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CHAPTER -8 Comparing Quantities | 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 - 8 Comparing Quantities

MCQs

Question 1.
The ratio of Rs 10 to 50 paise is
(a) 20 : 1
(b) 10 : 1
(c) 5 : 1
(d) 1 : 20

Answer

Answer: (a) 20 : 1
Hint:
Rs 10 : 50 paise = 10 × 100 paise: 50 paise
= 1000 : 50 = 20 : 1

Question 2.
The ratio of 6 kg to 400 g is
(a) 10 : 1
(b) 15 : 1
(c) 12 : 1
(d) 6 : 1

Answer

Answer: (b) 15 : 1
Hint:
6 kg : 400 g = 6 × 1000 g : 400 g
= 6000 : 400 = 15 : 1.

Question 3.
The ratio of 3 m to 60 cm is
(a) 4 : 1
(b) 3 : 1
(c) 5 : 1
(d) 2 : 1

Answer

Answer: (c) 5 : 1
Hint:
3 m : 60 cm = 3 × 100 cm : 60 cm
= 300 : 60 = 5 : 1.

Question 4.
The ratio of 15 days to 72 hours is
(a) 2 : 1
(b) 3 : 1
(c) 4 : 1
(d) 5 : 1

Answer

Answer: (d) 5 : 1
Hint:
15 days : 72 hours
= 15 × 24 hours : 72 hours
= 15 × 24 : 72 = 5 : 1.

Question 5.
The ages of father and son are 45 years and 10 years. The ratio of their ages is
(a) 3 : 2
(b) 5 : 2
(c) 9 : 2
(d) 15 : 2

Answer

Answer: (c) 9 : 2
Hint:
45 years : 10 years ⇒ 9 : 2

Question 6.
The cost of 3 envelopes is ₹ 15. Find the cost of 5 envelopes.
(a) ₹ 20
(b) ₹ 25
(c) ₹ 30
(d) ₹ 40

Answer

Answer: (b) ₹ 25
Hint:
3 : 5 = 15 : x
⇒ 35 = 15x ⇒ x = 25

Question 7.
The cost of 7 kg of potatoes is ₹ 42. How many kg of potatoes can be purchased for ₹ 96?
(a) 10 kg
(b) 12 kg
(c) 15 kg
(d) 16 kg

Answer

Answer: (d) 16 kg
Hint:
7. 7 : x = 42 : 96
⇒ 7x = 4296 ⇒ x = 16

Question 8.
The cost of 4 m of cloth is Rs 40. Find the cost of 9 m of cloth.
(a) ₹ 90
(b) ₹ 60
(c) ₹ 50
(d) ₹ 40

Answer

Answer: (a) ₹ 90
Hint:
8. 4 : 9 = 40 : x
⇒ 49 = 40x ⇒ x = 90

Question 9.
The cost of 8 pencils is ₹ 10. Find the cost of 20 pencils.
(a) ₹ 20
(b) ₹ 25
(c) ₹ 24
(d) ₹ 30

Answer

Answer: (b) ₹ 25
Hint:
8 : 20 = 10 : x
⇒ 820 = 10x ⇒ x = 25

Question 10.
A motorcycle goes 120 km in 3 l of petrol. How much petrol will be required to go 600 km?
(a) 10 l
(b) 12 l
(c) 15 l
(d) 20 l

Answer

Answer: (c) 15 l
Hint:
120 : 600 = 3 : x
⇒ 120600 = 3x ⇒ x = 15

Question 11.
14 as per cent is
(a) 20%
(b) 25%
(c) 30%
(d) 12 12%

Answer

Answer: (b) 25%
Hint:
14 = 14 × 100% = 25%

Question 12.
18 as per cent is
(a) 25%
(b) 121%
(c) 20%
(d) 16%

Answer

Answer: (b) 121%
Hint:
18 = 18 × 100% = 1212%

Question 13.
52 as per cent is
(a) 125%
(b) 150%
(c) 200%
(d) 250%

Answer

Answer: (d) 250%
Hint:
52 = 52 × 100% = 250%

Question 14.
0.1 as per cent is
(a) 1%
(b) 10%
(c) 100%
(d) 0.1%

Answer

Answer: (b) 10%
Hint:
0.1 = 110 = 110 × 100% = 10%

Question 15.
0.04 as per cent is
(a) 10%
(b) 20%
(c) 25%
(d) 4%

Answer

Answer: (d) 4%
Hint:
0.04 = 4100 = 4100 × 100% = 4%

Question 16.
22 as per cent is
(a) 19%
(b) 10%
(c) 100%
(d) none of these

Answer

Answer: (c) 100%
Hint:
22 = 22 × 100% = 100%

Question 17.
Out of 40 children in a class, 10 are boys. What is the percentage of boys?
(a) 10%
(b) 40%
(c) 4%
(d) 25%

Answer

Answer: (d) 25%
Hint:
Percentage of boys = 1040 × 100% = 25%

Important Questions

Question 1.
Find the ratio of:
(a) 5 km to 400 m
(b) 2 hours to 160 minutes
Solution:
(a) 5 km = 5 × 1000 = 5000 m
Ratio of 5 km to 400 m
= 5000 m : 400 m
= 25 : 2
Required ratio = 25 : 2
(b) 2 hours = 2 × 60 = 120 minutes
Ratio of 2 hours to 160 minutes
= 120 : 160
= 3 : 4
Required ratio = 3 : 4

Question 2.
State whether the following ratios are equivalent or not?
(a) 2 : 3 and 4 : 5
(b) 1 : 3 and 2 : 6
Solution:
(a) Given ratios = 2 : 3 and 4 : 5

Hence 2 : 3 and 4 : 5 are not equivalent ratios.
(b) Given ratios = 1 : 3 and 2 : 6
LCM of 3 and 6 = 6

Hence, 1 : 3 and 2 : 6 are equivalent ratios.

Question 3.
Express the following ratios in simplest form:
(a) 615 : 213
(b) 42 : 56
Solution:

Question 4.
Compare the following ratios:
3 : 4, 5 : 6 and 3 : 8
Solution:
Given: 3 : 4, 5 : 6 and 3 : 8
or 34 , 56 and 38
LCM of 4, 6 and 8 = 24

Hence, 3 : 8 < 3 : 4 < 5 : 6

Question 5.
State whether the following ratios are proportional or not:
(i) 20 : 45 and 4 : 9
(ii) 9 : 27 and 33 : 11
Solution:
(i) 20 : 45 and 4 : 9
Product of extremes = 20 × 9 = 180
Product of means = 45 × 4 = 180
Here, the product of extremes = Product of means
Hence, the given ratios are in proportion.
(ii) 9 : 27 and 33 : 11
Product of extremes = 9 × 11 = 99
Product of means = 27 × 33 = 891
Here, the product of extremes ≠ Product of means
Hence, the given ratios are not in proportion.

Question 6.
24, 36, x are in continued proportion, find the value of x.
Solution:
Since, 24, 36, x are in continued proportion.
24 : 36 :: 36 : x
⇒ 24 × x = 36 × 36
⇒ x = 54
Hence, the value of x = 54.

Question 7.
Find the mean proportional between 9 and 16.
Solution:
Let x be the mean proportional between 9 and 16.
9 : x :: x : 16
⇒ x × x = 9 × 16
⇒ x² = 144
⇒ x = √144 = 12
Hence, the required mean proportional = 12.

Question 8.
Find:
(i) 36% of 400
(ii) 1623% of 32
Solution:

Question 9.
Find a number whose 614% is 12.
Solution:
Let the required number be x.

Hence, the required number = 192.

Question 10.
What per cent of 40 kg is 440 g?
Solution:
Let x% of 40 kg = 440 g

Hence, the required Percentage = 1.1%

Comparing Quantities Class 7 Extra Questions Short Answer Type

Question 11.
Convert each of the following into the decimal form:
(а) 25.2%
(b) 0.15%
(c) 25%
Solution:

Question 12.
What per cent of
(a) 64 is 148.48?
(b) 75 is 1225?
Solution:

Question 13.
A machine costs ₹ 7500. Its value decreases by 5% every year due to usage. What will be its price after one year?
Solution:
The cost price of the machine = ₹ 7500
Decrease in price = 5%
Decreased price after one year

= 75 × 95
= ₹ 7125
Hence, the required price = ₹ 7125.

Question 14.
What sum of money lent out at 12 per cent p.a. simple interest would produce ₹ 9000 as interest in 2 years?
Solution:
Here, Interest = ₹ 9000
Rate = 12% p.a.
Time = 2 years
Principal = ?

Hence, the required principal amount = ₹ 37500.

Question 15.
Rashmi obtains 480 marks out of 600. Rajan obtains 560 marks out of 700. Whose performance is better?
Solution:
Rashmi obtains 480 marks out of 600
Marks Percentage = 480600 × 100 = 80%
Rajan obtains 560 marks out of 700
Marks Percentage = 560700 × 100 = 80%
Since, both of them obtained the same per cent of marks i.e. 80%.
So, their performance cannot be compared.

Question 16.
₹ 9000 becomes ₹ 18000 at simple interest in 8 years. Find the rate per cent per annum.
Solution:
Here, Principal = ₹ 9000
Amount = ₹ 18000
Interest = Amount – Principal = ₹ 18000 – ₹ 9000 = ₹ 9000

Hence, the required rate of interest = 1212%.

Question 17.
The cost of an object is increased by 12%. If the current cost is ₹ 896, what was its original cost?
Solution:
Here, rate of increase in cost = 12%
Increased Cost = ₹ 896
Original Cost = ?
Let the Original Cost be ₹ x

Hence, the required cost = ₹ 800.

Comparing Quantities Class 7 Extra Questions Long Answer Type

Question 18.
Radhika borrowed ₹ 12000 from her friends. Out of which ₹ 4000 were borrowed at 18% and the remaining at 15% rate of interest per annum. What is the total interest after 3 years? (NCERT Exemplar)
Solution:
Total amount borrowed by Radhika = ₹ 12,000
The amount borrowed by her at 18% p.a. = ₹ 4000

Total interest = ₹ 2160 + ₹ 3600 = ₹ 5760
Hence, the total interest = ₹ 5760.

Question 19.
Bhavya earns ₹ 50,000 per month and spends 80% of it. Due to pay revision, her monthly income increases by 20% but due to price rise, she has to spend 20% more. Find her new savings. (NCERT Exemplar)
Solution:
Monthly income of Bhavya = ₹ 50,000
Money spent by her = 80% of ₹ 50,000
= 80100 × 50,000 = ₹ 40,000
Due to pay revision, income is increased by 20%

So, the new savings = ₹ 60,000 – ₹ 48,000 = ₹ 12,000

Question 20.
The simple interest on a certain sum at 5% per annum for 3 years and 4 years differ by ₹ 82. Find the sum.
Solution:
Let the required sum be ₹ P.
Simple interest for 3 years

Alternate Method
Simple Interest gained from 3rd to 4th year = ₹ 82
Time (4th year – 3rd year) = 1 year

Required sum = ₹ 1640

Comparing Quantities Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 21.
Rajan’s monthly income is 20% more than the monthly income of Sarita. What per cent of Sarita’s income is less than Rajan’s monthly income?
Solution:
Let the monthly income of Sarita be ₹ 100.
Rajan’s monthly income

Now, Sarita’s monthly income is less than the monthly income of Raj an by = ₹ 120 – ₹ 100 = ₹ 20
Per cent of less in Rajan’s monthly income
= 20×100120 = 503% = 1623%
Hence, the required per cent = 1623%

Question 22.
If 10 apples are bought for ₹ 11 and sold at the rate of 11 apples for ₹ 10. Find the overall gain or loss per cent in these transactions.
Solution:
CP of 10 apples = ₹ 11
CP of 1 apple = ₹ 1110
SP of 11 apples = ₹ 10
SP of 1 apple = ₹ 1011

Question 23.
If 25 men can do a work in 36 hours, find the number of men required to do the same work in 108 hours.
Solution:
Let the number of men required to be x.
Men : Hours :: Men : Hours
25 : 36 :: x : 108
Product of extremes = 25 × 108
Product of means = 36 × x
Product of means = Product of extremes
36 × x = 25 × 108
⇒ x = 25 × 3 = 75
Hence, the required number of men = 75.

Question 24.
A machine is sold by A to B at a profit of 10% and then B sold it to C at a profit of 20%. If C paid ₹ 1200 for the machine, what amount was paid by A to purchase the machine?
Solution:
Cost price of machine for C = Selling price of the machine for B = ₹ 1200

Hence, the required cost price = ₹ 9091011 or ₹ 909.09 (approx)

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CHAPTER -7 Congruence of Triangles | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 7 Congruence of Triangles

MCQs

1. ‘Under a given correspondence, two triangles are congruent if the three sides of the one are equal to the three corresponding sides of the other.’
The above is known as
(a) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (a)

2. ‘Under a given correspondence, two triangles are congruent if two sides and the angle included between them in one of the triangles are equal to the corresponding sides and the angle included between them of the other triangle.’
The above is known as
(a) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (b)

3. ‘Under a given correspondence, two triangles are congruent if two angles and the side included between them in one of the triangles are equal to the corresponding angles and the side included between them of the other triangle.’
The above is known as
(а) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (c)

4. ‘Under a given correspondence, two right-angled triangles are congruent if the hypotenuse and a leg of one of the triangles are equal to the hypotenuse and the corresponding leg of the other triangle.’
The above is known as
(а) SSS congruence of two triangles
(b) SAS congruence of two triangles
(c) ASA congruence of two triangles
(d) RHS congruence of two right-angled triangles

Answer

Answer: (d)

5. For two given triangles ABC and PQR, how many matchings are possible?
(a) 2
(b) 4
(c) 6
(d) 3

Answer/Explanation

Answer: (c)
Explanation : ABC ↔ PQR, ABC ↔ PRQ,
ABC ↔ QRP, ABC ↔ QPR,
ABC ↔ RPQ, ABC ↔ RQP.

