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RD SHARMA SOLUTION CHAPTER –14 Compound Interest | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 14.1

Question 1.
Find the compound interest when principal = Rs 3,000, rate = 5% per annum and time = 2 years.
Solution:
Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 1
Amount after one year = Rs 3,000 + Rs 150 = 3,150
and principal for the second year = Rs 3,150
and interest for the second year

Compound interest for two years = Rs 150 + Rs 157.50 = Rs 307.50

Question 2.
What will be the compound interest on Rs 4,000 in two years when rate of interest is 5% per annum ?
Solution:
Principal (P) = Rs 4,000
Rate (R) = 5% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 3
Amount after one year = Rs 4,000 + Rs 200 = Rs 4,200
Principal for the second year = Rs 4,200
Interest for the second year

Compound interest for 2 years = Rs 200 + Rs 210 = Rs 410

Question 3.
Rohit deposited Rs 8,000 with a finance company for 3 years at an interest of 15% per annum. What is the compound interest that Rohit gets after 3 years ?
Solution:
Principal (P) = Rs 8,000
Rate (R) = 15% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 5
Amount after first year = Rs 8,000 + RS 1,200 = Rs 9,200
or Principal for the second year = Rs 9,200
Interest for the second year

Amount after 2 years = Rs 9,200 + Rs 1,380 = Rs 10,580
or Principal for the third year = Rs 10,580
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 7
Compound for the 3 years = Rs 1,200 + Rs 1,380 + Rs 1,587 = Rs 4,167

Question 4.
Find the compound interest on Rs 1,000 at the rate of 8% per annum for 112 years when interest is compounded half-yearly ?
Solution:
Principal (P) = Rs 1,000
Rate (R) = 8% p.a.
Period (T) = 112 years = 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 8
Amount after one half-year = Rs 1,000 + Rs 40 = 1,040
Or principal for the second half-year = Rs 1,040
Interest for the second half-year

Amount after second half-year = Rs 1,040 + 41.60 = Rs 1,081.60
Or principal for the third half-year = Rs 1081.60
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 10
Compound interest for the third half-years or 112 years
= Rs 40 + Rs 41.60 + Rs 43.264 = Rs 124.864

Question 5.
Find the compound interest on Rs 1,60,000 for one year at the rate of 20% per annum, if the interest is compounded quarterly.
Solution:
Principal (P) = Rs 1,60,000
Rate (R) = 20% p.a. or 5% quarterly
Period (T) = 1 year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 11
Amount after first quarter = Rs 1,60,000 + 8,000 = 1,68,000
Or principal for the second quarter = Rs 1,68,000
Interest for the second quarter

Amount after the second quarter = Rs 1,68,000 + Rs 8,400 = 1,76,400
Or principal for the third quarter = Rs 1,76,400
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 13
Amount after third quarter = Rs 1,76,400 + 8,820 = Rs 1,85,220
or Principal for the fourth quarter = Rs. 1,85,220
Interest for the fourth quarter

Total compound interest for the 4 quarters = Rs 8,000 + Rs 8,400 + Rs 8,820 + 9,261 = Rs 34,481

Question 6.
Swati took a loan of Rs 16,000 against her insurance policy at the rate of 1212 % per annum. Calculate the total compound interest payable by Swati after 3 years.
Solution:
Amount of loan or principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 15
Amount after first year = Rs 16,000 + Rs 2,000 = Rs 18,000
Principal for the second year = Rs 18,000
Interest for the second year

Amount after second year = Rs 18,000 + 2,250 = Rs 20,250
Principal for the third year = Rs 20,250

Compound for 3 years = Rs 2,000 + Rs 2,250 + 2531.25 = Rs 6,781.25

Question 7.
Roma borrowed Rs 64,000 from a bank for 112 years at the rate of 10% per annum. Compute the total compound interest payable by Roma after 112 years, if the interest is compounded half-yearly.
Solution:
Principal (sum borrowed) (P) = Rs 64,000
Rate (R) = 10% p.a. or 5% half-yearly
Period (T) = 112 years or 3 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 18
Amount after first half-year = Rs 64,000 + Rs 3,200 = Rs 67,200
Or principal for the second half-year = Rs 67,200
Interest for the second half-year

Amount after second half-year = Rs 6,7200 + 3,360 = Rs 70,560
Or principal for the third half-year = Rs 70,560
Interest for the third half-year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 20
Total compound interest for 3 half-years
or 112 years = Rs 3,200 + Rs 3,360 + Rs 3,528 = Rs 10,088

Question 8.
Mewa Lai borrowed Rs 20,000 from his friend RoopLal at 18% per annum simple interest. He lent it to Rampal at the same rate but compounded annually. Find his gain after 2 years.
Solution:
Principal (P) = Rs 20,000
Rate (R) = 18% p.a.
Period (T) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 21
In second case
Interest for the first year

Amount after one year = Rs 20,000 + Rs 3,600 = Rs 23,600
Or principal for the second year = Rs 23,600
Interest for the second year

Interest for two years = Rs 3,600 + 4,248 = Rs 7,848
Gain = Rs 7,848 – Rs 7,200 = Rs 648

Question 9.
Find the compound interest on Rs 8,000 for 9 months at 20% per annum compounded quarterly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 20% p.a. or 5% p.a. quarterly
Period (T) = 9 months or 3 quarters
Interest for the first quarterly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 24
Amount after first quarter = Rs 8,000 + Rs 400 = Rs 8,400
Or principal for second quarter = Rs 8,400
Interest for the second quarter

Amount after second quarter = Rs 8,400 + Rs 420 = Rs 8,820
Or principal for the third quarter = Rs 8,820
Interest for the third quarter
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 26
Compound interest for 9 months or 3 quarters = Rs 400 + Rs 420 + Rs 441 = Rs 1,261

Question 10.
Find the compound interest at the rate of 10% per annum for two years on that principal which in two years at the rate of 10% per annum gives Rs. 200 as simple interest.
Solution:
Simple interest = Rs 200
Rate (R) = 10% p.a.
Period (T) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 27
Now in second case,
Principal CP) = Rs 1,000
Rate (R) = 10% p.a.
Period (T) = 2 years.

Amount after one year = Rs 1,000+ Rs 100 = Rs 1,100
Or principal for the second year = Rs 1,100

Total interest for two years = Rs 100 + Rs 110 = Rs 210

Question 11.
Find the compound interest on Rs 64,000 for 1 year at the rate of 10% per annum compounded quarterly.
Solution:
Principal (P) = Rs 64,000
Rate (R) = 10% p.a. or 52 % quarterly
Period (T) = 1 year = 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 30
Amount after first quarter = Rs 64,000 + Rs 1,600 = Rs 65,600
Or principal for the second quarter = Rs 65,600
Interest for the second quarter

Amount after second quarter = Rs 65,600 + Rs 1,640 = Rs 67,240
Or principal for the third year = Rs 67,240
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 32
= Rs 1,681
Amount after third quarter = Rs 67,240 + Rs 1,681 = Rs 68,921
Or principal for the fourth quarter

Total compound interest for 4 quarters or one year
= Rs 1,600 + Rs 1,640 + Rs 1,681 + Rs 1723.025 = Rs 6644.025

Question 12.
Ramesh deposited Rs 7,500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months ?
Solution:
Principal (P) = Rs 7,500
Rate (R) = 12% p.a. or 3% quarterly
Time (T) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 34
Amount after one quarter = Rs 7,500 + Rs 225 = Rs 7,725
Or Principal for second quarter = Rs 7,725
Interest for the second quarter

Amount after second quarter = Rs 7,725 + Rs 231.75 = Rs 7956.75
Or principal for the third quarter

Total amount he received after 9 months = Rs 7956.75 + Rs 238.70 = Rs 8195.45

Question 13.
Anil borrowed a sum of Rs 9,600 to install a hand pump in his dairy. If the rate of interest is 512 % .per annum compounded annually, determine the compound interest which Anil will have to pay after 3 years.
Solution:
Principal (P) = Rs 9,600
Rate of interest (R) = 512 % = 112 % p.a.
Period (T) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 37
Amount after one year = Rs 9,600 + Rs 528 = Rs 10,128
Or principal for second year = Rs 10,128
Interest for the second year

Amount after second year = Rs 10,128 + Rs 557.04 = Rs 10685.04
or Principal for the third year = Rs 10685.04
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 39
Total compound interest = Rs 528 + Rs 557.04 + Rs 587.68 = Rs 1672.72

Question 14.
Surabhi borrowed a sum of Rs 12,000 from a finance company to purchase a refrigerator. If the rate of interest is 5% per annum compounded annually, calculate the compound interest that Surabhi has to pay to the company after 3 years.
Solution:
Sum of money borrowed (P) = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 40
Amount after one year = Rs 12,000 + Rs 600 = Rs 12,600
Or principal for the second year = Rs 12,600
Interest for the second year

Amount after second year = Rs 12,600 + Rs 630 = Rs 13,230
Or Principal for the third year = Rs 13,230
Interest for the third year
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 42
Total compound interest for 3 years = Rs 600 + Rs 630 + Rs 661.50 = Rs 1891.50

Question 15.
Daljit received a sum of Rs 40,000 as a loan from a finance company. If the rate of interest is 7% per annum compounded annually, calculate the compound interest that Daljit pays after 2 years.
Solution:
Amount of loan (P) = Rs 40,000
Rate (R) = 7% p.a.
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.1 43
Amount after one year = Rs 40,000 + Rs 2,800 = Rs 42,800
Or principal for the second year = Rs 42,800
Interest for the second year

Total interest paid after two years = Rs 2,800 + 2,996 = Rs 5,796

Exercise 14.2

Question 1.
Compute the amount and the compound interest in each of the following by using the formula when :
(i) Principal = Rs 3,000, Rate = 5%, Time = 2 years
(ii) Principal = Rs 3,000 Rate = 18%, Time = 2 years
(iii) Principal = Rs 5,000 Rate = 10 paise per rupee per annum, Time = 2 years
(iv) Principal = Rs 2,000, Rate = 4 paise per rupee per annum, Time = 3 years.
(v) Principal = Rs 12,800, Rate = 712 %, Time = 3 years
(vi) Principal = Rs 10,000, Rate = 20% per annum compounded half-yearly, time = 2 years
(vii) Principal = Rs 1,60,000, Rate = 10 paise per rupee per annum compounded half- yearly, Time = 2 years.
Solution:
(i) Principal (P) = Rs 3,000
Rate (R) = 5% p.a.
Time (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 1
and compound interest (C.I) = A – P = Rs 3307.50 – Rs 3,000 = Rs 307.50
(ii) Principal (P) = Rs 3,000
Rate (R) = 18% p.a.
Time (n) = 2 years

and compound interest (C.I.) = A – P = Rs 4177.20 – Rs 3,000 = Rs 1177.20
(iii) Principal (P) = Rs 5,000
Rate (R) =10 paise per rupee or 10% p.a.
Time (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 3
C.I. = A – P = Rs 6,050 – Rs 5,000 = Rs 1,050
(iv) Principal (P) = Rs 2,000
Rate (R) = 4 paise per rupee or 4% p.a.
Time (n) = 3 years

C.I. = A – P = Rs 2249.73 – Rs 2,000 = Rs 249.73
(v) Principal (P) = Rs 12,800

C.I. = A – P = Rs 15901.40 – Rs 12,800 = Rs 3101.40
(vi) Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Time = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 6
C.I. = A – P = Rs 14,641 – Rs 10,000 = Rs 4,641
(vii) Principal (P) = Rs 1,60,000
Rate (R) = 10 paise per rupee or 10% p.a. or 5% half-yearly
Time (n) = 2 years or 4 half-years.

C.I. = A – P = Rs 1,94,481 – Rs 1,60,000 = Rs 34,481

Question 2.
Find the amount of Rs 2,400 after 3 years, when the interest is compounded annually at the rate of 20% per annum.
Solution:
Principal (P) = Rs 2,400
Rate (R) = 20%
Time (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 8

Question 3.
Rahman lent Rs 16,000 to Rasheed at the rate of 1212 % per annum compound interest. Find the amount payable by Rasheed to Rahman after 3 years.
Solution:
Principal (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 9

Question 4.
Meera borrowed a sum of Rs 1,000 from Sita for two years. If the rate of interest is 10% compounded annually find the amount that Meera has to pay back.
Solution:
Amount of loan (P) = Rs 1,000
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 10

Question 5.
Find the difference between the compound interest and simple interest. On a sum of Rs 50,000 at 10% per annum for 2 years.
Solution:
Principal (P) = Rs 50,000
Rate (R) = 10% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 11
Difference between C.I. and S.I. = Rs 10,500 – Rs 10,000 = Rs 500

Question 6.
Amit borrowed Rs 16,000 at 1712 % per annum simple interest on the same day, he lent it to Ashu at the same rate but compounded annually. What does he gain at the end of 2 years ?
Solution:
Amount of loan (P) = Rs 16,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 12

C.I. = A – P = Rs 22,090 – Rs 16,000 = Rs 6,090
Now gain = C.I. – S.I = Rs 6,090 – 5,600 = Rs 490

Question 7.
Find the amount of Rs 4,096 for 18 months at 1212 % per annum, interest being compounded semi-annually ?
Solution:
Principal (P) = Rs 4,096
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 14

Question 8.
Find the amount and the compound interest on Rs 8,000 for 112 years at 10% per annum, compounded half-yearly.
Solution:
Principal (P) = Rs 8,000
Rate (R) = 10% p.a. or 5% half yearly
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 15
and C.I. = A – P = Rs 9,261 – Rs 8,000 = Rs 1,261

Question 9.
Kamal borrowed Rs 57,600 from LIC against her policy at 1212 % per annum to build a house. Find the amount that she pays LIC after 112 years if the interest is calculated half-yearly.
Solution:
Amount of loan (P) = Rs 57,600
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 16

Question 10.
Abha purchased a house from Avas Parishad on credit. If the cost of the house is Rs 64,000 and the rate of interest is 5% per annum compounded half-yearly, find the interest paid by Abha after one year and a half.
Solution:
Price of house (P) = Rs 64,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 17
Compound interest (C.I.) = A – P = Rs 68,921 – Rs 64,000 = Rs 4,921

Question 11.
Rakesh lent out Rs 10,000 for 2 years at 20% per annum compounded annually. How much more he could earn if the interest be compounded half-yearly ?
Solution:
Principal (P) = Rs 10,000
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 18

C.I. = Rs 14,400 – Rs 10,000 = Rs 4,400
Now difference in C.I. = Rs 4,641 – Rs 4,400 = Rs 241

Question 12.
Romesh borrowed a sum of Rs 2,45,760 at 12.5% per annum compounded annually. On the same day, he lent out his money to Ramu at the same rate of interest but compounded semi-annually. Find his gain after 2 years.
Solution:
In first case,
Principal (P) = Rs 2,45,760
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 20

C.I. = A – P = Rs 313203.75 – Rs 2,45,760 = Rs 67443.75
Gain = 67443.75 – Rs 65,280 = Rs2163.75

Question 13.
Find the amount that David would receive if he invests Rs 8,192 for 18 months at 1212 % per annum, the interest being compounded half-yearly.
Solution:
Principal (P) = Rs 8,192
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 22

Question 14.
Find the compound interest on Rs 15,625 for 9 months at 16% per annum, compounded quarterly.
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16% p.a. or 4% quarterly
Period (n) = 9 months or 3 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 24
Compound interest = A – P = Rs 17,576 – Rs 15,625 = Rs 1,951

Question 15.
Rekha deposited Rs 16,000 in a foreign bank which pays interest at the rate of 20% per annum compounded quarterly, find the interest received by Rekha after one year.
Solution:
Principal (P) = Rs 16,000
Rate (R) = 20% p.a. or 5% quarterly
Period (n) = one year or 4 quarters
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 25
C.I. = A – P = Rs 19448.10 – Rs 16,000 = Rs 3448.10

Question 16.
Find the amount of Rs 12,500 for 2 years compounded annually, the rate of interest being 15% for the first year and 16% for the second year.
Solution:
Principal (P) = Rs 12,500
Rate (R1) = 15% p.a. for first year
R2 = 16% p.a. for second year
Period = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 26

Question 17.
Ramu borrowed Rs 15,625 from a finance company to buy a scooter. If the rate of interest be 16% per annum compounded annually, what payment he will have to make after 214 years ?
Solution:
Principal (P) = Rs 15,625
Rate (R) = 16%
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 27

Question 18.
What will Rs 1,25,000 amount to at the rate of 6% if interest is calculated after every 4 months for one year ?
Solution:
Principal (P) = Rs 1,25,000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 29

Question 19.
Find the compound interest at the rate of 5% for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 12,000 as simple interest.
Solution:
In first case,
S.I. = Rs 12,000
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 30
= Rs 80,000
In second case,
Principal (P) = Rs 80,000
Rate (R) = 5% p.a.
Period (n) = 3 years

C.I. = A – P = Rs 92,610 – 80,000 = Rs 12,610

Question 20.
A sum of money was lent for 2 years at 20% compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is Rs 482 more. Find the sum.
Solution:
Let Sum (P) = Rs x
Rate (R) = 20% p.a. or 10% half-yearly
Period (n) = 2 years or 4 half years
In first case,
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 32
Interests = A – P = Rs 146.41 – Rs 100 = Rs 46.41
Now difference in interests = Rs 46.41 – Rs 44.00 = Rs 2.41
If difference is 2.41 then sum is 100 If difference is Rs 482, then sum

