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Page No 124:

Question 1:

Express each of the following ratios in simplest form:

(i) 24 : 40
(ii) 13.5 : 15
(iii) 623:712
(iv) 16:19
(v) 4:5:92
(vi) 2.5 : 6.5 : 8

Answer:

(i) HCF of 24 and 40 is 8.
∴ 24 : 40 = 2440= 24 ÷ 840 ÷ 8= 35 = 3 : 5

Hence, 24 : 40 in its simplest form is 3 : 5.

(ii) HCF of 13.5 and 15 is 1.5.

 13.515=135150The HCF of 135 and 150 is 15.=135 ÷ 15150 ÷ 15=910

Hence, 13.5 : 15 in its simplest form is 9 : 10.

(iii)  203 : 152=40 : 45
The HCF of 40 and 45 is 5.

∴ 40 : 45 = 4045=40 ÷ 545 ÷ 5=89  = 8 : 9

Hence, 623 : 712 in its simplest form is 8 : 9

(iv) 9 : 6
The HCF of 9 and 6 is 3.

9 : 6 = 96=9 ÷ 36 ÷ 3 = 3 : 2
Hence, 16:19 in its simplest form is 3 : 2.

(v) LCM of the denominators is 2.

4 : 5 : 92 = 8 : 10 : 9
The HCF of these 3 numbers is 1.


∴ 8 : 10 : 9 is the simplest form
.

(vi) 2.5 : 6.5 : 8 = 25 : 65 : 80

The HCF of 25, 65 and 80 is 5.
25 : 65 : 80 = 256580= 25 ÷ 565 ÷ 580 ÷ 5=51316 = 5 : 13 : 16

Page No 124:

Question 2:

Express each of the following ratios in simplest form:

(i) 75 paise : 3 rupees
(ii) 1 m 5 cm : 63 cm
(iii) 1 hour 5 minutes : 45 minutes
(iv) 8 months : 1 year
(v) (2 kg 250 g) : (3 kg)
(vi) 1 km : 750 m

Answer:

(i) Converting both the quantities into the same unit, we have:
   75 paise : (3 × 100) paise = 75 : 300

= 75300= 75 ÷ 75300 ÷ 75=14    (∵ HCF of 75 and 300 = 75)
= 1 paise : 4 paise

(ii)  Converting both the quantities into the same unit, we have:
105 cm : 63 cm = 10563=105÷2163÷21= 53   (∵ HCF of 105 and 63 = 21)
= 5 cm : 3 cm

(iii) Converting both the quantities into the same unit
65 min : 45 min = 6545= 65÷545÷5=139  (∵ HCF of 65 and 45 = 5)
= 13 min : 9 min

(iv) Converting both the quantities into the same unit, we get:
8 months : 12 months = 812=8÷412÷4=23  (∵ HCF of 8 and 12 = 4)
= 2 months : 3 months

(v) Converting both the quantities into the same unit, we get:

2250g : 3000 g = 22503000=2250÷7503000÷750= 34    (∵ HCF of 2250 and 3000 = 750)

= 3 g : 4 g

(vi)  Converting both the quantities into the same unit, we get:
1000 m : 750 m =  1000750=1000÷250750÷250 = 43    (∵ HCF of 1000 and 750 = 250)
= 4 m : 3 m

Page No 124:

Question 3:

If A : B = 7 : 5 and B : C = 9 : 14, find A : C.

Answer:

AB  = 75 and  BC = 914

Therefore, we have:

AB×BC = 75×914AC = 910

∴ A : C = 9 : 10

Page No 124:

Question 4:

If A : B = 5 : 8 and B : C = 16 : 25, find A : C.

Answer:

AB=58 and BC = 1625Now, we have:AB×BC = 58×1625AC = 25

∴ A : C = 2 : 5

Page No 124:

Question 5:

If A : B = 3 : 5 and B : C = 10 : 13, find A : B : C.

Answer:

A : B = 3 : 5

B : C = 10 : 13 =  10÷213÷2=5 :132

Now, A : B : C = 3 : 5 : 132

∴ A : B : C = 6 : 10 : 13

Page No 124:

Question 6:

If A : B = 5 : 6 and B : C = 4 : 7, find A : B : C.

Answer:

We have the following:

A : B = 5 : 6
B : C = 4 : 7  = 47 = 4×647×64= 6 : 212

A : B : C =  5 : 6 : 212 =  10 : 12 : 21

Page No 124:

Question 7:

Divide Rs 360 between Kunal and Mohit in the ratio 7 : 8.

