Yearly Archives: 2022

Exercise 10A

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:
Let x is the required number

Question 10.
Solution:
Let x be the required number

Question 11.
Solution:
10 % of Rs. 90 = 90 x 10100 = Rs. 9
Required amount = Rs. 90 + Rs. 9 = Rs. 99

Question 12.
Solution:
20 % of Rs. 60 = 60×20100 = 12
Required amount = Rs. 60 – 12 = Rs. 48

Question 13.
Solution:
3 % of x = 9

Question 14.
Solution:
12.5 % of x = 6
⇒ x x 12.5100 = 6

Question 15.
Solution:
Let x % of 84 = 14

Question 16.
Solution:

Exercise 10B

Question 1.
Solution:
Rupesh seemed 495 marks out of 750
Percentage of marks = 495750 x 100 = 66%

Question 2.
Solution:
Monthly salary = Rs. 15625
Increase = 12%
Amount of increase = 15625×12100 = Rs. 1875
New salary = Rs. 15625 + Rs. 1875 = Rs. 17500

Question 3.
Solution:
Excise duty in the beginning = Rs. 950
Reduced duty = Rs. 760
Reduction = Rs. 950 – Rs. 760 = Rs. 190
Reduction percent = 190×100950 = 20%

Question 4.
Solution:
Let x be the total cost of the T.V
96% of x = 10464

Total cost of T.V = Rs. 10900

Question 5.
Solution:
Let number of students = x
In a school boys = 70%
girls = 100 – 70 = 30%
Now 30% of x = 504

Number of boys = 1680 – 504 = 1176
and number of total students = 1680

Question 6.
Solution:
Copper required = 69 kg
copper in ore = 12%
Let quantity of ore = x kg
12% of x = 69

Quantity of ore = 575kg

Question 7.
Solution:
Pass marks = 36%
A students gets marks = 123
But failed by 39 marks
Pass marks = 123 + 39 = 162
Now, 36% of maximum marks = 162
Maximum marks = 162×10036 = 450 marks

Question 8.
Solution:
Let number of apples = x
Number of apples sold = 40% of x

Hence number of apples = 700

Question 9.
Solution:
Let total number of examinees = x
the numbers of examinees who passed = 72% of x
= xx72100
= 18×25

Question 10.
Solution:
Let the gross value of moped = x
Amount of commission = 5% of x

Question 11.
Solution:
Total gunpowder = 8 kg
Amount of nitre = 75%
amount of sulphur = 10%
Rest of powder which is charcoal = 100 – (75 + 10) = 100 – 85 = 15 = 15%
Amount of charcoal = 8 x 15100 = 120100
= 65 kg = 1 kg 200 grams = 1.2 kg

Question 12.
Solution:
Quantity of chalk = 1 kg or 1000 g

Question 13.
Solution:
Let total number of days on which the
school open = x
and Sonal’s attendance = 75%
x x 75% = 219

No. of days on which was school open = 292 days

Question 14.
Solution:
Rate of commission = 3%
Amount of commission = Rs. 42660
Let value of property = x
then 3% of x = 42660

Question 15.
Solution:
Total votes of the constituency = 60000
Votes polled = 80% of total votes
= 80100 x 60000 = 48000
Votes polled in favour of A = 60% of polled votes
= 60100 x 48000 = 28800
Votes polled in favour of B = 48000 – 28800 = 19200

Question 16.
Solution:
Let original price of shirt = Rs. x
Discount = 12%

Original price of shirt = Rs. 1350

Question 17.
Solution:
Let original price of sweater = x
Rate of increase = 8%
Increased price = x + 8% of x

Hence, original price of sweater = Rs. 1450

Question 18.
Solution:
Let total income = x

Question 19.
Solution:
Let the given number = 100
Then increase % = 20%

Decrease = 100 – 96 = 4
Decrease per cent = 4%

Question 20.
Solution:
Let original salary of the officer = Rs. 100
Increase = 20%

Question 21.
Solution:
Rate of commission = 2% on first Rs. 200000
1 % on next Rs. 200000 and 0.5% on remaining price
Sale price of property = Rs.200000 + 200000 +140000 = Rs. 540000
Now commission earned by the
= Rs. 200000 x 2% + Rs. 200000 x 1% + 140000 x 0.5%

Question 22.
Solution:
Let Akhil’s income = Rs. 100
Then income of Nikhil’s will be = Rs. 100 – 20 = Rs. 80
Amount which is more than that of Akhil’s = 100 – 80 = Rs. 20
% age = 20×10080 = 25%

Question 23.
Solution:
Let income of Mr Thomas = Rs. 100
then income of John = Rs. 100 + 20 = Rs. 120
Income of Mr Thomas is less than John = Rs. 120 – 100 = Rs. 20
% age = 20×100120
= 503 = 1623 %

Question 24.
Solution:
Present value of machine = Rs. 387000
Rate of depreciation = 10%
Let 1 year ago the value of machine was = x

1 year ago, value of machine = Rs. 430000

Question 25.
Solution:
Present value of car = Rs. 450000
Rate of decreasing of value = 20%
Value after 2 years

Question 26.
Solution:
Present population = 60000
Rate of increase = 10%
Increased population after 2 years

Question 27.
Solution:
Let the price of sugar = Rs. 100
and consumption = 100 kg.
Increase price of 100 kg = Rs. 100 + 25 = Rs. 125
Now increased amount on 100 kg = Rs. 125

Exercise 10C

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(b) 34 = 34 x 100 = 75 %

Question 2.
Solution:
(c)
2 : 5 = 25 = 25 x 100 = 40%

Question 3.
Solution:
(c)

Question 4.
Solution:
(c) x% of 75 = 9

Question 5.
Solution:
(d)

Question 6.
Solution:
(b)
Let x% of 1 day = 36 minutes

Question 7.
Solution:
(a)
Let x be the required number

Question 8.
Solution:
(b)
Let x be the required number, then

Question 9.
Solution:
(d)
Let ore = x, then
5% of x = 400g
xx5100 400

Question 10.
Solution:
(b)
Let gross value of T.V = x
Commission = 10%
After deducting commission, the value

Question 11.
Solution:
(b)
Increase in salary = 25%
Let original salary = x

Question 12.
Solution:
(c)
Let x be the number of total examinees
No. of examinees passed

Question 13.
Solution:
(c)
Let total number of apples = x

Question 14.
Solution:
(b)
Present value of machine = Rs. 25000
Rate of depreciation = 10% p.a.
Value of machine after one year

Question 15.
Solution:
(c) Let x be numbers

Question 16.
Solution:
(c) 60% of 450
= 60100 x 450 = 270

Question 17.
Solution:
(d) Rate of reduction = 6%
Price after reduction = Rs. 658

Question 18.
Solution:
(b) Boys = 70% of students
No. of girls = 240
Girls percentage = 100 – 70 = 30%

Question 19.
Solution:
(c) Let number = x
11% of x – 7% of x = 18

Question 20.
Solution:
(a) Let number = x
35% of x + 39 = x

Question 21.
Solution:
(c) Pass marks = 36%
A students get =145 marks
But failed by 3 5 marks
Then pass marks = 145 + 35 = 180
Maximum marks = 180×10036 = 500

Question 22.
Solution:
(d) Let number be = x
Then decreasing by 40%

Exercise 9A

Question 1.
Solution:
Cost of 15 oranges = Rs. 110
Cost of 1 orange = Rs. 11015
and cost of 39 oranges = Rs. 11015 x 39
= Rs. 22 x 13 = Rs. 286

Question 2.
Solution:
In Rs. 260, the sugar is bought = 8 kg
and in Re. 1, the sugar is bought = 8260 kg
Then in Rs. 877.50, the sugar will be bought = 8260 x 877.50 kg
= 8260 x 87750100
= 27 kg

Question 3.
Solution:
In Rs. 6290, silk is purchased = 37 m
and in Re. 1, silk is purchased = 376290 m
and in Rs. 4420, silk will be purchased 37
= 376290 x 4420 m = 26 m

Question 4.
Solution:
Rs. 1110 is wages for = 6 days.
Re. 1 will be wages for = 61110 days
and Rs. 4625 will be wages for

Question 5.
Solution:
In 42 litres of petrol, a car covers = 357 km
and in 1 litre, car will cover = 35742 km
and in 12 litres, car will cover = 35742 x 12 = 102 km

Question 6.
Solution:
Cost of travelling 900 km is = Rs. 2520
and cost of 1 km will be = Rs. 2520900
andcostof360kmwillbe = Rs. 2520900 x 360 = Rs. 1008

Question 7.
Solution:
To cover a distance of 51 km, time is taken = 45 minutes

Question 8.
Solution:
If weight is 85.5 kg, then length of iron rod = 22.5 m
If weight is 1 kg, then length of rod will be

Question 9.
Solution:
In 162 grams, sheets are = 6

Question 10.
Solution:
1152 bars of soap can be packed in 8 cartons
1 bar of soap coil be packed in

Question 11.
Solution:
In 44 mm of thickness, cardboards are = 16
In 1 mm of thickness, cardboards will be

Question 12.
Solution:
If length of shadow is 8.2 m, then
height of flag staff is = 7 m
If length of shadow is 1 m, then height will

Question 13.
Solution:
16.25 m long wall is build by = 15 men

Question 14.
Solution:
1350 litres of milk cm be consumed by = 60 patients
1 litres of milk can be consumed by = 601350 patients
and 1710 litres of milk can be consumed

Question 15.
Solution:
2.8 cm extension is produced by = 150 g.
1 cm extension will be produced by = 1502.8 g
and 19.6 cm extension will be produced by

Exercise 9B

Question 1.
Solution:
48 men can dig a trench in = 14 days
1 man will dig the trench in = 14 x 48 days (less men more days)
28 men will dig the trench m = 14×4828 (more men less days)
= 24 days

Question 2.
Solution:
In 30 days, a field is reaped by = 16 men
In 1 day, it will be reaped by = 16 x 30 (less days, more men)
and in 24 days, it will be reaped by = 16×3024 men (more days, less men)
= 48024
= 20 men

Question 3.
Solution:
In 13 days, a field is grazed by = 45 cows.
In 1 day, the field will be grazed by = 45 x 13 cows (less days, more cows)
and in 9 days the field will be grazed by = 45×139 cows (more days, less cows)
= 5 x 13 = 65 cows

Question 4.
Solution:
16 horses can consume corn in = 25 days
1 horse will consume it in = 25 x 16 days (Less horse, more days)
and 40 horses will consume it in = 25×1640 days (more horses, less days)
= 10 days

Question 5.
Solution:
By reading 18 pages a day, a book is finished in = 25 days
By reading 1 page a day, it will be finished = 25 x 18 days (Less page, more days)
and by reading 15 pages a day, it will be finished in = 25×1815 days (more pages, less days)
= 5 x 6 = 30 days

Question 6.
Solution:
Reeta types a document by typing 40 words a minute in = 24 minutes
She will type it by typing 1 word a minute in = 24 x 40 minutes (Less speed, more time)
Her friend will type it by typing 48 words 24 x 40 a minute in = 24×4048 minutes
(more speed, less time)
= 20 minutes

Question 7.
Solution:
With a speed of 45 km/h, a bus covers a distance in = 3 hours 20 minutes
= 313 = 103 hours
With a speed of 1 km/h it will cover the distance m = 10×453 h
(Less speed, more time)
and with a speed of 36 km/h, it will cover the distance in
= 10x453x36 hr (more speed, less time)
= 256 h
= 416 h
= 4 hr 10 minutes

Question 8.
Solution:
To make 240 tonnes of steel, material is sufficient in = 1 month or 30 days
To make 1 tonne of steel, it will be sufficient in = 30 x 240 days (Less steel, more days)
To make 240 + 60 = 300 tonnes of steel it will be sufficient in = 30×240300 days
= 24 days (more steel, less days)