Question 21.
Simple interest on a sum of money for 2 years at 612 % per annum is Rs 5,200. What will be the compound interest on the sum at the same rate for the same period ?
Solution:
In first case,
S.I. = Rs 5,200
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 34

Compound interest = A – P = Rs 45,369 – Rs 40,000 = Rs 5,369

Question 22.
Find the compound interest at the rate of 5% per annum for 3 years on that principal which in 3 years at the rate of 5% per annum gives Rs 1,200 as simple interest.
Solution:
In first case,
S.I. = Rs 1,200
Rate (R) = 5% p.a.
Period (T) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.2 36
In second case,
Principal (P) = Rs 8,000
Rate (R) = 5% p.a.
Period (n) = 3 years

= Rs 9,261
C.I. = A – P = Rs 9,261 – Rs 8000 = Rs 1,261

Exercise 14.3

Question 1.
On what sum will the compound interest at 5% p.a. annum for 2 years compounded annually be Rs 164 ?
Solution:
Let Principal (P) = Rs 100
Rate (R) = 5% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 1

Question 2.
Find the principal of the interest compounded annually at the rate of 10% for two years is Rs 210.
Solution:
Let principal (P) = Rs 100
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 3

Question 3.
A sum amounts to Rs 756.25 at 10% per annum in 2 years, compounded annually. Find the sum.
Solution:
Amount (A) = Rs 756.25
Rate (R) = 10% p.a.
Period (n) = 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 5

Question 4.
What sum will amount to Rs 4913 in 18 months, if the rate of interest is 1212 % per annum, compounded half-yearly.
Solution:
Amount (A) = Rs 4,913
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 6

Question 5.
The difference between the compound interest and simple interest on a certain sum at 15% per annum for 3 years is Rs 283.50. Find the sum.
Solution:
Let sum (P) = Rs 100
Rate (R) = 15% p.a.
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 7

Question 6.
Rachna borrowed a certain sum at the rate of 15% per annum. If she paid at the end of two years Rs 1,290 as interest compounded annually, find the sum she borrowed.
Solution:
C.I. = Rs 1,290
Rate (R) = 15% p.a.
Period (n) = 2 years
Let sum (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 10

Question 7.
The interest on a sum of Rs 2,000 is being compounded annually at the rate of 4% per annum. Find the period for which the compound interest is Rs 163.20.
Solution:
Sum (P) = Rs 2,000
C.I. = Rs 163.20
Amount (A) = P + C.I. = Rs 2000 + Rs 163.20 = Rs 2163.20
Rate (R) = 4%
Let period = n years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 11

Question 8.
In how much time would Rs 5,000 amount to Rs 6,655 at 10% per annum compound interest ?
Solution:
Principal (P) = Rs 5,000
Amount (A) = Rs 6,655
Rate (R) = 10%
Let period = n years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 13

Question 9.
In what time will Rs 4,400 becomes Rs 4,576 at 8% per annum interest compounded half-yearly ?
Solution:
Principal (P) = Rs 4,400
Amount (A) = Rs 4,576
Rate (R) = 8% or 4% half-yearly
Let period = n half-years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 14
n = 1
Period = 1 half year

Question 10.
The difference between the S.I. and C.I. on a certain sum of money for 2 years at 4% per annum is Rs 20. Find the sum.
Solution:
Difference between C.I. and S.I. = Rs 20
Rate (R) = 4% p.a.
Period (n) = 2 years
Let principal (P) = Rs 100
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 15

Question 11.
In what time will Rs 1,000 amount to Rs 1,331 at 10% per annum compound interest.
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1,331
Rate (R) = 10% p.a.
Let period = n year
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 17

Question 12.
At what rate percent compound interest per annum will Rs 640 amount to Rs 774.40 in 2 years ?
Solution:
Principal (P) = Rs 640
Amount (A) = Rs 774.40
Period (n) = 2 years.
Let R be the rate of interest p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 18

Question 13.
Find the rate percent per annum if Rs 2000 amount to Rs 2,662 in 112 years, interest being compounded half-yearly ?
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2,662
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 14.
Kamala borrowed from Ratan a certain rate for two years simple interest. She lent this sum at the same rate to Hari for two years compound interest. At the end of two years, she received Rs 210 as compound interest, but paid Rs 200 only as simple interest. Find the sum and the rate of interest.
Solution:
Simple interest = Rs 200
and compound interest = Rs 210.
Period = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 15.
Find the rate percent per annum, if Rs 2,000 amount to Rs 2,315.25, in an year and a half, interest being compounded six monthly.
Solution:
Principal (P) = Rs 2,000
Amount (A) = Rs 2315.25
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 22

Question 16.
Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.
Solution:
Let Principal (P) = Rs 100
then Amount (A) = Rs 200
Period (n) = 3 years
Let R be the rate % p.a.
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 24

Question 17.
Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half- yearly.
Solution:
Let Principal (P) = Rs 100
Then Amount (A) = Rs 400
Period (n) = 2 years or 4 half years
Let R be the rate % half-yearly, then
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 25
Rate % = 41.42% half yearly and 82.84% p.a.

Question 18.
A certain sum amounts to Rs 5,832 in 2 years at 8% compounded interest. Find the sum.
Solution:
Amount (A) = Rs 5,832
Let P be the sum
Rate (R) = 8% p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 26

Question 19.
The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs 360. Find the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 27

Question 20.
The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years is Rs 46. Determine the sum.
Solution:
Let sum (P) = Rs 100
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 29

Question 21.
Ishita invested a sum of Rs 12,000 at 5% per annum compound interest. She received an amount of Rs 13,230 after n years, Find the value of n.
Solution:
Principal (P) = Rs 12,000
Amount (A) = Rs 13,230
Rate (R) = 5% p.a.
Period = n years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 32

Question 22.
At what rate percent per annum will a sum of Rs 4,000 yield compound interest of Rs 410 in 2 years ?
Solution:
Principal (P) = Rs 4,060
C.I. = Rs 410
Amount (A) = Rs 4,000 + 410 = Rs 4,410
Let rate = R % p.a.
Period (n) = 2 years
lass 8 Solutions Chapter 14 Compound Interest Ex 14.3 33

Question 23.
A sum of money deposited at 2% per annum compounded annually becomes Rs 10404 at the end of 2 years. Find the sum deposited.
Solution:
Amount (A) = Rs 10,404
Rate (R) = 2% p.a.
Period (n) = 2 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 18

Question 24.
In how much time will a sum of Rs 1,600 amount to Rs 1852.20 at 5% per annum compound interest ?
Solution:
Principal (P) = Rs 1,600
Amount (A) = Rs 1852.20
Rate (R) = 5% p.a.
Let n be the time
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 20

Question 25.
At what rate percent will a sum of Rs 1,000 amount to Rs 1102.50 in 2 years at compound interest ?
Solution:
Principal (P) = Rs 1,000
Amount (A) = Rs 1102.50 .
Period (n) = 2 years
Let R be the rate of interest p.a.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 21

Question 26.
The compound interest on Rs 1,800 at 10% per annum for a certain period of time is Rs 378. Find the time in years.
Solution:
Principal (P) = Rs 1,800
C.I. = Rs 378
Amount (A) = P + C.I. = Rs 1,800 + 378 = Rs 2,178
Rate (R) = 10% p.a.
Let n be the period in years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 22
Comparing, we get:
n = 2
Period = 2 years

Question 27.
What sum of money will amount to Rs 45582.25 at 634 % per annum in two years, interest being compounded annually ?
Solution:
Amount (A) = Rs 45582.25
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 23

Question 28.
Sum of money amounts to Rs 4,53,690 in 2 years at 6.5% per annum compounded annually. Find the sum.
Solution:
Amount (A) = Rs 4,53,690
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.3 24

Exercise 14.4

Question 1.
The present population of a town is 28,000. If it increases at the rate of 5% per annum, what will be its population after 2 years ?
Solution:
Present population = 28000
Rate of increase (R) = 5% p.a.
Period (n) = 2 years
Population after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 1

Question 2.
The population of a city is 125000. If the annual birth rate and death rate are 5.5% and 3.5% respectively, calculate the population of city after 3 years.
Solution:
Present population = 125000
Rate of birth = 5.5%
and rate of death = 3.5%
Increase = 5.5 – 3.5 = 2% p.a.
Period = 3 years.
Population after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 2

Question 3.
The present population of a town is 25000. It grows at 4%, 5% and 8% during first year, second year and third year respectively. Find its population after 3 years.
Solution:
Present population = 25000
Increase in first year = 4%
in second year= 5% and
in third year = 8%
Population after 3 years =
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 4

Question 4.
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 4%, 5% and 3% respectively, find the present population.
Solution:
Three years ago,
Population of a town = 50000
Annual increase in population in first year = 4%
in second year = 5%
and in third year = 3%
Present population
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 5

Question 5.
There is a continuous growth in population of a village at the rate of 5% per annum. If its present population is 9261, what it was 3 years ago ?
Solution:
Let 3 years ago, population = P
Present population = 9261
Rate of increase (R) = 5% p.a.
Period (n) = 3 years.
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 6

Question 6.
In a factory, the production of scooters rose to 46305 from 40000 in 3 years. Find the annual rate of growth of the production of scooters.
Solution:
Production of scooters 3 years ago (P) = 40000
Present production (A) = 46305
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 7

Question 7.
The annual rate of growth in population of a certain city is 8%. If its present population is 196830, what it was 3 years ago ?
Solution:
Let 3 years ago, the population of a city = P
Rate of growth (R) = 8% p.a.
Present population = 196830
Period (n) = 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 9

Question 8.
The population of a town increases at the rate of 50 per thousand. Its population after 2 years will be 22050. Find its present population.
Solution:
Population after 2 years = 22050
Rate of increase = 50 per thousand
Period (n) = 2 years
Let present population = P, then
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 10

Question 9.
The count of bacteria in a culture grows by 10% in the first hour, decreases by 8% in the second hour and again increases by 12% in the third hour. If the count of bacteria in the sample is 13125000, what will be the count of bacteria after 3 hours ?
Solution:
Present count of bacteria = 13125000
In first hour increase = 10%
decrease in second hour = 8%
increase in third hour = 12%
Count of bacteria after 3 hours
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 11

Question 10.
The population of a certain city was 72000 on the last day of the year 1998. During next year it increased by 7% but due to an epidemic it decreased by 10% in the following year. What was its population at the end of the year 2000 ?
Solution:
On the last day of 1998,
Population of a town = 72000
In the first year, increase = 7%
In the second year, decrease = 10%
Population in the last day of 2000
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 12

Question 11.
6400 workers were employed to construct a river bridge in four years. At the end of the first year, 25% workers were retrenched. At the end of the second year, 25% of those working at that time were retrenched. However, to complete the project in time, the number of workers was increased by 25% at the end of the third year. How many workers were working during the fourth year ?
Solution:
Number of workers at the beginning = 6400
Period = 4 years.
At the end of 1st year, workers retrenched = 25%
At the end of second year, workers retrenched = 25%
At the end of third year, workers increased = 25%
Total number of workers during the 4 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 13

Question 12.
Aman started a factory with an initial investment of Rs 1,00,000. In the first year, he incurred a loss of 5%. However, during the second year, he earned a profit of 10% which in the third year rose to 12%. Calculate his net profit for the entire period of three years.
Solution:
Initial investment = Rs 100000
Loss in the first year = 5%
Profit in the second year = 10%
Profit in the third year = 12%
Investment at the end of 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 14

Question 13.
The population of a town increases at the rate of 40 per thousand annually. If the present population be 175760, what was the population three years ago ?
Solution:
Present population (A) = 175760
Increase rate = 40 per 1000
Period = 3 years
Let 3 years ago,
population was = P
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 15

Question 14.
The production of a mixi company in 1996 was 8000 mixies. Due to increase in demand it increases its production by 15% in the next two years and after two years its demand decreases by 5%. What will be its production after 3 years ?
Solution:
Production of Mixi in 1996 = 8000
Increase in next 2 years = 15%
Decrease in the third year = 5%
Production after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 16

Question 15.
The population of a city increases each year by 4% of what it had been at the beginning of each year. If the population in 1999 had been 6760000, find the population of the city in (i) 2001 (ii) 1997.
Solution:
Population of a city in 1999 = 6760000
Increase = 4%
(i) Population in 2001 is after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 17

Question 16.
Jitendra set up a factory by investing Rs 25,00,000. During the first two successive years his profits were 5% and f 10% respectively. If each year the profit was on previous year’s capital, compute his total profit.
Solution:
Investment in the beginning = Rs 25,00,000
Profit during the first 2 years = 5% and 10% respectively
Investment after 2 years will be
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.4 19
= Rs 28,87,500
Amount of profit = Rs 28,87,500 – Rs 25,00,000 = Rs 3,87,500

Exercise 14.5

Question 1.
Mr. Cherian purchased a boat for Rs 16,000. If the total cost of the boat is depreciating at the rate of 5% per annum, calculate its value after 2 years.
Solution:
Cost of boat = Rs 16,000
Rate of depreciating = 5% p.a.
Period = 2 years
Value of boat after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 1

Question 2.
The value of a machine depreciates at the rate of 10% per annum. What will be its value 2 years hence, if the present value is Rs 1,0,000 ? Also, find the total depreciation during this period.
Solution:
Present value of machine = Rs 1,00,000
Rate of depreciation = 10% p.a.
Period (n) = 2 years
Value of machine after 2 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 2

Question 3.
Pritam bought a plot of land for Rs 6,40,000. Its value is increasing by 5% of its previous value after every six months. What will be the value of the plot after 2 years ?
Solution:
Present value of plot = Rs 6,40,000
Increase = 5% per half year
Period (n) = 2 years or 4 half years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 3

Question 4.
Mohan purchased a house for Rs 30,000 and its value is depreciating at the rate of 25% per year. Find the value of the house after 3 years.
Solution:
Present value of the house (P) = Rs 30,000
Rate of depreciation = 25% p.a.
Period (n) = 3 years
Value of house after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 5

Question 5.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs 43,740, find its purchased price.
Solution:
Let the purchase price of machine = Rs P
Rate of depreciation = 10% p.a.
Period (n) = 3 years.
and present value = Rs 43,740
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 6

Question 6.
The value of a refrigerator which was purchased 2 years ago, depreciates at 12% per annum. If its present value is Rs 9,680, for how much was it purchased ?
Solution:
Let the refrigerator was purchased for = Rs P
Rate of depreciation (R) = 12% p.a.
Period (n) = 2 years
and present value (A) = Rs 9,680
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 8

Question 7.
The cost of a TV set was quoted Rs 17,000 at the beginning of 1999. In the beginning of2000, the price was hiked by 5%. Because of decrease in demand the cost was reduced by 4% in the beginning of 2001. What is the cost of the TV set in 2001 ?
Solution:
List price of TV set in 1999 = Rs 17,000
Rate of hike in 2000 = 5%
Rate of decrease in 2001 = 4%
Price of TV set in 2001
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 9

Question 8.
Ashish started the business with an initial investment of Rs 5,00,000. In the first year, he incurred a loss of 4%. However, during the second year he earned a profit of 5% which in third year, rose to 10%. Calculate the net profit for the entire period of 3 years.
Solution:
Initial investment = Rs 5,00,000
In the first year, rate of loss = 4%
In the second year, rate of gain = 5%
and in the third year, rate of gain = 10%
Investment after 3 years
RD Sharma Class 8 Solutions Chapter 14 Compound Interest Ex 14.5 10

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RD SHARMA SOLUTION CHAPTER –13 Profit, Loss, Discount and Value Added Tax (VAT) | CLASS 8TH MATHEMATICS-EDUGROWN

Exercise 13.1

Question 1.
A student buys a pen for Rs. 90 and sells it for Rs. 100. Find his gain and gain percent ?
Solution:
C.P. of a pen = Rs. 90
and S.P. = Rs. 100
Gain = S.P. – C.P. = Rs. 100 – 90 (S.P > C.P.)
= Rs. 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 1

Question 2.
Rekha bought a saree for Rs. 1240 and sold it for Rs. 1147. Find her loss and loss percent.
Solution:
C.P. of saree.= Rs. 1240 and
S.P. = Rs. 1147
Loss = C.P – S.P. = Rs. 1240 – Rs. 1147 (C.P. > S.P.)
= Rs. 93
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 2

Question 3.
A boy buys 9 apples for Rs. 9.60 and sells them at 11 for Rs. 12. Find his gain or loss percent.
Solution:
L.C.M. of 9 and 11 = 99
Let 99 apples were purchased.
C.P. of 99 apples at the rate of 9 apples for Rs. 9.60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 3

Question 4.
The cost price of 10 articles is equal to the selling price of 9 articles. Find the profit percent.
Solution:
C.P. of 10 articles = S.P. of 9 articles = 90 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 4

Question 5.
A retailer buys a radio for Rs. 225. His overhead expenses are Rs. 15. If he sells the radio for Rs. 300, determine his profit percent.
Solution:
Cost of radio = Rs. 225
Over head expenses = Rs. 15
Total C.P. of the ratio = Rs. 225 + 15 = Rs. 240
S.P. of radio = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 5