Answer:

Sum of the ratio terms  = 7 + 8 = 15

Now, we have the following:

Kunal's share = Rs 360 ×715= 24×7 = Rs 168

Mohit's share = Rs 360 ×815 = 24×8 = Rs 192



Page No 125:

Question 8:

Divide Rs 880 between Rajan and Kamal in the ratio 15:16.

Answer:

Sum of the ratio terms = 15+16=1130

Now, we have the following:
Rajan's share = Rs 880 ×151130 = Rs 880 ×611 = Rs 80×6  =  Rs 480
Kamal's share = Rs 880 ×161130= Rs 880 ×511= Rs 80 ×5 = Rs 400

Page No 125:

Question 9:

Divide Rs 5600 between A, B and C in the ratio 1 : 3 : 4.

Answer:

Sum of the ratio terms is (1 + 3 + 4) = 8

We have the following:

A's share =  Rs 5600 ×18 =Rs 56008 = Rs 700

B's share =  Rs 5600 ×38= Rs 700 × 3 = Rs 2100

C's share = Rs 5600 ×48 =Rs 700 ×4 = Rs 2800

Page No 125:

Question 10:

What number must be added to each term to the ratio 9 : 16 to make the ratio 2 : 3?

Answer:

Let x be the required number.
Then, (9 + x) : (16 + x) = 2 : 3

9+x16+x = 2327 + 3x = 32 + 2xx =5

Hence, 5 must be added to each term of the ratio 9 : 16 to make it 2 : 3.

Page No 125:

Question 11:

What number must be subtracted from each term of ratio 17 : 33 so that the ratio becomes 7 : 15?

Answer:

Suppose that x is the number that must be subtracted.
Then, (17 − x) : (33 − x) = 7 : 15

17 - x33 - x=715255 - 15x = 231 - 7x 8x = 255 - 231 = 24x = 3

Hence, 3 must be subtracted from each term of ratio 17 : 33 so that it becomes 7 : 15.

Page No 125:

Question 12:

Two numbers are in the ratio 7 : 11. If added to each of the numbers, the ratio becomes 2 : 3. Find the numbers.

Answer:

Suppose that the numbers are 7x and 11x.

Then, (7x + 7) : (11x + 7) = 2 : 3
7x + 711x + 7=23

⇒ 21x + 21 = 22x + 14

⇒ x = 7

Hence, the numbers are (7 × 7 =) 49 and (11 × 7 =) 77.

Page No 125:

Question 13:

Two numbers are in the ratio 5 : 9. On subtracting 3 from each, the ratio becomes 1 : 2. Find the numbers.

Answer:

Suppose that the numbers are 5x and 9x.
Then, (5x − 3) : (9x − 3) = 1 : 2

5x - 39x -3=12

⇒ 10x − 6 = 9x − 3
x = 3

Hence, the numbers are (5 × 3 =) 15 and (9 × 3 =) 27.

Page No 125:

Question 14:

Two numbers are in the ratio 3 : 4. If their LCM is 180, find the numbers.

Answer:

Let the numbers be 3x and 4x.
Their LCM is 12x.
Then, 12x = 180
x = 15

∴ The numbers are (3 × 15 =) 45 and (4 × 15 =) 60.

Page No 125:

Question 15:

The ages of A and B are in the ratio 8 : 3. Six years hence, their ages will be in the ratio 9 : 4. Find their present ages.

Answer:

Suppose that the present ages of A and B are 8x yrs and 3x yrs.
Then, (8x  + 6) : (3x + 6) = 9 : 4
8x+63x+6= 94
⇒ 32x + 24 = 27x + 54
⇒ 5x = 30
x = 6

Now, present age of A = 8 × 6 yrs = 48 yrs
Present age of  B = 3 × 6 yrs = 18 yrs

Page No 125:

Question 16:

The ratio of copper and zinc in an alloy is 9 : 5. If the weight of copper in the alloy is 48.6 grams, find the weight of zinc in the alloy.

Answer:

Suppose that the weight of zinc is x g.

Then, 48.6 : x = 9 : 5

x = 48.6×59=2439 = 27

Hence, the weight of zinc in the alloy is 27 g.

Page No 125:

Question 17:

The ratio of boys and girls in a school is 8 : 3. If the total number of girls be 375, find the number of boys in the school.

Answer:

Suppose that the number of boys is x.
Then, x : 375 = 8 : 3

⇒ x = 8×3753=8×125 = 1000

Hence, the number of girls in the school is 1000.

Page No 125:

Question 18:

The ratio of monthly income to the savings of a family is 11 : 2. If the savings be Rs 2500, find the income and expenditure.

Answer:

Suppose that the monthly income of the family is Rs x.

