ML Aggarwal Ratio and Proportion Exe-7.1 Class 10 ICSE Maths Solutions . We Provide Step by Step Answer of Exe-7.1 Questions for Ratio and Proportion as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-10.

## ML Aggarwal Ratio and Proportion Exe-7.1 Class 10 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 10th |

Chapter-7 | Ratio and Proportion |

Writer / Book | Understanding |

Topics | Solutions of Exe-7.1 |

Academic Session | 2024-2025 |

**Ch-7, Ratio and Proportion Exe-7.1**

ML Aggarwal Class 10 ICSE Maths Solutions

**Question- 1. ****An alloy consists of 27 ½ kg of copper and 2 ¾ kg of tin. Find the ratio by weight of tin to the alloy.**

**Answer :**

Copper = 27 ½ kg = 55/2 kg

Tin = 2 ¾ kg = 11/4 kg

Total alloy = 55/2 + 11/4

= (110 + 11)/ 4

= 121/4 kg

Ratio between= tin / alloy

= 11/4 kg : 121/4 kg

= 11: 121

= 1: 11

**Question -2. ****Find the compounded ratio of :**

**(i) 2 : 3 and 4 : 9**

**(ii) 4 : 5, 5 : 7 and 9 : 11**

**(iii) (a – b) : (a + b), (a + b) ^{2} : (a^{2} + b^{2}) and (a^{4} – b^{4}) : (a^{2} – b^{2})^{2}**

**Answer:**

**(i) 2: 3 and 4: 9**

Compound ratio = (2/3 )× (4/9)

= 8/27

= 8: 27

**(ii) 4: 5, 5: 7 and 9: 11**

Compound ratio = (4/5) × (5/7) × (9/11)

= 36 / 77

= 36: 77

**(iii) (a – b): (a + b), (a + b)**^{2}: (a^{2} + b^{2}) and (a^{4} – b^{4}): (a^{2} – b^{2})^{2}

^{2}: (a

^{2}+ b

^{2}) and (a

^{4}– b

^{4}): (a

^{2}– b

^{2})

