# What number must be subtracted from each term of the ratio 15 23 so that the ratio becomes 7:15

• Textbook Solutions
• Class 7
• Math
• ratio and proportion

Rs Aggarwal 2017 Solutions for Class 7 Math Chapter 8 Ratio And Proportion are provided here with simple step-by-step explanations. These solutions for Ratio And Proportion are extremely popular among Class 7 students for Math Ratio And Proportion Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2017 Book of Class 7 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2017 Solutions. All Rs Aggarwal 2017 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

#### Page No 124:

(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:

(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÷5 45÷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:

AB  = 75 and  BC = 914

Therefore, we have:

A B×BC = 75×914AC = 910

∴ A : C = 9 : 10

#### Page No 124:

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

∴ A : C = 2 : 5

#### Page No 124:

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:

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:

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:

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:

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:

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

⇒9+x 16+x = 23⇒27 + 3x  = 32 + 2x ⇒x =5

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

#### Page No 125:

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

⇒17 - x33 - x=715⇒255 -  15x = 231 -  7x ⇒8x =  255 - 231 = 24 ⇒x = 3

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

#### Page No 125:

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:

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:

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:

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:

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:

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:

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:

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

Total value of these coins = (5 x ×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:

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

⇒4x+ 53x + 11=1317⇒68x + 85 = 39x +  143⇒29x = 58 ⇒x = 2

#### Page No 125:

x y = 34⇒x=3y4

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

= 25 : 39

#### Page No 125:

xy = 611⇒x = 6y11

Now, we have:

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

∴ (8x − 3y) : (3x + 2y) = 3 : 8

#### Page No 125:

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:

(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=  8×5 13×5= 4065Clearly, 3965<4065

∴ (3 : 5) < (8 : 13)

#### Page No 125:

(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:

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:

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:

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:

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:

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:

(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:

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:

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:

(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:

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:

(i)  Supposethat 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 × x⇒x2 = 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 × x ⇒x2 = 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:

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 +  x2⇒60 + 17x = 63 + 16x⇒x = 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:

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 + x2⇒240 - 34x = 228 - 31x⇒3x = 12⇒x = 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:

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:

(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:

The correct option is (d).

ac= ab×bc = 34×89         = 23

Hence, a : c = 2 : 3

#### Page No 128:

(a) 15 : 8

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

#### Page No 128:

The correct option is (d).

A =  3B2C = 4B5∴ A : C = AC = 3B 24B5=  158
Hence, A : C = 15 : 8

#### Page No 128:

The correct option is (b).

15100A = 20100B⇒AB =43

Hence, A : B = 4 : 3

#### Page No 128:

(a)  1 : 3 : 6

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

#### Page No 129:

(b)  30 : 42 : 77

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

#### Page No 129:

(c)  6 : 4 : 3

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

#### Page No 129:

(a) 3 : 4 : 5

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

= 3 : 4 : 5

#### Page No 129:

(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:

xy= 34

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

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

#### Page No 129:

(c) 5 : 2

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

∴ a : b = 5 : 2

#### Page No 129:

(c)  9

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

#### Page No 129:

(b) 7

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

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

⇒3 +x5 + x= 56⇒18 + 6x = 25+ 5 x⇒x =7

#### Page No 129:

(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:

(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 - 3x⇒x = 3

#### Page No 129:

(a)  Rs 180

A's share = 37 × 420 = 180

#### Page No 129:

(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:

(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:

(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 = 144⇒x = 16

#### Page No 129:

(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 = 144⇒x = 12

#### Page No 129:

(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= 4 9⇒27x + 54 = 32x + 24⇒5x = 30⇒x = 6Hence, the present ages of A and B are 18 yrs and 48 yrs,  respectively.

#### Page No 131:

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= 3545 Clearly, 3545<3645Hence, (7 : 9 ) <(4 : 5)

#### Page No 131:

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:

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:

x : 18 :: 18 : 108

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

#### Page No 131:

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:

Suppose that the present ages of A and B are 4x yrs and 3xyrs, 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-8 3x - 8=107⇒28x - 56 = 30x - 80⇒2x = 24⇒x= 12Hence, present age of A = 4×12 yrs = 48 yrs  and present age of B = 3×12 yrs = 36 yrs

#### Page No 131:

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:

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:

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:

(d)  6 : 4 : 3

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

#### Page No 131:

(a)  2 : 3 : 4

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

#### Page No 131:

(c) 11 : 3

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

#### Page No 131:

(a) 3

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

⇒15 - x19 - x =34⇒ 60 - 4x = 57  - 3x⇒x =  3

#### Page No 131:

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B's share = Rs 840 ×37 = Rs 360

#### Page No 131:

(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:

(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:

(i) 15 : 8

AC  = AB×BC = 23× 45= 815

∴ C : A=15 : 8

(ii) 5 : 4

16100A = 20100B⇒AB= 2016 = 54

(iii) 1 : 3 : 6

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

(iv)  30 : 42 : 77

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

#### Page No 131:

(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 = 0.36 ⇒x = 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=51 ⇒3a + 5b = 15a - 25 b
⇒ 12a = 30b

⇒ a : b = 5 : 2

View NCERT Solutions for all chapters of Class 7

### What must be subtracted from each term of ratio 15 to 23 so that the ratio becomes 7 by 15?

7(23-x)=15(15-x).

### What must be subtracted from each term of the ratio 3/7 so that the ratio become 2 5?

Therefore, if 1/3 is subtracted from each term of the ratio 3:7, the ratio becomes 2:5.

### What number must be subtracted from each term of the ratio 19/21 to make 7 8?

This is Expert Verified Answer It means that we have to find out a number which should be subtracted from 19 : 21 to make it in the ratio 7 : 8. Ratio that make 7 : 8 are :- 14 : 16, 21 : 24, 28 : 32 etc. Condition is that if both the results are equal then only we can subtract them as told in question. = 5.

### What number should be subtracted from each of the numbers 18/24 28 38 so that the remainders may be in proportion?

Answer: The required number is 3. To find : What number should be subtracted from each of the number 18,24,28,38 so that remainders may be in proportion?