Then, x : 2500 = 11 : 2

x = 11×25002=11×1250
x = Rs 13750

Hence, the income is Rs 13,750.
∴ Expenditure = (monthly income − savings)
                         =Rs (13750 − 2500)
                          = Rs 11250

Page No 125:

Question 19:

A bag contains Rs 750 in the form of rupee, 50 P and 25 P coins in the ratio 5 : 8 : 4. Find the number of coins of each type.

Answer:

Let the numbers one rupee, fifty paise and twenty-five paise coins be 5x, 8x and 4x, respectively.

Total value of these coins = (5x ×100100+ 8x×50100 + 4x ×25100)

 5x + 8x2 + 4x4= 20x + 16x + 4x4=40x4=10x

However, the total value is Rs 750.
∴ 750 = 10x
x = 75

Hence, number of one rupee coins = 5 × 75 = 375
Number of fifty paise coins = 8 × 75 = 600
Number of twenty-five paise coins = 4 × 75 = 300

Page No 125:

Question 20:

If (4x + 5) : (3x + 11) = 13 : 17, find the value of x.

Answer:

(4x + 5) : (3x + 11) = 13 : 17

4x+ 53x + 11=131768x + 85 = 39x + 14329x = 58x = 2

Page No 125:

Question 21:

If x : y = 3 : 4, find (3x + 4y) : (5x + 6y).

Answer:

xy = 34x=3y4

Now, we have (3x + 4y) : (5x + 6y)
=3x +4y5x + 6y=3×3y4+4y5×3y4+6y= 9y+16y15y +24y  = 25y39y=2539

= 25 : 39

Page No 125:

Question 22:

If x : y = 6 : 11, find (8x − 3y) : (3x + 2y).

Answer:

xy = 611x = 6y11

Now, we have:

8x -3y3x + 2y = 8×6y11 -3y3×6y11+2y=48y-33y18y + 22y =15y40y=38

∴ (8x − 3y) : (3x + 2y) = 3 : 8

Page No 125:

Question 23:

Two numbers are in the ratio 5 : 7. If the sum of the numbers is 720, find the numbers.

Answer:

Suppose that the numbers are 5x and 7x.
The sum of the numbers is 720.
i.e., 5x + 7x = 720
⇒ 12x = 720
x = 60

Hence, the numbers are (5 × 60 =) 300 and (7 × 60 =) 420.

Page No 125:

Question 24:

Which ratio is greater?

(i) (5 : 6) or (7 : 9)
(ii) (2 : 3) or (4 : 7)
(iii) (1 : 2) or (4 : 7)
(iv) (3 : 5) or (8 : 13)

Answer:

(i) The LCM of 6 and 9 is 18.

56=5×36×3=151879=7×29×2=1418Clearly, 1418<1518

∴ (7 : 9) < (5 : 6)

(ii)  The LCM of 3 and 7 is 21.

23=2×73×7=1421 47=4×37×3=1221Clearly, 1221<1421

∴ (4 : 7) < (2 : 3)

(iii)  The LCM of 2 and 7 is 14.

  1×72×7=714   4×27×2= 814

Clearly, 714<814

∴ (1 : 2) < (4 : 7)

(iv) The LCM of 5 and 13 is 65.

 35=3×135×13= 3965 813=  8×513×5= 4065Clearly, 3965<4065

∴ (3 : 5) < (8 : 13)

Page No 125:

Question 25:

Arrange the following ratios in ascending order:

(i) (5 : 6), (8 : 9), (11 : 18)
(ii) (11 : 14), (17 : 21), (5 : 7) and (2 : 3)

Answer:

(i) We have 56, 89 and 1118.
 2  6,9, 18 3 3, 9, 9 3 1,3, 3     1,1, 1 

The LCM of 6, 9 and 18 is 18. Therefore, we have:

56= 5×36×3=151889= 8×29×2=1618 1118=1118Clearly, 1118<1518<1618

Hence, (11 : 18) < (5 : 6) < (8 : 9)

(ii) We have 1114, 1721, 57 and 23.
  2  14,21,7, 3  7 7,21,7, 3,   3 1,3,1, 3      1,1,1, 1 

The LCM of 14, 21, 7 and 3 is 42.

1114=11×314×3=33281721=17×221×2=344257=5×67×6=304223=2×143×14=2842Clearly, 2842<3042<3328<3442Hence, (2 : 3) < (5 : 7) < (11 : 14) < (17 : 21)



Page No 128:

Question 1:

Show that 30, 40, 45, 60 are in proportion.

Answer:

We have:

Product of the extremes = 30 ×  60 = 1800
Product of the means = 40 ×  45 = 1800
Product of extremes = Product of means

Hence, 30 : 40 :: 45 : 60

Page No 128:

Question 2:

Show that 36, 49, 6, 7 are not in proportion.