^{2}

Compound ratio

= 1/1

= 1: 1

**Question- 3. ****Find the duplicate ratio of :**

**(i) 2 : 3**

**(ii) √5 : 7**

**(iii) 5a : 6b**

**Answer :**

##### (i) Duplicate ratio of 2 : 3 = (2)^{2} : (3)^{2} = 4 : 9

##### (ii) Duplicate ratio of √5 : 7 = (√5)^{2} : (7)^{2} = 5 : 49

##### (iii) Duplicate ratio of 5a : 6b = (5a)^{2} : (6b)^{2} = 25a^{2} : 36b^{2}

**Question- 4. ****Find the triplicate ratio of :**

(i) 3: 4

(ii) ½: 1/3

(iii) 1^{3}: 2^{3}

**Answer- 4**

**(i) 3: 4**

Triplicate ratio of 3: 4 = 3^{3}: 4^{3}

= 27: 64

**(ii) ½: 1/3**

Triplicate ratio of ½: 1/3

= (1/2)^{3}: (1/3)^{3}

= 1/8: 1/27 = 27: 8

**(iii) 1**^{3}: 2^{3}

^{3}: 2

^{3}

Triplicate ratio of 1^{3}: 2^{3}

= (1^{3})^{3}: (2^{3})^{3}

= 1^{3}: 8^{3}

= 1: 512

**Question- 5. ****Find the sub-duplicate ratio of :**

**(i) 9 : 16**

**(ii) 1/4 : 1/9,**

**(iii) 9a ^{2} : 49b^{2}**

**Answer :**

**(i) 9: 16**

Sub-duplicate ratio of 9: 16

= √9: √16

= 3: 4

**(ii) ¼: 1/9**

Sub-duplicate ratio of ¼: 1/9

= √1/4: √1/9

= ½: 1/3

= 3: 2

**(iii) 9a**^{2}: 49b^{2}

^{2}: 49b

^{2}

Sub-duplicate ratio of 9a^{2}: 49b^{2}

= √9a^{2}: √49b^{2} = 3a: 7b

**Question- 6. ****Find the sub-triplicate ratio of :**

**(i) 1 : 216**

**(ii) 1/8 : 1/125**

**(iii) 27a ^{3} : 64b^{3}**

**Answer:**

##### (i) Sub-triplicate ratio of 1 : 216

=

**Question : ****Find the reciprocal ratio of :**

**(i) 4 : 7**

**(ii) 3 ^{2} : 4^{2}**

**(iii) 1/9 : 2**

**Answer :**

(i) Reciprocal ratio of 4 : 7 = 7 : 4

(ii) Reciprocal ratio of 3^{2} : 4^{2} = 4^{2} : 3^{2} = 16 : 9

(iii) Reciprocal ratio of 1/9 : 2 = 1/9 : 2 = 18 : 1

**Question -8. ****Arrange the following ratios in ascending order of magnitude:**

**2 : 3, 17 : 21, 11 : 14 and 5 : 7**

**Answer :**

Writing the given ratios in fraction

2/3, 17/21, 11/14, 5/7

Here the LCM of 3, 21, 14 and 7 is 42

By converting the ratio as equivalent

2/3 = (2 × 14)/ (3 × 14) = 28/42

17/21 = (17 × 2)/ (21 × 2) = 34/ 42

11/14 = (11 × 3)/ (14 × 3) = 33/42

5/7 = (5 × 6)/ (7 × 6) = 30/42

writing it in ascending order

28/42, 30/42, 33/42, 34/42

on simplification

2/3, 5/7, 11/14, 17/21

**Question -9**

**(i) If A : B = 2 : 3, B : C = 4 : 5 and C : D = 6 : 7, find A : D**

**(ii) If x : y = 2 : 3, and y : z = 4 : 7, find x : y : z**

**Answer:**

**(i) A: B = 2: 3, B: C = 4: 5 and C: D = 6: 7**

A/ B = 2/3, B/C = 4/5, C/D = 6/7

By multiplication

A/D = 16/35

A: D = 16: 35

**(ii) We know that the LCM of y terms 3 and 4 is 12**

Now making equals of y as 12

x/y = 2/3 = (2 × 4)/ (3 × 4) = 8/12 = 8: 12

y/z = 4 /7 = 12/21 mul by 3 in both N and D

= 12: 21

So x: y: z = 8: 12: 21

**Question- 10**

**(i) If A: B = 1/4: 1/5 and B: C = 1/7: 1/6, find A: B: C.**

**(ii) If 3A = 4B = 6C, find A: B: C**

**Answer :**

**(i) A: B = 1/4 × 5/1 = 5/4**

B: C = 1/7 × 6/1 = 6/7

making terms of B as 12

A/B = (5 × 3)/ (4 × 3) = 15/12 = 15: 12

B/C = (6 × 2)/ (7 × 2) = 12/14 = 12: 14

So A: B: C = 15: 12: 14

**(ii) 3A = 4B**

A/B = 4/3

A: B = 4: 3

Similarly 4B = 6C

B/C = 6/4 = 3/2

B: C = 3: 2

So we get

A: B: C = 4: 3: 2

**Question- 11**

(i) If (3x+5y)/(3x-5y)=7/3 , Find x : y

(ii) If a : b = 3 : 11, find (15a – 3b) : (9a + 5b).

**Answer- 11**

**(i) (3x+5y)/(3x-5y)=7/3**

⇒ 9x + 15y = 21x – 35y [By cross multiplication]

⇒ 21x – 9x = 15y + 35y

on simplification

21x – 9x = 15y + 35y

12x = 50y

x/y = 50/12 = 25/6

, x: y = 25: 6

**(ii) a: b = 3: 11**

a/b = 3/11

(15a – 3b)/ (9a + 5b)

dividing both numerator and denominator by b

= [15a/b – 3b/b]/ [9a/b + 5b/b]

= [15a/b – 3]/ [9a/b + 5]

Substituting the value of a/ b

= [15 × 3/11 – 3]/ [9 × 3/11 + 5]

= [45/11 – 3]/ [27/11 + 5]

Taking LCM

= [(45 – 33)/ 11]/ [(27 + 55)/ 11]

= (12/11)/( 82/11)

= 12/11 × 11/82

= 12/82

= 6/41

**Question- 12**

**(i) If (4x² + xy) : (3xy – y²) = 12 : 5, find (x + 2y) : (2x + y).**

**(ii) If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)².**

**Answer :**

**(i) (4x**^{2} + xy): (3xy – y^{2}) = 12: 5

^{2}+ xy): (3xy – y

^{2}) = 12: 5

(4x^{2} + xy)/ (3xy – y^{2}) = 12/ 5

20x^{2} + 5xy = 36xy – 12y^{2}

20x^{2} + 5xy – 36xy + 12y^{2} = 0

20x^{2} – 31xy + 12y^{2} = 0

divide by y^{2}

20x^{2}/y^{2} – 31xy/y^{2} + 12y^{2}/y^{2} = 0

20 (x/y)^{2} – 31 (x/y) + 12 = 0

20 (x/y)^{2} – 15(x/y) – 16 (x/y) + 12 = 0

5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0

[4 (x/y) – 3] [5 (x/y) – 4] = 0

4 (x/y) – 3 = 0

4 (x/y) = 3

x/y = ¾

Similarly 5 (x/y) – 4 = 0

5 (x/y) = 4

x/y = 4/5

dividing by y

(x + 2y)/ (2x + y) = (x/y + 2)/ (2 x/y + 1)