Question 9.
Solution:
In the beginning, number of men = 210
After 12 days, more men employed = 70
Total men = 210 + 70 = 280
Total period = 60 days.
After 12 days, remaining period = 60 – 12 = 48 days
Now 210 men can build the house in = 48 days
and 1 man can build the house in = 48 x 210 days (less men, more days) .
280 men can build the house in = 48×210280 days
(more men, less days)
= 36 days

Question 10.
Solution:
In 25 days, the food is sufficient for = 630 men
In 1 day, the food will be sufficient for = 630 x 25 men (less days, more men)
and in 30 days, the food will be sufficient for = 630×2530 hr
(more days less men)
= 525 men
Number of men to be transfered = 630 – 525 = 105 men

Question 11.
Solution:
Number of men in the beginning = 120
Number of men died = 30
Remaining = 120 – 30 = 90 men
Total period = 200 days
No. of days passed = 5
Remaining period = 200 – 5 = 195
Now, The food lasts for 120 men for = 195 days
The food will last for 1 man for = 195 x 120 days (Less men, more days)
The food will last for 90 men for = 195×12090
(more men less days)
= 65 x 4 = 260 days

Question 12.
Solution:
Period in the beginning = 28 days
No. of days passed = 4 days.
Remaining period = 28 – 4 = 24 days
The food is sufficient for 24 days for = 1200 soldiers
The food will be sufficient for 1 day for = 1200 x 24 soldiers (Less days, more men)
and the food will be sufficient for 32 days = 1200×2432
= 900 soldiers (more days, less men)
No. of soldiers who left the fort = 1200 – 900 = 300 soldiers

Exercise 9C

Objective Questions.
Marks (✓) against the correct answer in each of the following :
Question 1.
Solution:
(c)
Weight of 4.5 m rod = 17.1 kg

Question 2.
Solution:
(d) None of these 0.8 cm represent the map = 8.8 km

Question 3.
Solution:
(c) In 20 minutes, Raghu covers = 5 km
in 1 minutes, he will cover = 520 km
and in 50 minutes, he will cover

Question 4.
Solution:
(d)
No. of men in the beginning = 500
More men arrived = 300
No. of total men = 500 + 300 = 800
For 500 men, provision are for = 24 days
For 1 man, provision will be = 24 x 500 days (less men, more days)
and for 800 men, provision will be = 24800 x 500 days
(more men less days)
= 15 days

Question 5.
Solution:
(b) Total cistern = 1
Filled in 1 minute = 45
Unfilled = 1 – 45 = 15
45 of cistern is filled in = 1 minutes = 60 seconds
1 full cistern can be filled in = 60×54 = 75 seconds
More time = 75 – 60 = 15 seconds

Question 6.
Solution:
(a)
15 buffaloes can eat as much as = 21 cows
1 buffalo will eat as much as = 2115 cows
35 buffaloes will eat as much as
= 21×3515 cm = 49 cows

Question 7.
Solution:
(b) 4 m long shadow is of a tree of height = 6 m
1 m long shadow of flagpole will of height = 64 m
50 m long shadow, the height of pole 6 will be = 64 x 50 = 75 m

Question 8.
Solution:
(b) 8 men can finish the work in = 40 days
1 man will finish it in=40 x 8 days (less men, more days)
8 + 2 = 10 men will finish it in = 40×810 days
(more men, less days)
= 32 days

Question 9.
Solution:
(b)
16 men can reap a field in = 30 days
1 man will reap the field in = 30 x 16 days
and 20 men will reap the field in = 30×1620 = 24 days

Question 10.
Solution:
(c) 10 pipe can fill tank in = 24 minutes
1 pipe will fill it in = 24 x 10 minutes (less pipe, more time)
and 10 – 2 = 8 pipes will fill the tank in
= 24×108 = 30 minutes

Question 11.
Solution:
(d) 6 dozen or 6 x 12 = 72 eggs
Cost of 72 eggs is = Rs. 108
Cost of 1 egg will be = Rs. 10872
and cost of 132 eggs will be 108
= Rs. 10872 x 132 = Rs. 198

Question 12.
Solution:
(b) 12 workers take to complete the work = 4 hrs.
1 worker will take = 4 x 12 hrs. (less worker, more time)
15 workers will take = 4×1215 hrs. (more workers, less time)
= 165 hr. = 3 hrs. 12 min

Question 13.
Solution:
(a) 27 days – 3 days = 24 days
Men = 500 + 300 = 800
For 500 men, provision is sufficient = 24 days
For 1 man, provision will be = 24 x 500 (less man, more days)
and for 500 + 300 = 800 men provision
will be sufficient = 24×500800 = 15 days
(more men, less days)

Question 14.
Solution:
(c) No. of rounds of rope = 140
Radius of base of cylinder = 14 cm
Radius of second cylinder of cylinder = 20 cm
If radius is 14 cm, then rounds of rope are = 140
If radius is 1 cm, then round = 140 x 14 (less radius more rounds)
and if radius is 20 cm, then rounds will
be = 140×1420 = 98 (more radius less rounds)

Question 15.
Solution:
(d) A worker makes toy in 23 hr= 1
He will make toys in 1 hr = 1 x 32
and will make toys in 223 hrs. = 1 x 32 x 223
= 11 (more time more toys)

Question 16.
Solution:
(d) A wall is constructed in 8 days by = 10 men
It will be constructed in 1 day by = 10 x 8 men (less time, more men)
10 x 8

More men required = 160 – 10 = 150

Exercise 8A

Question 1.
Solution:
(i) 24 : 40
HCF of 24, 40 = 8
24 : 40 = 24 ÷ 8 : 40 ÷ 8 = 3 : 5 (Dividing by 8)
(ii) 13.5 : 15 or 135 : 150
HCF of 135 and 150 = 15
135 ÷ 15 : 150 ÷ 15 (Dividing by 15)
= 9 : 10

HCF of 25, 65, 80 = 5
Dividing by 5,
5 : 13 : 16

Question 2.
Solution:
(i) 75 paise : 3 rupees = 75 paise : 300 paise
(converting into same kind)
HCF of 75, 300 = 75
75 : 300 = 75 ÷ 75 : 300 ÷ 75 (Dividing by 75) = 1 : 4
(ii) 1 m 5 cm : 63 cm = 105 cm : 63 cm
(converting into same kind)
HCF of 105 and 63 = 21
105 ÷ 21 : 63 ÷ 21 (Dividing by 21)
= 5 : 3
(iii) 1 hour 5 minutes : 45 minutes = 65 minutes : 45 minutes
(converting into minutes)
13 : 9 (dividing by 5)
= 13 : 9
(iv) 8 months : 1 year = 8 months : 12 months
(converting into the same kind)
HCF of 8 and 12 = 4
Dividing by 4
8 ÷ 4 : 12 ÷ 4
= 2 : 3
(v) 2 kg 250 g : 3 kg = 2250 g : 3000 g (converting into the same kind)
HCF of 2250 and 3000 = 750
Dividing by 750,
2250 ÷ 750 : 3000 ÷ 750 = 3 : 4
(vi) 1 km : 750 m = 1000 m : 750 m
(converting into metre)
= 4 : 3 (dividing by 250)
= 4 : 3

Question 3.
Solution:

Question 4.
Solution:
A : B = 5 : 8, B : C = 16 : 25

Question 5.
Solution:
A : B = 3 : 5, B : C = 10 : 13

A : B : C = 6 : 10 : 13

Question 6.
Solution:
A : B = 5 : 6 and B : C = 4 : 7

Question 7.
Solution:
Total amount = Rs. 360
Sum of ratios = 7 + 8 = 15

Question 8.
Solution:
Total amount = Rs. 880

Question 9.
Solution:
Total amount = Rs. 5600
Ratio in A : B : C = 1 : 3 : 4

Question 10.
Solution:
Let x be added to each term Then
9 + x : 16 + x = 2 : 3

Question 11.
Solution:
Let x be subtracted from each term Then
(17 – x) : (33 – x) = 7 : 15

Question 12.
Solution:
Ratio in two numbers = 7 : 11
Let first number = 7x
Then second number = 11x
Then adding 7 to each number, the ratio is 2 : 3
7x+711x+7 = 23
By cross multiplying:
3 (7x + 7) = 2 (11x + 7)
⇒ 21x + 21 = 22x + 14
⇒ 21 – 14 = 22x – 21x
⇒ x = 7
First number = 7x = 7 x 7 = 49
and second number = 11x = 11 x 7 = 77
Hence numbers are 49, 77

Question 13.
Solution:
The ratio in two numbers = 5 : 9
Let the first number = 5x
Then second number = 9x
By subtracting 3 from each number the ratio is 1 : 2
5x–39x–3 = 12
By cross multiplication,
2 (5x – 3) = 1 (9x – 3)
⇒ 10x – 6 = 9x – 3
⇒ 10x – 9x = -3 + 6
⇒ x = 3
First number = 5x = 5 x 3 = 15
and second .number = 9x = 9 x 3 = 27
Hence numbers are 15, 27

Question 14.
Solution:
Ratio in two numbers = 3 : 4
LCM = 180
Let first number = 3x
Then second number = 4x
Now LCM = 3 x 4 x x = 12x
12x = 180
⇒ x = 15
Numbers will be 3 x 15 = 45 and 4 x 15 = 60

Question 15.
Solution:
Ratio in present ages of A and B = 8 : 3
Let A’s age = 8x
Then B’s age = 3x
6 years hence,
A’s age will be = 8x + 6
and B’s will be = 3x + 6
8x+63x+6 = 94
(By cross multiplication)
4 (8x + 6) = 9 (3x + 6)
⇒ 32x + 24 = 27x + 54
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s present age = 8x = 8 x 6 = 48 years
and B’s age = 3x = 3 x 6 = 18 years

Question 16.
Solution:
Ratio in copper and zinc = 9 : 5
Let alloy = x gm

Question 17.
Solution:
Ratio in boys and girls = 8 : 3
and total number of girls = 375
Let number of boys = 8x
Then number of girls = 3x
3x = 375
⇒ x = 125
Number of boys = 8x = 8 x 125 = 1000

Question 18.
Solution:
Ratio in income and savings = 11 : 2
Let income = 11x
Then savings = 2x
But savings = Rs. 2500
2x = 2500
⇒ x = 1250
Then income = 1250 x 11 = Rs. 13750
and expenditure = Total income – savings = 13750 – 2500 = Rs. 11250

Question 19.
Solution:
Total amount = Rs. 750
Ratio in rupee, 50 P and 25 P coins =5 : 8 : 4
Let number of rupees = 5x
Number of 50 P coins = 8x
and number of 25 coins = 4x
According to the condition,

Number of 1 Re coins = 5x = 5 x 75 = 375
Number of 50 P coins = 8x = 8 x 75 = 600
and number of 25 P coins = 4x = 4 x 75 = 300

Question 20.
Solution:
(4x + 5) : (3x + 11) = 13 : 17
4x+53x+11 = 1317
By cross multiplication,
68x + 85 = 39x + 143
⇒ 68x – 39x = 143 – 85
⇒ 29x = 58
x = 2
Hence x = 2

Question 21.
Solution:
x : y = 3 : 4

Question 22.
Solution:
x : y = 6 : 11
xy = 611
Now (8x – 3y) : (3x + 2y)

Question 23.
Solution:
Sum of two numbers = 720
Ratio of two numbers = 5 : 7
Let first number = 5x
Then second number = 7x
5x + 7x = 720
⇒ 12x = 720
⇒ x = 60
First number = 5x = 5 x 60 = 300
and second number = 7x = 7 x 60 = 420

Question 24.
Solution:
(i) (5 : 6) or (7 : 9)

Question 25.
Solution:
(i) (5 : 6), (8 : 9), (11 : 18)

Exercise 8B

Question 1.
Solution:
We know that a, b, c, d are in proportion if ad = bc
Now 30, 40, 45, 60 are in proportion
if 30 x 60 = 40 x 45
if 1800= 1800
which is true
30, 40, 45, 60 are in proportion.