Answer:

We have:
Product of the extremes = 36 × 7 = 252
Product of the means = 49 × 6 = 294
Product of the extremes Product of the means

Hence, 36, 49, 6 and 7 are not in proportion.

Page No 128:

Question 3:

If 2 : 9 :: x : 27, find the value of x.

Answer:

Product of the extremes = 2 ×  27 = 54
Product of the means  = 9 × x = 9x

Since 2 : 9 :: x : 27, we have:
Product of the extremes = Product of the means
⇒ 54 = 9x
⇒ x = 6

Page No 128:

Question 4:

If 8 : x :: 16 : 35, find the value of x.

Answer:

Product of the extremes = 8 ×  35 = 280
Product of the means = 16 × x = 16x

Since 8 : x :: 16 : 35, we have:
Product of the extremes = Product of the means
⇒ 280 = 16x
x = 17.5

Page No 128:

Question 5:

If x : 35 :: 48 : 60, find the value of x.

Answer:

Product of the extremes = x × 60 = 60x
Product of the means = 35 × 48 = 1680

Since x : 35 :: 48 : 60, we have:
Product of the extremes = Product of the means
⇒ 60x= 1680
x = 28

Page No 128:

Question 6:

Find the fourth proportional to the numbers:

(i) 8, 36, 6
(ii) 5, 7, 30
(iii) 2.8, 14, 3.5

Answer:

(i) Let the fourth proportional be x.
Then, 8 : 36 :: 6 : x

8 × x = 36 × 6                                   [Product of extremes = Product of means]
⇒ 8x = 216
x = 27
Hence, the fourth proportional is 27.

(ii) Let the fourth proportional be x.
Then, 5 : 7 :: 30 : x
5 ×x = 7 ×30                                      [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 5x = 210
x = 42

Hence, the fourth proportional is 42.

(iii) Let the fourth proportional be x.
Then, 2.8 × x =14 × 3.5                                [Product of extremes = Product of means]
⇒ 8x = 216
⇒ 2.8x = 49
x = 17.5
Hence, the fourth proportional is 17.5.

Page No 128:

Question 7:

If 36, 54, x are in continued proportion, find the value of x.

Answer:

36, 54 and x are in continued proportion.
Then, 36 : 54 :: 54 : x
36 × x =54 × 54                                 [Product of extremes = Product of means]
⇒ 36x = 2916
x = 81

Page No 128:

Question 8:

If 27, 36, x are in continued proportion, find the value of x.

Answer:

27, 36 and x are in continued proportion.
Then, 27 : 36 :: 36 : x
27×x = 36 ×36         [Product of extremes = Product of means]
⇒ 27x = 1296
x = 48

Hence, the value of x is 48.

Page No 128:

Question 9:

Find the third proportional to:

(i) 8 and 12
(ii) 12 and 18
(iii) 4.5 and 6

Answer:

(i)  Suppose that x is the third proportional to 8 and 12.
Then, 8 :12 :: 12 : x
⇒ 8 ×x = 12 × 12                                            (Product of extremes = Product of means )
⇒ 8x = 144
x = 18

Hence, the required third proportional is 18.

(ii) Suppose that x is the third proportional to 12 and 18.
Then, 12 : 18 :: 18 : x
12 × x = 18 ×18                                       (Product of extremes = Product of means )
⇒ 12x = 324
x = 27

Hence, the third proportional is 27.

(iii) Suppose that x is the third proportional to 4.5 and 6.
Then, 4.5 : 6:: 6 : x
4.5 × x= 6 × 6                                         (Product of extremes = Product of means )
⇒ 4.5x = 36
x = 8

Hence, the third proportional is 8.

Page No 128:

Question 10:

If the third proportional to 7 and x is 28, find the value of x.

Answer:

The third proportional to 7 and x is 28.
Then, 7 : x :: x : 28
⇒ 7 × 28 = x2           (Product of extremes = Product of means)
x = 14

Page No 128:

Question 11:

Find the mean proportional between:

(i) 6 and 24
(ii) 3 and 27
(iii) 0.4 and 0.9

Answer:

(i)  Suppose that x is the mean proportional.
 
Then, 6 : x :: x : 24

6 × 24 = x × x                                         (Product of extremes = Product of means)
x2  = 144
x = 12

Hence, the mean proportional to 6 and 24 is 12.

(ii)  Suppose that x is the mean proportional.

Then, 3 : x :: x : 27
3 × 27 = x × xx2 = 81                                         (Product of extremes =Product of means)
x = 9

Hence, the mean proportional to 3 and 27 is 9.

(iii)  Suppose that x is the mean proportional.