Question 6.
A retailer buys a cooler for Rs. 1200 and overhead expenses on it are Rs. 40. If he sells the cooler for Rs. 1550, determine his profit percent.
Solution:
Cost of cooler = Rs. 1200
Overhead expenses = Rs. 40
Total cost price of cooler = Rs. 1200 + Rs. 40 = Rs. 1240
Selling price = Rs. 1550
Gain = S.P. – C.P. = Rs. 1550 – Rs. 1240 = Rs. 310
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 6

Question 7.
A dealer buys a wristwatch for Rs. 225 and spends Rs. 15 on its repairs. If he sells the same for Rs. 300, find his profit percent.
Solution:
Cost of wristwatch = Rs. 225
Cost on repairs = Rs. 15
Total cost price = Rs. 225 + Rs. 15 = Rs. 240
Selling price = Rs. 300
Gain = S.P. – C.P. = Rs. 300 – Rs. 240 = Rs. 60
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 7

Question 8.
Ramesh bought two boxes for Rs. 1300. He sold one box at a profit of 20% and the other box at a loss of 12%. If the selling price of both boxes is the same, find the cost price of each box.
Solution:
Total cost price of two boxes = Rs. 1300
S.P. of each box is same.
Let S.P of each box = Rs. 100
S.P. of first box = Rs. 100
Gain = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 8

Question 9.
If the selling price of 10 pens is equal to cost price of 14 pens, find the gain percent.
Solution:
S.P. of 10 pens = C.P. of 14 pens = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 10

Question 10.
If the cost price of 18 chairs be equal to selling price of 16 chairs, And the gain or loss percent.
Solution:
C.P. of 18 chairs = S.P. of 16 chairs = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 11

Question 11.
If the selling price of 18 oranges is equal to the cost price of 16 oranges, And the loss percent.
Solution:
S.P. of 18 oranges = C.P. of 16 oranges = Rs. 100 (Suppose)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 13

Question 12.
Ravish sold his motorcycle to Vineet at a loss of 28%. Vineet spent Rs. 1680 on its repairs and sold the motor cycle to Rahul for Rs. 35910, thereby making a profit of 12.5%, find the cost price of the motor cycle for Ravish.
Solution:
Cost price of cycle for Rahul or Selling price for Vineet = Rs. 35910
Gain = 12.5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 14

Question 13.
By selling a book for Rs. 258, a book-seller gains 20%. For how much should he sell it to gain 30% ?
Solution:
S.P. of a book = Rs. 258
Gain = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 16

Question 14.
A defective briefcase costing Rs. 800 is being sold at a loss of 8%. If its price is further reduced by 5%, find its selling price.
Solution:
C.P. of a briefcase = Rs. 800
Loss = 8%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 17

Question 15.
By selling 90 ball pens for Rs. 160, a person losses 20%. How many ball pens should be sold for Rs. 96 so as to have a profit of 20%.
Solution:
S.P of 90 ball pens = Rs. 160
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 19

Question 16.
A man sells an article at a profit of 25%. If he had bought it at 20% less and sold it for Rs. 36.75 less, he would have gained 30%. Find the cost price of the article.
Solution:
Let C.P. of the article = Rs. 100
In first case gain = 25%
S.P. = 100 + 25 = Rs. 125
In second case,
C.P. = 20% less of Rs. 100 = 100 – 20 = Rs. 80
Gain = 30%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 20

Question 17.
A dishonest shopkeeper professes to sell pulses at his cost price but uses a false weight of 950 for each kilogram. Find his gain percent.
Solution:
Let C.P. of 1 kg of pulses = Rs. 100
S.P. of 950 gm = Rs. 100
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 21

Question 18.
A dealer bought two tables for Rs. 3120. He sold one of them at a loss of 15% and other at a gain of 36%. Then he found that each table was sold for the same price, find the cost price of each table.
Solution:
Cost price of two tables = Rs. 3120
Let S.P of each table = Rs. 100
Now S.P of one table = Rs. 100
Loss = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 23

Question 19.
Mariam bought two fans for Rs. 3605. She sold one at a profit of 15% and the other at a loss of 9%. If Mariam obtained the same amount for each fan, find the cost price of each fan.
Solution:
Total cost price of two fans = Rs. 3605
Let selling price of each fan = Rs. 100
Now S.P. of first fan = Rs. 100
Profit = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 25

Question 20.
Some toffees are bought at the rate of 11 for Rs. 10 and the same number at the rate of 9 for Rs. 10. If the whole lot is sold at one rupee per toffee, find the gain or loss percent on the whole transaction.
Solution:
L.C.M. of 11 and 9 = 99
Let each time 99 toffees are bought.
In first case, the C.P. of 99 toffees at the rate of 11 for Rs. 10
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 26

Question 21.
A tricycle is sold at a gain of 16%. Had it been sold for Rs. 100 more, the gain would have been 20%. Find the C.P. of the tricycle.
Solution:
Let the C.P. of tricycle = Rs. 100
In first case, gain = 16%
S.P. = Rs. 100 + 16 = Rs. 116
In second case, gain = 20%
S.P. = Rs. 100 + 20 = Rs. 120
Difference in S.P.’s = Rs. 120 – Rs. 116 = Rs. 4
If difference is Rs. 4,
then C.P. of the tricycle = Rs. 100
and if difference is Re. 1, then C.P.
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 27

Question 22.
Shabana bought 16 dozen ball pens and sold then at a loss equal to S.P. of 8 ball pens. Find
(i) her loss percent
(ii) S.P. of 1 dozen ball pens, if she purchased these 16 dozen ball pens for Rs. 576.
Solution:
C.P. of 16 dozen ball pens = S.P. of 16 dozen pens – loss
C.P. of 16 x 12 pens = S.P. of 16 dozen pens x S.P. of 8 pens
C.P. of 192 pens = S.P. of 16 x 12 pens x S.P. of 8 pens
S.P. of 192 pens + S.P. of 8 pens = S.P. of 200 pens
(i) Let C.P. of 1 pen = Re. 1
Then C.P. of 192 pens = Rs. 192
and S.P. of 200 pens = Rs. 192
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 28

Question 23.
The difference between two selling prices of a shirt at profits of 4% and 5% is Rs. 6. Find
(i) C.P. of the shirt
(ii) The two selling prices of the shirt
Solution:
The difference of two selling prices of shirt = Rs. 6
Difference in profits of 4% and 5% = 5 – 4 = 1%
(i) C.P. = 1 x 6 x 100 = Rs. 600
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 29

Question 24.
Toshiba bought 100 hens for Rs. 8000 and sold 20 of these at a gain of 5%. At what gain percent she must sell the remaining hens so as to gain 20% on the whole ?
Solution:
Total number of hens bought = 100
C.P. of 100 hens = Rs. 8000
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.1 30

Exercise 13.2

Question 1.
Find the S.P. If
(i) M.P. = Rs. 1300 and Discount = 10%
(ii) M.P. = Rs. 500 and Discount = 15%
Solution:
(i) M.P. = Rs. 1300, Discount = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 1

Question 2.
Find the M.P. If
(i) S.P. = Rs. 1222 and Discount = 6%
(ii) S.P. = Rs. 495 and Discount = 1%
Solution:
(i) S.P. = Rs. 1222, discount = 6%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 4

Question 3.
Find discount in percent when
(i) M.P. = Rs. 900 and S.P. = Rs. 873
(ii) M.P. = Rs. 500 and S.P. = Rs. 425
Solution:
(i) M.P. = Rs. 900
S.P. = Rs. 873
Discount = M.P. – S.P. = Rs. 900 – Rs. 873 = Rs. 27
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 5
(ii) M.P. = Rs. 500
S.P. = Rs. 425
Discount M.P. – S.P. = Rs. 500 – Rs. 425 = = Rs. 75

Question 4.
A shop selling sewing machines offers 3% discount on ail cash purchases. What cash amount does a customer pay for a sewing machine, the price of which is marked as Rs. 650.
Solution:
Marked price (M.P.) of one sewing machine = Rs. 650
Rate of discount = 3%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 7

Question 5.
The marked price of a ceiling fan is Rs. 720. During off season, it is sold for Rs. 684. Determine the discount percent.
Solution:
Marked price (M.P.) of fan = Rs. 720
Sale price (S.P.) = Rs. 684
Amount of discount = M.P. – S.P. = Rs. 720 – Rs. 684 = Rs. 36
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 8

Question 6.
On the eve of Gandhi Jayanti, a saree is sold for Rs. 720 after allowing 20% discount. What is the marked price ?
Solution:
S.P. of saree = Rs. 720
Rate of discount = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 9

Question 7.
After allowing a discount of 712 % on the marked price, an article is sold for Rs. 555. Find its marked price.
Solution:
Selling price (S.P.) = Rs. 555
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 10

Question 8.
A shopkeeper allows his customers 10% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 250?
Solution:
Marked price = Rs. 250
Discount allowed = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 11

Question 9.
A shopkeeper allows 20% off on the marked price of goods and still gets a profit of 25%. What is the actual cost to him of an article marked Rs. 500 ?
Solution:
Marked price (M.P.) of an article = Rs. 500
Discount allowed = 20%
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 13

Question 10.
A tradesman marks his goods at such a price that after allowing a discount of 15%, he makes a profit of 20%. What is the marked price of an article whose cost price is Rs 170 ?
Solution:
Rate of discount = 15% gain = 20%
Cost price (C.P.) of an article = Rs 170
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 14

Question 11.
A shopkeeper marks his goods in such a way that after allowing a discount of 25% on the marked price, he still makes a profit of 50%. Find the ratio of the C.P. to the M.P.
Solution:
Let cost price (C.P.) = Rs 100
Profit = 50%
Selling Price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 15

Question 12.
A cycle dealer offers a discount of 10% and still makes a profit of 26%. What is the actual cost to him of a cycle whose marked price is Rs 840 ?
Solution:
Rate of discount = 10%
Gain = 26%
Marked Price (M.P.) = Rs 840
Selling Price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 16

Question 13.
A shopkeeper allows 23% commission on his advertised price and still makes a profit of 10%. If he gains Rs 56 on one item, find his advertised price.
Solution:
Rate of commission = 23%
Profit = 10%
Total gain = Rs 56
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 17

Question 14.
A shopkeeper marks his goods at 40% above the cost price but allows a discount of 5% for cash payment to his customers. What actual profit does he make, if he receives Rs 1064 after paying the discount ?
Solution:
Let cost price (C.P.) = Rs 100
Marked price = Rs 100 + 40 = Rs 140
Rate of discount = 5%
Selling price (S.P)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 18

Question 15.
By selling a pair of ear rings at a discount of 25% on the marked price, a jeweller makes a profit of 16%. If the profit is Rs 48, what is the cost price ? What is the marked price and the price at which the pair was eventually bought ?
Solution:
Total profit = Rs 48
Profit percent = 16%
Cost price = 48×10016 = Rs 300
Selling Price = C.P. + profit = Rs 300 + Rs 48 = Rs 348
Rate of discount = 25%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 19

Question 16.
A publisher gives 32% discount on the printed price of a book to booksellers. What does a bookseller pay for a book whose printed price is Rs 275 ?
Solution:
Printed price of a book = Rs 275
Rate of discount = 32%
Selling price (S.P.)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 20

Question 17.
After allowing a discount of 20% on the marked price of a lamp, a trader loses 10%. By what percentage is the marked price above the cost price ?
Solution:
Rate of discount = 20%
Loss = 10%
Let the cost price of the lamp = Rs 100
Loss = 10%
Selling price = Rs 100 – 10 = Rs 90
Rate of discount = 20%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 21

Question 18.
The list price of a table fan is Rs 480 and it is available to a retailer at 25% discount. For how much should a retailer sell it to gain 15% ?
Solution:
List price of table fan (M.P.) = Rs 480
Rate of discount = 25%
Selling price (S.P)
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 22

Question 19.
Rohit buys an item at 25% discount on the marked price. He sells it for Rs 660, making a profit of 10%. What is the marked price of the item ?
Solution:
Rate of discount = 25%
Selling price (S.P.) for Rohit = Rs 660
Profit = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 23

Question 20.
A cycle merchant allows 20% discount on the marked price of the cycles and still makes a profit of 20%. If he gains Rs 360 over the sale of one cycle, find the marked price of the cycle.
Solution:
Rate of discount = 20%
Profit = 20%
Total gain = Rs 360
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 25

Question 21.
Jyoti and Meena run a ready-made garment shop. They mark the garments at such a price that even after allowing a discount of 12.5%, they make a profit of 10%. Find the marked price of a suit which costs them Rs 1470.
Solution:
Rate of discount = 12.5%
Profit = 10%
Cost price = Rs 1470
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 26

Question 22.
What price should Aslam mark on a pair of shoes ? Which costs him Rs 1200 so as to gain 12% after allowing a discount of 16% ?
Solution:
Cost price of shoes = Rs 1200
Gain = 12%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 28

Question 23.
Jasmine allows 4% discount on the marked price of her goods and still earns a profit of 20%. What is the cost price of a shirt for her marked at Rs 850 ?
Solution:
Marked price of a shirt = Rs 850
Discount = 4%
Selling price
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 29

Question 24.
A shopkeeper offers 10% off-season discount to the customers and still makes a profit of 26%. What is the cost price for the shopkeeper on a pair of shoes marked at Rs 1120 ?
Solution:
Marked price = Rs 1120
Rate of discount = 10%
S.P. of shoes
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 31

Question 25.
A lady shopkeeper allows her customers 10% discount on the marked price of the goods and still gets a profit of 25%. What is the cost price of a fan for her marked at Rs 1250 ?
Solution:
Marked price (M.P.) of fan = Rs 1250
Discount = 10%
S.P. of the fan
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.2 32

Exercise 13.3

Question 1.
The list price of a refrigerator is Rs 9700. If a value added tax of 6% is to be charged on it, how much one has to pay to buy the refrigerator ?
Solution:
List price of refrigerator = Rs 9700
Rate of VAT = 6%
Amount of VAT = Rs. 9700×6100 = Rs 582
Total price to be paid = Rs 9700 + 582 = Rs 10282

Question 2.
Vikram bought a watch for Rs 825. If this amount includes 10% VAT on the list price, what was the list price of the watch ?
Solution:
Price of watch including VAT = Rs 825
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 1

Question 3.
Aman bought a shirt for Rs 374.50 which includes 7% VAT. Find the list price of the shirt.
Solution:
Cost price of the shirt = Rs 374.50
Rate of VAT = 7%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 2

Question 4.
Rani purchases a pair of shoes whose sale price is Rs 175. If he pays a VAT at the rate of 7%, how much amount does he pays as VAT. Find the net value of the pair of shoes.
Solution:
Sale price of a pair of shoes = Rs 175
Rate of VAT = 7%
Total VAT paid = Rs. 175×7100 = Rs 12.25
and total amount paid = Rs 175 + Rs 12.25 = Rs 187.25https://googleads.g.doubleclick.net/pagead/ads?client=ca-pub-7601472013083661&output=html&h=280&adk=2523109437&adf=980553861&pi=t.aa~a.4089255474~i.16~rp.4&w=750&fwrn=4&fwrnh=100&lmt=1642576099&num_ads=1&rafmt=1&armr=3&sem=mc&pwprc=1894297687&psa=1&ad_type=text_image&format=750×280&url=https%3A%2F%2Fwww.learninsta.com%2Frd-sharma-class-8-solutions-chapter-13-profits-loss-discount-and-value-added-tax-vat-ex-13-3%2F&flash=0&fwr=0&pra=3&rh=188&rw=750&rpe=1&resp_fmts=3&wgl=1&fa=27&adsid=ChEIgN6ykAYQlqfqreX-0tatARI5AOa2GRQuFTRuP5gxgF63-8ooeukU8cZ7QCQZcMs0Bo08KBiUSCO_wUJ4DWFWZv_YD9YpnCl_xW2w&uach=WyJXaW5kb3dzIiwiMTAuMC4wIiwieDg2IiwiIiwiOTguMC40NzU4LjEwMiIsW10sbnVsbCxudWxsLCI2NCIsW1siIE5vdCBBO0JyYW5kIiwiOTkuMC4wLjAiXSxbIkNocm9taXVtIiwiOTguMC40NzU4LjEwMiJdLFsiR29vZ2xlIENocm9tZSIsIjk4LjAuNDc1OC4xMDIiXV1d&dt=1645025146248&bpp=2&bdt=6721&idt=2&shv=r20220214&mjsv=m202202090101&ptt=9&saldr=aa&abxe=1&cookie=ID%3D1be37a61a5e7cca6-2207ff21a8d0005c%3AT%3D1643370310%3ART%3D1645025060%3AS%3DALNI_MZlVwsQQNfk-mglMlxYK4Mv7pUwgg&prev_fmts=0x0%2C750x280%2C750x280&nras=3&correlator=3657959876739&frm=20&pv=1&ga_vid=528868594.1643370309&ga_sid=1645025145&ga_hid=193832627&ga_fc=1&u_tz=330&u_his=4&u_h=715&u_w=1270&u_ah=681&u_aw=1270&u_cd=24&u_sd=1.513&dmc=8&adx=57&ady=2221&biw=1254&bih=610&scr_x=0&scr_y=667&eid=42531397%2C44750773%2C21067496&oid=2&pvsid=1097847813229336&pem=972&tmod=1825076183&uas=3&nvt=1&ref=https%3A%2F%2Fwww.learninsta.com%2Frd-sharma-class-8-solutions%2F&eae=0&fc=1408&brdim=0%2C0%2C0%2C0%2C1270%2C0%2C1270%2C681%2C1270%2C610&vis=1&rsz=%7C%7Cs%7C&abl=NS&fu=128&bc=31&jar=2022-02-16-15&ifi=4&uci=a!4&btvi=3&fsb=1&xpc=BHm11P10xT&p=https%3A//www.learninsta.com&dtd=65