Question 2.
Solution:
We know that if a, b, c, d are in proportion if ad = bc
Now 36, 49, 6, 7 are in proportion
if 36 x 7 = 49 x 6
if 252 = 294
But 252 ≠ 294
36, 49, 6, 7 are not in proportion

Question 3.
Solution:
2 : 9 : : x : 27
9 x x = 2 x 27
x = 2×279 = 2 x 3 = 6

Question 4.
Solution:
8 : x : : 16 : 35
x x 16 = 8 x 35
x = 8×356 = 352 = 17.5

Question 5.
Solution:
x : 35 : : 48 : 60
x x 60 = 35 x 48
x = 7 x 4 = 28

Question 6.
Solution:
Let x be the fourth proportional, then

x = 352 = 17.5
Fourth proportional = 17.5

Question 7.
Solution:
36, 54, x are in continued proportion
36 : 54 : : 54 : x
⇒ 36 x x = 54 x 54

Question 8.
Solution:
27, 36, x are in continued proportion
27 : 36 :: 36 : x
27 x x = 36 x 36

Question 9.
Solution:
Let x be the third proportional, then
(i) 8 : 12 : : 12 : x

Question 10.
Solution:
Third proportional = 28 then
7 : x :: x : 28
⇒ 7 x 28 = x x x
⇒ x² = 28 x 7 = 196
⇒ x = √196 = 14
x = 14

Question 11.
Solution:
Let x be the mean proportional, then
(i) 6 : x :: x : 24
⇒ x² = 6 x 24 = 144
x = √144 = 12
Mean proportional = 12
(ii) 3 : x : : x : 27
⇒ x² = 3 x 27 = 81
x = √81 = 9
Mean proportional = 9
(iii) 0.4 : x :: x : 0.9
⇒ x² = 0.4 x 0.9
x = √o.36 = 0.6
Mean proportional = 0.6

Question 12.
Solution:
Let x be added to each of the given numbers then
5 + x, 9 + x, 7 + x, 12 + x are in proportion
5+x9+x = 7+x12+x
By cross multiplication :
(5 + x) (12 + x) = (7 + x) (9 + x)
⇒ 60 + 5x + 12x + x² = 63 + 7x + 9x + x²
⇒ 60 + 17x + x² = 63 + 16x + x²
⇒ 17x + x² – 16x – x² = 63 – 60
⇒ x = 3
Required number = 3

Question 13.
Solution:
Let x be subtracted from each of the given number, then
10 – x, 12 – x, 19 – x and 24 – x are in proportion
10–x12–x = 19–x24–x
By cross multiplication :
(10 – x) (24 – x) = (19 – x) (12 – x)
⇒ 240 – 10x – 24x + x² = 228 – 19x – 12x + x²
⇒ 240 – 34x + x² = 228 – 31x + x²
⇒ -34x + x² + 31x – x² = 228 – 240
⇒ -3x = -2
⇒ 3x = 12
⇒ x = 4
Required number = 4

Question 14.
Solution:
Scale of map = 1 : 5000000
Distance between two town on the map = 4 cm

Question 15.
Solution:
Height of a tree = 6 cm
and its shadow at same time = 8 m
Shadow of a pole = 20 m
Let height of pole = x m
6 : 8 = x : 20
⇒ x= 6×208 = 15 m
Height of pole = 15 m

Exercise 8C

Objective questions :
Mark (✓) against the correct answers in each of the following :
Question 1.
Solution:
(d) a : b = 3 : 4, b : c = 8 : 9

Question 2
Solution:
(a) A : B = 2 : 3, B : C = 4 : 5

Question 3.
Solution:
(d)

Question 4.
Solution:
(b) 15% of A = 20% of B

Question 5.
Solution:
(a)

Question 6.
Solution:
(b) A : B = 5 : 7, B : C = 6 : 11
LCM of 7, 6 = 42

Question 7.
Solution:
(c) 2A = 3B = 4C = x

Question 8.
Solution:
(a)
A3 = B4 = C5 = 1(suppose)
A = 3, B = 4, C = 5
A : B : C = 3 : 4 : 5

Question 9.
Solution:
(b)

Question 10.
Solution:
(c)

Question 11.
Solution:
(c) (3a + 5b) : (3a – 5b) = 5 : 1

Question 12.
Solution:
(c) 7 : x :: 35 : 45

x = 9

Question 13.
Solution:
(b) Let x to be added to each term of 3 : 5
Then 3+x5+x = 56
By cross multiplication
18 + 6x = 25 + 5x
6x – 5x = 25 – 18
x = 7
7 is to be added

Question 14.
Solution:
(d) Ratio in two numbers = 3 : 5
Let first number = 3x
Then second number = 5x
According to the condition,
3x+105x+10 = 57
(By cross multiplication)
25x + 50 = 21x + 70
25x – 21x = 70 – 50
4x = 20
x = 5
First number = 3 x 5 = 15
and second number = 5 x 5 = 25
Sum of numbers = 15 + 25 = 40

Question 15.
Solution:
(a)
Let x be subtracted from each of the term
15–x19–x = 34
⇒ 4 (15 – x) = 3 (19 – x)
⇒ 60 – 4x = 57 – 3x
⇒ -4x + 3x = 57 – 60
⇒ -x = -3
x = 3
Required number = 3

Question 16.
Solution:
(a)
Amount = Rs. 420
and ratio = 3 : 4
Sum of ratios = 3 + 4 = 7
A’s share = 420×37 = Rs. 60 x 3 = Rs. 180

Question 17.
Solution:
(d)
Let number of boys = x, then
x : 160 : : 8 : 5
⇒ x x 5 = 160 x 8
x = 160×85 = 32 x 8 = 256
Number of total students of the school = 256 + 160 = 416

Question 18.
Solution:
(a)

Question 19.
Solution:
(c)
Let x be the third proportional to 9 and 12 then
9 : 12 :: x : 12
⇒ 9 x x = 12 x 12
⇒ x = 12×129 = 1449 = 16
Third proportional = 16

Question 20.
Solution:
Answer = (b)
Mean proportional of 9 and 16 = √(9 x 16) = √144 = 12

Question 21.
Solution:
(a)
Let age of A = 3x
and age of B = 8x
6 years hence, their ages will be 3x + 6 and 8x + 6
3x+68x+6 = 49
⇒ 9 (3x + 6) = 4 (8x + 6)
⇒ 27x + 54 = 32x + 24
⇒ 32x – 27x = 54 – 24
⇒ 5x = 30
⇒ x = 6
A’s age = 3x = 3 x 6 = 18 years

Exercise 7A

Solve the following equations. Check your result in each case.
Question 1.
Solution:
3x – 5 = 0
Adding 5 to both sides
3x – 5 + 5 = 0 + 5
⇒ 3x = 5
⇒ x = 53
Check:
L.H.S. = 3x – 5
= 3 x 53 – 5
= 5 – 5
= 0
= R.H.S.
Hence x = 53

Question 2.
Solution:
8x – 3 = 9 – 2x
⇒ 8x + 2x = 9 + 3 (By transposing)
⇒ 10x = 12

Question 3.
Solution:
7 – 5x = 5 – 7x
⇒ – 5x + 7x = 5 – 7 (By transposing)
⇒ 2x = -2
x = -1
Check:
L.H.S. = 7 – 5x = 7 – 5(-1) = 7 + 5 = 12
R.H.S. = 5 – 7x = 5 – 7(-1) = 5 + 7 = 12
L.H.S. = R.H.S.
Hence x = -1

Question 4.
Solution:
3 + 2x = 1 – x
⇒ 2x + x = 1 – 3 (By transposing)
⇒ 3x = -2

Question 5.
Solution:
2(x – 2) + 3(4x – 1) = 0
⇒ 2x – 4 + 12x – 3 = 0
⇒ 2x + 12x = 4 + 3 (By transposing)
⇒ 14x = 7
⇒ x = 714 = 12
Check : L.H.S. = 2(x – 2) + 3 (4x -1)

Question 6.
Solution:
5 (2x – 3) – 3(3x – 7) = 5
⇒ 10x – 15 – 9x + 21 = 5
⇒ 10x – 9x – 15 + 21 = 5
⇒ 10x – 9x = 5 + 15 – 21 (By transposing)
⇒ x = 20 – 21 = -1
⇒ x = -1
Check:
L.H.S. = 5 (2x – 3) – 3(3x – 7)
= 5[2 x (-1) -3] -3[3 (-1) -7] = 5[-2 – 3] – 3[-3 – 7]
= 5 x (-5) -3 x (-10)
= -25 + 30
= 5 = R.H.S.
Hence x = -1

Question 7.
Solution:

Question 8.
Solution:

L.H.S. = R.H.S.
Hence x = 48

Question 9.
Solution:

Question 10.
Solution:
3x + 2(x + 2) = 20 – (2x – 5)
⇒ 3x + 2x + 4 = 20 – 2x + 5
⇒ 5x + 4 = 25 – 2x
⇒ 5x + 2x = 25 – 4 (By transposing)
⇒ 7x = 21
⇒ x = 3
Check:
L.H.S.= 3x + [2(x + 2)] = 3 x 3 + 2(3 + 2) = 9 + 2 x 5 = 9 + 10 = 19
R.H.S. = 20 – (2x – 5) = 20 – (2 x 3 – 5) = 20 – (6 – 5) = 20 – 1 = 19
L.H.S. = R.H.S.
Hence x = 3

Question 11.
Solution:
13(y – 4) – 3(y – 9) – 5(y + 4) = 0
⇒ 13y – 52 – 3y + 27 – 5y – 20 = 0
⇒ 13y – 3y – 5y – 52 + 27 – 20 = 0
⇒ 13y – 8y – 72 + 27 = 0
⇒ 5y – 45 = 0
⇒ 5y = 45 (By transposing)
⇒ y = 9
Check:
L.H.S. = 13(y – 4) – 3(y – 9) – 5(y + 4)
= 13(9 – 4) – 3(9 – 9) – 5(9 + 4)
= 13 x 5 – 3 x 0 – 5 x 13
= 65 – 0 – 65 = 0 = R.H.S.
Hence y = 9

Question 12.
Solution:
2m+53 = 3m – 10
⇒ 2m + 5 = 3 (3m – 10) (By cross multiplication)
⇒ 2m + 5 = 9m – 30
⇒ 2m – 9m = -30 – 5
⇒ -7m = -35
⇒ m = 5
m = 5
Check:

R.H.S. = 3m – 10 = 3 x 5 – 10 = 15 – 10 = 5
L.H.S. = R.H.S.
Hence m = 5

Question 13.
Solution:
6(3x + 2) – 5(6x – 1) = 3(x – 8) – 5(7x – 6) + 9x
⇒ 18x + 12 – 30x + 5 = 3x – 24 – 35x + 30 + 9x
⇒ 18x – 30x + 12 + 5 = 3x – 35x + 9x – 24 + 30
⇒ -12x + 17 = -23x + 6
⇒ – 12x + 23x = 6 – 17
⇒ 11x = -11
x = – 1
Check:
L.H.S. = 6(3x + 2) – 5(6x – 1)
= 6[3x (-1) + 2] – 5[6 x (-1) x -1]
= 6[-3 + 2] – 5[-6 – 1]
= 6 x (-1) – 5 x (-7)
= -6 + 35 = 29
R.H.S. = 3(x – 8) – 5 (7x – 6) + 9x
= 3[-1 – 8] -5 [7 x (-1) – 6] + 9 (-1)
= 3 x (-9) – 5 [-7 – 6] – 9
= -27 – 5(-13) – 9
= -27 + 65 – 9
= 65 – 36 = 29 .
L.H.S. = R.H.S.
Hence x = -1

Question 14.
Solution:
t – (2t + 5) – 5(1 – 2t) = 2(3 + 4t) – 3(t – 4)
⇒ t – 2t – 5 – 5 + 10t = 6 + 8t – 3t + 12t
⇒ t – 2t + 10t – 10 = 8t – 3t + 18
⇒ 9t – 10 = 5t + 18
⇒ 9t – 5t = 18 + 10 (By transposing)
⇒ 4t = 28
⇒ t = 7
Check:
L.H.S. = t – [2t + 5] -5[1 – 2t]
= 7 – [2 x 7 + 5] – 5[1 – 2 x 7]
= 7 – [14 + 5] – 5 [1 – 14]
= 7 – 19 – 5(-13)
= 7 – 19 + 65
= 72 – 19 = 53
R.H.S. = 2[3 + 4t) – 3(t – 4)
= 2 (3 + 4 x 7) – 3(7 – 4)
= 2(3 + 28) – 3(3)
= 2(31) – 9 = 62 – 9 = 53
L.H.S. = R.H.S.
Hence t = 7 Ans.