Then, 0.4 : x :: x : 0.9

0.4 × 0.9 = x × xx2 = 0.36                              (Product of extremes =Product of means)
⇒x = 0.6

Hence, the mean proportional to 0.4 and 0.9 is 0.6.

Page No 128:

Question 12:

What number must be added to each of the numbers 5, 9, 7, 12 to get the numbers which are in proportion?

Answer:

Suppose that the number is x.
 
Then, (5 + x) : ( 9 + x) :: (7 + x) : (12 + x)

(5 + x) ×(12 + x) = (9 + x) × (7 + x)            (Product of extremes = Product of means)60 +5 x + 12 x + x2 = 63 + 9x + 7x + x260 + 17x = 63 + 16xx = 3

Hence, 3 must be added to each of the numbers: 5, 9, 7 and 12, to get the numbers which are in proportion.

Page No 128:

Question 13:

What number must be subtracted from each of the numbers 10, 12, 19, 24 to get the numbers which are in proportion?

Answer:

Suppose that x is the number that is to be subtracted.

Then, (10 − x) : (12 − x) :: (19 − x) : (24 − x)

(10- x) ×(24 - x) =(12 - x) ×(19 - x)                   (Product of extremes =Product of means)240 - 10x -24x + x2 = 228 - 12x -19x + x2240 - 34x = 228 - 31x3x = 12x = 4
.
Hence, 4 must be subtracted from each of the numbers: 10, 12, 19 and 24, to get the numbers which are in proportion.

Page No 128:

Question 14:

The scale of a map is 1 : 5000000. What is the actual distance between two towns, if they are 4 cm apart on the map?

Answer:

Distance represented by 1 cm on the map = 5000000 cm = 50 km

Distance represented by 3 cm on the map = 50 × 4 km = 200 km

∴ The actual distance is 200 km.

Page No 128:

Question 15:

At a certain time a tree 6 m high casts a shadow of length 8 metres. At the same time a pole casts a shadow of length 20 metres. Find the height of the pole.

Answer:

(Height of tree) : (height of its shadow) = (height of the pole) : (height of its shadow)

Suppose that the height of pole is x cm.

Then, 6 : 8 = x : 20

x = 6×208 = 15
∴ Height of the pole = 15 cm

Page No 128:

Question 1:

Mark (✓) against the correct answer
If a : b = 3 : 4 and b : c = 8 : 9, then a : c = ?

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

Answer:

The correct option is (d).


ac= ab×bc = 34×89        = 23

Hence, a : c = 2 : 3

Page No 128:

Question 2:

Mark (✓) against the correct answer
If A : B = 2 : 3 and B : C = 4 : 5, then C : A = ?

(a) 15 : 8
(b) 6 : 5
(c) 8 : 5
(d) 8 : 15

Answer:

(a) 15 : 8

AB= 23BC= 45Then, AB×BC  = 23×45= 815Hence, C : A = 15 : 8

Page No 128:

Question 3:

Mark (✓) against the correct answer
If 2A = 3B and 4B = 5C, then A : C = ?

(a) 4 : 3
(b) 8 : 15
(c) 3 : 4
(d) 15 : 8

Answer:

The correct option is (d).


A = 3B2C = 4B5 A : C = AC = 3B24B5=  158
Hence, A : C = 15 : 8

Page No 128:

Question 4:

Mark (✓) against the correct answer
If 15% of A = 20% of B, then A : B = ?

(a) 3 : 4
(b) 4 : 3
(c) 17 : 16
(d) 16 : 17

Answer:

The correct option is (b).

15100A =20100BAB =43

Hence, A : B = 4 : 3

Page No 128:

Question 5:

Mark (✓) against the correct answer
If A=13B and B=12C, then A : B : C = ?

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

Answer:

(a)  1 : 3 : 6

A = 13BC = 2B A : B : C = 13B : B : 2B = 1 : 3 : 6



Page No 129:

Question 6:

Mark (✓) against the correct answer
If A : B = 5 : 7 and B : C = 6 : 11, then A : B : C = ?

(a) 30 : 42 : 55
(b) 30 : 42 : 77
(c) 35 : 49 : 66
(d) none of these

Answer:

(b)  30 : 42 : 77


AB = 57A = 5B7BC = 611C =11B6 A : B : C = 5B   7 : B : 11B6 = 30 : 42 : 77            

Page No 129:

Question 7:

Mark (✓) against the correct answer
If 2A = 3B = 4C, then A : B : C = ?

(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 6 : 4 : 3
(d) 3 : 4 : 6

Answer:

(c)  6 : 4 : 3

2A=3B = 4CThen, A = 3B2 and  C = 3B4 A : B : C = 3B2 : B : 3B4 = 6 : 4 : 3

Page No 129:

Question 8:

Mark (✓) against the correct answer
If A3=B4=C5, then A : B : C = ?