6. The symbol for congruence is
(a) ≡
(b) ≅
(c) ↔
(d) =

Answer

Answer: (b)

7. The symbol for correspondence is
(a) =
(b) ↔
(c) ≡
(d) ≅

Answer

Answer: (b)

8. If ∆ ABC = ∆ PQR, then AB¯¯¯¯¯¯¯¯ corresponds to
(a) PQ¯¯¯¯¯¯¯¯
(b) QR¯¯¯¯¯¯¯¯
(c) RP¯¯¯¯¯¯¯¯
(d) none of these

Answer

Answer: (a)

9. If ∆ ABC = ∆ PQR, then BC¯¯¯¯¯¯¯¯ corresponds to
(a) PQ¯¯¯¯¯¯¯¯
(b) QR¯¯¯¯¯¯¯¯
(c) RP¯¯¯¯¯¯¯¯
(d) none of these

Answer

Answer: (b)

10. If ∆ ABC = ∆ PQR, then CA¯¯¯¯¯¯¯¯ corresponds to
(a) PQ¯¯¯¯¯¯¯¯
(b) QR¯¯¯¯¯¯¯¯
(c) RP¯¯¯¯¯¯¯¯
(d) none of these

Answer

Answer: (c)

11. If ∆ ABC = ∆ PQR, then ∠A corresponds to
(a) ∠P
(b) ∠Q
(c) ∠R
(d) none of these

Answer

Answer: (a)

12. If ∆ ABC = ∆ PQR, then ∠B corresponds to
(a) ∠ P
(b) ∠ Q
(c) ∠ R
(d) none of these

Answer

Answer: (b)

13. If ∆ ABC= ∆ PQR, then ∠C corresponds to
(a) ∠ P
(b) ∠ Q
(c) ∠ R
(d) none of these

Answer

Answer: (c)

14. We want to show that ∆ ART = ∆ PEN and we have to use SSS criterion. We have AR = PE and RT = EN. What more we need to show?
(a) AT = PN
(b) AT = PE
(c) AT = EN
(d) none of these

Answer

Answer: (a)

15. We want to show that ∆ ART = ∆ PEN. We have to use SAS criterion. We have ∠ T = ∠ N, RT = EN. What more we need to show?
(a) PN = AT
(b) PN = AR
(c) PN = RT
(d) None of these

Answer

Answer: (a)

16. We want to show that ∆ ART = ∆ PEN. We have to use ASA criterion. We have AT = PN, ∠ A = ∠ P. What more we need to show?
(a) ∠T = ∠N
(b) ∠T = ∠E
(c) ∠T = ∠P
(d) None of these

Answer

Answer: (a)

17. Which congruence criterion do you use in the following?

Given AC = DF
AB = DE
BC = EF
So, ∆ ABC ≅ ∆ DEF
(a) SSS
(b) SAS
(c) ASA
(d) RHS

Answer

Answer: (a)

18. Which congruence criterion do you use in the following?

Given : ZX = RP
RQ = ZY
∠ PRQ = ∠ XZY
So, ∆ PRQ = ∆ XYZ
(a) SSS
(b) SAS
(c) ASA
(d) RHS

Answer

Answer: (b)

Important Questions

Question 1.
In the given figure, name
(a) the side opposite to vertex A
(b) the vertex opposite A to side AB
(c) the angle opposite to side AC
(d) the angle made by the sides CB and CA.

Solution:
(a) The side opposite to vertex A is BC.
(b) The vertex opposite to side AB is C.
(c) The angle opposite to side AB is ∠ACB.
(d) The angle made by the sides CB and CA is ∠ACB.

Question 2.
Examine whether the given triangles are congruent or not.
Solution:
Here,
AB = DE = 3 cm
BC = DF = 3.5 cm
AC = EF = 4.5 cm
ΔABC = ΔEDF (By SSS rule)

So, ΔABC and ΔEDF are congruent.

Question 3.
In the given congruent triangles under ASA, find the value of x and y, ΔPQR = ΔSTU.

Solution:
Given: ΔPQR = ΔSTU (By ASA rule)
∠Q = ∠T = 60° (given)
QR¯ = TU¯ = 4 cm (given)
∠x = 30° (for ASA rule)
Now in ΔSTU,
∠S + ∠T + ∠U = 180° (Angle sum property)
∠y + 60° + ∠x = 180°
∠y + 60° + 30° = 180°
∠y + 90° = 180°
∠y = 180° – 90° = 90°
Hence, x = 30° and y = 90°.

Question 4.
In the following figure, show that ΔPSQ = ΔPSR.

Solution:
In ΔPSQ and ΔPSR
PQ¯ = PR¯ = 6.5 cm (Given)
PS¯ = PS¯ (Common)
∠PSQ = ∠PSR = 90° (Given)
ΔPSQ = ΔPSR (By RHS rule)

Question 5.
Can two equilateral triangles always be congruent? Give reasons.
Solution:
No, any two equilateral triangles are not always congruent.
Reason: Each angle of an equilateral triangle is 60° but their corresponding sides cannot always be the same.

Question 6.
In the given figure, AP = BQ, PR = QS. Show that ΔAPS = ΔBQR

Solution:
In ΔAPS and ΔBQR
AP = BQ (Given)
PR = QS (Given)
PR + RS = QS + RS (Adding RS to both sides)
PS = QR
∠APS = ∠BQR = 90° (Given)
ΔAPS = ΔBQR (by SAS rule)

Question 7.
Without drawing the figures of the triangles, write all six pairs of equal measures in each of the following pairs of congruent triangles.
(i) ΔABC = ADEF
(ii) ΔXYZ = ΔMLN
Solution:
(i) Given: ΔABC = ΔDEF
Here AB = DE
BC = EF
AC = DF
∠A = ∠D, ∠B = ∠E and ∠C = ∠F
(ii) Given ΔXYZ = ΔMLN
Here XY = ML
YZ = LN
XZ = MN
∠X = ∠M, ∠Y = ∠L and ∠Z = ∠N

Question 8.
Lengths of two sides of an isosceles triangle are 5 cm and 8 cm, find the perimeter of the triangle.
Solution:
Since the lengths of any two sides of an isosceles triangle are equal, then
Case I: The three sides of the triangle are 5 cm, 5 cm and 8 cm.
Perimeter of the triangle = 5 cm + 5 cm + 8 cm = 18 cm
Case II: The three sides of the triangle are 5 cm, 8 cm and 8 cm.
Perimeter of the triangle = 5 cm + 8 cm + 8 cm = 21 cm
Hence, the required perimeter is 18 cm or 21 cm.

Question 9.
Write the rule of congruence in the following pairs of congruent triangles.

Solution:
(i) Here, AB = ST = 3 cm
BC = TU = 4.5 cm
∠ABC = ∠STU = 110°
ΔABC = ΔSTU (By SAS rule)
(ii) Here ∠PQR = ∠MNL = 90°
hypt. PR = hypt. ML
QR = NL = 3 cm
ΔPQR = ΔMNL (By RHS rule)

Question 10.
In the given figure, state the rule of congruence followed by congruent triangles LMN and ONM.

Solution:
In ΔLMN and ΔONM
LM = ON
LN = OM
MN = NM
ΔLMN = ΔONM

Congruence of Triangles Class 7 Extra Questions Short Answer Type

Question 11.
In the given figure, PQR is a triangle in which PQ = PR. QM and RN are the medians of the triangle. Prove that
(i) ΔNQR = ΔMRQ
(ii) QM = RN
(iii) ΔPMQ = ΔPNR

Solution:
ΔPQR is an isosceles triangle. [∵ PQ = PR]
⇒ 12 PQ = 12 PR
⇒ NQ = MR and PN = PM
(i) In ΔNQR and ΔMRQ
NQ = MR (Half of equal sides)
∠NQR = ∠MRQ (Angles opposite to equal sides)
QR = RQ (Common)
ΔNQR = ΔMRQ (By SAS rule)
(ii) QM = RN (Congruent parts of congruent triangles)
(iii) In ΔPMQ and ΔPNR
PN = PM (Half of equal sides)
PR = PQ (Given)
∠P = ∠P (Common)
ΔPMQ = ΔPNR (By SAS rule)

Question 12.
In the given figure, PQ = CB, PA = CR, ∠P = ∠C. Is ΔQPR = ΔBCA? If yes, state the criterion of congruence.

Solution:
Given:
PQ = CB, PA = CR
and ∠P = ∠C
In ΔQPR and ΔBCA,
PQ = CB (Given)
∠QPR = ∠BCA (Given)
PA = CR (Given)
PA + AR = CR + AR (Adding AR to both sides)
or PR = CA
ΔQPR = ΔBCA (By SAS rule)

Question 13.
In the given figure, state whether ΔABC = ΔEOD or not. If yes, state the criterion of congruence.

Solution:
In ΔABC and ΔEOD
AB = OE
∠ABC = ∠EOD = 90°
AC = ED
ΔABC = ΔEOD
Hence, ΔABC = ΔEOD
RHS is the criterion of congruence.

Question 14.
In the given figure, PQ || RS and PQ = RS. Prove that ΔPUQ = ΔSUR.

Solution:
In ΔPUQ and ΔSUR
PQ = SR = 4 cm
∠UPQ = ∠USR (Alternate interior angles)
∠PQU = ∠SRU (Alternate interior angles)
ΔPUQ = ΔSUR (By ASA rule)

Congruence of Triangles Class 7 Extra Questions Long Answer Type

Question 15.
In the given figure ΔBAC = ΔQRP by SAS criterion of congruence. Find the value of x and y.

Solution:
Given: ΔBAC = ΔQRP (By SAS rule)
So, BA = QR
⇒ 3x + 10 = 5y + 15 ……(i)
∠BAC = ∠QRP
⇒ 2x + 15° = 5x – 60° ……(ii)
From eq. (ii), we have
2x + 15 = 5x – 60
⇒ 2x – 5x = -15 – 60
⇒ -3x = -7 5
⇒ x = 25
From eq. (i), we have
3x + 10 = 5y + 15
⇒ 3 × 25 + 10 = 5y + 15
⇒ 75 + 10 = 5y + 15
⇒ 85 = 5y + 15
⇒ 85 – 15 = 5y
⇒ 70 = 5y
⇒ y = 14
Hence, the required values of x andy are 25 and 14 respectively.

Question 16.
Observe the figure and state the three pairs of equal parts in triangles ABC and DCB.
(i) Is ΔABC = ΔDCB? Why?
(ii) Is AB = DC? Why?
(iii) Is AC = DB? Why? (NCERT Exemplar)

Solution:
(i) In ΔABC and ΔDCB
∠ABC = ∠DCB = 70° (40° + 30° = 70°) (Given)
∠ACB = ∠DCB = 30° (Given)
BC = CB (Common)
ΔABC = ΔDCB (By ASA rule)
(ii) Yes,
AB = DC (Congruent parts of congruent triangles)
(iii) Yes,
AC = DB (Congruent parts of congruent triangles)

Question 17.
In the given figure, ΔQPS = ΔSRQ. Find each value.
(a) x
(b) ∠PQS
(c) ∠PSR

Solution:
(a) ΔQPS = ΔSRQ
∠QPS = ∠SRQ (Congruent part of congruent triangles)
106 = 2x + 12
⇒ 106 – 12 = 2x
⇒ 94 = 2x
⇒ x = 47
∠QRS = 2 × 47 + 12 = 94 + 12 = 106°
So, PQRS is a parallelogram.
∠QSR = 180° – (42° + 106°) = 180° – 148° = 32°
(b) ∠PQS = 32° (alternate interior angles)
(c) ∠PSQ = 180° – (∠QPS + ∠PQS) = 180° – (106° + 32°) = 180° – 138° = 42°
∠PSR = 32° + 42° = 74°

Congruence of Triangles Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 18.
In ΔABC, medians BD and CE are equal and intersect each other at O. Prove that ΔABC is an isosceles triangle.

Solution:
We know that the medians of a triangle intersect each other in the ratio 2 : 1.
BD = CE (Given)
23 BD = 23 CE
⇒ OB = OC
13 BD = 13 CE
⇒ OE = OD
In ΔBOE and ΔCOD,
OB = OC
OE = OD
∠BOE = ∠COD (Vertically opposite angles)
ΔBOE = ΔCOD (By SAS rule)
BE = CD (Congruent parts of congruent triangles)
2BE = 2CD
⇒ AB = AC
Hence ΔABC is an isosceles triangle.

Question 19.
Prove that the lengths of altitudes drawn to equal sides of an isosceles triangle are also equal.
(i) ∠TRQ = ∠SQR?
(ii) If ∠TRQ = 30°, find the base angles of the ΔPQR.
(iii) Is ΔPQR an equilateral triangle?

Solution:
In ΔQTR and ΔRSQ
∠QTR = ∠RSQ = 90° (Given)
∠TQR = ∠SRQ (Base angle of an isosceles triangle)
∠QRT = ∠RQS (Remaining third angles)
QR = QR (Common)
ΔQTR = ΔRSQ (By ASA rule)
QS = RT (Congruent parts of congruent triangles)
Hence proved.
(i) ∠TRQ = ∠SQR (Congruent parts of congruent triangles)
(ii) In ΔQTR,
∠TRQ = 30° (Given)
∠QTR + ∠TQR + ∠QRT = 180° (Angle sum property)
⇒ 90° + ∠TQR + 30° = 180°
⇒ 120° + ∠TQR = 180°
⇒ ∠TQR = 180° – 120° = 60°
⇒ ∠TQR = ∠SRQ = 60°
Each base angle = 60°
(iii) In ΔPQR,
∠P + ∠Q + ∠R = 180° (Angle sum property)
⇒ ∠P + 60° + 60° = 180° (From ii)
⇒ ∠P + 120° = 180°
⇒ ∠P = 180° – 120° = 60°
Hence, ΔPQR is an equilateral triangle.