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
0.18 (5x – 4) = 0.5x + 0.8

Question 29.
Solution:
2.4 (3 – x) – 0.6 (2x – 3) = 0
⇒ 7.2 – 2.4x – 1.2x + 1.8 = 0
⇒ -2.4x – 1.2x = – (7.2 + 1.8).
L.H.S. = 2.4 (3 – x) – 0.6 (2x – 3)
⇒ 2.4 (3 – 2.5) – 0.6 (2 x 2.5 – 3)
⇒ 2.4 (0.5) – 0.6 (5 – 3)
⇒ 1.2 – 0.6 x 2 = 1.2 – 1.2 = 0 = R.H.S.
Hence x = 2.5

Question 30.
Solution:
0.5x – (0.8 – 0.2x) = 0.2 – 0.3x
⇒ 0.5x – 0.8 + 0.2x = 0.2 – 0.3x
⇒ 0.5x + 0.2x + 0.3x = 0.2 + 0.8
⇒ 1.0x = 1.0
⇒ x = 1
Check :
L.H.S. = 0.5x – (0.8 – 0.2x)
= 0.5 x 1 – (0.8 – 0.2 x 1)
= 0.5 – (0.8 – 0.2) = 0.5 – 0.6 = -0.1
R.H.S. = 0.2 – 0.3x = 0.2 – 0.3 x 1 = 0.2 – 0.3 = -0.1
L.H.S. = R.H.S.
Hence x = 1

Question 31.
Solution:

Question 32.
Solution:

Exercise 7B

Question 1.
Solution:
Let the required number = x
Then 2x – 7 = 45
2x = 45 + 7 = 52
x = 26
Required number = 26

Question 2.
Solution:
Let the required number = x Then
3x + 5 = 44
⇒ 3x = 44 – 5 = 39
x = 13
Required number = 13

Question 3.
Solution:
Let the required fraction = x
then 2x + 4 = 265

Question 4.
Solution:
Let the required number = x
and half of .the number = x2

Question 5.
Solution:
Let the required number = x
Two third of the number = 23 x

Question 6.
Solution:
Let the required number = x
Then, 4x = x + 45
⇒ 4x – x = 45
⇒ 3x = 45
⇒ x = 15
Required number = 15

Question 7.
Solution:
Let the required number = x
Then x – 21 = 71 – x
⇒ x + x = 71 + 21
⇒ 2x = 92
⇒ x = 46

Question 8.
Solution:
Let the original number = x
Then 23 of the number = 23 x

Question 9.
Solution:
Let the second number = x
then first number = 25 x
their sum = 70

Question 10.
Solution:
Let the required number = x

Question 11.
Solution:
Let the required number = x
Fifth part of the number = x5
Fourth part of the number = x4

Question 12.
Solution:
Let first natural number = x then
next number = x + 1
x + x + 1 = 63
⇒ 2x = 63 – 1 = 62
x = 31
first number = 31
and second number = 31 + 1 = 32
Numbers are 31, 32

Question 13.
Solution:
Let first odd number = 2x + 1
second odd number = 2x + 3
2x + 1 + 2x + 3 = 76
⇒ 4x + 4 = 76
⇒ 4x = 76 – 4 = 72
x = 18
First number = 2x + 1 = 2 x 18 + 1 = 36 + 1 = 37
Second number = 2x + 3 = 2 x 18 + 3 = 36 + 3 = 39
Numbers are 37, 39

Question 14.
Solution:
Let first positive even number = 2x
Second number = 2x + 2
Third number = 2x + 4
2x + 2x +2 + 2x + 4 = 90
⇒ 6x + 6 = 90
⇒ 6x = 90 – 6 = 84
x = 14
First even number = 2x = 2 x 14 = 28
Second number = 2x + 2 = 2 x 14 + 2 = 28 + 2 = 30
Third number = 30 + 2 = 32
Required numbers are 28, 30, 32

Question 15.
Solution:
Sum of two numbers = 184
Let first number = x
Then second number = 184 – x

First part = 72
Second part = 184 – 72 = 112
Hence parts are 72, 112

Question 16.
Solution:
Total number of notes = 90
Let number of notes of Rs. 5 = x
Then number of notes of Rs.10 = 90 – x
Then x x 5 + (90 – x) x 10 = 500
⇒ 5x + 900 – 10x = 500
⇒ -5x = 500 – 900 = -400
x = 8
Number of 5 rupees notes = 80
and ten rupees notes = 90 – 80 = 10

Question 17.
Solution:
Amount of coins = Rs. 34
Let 50 paisa coins = x
then 25 paisa coins = 2x

Number of 50 paisa coins = 34
and number of 25 paisa coins = 2x = 2 x 34 = 68

Question 18.
Solution:
Let present age of Raju’s cousin = x years
then age of Raju = (x – 19) years
After 5 years,
Raju’s age = x – 19 + 5 = (x – 14) years
and his cousin age = x + 5
(x – 14) : (x + 5) = 2 : 3
⇒ x–14x+5 = 23
⇒ 3(x – 14) = 2 (x + 5) (By cross multiplication)
⇒ 3x – 42 = 2x + 10
⇒ 3x – 2x = 10 + 42
⇒ x = 52
Raju’s age = x – 19 = 52 – 19 = 33 years
and his cousin age = 52 years.

Question 19.
Solution:
Let present age of son = x years
Age of father = (x + 30) years
12 years after,
Father’s age = x + 30 + 12 = (x + 42) years
and son’s age = (x + 12) years
(x + 42) = 3(x + 12)
⇒ x + 42 = 3x + 36
⇒ 3x + 36 = x + 42
⇒ 3x – x = 42 – 36
⇒ 2x = 6
⇒ x = 3
Son’s age = 3 years
Father’s age = 3 + 30 = 33 years

Question 20.
Solution:
Ratio in present ages of Sonal and Manoj = 7 : 5
Let Sonal’s age = 7x
then Manoj’s age = 5x
10 years hence,
Sonal’s age will be = 7x + 10
and Manoj’s age = 5x + 10

Sonal’s present age = 7x = 7 x 5 = 35 years
and Manoj’s age = 5x = 5 x 5 = 25 years

Question 21.
Solution:
Five years ago,
Let Son’s age = x years
and father’s age = 7x years
Present age of son = (x + 5) years
and age of father = (7x + 5) years
5 years hence,
father’s age = 7x + 5 + 5 = 7x + 10
and Son’s age = x + 5 + 5 = x + 10
(7x + 10) = 3(x + 10)
⇒ 7x + 10 = 3x + 30
⇒ 7x – 3x = 30 – 10
⇒ 4x = 20
⇒ x = 5
Father present age = 7x + 5 = 7 x 5 + 5 = 35 + 5 = 40 years
and son’s age = x + 5 = 5 + 5 = 10 years

Question 22.
Solution:
Let age of Manoj 4 years ago = x
then his present age = x + 4
After 12 years his age will be = x + 4 + 12 = x + 16
x + 16 = 3(x)
x + 16 = 3x
⇒ 16 = 3x – x
⇒ 2x = 16
x = 8
His present age = 8 + 4 = 12 years

Question 23.
Solution:
Let total marks = x
Pass marks = 40% of x = 40×100 = 25 x
No. of marks got by Rupa = 185
No. of marks by which she failed = 15
Pass marks = 185 + 15 = 200
25 x = 200
⇒ x = 200×52 x
⇒ x = 500
Hence total marks = 500

Question 24.
Solution:
Sum of digits = 8
Let units digit = x
Then tens digit = 8 – x
and number will be x + 10 (8 – x) ….(i)
By adding 18, the digits are reversed then
units digit = 8 – x
and tens digit = x
Number = (8 – x) = 10x
According to the condition,
(8 – x) + 10x = 18 + x + 10 (8 – x)
⇒ 8 – x + 10x = 18 + x + 80 – 10x
⇒ 10x – x – x + 10x = 18 + 80 – 8
⇒ 18x = 90
⇒ x = 5
Number is
x + 10(8 – x) = 5 + 10(8 – 5) = 5 + 10 x 3 = 35

Question 25.
Solution:
Cost of 3 tables and 2 chairs = 1850
Cost of table = Rs. 75 + cost of a chair
Let cost of chair = Rs. x,
then Cost of table = Rs. 75 + x
According to the condition,
3 (75 + x) + 2x = 1850
⇒ 225 + 3x + 2x = 1850
⇒ 225 + 5x = 1850
⇒ 5x = 1850 – 225 = 1625
x = 325
Cost of chair = Rs. 325
and cost of table = Rs. 325 + 75 = Rs. 400

Question 26.
Solution:
S.P of article = Rs. 495
gain = 10%
Let cost price = Rs. x

Question 27.
Solution:
Perimeter of field = 150 m
Length + Breadth = 1502 = 75 m
[Perimeter = 2(l + b)]
Let length = x Then breadth = 75 – x
Then x = 2(75 – x)
⇒ x = 150 – 2x
⇒ x + 2x = 150
⇒ 3x = 150
⇒ x = 1503 = 50
Length = 50 m
and breadth = 75 – 50 = 25 m

Question 28.
Solution:
Perimeter of an isosceles triangle = 55 m
Let the third side of an isosceles triangle = x
Then each equal side = (2x – 5) m
According to the condition,
x + 2 (2x – 5) = 55
⇒ x + 4x – 10 = 55
⇒ 5x = 55 + 10
⇒ 5x = 65
⇒ x = 13
and 2x – 5 = 2 x 13 – 5 = 21 m
Sides will be 13m, 21m, 21m

Question 29.
Solution:
Sum of two complementary angles = 90°
Let first angle = x
then second = 90° – x
x – (90 – x) = 8
⇒ x – 90 + x = 8
⇒ 2x = 8 + 90
⇒ 2x = 98
⇒ x = 49
first angle = 49°
and second angle = 90° – 49° = 41°
Hence angles are 41°, 49°

Question 30.
Solution:
Sum of two supplementary angles = 180°
Let first angle = x
Then second angle = 180° – x
x – (180° – x) = 44°
⇒ x – 180° + x = 44°
⇒ 2x = 44° + 180° = 224°
⇒ 2x = 224°
⇒ x = 112°
First angle = 112°
and second angle = 180° – 112° = 68°
Hence angles are 68°, 112°

Question 31.
Solution:
In an isosceles triangle
Let each equal base angles = x
Then vertex angle = 2x
According to the condition,
x + x + 2x = 180° (sum of angles of a triangle)
⇒ 4x = 180°
⇒ x = 45°
Then vertex angle = 2x = 2 x 45° = 90°
Angles of the triangle are 45°, 45° and 90°

Question 32.
Solution:
Let length of total journey = x km
According to the condition,