Question 5.
Swarna paid Rs 20 as VAT on a pair of shoes worth Rs 250. Find the rate of VAT.
Solution:
Amount of VAT paid = Rs 20
Price of the pair of shoes = Rs 250
Rate of VAT = 20×100250 = 8%

Question 6.
Sarita buys goods worth Rs 5,500. She gets a rebate of 5% on it. After getting the rebate if VAT at the rate of 5% is charged, find the amount she will have to pay for the goods.
Solution:
Price of goods = Rs 5,500
Rate of rebate = 5%
Sales price after rebate
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 3
= Rs 5,225
Rate of VAT = 5%
Amount of VAT

Amount to be paid after paying VAT = Rs 5,225 + 261.25 = Rs 5486.25

Question 7.
The cost of furniture inclusive of VAT is Rs 7,150. If the rate of VAT is 10%, find the original cost of the furniture.
Solution:
Cost of furniture including VAT = Rs 7,150
Rate of VAT = 10%
Original cost of furniture
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 5

Question 8.
A refrigerator is available for Rs 13,750 including VAT. If the rate of VAT is 10%, find the original cost of refrigerator.
Solution:
Cost of refrigerator including VAT = Rs 13,750
Rate of VAT = 10%
Actual price of refrigerator
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 6

Question 9.
A colour TV is available for Rs 13,440 inclusive of VAT. If the original cost of TV is Rs 12,000, find the rate of VAT.
Solution:
Cost of TV including VAT = Rs 13,440
Actual cost = Rs 12,000
Amount of VAT = Rs 13,440 – Rs 12,000 = Rs 1,440
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 7

Question 10.
Reena goes to a shop to buy a radio, costing Rs 2,568. The rate of VAT is 7%. She tells the shopkeeper to reduce the price of the radio such that she has to pay Rs 2,568 inclusive of VAT. Find the reduction needed in the price of radio.
Solution:
Price of radio inclusive of VAT = Rs 2,568
Rate of VAT = 7%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 8
But the cost of radio in the beginning = Rs 2,568
Reduction = Rs 2568 – Rs 2,400 = Rs 168

Question 11.
Rajat goes to a departmental store and buys the following articles :
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 9
Calculate the total amount he has to pay to the store.
Solution:
Price of 2 pairs of shoes @ Rs 800 = Rs800 x 2 = Rs 1,600
Rate of VAT = 5%

Total amount to be paid = Rs 1,680 + Rs 1,590 + Rs 1,352 = Rs 4,622

Question 12.
Ajit buys a motorcycle for Rs 17,600 including value added tax. If the rate of VAT is 10%, what is the sale price of the motorcycle ?
Solution:
Cost price of motorcycle (including VAT) = Rs 17,600
Rate of VAT = 10%
Sale price of motorcycle
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 12

Question 13.
Manoj buys a leather coat costing Rs 900 at Rs 990 after paying the VAT. Calculate the VAT charged on the cost.
Solution:
Cost price of coat = Rs 900
And sale price including VAT = Rs 990
Amount of VAT = Rs 990 – Rs 900 = Rs 90
Rate of VAT = 90×100900 = 10%

Question 14.
Rakesh goes to a departmental store and purchases the following articles:
(i) biscuits and bakery products costing Rs 50, VAT @ 5%.
(ii) medicines costing Rs 90. VAT @ 10%,
(iii) clothes costing Rs 400, VAT @ 1% and
(iv) cosmetics costing Rs 150, VAT @ 10% Calculate the total amount to be paid by Rakesh to the store.
Solution:
(i) Cost of biscuits and bakery product = Rs 50
VAT = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 13
Total amount of the bill = Rs 52.50 + Rs 99 + Rs 404 + Rs 165 = Rs 720.50

Question 15.
Rajeeta purchased a set of cosmetics. She paid Rs 165 for it including VAT. If the rate of VAT is 10%, find the sale price of the set.
Solution:
Total price of set including VAT = Rs 165
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 14

Question 16.
Sunita purchases a bicycle for Rs 660. She has paid a VAT of 10%. Find the list price of bicycle.
Solution:
Cost price of bicycle including VAT = Rs. 660
Rate of VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 15

Question 17.
The sales price of a television inclusive of VAT is Rs 13,500. If VAT is charged at the rate of 8% of the list price, find the list price of the television.
Solution:
Sale price of television including VAT = Rs 13,500
Rate of VAT = 8%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 16

Question 18.
Shikha purchased a car with a marked price of Rs 2,10,000 at a discount of 5%. If VAT is charged at the rate of 10%, find the amount Shikha had paid for purchasing the car.
Solution:
Marked price of car = Rs 2,10,000
Rate of discount = 5%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 18

Question 19.
Sharuti bought a set of cosmetic items for Rs 345 including 15% value added tax and a purse for Rs 110 including 10% VAT. What percent is the VAT charged on the whole transactions ?
Solution:
Cost price of set = Rs 345
VAT = 15%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 19

Question 20.
List price of a cooler is Rs 2,563. The rate of VAT is 10%. The customer requests the shopkeeper to allow a discount in the price of the cooler to such an extent that the price remains Rs 2,563 inclusive of VAT. Find the discount in the price of the cooler.
Solution:
List price of cooler = Rs. 2,563
On request the price of cooler is paid Rs 2,563 including VAT
VAT = 10%
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 21
Amount of discount = Rs 2,563 – Rs 2,330 = Rs 233

Question 21.
List price of a washing machine is Rs 9,000. If the dealer allows a discount of 5% on the cash payment, how much money will a customer pay to the dealer in cash, if the rate of VAT is 10%.
Solution:
List price of washing machine = Rs 9,000
Rate of discount = 5%
Amount after giving discount
RD Sharma Class 8 Solutions Chapter 13 Profits, Loss, Discount and Value Added Tax (VAT) Ex 13.3 22

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CHAPTER – 1 Nutrition in Plants | CLASS 7TH |NCERT SCIENCE IMPORTANT QUESTIONS & MCQS | EDUGROWN

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Get Chapter Wise MCQ Questions for Class 7 Science with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Science MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7Science. Every question of the textbook has been answered here.

Chapter - 1 Nutrition in Plants

MCQs

Question 1.
Which of the following is a nutrient?
(a) Protein
(b) Fat
(c) Vitamin
(d) All of these

Answer

Answer: (d) All of these

Question 2.
Human beings can be categorised as
(a) heterotrophs
(b) autotrophs
(c) parasites
(d) saprotrophs

Answer

Answer: (a) heterotrophs

Question 3.
The food making process in plants is called as
(a) glycolysis
(b) photosynthesis
(c) photolysis
(d) chemosynthesis

Answer

Answer: (b) photosynthesis

Question 4.
Which part of the plant is called its food factory ?
(a) Fruits
(b) Seeds
(c) Leaves
(d) Flowers

Answer

Answer: (c) Leaves

Question 5.
Tiny pores present on the surface of leaves through which gaseous exchange occurs are called
(a) stomata
(b) guard cells
(c) food holes
(d) gas holes

Answer

Answer: (a) stomata

Question 6.
What is the ultimate source of energy for all living organisms?
(a) Water energy
(b) Wind energy
(c) Solar energy
(d) Chemical energy

Answer

Answer: (c) Solar energy

Question 7.
Green pigment present in the leaves is called
(a) haemoglobin
(b) globulin
(c) albumin
(d) chlorophyll

Answer

Answer: (d) chlorophyll

Question 8.
During photosynthesis plants
(a) take oxygen and release carbon dioxide
(b) take carbon dioxide and release oxygen
(c) take carbon dioxide but do not release oxygen
(d) take oxygen but do not release carbon dioxide

Answer

Answer: (b) take carbon dioxide and release oxygen

Question 9.
During photosynthesis
(а) solar energy is converted into chemical energy
(b) solar energy is converted into mechanical energy
(c) chemical energy is converted into mechanical energy
(d) bioenergy is converted into chemical energy

Answer

Answer: (а) solar energy is converted into chemical energy

Question 10.
The raw materials for photosynthesis are
(а) CO₂
(b) CO₂, O₂, H₂
(c) N₂ water
(d) O₂ water

Answer

Answer: (а) CO₂

Question 11.
The end products of photosynthesis are
(a) carbohydrates, oxygen
(b) carbohydrates, hydrogen
(c) carbohydrates, water vapours
(d) carbohydrates, oxygen and water vapours

Answer

Answer: (a) carbohydrates, oxygen

Question 12.
Which one of the following is a parasite?
(a) Lichen
(b) Cuscuta
(c) Pitcher plant
(d) Rhizobium

Answer

Answer: (b) Cuscuta

Question 13.
Which of the following class of organisms belongs to saprotrophs?
(a) Fungi
(b) Algae
(c) Lichens
(d) Bryophytes

Answer

Answer: (a) Fungi

Question 14.
Which one of the following is a pair of symbiotic organisms?
(a) Lichens
(b) Rhizobium and a legume
(c) None of these
(d) Both (a) and (b)

Answer

Answer: (d) Both (a) and (b)

Question 15.
Which of the following is an insectivorous plant?
(a) Pitcher plant
(b) Indian telegraph plant
(c) 4 ‘O’clock plant
(d) Cuscuta

Answer

Answer: (a) Pitcher plant

Match the following:

Column A	Column B
(i) Autotrophs	(a) Fungi
(ii) Heterotroph	(b) Lichen
(iii) Parasite	(c) Pitcher plant
(iv) Saproptroph	(d) Algae
(v) Symbiont	(e) Man
(vi) Insectivorous	(f) Cuscuta

Answer

Answer:

Column A	Column B
(i) Autotrophs	(d) Algae
(ii) Heterotroph	(e) Man
(iii) Parasite	(f) Cuscuta
(iv) Saproptroph	(a) Fungi
(v) Symbiont	(b) Lichen
(vi) Insectivorous	(c) Pitcher plant

Fill in the blanks:

1. All organisms take ……………… and utilize it to get energy for the growth and the maintenance of their bodies.

Answer

Answer: food

2. Green plants synthesise their food themselves by the process of ………….. they are called ……………..

Answer

Answer: photosynthesis, autotrophs

3. …………………. energy is stored by the leaves with the help of chlorophyll.

Answer

Answer: Solar

4. …………………. derive nutrition from, dead, decaying matter.

Answer

Answer: Fungi/s aprotrophs

5. Plants like cuscuta take food from ………………… plant.

Answer

Answer: host

6. All animals are categorised as ……………..

Answer

Answer: heterotrophs

7. …………….. is produced and …………….. is utilized during photosynthesis.

Answer

Answer: Oxygen, carbon dioxide

8. …………….. is the site of reception of light energy in leaves.

Answer

Answer: Chlorophyll

Choose the true and false statements from the following:

1. Food is essential for all living organisms.

Answer

Answer: True

2. Leaves are the food factories of plant.

Answer

Answer: True

3. Water comes into leaves through stomata in the form of vapours.

Answer

Answer: False

4. Plants utilize the carbon dioxide dissolved in the water absorbed by the roots for photosynthesis.

Answer

Answer: False

Important Questions

Question 1.
Potato and ginger are both underground parts that store food. Where is the food prepared in these plants? [NCERT Exemplar]
Anwer:
In both the plants, shoot system and leaves are above ground. They prepare food through photosynthesis and transport it to the underground part for storage.

Question 2.
Plants prepare their food using a different mode of nutrition than us. What is it?
Answer:
The mode of nutrition in plant is autotrophic, i.e. they synthesise their own food.

Question 3.
Photosynthesis requires chlorophyll and a few other raw materials. Add the missing raw materials to the list given below:
Water, minerals, (a) …… (b) …….
Answer:
(a) Sunlight
(b) Carbon dioxide

Question 4.
The tiny openings present on the leaf surface. What are they called?
Answer:
Stomata are the tiny pores present on the surface of leaves through which gaseous exchange takes place in plants.

Question 5.
What is the function of guard cells of stomata?
Answer:
Guard cells help in controlling the opening and closing of stomata for gaseous exchange.

Question 6.
Which parts of the plant are called food factories of the plant?
Answer:
Leaves are referred to as food factories of plants. This is because, leaves synthesise food by the process of photosynthesis.

Question 7.
A carbohydrate is produced by plants as food source. It is constituted from which molecules?
Answer:
Carbohydrates are composed of carbon, hydrogen and oxygen.

Question 8.
Why do some plants feed on insects?
Answer:
Insectivorous plants grow in soil which lack nitrogen, therefore they eat insects to fulfill their need of nitrogen.

Question 9.
Define parasites.
Answer:
Parasites they are those organisms which grow on other plants or animals for their food, e.g. Cuscuta.

Question 10.
Name the bacteria that can fix atmospheric nitrogen.
Answer:
Rhizobium is the bacterium which can fix atmospheric nitrogen.

Question 11.
Except plants, why can’t other living organisms prepare their food using CO₂, water and minerals? [HOTS]
Answer:
Our body does not contain chlorophyll for absorbing solar energy which is necessary for preparing food using air, water, etc.

Question 12.
A leguminous plant can restore the soil’s concentration of mineral nutrients. Can you give examples of some such plants?
Answer:
Plants such as gram, pulses and beans are leguminous.

Question 13.
Algae are green in colour. Why?
Answer:
Algae contain chlorophyll which imparts green colour to them.

Question 14.
what do you understand by nutrition?
Answer:
The process of utilising nutrients like carbohydrates, proteins, fats, etc., to generate energy is called nutrition.

Question 15.
Fungus can be harmful and useful. Give an example showing both of these traits of fungus.
Answer:
Fungus produces antibiotics like penicillin used to treat diseases and fungus can also harm us by causing fungal infections on skin and hair.

Question 16.
A unique feature in leaves allows them to prepare the food while other parts of plants cannot. Write the possible reason for this. [HOTS]
Answer:
Leaves contain chlorophyll which is essential for food preparation and is absent in other parts of plant.

Question 17.
Algae and fungi form a unique association sharing benefits from each other. What is the name of association between them?
Answer:
Lichens.

Question 18.
In a plant, photosynthesis occurs in a part other than leaf. Name that plant and the part where photosynthesis occurs.
Answer:
Cactus, the part where photosynthesis occurs are stem and branches which are green.

Question 19.
Why is Cuscuta, categorised as a parasite?
Answer:
Cuscuta derives its nutrition using an association where it deprives its host of all valuable nutrients and absorbs them itself. Hence, it is called a parasitic plant.

Question 20.
Plant cannot use the nitrogen present in the soil directly. Why?
Answer:
Plants can use nitrogen only in soluble form while in soil nitrogen is present in inorganic form.

Question 21.
Why are insectivorous plants called partial heterotrophs?
Answer:
Insectivorous plants are autotrophs, i.e. they prepare their own food. They are partial heterotrophs as they eat insects for obtaining nitrogen.

Question 22.
What is the stored food form in sunflower seeds?
Answer:
In sunflower seeds, glucose is stored in the form of oils (fats).

Question 23.
What do you understand by saprotrophic mode of nutrition?
Answer:
The mode of nutrition in which organisms take their nutrients from dead and decaying matter is called saprotrophic mode of nutrition.

Question 24.
A mutually beneficial relationship that occurs between two plants. It is known by what name? Give an example.
Answer:
Symbiosis is the mutually benefitting association between two plants, e.g. lichens.

Question 25.
For testing the presence of starch in leaves, a boiled leaf is used. Why?
Answer:
Boiling the leaf remove chlorophyll/green colour from the leaves.

Question 26.
Mosquitoes, bed bugs, lice and leeches suck our blood. Can they be called as parasites? [HOTS]
Answer:
Yes, these animals/insects are parasites as they harm the hosts while they suck blood.

Question 27.
Insectivorous plants have one or the other specialised organs to catch their prey. What is that organ?
Answer:
Leaves of insectivorous plants catches the prey.

Question 28.
Farmers spread manure of fertilisers in the field or in gardens, etc. Why are these added to the soil?
Answer:
Plants absorb mineral nutrients from soil. Thus, declining their concentration in soil fertilisers and manures enhance or add these essential nutrients back in soil.

Question 29.
A cell is formed of many sub-components. Identify different constituents of the cell. Are animal and plant cells similar?
Answer:
A cell contains nucleus, cytoplasm, vacuole, cell organelles like chloroplast, mitochondria, etc. No, animal cells are different from plant cells.

Question 30.
A goat eats away all the leaves of a small plant (balsam). However, in a few days, new leaves could be seen sprouting in the plant again. How did the plant survive without leaves? [NCERT Exemplar; HOTS]
Answer:
The plant of balsam survived on the food stored in the stem and roots.