(a) If x/y = 3/4, then

= (x/y + 2)/ (2 x/y + 1)

Substituting the values

= (3/4 + 2)/ (2 × 3/4 + 1)

= 11/4/ (3/2 + 1)

= 11/4/ 5/2

= 11/4 × 2/5

= 11/10

(x + 2y): (2x + y) = 11: 10

(b) If x/y = 4/5 then

(x + 2y)/ (2x + y) = [x/y + 2]/ [2 x/y + 1]

Substituting the value of x/y

= [4/5 + 2]/ [2 × 4/5 + 1]

= 14/5/ [8/5 + 1]

= 14/5/ 13/5

= 14/5 × 5/13

= 14/13

(x + 2y)/ (2x + y) = 11/10 or 14/13

(x + 2y): (2x + y) = 11: 10 or 14: 13

**(ii) y (3x – y): x (4x + y) = 5: 12**

(3xy – y^{2})/ (4x^{2} + xy) = 5/12

36xy – 12y^{2} = 20x^{2} + 5xy

20x^{2} + 5xy – 36xy + 12y^{2} = 0

20x^{2} – 31xy + 12y^{2} = 0

Divide by y^{2}

20x^{2}/y^{2} – 31 xy/y^{2} + 12y^{2}/y^{2} = 0

20(x^{2}/y^{2}) – 31 (xy/y^{2}) + 12 = 0

20(x^{2}/y^{2}) – 15 (x/y) – 16 (x/y) + 12 = 0

5 (x/y) [4 (x/y) – 3] – 4 [4 (x/y) – 3] = 0

[4 (x/y) – 3] [5 (x/y) – 4] = 0

4 (x/y) – 3 = 0

4 (x/y) = 3

x/y = 3/4

5 (x/y) – 4 = 0

5 (x/y) = 4

x/y = 4/5

(a) x/y = 3/4

(x^{2} + y^{2}): (x + y)^{2} = (x^{2} + y^{2})/ (x + y)^{2}

Dividing both numerator and denominator by y^{2}

= (x^{2}/y^{2} + y^{2}/y^{2})/ [1/y^{2} (x + y)^{2}]

= (x^{2}/ y^{2} + 1) (x/y + 1)^{2}

Substituting the value of x/y

= [(3/4)^{2} + 1]/ [3/4 + 1]^{2}

= (9/16 + 1)/ (7/4)^{2}

= 25/16/ 49/16

= 25/16 × 16/49

= 25/49

(x^{2} + y^{2}): (x + y)^{2} = 25: 49

(b) x/y = 4/5

(x^{2} + y^{2}): (x + y)^{2} = (x^{2} + y^{2})/ (x + y)^{2}

Dividing both numerator and denominator by y^{2}

= (x^{2}/y^{2} + y^{2}/y^{2})/ [1/y^{2} (x + y)^{2}]

= (x^{2}/ y^{2} + 1) (x/y + 1)^{2}

Substituting the value of x/y

= [(4/5)^{2} + 1]/ [4/5 + 1]^{2}

= (16/25 + 1)/ (9/5)^{2}

= 41/25/ 81/25

= 41/25 × 25/81

= 41/81

(x^{2} + y^{2}): (x + y)^{2} = 41: 81

Ch-7, Ratio and Proportion Exe-7.1

**ML Aggarwal Class 10 ICSE Maths Solutions**

Page-117

**Question- 13**

**(i) If (x – 9) : (3x + 6) is the duplicate ratio of 4 : 9, find the value of x.**

**(ii) If (3x + 1) : (5x + 3) is the triplicate ratio of 3 : 4, find the value of x.**

**(iii) If (x + 2y) : (2x – y) is equal to the duplicate ratio of 3 : 2, find x : y.**

**Answer:**

**(i) (x – 9)/ (3x + 6) = (4/9)**^{2}

^{2}

(x – 9)/ (3x + 6) = 16/81

81x – 729 = 48x + 96

81x – 48x = 96 + 729

33x = 825

x = 825/33

= 25

**(ii) (3x + 1)/ (5x + 3) = 3**^{3}/ 4^{3}

^{3}/ 4

^{3}

(3x + 1)/ (5x + 3) = 27/64

64 (3x + 1) = 27 (5x + 3)