⇒ 39x + 80 = 40x
⇒ 40x – 39x = 80
⇒ x = 80
Total journey = 80km

Question 33.
Solution:
No. of days = 20 Let no. of days he worked = x
Then he will receive amount = x x Rs. 120 = Rs. 120x
No. of days he did not work = 20 – x
Fine paid = (20 – x) x Rs. 10 = Rs. 10(20 -x)
120x – 10 (20 – x) = 1880
⇒ 120x – 200 + 10x = 1880
⇒ 130x = 1880 + 200 = 2080
x = 16
No. of days he remained absent = 20 – x = 20 – 16 = 4 days

Question 34.
Solution:
Let value of property = x

Question 35.
Solution:
Solution = 400 mL
Quantity of alcohol = 15% of 400 mL
= 400×15100 = 60 mL
Let pure alcohol added = x mL
Total solution = 400 + x
and total alcohol = (x + 60)
Now (400 + x) x 32% = x + 60
⇒ (400 + x) x 32100 = x + 60
⇒ 32 (400 + x) = 100 (x + 60)
⇒ 12800 + 32x = 100x + 6000
⇒ 12800 – 6000 = 100x – 32x
⇒ 6800 = 68x
⇒ x = 6800
Pure alcohol added = 100 mL

Exercise 7C

Objective Questions :
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(d)

Question 2.
Solution:
(d)

Question 3.
Solution:
(a)
2n + 5 = 3 (3n – 10)
⇒ 2n + 5 = 9n – 30
⇒ 9n – 2n = 5 + 30
⇒ 7n = 35
⇒ n = 5

Question 4.
Solution:
(c)

Question 5.
Solution:
(c)
8 (2x – 5) – 6 (3x – 7) = 1
⇒ 16x – 40 – 18x + 42 = 1
⇒ -2x + 2 = 1
⇒ -2x = 1 – 2 = -1
x = 12

Question 6.
Solution:
(d)

Question 7.
Solution:
(a)

Question 8.
Solution:
(b)
Let first whole number=x
Then second number = x + 1
and sum = 53
x + x + 1 = 53
⇒ 2x = 53 – 1
⇒ 2x = 52
⇒ x = 26
Smaller number = 26

Question 9.
Solution:
Let first even number = 2x
Then second number = 2x + 2
and sum = 86
2x + 2x + 2 – 86
⇒ 4x = 86 – 2 = 84
⇒ x = 21
Larger even number = 2x + 2 = 2 x 21 + 2 = 42 + 2 = 44

Question 10.
Solution:
(b)
Let first odd number = 2x + 1
Second number = 2x + 3
2x + 1 + 2x + 3 = 36
⇒ 4x + 4 = 36
⇒ 4x = 36 – 4 = 32
⇒ x = 8
Smaller number = 2x + 1 = 2 x 8 + 1 = 16 + 1 = 17

Question 11.
Solution:
(d)
Let number = x
2x + 9 = 31
⇒ 2x = 31 – 9 = 22
⇒ x = 11

Question 12.
Solution:
(a)
Let number = x then
3x + 6 = 24
⇒ 3x = 24 – 6 = 18
⇒ x = 6
Number = 6

Question 13.
Solution:
(a)

Question 14.
Solution:
(b)
Let first angle = x
Then second = 90° – x
x – (90° – x) = 10
⇒ x – 90° + x = 10°
⇒ 2x = 10° + 90° = 100°
x = 50°
Second angle = 90° – 50° = 40°
Larger angle = 50°

Question 15.
Solution:
(b)
Let first angle = x
Then second = 180° – x
x – (180° – x) = 20°
⇒ x – 180° + x = 20°
⇒ 2x = 20° + 180° = 200°
x = 100°
Second angle = 180° – 100° = 80°
Smaller angle = 80°

Question 16.
Solution:
(c)
Let age of A = 5x
Then age of B = 3x
After 6 years,
A’s age = 5x + 6
and B’s age = 3x + 6
5x+63x+6 = 75
⇒ 25x + 30 = 21x + 42
⇒ 25x – 21x = 42 – 30
⇒ 4x = 12
⇒ x = 3
A’s age = 5x = 5 x 3 = 15 years

Question 17.
Solution:
(b)
Let the number = x
According to the condition,
5x = 80 + x
⇒ 5x – x = 80
⇒ 4x = 80
⇒ x = 20
Number = 20

Question 18.
Solution:
(c)
Let width of rectangle = x m
Then length = 3x m
Perimeter = 96 m
2 (x + 3x) = 96
⇒ x + 3x = 962 = 48
⇒ 4x = 48
⇒ x = 12
Length = 3x = 12 x 3 = 36 m

Exercise 6A

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:
Sum of (a + 3b – 4c), (4a – b + 9c) and (-2b + 3c – a)

Now subtract (2a – 3b + 4c) from 4a + 8c
= 4a + 8c – (2a – 3b + 4c)
= 4a + 8c – 2a + 3b – 4c
= 4a – 2a + 3b + 8c – 4c
= 2a + 3b + 4c

Question 4.
Solution:

Question 5.
Solution:
Sum of (8a – 6a² + 9) and (-10a – 8 + 8a²)
= 8a – 6a² + 9 + (-10a) – 8 + 8a²
= 8a – 10a – 6a² + 8a² + 9 – 8
= -2a + 2a² + 1
Now -3 – (-2a + 2a² + 1)
= (-3 + 2a – 2a² – 1)
= -4 + 2a – 2a²

Question 6.
Solution:

Exercise 6B

Find the products:
Question 1.
Solution:
3 x 8 a²⁺⁴ = 24a⁶

Question 2.
Solution:
(-6x³) x (5x²) = -6 x 5x²⁺³ = -30x⁵

Question 3.
Solution:
(-4ab) x (-3a²bc)
= (-4) x (-3) a. a².b. b. c
= 12.a²⁺¹. b¹⁺¹.c
= 12a³b²c

Question 4.
Solution:
= (2a²b³) x (-3a³b)
= 2 x (-3) a². a³. b³. b. b
= -6a²⁺³.b³⁺¹
= -6a⁵.b⁴

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Find the products given below and in each case verify the result for a = 1, b = 2 and c = 3 :
Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:

Exercise 6C

Find each of the following products:
Question 1.
Solution:
4a(3a + 7b) = 4a x 3a + 4a x 7b = 12a² + 28ab

Question 2.
Solution:
5a(6a – 3b) = 5a x 6a – 5a x 3b = 30a² – 15ab

Question 3.
Solution:
8a² (2a + 5b) = 8a² x 2a + 8a² x 5b = 16a³ + 40a²b

Question 4.
Solution:
9x² (5x + 7) = 9x² x 5x + 9x² x 7 = 45x³ + 63x²

Question 5.
Solution:
ab(a² – b²) = ab x a² – ab x b² = a³b – ab³

Question 6.
Solution:
2x² (3x – 4x²) = 2x² x 3x – 2x² x 4x² = 6x³ – 8x⁴

Question 7.
Solution:

Question 8.
Solution:
-1 7x² (3x – 4) = -17x² x 3x – 17x² x (-4) = -51x³ + 68x²

Question 9.
Solution:

Question 10.
Solution:
-4x²y (3x² – 5y)
= -4x² y x 3x² – 4x² y x (-5y)
= -12x² y + 20x² y²

Question 11.
Solution:

Question 12.
Solution:
9t² (t + 7t³) = 9t² x t + 9t² x 7t³ = 9t³ + 63t⁵

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:
24x² (1 – 2x)
= 24x² x 1 – 24x² x 2x
= 24x² – 48x³
If x = 2, then
24x² – 48x³
= 24(2)² – 48(2)³
= 24 x 4 – 48 x 8
= 96 – 384
= -288

Question 17.
Solution:

Question 18.
Solution:
s (s² – st) = s x s² – s x st = s³ – s²t
If s = 2, t = 3, then
s³ – s²t = (2)³ – (2)² x 3 = 8 – 4 x 3 = 8 – 12 = -4

Question 19.
Solution:
-3y (xy + y²) = -3y x xy + (-3y) x y² = -3xy² – 3y³
if x = 4, y = 5, then
-3xy² – 3y³
= -3(4)(5)² – 3(5)³ = -3 x 4 x 25 – 3 x 125 = -300 – 375 = -675

Simplify each of the following:
Question 20.
Solution:
a(b – c) + b(c – a) + c(a – b) = ab – ac + bc – ab + ac – bc = 0

Question 21.
Solution:
a(b – c) – b(c – a) – c(a – b) = ab – ac – bc + ab – ac + bc = 2ab – 2ac

Question 22.
Solution:
3x² + 2(x + 2) – 3x (2x + 1)
= 3x² + 2x + 4 – 6x² – 3x = 3x² – 6x² + 2x – 3x + 4 = -3x² – x + 4

Question 23.
Solution:
x (x + 4) + 3x (2x² – 1) + 4x² + 4
= x² + 4x + 6x³ – 3x + 4x² + 4
= 6x³ + x² + 4x² + 4x – 3x + 4
= 6x³ + 5x² + x + 4

Question 24.
Solution:
2x² + 3x (1 – 2x³) + x (x + 1)
= 2x² + 3x – 6x⁴ + x² + x
= – 6x⁴ + 2x² + x² + 3x + x
= – 6x⁴ + 3x² + 4x

Question 25.
Solution:
a²b (a – b²) + ab(4ab – 2a²) – a³b (1 – 2b)
= a³b – a²b³ – 2a³b² + 4a²b³ – 2a³b² – a³b + 2a³b²
= a³b – a³b + 2a³b² – a²b³ + 4a²6³
= 3a²b³

Question 26.
Solution:
4st (s – t) – 6s² (t – t²) – 3t² (2s² – s) + 2st(s – t)
= 4s²t – 4st² – 6s²t + 6s²t² – 6s²t² + 3st² + 2s²t – 2st²
= 4s²t – 6s²t + 2s²t – 4st² + 3st² – 2st² + 6s²t² – 6s²t²
= 6s²t – 6s²t – 6st² + 3st² + 6s²t² – 6s²t²
= – 3st²

Exercise 6D

Find each of the following products.
Question 1.
Solution:
(5x + 7) (3x + 4)
= 5x (3x + 4) + 7 (3x + 4)
= 5x x 3x + 5x x 4 + 7 x 3x + 7 x 4
= 15x² + 20x + 21x + 28
= 15x² + 41x + 28

Question 2.
Solution:
(4x – 3) (2x + 5)
= 4x (2x + 5) – 3 (2x + 5)
= 4x x 2x + 4x x 5 – 3 x 2x -3 x 5
= 8x² + 20x – 6x – 15
= 8x² + 14x – 15

Question 3.
Solution:
(x – 6) (4x + 9)
= x (4x + 9) – 6 (4x + 9)
= x x 4x + x x 9 – 6 x 4x – 6 x 9
= 4x² + 9x – 24x – 54
= 4x² – 15x – 54

Question 4.
Solution:
(5y – 1) (3y – 8)
= 5y x 3y – 5y x 8 – 1 x 3y – 8 x (-1)
= 15y² – 40y – 3y + 8
= 15y² – 43y + 8

Question 5.
Solution:
(7x + 2y) (x + 4y)
= 7x (x + 4y) + 2y (x + 4y)
= 7x x x + 7x x 4y + 2y x x + 2y x 4y
= 7x² + 28xy + 2xy + 8y²
= 7x² + 30xy + 8y²

Question 6.
Solution:
(9x + 5y) (4x + 3y)
= 9x (4x + 3y) + 5y (4x + 3y)
= 9x x 4x + 9x x 3y + 5y x 4x + 5y x 3y
= 36x² + 27xy + 20xy + 15y²
= 36x² + 47xy +15y²