(a) 3 : 4 : 5
(b) 4 : 3 : 5
(c) 5 : 4 : 3
(d) 20 : 15 : 12

Answer:

(a) 3 : 4 : 5

A =3B4C = 5B4 A : B : C = 3B4  : B : 5B4

                 = 3 : 4 : 5

Page No 129:

Question 9:

Mark (✓) against the correct answer
If1x:1y:1z=2:3:5, then, x : y : z = ?

(a) 2 : 3 : 5
(b) 15 : 10 : 6
(c) 5 : 3 : 2
(d) 6 : 10 : 15

Answer:

(b)  15 : 10 : 6

1x :1y=2 : 3Then, y : x = 2 : 3 and y = 23x1y:1z = 3 : 5Then, z : y =3 : 5  and z = 35y x : y : z = x : 23x : 35y  = x : 23x: 35×23x =x : 23x : 25x  =  15 : 10 : 6

Page No 129:

Question 10:

Mark (✓) against the correct answer
If x : y = 3 : 4, then (7x + 3y) : (7x − 3y) = ?

(a) 4 : 3
(b) 5 : 2
(c) 11 : 3
(d) 37 : 39

Answer:

xy= 34

x = 3y4 7x + 3y7x - 3y = 73y4+3y73y4 - 3y=21y + 12y21y - 12y = 33y9y= 113

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

Page No 129:

Question 11:

Mark (✓) against the correct answer
If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = ?

(a) 2 : 1
(b) 3 : 2
(c) 5 : 2
(d) 5 : 3

Answer:

(c) 5 : 2

3a + 5b3a - 5b=513a + 5b = 15a - 25b12a = 30bab= 3012=52

a : b = 5 : 2

Page No 129:

Question 12:

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If 7 : x :: 35 : 45, then x = ?

(a) 11
(b) 15
(c) 9
(d) 5

Answer:

(c)  9

7 × 45 = x × 35      (Product of extremes = Product of means)35x = 315x = 9

Page No 129:

Question 13:

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What number has to be added to each term of 3 : 5 to make the ratio 5 : 6?

(a) 6
(b) 7
(c) 12
(d) 11

Answer:

(b) 7

Suppose that x is the number that is to be added.

Then, (3 + x) : (5 + x) = 5 : 6

3 +x5 + x= 5618 + 6x = 25+ 5xx =7

Page No 129:

Question 14:

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Two numbers are in the ratio 3 : 5. If each number is increased by 10, the ratio becomes 5 : 7. The sum of the numbers is

(a) 8
(b) 16
(c) 35
(d) 40

Answer:

(d) 40

Suppose that the numbers are x and y.

Then, x : y = 3 : 5 and (x + 10) : (y + 10) = 5 : 7

xy=35x = 3y5=>x + 10y + 10=57=>7x+70 = 5y + 50=>7×3y5 + 70 =5y + 50=>5y -21y5 = 20=>4y5= 20=>y = 25Therefore, x =3 ×255= 15

Hence, sum of numbers = 15 + 25 = 40

Page No 129:

Question 15:

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What least number is to be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4?

(a) 3
(b) 5
(c) 6
(d) 9

Answer:

(a)  3
Suppose that x is the number that is to be subtracted.
Then, (15 − x) : (19 − x) = 3 : 4

15 - x19- x=34Cross multiplying, we get:60 - 4x = 57 - 3xx = 3 

Page No 129:

Question 16:

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If Rs 420 is divided between A and B in the ratio 3 : 4, then A's share is

(a) Rs 180
(b) Rs 240
(c) Rs 270
(d) Rs 210

Answer:

(a)  Rs 180

A's share = 37 × 420 = 180

Page No 129:

Question 17:

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The boys and girls in a school are in the ratio 8 : 5. If the number of girls is 160, what is the total strength of the school?

(a) 250
(b) 260
(c) 356
(d) 416

Answer:

(d) 416

Let x be the number of boys.
Then, 8 : 5 = x : 160

85= x160 x=8×1605= 256 Total strength of the school =256 + 160 = 416

Page No 129:

Question 18:

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Which one is greater out of (2 : 3) and (4 : 7)?

