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

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

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What number must be subtracted from each of the numbers $10,12,19,24$ to get the numbers which are in proportion?
What must be subtracted from each term of the ratio 4 is to 7 so that the ratio becomes 2 is to 5?
What number must be subtracted from each term of the ratio 7 ratio 33 so that the ratio becomes 7 ratio 15?
What must be subtracted from each term of the ratio 3 is to 7 so that the ratio becomes 2 is to 5?
What must be subtract from 8x4 14x3 2 x2 7x 8 so that resulting polynomial is exactly divisible by 4x2 3x 2?

Answer

Verified

Hint: The number subtracting from the given numbers is the same. So we can form an equation using the definition of proportion. Proportionality can be expressed using fractions. Solving this we get the answer.

Formula used: If we say four numbers $a,b,c,d$ are in proportion we mean $a:b = c:d$ or $\dfrac{a}{b} = \dfrac{c}{d}$.

Complete step-by-step solution:
We are given four numbers $10,12,19,24$.
It is said that the same number when subtracted from these numbers make a proportion.
If we say four numbers $a,b,c,d$ are in proportion we mean $a:b = c:d$ or $\dfrac{a}{b} = \dfrac{c}{d}$.
Let the number subtracting be $x$.
So we have,
$\Rightarrow$$\dfrac{{10 - x}}{{12 - x}} = \dfrac{{19 - x}}{{24 - x}}$
Cross multiplying we have,
$\Rightarrow$$(10 - x)(24 - x) = (12 - x)(19 - x)$
Opening the brackets we get,
$\Rightarrow$$240 - 10x - 24x + {x^2} = 228 - 12x - 19x + {x^2}$
Simplifying we get,
$\Rightarrow$$240 - 34x + {x^2} = 228 - 31x + {x^2}$
Cancelling ${x^2}$ from both sides we get,
$\Rightarrow$$240 - 34x = 228 - 31x$
Rearranging the terms we have,
$\Rightarrow$$240 - 228 = - 31x + 34x$
$ \Rightarrow 3x = 12$
Dividing both sides by $3$ we get,
$\Rightarrow$$x = \dfrac{{12}}{3} = 4$
That is, $4$ has to be subtracted from the numbers to make them proportion.

$\therefore $ The answer is $4$.

Note: Proportion says that two ratios (or fractions) are equal.
If $a:b = c:d$, then the quantity $d$ is called the fourth proportional to $a,b$ and $c$.
Ratio and proportion is a widely used concept in Mathematics as well as in day to day life.
If the number subtracting from each number is different we cannot form an equation like this. So we need more information for those problems. But here this works, since the same number is subtracting from each. So we got an equation with a single variable which could be solved easily.

Rs Aggarwal 2018 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 2018 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 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 7 Math are prepared by experts and are 100% accurate.

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 = 58×1625⇒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.

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 = 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?

Answer:

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.

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 - 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.

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

Page No 125:

Question 21:

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

Answer:

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).

Answer:

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.

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=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.

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) 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?

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 + 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?

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 + 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?

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 =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

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 = 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

Answer:

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

Answer:

(a) 3 : 4 : 5

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

= 3 : 4 : 5

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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 : 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

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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=21 y + 12y21y - 12y = 33y9y= 113

Hence, (7x + 3y) : (7x − 3y) = 11 : 3

The correct option is (c).

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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:

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

∴ a : b = 5 : 2

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Question 12:

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

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

Answer:

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

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Question 13:

Mark (✓) against the correct answer
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= 56⇒18 + 6x = 25+ 5x⇒x = 7

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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

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

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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

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 - 3x⇒x = 3

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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

Answer:

(a) Rs 180

A's share = 37 × 420 = 180

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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

Answer:

(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

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Question 18:

Mark (✓) against the correct answer
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

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

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Question 19:

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

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

Answer:

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

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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 = 144⇒x = 12

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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= 49⇒27x + 54 = 32x + 24 ⇒5x = 30⇒x = 6Hence, the present ages of A and B are 18 yrs and 48 yrs, respectively.

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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<3645 Hence, (7 : 9 ) <(4 : 5)

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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

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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.

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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 = 324⇒x =3Hence, the value of x is 3.

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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.

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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 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

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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

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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 .

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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.

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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

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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

Answer:

(a) 2 : 3 : 4

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

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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:

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

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Question 13:

Mark (✓) against the correct answer
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 - x 19 - x =34⇒60 - 4x = 57 - 3x ⇒x = 3

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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

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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:

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.

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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

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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 = 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

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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 = 0.36⇒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

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What must be subtracted from each term of the ratio 3 is to 7 so that the ratio becomes 2 is to 5?

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

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Hence, we have to subtract `11x-8` so that `8x^4+14 x^3-2x^2+7x-8` is exactly divisible by `4x^2+3x-2`.