Question 7.
Solution:
(3m – 4n) (2m – 3n)
= 3m (2m – 3n) – 4n (2m – 3n)
= 3m x 2m – 3m x 3n – 4n x 2m – 4n x (-3n)
= 6m² – 9mn – 8mn + 12n²
= 6m² – 17mn + 12n²

Question 8.
Solution:
(0.8x – 0.5y) (1.5x – 3y)
= 0.8x (1.5x – 3y) – 0.5y (1.5x – 3y)
= 0.8x x 1.5x – 0.8x x 3y – 0.5y x 1.5x – 0.5y x (-3y)
= 1.20x² – 2.4xy – 0.75xy + 1.5y²
= 1.2x² – 3.15xy + 1.5y²

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
(x² – a²) (x – a)
= x² (x – a) – a² (x – a)
= x² x x – x² x a – a² x x – a² (-a)
= x³ – x² a – xa² + a³

Question 13.
Solution:
(3p² + q²) (2p² – 3q²)
= 3p² (2p² – 3q²) + q² (2p² – 3q²)
= 3p² x 2p² – 3p² x 3q² + q² x 2p² – q² x 3q²
= 6q⁴ – 9p²q² + 2p²q² – 3q⁴
= 6p⁴ – 7p²q² – 3q⁴

Question 14.
Solution:
(2x² – 5y²) (x² + 3y²)
= 2x² (x² + 3y²) – 5y² (x² + 3y²)
= 2x² x x² + 2x² x 3y² – 5y² x x² – 5y² x 3y²
= 2x⁴ + 6x² y² – 5x² y² – 15y⁴
= 2y⁴ + x² y² – 15y⁴

Question 15.
Solution:
(x³ – y³) (x² + y²)
= x³ (x² + y²) – y³ (x² + y²)
= x³ x x² + x³ x y² – y³ x x² – y³ x y²
= x⁵ + x³ y² – x² y³ – y⁵

Question 16.
Solution:
(x⁴ + y⁴) (x² – y²)
= x⁴ (x² – y²) + y⁴ (x² – y²)
= x⁴ x x² – x⁴ x y² + y⁴ x x² – y⁴ x y²
= x⁶ – x⁴ y² + x² y⁴ – y⁶

Question 17.
Solution:

Question 18.
Solution:
(x² – y²) (x + 2y)
= x² (x + 2y) – y² (x + 2y)
= x² x x + x² x 2y – y² x x – y² x 2y
= x³ + 2x² y – xy² – 2y³

Question 19.
Solution:
(2x + 3y – 5) (x + y)
= 2x (x + y) + 3y (x + y) – 5 (x + y)
= 2x x x + 2x x y + 3y x x + 3y x y – 5 x x – 5 x y
= 2x² + 2xy + 3xy + 3y² – 5x – 5y
= 2x² + 5xy + 3y² – 5x – 5y

Question 20.
Solution:
(3x + 2y – 4) (x – y)
= 3x (x – y) + 2y (x – y) – 4 (x – y)
= 3x x x – 3x x y + 2y x x – 2y x y – 4 x x – 4 x (-y)
= 3x² – 3xy + 2xy – 2y² – 4x + 4y
= 3x² – xy – 2y² – 4x + 4y

Question 21.
Solution:
(x² – 3x + 7) (2x + 3)
= x² (2x + 3) – 3x (2x + 3) + 7 (2x + 3)
= x² x 2x + x² x 3 – 3x x 2x – 3x x 3 + 7 x 2x + 7 x 3
= 2x³ + 3x² – 6x² – 9x + 14x + 21
= 2x³ – 3x² + 5x + 21

Question 22.
Solution:
(3x² + 5x – 9) (3x – 9)
= 3x² (3x – 9) + 5x (3x – 9) – 9 (3x – 9)
= 3x² x 3x – 3×2 x 9 + 5x x 3x + 5x x (-9) – 9 x 3x – 9 x (-9)
= 9x³ – 27x² + 15x² – 45x – 27x + 81
= 9x³ – 12x² – 72x + 81

Question 23.
Solution:
(9x² – x + 15) (x² – 3)
= 9x² (x² – 3) – x (x² – 3) + 15 (x² – 3)
= 9x² x x² – 9x² x 3 – x x x² + x x 3 + 15 x x² – 15 x 3
= 9x⁴ – 27x² – x³ + 3x + 15x² – 45
= 9x⁴ – x³ – 12x² + 3x – 45

Question 24.
Solution:
(x² + xy + y²) (x – y)
= x² (x – y) + xy (x – y) + y² (x – y)
= x² x x – x² x y + xy x x – xy x y + y² x x – y² x y
= x³ – x²y + x²y – xy² + xy² – y²
= x³ – y³

Question 25.
Solution:
(x² – xy + y²) (x + y)
= x² (x + y) – xy (x + y) + y² (x + y)
= x³ + x² y – x² y – xy² + xy² + y³
= x³ + y³

Question 26.
Solution:
(x² – 5x + 8) (x² + 2)
= x² (x² + 2) – 5x (x² + 2) + 8 (x² + 2)
= x² x x² + x² x 2 – 5x x x² – 5x x 2 + 8 x x² + 8 x 2
= x⁴ + 2x² – 5x³ – 10x + 8x² + 16
= x⁴ – 5x³ + 2x² + 8x² – 10x + 16
= x⁴ – 5x³ + 10x² – 10x + 16

Simplify:
Question 27.
Solution:
(3x + 4) (2x – 3) + (5x – 4) (x + 2)
= [3x (2x – 3) + 4 (2x – 3)] + [5x (x + 2) – 4 (x + 2)]
= [3x x 2x – 3x x 3 + 4 x 2x – 4 x 3] + [5x x x + 5x x 2 – 4 x x – 4 x 2]
= [6x² – 9x + 8x – 12] + [5x² + 10x – 4x – 8]
= (6x² – x – 12) + (5x² + 6x – 8)
= 6x² – x – 12 + 5x² + 6x – 8
= 6x² + 5x² – x + 6x – 12 – 8
= 11x² + 5x – 20

Question 28.
Solution:
(5x – 3) (x + 4) – (2x + 5) (3x – 4)
= [5x x x + 5x x 4 – 3 x x – 3 x 4] – [2x x 3x + 2x x (-4) + 5 x 3x + 5x (-4)]
= [5x² + 20x – 3x – 12] – [6x² – 8x + 15x – 20]
= (5x² + 17x – 12) – (6x² + 7x – 20)
= 5x² + 17x – 12 – 6x² – 7x + 20
= 5x² – 6x² + 17x – 7x – 12 + 20
= -x² + 10x + 8

Question 29.
Solution:
(9x – 7) (2x – 5) – (3x – 8) (5x – 3)
= [9x (2x – 5) -7 (2x – 5)] – [3x (5x – 3) -8 (5x – 3)]
= [9x x 2x – 9x x 5 – 7 x 2x – 7 x (-5)] – [3x x 5x – 3x x 3 – 8 x 5x – 8 x (-3)]
= [18x² – 45x – 14x + 35] – [15x² – 9x – 40x + 24]
= 18x² – 45x – 14x + 35 – 15x² + 9x + 40x – 24
= 18x² – 15x² – 45x – 14x + 9x + 40x + 35 – 24
= 3x² – 59x + 49x + 11
= 3x² – 10x + 11

Question 30.
Solution:
(2x + 5y) (3x + 4y) – (7x + 3y) (2x + y)
= [2x (3x + 4y) + 5y (3x + 4y)] – [7x (2x + y) + 3y (2x + y)]
= [2x x 3x + 2x x 4y + 5y x 3x + 5y x 4y] – [7x x 2x + 7x x y + 3y x 2x + 3y x y]
= [6x² + 8xy + 15xy + 20y²] – [14x² + 7xy + 6xy + 3y²]
= [6x² + 23xy + 20y²] – [14x² + 13xy + 3y^2]
= 6x² + 23xy + 20y² – 14x² – 13xy – 3y²
= 6x² – 14x² + 23xy – 13xy + 20y² – 3y²
= -8x² + 10xy + 11y²

Question 31.
Solution:
(3x² + 5x – 7) (x – 1) – (x² – 2x + 3) (x + 4)
= [3x² (x – 1) + 5x (x – 1) – 7 (x – 1)] – [x² (x + 4) – 2x (x + 4) + 3 (x + 4)]
= [3x² x x – 3x² x 1 + 5x x x – 5x x (-1) – 7 x x -7 x (-1)] -[x² x x + x² x 4 – 2x x x – 2x x 4 + 3 x x + 3 x 4]
= [3x³ – 3x² + 5x² – 5x – 7x + 7] – [x³ + 4x² – 2x² – 8x + 3x + 12]
= [3x³ + 2x² – 12x + 7] – [x³ + 2x² – 5x + 12]
= 3x³ + 2x² – 12x + 7 – x³ – 2x² + 5x – 12
= 3x³ – x³ + 2x² – 2x² – 12x + 5x – 12 + 7
= 2x³ – 7x – 5

Exercise 5A

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
Product of two numbers = 4

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Exercise 5B

Question 1.
Solution:
We can write in standard form
(i) 538 = 5.38 x 10²
(ii) 6428000 = 6.428000 x 10⁶ = 6.428 x 10⁶
(iii) 82934000000 = 8.2934000000 x 10¹⁰ = 8.2934 x 10¹⁰
(iv) 940000000000 = 9.40000000000 x 10¹¹ = 9.4 x 10¹¹
(v) 23000000 = 2.3000000 x 10⁷ = 2.3 x 10⁷

Question 2.
Solution:
(i) Diameter of Earth = 12756000 m = 1.2756000 x 10⁷m = 1.2756 x 10⁷ m
(ii) Distance between Earth and Moon = 384000000 m = 3.84000000 x 10⁸ = 3.84 x 10⁸
(iii) Population of India in March 2001 = 1027000000 = 1.027000000 x 10⁹ = 1.027 x 10⁹
(iv) Number of stars in a galaxy = 100000000000 = 1.00000000000 x 10¹¹ = 1 x 10¹¹
(v) The present age of universe = 12000000000 years = 1.2000000000 x 10¹⁰ years = 1.2 x 10¹⁰ years

Question 3.
Solution:
The expanded form will be
(i) 684502 = 6 x 10⁵ + 8 x 10⁴ + 4 x 10³ + 5 x 10² + 2
(ii) 4007185 = 4 x 10⁶ + 0 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 1 x 10² + 8 x 10¹ + 5 x 10⁰
(iii) 5807294 = 5 x 10⁶ + 8 x 10⁵ + 0 x 10⁴ + 7 x 10³ + 2 x 10² + 9 x 10¹ + 4 x 10⁰
(iv) 50074 = 5 x 10⁴ + 0 x 10³ + 0 x 10² + 7 x 10¹ + 4 x 10⁰

Question 4.
Solution:
(i) 6 x 10⁴ + 3 x 10³ + 0 x 10² + 7 x 10¹ + 8 x 10⁰
= 60000 + 3000 + 0 + 70 + 8
= 63078
(ii) 9 x 10⁶ + 7 x 10⁵ + 0 x 10⁴ + 3 x 10³ + 4 x 10² + 6 x 10¹ + 2 x 10⁰
= 9000000 + 700000 + 0 + 3000 + 400 + 60 + 2
= 9703462
(iii) 8 x 10⁵ + 6 x 10⁴ + 4 x 10³ + 2 x 10² + 9 x 10¹ + 6 x 10⁰
= 800000 + 60000 + 4000 + 200 + 90 + 6
= 864296

Exercise 5C

OBJECTIVE QUESTIONS
Mark (✓) tick against the correct answer in each of the following :
Question 1.
Solution:
(d)

Question 2.
Solution:
(c)

Question 3.
Solution:
(c)

Question 4.
Solution:
(b)

Question 5.
Solution:
(c)

Question 6.
Solution:
(b)

Question 7.
Solution:
(b)

Question 8.
Solution:
(a)

Question 9.
Solution:
(c)

Question 10.
Solution:
(b)

Question 11.
Solution:
(b)

Question 12.
Solution:
(b)

Question 13.
Solution:
(d)

Question 14.
Solution:
(a)

Question 15.
Solution:
(c)

Question 16.
Solution:
(a)

Question 17.
Solution:
(c)

Question 18.
Solution:
(b)

Question 19.
Solution:
(c)

Question 20.
Solution:
(c)
Required number = 10^-1 ÷ (-8)^-1

Question 21.
Solution:
(c)
The number which is in standard form is 2.156 x 10⁶

Exercise 4A

Question 1.
Solution:
(i) Rational numbers: The numbers of the form pq where p and q are integers and q ≠ 0, are called rational numbers.