(a) 2 : 3
(b) 4 : 7
(c) both are equal

Answer:

(a) (2 :3)

LCM of 3 and 7 = 7×3=21

2×73×7 = 1421 and 4×37×3= 1221Clearly, 1221<1421Hence, (4 : 7) < (2 : 3)

Page No 129:

Question 19:

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The third proportional to 9 and 12 is

(a) 10.5
(b) 8
(c) 16
(d) 21

Answer:

(c) 16

Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x

9 × x = 12 × 12              (Product of extremes= Product of means)9x = 144x = 16                               

Page No 129:

Question 20:

Mark (✓) against the correct answer
The mean proportional between 9 and 16 is

(a) 12.5
(b) 12
(c) 5
(d) none of these

Answer:

(b) 12

Suppose that the mean proportional is x.

Then, 9 : x :: x : 16

9 ×16 = x × x                 (Product of extremes =Product of means)x2 = 144x = 12

Page No 129:

Question 21:

Mark (✓) against the correct answer
The ages of A and B are in the ratio 3 : 8. Six years hence, their ages will be in the ratio 4 : 9. The present age of A is

(a) 18 years
(b) 15 years
(c) 12 years
(d) 21 years

Answer:

(a)  18 years

Suppose that the present ages of A and B are 3x yrs and 8x yrs, respectively.
After six years, the age of A will be (3x+6) yrs and that of B will be (8x+6) yrs.
Then, (3x +6) : (8x + 6) = 4 : 9

3x + 68x+6= 4927x + 54 = 32x + 245x = 30x = 6Hence, the present ages of A and B are 18 yrs and 48 yrs, respectively. 



Page No 131:

Question 1:

Compare 4 : 5 and 7 : 9.

Answer:

The given fractions are 45 and 79.
LCM of 5 and 9 = 5 × 9 = 45

Now, we have:4×95×9=3645 and  7×59×5= 3545Clearly, 3545<3645Hence, (7 : 9 ) <(4 : 5)

Page No 131:

Question 2:

Divide Rs 1100 among A, B and C in the ratio 2 : 3 : 5.

Answer:

The sum of ratio terms is 10.

Then, we have:

A's share = Rs 210×1100 = Rs 220

B's share =Rs 310× 1100 =Rs 330 

C's share =Rs 510× 1100 = Rs 550

Page No 131:

Question 3:

Show that the numbers 25, 36, 5, 6 are not in proportion.

Answer:

Product of the extremes = 25 × 6 = 150
Product of the means = 36 × 5 = 180

The product of the extremes is not equal to that of the means.

Hence, 25, 36, 5 and 6 are not in proportion.

Page No 131:

Question 4:

If x,18,108 are in continued proportion, find the value of x.

Answer:

x : 18 :: 18 : 108

 x × 108 = 18×18            (Product of extremes =Product of means)108x = 324x =3Hence, the value of x is 3.

Page No 131:

Question 5:

Two numbers are in the ratio 5 : 7. If the sum of these numbers is 84, find the numbers.

Answer:

Suppose that the numbers are 5x and 7x.
Then, 5x + 7x = 84
⇒ 12x = 84
x = 7

Hence, the numbers are (5 × 7 =) 35 and (7 × 7 =) 49.

Page No 131:

Question 6:

The ages of A and B are in the ratio 4 : 3. Eight years ago, their ages were in the ratio 10 : 7. Find their present ages.

Answer:

Suppose that the present ages of A and B are 4x yrs and 3x yrs, respectively.
Eight years ago, age of A = (4x − 8) yrs
Eight years ago, age of B = (3x − 8) yrs
Then, (4x − 8) : (3x − 8) = 10 : 7

4x-83x - 8=10728x - 56 = 30x - 802x = 24x= 12Hence, present age of A = 4×12 yrs = 48 yrs and present age of B = 3×12 yrs = 36 yrs 

Page No 131:

Question 7:

If a car covers 54 km in an hour, how much distance will it cover in 40 minutes?

Answer:

Distance covered in 60 min = 54 km
Distance covered in 1 min = 5460 km

∴ Distance covered in 40 min = 5460×40 = 36 km

Page No 131:

Question 8:

Find the third proportional to 8 and 12.

Answer:

Suppose that the third proportional to 8 and 12 is x.
Then, 8 : 12 :: 12 : x

⇒ 8x = 144  (Product of extremes = Product of means)
x = 18

Hence, the third proportional is 18 .

Page No 131:

Question 9:

If 40 men can finish a piece of work in 60 days, in how many days will 75 men finish the same work?

Answer:

40 men can finish the work in 60 days.
1 man can finish the work in 60 × 40 days.     [Less men, more days]
75 men can finish the work in 60×4075 = 32 days 

Hence, 75 men will finish the same work in 32 days.

Page No 131:

Question 10:

Mark (✓) against the correct answer
If 2A = 3B = 4C then A : B : C = ?

