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## Thủ Thuật về What must be added to each of the number 10 17 24 and 38 to get the number which are in proportion? Mới Nhất

Quý khách đang tìm kiếm từ khóa What must be added to each of the number 10 17 24 and 38 to get the number which are in proportion? được Update vào lúc : 2022-10-05 00:20:08 . Với phương châm chia sẻ Thủ Thuật về trong nội dung bài viết một cách Chi Tiết 2022. Nếu sau khi tìm hiểu thêm Post vẫn ko hiểu thì hoàn toàn có thể lại Comments ở cuối bài để Ad lý giải và hướng dẫn lại nha.

Rs Aggarwal 2022 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 2022 Book of Class 7 Math Chapter 8 are provided here for you for không lấy phí. You will also love the ad-không lấy phí experience on
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What must be added to each of the number 10 17 24 and 38 to get the numbers which are in proportion?What must be added to the numbers 10/18 22 and 38 to get the numbers which are in proportion?How do you make a number proportional?What must be added to number 13 10 and 18 so that they are in proportion?What least number must be added to each of 6 14 18 and 38 to make them in proportion?

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

(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

(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.

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.

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

∴ A : C = 2 : 5

Page No 124:

Question 5:

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

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.

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.

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.

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.

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?

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

⇒9+x16+
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:

Question 11:

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

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:

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.

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.

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

⇒ 5x – 3
9x -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.

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.

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.

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.

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.

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.

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 + 8×2 +
4×4= 20x + 16x + 4×4=40×4=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.

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

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

Page No 125:

Question
21:

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

xy = 34⇒x=3y4

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

= 25 : 39

Page No 125:

Question 22:

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

xy =
611⇒x = 6y11

Now, we have:

8x -3y3x + 2y = 8×6y11 -3y3×6y11+2y=48y-33
y18y + 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.

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)

(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)

(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=1518
89= 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=11×3
14×3=33281721=17×221×2=344257=5×67×
6=304223=2×143×14=2842Clearly, 2842<3042<33
28<3442Hence, (2 : 3) < (5 : 7) < (11 : 14) <
(17 : 21)

Page No 128:

Question 1:

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

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.

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.

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.

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.

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

(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.

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.

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

(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.

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

(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:

Question 12:

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

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:

Question 13:

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

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:

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?

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.

(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

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

(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

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

The correct option is (b).

15100A =20100
B⇒AB =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

(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

(b)  30 : 42 : 77

AB = 57⇒A =
5B7BC = 611
⇒C =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

(c)  6 : 4 : 3

2A=3
B = 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

(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

(b)  15 : 10 : 6

1x :1y=2 : 3
Then, 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

xy
= 34

x = 3y4∴ 7x + 3y7x – 3y
= 73y4+3y73y4 – 3y=21
y + 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

(c) 5 : 2

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

∴ a : b = 5 : 2

Page No 129:

Question 12:

Mark (✓) against the correct answer
If 7 :
x :: 35 : 45, then x = ?

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

(c)  9

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

Page No 129:

Question 13:

Mark
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

(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+ 5x⇒x =
7

Page No 129:

Question 14:

Mark (✓) against the correct answer
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

(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:

Mark (✓) against the correct answer
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

(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:

Question 16:

Mark (✓) against the correct answer
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

(a)  Rs 180

A’s share = 37 × 420 = 180

Page No 129:

Question 17:

Mark (✓) against the correct answer
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

(d) 416

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

⇒85= x160 ⇒x=8×160
5= 256∴ Total strength of the school =256 + 160 = 416

Page No 129:

Question 18:

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

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

(a) (2 :3)

LCM of 3 and 7 = 7×3=21

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

Page No 129:

Question 19:

Mark (✓) against the correct answer
The third proportional to 9 and 12 is

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

(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:

Question 20:

Mark
The mean proportional between 9 and 16 is

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

(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:

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

(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= 49⇒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:

Question 1:

Compare 4 : 5 and 7 : 9.

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<3645
Hence, (7 : 9 ) <(4 : 5)

Page No 131:

Question 2:

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

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.

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.

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:

Question 5:

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

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.

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-83x – 8
=107⇒28x – 56 = 30x – 80⇒2x = 24⇒x= 12
Hence, 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?

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.

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?

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

(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
If
A2=B3=C4 then A : B : C = ?
(a) 2 : 3 : 4
(b) 4 : 3 : 2
(c) 3 : 2 : 4
(d) none of these

(a)  2 : 3 : 4

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

(c)
11 : 3

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

Page No 131:

Question 13:

Mark (✓) against the correct

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

(a) 3

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

⇒15 – x
19 – x =34⇒60 – 4x
= 57 – 3x
⇒x = 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

(b) 360

Sum of the ratio terms = 4 + 3 = 7

∴ B’s share = Rs 840 ×37 = Rs 360

Page No 131:

Question 15:

Mark
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

(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

(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 = …… .

(i) 15 : 8

AC  = AB×BC = 23×45=
815

∴ C : A=15 : 8

(ii) 5 : 4

16100A = 20100B⇒AB= 2022= 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:

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

(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 added to each of the number 10 17 24 and 38 to get the numbers which are in proportion?

Hence, 4 must be added to each of the numbers.

### How do you make a number proportional?

Note: The numbers are proportional when the ratio of the LHS of the proportions is equal to the RHS of the proportion. To check if the numbers are in proportion we just find their ratio of both the sides. If ab and c are unit vectors then left ab2 right+bc2+ca…

### What must be added to number 13 10 and 18 so that they are in proportion?

What must be added to numbers 1,3,10,18 so that they are in proportion? 1+3+10+18. = 32. =2.

### What least number must be added to each of 6 14 18 and 38 to make them in proportion?

Answer: The least number which should be added to make the numbers proportional is 2.
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