(iv) Yes, there is one rational number (0) which is neither positive nor negative.

Question 2.
Solution:

(viii) 01 are all rational number but 10 and 00 are not rational number as their denominator is zero.

Question 3.
Solution:
(i) Numerator = 8, denominator =19
(ii) Numerator = 5, denominator = – 8
(iii) Numerator =-13, denominator =15
(iv) Numerator = – 8, denominator = -11
(v) Numerator = 9, denominator = 1

Question 4.
Solution:

Question 5.
Solution:
According to the definition, a rational number is positive if both of numerator and denominator have same signs. Therefore
(iii), (iv) and (vi) 8 are positive rational numbers.

Question 6.
Solution:
According to the definition, a rational number is negative if numerator and denominator have opposite sign. Therefore.
(iii), (iv), (v), (vi) are all negative rational numbers.

Question 7.
Solution:
Equivalent rational numbers of each are given below:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Exercise 4B

Question 1.
Solution:
(i) Draw a number line and locate a point O on it. Let it represent 0 Now 13 has been presented on the number line given below.

(ii) Draw a number line and locate a point O on it. Let it represent 0. The number 27 has been represented on the number line given below:

(iii) Draw a number line and locate a point O on it. Let it represent 0. The number 73 has been represented on the number line given below:

(iv) Draw a number line and locate a point O on it. Let it represent 0. The number 73 has been represented on it as given below:

(v) Draw a number line and locate a point O on it. Let it represented 0. The number 378 has been represented on it as given below:
378 = 458

(vi) Draw a number line and locate a point O on it. Let it represent 0. The number −13 has been represented on it as given below:

(vii) Draw a number line and locate a point O on it. Let it represent 0. The number −34 has been represented on it is as given below:

(viii) Draw a number line and locate a point on it. Let it represent 0. The number −127 has been represented on it as given below:

(ix) Draw a number line and locate a point O on it. Let it represent 0. The number 36−5 has been represented on it as given below:

(x) Draw a number line and locate is point O on it. Let is represent 0. The number −439 has been represented on it as given below:

Question 2.
Solution:
(i) 56 or 0, 56 is greater as any positive number is always greater than 0.

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
(i) True: All negative numbers lie on the left of 0.
(ii) False: All negative numbers lie on the left of 0.
(iii) True: All positive numbers lie on the right of 0 and all negative numbers on the left of 0.
(iv) False: −18−13 = 1813 which is positive and positive number lie on the left of 0.
(v) True: −5−8 = 58 which is positive and all positive number lie on the right of negative numbers.
(i), (iii) and (iv) are true.

Question 8.
Solution:
5 rational numbers between -3 and -2.

Question 9.
Solution:

Question 10.
Solution:
L.C.M. of 5 and 2 = 10

Exercise 4C

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Exercise 4D

Question 1.
Solution:
(i) Additive inverse of 5 = -5
(ii) Additive inverse of -9 = – (-9) = 9

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:
The required number = −11 – 29

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Exercise 4E

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Cost of 1 metre of cloth

Question 6.
Solution:

Exercise 4F

Question 1.
Solution:

(vii) Reciprocal of -1 = -1
(viii) Reciprocal of 0 does not exist.

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:
Product of two number = 10
One number = -8
Second number = 10 ÷ (-8)

Question 8.
Solution:
Product of two rational numbers = – 9
One number = -12
Second number = (-9) ÷ (-12)

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Cloth required for 24 pairs of trousers =54 m
Cloth required for one pair = (54 ÷ 24) m

Question 12.
Solution:
Total length of rape = 30 m

Question 13.
Solution:

Exercise 4G

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)

Question 2.
Solution:
(b)

Question 3.
Solution:
(a)

Question 4.
Solution:
(c)

Question 5.
Solution:
(b)

Question 6.
Solution:
(a)

Question 7.
Solution:
(a)

Question 8.
Solution:
(c)

Question 9.
Solution:
(b)

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:

Question 19.
Solution:

Question 20.
Solution:

Exercise 3A

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
(i) 6.5, 16.03, 0.274, 119.4
In these decimals, the greatest places of decimal is 3
6.5 = 6.500
16.03 = 16.030
0. 274 = 0.274
119.4 = 119.400 are like decimals.
(ii) 3.5, 0.67, 15.6, 4
In these decimal, the greatest place of decimal is 2
3.5 = 3.50
0.67 = 0.67
15.6 = 15.60
4 = 4.00 are the like decimals

Question 5.
Solution:
(i) Among 78.23 and 69.85,
78.23 is greater than 69.85 (78 > 69)
78.23 > 69.85
(ii) Among 3.406 and 3.46,
3.406 is less than 3.46 (40 < 46)
3.406 < 3.46
(iii) Among 5.68 and 5.86,
5.68 is less than 5.86 (68 < 86)
5.68 < 5.86
(iv) Among 14.05 and 14.005
14.5 is greater than 14.005 (05 > 00)
14.5 >14.005
(v) Among 1.85 and 1.805,
1.85 is greater than 1.805 (85 > 80)
1.85 > 1.805
(vi) Among 0.98 and 1.07,
0.98 is less than 1.07 (0 < 1)
0.98 < 1.07

Question 6.
Solution:
(i) 4.6, 7.4, 4.58, 7.32, 4.06
Converting the given decimals into like decimals, we get:
4.60, 7.40, 4.58, 7.32, 4.06.
We see that 4.06 < 4.58 < 4.60 < 7.32 < 7.40.
Writing in ascending order, 4.06, 4.58, 4.6, 7.32, 7.4
(ii) 0.5, 5.5, 5.05, 0.05, 5.55
Converting the given decimals into like decimals, we get:
0. 50, 5.50, 5.05, 0.05, 5.55
We see that 0.05 < 0.50 < 5.05 < 5.50 < 5.55.
Writing in ascending order, 0.05, 0.50, 5.05, 5.5, 5.55
(iii) 6.84, 6.84, 6.8, 6.4, 6.08
Converting the given decimals into like decimals
6.84, 6.48, 6.80, 6.40, 6.08
We see that 6.08 < 6.40 < 6.48 < 6.80 < 6.84
Writing in ascending order,
6.08, 6.4, 6.48, 6.8, 6.84
(iv) 2.2, 2.202, 2.02, 22.2, 2.002
Converting them into like decimals
2.200, 2.202, 2.020, 22.200, 2.002 we see that
2.002 < 2.020 < 2.200 < 2.202 < 22.200
Now writing in ascending order,
2.002, 2.020, 2.2, 2.202, 22.2

Question 7.
Solution:
(i) 7.4, 8.34, 74.4, 7.44, 0.74
Converting them into like decimals,
7.40, 8.34, 74.40, 7.44, 0.74
we see that
74.40 > 8.34 > 7.44 > 7.40 > 0.74
Writing in descending order,
74.4, 8.34, 7.44, 7.4, 0.74
(ii) 2.6, 2.26, 2.06, 2.007, 2.3
Converting them into like decimals,
2.600, 2.260, 2.060, 2.007, 2.300
We see that
2.600 > 2.300 > 2.260 > 2.060 > 2.007
Writing in descending order,
2.6, 2.3, 2.26, 2.06, 2.007

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Exercise 3B

Question 1.
Solution:
Converting them into like decimals 16.00, 8.70, 0.94, 6.80 and 7.77
Now, adding them,
16.0 + 8.70 + 0.94 + 6.80 + 7.77 = 40.21

Question 2.
Solution:
Converting them into like decimals 18.600, 206.370, 8.008, 26.400, 6.900
Adding we get
18.600 + 206.370 + 8.008 + 26.400 + 6.900 = 266.278

Question 3.
Solution:
Converting them into like decimals, 63.50, 9.70, 0.80, 26.66, 12.17
Adding we get:
63.50 + 9.70 + 0.80 + 26.66 + 12.17 = 112.83

Question 4.
Solution:
Converting them into like decimals 17.400, 86.390, 9.435, 8.800, 0.060
Adding we get:
17.400 + 86.390 + 9.435 + 8.800 + 0.060 = 122.085

Question 5.
Solution:
Converting them into like decimals 26.900, 19.740, 231.769, 0.048
Now adding we get:
26.900 + 19.740 + 231.769 + 0.048 = 278.457

Question 6.
Solution:
Converting them into like decimals 23.800, 8.940, 0.078 and 214.600
Now adding we get:
23.800 + 8.940 + 0.078 + 214.600 = 247.418

Question 7.
Solution:
Converting them into like decimals.
6.606, 66.600, 666.000,0.066, 0.660
Now adding we get:
6.606 + 66.600 + 666.000 + 0.066 + 0,660 = 739.932

Question 8.
Solution:
9.090, 0.909, 99.900, 9.990, 0.099
Now adding we get:
9.090 + 0.909 + 99.900 + 9.990 + 0.099 = 119.988

Subtract:
Question 9.
Solution:
14.79 from 72.43
72.43 – 14.79 = 57.64

Question 10.
Solution:
Converting them into like decimals, We get
36.74 and 52.60
Now 52.60 – 36.74 = 15.86

Question 11.
Solution:
Converting them into like decimals, We get
13.876 and 22.000
22.000 – 13.876 = 8.124

Question 12.
Solution:
Converting them into like decimals, We get
15.079 and 24.160
24.160 – 15.079 = 9.081

Question 13.
Solution:
Converting them into like decimals We get
0.680 and 1.007
1.007 – 0.680 = 0.327

Question 14.
Solution:
Converting them into like decimals,
We get 0.4678 and 5.0500
5.0500 – 0.4678 = 4.5822

Question 15.
Solution:
Converting them into like decimals,
We get 2.5307 and 8.0000
8.0 – 2.5307 = 5.4693

Question 16.
Solution:
There are like decimals
9.1 – 6.732 = 2.269

Question 17.
Solution:
Converting them into like decimals,
We get 5.746 and 9.100
9.100 – 5.746 = 3.354

Question 18.
Solution:
Converting into like decimals, we get,
63.59 and 92.00
Required number = 92.00 – 63.58 = 28.42

Question 19.
Solution:
Converting into like decimals, we get:
8.100 and 0.813
Required number = 8.100 – 0.813 = 7.287

Question 20.
Solution:
Converting them into like decimals, we get: 32.67 and 60.10
Required number = 60.10 – 32.67 = 27.43

Question 21.
Solution:
Converting into like decimals, we get 74.3 and 26.87
Required number = 74.30 – 26.87 = 47.43

Question 22.
Solution:
Cost of notebook = Rs. 23.75
Cost ofpencil = Rs. 2.85
Costofpen =Rs. 15.90
Total cost = Rs. 42.50
Amount gave to the shop keeper = 50 rupees
Balance amount got = Rs 50.00 – Rs 42.50 = 7.50

Exercise 3C

Question 1.
Solution:
We know that by multiplying by 10, the decimal point is shifted one place to its right side.
(i) 73.92 x 10 = 739.2
(ii) 7.54 x 10 = 75.4
(iii) 84.003 x 10 = 840.03
(iv) 0.83 x 10 = 8.3
(v) 0.7 x 10 = 7.0
(vi) 0.032 x 10 = 0.32