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CHAPTER -6 The Triangle and its Properties | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 6 The Triangle and its Properties

MCQs

1. How many elements are there in a triangle?
(a) 3
(b) 6
(c) 4
(d) None of these.

Answer/Explanation

Answer: (b)
Explanation : See a triangle.

2. How many vertices does a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : See a triangle.

3. How many sides are there in a triangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : See a triangle.

4. How many angles are there in a triangle?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : See a triangle.

5. If two sides of a triangle are not equal, the triangle is called
(a) scalene
(b) isosceles
(c) equilateral
(d) right-angled

Answer/Explanation

Answer: (a)
Explanation : Definition of a scalene triangle.

6. If two sides of a triangle are equal, the triangle is called
(a) isosceles
(b) equilateral
(c) scalene
(d) right-angled

Answer/Explanation

Answer: (a)
Explanation : Definition of an isosceles triangle.

7. If all the three sides of a triangle are equal, the triangle is called
(a) equilateral
(b) right-angled
(c) isosceles
(d) scalene

Answer/Explanation

Answer: (a)
Explanation : Definition of an equilateral triangle.

8. If all the angles of a triangle are acute, the triangle is called
(a) obtuse-angled
(b) acute-angled
(c) right-angled
(d) none of these

Answer/Explanation

Answer: (b)
Explanation : Definition of an acute-angled triangle.

9. If one angle of a triangle measures 90°, the triangle is called
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) none of these

Answer/Explanation

Answer: (c)
Explanation : Definition of a right triangle.

10. If one angle of a triangle is obtuse, the triangle is called
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) none of these

Answer/Explanation

Answer: (b)
Explanation : Definition of an obtuse angled triangle.

11, How many medians can a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : Draw medians and count.

12. How many altitudes can a triangle have?
(a) 1
(b) 2
(c) 3
(d) 4

Answer/Explanation

Answer: (c)
Explanation : Draw altitudes and count.

13. The total measure of the three angles of a triangle is
(a) 360°
(b) 90°
(c) 180°
(d) none of these

Answer/Explanation

Answer: (c)
Explanation : Angle Sum Property of a triangle.

14. The measure of each angle of an equilateral triangle is
(a) 30°
(b) 45°
(c) 90°
(d) 60°

Answer/Explanation

Answer: (d)
Explanation : x° + x° + x° = 180° ⇒ x° = 60°.

15. Which of the following statements is true?
(a) A triangle can have two right angles
(b) A triangle can have two obtuse angles
(c) A triangle can have two acute angles
(d) A triangle can have all the three angles less than 60°

Answer/Explanation

Answer: (c)
Explanation :

Important Questions

Question 1.
In ∆ABC, write the following:
(a) Angle opposite to side BC.
(b) The side opposite to ∠ABC.
(c) Vertex opposite to side AC.

Solution:
(a) In ∆ABC, Angle opposite to BC is ∠BAC
(b) Side opposite to ∠ABC is AC
(c) Vertex opposite to side AC is B

Question 2.
Classify the following triangle on the bases of sides

Solution:
(i) PQ = 5 cm, PR = 6 cm and QR = 7 cm
PQ ≠ PR ≠ QR
Thus, ∆PQR is a scalene triangle.
(ii) AB = 4 cm, AC = 4 cm
AB = AC
Thus, ∆ABC is an isosceles triangle.
(iii) MN = 3 cm, ML = 3 cm and NL = 3 cm
MN = ML = NL
Thus, ∆MNL is an equilateral triangle.

Question 3.
In the given figure, name the median and the altitude. Here E is the midpoint of BC.
Solution
In ∆ABC, we have

AD is the altitude.
AE is the median.

Question 4.
In the given diagrams, find the value of x in each case.

Solution:
(i) x + 45° + 30° = 180° (Angle sum property of a triangle)
⇒ x + 75° – 180°
⇒ x = 180° – 75°
x = 105°
(ii) Here, the given triangle is right angled triangle.
x + 30° = 90°
⇒ x = 90° – 30° = 60°
(iii) x = 60° + 65° (Exterior angle of a triangle is equal to the sum of interior opposite angles)
⇒ x = 125°

Question 5.
Which of the following cannot be the sides of a triangle?
(i) 4.5 cm, 3.5 cm, 6.4 cm
(ii) 2.5 cm, 3.5 cm, 6.0 cm
(iii) 2.5 cm, 4.2 cm, 8 cm
Solution:
(i) Given sides are, 4.5 cm, 3.5 cm, 6.4 cm
Sum of any two sides = 4.5 cm + 3.5 cm = 8 cm
Since 8 cm > 6.4 cm (Triangle inequality)
The given sides form a triangle.

(ii) Given sides are 2.5 cm, 3.5 cm, 6.0 cm
Sum of any two sides = 2.5 cm + 3.5 cm = 6.0 cm
Since 6.0 cm = 6.0 cm
The given sides do not form a triangle.

(iii) 2.5 cm, 4.2 cm, 8 cm
Sum of any two sides = 2.5 cm + 4.2 cm = 6.7 cm
Since 6.7 cm < 8 cm
The given sides do not form a triangle.

Question 6.
In the given figure, find x.

Solution:
In ∆ABC, we have
5x – 60° + 2x + 40° + 3x – 80° = 180° (Angle sum property of a triangle)
⇒ 5x + 2x + 3x – 60° + 40° – 80° = 180°
⇒ 10x – 100° = 180°
⇒ 10x = 180° + 100°
⇒ 10x = 280°
⇒ x = 28°
Thus, x = 28°

Question 7.
One of the equal angles of an isosceles triangle is 50°. Find all the angles of this triangle.
Solution:
Let the third angle be x°.
x + 50° + 50° = 180°
⇒ x° + 100° = 180°
⇒ x° = 180° – 100° = 80°
Thus ∠x = 80°

Question 8.
In ΔABC, AC = BC and ∠C = 110°. Find ∠A and ∠B.

Solution:
In given ΔABC, ∠C = 110°
Let ∠A = ∠B = x° (Angle opposite to equal sides of a triangle are equal)
x + x + 110° = 180°
⇒ 2x + 110° = 180°
⇒ 2x = 180° – 110°
⇒ 2x = 70°
⇒ x = 35°
Thus, ∠A = ∠B = 35°

The Triangles and its Properties Class 7 Extra Questions Short Answer Type

Question 9.
Two sides of a triangle are 4 cm and 7 cm. What can be the length of its third side to make the triangle possible?
Solution:
Let the length of the third side be x cm.
Condition I: Sum of two sides > the third side
i.e. 4 + 7 > x ⇒ 11 > x ⇒ x < 11
Condition II: The difference of two sides less than the third side.
i.e. 7 – 4 < x ⇒ 3 < x ⇒ x > 3
Hence the possible value of x are 3 < x < 11
i.e. x < 3 < 11

Question 10.
Find whether the following triplets are Pythagorean or not?
(a) (5, 8, 17)
(b) (8, 15, 17)
Solution:
(a) Given triplet: (5, 8, 17)
17² = 289
8² = 64
5² = 25
8² + 5² = 64 + 25 = 89
Since 89 ≠ 289
5² + 8² ≠ 17²
Hence (5, 8, 17) is not Pythagorean triplet.

(b) Given triplet: (8, 15, 17)
17² = 289
15² = 225
8² = 64
15² + 8² = 225 + 64 = 289
17² = 15² + 8²
Hence (8, 15, 17) is a Pythagorean triplet.

Question 11.
In the given right-angled triangle ABC, ∠B = 90°. Find the value of x.

Solution:
In ΔABC, ∠B = 90°
AB² + BC² = AC² (By Pythagoras property)
(5)² + (x – 3)² = (x + 2)²
⇒ 25 + x² + 9 – 6x = x² + 4 + 4x
⇒ -6x – 4x = 4 – 9 – 25
⇒ -10x = -30
⇒ x = 3
Hence, the required value of x = 3

Question 12.
AD is the median of a ΔABC, prove that AB + BC + CA > 2AD (HOTS)

Solution:
In ΔABD,
AB + BD > AD …(i)
(Sum of two sides of a triangle is greater than the third side)
Similarly, In ΔADC, we have
AC + DC > AD …(ii)
Adding (i) and (ii), we have
AB + BD + AC + DC > 2AD
⇒ AB + (BD + DC) + AC > 2AD
⇒ AB + BC + AC > 2AD
Hence, proved.

Question 13.
The length of the diagonals of a rhombus is 42 cm and 40 cm. Find the perimeter of the rhombus.

Solution:
AC and BD are the diagonals of a rhombus ABCD.
Since the diagonals of a rhombus bisect at the right angle.
AC = 40 cm
AO = 402 = 20 cm
BD = 42 cm
OB = 422 = 21 cm
In right angled triangle AOB, we have
AO² + OB² = AB²
⇒ 20² + 21² = AB²
⇒ 400 + 441 = AB²
⇒ 841 = AB²
⇒ AB = √841 = 29 cm.
Perimeter of the rhombus = 4 × side = 4 × 29 = 116 cm
Hence, the required perimeter = 116 cm

Question 14.
The sides of a triangle are in the ratio 3 : 4 : 5. State whether the triangle is right-angled or not.
Solution:
Let the sides of the given triangle are 3x, 4x and 5x units.
For right angled triangle, we have
Square of the longer side = Sum of the square of the other two sides
(5x)² = (3x)² + (4x)²
⇒ 25x² = 9x² + 16x²
⇒ 25x² = 25x²
Hence, the given triangle is a right-angled.

Question 15.
A plane flies 320 km due west and then 240 km due north. Find the shortest distance covered by the plane to reach its original position.
Solution:
Here, OA = 320 km
AB = 240 km
OB = ?

Clearly, ∆OBA is right angled triangle
OB² = OA² + AB² (By Pythagoras property)
⇒ OB² = 320² + 240²
⇒ OB² = 102400 + 57600
⇒ OB² = 160000
⇒ OB = √160000 = 400 km.
Hence the required shortest distance = 400 km.

The Triangles and its Properties Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type

Question 16.
In the following figure, find the unknown angles a and b, if l || m.

Solution:
Here, l || m
∠c = 110° (Corresponding angles)
∠c + ∠a = 180° (Linear pair)
⇒ 110° + ∠a = 180°
⇒ ∠a = 180° – 110° = 70°
Now ∠b = 40° + ∠a (Exterior angle of a triangle)
⇒ ∠b = 40° + 70° = 110°
Hence, the values of unknown angles are a = 70° and b = 110°

Question 17.
In figure (i) and (ii), Find the values of a, b and c.

Solution:
(i) In ∆ADC, we have
∠c + 60° + 70° = 180° (Angle sum property)
⇒ ∠c + 130° = 180°
⇒ ∠c = 180° – 130° = 50°
∠c + ∠b = 180° (Linear pair)
⇒ 50° + ∠b = 180°
⇒ ∠ b = 180° – 50° = 130°
In ∆ABD, we have
∠a + ∠b + 30° = 180° (Angle sum property)
⇒ ∠a + ∠130° + 30° = 180°
⇒ ∠a + 160° = 180°
⇒ ∠a = 180° – 160° = 20°
Hence, the required values are a = 20°, b = 130° and c = 50°

(ii) In ∆PQS, we have
∠a + 60° + 55° = 180°(Angle sum property)
⇒ ∠a + 115° = 180°
⇒ ∠a = 180° – 115°
⇒ ∠a = 65°
∠a + ∠b = 180° (Linear pair)
⇒ 65° + ∠b = 180°
⇒ ∠b = 180° – 65° = 115°
In ∆PSR, we have
∠b + ∠c + 40° = 180° (Angle sum property)
⇒ 115° + ∠c + 40° = 180°
⇒ ∠c + 155° = 180°
⇒ ∠c = 180° – 155° = 25°
Hence, the required angles are a = 65°, b = 115° and c = 25°

Question 18.
I have three sides. One of my angle measure 15°. Another has a measure of 60°. What kind of a polygon am I? If I am a triangle, then what kind of triangle am I? [NCERT Exemplar]
Solution:
Since I have three sides.
It is a triangle i.e. three-sided polygon.
Two angles are 15° and 60°.
Third angle = 180° – (15° + 60°)
= 180° – 75° (Angle sum property)
= 105°
which is greater than 90°.
Hence, it is an obtuse triangle.

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CHAPTER -5 Lines and Angles | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 5 Lines and Angles

MCQs

1. When the sum of the measures of two angles is 90°, the angles are called
(а) supplementary angles
(b) complementary angles
(c) adjacent angles
(d) vertically opposite angles

Answer/Explanation

Answer: (b)
Explanation : Definition of complementary angles

2. The sum of the measures of two complementary angles is
(a) 180°
(b) 60°
(c) 45°
(d) 90°

Answer/Explanation

Answer: (d)
Explanation : Definition of complementary angles

3. The measure of the complement of the angle 30° is
(а) 30°
(b) 16°
(c) 60°
(d) 160°

Answer/Explanation

Answer: (c)
Explanation : 90° – 30° = 60°.

4. Which of the following statements is true?
(a) Two acute angles can be complementary to each other
(b) Two obtuse angles can be complementary to each other
(c) Two right angles can be complementary to each other
(d) One obtuse angle and one acute angle can be complementary to each other

Answer

Answer: (a)

5. The measure of the complement of the angle 46° is
(a) 90°
(b) 46°
(c) 16°
(d) 136°

Answer/Explanation

Answer: (b)
Explanation : 90°- 45° = 45°.

6. What is the measure of the complement of the angle 80°?
(a) 10°
(b) 100°
(c) 36°
(d) 20°

Answer/Explanation

Answer: (a)
Explanation : 90° – 80° = 10°.

7. Which pair of the following angles are complementary?

Answer/Explanation

Answer: (a)
Explanation : 60°+ 30° = 90°.

