Ratio and Proportion Class 10 RS Aggarwal Exe-7A Goyal Brothers ICSE Maths Solutions Ch-7. Step by step solutions of exercise-7A questions as latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ICSE Board Class-10.
Ratio and Proportion Class 10 RS Aggarwal Exe-7A Goyal Brothers ICSE Maths Solutions Ch-7
| Board | ICSE |
| Subject | Maths |
| Class | 10th |
| Chapter-7 | Ratio and Proportion |
| Writer/ Book | RS Aggarwal |
| Topics | Solution of Exe-7A Questions |
| Academic Session | 2024-2025 |
Solution of Exe-7A Questions
Ratio and Proportion Class 10 RS Aggarwal Goyal Brothers ICSE Maths Solutions Ch-7
In this exercise you will learn the important property and comparison of ratio with practice questions and their step by step solutions.
Imp Features of Ratio
- The ratio should exist between the quantities of the same kind.
- While comparing two things, the units should be similar.
- There should be significant order of terms.
- The comparison of two ratios can be performed,
- multiplying / dividing same term the ratio does not change
Que-1: Find the ratio between :
(i) 60 paise and Rs 1.35 (ii) 1.8m and 75cm (iii) 1.5kg and 600g (iv) 35 min and 1*(3/4) hrs.
Sol: (i) 60p ; 1.35 Rs = 135p
60 p / 135 p
12 / 27
4 / 9
4 : 9.
(ii) 1.8m = 180cm ; 75cm
= 180 : 75
= 12 : 5.
(iii) 1.5kg = 1500g ; 600g
= 1500 : 600
= 5 : 2.
(iv) 1*(3/4) hrs = 7/4 hrs
7/4 hrs = 1.75 hrs.
35 min = 0.58 hrs.
= 0.58 : 1.75
= 1 : 3.
Que-2: If A:B = 5:6 and B:C = 9:11, find (i) A:C (ii) A:B:C
Sol: (i) Given
A/B = 5/6 …(1)
B/C = 9/11 …(2)
On multiplying (1) & (2)
A/C = 45/66 = 15/22 …(3)
(ii) Let A = x
By (1) B = 6x/5 & by (3) C = 22x/15
Therefore
A:B:C :: 1:6/5:22/15
=> A:B:C :: 15:18:22
Que-3: If P:Q = 7:4 and Q:R = 5:14, find (i) R:P (ii) P:Q:R
Sol: 5:14. lcm of 4and 5 is 20
7×5:4×5
5×4:14×4
P:Q:R
35:20:56
(i) R : P = 56 : 35 = 8:5
(ii) P : Q : R = 35:20:56.
Que-4: If 3A = 5B = 6C, find A:B:C
Sol: Given information: 3A = 5B = 6C
Consider we need to find the ratio A:B:C.
Coefficients of A,B and C are 3, 5 and 6 respectively.
LCM (3,6,5) = 30.
Divide the given equation by 30.
3A/30 = 5B/30 = 6C/30
Cancel out common factors.
A/10 = B/6 = C/5.
Therefore, the ratio A:B:C is 10:6:5.
Que-5: If A/2 = B/3 = C/6, find A:B:C.
Sol: let’s set each of these equal to a common variable k. This gives us:
A/2 = k
A = 2k
B/3 = k
B = 3k
C/6 = k
C = 6k
A:B:C = 2k:3k:6k
The k cancels out, leaving us with the ratio:
A:B:C = 2:3:6.
Que-6: If a:b = 8:5, find (7a+5b) : (8a-9b)
Sol: a:b = 8:5
a = 8
b = 5
(7a + 5b):(8a – 9b)
(7(8) + 5(5)):(8(8) – 9(5)
(56 + 25):(64 – 45)
81:19.
Que-7: If x:y = 3:2, find (5x-3y) : (7x+2y)
Sol: x:y = 3:2
x = 3, y = 2
(5x-3y) : (7x+2y)
(5×3-3×2) : (7×3+2×2)
(15-6) : (21+4)
9 : 25.
Que-8: If x:y = 10:3, find (3x²+2y²) : (3x²-2y²)
Sol: x:y = 10:3
x = 10 and y = 3.
(3×10²+2×3²) : (3×10²-2×3²)
(300+18) : (300-18)
318 : 282
= 318/282
= 53/47
= 53 : 47.
Que-9: If a:b = 2:5, find (3a²-2ab+5b²) : (a²+7ab-2b²)
Sol: a:b = 2:5
a = 2 and b = 5
(3×2²-2x2x5+5×5²) : (2²+7x2x5-2×5²)
(12-20+125) : (4+70-50)
117 : 24
39 : 8.