(a) 2 : 3 : 4
(b) 3 : 4 : 6
(c) 4 : 3 : 2
(d) 6 : 4 : 3

Answer:

(d)  6 : 4 : 3

A = 32BC = 34B A : B : C = 32B : B : 34B                         =  6 : 4 : 3

Page No 131:

Question 11:

Mark (✓) against the correct answer
IfA2=B3=C4 then A : B : C = ?
(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 3 : 2 : 4
(d) none of these

Answer:

(a)  2 : 3 : 4


A = 23BC = 43B A : B : C =  23B : B : 43 B                         = 2 : 3 : 4

Page No 131:

Question 12:

Mark (✓) against the correct answer
If (x : y) = 3 : 4, then (7x + 3y) : (7x − 3y) = ?

(a) 7 : 3
(b) 5 : 2
(c) 11 : 3
(d) 14 : 9

Answer:

(c) 11 : 3

We have x = 34yNow, 7x + 3y7x - 3y = 7×34y +3y7×34y - 3y = 21y + 12y21y -12y=33y9y=113

Page No 131:

Question 13:

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What least number must be subtracted from each term of the ratio 15 : 19 to make the ratio 3 : 4?

(a) 3
(b) 5
(c) 6
(d) 9

Answer:

(a) 3
 
Suppose that the number to be subtracted is x.
Then, (15 − x) : (19 − x) = 3 : 4

15 - x19 - x =3460 - 4x = 57 - 3xx = 3

Page No 131:

Question 14:

Mark (✓) against the correct answer
If Rs 840 is divided between A and B in the ratio 4 : 3, then B's share is

(a) Rs 480
(b) Rs 360
(c) Rs 320
(d) Rs 540

Answer:

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B's share = Rs 840 ×37 = Rs 360

Page No 131:

Question 15:

Mark (✓) against the correct answer
The ages of A and B are in the ratio 5 : 2. After 5 years, their ages will be in the ratio 15 : 7. The present age of A is

(a) 48 years
(b) 36 years
(c) 40 years
(d) 35 years

Answer:

(c) 40 years

Suppose that the present ages of A and B are 5x yrs and 2x yrs, respectively.
After 5 years, the ages of A and B will be (5x+5) yrs and (2x+5) yrs, respectively.

Then, (5x + 5) : (2x + 5) = 15 : 7

5x+52x +5=157

Cross multiplying, we get:

35x + 35 = 30x + 75
⇒ 5x = 40
x = 8

Hence, the present age of A is 5 × 8 = 40 yrs.

Page No 131:

Question 16:

Mark (✓) against the correct answer
The boys and girls in a school are in the ratio 9 : 5. If the number of girls is 320, then the total strengh of the school is

(a) 840
(b) 896
(c) 920
(d) 576

Answer:

(b)  896

Suppose that the number of boys in the school is x.
Then, x : 320 = 9 : 5
⇒ 5x = 2880
x = 576

Hence, total strength of the school = 576 + 320 = 896

Page No 131:

Question 17:

Fill in the blanks.

(i) If A : B = 2 : 3 and B : C = 4 : 5, then C : A = ...... .
(ii) If 16% of A = 20% of B, then A : B = ...... .
(iii) If A=13B and B=12C, then A : B : C = ...... .
(iv) If A : B = 5 : 7 and B : C = 6 : 11, then A : B : C = ...... .

Answer:

(i) 15 : 8

AC  = AB×BC = 23×45= 815

∴ C : A=15 : 8

(ii) 5 : 4

16100A = 20100BAB= 2016= 54

(iii) 1 : 3 : 6

A : B : C = 13 B : B :  2B = 1 : 3 : 6

(iv)  30 : 42 : 77

AB= 5×67×6=3042BC= 6×711×7= 4277A : B : C = 30 : 42 : 77

Page No 131:

Question 18:

Write 'T' for true and 'F' for false

(i) Mean proportional between 0.4 and 0.9 is 6.
(ii) The third proportional to 9 and 12 is 10.5.
(iii) If 8 : x :: 48 : 18, then x = 3.
(iv) If (3a + 5b) : (3a − 5b) = 5 : 1, then a : b = 5 : 2

Answer:

(i) F
Suppose that the mean proportional is x.
Then, 0.4 : x :: x : 0.9
0.9 × 0.4 =x × x       (Product of extremes =Product of means) 
x2 = 0.36x = 0.6


(ii)  F
Suppose that the third proportional is x.
Then, 9 : 12 :: 12 : x
⇒ 9x = 144                      (Product of extremes = Product of means)
x = 16

(iii)  T
8 : x :: 48 : 18
⇒ 144 = 48x               (Product of extremes = Product of means)        
x = 3

(iv) T

3a + 5b3a-5b=513a + 5b = 15a - 25 b
⇒ 12a = 30b

a : b = 5 : 2



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