Question 2.
Solution:
We know that by multiplying a decimal by 100, two decimal points are shifted to it right side
(i) 2.397 x 100 = 239.7
(ii) 6.83 x 100 = 683.0
(iii) 2.9 x 100 = 290
(iv) 0.08 x 100 = 8
(v) 0.6 x 100 = 60
(vi) 0.003 x 100 = 0.3

Question 3.
Solution:
We know that by multiplying a decimal by 1000, three places of decimal are shifted to its right.
(i) 6.7314 x 1000 = 6731.4
(ii) 0.182 x 1000 = 182
(iii) 0.076 x 1000 = 76
(iv) 6.25 x 1000 = 6250
(v) 4.8 x 1000=4800
(vi) 0.06 x 1000 = 60

Question 4.
Solution:
(i) 5.4 x 16 = 86.4 (One place of decimal)
(ii) 3.65 x 19 = 69.35 (Two place of decimal)
(iii) 0.854 x 12 = 10.2468 (Three place of decimal)
(iv) 36.73 x 48 = 1763.04 (Two places of decimal)

(v) 4.125 x 86=354.750 (Three places of decimal)
= 354.75

Question 5.
Solution:
(i) 7.6 x 2.4= 18.24
{Sum of decimal places = 1 + 1 = 2}

Question 6.
Solution:
(i) 13 x 1.3 x 0.13 = 2.197
{Sum of decimal places = 1 + 2 = 3}

(ii) 2.4 x 1.5 x 2.5 = 9.000 = 9
{Sum of decimal places = 1 + 1 + 1 = 3}
(iii) 0.8 x 3.5 x 0.05 = 0.1400 = 0.14
{Sum of decimal places = 1 + 1 + 2 = 4}
(iv) 0.2 x 0.02 x 0.002 = 0.000008
{Sum of decimal places = 1 + 2 + 3 = 6}
(v) 11.1 x 1.1 x 0.11 = 1.3431
{Sum of decimal places = 1 + 1 + 2 = 4}

(vi) 2.1 x 0.21 x 0.021 = 0.00926
21 x 21 = 441
441 x 21 = 9261
{Sum of decimal places = 1 + 2 + 3 = 6}

Question 7.
Solution:
(i) (1.2)²= 1.2 x 1.2 = 1.44
{Sum of decimal places = 1 + 1 = 2}
(ii) (0.7)² = 0.7 x 0.7 = 0.49
{Sum of decimal places = 1 + 1 = 2}
(iii) (0.04)² = 0.04 x 0.04 = 0.0016
{Sum of decimal places = 2 + 2 = 4}
(iv) (0.11)² = 0.11 x 0.11 =0.0121
{Sum of decimal places = 2 + 2 = 4}

Question 8.
Solution:
(i) (0.3)³ = 0.3 x 0.3 x 0.3 = 0.027
{Sum of decimal places = 1 + 1 + 1 = 3}
(ii) (0.05)³= 0.05 x 0.05 x 0.05 = 0.000125
{Sum of decimal places = 2 + 2 + 2 = 6}
(iii) (1.5)³ = 1.5 x 1.5 x 1.5 = 3.375
{Sum of decimal places = 1 + 1 + 1 = 3}

Question 9.
Solution:
Distance covered in one hour = 62.5 km
Distance covered in 18 hours = 62.5 x 18 km = 1125.0 km

Question 10.
Solution:
Weight of one tin of oil = 16.8 kg
Weight of 45 tins = 16.8 x 45 kg = 756.0 kg = 756 kg

Question 11.
Solution:
Weight of wheat in one bag = 97.8 kg
weight of wheat in 500 bags = 97.8 x 500 kg = 48900.0 kg = 48900 kg

Question 12.
Solution:
Weight of one bag = 48.450 kg
Weight of 16 bags = 48.450 x 16 = 775.200 kg

Question 13.
Solution:
Quantity of sauce in one bottle = 0.845 kg
quantity of sauce in 72 bottles = 0.845 x 72 kg = 60.840 kg

Question 14.
Solution:
Quantity of jam in one bottle = 925 .
Quantity of jam in 25 bottles = 925 x 25 g = 23135 g = 23.125 kg

Question 15.
Solution:
Oil in one drum = 16.850 litres
Oil in 48 drums = 16.850 x 48 = 808.800 = 808.800 litres

Question 16.
Solution:
Cost of 1 kg rice = Rs 56.80
Cost of 16.25 kg of rice = Rs 56.80 x 16.25 = Rs 923.0000 = Rs 923

Question 17.
Solution:
Cost of one metre of cloth = Rs 108.5 0
Costof 18.5 metres of cloth = Rs 108.50 x 18.5 = Rs 2007.250 = Rs 2007.25

Question 18.
Solution:
Distance covered in one litre = 8.6 km
Distance covered in 36.5 litres = 8.6 x 36.5 km = 313.90 km = 313.9 km

Question 19.
Solution:
Charges for 1 km = Rs 9.80
Charges for 106.5 km = Rs 9.80 x 106.5 = Rs 1043.700 = Rs 1043.70

Exercise 3D

Question 1.
Solution:
We know that a decimal divided by 10, the decimal point is shifted to the left by one place. Therefore
(i) 131.6 ÷ 10 = 13.16
(ii) 32.56 ÷ 10 = 3.256
(iii) 4.38 ÷ 10 = 0.438
(iv) 0.34 ÷ 10 = 0.034
(v) 0.08 ÷ 10 = 0.008
(vi) 0.062 ÷ 10 = 0.0062

Question 2.
Solution:
We know that decimal divided by 100, the decimal point is shifted to the left by two place. Therefore
(i) 137.2 ÷ 100 = 1.372
(ii) 23.4 ÷ 100= 0.234
(iii) 4.1 ÷ 100 = 0.047
(iv) 0.3 ÷ 100 = 0.003
(v) 0.58 ÷ 100 = 0.0058
(vi) 0.02 ÷ 100 = 0.0002

Question 3.
Solution:
We know that a decimal divided by 1000, the decimal point is shifted to the left by three places. Therefore:
(i) 1286.5 ÷ 1000= 1.2865
(ii) 354.16 ÷ 1000 = 0.35416
(iii) 38.9 ÷ 1000 = 0.0389
(iv) 4.6 ÷ 1000 = 0.0046
(v) 0.8 ÷ 1000 = 0.0008
(vi) 2 ÷ 1000 = 0.002

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:
Cost of 24 chairs = Rs. 9255.60
Cost of one chair = Rs. 9255.6024 = Rs. 385.65

Question 9.
Solution:
Length of cloth for one shirt = 1.8 m
Total length of piece of cloth = 45 m
Number of shirts will be = 45 ÷ 1.8

Question 10.
Solution:
A car covers in 2.4 litre = 22.8 km
It will cover in 1 litre = 22.82.4 km = 9.5 km

Question 11.
Solution:
Oil in one tin = 16.5 l
Total oil = 478.5 l

Question 12.
Solution:
Weight of 37 bags of sugar=3644.5 kg
Weight of one bag of sugar = 3644.5 ÷ 37

Question 13.
Solution:
Capacity of 69 buckets = 586.5 litres
Capacity of 1 bucket = 586.5 ÷ 69

Question 14.
Solution:
Number of pieces in 1.15 m = 1

Question 15.
Solution:
Total weight of cement = 1792.8 kg
Cement in one bag = 49.8 kg
Number of bags = 1792.8 ÷ 49.8

Question 16.
Solution:
Total thickness = 1.89 m = 189 cm
Thickness of one piece = 0.3 5 cm
Number of pieces = 189 ÷ 0.35

Question 17.
Solution:
Product of two decimals = 261.36
One decimal = 17.6
Second decimal = 261.36 ÷ 17.6

Exercise 3E

OBJECTIVE QUESTIONS
Mark (✓) against the correct answer in each of the following:
Question 1.
Solution:
(b)

Question 2.
Solution:
(c)

Question 3.
Solution:
(b)

= 208100 = 2.08

Question 4.
Solution:

Question 5.
Solution:
(b) 70g = 701000 = 0.07 kg

Question 6.
Solution:
(c) 5 kg 6 g = 561000 kg = 5.006 kg

Question 7.
Solution:
(c) 2 km 5 m = 251000 km = 2.005 km

Question 8.
Solution:
(c)
1.007 – 0.7 = 1.007 – 0.700 = 0.307

Question 9.
Solution:
(b)
0.1 – 0.03 = 0.10 – 0.03 = 0.07

Question 10.
Solution:
(c)
3.5 – 3.07 = 3.50 – 3.07 = 0.43

Question 11.
Solution:
(c)
0.23 x 0.3 = 0.069

Question 12.
Solution:
(b)
0.02 x 30 = .60 = .6

Question 13.
Solution:
(b)
0.25 x 0.8 = 0.200 = 0.2

Question 14.
Solution:
(c)
0.4 x 0.4 x 0.4 = 0.064

Question 15.
Solution:
(b)
1.1 x .1 x .01 = .0011

Question 16.
Solution:
(a)

Question 17.
Solution:
(b)
1.02 ÷ 6 = 1.026 = 0.17

Question 18.
Solution:

Question 19.
Solution:
(b)

Question 20.
Solution:
(a)

Question 21.
Solution:
(c)

Exercise 2A

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:
Reenu get 27 of an apple while Sonal gets 45 part of it

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
For finding the required fraction, we have to subtract 735 from 18

Question 12.
Solution:
For finding the required fraction we should subtract 7415 from 825

Question 13.
Solution:

Question 14.
Solution:

Question 15.
Solution:

Question 16.
Solution:

Exercise 2B

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Cost of 1 metre pf cloth

Question 6.
Solution:
Distance covered in 1 hour

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:
Weight of Amit = 35 kg.

Question 12.
Solution:
Total number of students in a class = 42
Number of boys = 57 of 42 = 5 x 6 = 30
Number of girls = 42 – 30 = 12

Question 13.
Solution:
Sapna total income for one month = Rs 24000
Amount spent = 78 of her income
= 78 x 24000
= Rs (7 x 3000) = Rs 21000
Amount deposited in the bank per month = Rs 24000 – 21000 = Rs 3000

Question 14.
Solution:
Length of each side of a square

Question 15.
Solution:
Length of rectangular field (l)

Exercise 2C

Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:
Total weight of 18 boxes

Question 6.
Solution:
Total amount of oranges=Rs 378

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:
Total quantity of milk = 24 litres
and quantity of milk got by one student = 25 litres

Question 14.
Solution:
Quantity of water in a bucket

Question 15.
Solution:

Question 16.
Solution:
Product of two numbers = 42

Question 17.
Solution:

Exercise 2D

OBJECTIVE QUESTIONS
Mark (√) against the correct answer in each of the following:
Question 1.
Solution:
(c)
Denominator in (a) and (b) is 10
These are decimal fractions
But denominator of (c) is 3
103 is a vulgar fraction

Question 2.
Solution:
(c)
710 and 79 are proper fractions as each of these have numerator less than its denominator
97 is improper fraction

Question 3.
Solution:
(a)
105112 is reducible fraction because HCF 112 of 105 and 112 is 7

Question 4.
Solution:
(c)

Question 5.
Solution:
(c)

Question 6.
Solution:
(d)

Question 7.
Solution:
(c)

Question 8.
Solution:
(d)

Question 9.
Solution:
(d)

https://googleads.g.double

Question 10.
Solution:
(b)

Question 11.
Solution:
(d)

Question 12.
Solution:
(c)

Question 13.
Solution:
(d)

Question 14.
Solution:
(d)

Question 15.
Solution:
(b)
The correct statement will be

Question 16.
Solution:
(c)
A car runs in 1 litre of petrol = 16 km

Question 17.
Solution:
(a)