8. The measure of the angle which is equal to its complement is
(a) 30°
(b)60°
(c) 46°
(d) 90°

Answer/Explanation

Answer: (c)
Explanation : x° + x° = 90° ⇒ x° = 45°.

9. Which of the following pairs of angles is not a pair of complementary angles?
(a) 60°, 30°
(b) 66°, 34°
(c) 0°, 90°
(d) 160°, 30°

Answer/Explanation

Answer: (d)
Explanation : 150° + 30° = 180° ≠ 90°.

10. What is the measure of the complement of the angle 90°?
(а) 90°
(b) 0°
(c) 180°
(d) 46°

Answer/Explanation

Answer: (b)
Explanation : 90° – 90° = 0°.

11. When the sum of the measures of two angles is 180°, the angles are called
(a) adjacent angles
(b) complementary angles
(c) vertically opposite angles
(d) supplementary angles

Answer/Explanation

Answer: (d)
Explanation : Definition of supplementary angles.

12. The sum of the measures of two supplementary angles is
(a) 90°
(b) 180°
(c) 360°
(d) none of these

Answer/Explanation

Answer: (b)
Explanation : Definition of supplementary angles.

13. The measure of the supplement of the angle 120° is
(a) 30°
(b)45°
(c) 60°
(d) 90°

Answer/Explanation

Answer: (c)
Explanation : 180° – 120° = 60°.

14. Which of the following statements is true?
(a) Two acute angles can be supplementary.
(b) Two right angles can be supplementary.
(c) Two obtuse angles can be supplementary.
(d) One obtuse angle and one acute angle cannot be supplementary

Answer

Answer: (b)

15. The measure of the supplement of the angle 90° is
(a) 45°
(b) 60°
(c) 30°
(d) 90°

Answer/Explanation

Answer: (d)
Explanation : 180° – 90° = 90°.

16. The measure of the angle which is equal to its supplement is
(a) 30°
(b) 45°
(c) 90°
(d) 60°

Answer/Explanation

Answer: (c)
Explanation : x° + x° = 180° ⇒ x° = 90°.

Important Questions

Question 1.
Find the angles which is 15 of its complement.
Solution:
Let the required angle be x°
its complement = (90 – x)°
As per condition, we get

Question 2.
Find the angles which is 23 of its supplement.
Solution:
Let the required angle be x°.
its supplement = (180 – x)°
As per the condition, we get
23 of (180 – x)° = x°

Question 3.
Find the value of x in the given figure.

Solution:
∠POR + ∠QOR = 180° (Angles of linear pair)
⇒ (2x + 60°) + (3x – 40)° = 180°
⇒ 2x + 60 + 3x – 40 = 180°
⇒ 5x + 20 = 180°
⇒ 5x = 180 – 20 = 160
⇒ x = 32
Thus, the value of x = 32.

Question 4.
In the given figure, find the value of y.

Solution:
Let the angle opposite to 90° be z.
z = 90° (Vertically opposite angle)
3y + z + 30° = 180° (Sum of adjacent angles on a straight line)
⇒ 3y + 90° + 30° = 180°
⇒ 3y + 120° = 180°
⇒ 3y = 180° – 120° = 60°
⇒ y = 20°
Thus the value of y = 20°.

Question 5.
Find the supplements of each of the following:
(i) 30°
(ii) 79°
(iii) 179°
(iv) x°
(v) 25 of right angle
Solution:
(i) Supplement of 30° = 180° – 30° = 150°
(ii) Supplement of 79° = 180° – 79° = 101°
(iii) Supplement of 179° = 180° – 179° = 1°
(iv) Supplement of x° = (180 – x)°
(v) Supplement of 25 of right angle
= 180° – 25 × 90° = 180° – 36° = 144°

Question 6.
If the angles (4x + 4)° and (6x – 4)° are the supplementary angles, find the value of x.
Solution:
(4x + 4)° + (6x – 4)° = 180° (∵ Sum of the supplementary angle is 180°)
⇒ 4x + 4 + 6x – 4 = 180°
⇒ 10x = 180°
⇒ x = 18°
Thus, x = 18°

Question 7.
Find the value of x.

Solution:
(6x – 40)° + (5x + 9)° + (3x + 15) ° = 180° (∵ Sum of adjacent angles on straight line)
⇒ 6x – 40 + 5x + 9 + 3x + 15 = 180°
⇒ 14x – 16 = 180°
⇒ 14x = 180 + 16 = 196
⇒ x = 14
Thus, x = 14

Question 8.
Find the value of y.

Solution:
l || m, and t is a transversal.
y + 135° = 180° (Sum of interior angles on the same side of transversal is 180°)
⇒ y = 180° – 135° = 45°
Thus, y = 45°

Lines and Angles Class 7 Extra Questions Short Answer Type

Question 9.
Find the value ofy in the following figures:

Solution:
(i) y + 15° = 360° (Sum of complete angles round at a point)
⇒ y = 360° – 15° = 345°
Thus, y = 345°
(ii) (2y + 10)° + 50° + 40° + 130° = 360° (Sum of angles round at a point)
⇒ 2y + 10 + 220 = 360
⇒ 2y + 230 = 360
⇒ 2y = 360 – 230
⇒ 2y = 130
⇒ y = 65
Thus, y = 65°
(iii) y + 90° = 180° (Angles of linear pair)
⇒ y = 180° – 90° = 90°
[40° + 140° = 180°, which shows that l is a straight line]

Question 10.
In the following figures, find the lettered angles.

Solution:
(i) Let a be represented by ∠1 and ∠2
∠a = ∠1 + ∠2
∠1 = 35° (Alternate interior angles)
∠2 = 55° (Alternate interior angles)
∠1 + ∠2 = 35° + 55°
∠a = 90°
Thus, ∠a = 90°

Question 11.
In the given figure, prove that AB || CD.

Solution:
∠CEF = 30° + 50° = 80°
∠DCE = 80° (Given)
∠CEF = ∠DCE
But these are alternate interior angle.
CD || EF ……(i)
Now ∠EAB = 130° (Given)
∠AEF = 50° (Given)
∠EAB + ∠AEF = 130° + 50° = 180°
But these are co-interior angles.
AB || EF …(ii)
From eq. (i) and (ii), we get
AB || CD || EF
Hence, AB || CD
Co-interior angles/Allied angles: Sum of interior angles on the same side of transversal is 180°.

Question 12.
In the given figure l || m. Find the values of a, b and c.

Solution:
(i) We have l || m
∠b = 40° (Alternate interior angles)
∠c = 120° (Alternate interior angles)
∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight angle)
⇒ ∠a + 40° + 120° = 180°
⇒ ∠a + 160° = 180°
⇒ ∠a = 180° – 160° = 20°
Thus, ∠a = 20°, ∠b = 40° and ∠c = 120°.
(ii) We have l || m
∠a = 45° (Alternate interior angles)
∠c = 55° (Alternate interior angles)
∠a + ∠b + ∠c = 180° (Sum of adjacent angles on straight line)
⇒ 45 + ∠b + 55 = 180°
⇒ ∠b + 100 = 180°
⇒ ∠b = 180° – 100°
⇒ ∠b = 80°

Question 13.
In the adjoining figure if x : y : z = 2 : 3 : 4, then find the value of z.

Solution:
Let x = 2s°
y = 3s°
and z = 4s°
∠x + ∠y + ∠z = 180° (Sum of adjacent angles on straight line)
2s° + 3s° + 4s° = 180°
⇒ 9s° = 180°
⇒ s° = 20°
Thus x = 2 × 20° = 40°, y = 3 × 20° = 60° and z = 4 × 20° = 80°

Question 14.
In the following figure, find the value of ∠BOC, if points A, O and B are collinear. (NCERT Exemplar)

Solution:
We have A, O and B are collinear.
∠AOD + ∠DOC + ∠COB = 180° (Sum of adjacent angles on straight line)
(x – 10)° + (4x – 25)° + (x + 5)° = 180°
⇒ x – 10 + 4x – 25 + x + 5 = 180°
⇒ 6x – 10 – 25 + 5 = 180°
⇒ 6x – 30 = 180°
⇒ 6x = 180 + 30 = 210
⇒ x = 35
So, ∠BOC = (x + 5)° = (35 + 5)° = 40°

Question 15.
In given figure, PQ, RS and UT are parallel lines.
(i) If c = 57° and a = c3, find the value of d.
(ii) If c = 75° and a = 25c , find b.

Solution:
(i) We have ∠c = 57° and ∠a = ∠c3
∠a = 573 = 19°
PQ || UT (given)
∠a + ∠b = ∠c (Alternate interior angles)
19° + ∠b = 57°
∠b = 57° – 19° = 38°
PQ || RS (given)
∠b + ∠d = 180° (Co-interior angles)
38° + ∠d = 180°
∠d = 180° – 38° = 142°
Thus, ∠d = 142°
(ii) We have ∠c = 75° and ∠a = 25 ∠c
∠a = 25 × 75° = 30°
PQ || UT (given)
∠a + ∠b = ∠c
30° + ∠b = 75°
∠b = 75° – 30° = 45°
Thus, ∠b = 45°

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CHAPTER -4 Simple Equations | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 4 Simple Equations

MCQs

1. Write the following statement in the form of an equation:
The sum of three times x and 10 is 13.
(a) 3x + 10 = 13
(b) 3x – 10 = 13
(c) 3x + 13 = 10
(d) none of these

Answer

Answer: (a)

2. Write the following statement in the form of an equation:
If you subtract 3 from 6 times a number, you get 9
(a) 3x – 6 = 9
(b) 6x – 3 = 9
(c) 6x + 3 = 9
(d) 3x + 6 = 9

Answer

Answer: (b)

3. Write the following statement in the form of an equation:
One fourth of n is 3 more than 2
(a) n4 – 2 = 3
(b) n4 + 2 = 3
(c) n2 – 4 = 3
(d) n2 + 4 = 3

Answer

Answer: (a)

4. Write the following statement in the form of an equation:
One third of a number plus 2 is 3
(a) m3 – 2 = 3
(b) m3 + 2 = 3
(c) m2 – 3 = 3
(d) m2 + 3 = 3

Answer

Answer: (b)

5. Write the following statement in the form of an equation:
Taking away 5 from x gives 10
(a) x – 5 = 10
(b) x + 5 = 10
(c) x – 10 – 5
(d) none of these

Answer

Answer: (a)

6. Write the following statement in the form of an equation:
Four times a number p is 8.
(a) 4P = 8
(b) P + 4 = 8
(c) p – 4 = 8
(d) p ÷ 4 = 8

Answer

Answer: (a)

7. Write the following statement in the form of an equation:
Add 1 to three times n to get 7
(a) 3n + 1 = 7
(b) 3n – 1 = 7
(c) 3n + 7 = 1
(d) none of these

Answer

Answer: (a)

8. Write the following statement in the form of an equation:
The number b divided by 6 gives 5.
(a) b6 = 5
(b) b – 5 = 6
(c) 5b = 6
(d) b + 5 = 6

Answer

Answer: (a)

9. The solution of the equation x + 3 = 0 is
(a) 3
(b) – 3
(c) 0
(d) 1

Answer

Answer: (b)

10. The solution of the equation x – 6 = 1 is
(a) 1
(b) 6
(c) – 7
(b) 7

Answer

Answer: (d)

11. The solution of the equation 5x = 10 is
(a) 1
(b) 2
(c) 5
(d) 10

Answer

Answer: (b)

12. The solution of the equation m2 = 3 is
(a) 2
(b) 3
(c) 12
(d) 6

Answer

Answer: (d)

13. The solution of the equation 7n + 5 = 12 is
(a) 0
(b) – 1
(c) 1
(d) 5

Answer

Answer: (c)

14. The solution of the equation 4p – 3 = 9 is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (c)

15. The solution of the equation 5p + 2 = 7 is
(a) 0
(b) 1
(c) – 1
(d) 2

Answer

Answer: (b)

16. The solution of the equation 3p – 2 = 4 is
(a) 0
(b) 1
(c) 2
(d) 3

Answer

Answer: (c)

Important Questions

Question 1.
Write the following statements in the form of equations.
(a) The sum of four times a number and 5 gives a number five times of it.
(b) One-fourth of a number is 2 more than 5.
Solution:
(a) Let the number be x.
Sum of 4x and 5 = 4x + 5
The sum is 5x.
The equation is 4x + 5 = 5x as required.
(b) Let the number be x.
14x = 5 + 2
⇒ 14x = 7 as required.

Question 2.
Convert the following equations in statement form:
(a) 5x = 20
(b) 3y + 7 = 1
Solution:
(a) Five times a number x gives 20.
(b) Add 7 to three times a number y gives 1.

Question 3.
If k + 7 = 10, find the value of 9k – 50.
Solution:
k + 7 = 10
⇒ k = 10 – 7 = 3
Put k = 3 in 9k – 50, we get
9 × 3 – 50 = 27 – 50 = -23
Thus the value of k = -23

Question 4.
Solve the following equations and check the answers.

Solution:

Question 5.
Solve the following equations:
3(y – 2) = 2(y – 1) – 3
Solution:
3(y – 2) = 2(y – 1) – 3
⇒ 3y – 6 = 2y – 2 – 3 (Removing the brackets)
⇒ 3y – 6 = 2y – 5
⇒ 3y – 2y = 6 – 5 (Transposing 6 to RHS and 2y to LHS)
⇒ y = 1
Thus y = 1

Question 6.
If 5 is added to twice a number, the result is 29. Find the number.
Solution:
Let the required number be x.
Step I: 2x + 5
Step II: 2x + 5 = 29
Solving the equation, we get
2x + 5 = 29
⇒ 2x = 29 – 5 (Transposing 5 to RHS)
⇒ 2x = 24
⇒ x = 12 (Dividing both sides by 2)
⇒ x = 12
Thus the required number is 12.