192x + 64 = 135x + 81

192x – 135x = 81 – 64

57x = 17

x = 17/57

**(iii) (x + 2y)/ (2x – y) = 3**^{2}/ 2^{2}

^{2}/ 2

^{2}

(x + 2y)/ (2x – y) = 9/4

9 (2x – y) = 4 (x + 2y)

18x – 9y = 4x + 8y

18x = 4x = 8y + 9y

14x = 17y

x/y = 17/14

x: y = 17: 14

**Question- 14**

**(i) Find two numbers in the ratio of 8 : 7 such that when each is decreased by , they are in the ratio 11 : 9.**

**(ii) The income of a man is increased in the ratio of 10 : 11. If the increase in his income is Rs 600 per month, find his new income.**

**Answer- 14**

**(i) The ratio is 8 : 7**

Let the numbers be 8x and 7x,

According to condition,

[8x – 25/2]/ [7x – 25/2] = 11/9

Taking LCM

[(16x – 25)/ 2]/ [(14x – 25)/ 2] = 11/9

[(16x – 25) × 2]/ [2 (14x – 25)] = 11/9

(16x – 25)/ (14x – 25) = 11/9

154x – 144x = 275 – 225

10x = 50

x = 50/10 = 5

8x = 8 × 5 = 40

7x = 7 × 5 = 35

**(ii) let present income = 10x**

and Increased income = 11x

the increase/ month = 11x – 10x = x

but x = Rs 600 given

New income = 11x = 11 × 600 = Rs 6600

**Question -15**

**(i) A woman reduces her weight in the ratio 7 : 5. What does her weight become if originally it was 91 kg.**

**(ii) A school collected Rs 2100 for charity. It was decided to divide the money between an orphanage and a blind school in the ratio of 3 : 4. How much money did each receive?**

**Answer :**

**(i) Ratio between the original weight and reduced weight = 7 : 5**

Let original weight = 7x

then reduced weight = 5x

If original weight = 91 kg.

So the reduced weight = (91 × 5x)/ 7x = 65 kg

**(ii) Amount collected for charity = Rs 2100**

Here the ratio between orphanage and a blind school = 3: 4

Sum of ratios = 3 + 4 = 7

We know that

Orphanage schools share = 2100 × 3/7 = Rs 900

Blind schools share = 2100 × 4/7 = Rs 1200

**Question -16**

**(i) The sides of a triangle are in the ratio 7 : 5 : 3 and its perimeter is 30 cm. Find the lengths of sides.**

**(ii) If the angles of a triangle are in the ratio 2 : 3 : 4, find the angles.**

**Answer:**

**(i) Perimeter of a triangle = 30 cm.**

Ratio among sides = 7 : 5 : 3

Sum of ratios 7 + 5 + 3 = 15

Length of first side = 30 × 7/15 = 14 cm

Length of second side = 30 × 5/15 = 10 cm

Length of third side = 30 × 3/15 = 6 cm

Therefore, the sides are 14 cm, 10 cm and 6 cm.

**(ii) Sum of all the angles of a triangle = 180**^{0}

^{0}

Here the ratio among angles = 2: 3: 4

Sum of ratios = 2 + 3 + 4 = 9

First angle = 180 × 2/9 = 40^{0}

Second angle = 180 × 3/9 = 60^{0}

Third angle = 180 × 4/9 = 80^{0}

Hence, the angles are 40^{0}, 60^{0} and 80^{0}.

**Question- 17**

**Three numbers are in the ratio 1/2 : 1/3 : 1/4 If the sum of their squares is 244, find the numbers.**

**Answer:**

Ratio of three numbers = 1/2: 1/3: 1/4

= (6: 4: 3)/ 12

= 6: 4: 3

r first number = 6x

Second number = 4x

Third number = 3x

on the condition

(6x)^{2} + (4x)^{2} + (3x)^{2} = 244

36x^{2} + 16x^{2} + 9x^{2} = 244

61x^{2} = 244

x^{2} = 244/61 = 4 = 2^{2}

x = 2

First number = 6x = 6 × 2 = 12

Second number = 4x = 4 × 2 = 8

Third number = 3x = 3 × 2 = 6

**Question- 18**

**(i) A certain sum was divided among A, B and C in the ratio 7 : 5 : 4. If B got Rs 500 more than C, find the total sum divided.**

**(ii) In a business, A invests Rs 50000 for 6 months, B Rs 60000 for 4 months and C, Rs 80000 for 5 months. If they together earn Rs 18800 find the share of each.**