Nutrition in Plants Class 7 Science Extra Questions Short Answer Type

Question 1.
Different modes of nutrition has been observed in plants. What are they? Give example of each.
Answer:
Plants show two major modes of nutrition, i.e.
(i) Autotrophs are those which can synthesise their own food.
(ii) Heterotrophs are those which are dependent on other plants and animals for their food. They are of following types:
(a) Parasites, e.g. Cuscuta
(b) Saprotrophs, e.g. fungi.

Question 2.
Sunlight, chlorophyll, carbon dioxide, water and minerals are raw materials essential for photosynthesis. Do you know where they are available? Fill in the blanks with the appropriate raw materials.
(a) Available in the plant: ………
(b) Available in the soil: ………
(c) Available in the air: ………
(d) Available during day : ……… [NCERT Exemplar]
Answer:
(a) Available in the plant: chlorophyll
(b) Available in the soil : water, minerals
(c) Available in the air : carbon dioxide
(d) Available during day : sunlight

Question 3.
Plants are considered an essential part of earth as they keep a check on lot of process occurring all over. What would happen if all the green plants are wiped from earth? [HOTS]
Answer:
Green plants are the source of energy for all the living organisms so that they can perform their normal functions. If all green plants and trees disappear, all the organism depending on them for food and shelter will also die.

The lack of gaseous exchange will lead to increase in amount of CO₂, causing death in humans and other animals also. The cycle of life will gradually disappear.

Question 4.
Autotrophs and heterotrophs are two different organisms with distinct modes of nutrition state. How are they different from each other?
Answer:
The difference between autotrophs and heterotrophs are as follows:

Autotrophs	Heterotrophs
They can prepare their own food.	They cannot prepare their own food.
Autotrophs take simple inorganic substances and change it into complex organic food, e.g. green plants.	They take in complex food and breakdown it into simple compounds, e.g. all animals, fungi and non-green plants.

Question 5.
Wheat dough if left in the open, after a few days, starts to emit a foul smell and becomes unfit for use. Give reason. [NCERT Exemplar; HOTS]
Answer:
Carbohydrates in wheat dough encourage the growth of yeast and other saprophytic fungi which breakdown carbohydrates into simpler compounds like CO₂ and alcohol and emit a foul smell.

Question 6.
What are the various raw materials for photosynthesis?
Answer:
Plants utilise carbon dioxide from air and water and minerals are derived from soil (through roots) as raw material for photosynthesis. Besides these chlorophyll present in green leaf is necessary for the process and sunlight is the source of energy which is converted into chemical energy during the process of photosynthesis.

Question 7.
Observe the given figure and label the following terms given in the box. Stomatal opening, guard cell

Answer:
Labelled figure is given below:

Question 8.
Nitrogen is an essential nutrient for plants growth. But farmers who cultivate pulses as crops like green gram, bengal gram, black gram, etc., do not apply nitrogenous fertilisers during t cultivation. Why? [NCERT Exemplar; HOTS]
Answer:
Roots of pulses (leguminous plants) have a symbiotic association with a bacterium called Rhizobium. This bacteria convert gaseous nitrogen of air into water soluble nitrogen compounds and give them to the leguminous plants for their growth. Hence, farmers need not use nitrogenous fertilisers.

Question 9.
Pooja is worried about her new shoes which she wore on special occassions that they were spoiled by fungus during rainy season. Is she right to worry, if yes, then tell why does fungi suddenly appears during the rainy season? [HOTS]
Answer:
Yes, the fungi reproduces by spores which are generally present in the air and grow on any article that are left in hot and humid weather for a long time. During rainy season they land on wet and warm things and begin to germinate and grow.

Question 10.
In what unique manner does a pitcher plant derive its nutrition?
Answer:
Nepenthes or pitcher plant modifies its leaf axis into a long tubular pitcher to form a pitfall trap. Inside the pitcher sticky liquid is present. When any insect comes in contact with the leaf, the lid present on it is closed and insect is trapped. The liquid contains digestive enzymes which slowly digest the trapped insects.

Question 11.
Water and minerals are absorbed by the roots and then transported to leaves. How?
Answer:
Water and minerals are transported to the leaves by the vessels which run like pipes throughout the root, stem, branches and the leaves. These vessels are xylem and phloem, forming a continuous path or passage for the nutrients to make them reach the leaf.

Question 12.
Some plants have deep red, violet or brown coloured leaves. Can these leaves perform the photosynthesis process? [HOTS]
Answer:
Yes, plants having deep red, violet or brown coloured leaves can also carry out photosynthesis because they contain chlorophyll. But their green colour of chlorophyll is masked by the large amount of all other coloured pigments.

Question 13.
If plant has a requirement for nitrogen, then from where will they obtain it?
Answer:
Soil contains nitrogen in the form that is not usable by plants. Bacteria like Rhizobium converts nitrogen into soluble form that can be easily used by plants. So, if plant has a requirement for nitrogen, then it will obtain that which the help of bacteria.

Question 14.
In the absence of photosynthesis, life would be impossible on earth. Is it true or false?
Answer:
True, because photosynthesis is important for the existence of life on the earth. Photosynthesis is important process as it is provides food to all living organisms and maintains CO₂ – O₂ balance of nature.

Nutrition in Plants Class 7 Science Extra Questions Long Answer Type

Question 1.
Describe the process by which plants prepare their food using different raw materials.
Answer:
The process by which green plants can prepare their own food is called photosynthesis. Green plants possess chlorophyll in their leaf and utilises carbon dioxide (from air) water, minerals (from soil, through root) as raw material and sunlight as source of energy and convert light energy into chemical energy. The food thus synthesised is in the form of starch (carbohydrate). The overall reaction for photosynthesis can be given as follows:

Question 2.
Describe the method for replinishing the soils with minerals and other essential constituents used by plants growing in those soil by farmers.
Answer:
Replenishment of Nutrients in Soil

Crops require a lot of nitrogen to make proteins. After the harvest, the soil becomes deficient in nitrogen. Plants cannot use the nitrogen gas available in atmosphere directly. Action of certain bacteria can convert this nitrogen into form readily used by plants. Rhizobium bacteria live in the root nodules of leguminous plants. These bacteria take nitrogen gas from the atmosphere and convert it into water soluble nitrogen compounds making it available to the leguminous plants for their growth.

In return, leguminous plants provide food and shelter to the bacteria as Rhizobium cannot prepare its food. They, thus have a symbiotic relationship. This association is very important for the farmers, as they do not need to add nitrogen fertilisers to the soil in which leguminous plants are grown.

Question 3.
Harish went to visit his grandfather in village where he saw that his grandfather’s field of wheat are infected with fungus but no one is aware of this. Harish rushed to his grandfather’s side and told him that the field have been infected with fungi. He should use an antifungal agent in his fields to stop this infection.
(a) What is fungus?
(b) Can fungus only cause diseases or can it be helpful also?
(c) What values are shown by Harish? [Value Based Question]
Answer:
(a) Fungus are saprophytic organisms usually present as spores in atmosphere which can germinate on any substrate in optimal conditions.
(b) Fungus are also useful in that they produce many antibiotics which can cure different types of infections like penicillin.
(c) Harish is sincere, curious and knowledgeable with a keen sense of applying it where necessary.

Question 4.
Wild animals like tiger, wolf, lion and leopard do not eat plants. Does this mean that they can survive without plants? Can you provide a suitable explanation? [HOTS]
Answer:
Animals like tiger, wolf, lion and leopard are carnivores and do not eat plants. They hunt and eat herbivorous animals like deer, gaur, bison, zebra, giraffe, etc., which are dependent on plants for food.

If there are no plants, herbivorous animals will not survive and ultimately animals like tiger, wolf, lion and leopard will have nothing to eat.

Question 5.
Asha went to visit her grandfather in his village. He was having a serious discussion with his fellow members regarding the productivity level of crops for present year. They all were worried about how to increase the productivity of crop Asha listened to this and then suggested to the group that the reason may be decreased level of minerals in soil.
She told her grandfather to plant crops like pulses, gram, beans, etc., for a year then follow with regular crops. This will increase the crop productivity?
(a) What will you name the process suggested by Asha? Why is there decrease in crop productivity?
(b) What are noted benefits of this process? Will the results be as what Asha expressed?
(c) What values are shown by Asha? [Value Based Question]
Answer:
(a) This process is known as crop rotation. All the plants/crops grown in soil use the minerals present in soil for their own use. This continuous usage depletes the concentration of mineral in soil.
(b) After growing leguminous plants, the mineral content of soil is restored and enriched to new level. Yes, the benefit of leguminous plant is the re-enrichment of soil minerals.
(c) Asha is observant, sincere and interested in applying her knowledge to situation

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CHAPTER – 15 Visualising Solid Shapes | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Get Chapter Wise MCQ Questions for Class 7 Maths with Answers prepared here according to the latest CBSE syllabus and NCERT curriculum. Students can practice CBSE Class 7 Maths MCQs Multiple Choice Questions with Answers to score good marks in the examination. Students can also visit the most accurate and elaborate NCERT Solutions for Class 7 Maths. Every question of the textbook has been answered here.

Chapter - 15 Visualising Solid Shapes

MCQs

Question 1.
The name of the solid shape is

(a) cone
(b) cylinder
(c) sphere
(d) cube

Answer

Answer: (b) cylinder

Question 2.
The name of the solid shape is

(a) cuboid
(b) cube
(c) pyramid
(d) cone

Answer

Answer: (a) cuboid

Question 3.
The name of the solid shape is

(a) cylinder
(b) cone
(c) sphere
(d) cube

Answer

Answer: (c) sphere

Question 4.
The name of the solid shape is

(a) cube
(b) cylinder
(c) cone
(d) sphere

Answer

Answer: (a) cube

Question 5.
The name of the solid shape is

(a) cylinder
(b) cone
(c) cuboid
(d) sphere

Answer

Answer: (b) cone

Question 6.
The name of the solid shape is

(a) cylinder
(b) cone
(c) sphere
(d) pyramid

Answer

Answer: (d) pyramid

Question 7.
The number of vertices of a cube is
(a) 8
(b) 12
(c) 6
(d) 3

Answer

Answer: (a) 8

Question 8.
The number of edges of a cube is
(a) 8
(b) 12
(c) 6
(d) 3

Answer

Answer: (b) 12

Question 9.
The number of faces of a cube is
(a) 8
(b) 12
(c) 6
(d) 3

Answer

Answer: (c) 6

Question 10.
The number of vertices of the solid shape is

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 11.
The number of faces of the solid shape is

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 12.
The number of edges of the solid shape is

(a) 1
(b) 2
(c) 3
(d) 6

Answer

Answer: (d) 6

Question 13.
The number of vertices of the solid shape is

(a) 9
(b) 4
(c) 6
(d) 8

Answer

Answer: (a) 9

Question 14.
The number of faces of the solid shape is

(a) 4
(b) 6
(c) 9
(d) 8

Answer

Answer: (c) 9

Question 15.
The number of edges of the solid shape is

(a) 16
(b) 9
(c) 6
(d) 4

Answer

Answer: (a) 16

Question 16.
Two cubes of edge length 2 cm are placed side by side. The length of the resulting cuboid is
(a) 2 cm
(b) 4 cm
(c) 1 cm
(d) none of these

Answer

Answer: (b) 4 cm

Important Questions

Question 1.
If three cubes of dimensions 2 cm × 2 cm × 2 cm are placed end to end, what would be the dimension of the resulting cuboid?
Solution:
Length of the resulting cuboid = 2 cm + 2 cm + 2 cm = 6 cm
Breadth = 2 cm
Height = 2 cm

Hence the required dimensions = 6 cm × 2 cm × 2 cm.

Question 2.
Answer the following:
(i) Why a cone is not a pyramid?
(ii) How many dimension a solid have?
(iii) Name the solid having one curved and two flat faces but no vertex.
Solution:
(i) Cone is not a pyramid because its base is not a polygon.
(ii) Three.
(iii) Cylinder

Question 3.
Write down the number of edges on each of the following solid figures:
(i) Cube
(ii) Tetrahedron
(iii) Sphere
(iv) Triangular prism
Solution:
(i) 12
(ii) 6
(iii) 0
(iv) 9

Question 4.
What cross-section do you get when you give a horizontal cut to an ice cream cone?
Solution:
Circle.

Question 5.
Determine the number of edges, vertices and faces in the given figure.

Solution:
Edges = 8
Vertices = 5
Faces = 5

Question 6.
Draw the sketch of two figure that has no edge.
Solution:

Question 7.
Draw the sketches of two figures that have no vertex.
Solution:

Question 8.
Name any three objects which resemble a sphere and cone.
Solution:
Sphere: Football, Earth, Round table
Cone: Conical funnel, ice cream cone, conical cracker

Question 9.
What shape would we get from the given figure?

Solution:
From the given net, we get a rectangular pyramid.

Visualising Solid Shapes Class 7 Extra Questions Short Answer Type

Question 10.
For the solids given below sketch the front, side and top view

Solution:

Question 11.
Match the following:

Solution:
(i) → (e)
(ii) → (a)
(iii) → (b)
(iv) → (c)
(v) → (d)

Question 12.
Complete the following table:

Solution:

Question 13.
Draw a plan, front and side elevations of the following solids.

Solution:

Question 14.
Name the solid that would be formed by each net:

Solution:
(i) Triangular pyramid
(ii) Square pyramid
(iii) Hexagonal pyramid

Question 15.
Name the solids that have:
(i) 1 curved surface
(ii) 4 faces
(iii) 6 faces
(iv) 5 faces and 5 vertices
(v) 8 triangular faces
(vi) 6 triangular faces and 2 hexagonal faces.
Solution:
(i) Cylinder
(ii) Tetrahedron
(iii) Cube and cuboid
(iv) Square pyramid or rectangular pyramid
(v) Regular octahedron
(vi) Hexagonal prism.

Question 16.
Draw the top, side and front views of the given solids:

Solution:

Question 17.
Draw the net of a cuboid having same breadth and height, but length double the breadth.
Solution:

Question 18.
Draw the nets of the following:
(i) Triangular prisms
(ii) Tetrahedron
(iii) Cuboid.
Solution:

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CHAPTER – 14 Symmetry | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 14 Symmetry

MCQs

Question 1.
How many lines of symmetry are there in an equilateral triangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 2.
How many lines of symmetry are there in a square?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 3.
How many lines of symmetry are there in a rectangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 4.
How many lines of symmetry are there in a regular pentagon?
(a) 1
(b) 2
(c) 3
(d) 5

Answer

Answer: (d) 5

Question 5.
How many lines of symmetry are there in a regular hexagon?
(a) 2
(b) 4
(c) 6
(d) 3

Answer

Answer: (c) 6

Question 6.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 7.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 8.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 9.
How many lines of symmetry are there in the following figure?

(a) 4
(b) 3
(c) 2
(d) 1

Answer

Answer: (d) 1

Question 10.
How many lines of symmetry are there in the following figure?

(a) 2
(b) 1
(c) 4
(d) 3

Answer

Answer: (b) 1

Question 11.
How many lines of symmetry are there in following figure?

(a) 1
(b) 2
(c) 3
(d) None of these

Answer

Answer: (a) 1

Question 12.
How many lines of symmetry are there in the following figure?

(a) 1
(b) 2
(c) 3
(d) Infinitely many

Answer

Answer: (d) Infinitely many

Question 13.
How many lines of symmetry are there in an isosceles triangle?
(a) 4
(b) 3
(c) 1
(d) 2

Answer

Answer: (c) 1

Question 14.
How many lines of symmetry are there in a scalene triangle?
(a) 1
(b) 0
(c) 2
(d) 4

Answer

Answer: (b) 0

Question 15.
How many lines of symmetry are there in a rhombus?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 16.
How many lines of symmetry are there in a parallelogram?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer

Answer: (a) 0

Question 17.
How many lines of symmetry are there in a quadrilateral?
(a) 0
(b) 2
(c) 4
(d) None of these

Answer

Answer: (a) 0

Question 18.
The order of rotational symmetry of an equilateral triangle is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 19.
The order of rotational symmetry of a square is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 20.
What is the order of the rotational symmetry of the following figure?

(a) 4
(b) 3
(c) 2
(d) 1

Answer

Answer: (a) 4

Important Questions

Question 1.
Draw any two English alphabets having an only a vertical line of symmetry.
Solution:

Question 2.
Draw any two English alphabets having a horizontal line of symmetry.
Solution:

Question 3.
Draw any two English alphabets having both horizontal and vertical line of symmetry.
Solution:

Question 4.
Dray any two English alphabets which has no line of symmetry.
Solution:

Question 5.
Draw any two figures which have the order of rotational symmetry 4.
Solution:

Question 6.
Draw any two figures which have the order of rotational symmetry 2.
Solution:

Question 7.
State the order of rotational symmetry of the following figures.

Solution:
(i) Order of equilateral triangle = 3
(ii) Order of regular pentagon = 5.

Question 8.
Draw a figure having an infinite number of lines of symmetry.
Solution:

A circle has the infinite number of lines of symmetry.

Question 9.
Draw any two figure having no lines of symmetry.
Solution:
English alphabet R and P have no lines of symmetry.