Question 7.
If one-third of a number exceeds its one-fourth by 1, find the number.
Solution:
Let the required number be x.

Simple Equations Class 7 Extra Questions Short Answer Type

Question 8.
The length of a rectangle is twice its breadth. If its perimeter is 60 cm, find the length and the breadth of the rectangle.
Solution:
Let the breadth of the rectangle be x cm.
its length = 2x
Perimeter = 2 (length + breadth) = 2(2x + x) = 2 × 3x = 6x
As per the condition of the question, we have
6x = 60 ⇒ x = 10
Thus the required breadth = 10 cm
and the length = 10 × 2 = 20 cm.

Question 9.
Seven times a number is 12 less than thirteen times the same number. Find the number.
Solution:
Let the required number be x.
7x = 13x – 12
⇒ 7x – 13x = -12 (Transposing 13x to LHS)
⇒ -6x = -12
⇒ x = 2
Thus, the required number is 2.

Question 10.
The present age of a son is half the present age of his father. Ten years ago, the father was thrice as old as his son. What are their present age?
Solution:
Let the present age of a father be x years.
Son’s age = 12x years
10 years ago, father’s age was (x – 10) years
10 years ago, son’s age was (x2 – 10) years
As per the question, we have

Question 11.
The sum of three consecutive multiples of 2 is 18. Find the numbers.
Solution:
Let the three consecutive multiples of 2 be 2x, 2x + 2 and 2x + 4.
As per the conditions of the question, we have
2x + (2x + 2) + (2x + 4) = 18
⇒ 2x + 2x + 2 + 2x + 4 = 18
⇒ 6x + 6 = 18
⇒ 6x = 18 – 6 (Transposing 6 to RHS)
⇒ 6x = 12
⇒ x = 2
Thus, the required multiples are
2 × 2 = 4, 4 + 2 = 6, 6 + 2 = 8 i.e., 4, 6 and 8.

Question 12.
Each of the 2 equal sides of an isosceles triangle is twice as large as the third side. If the perimeter of the triangle is 30 cm, find the length of each side of the triangle. [NCERT Exemplar]
Solution:
Let the length of the third side be x cm.
Each equal side = 2x cm.
As per the condition of the question, we have
Perimeter = x + 2x + 2x = 30
⇒ 5x = 30
⇒ x = 6
Thus, the third side of the triangle = 6 cm
and other two equal sides are 2 × 6 = 12 cm each

Question 13.
A man travelled two-fifth of his journey by train, one-third by bus, one-fourth by car and the remaining 3 km on foot. What is the length of his total journey? [NCERT Exemplar]
Solution:
Let the total length of total journey be x km.
Distance travelled by train = 25x km
Distance travelled by bus = 13x km
Distance travelled by car = 14x km
Remaining distance = 3 km
As per the question, we have

Thus, the required journey = 180 km.

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CHAPTER -3 Data Handling | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 3 Data Handling

MCQs

1. On which day is the number of tourists maximum?
(a) Sunday
(b) Wednesday
(c) Tuesday
(d) Saturday

Answer

Answer: (a)

2. On which day is the number of tourists minimum?
(a) Friday
(b) Monday
(c) Thursday
(d) Saturday

Answer

Answer: (a)

3. On which day 60 tourists visit?
(a) Monday
(b) Tuesday
(c) Friday
(d) Sunday

Answer

Answer: (c)

4. What is the difference between the num-ber of tourists visiting on Friday and Monday?
(a) 10
(b) 24
(c) 38
(d) 5

Answer

Answer: (d)

5. The sum of the number of tourists visit-ing on Sunday and Friday is
(a) 160
(b) 60
(c) 220
(d) 100

Answer

Answer: (c)

6. The difference between the maximum and minimum number of tourists is
(a) 50
(b) 80
(c) 90
(d) 100

Answer

Answer: (d)

7. A batsman scored the following number of runs in six innings:
35, 30, 45, 65, 39, 20
The mean runs scored by him in an inning is
(a) 39
(b) 38
(c) 37
(d) 40

Answer

Answer: (a)

8. The mean of the numbers 10,20, 30 and 40 is
(a) 20
(b) 25
(c) 30
(d) 50

Answer

Answer: (b)

9. The range of the weights (in kg) of a students of a class given below is:
49, 60, 47, 50, 47, 59, 58, 45, 53
(a) 10
(b) 15
(c) 20
(d) 2

Answer

Answer: (b)

10. The marks of 11 students of a class are as given below:
78, 11, 99, 63, 94, 6, 78, 36, 30, 55, 22
The range of marks is
(a) 90
(b) 91
(c) 92
(d) 93

Answer

Answer: (d)

11. The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

On how many days was the rainfall less than 6 mm?
(a) 0
(b) 3
(c) 6
(d) 7

Answer

Answer: (d)

12. The mode of the distribution 3,5, 7, 4, 2, 1, 4, 3, 4 is
(a) 7
(b) 4
(c) 3
(d) 1

Answer

Answer: (b)

13. The marks of some students are as given below:
30, 31, 32, 32, 33, 32, 34, 35, 30, 31, 33, 32
Find the mode of their marks.
(a) 30
(b) 31
(c) 32
(d) 33

Answer

Answer: (c)

14. The median of the distribution 2, 3, 4, 7, 5, 1, 6 is
(a) 1
(b) 2
(c) 3
(d) 4

Answer

Answer: (d)

15. The median of the data 20, 30, 40, 10, 15, 25, 35 is
(a) 20
(b) 25
(c) 30
(d) 40

Answer

Answer: (b)

16. Which of the following statements is true?
(a) The mode is always one of the numbers in a data
(b) The mean is always one of the numbers in a data
(c) Mean < Mode in a data
(d) Median < Mode in a data

Answer

17. A coin is tossed. What is the probability of getting head?
(a) 0
(b) 1
(c) $\frac { 1 }{ 2 }$
(d) 2

Answer

Answer: (c)

18. A coin is tossed. What is the probability of getting tail?
(a) 1
(b) $\frac { 1 }{ 2 }$
(c) 2
(d) 0

Answer

Answer: (b)

19. A die is thrown. What is the probability of getting 1?
(a) 0
(b) 1
(c) $\frac { 1 }{ 2 }$
(d) $\frac { 1 }{ 6 }$

Answer

Answer: (d)

20. A die is thrown. What is the probability of getting 6?
(a) 0
(b) $\frac { 1 }{ 6 }$
(c) $\frac { 1 }{ 2 }$
(d) 1

Answer

Answer: (b)

Read the following bar graph and answer the following related questions :

21. Who got the maximum marks?
(a) Apala
(b) Meenu
(c) Saroj
(d) Indu

Answer

Answer: (a)

22. Who got the minimum marks?
(a) Apala
(b) Meenu
(c) Vimla
(d) Indu

Answer

Answer: (b)

23. The difference between maximum and minimum marks is
(a) 100
(d) 200
(c) 400
(d) 500

Answer

Answer: (d)

24. The ratio between the marks obtained by Saroj and Vimla is
(a) 1 : 2
(b) 2 : 3
(c) 3 : 4
(d) 1 : 6

Answer

Answer: (a)

Important Questions

Question 1.
Find the range of the following data:
21, 16, 30, 15, 16, 18, 10, 24, 26, 20
Solution:
Greatest number 30
Smallest number = 10
Range = 30 – 10 = 20

Question 2.
Find the mode of the following data:
24, 26, 23, 26, 22, 25, 26, 28
Solution:
Arranging the given data with the same value together, we get
22, 23, 24, 25, 26, 26, 26, 28
Here, 26 occurs the greatest number of times i.e. 3 times
Thus, the required mode = 26.

Question 3.
Find the average of the numbers 8, 13, 15.
Solution:

Question 4.
Find the median of the following data:
8, 6, 10, 12, 14
Solution:
Let us arrange the given data in increasing order,
6, 8, 10, 12, 14
n = 5 (odd)
Median = (n+12)th term = 3rd term = 10
Thus, the required median = 10.

Question 5.
Find the median of the following data:
20, 14, 6, 25, 18, 13, 19, 10, 9, 12
Solution:
Arranging the given data in increasing order, we get
6, 9, 10, 12, 13, 14, 18, 19, 20, 25
n = 10 (even)

Thus, the required median = 13.5

Question 6.
A fair die is rolled, find the probability of getting a prime number.
Solution:
Number on a die = 1, 2, 3, 4, 5, 6
n(S) = 6
Prime numbers = 2, 3, 5
n(E) = 3
Probability = n(E)n(S) = 36 = 12
Thus the required probability = 12.

Question 7.
If the averages of the given data 6, 10, 12, x, 16 is 14, find the value of x.
Solution:
Average of the given numbers

Thus, the required value of x is 26.

Question 8.
Find the mean of the first 5 multiples of 3.
Solution:
Five multiples of 3 are 3, 6, 9, 12 and 15

Hence, the required mean = 9.

Data Handling Class 7 Extra Questions Short Answer Type

Question 9.
The following bar graph shows the number of books sold by a publisher during the five consecutive years. Read the bar graph and answer the following questions:
(i) About how many books were sold in 2008, 2009 and 2012 years?
(ii) In which years were 575 books were sold?
(iii) In which years were the minimum number of books sold?

Solution:

(ii) In the year of 2012, maximum number of books i.e. 575 were sold.
(iii) Minimum number of books i.e. 150 were sold in the year 2008.

Question 10.
Find the mean and median of first five prime numbers.
Solution:
First five prime numbers are: 2, 3, 5, 7 and 11

Here, n = 5
Median is the middle term, i.e., 5.

Question 11.
The marks obtained (out of 10) by 80 students in a class test are given below:

Find the mode of the above data.
Solution:
In the given frequency distribution table, we find that the observation 7 has maximum frequency, i.e., 20
Hence, the required mode = 7.

Question 12.
A bag contains 5 white and 9 red balls. One ball is drawn at random from the bag. Find the probability of getting
(a) a white ball
(b) a red ball
Solution:
Total number of balls = 5 + 9 = 14 balls
n(S) = 14
(i) Number of white ball = 5
n(E) = 5
Probability of getting white ball = n(E)n(S) = 514
(ii) Number of red balls = 9
n(E) = 9
Probability of getting white ball = n(E)n(S) = 914

Question 13.
A dice is tossed once. Find the probability of getting
(i) a number 5
(ii) a number greater than 5
(iii) a number less than 5
(iv) an odd number
(v) an even number
(vi) a number greater than 6
Solution:
Total number of outcomes = 6
n(S) = 6
(i) An event of getting a number 5
n(E) = 1
Probability = n(E)n(S) = 16
(ii) An event of getting a number 5 greater than 5, i.e., 6
n(E) = 1
Probability = n(E)n(S) = 16
(iii) An event of getting a number less than 5, i.e., 1, 2, 3 and 4.
n(E) = 4
Probability = n(E)n(S) = 46 = 23
(iv) An event of getting an odd number, i.e., 1, 3 and 5.
n(E) = 3
Probability = n(E)n(S) = 36 = 12
(v) An event of getting an even number, i.e., 2, 4 and 6.
n(E) = 3
Probability = n(E)n(S) = 36 = 12
(vi) An event of getting a number greater than 6, i.e., Nil.
n(E) = 0
Probability = n(E)n(S) = 06 = 0

Data Handling Class 7 Extra Questions Long Answer Type

Question 14.
The data given below shows the production of motorbikes in a factory for some months of two consecutive years.

Study the table given above and the answer the following questions:
(a) Draw a double bar graph using an appropriate scale to depict the above information and compare them.
(b) In which year was the total output maximum?
(c) Find the mean production for the year 2007.
(d) For which month was the difference between the production for the two years is the maximum?
(e) In which month for the year 2008, the production was the maximum?
(f) In which month for the year 2007, the production was the least? [NCERT Exemplar]
Solution:
(a) Double bar graph
Scale : 1 cm = 100 Motor Bikes

The above bar graph depicts the total production of motorbikes in two consecutive years.
Total production in 2007 was 22100 whereas in 2008 it was 21100.
(b) In the year 2007, the total production was maximum (22100)
(c) Mean production in the year 2007 is

(d) Production of motorbikes in the May 2007 = 4500 and in May 2008 = 3200
Difference = 4500 – 3200 = 1300 which is the maximum
(e) In the month of August 2008, production was maximum i.e., 6000
(f) In the month of Feb. 2007 the production was least i.e., 2800.

Question 15.
A coin and a die are tossed once together. Find the total number of outcomes.
Solution:
A coin has two faces, Head (H) and Tail (T)
A die has six faces marked with numbers 1, 2, 3, 4, 5, 6
Possible outcomes are:
H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6
Total number of outcomes = 2 × 6 = 12

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CHAPTER -2 Fractions and Decimals | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 2 Fractions and Decimals

MCQs

1. Which of the following fraction?

Answer

Answer: (d)

2. Which of the following fraction?

Answer

Answer: (d)

3. Which of the following is an improper fraction?

Answer

Answer: (d)

4. Which of the following is an improper fraction?

Answer

Answer: (d)

5. Which of the following is a mixed fraction?

Answer

Answer: (d)

6. The improper fraction $\frac { 33 }{ 4 }$ in the form of a mixed fraction is

Answer

Answer: (a)

7. Which of the following is not an equiva-lent fraction of $\frac { 3 }{ 5 }$ ?

Answer

Answer: (d)

8. Which of the following is an equivalent fraction of $\frac { 2 }{ 3 }$ ?

Answer

Answer: (d)

9. 1 – $\frac { 1 }{ 5 }$ is equal to

Answer

Answer: (c)

10. 2 + $\frac { 1 }{ 4 }$ is equal to

Answer

Answer: (b)

11. $\frac { 1 }{ 2 }$ + $\frac { 1 }{ 3 }$ is equal to

Answer

Answer: (b)

12. $\frac { 1 }{ 2 }$ – $\frac { 1 }{ 4 }$ is equal to

Answer

Answer: (d)

13. Apala ate $\frac { 3 }{ 5 }$ of an orange. The remaining orange was eaten by Meenu. What part of the orange was eaten by Meenu?