**Answer :**

**(i) Ratio between A, B and C = 7 : 5 : 4**

Let A’s share = 7x

B’s share = 5x

and C’s share = 4x

Total sum = 7x + 5x + 4x = 16x

5x – 4x = 500

x = 500

So the total sum = 16x = 16 × 500 = Rs 8000

**(ii) 6 months investment of A = Rs 50000**

1 month investment of A = 50000 × 6 = Rs 300000

4 months investment of B = Rs 60000

1 month investment of B = 60000 × 4 = Rs 240000

5 months investment of C = Rs 80000

1 month investment of C = 80000 × 5 = Rs 400000

the ratio between their investments = 300000: 240000: 400000

= 30: 24: 40

Sum of ratio = 30 = 24 + 40 = 94

Total earnings = Rs 18800

So we get

A share = 30/94 × 18800 = Rs 6000

B share = 24/94 × 18800 = Rs 4800

C share = 40/94 = 18800 = Rs 8000

**Question -19**

**(i) In a mixture of 45 litres, the ratio of milk to water is 13 : 2. How much water must be added to this mixture to make the ratio of milk to water as 3 : 1 ?**

**(ii) The ratio of the number of boys to the number of girls in a school of 560 pupils is 5 : 3. If 10 new boys are admitted, find how many new girls may be admitted so that the ratio of the number of boys to the number of girls may change to 3 : 2.**

**Answer :**

**(i) Mixture of milk and water = 45 litres**

Ratio of milk and water =13 : 2

Sum of ratio = 13 + 2 = 15

quantity of milk = (45 × 13)/ 15 = 39 litres

Quantity of water = 45 × 2/15 = 6 litres

let x litre of water to be added,

then water = (6 + x) litres

So the new ratio = 3: 1

39: (6 + x) = 3: 1

39/ (6 + x) = 3/1

39 = 18 + 3x

3x = 39 – 18 = 21

x = 21/3 = 7 litres

**(ii) Ratio between boys and girls = 5: 3**

Number of pupils = 560

So the sum of ratios = 5 + 3 = 8

Number of boys = 5/8 × 560 = 350

Number of girls = 3/8 × 560 = 210

Number of new boys admitted = 10

So the total number of boys = 350 + 10 = 360

Let x as the number of girls admitted

Total number of girls = 210 + x

360: 210 + x = 3: 2

360/ 210 + x = 3/2

630 + 3x = 720

3x = 720 – 630 = 90

x = 90/3 = 30

**Question- 20**

**(i) The monthly pocket money of Ravi and Sanjeev are in the ratio 5 : 7. Their expenditures are in the ratio 3 : 5. If each saves Rs 80 every month, find their monthly pocket money.**

**(ii) In class X of a school, the ratio of the number of boys to that of the girls is 4 : 3. If there were 20 more boys and 12 less girls, then the ratio would have been 2 : 1, How many students were there in the class?**

**Answer:**

**(i) Let the monthly pocket money of Ravi and Sanjeev be 5x and 7x respectively.**

Also, let their expenditure be 3y and 5y respectively.

5x – 3y = 80 …… (1)

7x – 5y = 80 …… (2)

multiply equation (1) by 7 and (2) by 5

35x – 21y = 560 …… (1)

35x – 25y = 400 …… (2)

Sub (2) from (1) equation

4y = 160

y = 40

In equation (1)

5x = 80 + 3 × 40 = 200

x = 40

monthly pocket money of Ravi = 5 × 40 = 200

**(ii) Let x number of students in class**

Ratio of boys and girls = 4: 3

Number of boys = 4x/7

Number of girls = 3x/7

(4x/7 + 20): (3x/7 – 12) = 2: 1

(4x + 140)/ 7: (3x – 84)/ 7 = 2: 1

(4x + 140)/ 7 × 7/ (3x – 84) = 2/1

(4x + 140)/ (3x – 84) = 2/1

6x – 168 = 4x + 140

6x – 4x = 140 + 168

2x = 308

x = 308/2 = 154

**Question- 21. ****In an examination, the ratio of passes to failures was 4 : 1. If 30 less had appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. How many students appeared for the examination**

**Answer :**

Let the number of passes = 4x

and number of failures = x

The total number of students appeared = 4x + x = 5x

In the second case, the number of students appeared = 5x – 30

and number of passes = 4x – 20

So the number of failures = (5x – 30) – (4x – 20)

= 5x – 30 – 4x + 20

= x – 10

(4x – 20)/ (x – 10) = 5/1

5x – 50 = 4x – 20

5x – 4x = – 20 + 50

x = 30

Hence Number of students appeared = 5x = 5 × 30 = 150

— : End of ML Aggarwal Ratio and Proportion Exe-7.1 Class 10 ICSE Maths Solutions: –

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