Question 10.
State the English alphabet which has only the horizontal line of symmetry.
Solution:
B, C, D, E, H, I, O, X and K are the English alphabets having an only horizontal line of symmetry.

Symmetry Class 7 Extra Questions Short Answer Type

Question 11.
Give the order of rotational symmetry of each of the following figures:

Solution:
(a) Order of rotational symmetry = 4
(b) Order of rotational symmetry = 5
(c) Order of rotational symmetry = 3
(d) Order of rotational symmetry = 6
(e) Order of rotational symmetry = 3
(f) Order of rotational symmetry = 4

Question 12.
How many lines of symmetry do the following have:
(a) a parallelogram
(b) an equilateral triangle
(c) a right angle with equal legs
(d) an angle with equal arms
(e) a semicircle
(f) a rhombus
(g) a square
(h) scalene triangle
Solution:

Question 13.
What letters of the English alphabet have reflectional symmetry about
(а) a vertical mirror
(b) a horizontal mirror
(c) both horizontal and vertical mirrors
Solution:
(a) The letters of English alphabet having reflectional symmetry about a vertical mirror are:
A, H, I, M, O, T, U, V, W, X, Y
(b) Letters of the English alphabets having reflectional symmetry about a horizontal mirror are:
B, C, D, E, H, I, O, X
(c) Letters of the English alphabet having reflectional of symmetry about vertical and horizontal mirror are:
O, X, I, H.

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CHAPTER – 13 Exponents and Powers | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 13 Exponents and Powers

MCQs

Question 1.
The exponential form of 10000 is
(a) 10³
(b) 10⁴
(c) 10⁵
(d) none of these

Answer

Answer: (b) 10⁴
Hint:
10000 = 10 × 10 × 10 × 10 = 10⁴

Question 2.
The exponential form of 100000 is
(a) 10³
(b) 10⁴
(c) 10⁵
(d) none of these

Answer

Answer: (c) 10⁵
Hint:
100000 = 10 × 10 × 10 × 10 × 10 = 10⁵

Question 3.
The exponential form of 81 is
(a) 3⁴
(b) 3³
(c) 3²
(d) none of these

Answer

Answer: (a) 3⁴
Hint:
81 = 3 × 3 × 3 × 3 = 3⁴

Question 4.
The exponential form of 125 is
(a) 5⁴
(b) 5³
(c) 5²
(d) none of these

Answer

Answer: (b) 5³
Hint:
125 = 5 × 5 × 5 = 5³

Question 5.
The exponential form of 32 is
(a) 2³
(b) 2⁴
(c) 2⁵
(d) none of these

Answer

Answer: (c) 2⁵
Hint:
32 = 2 × 2 × 2 × 2 × 2 = 2⁵

Question 6.
The exponential form of 243 is
(a) 3⁵
(b) 3⁴
(c) 3³
(d) 3²

Answer

Answer: (a) 3⁵
Hint:
243 = 3 × 3 × 3 × 3 × 3 = 3⁵

Question 7.
The exponential form of 64 is
(a) 2⁵
(b) 2⁶
(c) 2⁷
(d) 2⁸

Answer

Answer: (b) 2⁶
Hint:
64 = 2 × 2 × 2 × 2 × 2 × 2 = 2⁶

Question 8.
The exponential form of 625 is
(a) 5²
(b) 5³
(c) 5⁴
(d) 5⁵

Answer

Answer: (c) 5⁴
Hint:
625 = 5 × 5 × 5 × 5 = 5⁴

Question 9.
The exponential form of 1000 is
(a) 10¹
(b) 10²
(c) 10³
(d) 10⁴

Answer

Answer: (c) 10³
Hint:
1000 = 10 × 10 × 10 = 10³

Question 10.
The value of (- 2)³ is
(a) 8
(b) -8
(c) 16
(d) -16

Answer

Answer: (b) -8
Hint:
(-2)³ = (-2) × (-2) × (-2) = -8

Question 11.
The value of (- 2)⁴ is
(a) 8
(b) -8
(c) 16
(d) -16

Answer

Answer: (c) 16
Hint:
(-2)⁴ = (-2) × (-2) × (-2) × (-2) = 16

Question 12.
What is the base in 8²?
(a) 8
(b) 2
(c) 6
(d) 10

Answer

Answer: (a) 8

Question 13.
What is the exponent in 8²?
(a) 8
(b) 2
(c) 16
(d) 6

Answer

Answer: (b) 2

Question 14.
(-1)^{even number} =
(a) -1
(b) 1
(c) 0
(d) none of these

Answer

Answer: (b) 1

Question 15.
(-1)^{odd number} =
(a) -1
(b) 1
(c) 0
(d) none of these

Answer

Answer: (a) -1

Question 16.
0 × 10⁴ =
(a) 0
(b) 10⁴
(c) 1
(d) none of these

Answer

Answer: (a) 0

Question 17.
If 2³ × 2⁴ = 2^?, then ? =
(a) 3
(b) 4
(c) 1
(d) 7

Answer

Answer: (d) 7
Hint:
2³ × 2⁴ = 2³⁺⁴ = 2⁷

Important Questions

Question 1.
Express 343 as a power of 7.
Solution:
We have 343 = 7 × 7 × 7 = 7³
Thus, 343 = 7³

Question 2.
Which is greater 3² or 2³?
Solution:
We have 3² = 3 × 3 = 9
2³ = 2 × 2 × 2 = 8
Since 9 > 8
Thus, 3² > 2³

Question 3.
Express the following number as a powers of prime factors:
(i) 144
(ii) 225
Solution:
(i) We have
144 = 2 × 2 × 2 × 2 × 3 × 3 = 2⁴ × 3²
Thus, 144 = 2⁴ × 3²

(ii) We have
225 = 3 × 3 × 5 × 5 = 3² × 5²
Thus, 225 = 3² × 5²

Question 4.
Find the value of:
(i) (-1)¹⁰⁰⁰
(ii) (1)²⁵⁰
(iii) (-1)¹²¹
(iv) (10000)⁰
Solution:
(i) (-1)¹⁰⁰⁰ = 1 [∵ (-1)^{even number} = 1]
(ii) (1)²⁵⁰ = 1 [∵ (1)^{even number} = 1]
(iii) (-1)¹²¹ = -1 [∵ (-1)^{odd number} = -1]
(iv) (10000)⁰ = 1 [∵ a⁰ = 1]

Question 5.
Express the following in exponential form:
(i) 5 × 5 × 5 × 5 × 5
(ii) 4 × 4 × 4 × 5 × 5 × 5
(iii) (-1) × (-1) × (-1) × (-1) × (-1)
(iv) a × a × a × b × c × c × c × d × d
Solution:
(i) 5 × 5 × 5 × 5 × 5 = (5)⁵
(ii) 4 × 4 × 4 × 5 × 5 × 5 = 4³ × 5³
(iii) (-1) × (-1) × (-1) × (-1) × (-1) = (-1)⁵
(iv) a × a × a × b × c × c × c × d × d = a³b¹c³d²

Question 6.
Express each of the following as product of powers of their prime factors:
(i) 405
(ii) 504
(iii) 500
Solution:
(i) We have
405 = 3 × 3 × 3 × 3 × 5 = 3⁴ × 5¹
Thus, 405 = 3⁴ × 5¹

(ii) We have
504 = 2 × 2 × 2 × 3 × 3 × 7 = 2³ × 3² × 7¹
Thus, 504 = 2³ × 3² × 7¹

(iii) We have
500 = 2 × 2 × 5 × 5 × 5 = 2² × 5³
Thus, 500 = 2² × 5³

Question 7.
Simplify the following and write in exponential form:
(i) (5²)³
(ii) (2³)³
(iii) (a^b)^c
(iv) [(5)²]²
Solution:
(i) (5²)³ = 5^2×3 = 5⁶
(ii) (2³)³ = 2^3×3 = 2⁹
(iii) (a^b)^c = a^b×c = a^bc
(iv) [(5)²]² = 5^2×2 = 5⁴

Question 8.
Verify the following:

Solution:

Question 9.
Simplify:

Solution:

Question 10.
Simplify and write in exponential form:

Solution:

Exponents and Powers Class 7 Extra Questions Short Answer Type

Question 11.
Express each of the following as a product of prime factors is the exponential form:
(i) 729 × 125
(ii) 384 × 147
Solution:
(i) 729 × 125 = 3 × 3 × 3 × 3 × 3 × 3 × 5 × 5 × 5 = 3⁶ × 5³
Thus, 729 × 125 = 3⁶ × 5³

(ii) 384 × 147 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 7 × 7 = 2⁷ × 3² × 7²
Thus, 384 × 147 = 2⁷ × 3² × 7²

Question 12.
Simplify the following:
(i) 10³ × 9⁰ + 3³ × 2 + 7⁰
(ii) 6³ × 7⁰ + (-3)⁴ – 9⁰
Solution:
(i) 10³ × 9⁰ + 3³ × 2 + 7⁰
= 1000 + 54 + 1
= 1055
(ii) 6³ × 7⁰ + (-3)⁴ – 9⁰
= 216 × 1 + 81 – 1
= 216 + 80
= 296

Question 13.
Write the following in expanded form:
(i) 70,824
(ii) 1,69,835
Solution:
(i) 70,824
= 7 × 10000 + 0 × 1000 + 8 × 100 + 2 × 10 + 4 × 10⁰
= 7 × 10⁴ + 8 × 10² + 2 × 10¹ + 4 × 10⁰
(ii) 1,69,835
= 1 × 100000 + 6 × 10000 + 9 × 1000 + 8 × 100 + 3 × 10 + 5 × 10⁰
= 1 × 10⁵ + 6 × 10⁴ + 9 × 10³ + 8 × 10² + 3 × 10¹ + 5 × 10⁰

Question 14.
Find the number from each of the expanded form:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
(ii) 4 × 10⁷ + 6 × 10³ + 5
Solution:
(i) 7 × 10⁸ + 3 × 10⁵ + 7 × 10² + 6 × 10¹ + 9
= 7 × 100000000 + 3 × 100000 + 7 × 100 + 6 × 10 + 9
= 700000000 + 300000 + 700 + 60 + 9
= 700300769
(ii) 4 × 10⁷ + 6 × 10³ + 5
= 4 × 10000000 + 6 × 1000 + 5
= 40000000 + 6000 + 5
= 40006005

Question 15.
Find the value of k in each of the following:

Solution:

Question 16.
Find the value of
(a) 3⁰ ÷ 4⁰
(b) (8⁰ – 2⁰) ÷ (8⁰ + 2⁰)
(c) (2⁰ + 3⁰ + 4⁰) – (4⁰ – 3⁰ – 2⁰)
Solution:
(a) We have 3⁰ ÷ 4⁰ = 1 ÷ 1 = 1 [∵ a⁰ = 1]
(6) (8⁰ – 2⁰) ÷ (8⁰ + 2⁰) = (1 – 1) ÷ (1 + 1) = 0 ÷ 2 = 0
(c) (2⁰ + 3⁰ + 4⁰) – (4⁰ – 3⁰ – 2⁰)
= (1 + 1 + 1) – (1 – 1 – 1) [∵ a⁰ = 1]
= 3 – 1
= 2

Question 17.
Express the following in standard form:
(i) 8,19,00,000
(ii) 5,94,00,00,00,000
(iii) 6892.25
Solution:
(i) 8,19,00,000 = 8.19 × 10⁷
(ii) 5,94,00,00,00,000 = 5.94 × 10¹¹
(iii) 6892.25 = 6.89225 × 10³

Question 18.
Evaluate:

Solution:

Exponents and Powers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 19.
Find the value of x, if

Solution:

Question 20.

Solution:

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CHAPTER – 12 Algebraic Expressions | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 12 Algebraic Expressions

MCQs

Question 1.
How many terms are there in the expression 2x²y?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (a) 1

Question 2.
How many terms are there in the expression 2y + 5?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (b) 2

Question 3.
How many terms are there in the expression 1.2ab – 2.4b + 3.6a?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 4.
How many terms are there in the expression – 2p³ – 3p² + 4p + 7?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 5.
What is the coefficient of x in the expression 4x + 3y?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d) 4

Question 6.
What is the coefficient of x in the expression 2 – x + y?
(a) 2
(b) 1
(c) -1
(d) None of these

Answer

Answer: (c) -1

Question 7.
What is the coefficient of x in the expression y²x + y?
(a) y²
(b) y
(c) 1
(d) 0

Answer

Answer: (a) y²

Question 8.
What is the coefficient of x in the expression 2z – 3xz?
(a) 3
(b) z
(c) 3z
(d) -3z

Answer

Answer: (d) -3z

Question 9.
What is the coefficient of x in the expression 1 + x + xz?
(a) z
(b) 1 + z
(c) 1
(d) 1 + x

Answer

Answer: (b) 1 + z

Question 10.
What is the coefficient of x in the expression 2x + xy²?
(a) 2 + y²
(b) 2
(c) y²
(d) None of these

Answer

Answer: (a) 2 + y²

Question 11.
What is the coefficient of y² in the expression 4 – xy²?
(a) 4
(b) x
(c) -x
(d) None of these

Answer

Answer: (c) -x

Question 12.
What is the coefficient of y² in the expression 3y² + 4x?
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c) 3

Question 13.
What is the coefficient of y² in the expression 2x²y – 10xy² + 5y²?
(a) 5 – 10x
(b) 5
(c) -10 x
(d) None of these

Answer

Answer: (a) 5 – 10x

Question 14.
What is the coefficient of x in the expression ax³ + bx² + d?
(a) a
(b) b
(c) d
(d) 0

Answer

Answer: (d) 0

Question 15.
What is the coefficient of x² in the expression ax + b?
(a) a
(b) b
(c) a + b
(d) 0

Answer

Answer: (d) 0

Imporatant Questions

Question 1.
Identify in the given expressions, terms which are not constants. Give their numerical coefficients.
(i) 5x – 3
(ii) 11 – 2y²
(iii) 2x – 1
(iv) 4x²y + 3xy² – 5
Solution:

Question 2.
Group the like terms together from the following expressions:
-8x²y, 3x, 4y, −32x , 2x²y, -y
Solution:
Group of like terms are:
(i) -8x²y, 2x²y
(ii) 3x, −32x
(iii) 4y, -y

Question 3.
Identify the pairs of like and unlike terms:
(i) −32x, y
(ii) -x, 3x
(iii) −12y2x, 32xy²
(iv) 1000, -2
Solution:
(i) −32x, y → Unlike Terms
(ii) -x, 3x → Like Terms
(iii) −12y2x, 32xy² → Like Terms
(iv) 1000, -2 → Like Terms

Question 4.
Classify the following into monomials, binomial and trinomials.
(i) -6
(ii) -5 + x
(iii) 32x – y
(iv) 6x² + 5x – 3
(v) z² + 2
Solution:
(i) -6 is monomial
(ii) -5 + x is binomial
(iii) 32x – y is binomial
(iv) 6x² + 5x – 3 is trinomial
(v) z² + z is binomial

Question 5.
Draw the tree diagram for the given expressions:
(i) -3xy + 10
(ii) x² + y²
Solution:

Question 6.
Identify the constant terms in the following expressions:
(i) -3 + 32x
(ii) 32 – 5y + y²
(iii) 3x² + 2y – 1
Solution:
(i) Constant term = -3
(ii) Constant term = 32
(iii) Constant term = -1

Question 7.
Add:
(i) 3x²y, -5x²y, -x²y
(ii) a + b – 3, b + 2a – 1
Solution:
(i) 3x²y, -5x²y, -x²y
= 3x²y + (-5x²y) + (-x²y)
= 3x²y – 5x²y – x²y
= (3 – 5 – 1 )x²y
= -3x²y
(ii) a + b – 3, b + 2a – 1
= (a + b – 3) + (b + 2a – 1)
= a + b – 3 + b + 2a – 1
= a + 2a + b + b – 3 – 1
= 3a + 2b – 4

Question 8.
Subtract 3x² – x from 5x – x².
Solution:
(5x – x²) – (3x² – x)
= 5x – x² – 3x² + x
= 5x + x – x² – 3x²
= 6x – 4x²

Question 9.
Simplify combining the like terms:
(i) a – (a – b) – b – (b – a)
(ii) x² – 3x + y² – x – 2y²
Solution:
(i) a – (a – b) – b – (b – a)
= a – a + b – b – b + a
= (a – a + a) + (b – b – b)
= a – b
(ii) x² – 3x + y² – x – 2y²
= x² + y² – 2y² – 3x – x
= x² – y² – 4x

Algebraic Expressions Class 7 Extra Questions Short Answer Type

Question 10.
Subtract 24xy – 10y – 18x from 30xy + 12y – 14x.
Solution:
(30xy + 12y – 14x) – (24xy – 10y – 18x)
= 30xy + 12y – 14x – 24xy + 10y + 18x
= 30xy – 24xy + 12y + 10y – 14x + 18x
= 6xy + 22y + 4x

Question 11.
From the sum of 2x² + 3xy – 5 and 7 + 2xy – x² subtract 3xy + x² – 2.
Solution:
Sum of the given term is (2x² + 3xy – 5) + (7 + 2xy – x²)
= 2x² + 3xy – 5 + 7 + 2xy – x²
= 2x² – x² + 3xy + 2xy – 5 + 7
= x² + 5xy + 2
Now (x² + 5xy + 2) – (3xy + x² – 2)
= x² + 5xy + 2 – 3xy – x² + 2
= x² – x² + 5xy – 3xy + 2 + 2
= 0 + 2xy + 4
= 2xy + 4

Question 12.
Subtract 3x² – 5y – 2 from 5y – 3x² + xy and find the value of the result if x = 2, y = -1.
Solution:
(5y – 3x² + xy) – (3x² – 5y – 2)
= 5y – 3x² + xy – 3x² + 5y + 2
= -3x² – 3x² + 5y + 5y + xy + 2
= -6x² + 10y + xy + 2
Putting x = 2 and y = -1, we get
-6(2)² + 10(-1) + (2)(-1) + 2
= -6 × 4 – 10 – 2 + 2
= -24 – 10 – 2 + 2
= -34

Question 13.
Simplify the following expressions and then find the numerical values for x = -2.
(i) 3(2x – 4) + x² + 5
(ii) -2(-3x + 5) – 2(x + 4)
Solution:
(i) 3(2x – 4) + x² + 5
= 6x – 12 + x² + 5
= x² + 6x – 7
Putting x = -2, we get
= (-2)² + 6(-2) – 7
= 4 – 12 – 7
= 4 – 19
= -15
(ii) -2(-3x + 5) – 2(x + 4)
= 6x – 10 – 2x – 8
= 6x – 2x – 10 – 8
= 4x – 18
Putting x = -2, we get
= 4(-2) – 18
= -8 – 18
= -26

Question 14.
Find the value of t if the value of 3x² + 5x – 2t equals to 8, when x = -1.
Solution:
3x² + 5x – 2t = 8 at x = -1
⇒ 3(-1)² + 5(-1) – 2t = 8
⇒ 3(1) – 5 – 2t = 8
⇒ 3 – 5 – 2t = 8
⇒ -2 – 2t = 8
⇒ 2t = 8 + 2
⇒ -2t = 10
⇒ t = -5
Hence, the required value of t = -5.