Answer

Answer: (b)

14. The side of an equilateral triangle is $\frac { 1 }{ 2 }$ cm. The perimeter of the triangle is

Answer

Answer: (c)

15. The side of a square is $\frac { 1 }{ 2 }$ cm. The perimeter of the square is

Answer

Answer: (b)

16. Manish worked for $\frac { 1 }{ 2 }$ an hour. Yash worked for $\frac { 1 }{ 4 }$ of an hour. For how much time did both work together?

Answer

Answer: (b)

Important Questions

Question 1.
If 23 of a number is 6, find the number.
Solution:
Let x be the required number.

Hence, the required number is 9.

Question 2.
Find the product of 67 and 223.
Solution:

Question 3.
Solve the following:

Solution:

Question 4.
Multiply 2.05 and 1.3.
Solution:

Fractions and Decimals Class 7 Extra Questions Short Answer Type

Question 5.
Solve the following:
(a) 3 – 23
(b) 4 + 25
Solution:

Question 6.
Arrange the following in ascending order:

Solution:

Question 7.
Find the products:
(i) 2.4 × 100
(ii) 0.24 × 1000
(iii) 0.024 × 10000
Solution:

Question 8.
Arnav spends 134 hours in studies, 212 hours in playing cricket. How much time did he spend in all?
Solution:
Time spent by Arnav in studies = 134 hours
Time spent by Arnav in playing cricket = 212 hours
Total time spent by Arnav = 134 hours + 212 hours

Question 9.
A square paper sheet has 1025 cm long side. Find its perimeter and area.
Solution:

Question 10.

Solution:

Question 11.
The product of two numbers is 2.0016. If one of them is 0.72, find the other number.
Solution:
Product of two numbers = 2.0016
One number = 0.72
Other number = 2.0016 ÷ 0.72

Hence, the required number = 2.78.

Question 12.
Reemu reads 15th pages of a book. If she reads further 40 pages, she would have read 710th
page of the book. How many pages are left to be read? [NCERT Exemplar]
Solution:
Let the total number of pages be x.
Number of pages read by Reemu = 15x
If she reads 40 more pages,
Total number of pages read by her = 15x + 40

Question 13.
18 of a number equals 25 ÷ 120. What is the number? (NCERT Exemplar)
Solution:
Let the number be x.

Hence, the required number = 64.

Question 14.
Simplify the following:

Solution:

Question 15.
The weight of an object on the Moon is 16 its weight on the Earth. If an object weight 535 kg on the Earth. How much would it weight on the Moon? [NCERT Exemplar]
Solution:

Question 16.
A picture hall has seats for 820 persons. At a recent film show, one usher guessed it was 34 full, another that it was 23 full. The ticket office reported 648 sales. Which usher (first or second) made the better guess? [NCERT Exemplar]
Solution:
Total number of seats = 820
Number of ticket sold = 648
For first usher = 34 × 648 = 3 × 162 = 486
For second usher = 23 × 648 = 2 × 216 = 432
Since 432 < 486
Hence, the first usher guessed better.

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CHAPTER -1 INTEGERS | CLASS 7TH |NCERT MATHS IMPORTANT QUESTIONS & MCQS | EDUGROWN

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter -1 Integers Important Questions & MCQs

MCQs

1. On a number line, when we add a positive integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these

Answer

Answer: (a)

2. On a number line, when we add a negative integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these

Answer

Answer: (b)

3. On a number line, when we subtract a positive integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these

Answer

Answer: (b)

4. On a number line, when we subtract a negative integer, we
(a) move to the right
(b) move to the left
(c) do not move at all
(d) none of these

Answer

Answer: (a)

5. When two positive integers are added, we get
(a) a positive integer
(b) a negative integer
(c) sometimes a positive integer, sometimes a negative integer
(d) none of these

Answer

Answer: (a)

6. When two negative integers are added, we get
(a) a positive integer
(b) a negative integer
(c) sometimes a positive integer, sometimes a negative integer
(d) none of these

Answer

Answer: (b)

7. Which of the following statements is wrong?
(а) When a positive integer and a negative integer are added, we always get a negative integer
(b) Additive inverse of 8 is (- 8)
(c) Additive inverse of (- 8) is 8
(d) For subtraction, we add the additive inverse of the integer that is being subtracted, to the other integer

Answer

Answer: (a)

8. Which of the following is true?
(а) (- 8) + (- 4) > (- 8) – (- 4)
(b) (- 8) + (- 4) < (- 8) – (- 4)
(c) (- 8) + (- 4) = (- 8) – (- 4)
(d) none of these

Answer

Answer: (b)

9. The product of two negative integers is
(a) a positive integer
(b) a negative integer
(c) either a positive integer or a negative integer
(d) none of these

Answer

Answer: (a)

10. The product of three negative integers is
(a) a positive integer
(b) a negative integer
(c) either a positive integer or a nega¬tive integer
(d) none of these

Answer

Answer: (b)

11. (- 1) × (- 1) × (- 1) × ……. 10 times is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these

Answer

Answer: (a)

12. (- 1) × (- 1) × (- 1) × …… 5 times is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these

Answer

Answer: (b)

13. (- 1) × (- 1) × (- 1) × …… 2m times, where m is a natural number, is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these

Answer

Answer: (a)

14. (- 1) x (- 1) x (- 1) × …………….. (2m + 1) times, where m is a natural number, is equal to
(a) 1
(b) – 1
(c) 1 or – 1
(d) none of these

Answer

Answer: (b)

15. (- 20) × (- 5) is equal to
(a) 100
(b) – 100
(c) 20
(d) 5

Answer

Answer: (a)

16. (- 30) × 20 is equal to
(a) 600
(b) – 600
(c) 50
(d) 10

Answer

17. 10 × (- 20) is equal to
(a) 200
(b) -200
(c) 30
(d) 10

Answer

Answer: (b)

18. 3 × 0 is equal to
(a) 0
(b) 3
(c) 1
(d) -3

Answer

Answer: (a)

19. 0 × (- 5) is equal to
(a) 0
(b) 5
(c) – 5
(d) 1

Answer

Answer: (a)

20. (- 2) × 1 is equal to
(a) 2
(b) – 2
(c) 1
(d) – 1

Answer

Answer: (b)

IMPORTANT QUESTIONS

Question 1.
Fill in the blanks using < or >.
(a) -3 …… -4
(b) 6 ……. -20
(c) -8 …… -2
(d) 5 …… -7
Solution:
(a) -3 > -4
(b) 6 > -20
(c) -8 < -2
(d) 5 > -7

Question 2.
Solve the following:
(i) (-8) × (-5) + (-6)
(ii) [(-6) × (-3)] + (-4)
(iii) (-10) × [(-13) + (-10)]
(iv) (-5) × [(-6) + 5]
Solution:
(i) (-8) × (-5) + (-6)
= (-) × (-) × [8 × 5] + (-6)
= 40 – 6
= 34

(ii) [(-6) × (-3)] + (-4)
= (-) × (-) × [6 × 3] + (-4)
= 18 – 4
= 14

(iii) (-10) × [(-13) + (-10)]
= (-10) × (-23)
= (-) × (-) × [10 × 23]
= 230

(iv) (-5) × [(-6) + 5]
= (-5) × (-1)
= (-) × (-) × 5 × 1
= 5

Question 3.
Starting from (-7) × 4, find (-7) × (-3)
Solution:
(-7) × 4 = -28
(-7) × 3 = -21 = [-28 + 7]
(-7) × 2 – -14 = [-21 + 7]
(-7) × 1 = -7 = [-14 + 7]
(-7) × 0 = 0 = [-7 + 7]
(-7) × (-1) = 7 = [0 + 7]
(-7) × (-2) = 14 = [7 + 7]
(-7) × (-3) = 21 = [14 + 7]

Question 4.
Using number line, find:
(i) 3 × (-5)
(ii) 8 × (-2)
Solution:
(i) 3 × (-5)
Integers Class 7 Extra Questions Maths Chapter 1 Q4
From the number line, we have
(-5) + (-5) + (-5) = 3 × (-5) = -15

(ii) 8 × (-2)
Integers Class 7 Extra Questions Maths Chapter 1 Q4.1
From the number line, we have
(-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) + (-2) = 8 × (-2) = -16

Question 5.
Write five pair of integers (m, n ) such that m ÷ n = -3. One of such pair is (-6, 2).
Solution:
(i) (-3, 1) = (-3) ÷ 1 = -3
(ii) (9, -3) = 9 ÷ (-3) = -3
(iii) (6, -2) = 6 ÷ (-2) = -3
(iv) (-24, 8) = (-24) ÷ 8 = -3
(v) (18, -6) = 18 ÷ (-6) = -3

Integers Class 7 Extra Questions Short Answer Type

Question 6.
Solve the following:
(i) (-15) × 8 + (-15) × 4
(ii) [32 + 2 × 17 + (-6)] ÷ 15
Solution:
(i) (-15) × 8 + (-15) × 4
= (-15) × [8 + 4]
= (-15) × 12
= -180

(ii) [32 + 2 × 17 + (-6)] ÷ 15
= [32 + 34 – 6] ÷ 15
= [66 – 6] ÷ 15
= 60 ÷ 15
= 4

Question 7.
The sum of two integers is 116. If one of them is -79, find the other integers.
Solution:
Sum of two integers = 116
One integer = -79
Other integer = Sum of integer – One of integer = 116 – (-79) = 116 + 79 = 195

Question 8.
If a = -35, b = 10 cm and c = -5, verify that:
(i) a + (b + c) = (a + b) + c
(ii) a × (b + c) = a × b + a × c
Solution:
(i) Given that a = -35, b = 10, c = -5
LHS = a + (b + c) = (-35) + [10 + (-5)] = (-35) + 5 = -30
RHS = (a + b) + c = [(-35) + 10] + (-5) = (-25) + (-5) = -(25 + 5) = -30
LHS = RHS
Hence, verified.

(ii) a × (b + c) = a × b + a × c
LHS = a × (b + c) = (-35) × [10 + (-5)] = (-35) × 5 = -175
RHS = a × b + a × c = (-35) × 10 + (-35) × (-5) = -350 + (-) × (-) × (35 × 5) = -350 + 175 = -175
LHS = RHS
Hence, verified.

Question 9.
Write down a pair of integers whose
(i) sum is -5
(ii) difference is -7
(iii) difference is -1
(iv) sum is 0
Solution:
(i) (-2) + (-3) = -5
Hence, the required pair of integers = (-2, -3)
(ii) -10 – (-3) = -10 + 3 = -7
Hence, the required pair of integers = (-10, -3)
(iii) (-3) – (-2) = -1
Hence, the required pair of integers = (-3, -2)
(iv) (-4) + (4) = 0
Hence, the required pair of integers = (-4, 4)

Question 10.
You have ₹ 500 in your saving account at the beginning of the month. The record below shows all of your transactions during the month. How much money is in your account after these transactions?
Integers Class 7 Extra Questions Maths Chapter 1 Q10
Solution:
Amount in the beginning of the month in the account = ₹ 500
Amount deposited in the account for Jal Board = ₹ 200
Amount paid to Jal Board = ₹ 120
Amount left in the account after the above transactions = ₹ (500 + 200 – 120) = ₹ (700 – 120) = ₹ 580
Amount deposited for LIC India = ₹ 150
Amount paid to LIC India = ₹ 240
Amount left after this transactions = ₹ (580 + 150 – 240) = ₹ (730 – 240) = ₹ 490

Question 11.
The given table shows the freezing points in °F of different gases at sea level. Convert each of these into °C to the nearest integral value using the relations and complete the table
C = 59 [F – 32]
Integers Class 7 Extra Questions Maths Chapter 1 Q11
Solution:
Freezing point of Hydrogen = -435°F
C = 59 [-435 – 32]
= 59 [-467]
= 5 × (-51.9)
= 259.5°C or 259°C
For Krypton, freezing point = -251°F
C = 59 [-251 – 32]
= 59 [-283]
= 5 × (-31.4)]
= -157°C
For Oxygen, freezing point = -369°F
C = 59 [-369 – 32]
= 59 [-401]
= 5 × (44.56)
= 222.80° C or 223°C
Hence, the required freezing points at sea level in °C for Hydrogen = -259°C, Krypton = -157°C, Oxygen = -223°C.

Question 12.
Taking today as zero on the number line, if the day before yesterday is 17 January, what is the date on 3 days after tomorrow? [NCERT Exemplar]
Solution:
Integers Class 7 Extra Questions Maths Chapter 1 Q12
The date before yesterday = 17 January
The date of yesterday = 17 + 1 = 18 January
Today’s date = 18 + 1 = 19 January
Tomorrow’s date = 19 + 1 = 20 January
Date on 3 days after tomorrow = (20 + 3) = 23rd January

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Chapter – 20 विप्लव गायन | Class 7th | NCERT Hindi Vasant 2 Solutions | Edugrown

NCERT Solutions for Class 7th Hindi Vasant Part 2

Here students can get info about NCERT Solutions for Class 7 Hindi Vasant Bhag 2 so that they can get to know the answers to the questions in case they are not able to find it. You can find the best Class 7th Hindi NCERT Solutions वसंत भाग 2 explained in conformance with the CBSE curriculum for Class 7.