Que-10: If (5x+2y) : (7x+4y) = 13:20, find x:y
Sol: (5x+2y) : (7x+4y) = 13 : 20
(5x+2y)/(7x+4y) = 13/20
20(5x+2y) = 13(7x+4y)
100x+40y = 91x+52y
9x = 12y
x/y = 12/9
x : y = 4 : 3.
Que-11: If (3x+5y) : (3x-5y) = 7:3, find x:y.
Sol: (3x+5y) : (3x-5y) = 7:3
(3x+5y)/(3x-5y) = 7/3
⇒ 9x + 15y = 21x – 35y …[By cross multiplication]
⇒ 21x – 9x = 15y + 35y
⇒ 12x = 50y
x/y = 50/12 = 25/6
x : y = 25 : 6.
Que-12: If (6x²-xy) : (2xy-y²) = 6:1, find x:y
Sol: (6x² – xy) : (2xy – y²) = 6 : 1
⇒ (6x² – xy) / (2xy – y²) = 6 / 1
⇒ 1 (6x² – xy) = 6 (2xy – y²)
⇒ 6x² – xy = 12xy – 6y²
⇒ 6x² – 13xy + 6y² = 0
⇒ 6x² – (9 + 4)xy + 6y² = 0
⇒ 6x² – 9xy – 4xy + 6y² = 0
⇒ 3x (2x – 3y) – 2y (2x – 3y) = 0
⇒ (2x – 3y) (3x – 2y) = 0
So, either 2x – 3y = 0 or, 3x – 2y = 0
⇒ 2x = 3y & 3x = 2y
⇒ x/y = 3/2 & x/y = 2/3
⇒ x : y = 3 : 2 & x : y = 2 : 3
Que-13: If (4x²-3y²) : (2x²+5y²) = 12:19, find x:y.
Sol: ( 4x² – 3y²) : ( 2x² + 5y² ) = 12 : 19
( 4x² – 3y² ) 19 = ( 2x² + 5y² )12
76x² – 57y² = 24x² + 60y²
76x² – 24x² = 60y² +57y²
52x² = 117y²
x² / y² = 117 / 52
( x/ y )² = ( 13 × 9 ) / ( 13 × 4 )
= 9 / 4
( x / y )² = ( 3 / 2 )²
Therefore ,
x / y = 3 / 2
x : y = 3 : 2
Que-14: If x²+4y² = 4xy, find x:y.
Sol: x²+4y² = 4xy
x²−4xy+4y² = 0
⟹ (x−2y)² = 0
⟹ x = 2y
⟹ x/y = 2/1
x : y = 2 : 1.
Que-15: If 10x²-23xy+9y² = 0, find x:y.
Sol: 10x²-23xy+9y² = 0
=> 10x²-5xy-18xy+9y² = 0
=> 5x(2x-1y)-9y(2x-1y) = 0
=> (2x-y)(5x-9y) = 0
2x-y = 0 (or) 5x-9y = 0
2x = 1y 5x = 9y
x/y=1/2 x/y=9/5
=> x : y = 1:2 or 9:5
Que-16: A ratio is equal to 3:4. If its consequent is 144, what is its antecedents?
Sol: Let the antecedent be ‘x’
ATQ. 3/4 = x/144
x = (144×3)/4
x = 108.
Que-17: Two numbers are in the ratio 8:13. If 14 is added to each of the numbers, the ratio becomes 2:3. Find the numbers.
Sol: Ratio of the numbers = 8 : 13
Let the two numbers be 8x and 13x .
When 14 is added to these numbers the ratio becomes 2 : 3 .
(8x+14)/(3x+14) = 2/3
3(8x+14) = 2(3x+14)
24x+42 = 6x+28
2x = 14
x = 7
Now, the number are
8x = 8 x 7 = 56
13x = 13 x 7 = 91.
Que-18: Two numbers are in the ratio 5:7. If 8 is subtracted from each, the ratio becomes 3:5. Find the numbers.
Sol: Ratio between the original numbers= 5:7
Let the original numbers be 5x and 7x
8 is subtracted from each and ratio becomes 3:5
i.e 5x -8 : 7x – 8 = 3:5
=》 5x – 8 / 7x – 8 = 3/5
=》 5( 5x- 8) = 3(7x -8) ( cross multiplication)
=》 25x – 40 = 21x – 24
=》 25x- 21x = -24 + 40
=》 4x = 16
=》 x = 16/4
=》 x = 4
Now the number are :
5x = 5 x 4 = 20.