Question 15.
Subtract the sum of -3x³y² + 2x²y³ and -3x²y³ – 5y⁴ from x⁴ + x³y² + x²y³ + y⁴.
Solution:
Sum of the given terms:

Required expression

Question 16.
What should be subtracted from 2x³ – 3x²y + 2xy² + 3y² to get x³ – 2x²y + 3xy² + 4y²? [NCERT Exemplar]
Solution:
We have

Required expression

Question 17.
To what expression must 99x³ – 33x² – 13x – 41 be added to make the sum zero? [NCERT Exemplar]
Solution:
Given expression:
99x³ – 33x² – 13x – 41
Negative of the above expression is
-99x³ + 33x² + 13x + 41
(99x³ – 33x² – 13x – 41) + (-99x³ + 33x² + 13x + 41)
= 99x³ – 33x² – 13x – 41 – 99x³ + 33x² + 13x + 41
= 0
Hence, the required expression is -99x³ + 33x² + 13x + 41

Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 18.
If P = 2x² – 5x + 2, Q = 5x² + 6x – 3 and R = 3x² – x – 1. Find the value of 2P – Q + 3R.
Solution:
2P – Q + 3R = 2(2x² – 5x + 2) – (5x² + 6x – 3) + 3(3x² – x – 1)
= 4x² – 10x + 4 – 5x² – 6x + 3 + 9x² – 3x – 3
= 4x² – 5x² + 9x² – 10x – 6x – 3x + 4 + 3 – 3
= 8x² – 19x + 4
Required expression.

Question 19.
If A = -(2x + 3), B = -3(x – 2) and C = -2x + 7. Find the value of k if (A + B + C) = kx.
Solution:
A + B + C = -(2x + 13) – 3(x – 2) + (-2x + 7)
= -2x – 13 – 3x + 6 – 2x + 7
= -2x – 3x – 2x – 13 + 6 + 7
= -7x
Since A + B + C = kx
-7x = kx
Thus, k = -7

Question 20.
Find the perimeter of the given figure ABCDEF.

Solution:
Required perimeter of the figure
ABCDEF = AB + BC + CD + DE + EF + FA
= (3x – 2y) + (x + 2y) + (x + 2y) + (3x – 2y) + (x + 2y) + (x + 2y)
= 2(3x – 2y) + 4(x + 2y)
= 6x – 4y + 4x + 8y
= 6x + 4x-4y + 8y
= 10x + 4y
Required expression.

Question 21.
Rohan’s mother gave him ₹ 3xy² and his father gave him ₹ 5(xy² + 2). Out of this total money he spent ₹ (10 – 3xy²) on his birthday party. How much money is left with him? [NCERT Exemplar]
Solution:
Money give by Rohan’s mother = ₹ 3xy²
Money given by his father = ₹ 5(xy² + 2)
Total money given to him = ₹ 3xy² + ₹ 5 (xy² + 2)
= ₹ [3xy² + 5(xy² + 2)]
= ₹ (3xy² + 5xy² + 10)
= ₹ (8xy² + 10).
Money spent by him = ₹ (10 – 3xy)²
Money left with him = ₹ (8xy² + 10) – ₹ (10 – 3xy²)
= ₹ (8xy² + 10 – 10 + 3x²y)
= ₹ (11xy²)
Hence, the required money = ₹ 11xy²

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CHAPTER – 11 Perimeter and Area | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 11 Perimeter and Area

MCQs

Question 1.
Perimeter of a square =
(a) side × side
(b) 3 × side
(c) 4 × side
(d) 2 × side

Answer

Answer: (c) 4 × side
Hint:
Formula

Question 2.
Perimeter of a rectangle of length Z and breadth 6 is
(a) l + b
(b) 2 × (l + b)
(c) 3 × (l + b)
(d) l × b

Answer

Answer: (b) 2 × (l + b)
Hint:
Formula

Question 3.
Area of a square =
(a) side × side
(b) 2 × side
(c) 3 × side
(d) 4 × side

Answer

Answer: (a) side × side
Hint:
Formula

Question 4.
Area of a rectangle of length l and breadth b is
(a) l × b
(b) l + b
(c) 2 × (l + b)
(d) 6 × (l + b)

Answer

Answer: (a) l × b
Hint:
Formula

Question 5.
Area of a parallelogram =
(а) base × height
(b) 12 × base × height
(c) 13 × base × height
(d) 14 × base × height

Answer

Answer: (а) base × height
Hint:
Formula

Question 6.
Area of a triangle =
(а) base × height
(b) 12 × base × height
(c) 13 × base × height
(d) 14 × base × height

Answer

Answer: (b) 12 × base × height
Hint:
Formula

Question 7.
The circumference of a circle of radius r is
(a) πr
(b) 2πr
(c) πr²
(d) 14 πr²

Answer

Answer: (b) 2πr
Hint:
Formula

Question 8.
The circumference of a circle of diameter d is
(a) πd
(b) 2πd
(c) 12 πd
(d) πd²

Answer

Answer: (a) πd
Hint:
Formula

Question 9.
If r and d are the radius and diameter of a circle respectively, then
(a) d = 2 r
(b) d = r
(C) d = 12 r
(d) d = r²

Answer

Answer: (a) d = 2 r
Hint:
Formula

Question 10.
The area of a circle of radius r is
(a) πr²
(b) 2πr²
(c) 2πr
(d) 4πr²

Answer

Answer: (a) πr²
Hint:
Formula

Question 11.
The area of a circle of diameter d is
(a) πd²
(b) 2πd²
(c) 14 πd²
(d) 2πd

Answer

Answer: (c) 14 πd²
Hint:
Formula

Question 12.
1 cm² =
(a) 10 mm²
(b) 100 mm²
(c) 1000 mm²
(d) 10000 mm²

Answer

Answer: (b) 100 mm²
Hint:
Formula

Question 13.
1 m² =
(a) 10 cm²
(b) 100 cm²
(c) 1000 cm²
(d) 10000 cm²

Answer

Answer: (d) 10000 cm²
Hint:
Formula

Question 14.
1 hectare =
(a) 10 m²
(b) 100 m²
(c) 1000 m²
(d) 10000 m²

Answer

Answer: (d) 10000 m²
Hint:
Formula

Important Questions

Question 1.
The side of a square is 2.5 cm. Find its perimeter and area.
Solution:
Side of the square = 2.5 cm
Perimeter = 4 × Side = 4 × 2.5 = 10 cm
Area = (side)² = (4)² = 16 cm²

Question 2.
If the perimeter of a square is 24 cm. Find its area.
Solution:
Perimeter of the square = 24 cm
Side of the square = 244 cm = 6 cm
Area of the square = (Side)² = (6)² cm² = 36 cm²

Question 3.
If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find
(i) Perimeter
(ii) Area of the rectangle.
Solution:
Length = 36 cm, Breadth = 24 cm
(i) Perimeter = 2(l + b) = 2(36 + 24) = 2 × 60 = 120 cm
(ii) Area of the rectangle = l × b = 36 cm × 24 cm = 864 cm²

Question 4.
The perimeter of a rectangular field is 240 m. If its length is 90 m, find:
(i) it’s breadth
(ii) it’s area.
Solution:
(i) Perimeter of the rectangular field = 240 m
2(l + b) = 240 m
l + b = 120 m
90 m + b = 120 m
b = 120 m – 90 m = 30 m
So, the breadth = 30 m.
(ii) Area of the rectangular field = l × b = 90 m × 30 m = 2700 m²
So, the required area = 2700 m²

Question 5.
The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of ₹ 2.50 per m².
Solution:
Length = 600 m, Breadth = 400 m
Area of the field = l × b = 600 m × 400 m = 240000 m²
Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000
Hence, the required cost = ₹ 6,00,000.

Question 6.
The perimeter of a circle is 176 cm, find its radius.
Solution:
The perimeter of the circle = 176 cm

Question 7.
The radius of a circle is 3.5 cm, find its circumference and area.
Solution:
Radius = 3.5 cm
Circumference = 2πr

Question 8.
Area of a circle is 154 cm², find its circumference.
Solution:
Area of the circle = 154 cm²

Question 9.
Find the perimeter of the figure given below.

Solution:
Perimeter of the given figure = Circumference of the semicircle + diameter
= πr + 2r
= 227 × 7 + 2 × 7
= 22 + 14
= 36 cm
Hence, the required perimeter = 36 cm.

Question 10.
The length of the diagonal of a square is 50 cm, find the perimeter of the square.

Solution:
Let each side of the square be x cm.
x² + x² = (50)² [Using Pythagoras Theorem]
2x² = 2500
x² = 1250

x = √1250 = 2×5×5×5×5−−−−−−−−−−−−−√
x = 5 × 5 × √2 = 25√2
The side of the square = 25√2 cm
Perimeter of the square = 4 × side = 4 × 25√2 = 100√2 cm

Perimeter and Area Class 7 Extra Questions Short Answer Type

Question 11.
A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?
Solution:
Length of the wire = 176 cm
Side of the square = 176 ÷ 4 cm = 44 cm
Area of the square = (Side)² = (44)² cm² = 1936 cm²
Circumference of the circle = 176 cm

Since 2464 cm² > 1936 cm²
Hence, the circle will have more area.

Question 12.
In the given figure, find the area of the shaded portion.

Solution:
Area of the square = (Side)² = 10 cm × 10 cm = 100 cm²
Area of the circle = πr²
= 227 × 3.5 × 3.5
= 772 cm²
= 38.5 cm²
Area of the shaded portion = 100 cm² – 38.5 cm² = 61.5 cm²

Question 13.
Find the area of the shaded portion in the figure given below.

Solution:
Area of the rectangle = l × b = 14 cm × 14 cm = 196 cm²
Radius of the semicircle = 142 = 7 cm
Area of two equal semicircle = 2 × 12 πr²
= πr²
= 227 × 7 × 7
= 154 cm²
Area of the shaded portion = 196 cm² – 154 cm² = 42 cm²

Question 14.
A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.

Solution:
Length of the rectangular park = 45 m
Breadth of the park = 30 m
Area of the park = l × 6 = 45m × 30m = 1350 m²
Length of the park including the path = 45 m + 2 × 2.5 m = 50 m
Breadth of the park including the path = 30 m + 2 × 2.5 m = 30m + 5m = 35m
Area of the park including the path = 50 m × 35 m = 1750 m²
Area of the path = 1750 m² – 1350 m² = 400 m²
Hence, the required area = 400 m².

Question 15.
In the given figure, calculate:
(а) the area of the whole figure.
(b) the total length of the boundary of the field.

Solution:
Area of the rectangular portions = l × b = 80 cm × 42 cm = 3360 cm²
Area of two semicircles = 2 × 12 πr² = πr²
= 227 × 21 × 21
= 22 × 3 × 21
= 1386 cm²
Total area = 3360 cm² + 1386 cm² = 4746 cm²
Total length of the boundary of field = (2 × 80 + πr + πr) cm
= (160 + 227 × 21 + 227 × 21)
= (160 + 132) cm
= 292 cm
Hence, the required (i) area = 4746 cm² and (ii) length of boundary = 292 cm.

Question 16.
How many times a wheel of radius 28 cm must rotate to cover a distance of 352 m?
(Take π = 227)
Solution:
Radius of the wheel = 28 cm
Circumference = 2πr = 2 × 227 × 28 = 176 cm
Distance to be covered = 352 m or 352 × 100 = 35200 m
Number of rotation made by the wheel to cover the given distance = 35200176 = 200
Hence, the required number of rotations = 200.

Perimeter and Area Class 7 Extra Questions Long Answer Type

Question 17.
A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)

Solution:
Area of the playground = l × b = 160 m × 80 m = 12800 m²
Area left for swings = l × b = 80m × 80m = 6400 m²
Area of the remaining portion = 12800 m² – 6400 m² = 6400 m²
Area of the vertical road = 80 m × 1.5 m = 120 m²
Area of the horizontal road = 80 m × 1.5 m = 120 m²
Area of the common portion = 1.5 × 1.5 = 2.25 m²
Area of the two roads = 120 m² + 120 m² – 2.25 m² = (240 – 2.25) m² = 237.75 m²
Area of the portion to be planted by grass = 6400 m² – 237.75 m² = 6162.25 m²
Hence, the required area = 6162.25 m².

Question 18.
Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)

Solution:
DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm
DB = AC = 13 cm (Diagonal of a rectangle are equal)
In right ∆ADC,
AD² + DC² = AC² (By Pythagoras Theorem)
⇒ (5)² + DC² = (13)²
⇒ 25 + DC² = 169
⇒ DC² = 169 – 25 = 144
⇒ DC = √144 = 12 cm
Perimeter of rectangle ABCD = 2(AD + DC)
= 2(5 cm + 12 cm)
= 2 × 17 cm
= 34 cm

Question 19.
Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?

Solution:
Here, AB = 10 cm
AF = 4 cm
FB = 10 cm – 4 cm = 6 cm
Area of the parallelogram = Base × Height = FB × AD = 6 cm × 6 cm = 36 cm²
Hence, the required area of shaded region = 36 cm².
Area ∆DEF = 12 × b × h
= 12 × AF × AD
= 12 × 4 × 6
= 12 cm²
Area ∆BEC = 12 × b × h
= 12 × GC × BC
= 12 × 4 × 6
= 12 cm²
Area of Rectangle ABCD = l × b = 10 cm × 6 cm = 60 cm²
Remaining area of Rectangle = 60 cm² – 36 cm² = 24 cm²
Required Ratio = 36 : 24 = 3 : 2

Question 20.
A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)

Solution:
Length of the rectangular piece = 6 cm
Breadth = 5 cm
Area of the sheet = l × b = 6 cm × 5 cm = 30 cm²
Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm²
Ratio of areas of two rectangles = 30 cm² : 6 cm² = 5 : 1

Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 21.
In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
(Take π = 227)

Solution:
PQ = 12 AB = 12 × 14 = 7 cm
PQRS is a square with each side 7 cm
Radius of each circle = 72 cm
Area of the quadrants of each circle = 14 × πr²
Area of the four quadrants of all circles

Area of the square PQRS = Side × Side = 7 cm × 7 cm = 49 cm²
Area of the shaded portion = 49 cm² – 38.5 cm² = 10.5 cm²
Hence, the required area = 10.5 cm².

Question 22.
Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.

Solution:
BE = AB – AE
= 12 cm – (AC + CE)
= 12 cm – (2.4 cm + 6 cm)
= 12 cm – 8.4 cm
= 3.6 cm

Area of the polygon AFGBH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH
= 3.6 cm² + 4.32 cm² + 21.6 cm² + 6.48 cm² + 14.4 cm²
= 50.40 cm²
Hence, the required area = 50.40 cm².

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CHAPTER – 10 Practical Geometry | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 10 Practical Geometry

MCQs

Question 1.
In ΔRST, R = 5 cm, and ∠SRT = 45° and ∠RST = 45°. Which criterion can be used to construct ΔRST?
(a) A.S.A. criterion
(b) S.A.S. criterion
(c) S.S.S. criterion
(d) R.H.S. criterion

Answer

Answer: (a) A.S.A. criterion
Hint:

Clearly, from the figure two angles and the included side are given. So, A.S.A. criterion can be used to construct ARST.