Chapter -20 विप्लव गायन

पाठ्यपुस्तक के प्रश्न-अभ्यास

कविता से
प्रश्न 1.
‘कण-कण में है व्याप्त वही स्वर ……… कालकूट फणि को चिंतामणि।’
(क) ‘वही स्वर’, ‘वह ध्वनि’ एवं ‘वही तान’ आदि वाक्यांश किसके लिए किस भाव के लिए प्रयुक्त हुए हैं?
(ख) वही स्वर, वह ध्वनि एवं वही तान से संबंधित भाव का ‘रुद्ध-गीत की क्रुद्ध तान है/निकली मेरी अंतरतर से-पंक्तियों से क्या कोई संबंध बनता है?
उत्तर-
(क) ‘वही स्वर’ ‘वह ध्वनि’ एवं ‘वही तान’ नवनिर्माण का रास्ते खोलने के लिए तथा जनजागृति का आहवान करने के लिए प्रयोग किया गया है। इसके अलावे इन वाक्यांशों का प्रयोग क्रांति की उस भावना के लिए प्रयोग किए हुए हैं, जो हर ओर व्याप्त है।
(ख) हाँ, वही स्वर, वही ध्वनि एवं वही तान से संबंधित भाव रुद्ध-गीत की क्रुद्ध तान है। निकली मेरे अंतरतर से पंक्तियों में सही संबंध बनता है क्योंकि कवि इनकी पंक्तियों में वर्तमान व्यवस्था के प्रति आक्रोश है, वही क्रांति गीत के रूप में निकल रहा है। वही क्रांति रोम-रोम में घुलकर प्रति ध्वनित होने लगती है।

प्रश्न 2.
नीचे दी गई पंक्तियों का भाव स्पष्ट कीजिए-
‘सावधान! मेरी वीणा में ” दोनों मेरी ऐंठी हैं।’
उत्तर
इन पंक्तियों का भाव यह है कि कवि लोगों को परिवर्तन के प्रति सावधान करता है और वीणा से कोमल स्वर निकालने की बजाय कठोर स्वर निकालने के कारण उसकी उँगलियों की मिज़राबें टूटकर गिर गईं, जिससे उसकी उँगलियाँ ऐंठकर घायल हो जाती हैं।

कविता से आगे

प्रश्न 1.
स्वाधीनता संग्राम के दिनों में अनेक कवियों ने स्वाधीनता को मुखर करने वाली ओजपूर्ण कविताएँ लिखीं। माखनलाल चतुर्वेदी, मैथिलीशरण गुप्त और सूर्यकांत त्रिपाठी ‘निराला’ की ऐसी कविताओं की चार-चार पंक्तियाँ इकट्ठा कीजिए जिनमें स्वाधीनता के भावे ओज से मुखर हुए हैं।
उत्तर-
(क) द्वार बालिका खोल, चल, भूडोल कर दें,
एक हिमगिरि, एक सिर का मोल कर दें,
मसल कर अपने इरादों-सी उठाकर,
दो हथेली है कि पृथ्वी गोलकर दें। -माखनलाल चर्तुवेदी

(ख) विचार लो कि मर्त्य हो न मृत्यु से डरो कभी,
मरो परंतु यो मरो कि याद जो करे सभी।
हुई न यो सु-मृत्यु तो वृथा मरे वृथा जिए।
मरा नहीं वही कि जो जिया न आपके लिए।
यही पशु-प्रवृत्ति है कि आप-आप ही चरे
वही मनुष्य है कि जो मनुष्य के लिए मरे। -मैथिलीशरण गुप्त
बाधाएँ आएँ तन पर,
देखें तुझे नयन मन भर,
मुझे देख तू सजल दृगों से
अपलक उर के शतदल पर;
क्लेद-युक्त, अपना तने दूंगा,
मुक्त करूंगा तुझे अटल,
तेरे चरणों पर देकर बलि,
सकल श्रेय-श्रम संचित फल । -सूर्यकांत त्रिपाठी निराला

अनुमान और कल्पना

प्रश्न 1.
कविता के मूलभाव को ध्यान में रखते हुए बताइए कि इसका शीर्षक ‘विप्लव-गायन’ क्यों रखा गया होगा?
उत्तर
कविता का मूल भाव है गलत रीति-रिवाजों, रूढ़िवादी विचारों व परस्पर भेदभाव त्यागकर नवनिर्माण के लिए जनता को प्रेरित करना। इसीलिए इस कविता का शीर्षक ‘विप्लव-गायन’ रखा गया है जिसका अर्थ है क्रांति के लिए आह्वान करना।

भाषा की बात

प्रश्न 1.
कविता में दो शब्दों के मध्य (-) का प्रयोग किया गया है, जैसे- ‘जिससे उथल-पुथल मच जाए’ एवं ‘कण-कण में है व्याप्त वही स्वर’। इन पंक्तियों को पढ़िए और अनुमान लगाइए कि कवि ऐसा प्रयोग क्यों करते हैं?
उत्तर-
कवि ऐसा प्रयोग इसलिए करते हैं, क्योंकि शब्द की पुनरुक्ति करके चमत्कार उत्पन्न करने के लिए जैसे रोम-रोम गाता है। काव्य को प्रभावशाली बनाने व शब्दों में प्रवाह लाने के लिए (-) योजक चिह्न का प्रयोग किया जाता है।

प्रश्न 2.
कविता (में, -। आदि) विराम चिह्नों का उपयोग रुकने, आगे-बढ़ने अथवा किसी खास भाव को अभिव्यक्त करने के लिए किया जाता है। कविता पढ़ने में इन विराम चिह्नों का प्रभावी प्रयोग करते हुए काव्य पाठ कीजिए। गद्य में आमतौर पर है शब्द का प्रयोग वाक्य के अंत में किया जाता है, जैसे-देशराज जाता है। अब कविता की निम्न पंक्तियों को देखिए-
‘कण-कण में है व्याप्त … वही तान गाती रहती है.’
इन पंक्तियों में है शब्द का प्रयोग अलग-अलग जगहों पर किया गया है। कविता में अगर आपको ऐसे अन्य प्रयोग मिलें तो उन्हें छाँटकर लिखिए।
उत्तर
कंठ रुका है महानाश का मारक गीत रुद्ध होता है।

प्रश्न 3.
निम्न पंक्तियों को ध्यान से देखिए
‘कवि कुछ ऐसी तान सुनाओ ……. एक हिलोर उधर से आए’,
इन पंक्तियों के अंत में आए, जाए जैसे तुक मिलानेवाले शब्दों का प्रयोग किया गया है। इसे तुकबंदी या अंत्यानुप्रास कहते हैं। कविता से तुकबंदी के अन्य शब्दों को छाँटकर लिखिए। छाँटे गए शब्दों से अपनी कविता बनाने की कोशिश कीजिए।
उत्तर-
तुकबंदी वाले शब्द/पद
बैठी हैं- ऐंठी हैं गाती-रहती कुद्ध-युद्ध छात्र इन शब्दों के आधार पर कविता लिखने का प्रयास करें।

अन्य पाठेतर हल प्रश्न

बहुविकल्पी प्रश्नोत्तर
(क) कवि अपनी कविता के माध्यम से आह्वान कर रहा है
(i) स्वतंत्रता सेनानियों
(ii) देशवासियों से
(iii) नवयुवकों से
(iv) सेना से।

(ख) ‘उथल-पुथल मचने’ से कवि का क्या अभिप्राय है?
(i) विद्रोह का होना
(ii) क्रांति का आगमन होना
(iii) आँधी का आना
(iv) समाज में परिवर्तन का होना।

(ग) कवि देशवासियों को कैसी तान सुनाना चाहता है?
(i) प्राचीन परंपराओं को समाप्त करने की
(ii) परिवर्तन एवं नवनिर्माण करना
(iii) बदलाव की
(iv) उपर्युक्त सभी।

(घ) इस कविता के रचयिता कौन हैं?
(i) रामधारी सिंह दिनकर
(ii) बालकृष्ण शर्मा ‘नवीन’।
(iii) सुमित्रानंदन पंत
(iv) सूर्यकांत त्रिपाठी “निराला’।

(ङ) कवि कैसा गीत नहीं लिख पा रहा है
(i) रुद्र गीत
(ii) क्रांति गीत
(iii) मारक गीत
(iv) प्रेम गीत।

(च) कवि की वीणा में कैसी चिनगारियाँ आ बैठी हैं?
(i) शांति की
(ii) भ्रांति की
(iii) क्रांति की
(iv) उपर्युक्त सभी।

(छ) यह गीत कैसा गीत है?
(i) वीरतापूर्ण
(ii) ओजस्वी
(iii) रौद्र
(iv) हास्य।

उत्तर
(क) (iii)
(ख) (ii)
(ग) (iv)
(घ) (ii)
(ङ) (iii)
(च) (iii)
(छ) (ii)

अतिलघु उत्तरीय प्रश्न

(क) यह कविता किस वाद से प्रभावित है?
उत्तर-
यह कविता प्रगतिवाद से प्रभावित है।

(ख) कवि अन्य कवियों से क्या आह्वान करता है?
उत्तर-
क्रांतिकारी गीत की रचना के लिए आह्वान करता है।

(ग) क्रांति लाने के लिए कवि किसका सहारा लेता है?
उत्तर-
क्रांति लाने के लिए कवि गीत का सहारा लेता है।

(घ) कवि के कंठ से निकले गीत का क्या प्रभाव पड़ेगा?
उत्तर-
कवि के कंठ से निकले गीत जीर्ण-शीर्ण विचारधाराओं और रुढ़िवादी विचारों का नाश हो जाएगा।

लघु उत्तरीय प्रश्न

(क) कवि कैसी तान सुनाना चाहते हैं?
उत्तर-
कवि ऐसी तान सुनाना चाहते हैं जिससे चारों तरफ़ हलचल मच जाए।

(ख) कवि विप्लव गान क्यों गाना चाहता है?
उत्तर-
कवि का मानना है कि विप्लव गान द्वारा ही वह लोगों को समाज के नवनिर्माण के लिए जाग्रत कर सकता है, क्योंकि सुंदर राष्ट्र की नींव पुराने, गले-सड़े रीति-रिवाजों व रूढ़िवादी विचारों पर नहीं रखा जा सकता।

(ग) कविता में कालकूट फणि की चिंतामणि शब्दों का अर्थ क्या है?
उत्तर-
विष से परिपूर्ण शेषनाग को अपनी सबसे प्रिय मणि की चिंता हरदम रहती है, वैसे ही कवि चाहता है कि प्रत्येक मनुष्य के मन में नवनिर्माण की चिंता जाग्रत कर सके।

दीर्घ उत्तरीय प्रश्न

(क) कवि के अनुसार जीवन का रहस्य क्या है?
उत्तर-
कवि के अनुसार, कवि विप्लव के माध्यम से परिवर्तन की हिलोर लाना चाहता है। इस कविता का भाव है जीवन का रहस्य है। विकास और गतिशीलता में रुकावट पैदा करने वाली प्रवृत्ति से संघर्ष करके नया निर्माण करना। नव-निर्माण के लिए कवि विध्वंस और महानाश को आवश्यक मानता है। यह विनाश सदियों से चली आ रही रुढ़िवादी मानसिकता, जड़ता तथा अंधविश्वास को काटकर दूर फेंक देगा। सारी रुकावट समाप्त कर नए सृजन तथा नए राष्ट्र को निर्माण का रास्ता साफ़ हो जाएगा। इसके लिए कवि द्वारा एक क्रांति की चिंगारी जलाने की जरूरत है।

मूल्यपरक प्रश्न

(क) क्या आप क्रांति के समर्थक हैं? क्या क्रांति के द्वारा समाज में व्याप्त अव्यवस्था को दूर किया जा सकता है? कैसे तर्क सहित उत्तर लिखिए।
उत्तर-
हाँ, मैं समाज में कुरीतियों को दूर करने के लिए क्रांति का समर्थक हूँ। लेकिन जब तक इस समाज में क्रांति की आवश्यकता बनी हुई है तब तक मैं क्रांति का बिगुल बजाता रहता हूँ। लक्ष्य पाने के बाद क्रांति की आवश्यकता होती है। हम ऐसी क्रांति का समर्थन करते

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Chapter – 19 आश्रम का अनुमानित व्यय | Class 7th | NCERT Hindi Vasant 2 Solutions | Edugrown

NCERT Solutions for Class 7th Hindi Vasant Part 2

Chapter -19 आश्रम का अनुमानित व्यय

पाठ्यपुस्तक के प्रश्न-अभ्यास

लेखा-जोखा
प्रश्न 1.
हमारे यहाँ बहुत से काम लोग खुद नहीं करके किसी पेशेवर कारीगर से करवाते हैं। गांधी जी छेनी, हथौड़े, बसूले क्यों खरीदना चाहते होंगे?
उत्तर
यह सत्य है कि हमारे यहाँ अर्थात् भारत में बहुत से काम लोग खुद न करके किसी पेशेवर कारीगर से करवाते हैं। गांधी जी छेनी, हथौड़े, वसूले इसलिए खरीदना चाहते होंगे ताकि लोग कुटीर उद्योग, लुहार व बढ़ईगिरी आदि को बढ़ावा दें। आत्मनिर्भर बनें व छोटे-छोटे कामों के लिए दूसरों का मुँह न ताकें।।

प्रश्न 2.
गांधी जी ने अखिल भारतीय कांग्रेस सहित कई संस्थाओं व आंदोलनों का नेतृत्व किया। उनकी जीवनी या उन पर लिखी गई किताबों से उन अंशों को चुनिए जिनसे हिसाब-किताब के प्रति गांधी जी की चुस्ती का पता चलता है।
उत्तर
गांधी जी कोई भी कार्य बिना हिसाब किताब के नहीं करते थे। वे प्रत्येक विषय के प्रति नकारात्मक व सकारात्मक सोच बराबर रखते थे। निम्ने उदाहरणों द्वारा इस वक्तव्य को स्पष्टता दे सकते हैं-

‘दांडी यात्रा’ के लिए गाँधी जी जब ‘रास’ नामक स्थान पर पहुँचे तो वहाँ निषेधाज्ञा लागू थी अर्थात कोई भी नेता किसी प्रकार के विचार जलूस-जलसे के रूप में नहीं प्रकट कर सकता था। गांधी जी तो लोगों को संबोधित किए बिना रह नहीं सकते थे तो पहले ही यह योजना बना ली गई कि यदि उन्हें गिरफ्तार कर लिया गया तो अब्बास तैयबजी दांडी यात्रा का नेतृत्व करेंगे।
असहयोग आंदोलन के समय भी वे यह हिसाब लगाने में पूर्णतया सक्ष्म थे कि किस स्थान पर किस तरह से ब्रिटिश शासन पर प्रहार करना है। यही कारण था कि लोग उनके हर विचार की कद्र करते थे और उनका कहा पूरी तरह से मानते थे। |
वे बिल्कुल भी फिजूल खर्च न करते थे एक-एक पैसा सोच समझकरे खर्च करते थे यहाँ तक कि कई बार तो पच्चीस-पच्चीस किलोमीटर एक दिन में पैदल चलते थे। उनका मानना था कि धन को जरूरी कामों के लिए ही खर्च करना चाहिए। शानो-शौकत या वैभवपूर्व जीवन जीने के लिए नहीं।
किसी भी आश्रम या सभा का हिसाब-किताब वे बहुत कुशलता से लगाते थे। साबरमती आश्रम में भी उन्होंने ऐसा बजट बनाया कि आने वाले मेहमानों के खर्च भी उसमें शामिल किए गए।

प्रश्न 3.
मान लीजिए, आपको कोई बाल आश्रम खोलना है। इस बजट से प्रेरणा लेते हुए उसको अनुमानित बजट बनाइए। इस बजट में दिए गए किन-किन मदों पर आप कितना खर्च करना चाहेंगे। किन नई मदों को जोड़ना-हटाना चाहेंगे?
उत्तर-
छात्र इस पाठ से उदाहरण लेकर बाल आश्रम के लिए आवश्यक चीज़ों और उनके अनुमानित-खर्च का बजट तैयार करें।

प्रश्न 4.
आपको कई बार लगता होगा कि आप कई छोटे-मोटे काम ( जैसे- घर की पुताई, दूध दुहना, खाट बुनना ) करना चाहें तो कर सकते हैं। ऐसे कामों की सूची बनाइए जिन्हें आप चाहकर भी नहीं सीख पाते। इसके क्या कारण रहे होंगे उन कामों की सूची भी बनाइए, जिन्हें आप सीख कर ही छोड़ेंगे?
उत्तर-
हमारे जीवन में ऐसे बहुत से काम होते हैं जिसे हम चाहकर भी नहीं सीख पाते; जैसे- घर पुताई सफ़ेदीवाला करता है, दूधवाला दूध देता है और खाट (चारपाई) बुननेवाले से बुनवाई जाती है। कुछ ऐसे ही निम्न कार्य हैं, मैं चाहकर भी सीख नहीं पाता; जैसे

कार्य	कारण
रोटी बनाने का कार्य	लगन की कमी
सिलाई करने का काम	सिखानेवाला नहीं मिला

चप्पल जूते में टाँका लगाना – जानकारी का अभाव एवं औजारों की कमी पर मैं इन कामों को सीखने का पूरा प्रयास कर रहा हूँ। मैं इन कामों को सीखकर ही दम लूंगा।

मैं इन कामों को सिखाने वाले प्रशिक्षित व्यक्ति के तालाश में हूँ। मैं इस काम को सीखकर ही दम लूंगा।

प्रश्न 5.
इस अनुमानित बजट को गहराई से पढ़ने के बाद आश्रम के उद्देश्यों और कार्यप्रणाली के बारे में क्या-क्या अनुमान लगाए जा सकते हैं?
उत्तर-
अनुमानित बजट को गहराई से अध्ययन करने के बाद हम आश्रम के उद्देश्यों को भलीभाँति समझ सकते हैं स्वावलंबन की भावना का विकास करना, अतिथि सत्कार करना, जरूरतमंदों को आवश्यक सुविधाएँ प्रदान करना, बेकार लोगों को आजीविका प्रदान करना, श्रम का महत्त्व समझना, कुटीर उद्योगों को बढ़ावा देना, चरखे खादी आदि से स्वदेशी आंदोलन को बढ़ावा देना। सहयोग की भावना का विकास। इस आश्रम की कार्य प्रणाली का मुख्य आधार आत्मनिर्भरता है।

भाषा की बात

प्रश्न 1.
अनुमानित शब्द अनुमान में इत प्रत्यय जोड़कर बना है। इत प्रत्यय जोड़ने पर अनुमान का ‘न’ नित में परिवर्तित हो जाता है। नीचे इत प्रत्यय वाले कुछ और शब्द लिखे हैं। उनमें मूल शब्द पहचानिए और देखिए कि क्या परिवर्तन हो रहा है

प्रमाणित	व्यथित	द्रवित	मुखरित
झंकृत	शिक्षित	मोहित	चर्चित

इत प्रत्यय की भाँति इक प्रत्यय से भी शब्द बनते हैं और तब शब्द के पहले अक्षर में भी परिवर्तन हो जाता है; जैसे सप्ताह के इक + साप्ताहिक। नीचे इक प्रत्यय से बनाए गए शब्द दिए गए हैं। इनमें मूल शब्द पहचानिए और देखिए कि क्या परिवर्तन हो रहा है

मौखिक	संवैधानिक	प्राथमिक
नैतिक	पौराणिक	दैनिक

उत्तर-
इत प्रत्यय युक्त शब्द

मूल शब्द	प्रत्यय
प्रमाणित झंकृत व्यथित द्रवित मुखरित शिक्षित द्रवित मोहित मुखरित चर्चित मौखिक नैतिक संवैधानिक पौराणिक प्राथमिक दैनिक	प्रमाण + इत झंकार + इत व्यथा + इत द्रव + इत मुखर + इत शिक्षा + इत द्रव + इत मोह + इत मुखर + इत चर्चा + इत मुख + इक नीति + इक संविधान + इक पुराण + इक प्रथम + इक दिन + इक

प्रश्न 2.
बैलगाड़ी और घोड़ागाड़ी शब्द दो शब्दों को जोड़ने से बने हैं। इसमें दूसरा शब्द प्रधान है, यानी शब्द का प्रमुख अर्थ दूसरे शब्द पर टिका है। ऐसे समास को तत्पुरुष समास कहते हैं। ऐसे छह शब्द और सोचकर लिखिए और समझिए कि उनमें दूसरा शब्द प्रमुख क्यों है?
उत्तर
राहखर्च क्रीडाक्षेत्र
तुलसीकृत घुड़सवार
गंगाजल वनवास
इन शब्दों में दूसरा शब्द प्रमुख है क्योंकि दूसरा शब्द पहले शब्द की सार्थकता को स्पष्ट कर रहा है।

जैसे-
राहखर्च राह के लिए खर्च
तुलसीकृत तुलसी द्वारा कृत
गंगाजल गंगा का जल
क्रीडाक्षेत्र क्रीड़ा के लिए क्षेत्र
घुड़सवार घोड़े पर सवार
वनवास वन में वास

अन्य पाठेतर हल प्रश्न

बहुविकल्पी प्रश्नोत्तर
(क) गांधी जी क्या बना रहे थे?
(i) आश्रम
(ii) अहमदाबाद के आश्रम का होने वाला खर्च का ब्यौरा
(iii) अंग्रेजों के विरुद्ध योजनाएँ
(iv) उपर्युक्त सभी।

(ख) कुछ समय बाद आगंतुकों की संख्या आश्रम में किंतने होने वाली थी?
(i) 30
(ii) 40
(iii) 50
(iv) 60

(ग) सपरिवार रहने वाले अतिथि की संख्या आश्रम में कितनी होगी?
(i) 2
(ii) 3
(iii) 5
(iv) 3 से 5

(घ) आश्रम में कितनी पुस्तकें रखने की बात हो रही थी?
(i) 1000
(ii) 1500
(iii) 2000
(iv) 3000

(ङ) स्टेशन से अतिथि और सामान को लाने के लिए किस साधन का प्रयोग करने की बात हो रही थी?
(i) कार
(ii) ओटो रिक्शा
(iii) बैलगाड़ी
(iv) रिक्शा।

(च) आश्रम में औज़ारों की आवश्यकता क्यों महसूस हो रही थी?
(i) ताकि लोग आत्मनिर्भर बनें
(ii) ताकि लोग काम करना सीखें
(iii), ताकि आश्रम के छोटे-मोटे काम स्वयं करें
(iv) दिए गए उपर्युक्त सभी।

(छ) आश्रम में हर महीने कितने अतिथियों के आने की संभावना थी?
(i) पाँच
(ii) आठ
(iii) दस
(iv) बारह

(ज) आश्रम में कितने रसोईघर बनाने का लेखा-जोखा था?
(i) दो
(ii) तीन
(iii) चार
(iv) पाँच

उत्तर
(क) (ii)
(ख) (iii)
(ग) (iv)
(घ) (iv)
(ङ) (ii)
(च) (iii)
(छ) (iii)
(ज) (iii)

अतिलघु उत्तरीय प्रश्न

(क) गांधी जी कौन-सा आश्रम बना रहे थे?
उत्तर-
गांधी जी अहमदाबाद में साबरमती आश्रम बना रहे थे।

(ख) आश्रम में शुरुआत में कितने लोग थे?
उत्तर-
आश्रम में शुरुआत में चालीस लोग थे

(ग) पुस्तकालय में कितनी पुस्तकें रखी जाती थीं?
उत्तर-
पुस्तकालय में तीन हजार पुस्तकें रखी जाती थीं।

(घ) शिक्षण के सामान में कितने हथकरघों की आवश्यकता होगी?
उत्तर-
पाँच-छह देशी हथकरघों की आवश्यकता होगी।

(ङ) गांधी जी ने आश्रम की स्थापना कब की थी?
उत्तर-
गांधी जी ने आश्रम की स्थापना दक्षिण अफ्रीका से लौटने के बाद की थी।

लघु उत्तरीय प्रश्न

(क) गांधी जी ने आश्रम की स्थापना कब करनी चाही?
उत्तर-
गांधी ने सन् 1915 में, जब वे दक्षिण अफ्रीका से लौटे तो अहमदाबाद में आश्रम बनाने की योजना बनाई।

(ख) गांधी जी को आश्रम के लिए कितने स्थान की ज़रूरत थी और क्यों?
उत्तर-
साबरमती आश्रम में लगभग 40-50 लोगों के रहने, इनमें हर महीने दस अतिथियों के आने की संभावना, जिनमें तीन या पाँच सपरिवार आने की उम्मीद थी। अतः आश्रम में तीन रसोईघर तथा रहने के मकान के लिए 50,000 फुट क्षेत्रफल में बने मकान की आवश्यकता थी। इसके अलावे-खेती के लिए पाँच एकड़ जमीन की ज़रूरत थी, क्योंकि इतने लोगों के भोजन का सामान खरीदना कठिन था।

दीर्घ उत्तरीय प्रश्न

(क) गांधी जी ने आश्रम के अनुमानित खर्च का ब्यौरा क्यों तैयार किया?
उत्तर-
गांधी जी द्वारा लिखे गए पाठ ‘आश्रम का अनुमानित व्यय’ से हमें सीख मिलती है कि यदि हम कोई भी कार्य करना चाहें तो सोच-समझकर पहले ब्यौरा बना लेना चाहिए ताकि उसके अनुमानित खर्च को भी जाना जा सके तथा इस हिसाब से आगे बढ़ने का रास्ता भी साफ़ दिखाई देने लगता है। गांधी जी एक ऐसे आश्रम की स्थापना कर रहे थे। इसके लिए स्थान की ज़रूरत थी, आवश्यक वस्तुओं, पुस्तकों, भोजन की व्यवस्था करने की जरूरत थी। वहाँ सत्याग्रह तथा स्वदेशी आंदोलन की योजनाएँ तैयार करनी थीं। वह आश्रम एक दो दिन के लिए नहीं, लंबे समय के लिए बनाया जा रहा था। अतः स्थायी व्यवस्था के लिए गांधी ने खर्च का लेखा-जोखा तैयार दिया।

(ख) गांधी जी के अनुसार आश्रम में कौन-कौन से खर्च थे? वह उसे कहाँ से जुटाना चाहते थे?
उत्तर-
गांधी जी के अनुसार यदि अन्य खर्च अहमदाबाद उठा ले, तो वह खाने का खर्च जुटा लेंगे। उनके अनुसार आश्रम के मद में निम्नलिखित खर्च थे।

मकान और जमीन का किराया।
किताबों की अलमारियों का खर्च ।
बढ़ई के औजार।।
मोची के औजार।
चौके के सामान।
एक बैलगाड़ी या घोडागाड़ी।
एक वर्ष में भोजन का खर्च- 6000 रु०।।

मूल्यपूरक प्रश्न

(क) क्या आप इस बात से सहमत हैं कि गांधी जी द्वारा आश्रम संबंधी दृष्टिकोण व्यावहारिक था?
उत्तर-
हाँ, मैं इस बात से सहमत हूँ कि गांधी का आश्रम संबंधी दृष्टिकोण व्यावहारिक था। वे स्वावलंबन पर जोर देते थे, अतः खर्च को न्यूनतम बनाने का प्रयास किया गया है। ऐसा उन्होंने संभव कर दिया था।

Yearly Archives: 2022

MATHS IMPORTANT QUESTIONS & MCQS FOR CLASS 7th

Chapter - 8 Comparing Quantities

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Chapter - 7 Congruence of Triangles

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Chapter - 6 The Triangle and its Properties

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Chapter - 5 Lines and Angles

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Chapter - 4 Simple Equations

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Chapter - 3 Data Handling

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Chapter - 2 Fractions and Decimals

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Chapter -1 Integers Important Questions & MCQs

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Chapter -20 विप्लव गायन

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