Question 2.
Identify the criterion of construction of the equilateral triangle LMN given LM = 6 cm.
(a) S.A.S. criterion
(b) R.H.S. criterion
(c) A.S.A. criterion
(d) S.S.S. criterion

Answer

Answer: (d) S.S.S. criterion
Hint:
Since ALMN is equilateral the measurement of one side is used for the other two sides of the triangle. Hence ALMN can be constructed by S.S.S. criterion.

Question 3.
The idea of equal alternate angles is used to construct which of the following?
(a) A line parallel to a given line
(b) A triangle
(c) A square
(d) Two triangles

Answer

Answer: (a) A line parallel to a given line.

Question 4.
A Given AB = 3 cm, AC = 5 cm,and ∠B = 30°, ΔABC cannot be uniquely constructed, with AC as base, why?
(a) Two sides and included angle are given.
(b) The other two angles are not given.
(c) The vertex B cannot be uniquely located.
(d) The vertex A coincides with the vertex C.

Answer

Answer: (c) The vertex B cannot be uniquely located.

Question 5.
A line panda point X not on it are given. Which of the following is used to draw a line parallel to p through X?
(a) Equal corresponding angles.
(b) Congruent triangles.
(c) Angle sum property of triangles.
(d) Pythagoras’ theorem.

Answer

Answer: (a) Equal corresponding angles.
Hint:
Corresponding angles of parallel lines are equal.

Question 6.
Δ PQR is such that ∠P = ∠Q = ∠R = 60° which of the following is true?
(a) Δ PQR is equilateral.
(b) Δ PQR is acute angled.
(c) Both [a] and [b]
(d) Neither [a] nor [b]

Answer

Answer: (c) Both [a] and [b]
Hint:
In ΔPQR since all the angles are acute, it is acute angled. Also since all the angles are equal, it is equilateral.

Question 7.
Which vertex of ΔABC is right angled if AB¯¯¯¯¯¯¯¯ = 8 cm, AC¯¯¯¯¯¯¯¯ = 6 cm,and BC¯¯¯¯¯¯¯¯ = 10 cm,?
(a) ∠C
(b) ∠A
(c) ∠B
(d) A or C

Answer

Answer: (b) ∠A
Hint:
From the given measurements, BC¯¯¯¯¯¯¯¯ is the hypotenuse. The angle opposite to BC¯¯¯¯¯¯¯¯ is ∠A which is a right angle.

Question 8.
An isosceles triangle is constructed as shown in the figure.

Which of the given statements is incorrect?
(a) PR¯¯¯¯¯¯¯¯ is the hypotenuse of ΔPQR.
(b) ΔPQR is an equilateral triangle.
(c) ΔPQR is a right angled triangle.
(d) If right angled ΔPQR has its equal angles measuring 45° each.

Answer

Answer: (b) ΔPQR is an equilateral triangle.

Question 9.
ΔPQR is constructed with all its angles measuring 60° each. Which of the following is correct?
(a) ΔPQR is an equilateral triangle.
(b) ΔPQR is isosceles triangle.
(c) ΔPQR is a scalene triangle.
(d) ΔPQR is a right angled triangle.

Answer

Answer: (a) ΔPQR is an equilateral triangle.

Question 10.
How many perpendicular lines can be drawn to a line from a point not on it?
(a) 1
(b) 2
(c) 0
(d) Infinite

Answer

Answer: (a) 1
Hint:

As can be seen from the given figure, one and only one perpendicular line can be drawn to a given line from a point not on it.

Question 11.
Identify the false statement.
(a) A triangle with three equal sides is called an equilateral triangle.
(b) A triangle with a right angle is called a right angled triangle.
(c) A triangle with two equal sides is called a scalene triangle.
(d) A right angled triangle has two acute angles and a right angle.

Answer

Answer: (c) A triangle with two equal sides is called a scalene triangle.

Question 12.
ΔPQR is constructed such that PQ = 5 cm, PR = 5 cm and ∠RPQ = 50° Identify the type of triangle constructed.
(a) An isosceles triangle
(b) An acute angled triangle
(c) An obtuse angled triangle
(d) Both [a] and [b]

Answer

Answer: (d) Both [a] and [b]

Question 13.
Which of the following is NOT constructed using a ruler and a set square?
(a) A perpendicular to a line from a point not on it.
(b) A perpendicular bisector of a line segment.
(c) A perpendicular to a line at a point on the line.
(d) A line parallel to a given line through a given point.

Answer

Answer: (b) A perpendicular bisector of a line segment.

Important Questions

Question 1.
State whether the triangle is possible to construct if
(a) In ΔABC, m∠A = 80°, m∠B = 60°, AB = 5.5 cm
(b) In ΔPQR, PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
Solution:
(a) m∠A = 80°, m∠B = 60°
m∠A + m∠B = 80° + 60° = 140° < 180°
So, ΔABC can be possible to construct.
(b) PQ = 5 cm, QR = 3 cm, PR = 8.8 cm
PQ + QR = 5 cm + 3 cm = 8 cm < 8.8 cm
or PQ + QR < PR
So, the ΔPQR can not be constructed.

Question 2.
Draw an equilateral triangle whose each side is 4.5 cm.

Solution:
Steps of construction:
(i) Draw AB = 4.5 cm.
(ii) Draw two arcs with centres A and B and same radius of 4.5 cm to meet each other at C.
(iii) Join CA and CB.
(iv) ΔCAB is the required triangle.

Question 3.
Draw a ΔPQR, in which QR = 3.5 cm, m∠Q = 40°, m∠R = 60°.

Solution:
Steps of construction:
(i) Draw QR = 3.5 cm.
(ii) Draw ∠Q = 40°, ∠R = 60° which meet each other at P.
(iii) ΔPQR is the required triangle.

Question 4.
There are four options, out of which one is correct. Choose the correct one:
(i) A triangle can be constructed with the given measurement.
(a) 1.5 cm, 3.5 cm, 4.5 cm
(b) 6.5 cm, 7.5 cm, 15 cm
(c) 3.2 cm, 2.3 cm, 5.5 cm
(d) 2 cm, 3 cm, 6 cm
(ii) (a) m∠P = 40°, m∠Q = 60°, AQ = 4 cm
(b) m∠B = 90°, m∠C = 120° , AC = 6.5 cm
(c) m∠L = 150°, m∠N = 70°, MN = 3.5 cm
(d) m∠P = 105°, m∠Q = 80°, PQ = 3 cm
Solution:
(i) Option (a) is possible to construct.
1.5 cm + 3.5 cm > 4.5 cm
(ii) Option (a) is correct.
m∠P + m∠Q = 40° + 60° = 100° < 180°

Question 5.
What will be the other angles of a right-angled isosceles triangle?

Solution:
In right angled isosceles triangle ABC, ∠B = 90°
∠A + ∠C = 180° – 90° = 90°
But ∠A = ∠B
∠A = ∠C = 902 = 45°
Hence the required angles are ∠A = ∠C = 45°

Question 6.
What is the measure of an exterior angle of an equilateral triangle?

Solution:
We know that the measure of each interior angle = 60°
Exterior angle = 180° – 60° = 120°

Question 7.
In ΔABC, ∠A = ∠B = 50°. Name the pair of sides which are equal.

Solution:
∠A = ∠B = 50°
AC = BC [∵ Sides opposite to equal angles are equal]
Hence, the required sides are AC and BC.

Question 8.
If one of the other angles of a right-angled triangle is obtuse, whether the triangle is possible to construct.
Solution:
We know that the angles other than right angle of a right-angled triangle are acute angles.
So, such a triangle is not possible to construct.

Question 9.
State whether the given pair of triangles are congruent.

Solution:
Here, AB = PQ = 3.5 cm
AC = PR = 5.2 cm
∠BAC = ∠QPR = 70°
ΔABC = ΔPQR [By SAS rule]

Practical Geometry Class 7 Extra Questions Short Answer Type

Question 10.
Draw a ΔABC in which BC = 5 cm, AB = 4 cm and m∠B = 50°.

Solution:
Steps of construction:
(i) Draw BC = 5 cm.
(ii) Draw ∠B = 50° and cut AB = 4 cm.
(iii) Join AC.
(iv) ΔABC is the required triangle.

Question 11.
Draw ΔPQR in which QR = 5.4 cm, ∠Q = 40° and PR = 6.2 cm.

Solution:
Steps of construction:
(i) Draw QR = 5.4 cm.
(ii) Draw ∠Q = 40°.
(iii) Take R as the centre and with radius 6.2 cm, draw an arc to meet the former angle line at P.
(iv) Join PR.
(v) ΔPQR is the required triangle.

Question 12.
Construct a ΔPQR in which m∠P = 60° and m∠Q = 30°, QR = 4.8 cm.

Solution:
m∠Q = 30°, m∠P = 60°
m∠Q + m∠P + m∠R = 180° (Angle sum property of triangle)
30° + 60° + m∠R = 180°
90° + m∠R = 180°
m∠R = 180° – 90°
m∠R = 90°
Steps of construction:
(i) Draw QR = 4.8 cm.
(ii) Draw ∠Q = 30°.
(iii) Draw ∠R = 90° which meets the former angle line at P.
(iv) ∠P = 180° – (30° + 90°) = 60°
(v) ΔPQR is the required triangle.

Practical Geometry Class 7 Extra Questions Higher Order Thinking Skills [HOTS] Type

Question 13.
Draw an isosceles right-angled triangle whose hypotenuse is 5.8 cm.
Solution:
Right angled triangle is an isosceles triangle
Each of its acute angles = 902 = 45°

Steps of construction:
(i) Draw AB = 5.8 cm.
(ii) Construct ∠A = 45° and ∠B = 45° to meet each other at C.
(iii) ∠C = 180° – (45° + 45°) = 90°
(iv) ΔACB is the required isosceles right angle triangle.

Question 14.
Construct a ΔABC such that AB = 6.5 cm, AC = 5 cm and the altitude AP to BC is 4 cm.

Solution:
Steps of construction:
(i) Draw a line l and take any point P on it.
(ii) Construct a perpendicular to l at P.
(iii) Cut AP = 4 cm.
(iv) Draw two arcs with centre A and radii 6.5 cm and 5 cm to cut the line l at B and C respectively.
(v) Join AB and AC.
(vi) ΔABC is the required triangle.

Question 15.
Construct an equilateral triangle whose altitude is 4.5 cm.

Solution:
Steps of construction:
(i) Draw any line l and take a point D on it.
(ii) Construct a perpendicular to l at D and cut AD = 4.5 cm.
(iii) Draw the angle of 30° at on either side of AD to meet the line l at B and C.
(iv) ΔABC is the required equilateral triangle.

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CHAPTER -9 Rational Numbers | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 9 Rational Numbers

MCQs

Question 1.
The numerator of the rational number 35 is
(a) 3
(b) 5
(c) 2
(d) 8

Answer

Answer: (a) 3

Question 2.
The numerator of the rational number 1100 is
(a) 100
(b) 1
(c) 10
(d) 99

Answer

Answer: (b) 1

Question 3.
The denominator of the rational number 47 is
(a) 7
(b) 4
(c) 3
(d) 11

Answer

Answer: (a) 7

Question 4.
The denominator of the rational number 713 is
(a) 13
(b) 7
(c) 6
(d) 91

Answer

Answer: (a) 13

Question 5.
The numerator of the rational number −34 is
(a) -3
(b) 3
(c) 4
(d) -4

Answer

Answer: (a) -3

Question 6.
The numerator of the rational number −29 is
(a) -2
(b) 2
(c) -9
(d) 9

Answer

Answer: (a) -2

Question 7.
The denominator of the rational number 5−3 is
(a) 5
(b) -3
(c) 3
(d) 8

Answer

Answer: (b) -3

Question 8.
The denominator of the rational number 3−7 is
(a) 7
(b) -7
(c) 3
(d) -3

Answer

Answer: (b) -7

Question 9.
The numerator of the rational number −2−5 is
(a) 2
(b) -2
(c) 5
(d) -5

Answer

Answer: (b) -2

Question 10.
the numerator of the rational number −5−3 is
(a) -5
(b) 5
(c) -3
(d) 3

Answer

Answer: (a) -5

Question 11.
The denominator of the rational number −2−9 is
(a) -2
(b) 2
(c) 9
(d) -9

Answer

Answer: (d) -9

Question 12.
The denominator of the rational number −6−5 is
(a) 6
(b) -6
(c) 5
(d) -5

Answer

Answer: (d) -5

Question 13.
The denominator of the rational number −13−11 is
(a) -13
(b) 13
(c) 11
(d) -11

Answer

Answer: (d) -11

Question 14.
The numerator of the rational number 0 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: (a) 0

Question 15.
The denominator of the rational number 0 is
(a) 0
(b) 1
(c) -1
(d) any non-zero integer

Answer

Answer: (d) -1

Question 16.
The numerator of a rational number 8 is
(a) 2
(b) 4
(c) 6
(d) 8

Answer

Answer: (d) 8

Question 17.
The denominator of the rational number 8 is
(a) 8
(b) 2
(c) 4
(d) 1

Answer

Answer: (d) 1

Question 1.
Find three rational numbers equivalent to each of the following rational numbers.
(i) −25
(ii) 37
Solution:

Question 2.
Reduce the following rational numbers in standard form.
(i) 35−15
(ii) −36−216
Solution:

Question 3.
Represent 32 and −34 on number lines.
Solution:

Question 4.
Which of the following rational numbers is greater?
(i) 34, 12
(ii) −32, −34
Solution:

Question 5.
Find the sum of

Solution:

Question 6.
Subtract:

Solution:

Question 7.
Find the product:

Solution:

Rational Numbers Class 7 Extra Questions Short Answer Type

Question 8.
If the product of two rational numbers is −916 and one of them is −415, find the other number.
Solution:
Let the required rational number be x.

Question 9.
Arrange the following rational numbers in ascending order.

Solution:

Question 10.
Insert five rational numbers between:

Solution:

Rational Numbers Class 7 Extra Questions Long Answer Type

Question 11.
Evaluate the following:

Solution:

Question 12.
Subtract the sum of −56 and -135 from the sum 223 and -625.
Solution:

Question 13.

Solution:
We have

Question 14.
Divide the sum of -21517 and 3534 by their difference.
Solution:

Question 15.
During a festival sale, the cost of an object is ₹ 870 on which 20% is off. The same object is available at other shops for ₹ 975 with a discount of 623 %. Which is a better deal and by how much?
Solution:
The cost of the object = ₹ 870
Discount = 20% of ₹ 870 = 20100 × 870 = ₹ 174
Selling price = ₹ 870 – ₹ 174 = ₹ 696
The same object is available at other shop = ₹ 975

Selling price = ₹ 975 – ₹ 65 = ₹ 910
Since ₹ 910 > ₹ 696
Hence, deal at first shop is better and by ₹ 910 – ₹ 696 = ₹ 214

Rational Numbers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 16.
Simplify:
21.5 ÷ 5 – 15 of (20.5 – 5.5) + 0.5 × 8.5
Solution:
Using BODMAS rule, we have
21.5 ÷ 5 – 15 of (20.5 – 5.5) + 0.5 × 8.5
= 21.5 ÷ 5 – 15 of 15 + 0.5 × 8.5
= 21.5 × 15 – 15 × 15 + 0.5 × 8.5
= 4.3 – 3 + 4.25
= 4.3 + 4.25 – 3
= 8.55 – 3
= 5.55

Question 17.
Simplify:

Solution:
Using BODMAS rule, we have
2.3 – [1.89 – {3.6 – (2.7 – 0.77)}]
= 2.3 – [1.89 – {3.6 – 1.93}]
= 2.3 – [1.89 – 1.67]
= 2.3 – 0.22
= 2.08

Question 18.

Solution:
Using BODMAS rule, we have

Yearly Archives: 2022

Exercise 14.1

Exercise 14.2

Exercise 14.3

Exercise 14.4

Exercise 14.5

Exercise 13.1

Exercise 13.2

Exercise 13.3

SCIENCE IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 1 Nutrition in Plants

MCQs

Nutrition in Plants Class 7 Science Extra Questions Short Answer Type

Nutrition in Plants Class 7 Science Extra Questions Long Answer Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 15 Visualising Solid Shapes

MCQs

Visualising Solid Shapes Class 7 Extra Questions Short Answer Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 14 Symmetry

MCQs

Symmetry Class 7 Extra Questions Short Answer Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 13 Exponents and Powers

MCQs

Exponents and Powers Class 7 Extra Questions Short Answer Type

Exponents and Powers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 12 Algebraic Expressions

MCQs

Algebraic Expressions Class 7 Extra Questions Short Answer Type

Algebraic Expressions Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 11 Perimeter and Area

MCQs

Perimeter and Area Class 7 Extra Questions Short Answer Type

Perimeter and Area Class 7 Extra Questions Long Answer Type

Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 10 Practical Geometry

MCQs

Practical Geometry Class 7 Extra Questions Short Answer Type

Practical Geometry Class 7 Extra Questions Higher Order Thinking Skills [HOTS] Type

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 9 Rational Numbers

MCQs

Rational Numbers Class 7 Extra Questions Short Answer Type

Rational Numbers Class 7 Extra Questions Long Answer Type

Rational Numbers Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

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