7x = 7 x 4 = 28.
Que-19: What least number must be added to each term of the ratio 5:7 to make it 8:9 ?
Sol: Let the number be x.
ATQ, the equation will be
(5+x)/(7+x) = 8/9
45+9x = 56+8x
x = 11.
Que-20: Out of the monthly income of Rs 45000, Rahul spends Rs 31500 and the rest he saves. Find the ratio of his
(i) income to expenditure (ii) income to saving (iii) saving to expenditure.
Sol: monthly income = ₹ 45000
Rahul’s expenditure = ₹ 31500
Rahul’s savings = ₹ 13,500
(i) income to expenditure = 45000/31500
= 10/7
(ii) income to savings = 45000/13500
= 10/3
(iii) savings to expenditure = 13500/31500
= 3/7
Que-21: The cost of making an umbrella is divided between material, labour and overheads in the ratio 6:4:1. If the material costs Rs 132, find the cost of production of an umbrella.
Sol: The costs 6x, 4x, x will be in the same ratio as 6:4:1.
The sum of the cost is the cost of making an umbrella which is
6x + 4x + x = 11x
We are given in the question that the material cost is 132 rupees. So we have,
6x = 132
⇒ x = 132/6 = 22
We are asked to find the total cost. The total cost in rupees is
11x = 11×22 = 242
Therefore, the total cost is Rs.242.
Que-22: Divide Rs 6720 in the ratio 5:3 .
Sol: The ratio is 5 : 3
Sum of the ratio = 5+3 = 8
First part = 6720 x 5/8 = 4200
Second part = 6720 x 3/8 = 2520.
Que-23: Divide Rs 11620 among A,B and C in the ratio 35:28:20.
Sol: Sum of ratio terms = (35 + 28 + 20) = 83
A’s share = Rs.(11620×(35/83)) = Rs.4900
B’s share = Rs. (11620×(28/83)) = Rs.3920
C’s share = Rs.(11620×(20/83)) = Rs.2800
Que-24: Divide Rs 782 among P,Q and R in the ratio (1/2):(2/3):(3/4).
Sol: Given number = 782
{1/2+2/3+3/4}⋅x = 782
{(12+16+18)/24} × x = 782
{46/24}⋅x = 782
{23/12}⋅x = 782
x = 408
now the first part = 1/2⋅x = (1/2⋅408) = 204
second part = 2/3⋅x = (2/3⋅408) = 272
third part = 3/4⋅x = (3/4⋅408) = 306
Que-25: If Rs 5100 be divided among A,B,C in such a way that A gets (2/3) of what B gets and B gets (1/4) of what C gets, find their respective shares.
Sol: Share of C is x
(2/2)×(1/4x)+(x/4)+x = 5100
{(2x×3x)+12x}/12 = 5100
17x/12 = 5100
x = (5100 x 12)/17
x = 3600
C = 3600
B = 360/4 = 900
A = (2×900)/3 = 600.
Que-26: Divide Rs 8300 among A,B and C such that 4 times A’s share, 5 times B’s share and 7 times C’s share may all be equal.
Sol: Let A’s share be x
B’s share be y
C’s share be z
ATQ, 4x = 5y = 7z
y = 4x/5 and z = 4x/7
x+y+z = 8300
(x)+(4x/5)+(4x/7) = 8300
(35x+28x+20x)/35 = 8300
83x/35 = 8300
x = (8300 x 35)/83
x = 3500
y = (4 x 3500)/5 = 2800
z = (4 x 3500)/7 = 2000.
Que-27: A sum of money is divided between A and B in the ratio 6:11. If B’s share is Rs 7315, find (i) A’ share (ii) the total amount of money.
Sol: The ratio of money between A and B = 6 : 11
Let share of A = 6x
Share of B = 11x
According to question- B’s share is 7315
(i) So, 11x = 7315
x = 7315//11
x = 665
Hence, share of A = 6x
= 6×665
Share of A = 3,990 Rs
(ii) Share of B = 7,315 Rs
Total amount = Share of A + Share of B
= 7315 + 3990
Total amount = 11,305 Rs
Que-28: The ages of Tanvy and Divya are in the ratio 5:7. Five years hence, their ages will be in the ratio 3:4. Find their present ages.
Sol: Age of Tanvi: Age of Divya = 5:7
Therefore, age of Tanvi = 5x
And, age of Divya = 7x
5 years hence, their ages of Tanvi and Divya will be (5x+5) and (7x+5) respectively.
Now, according to the question,
⇒3/4 = 5x+5/7x+5
⇒3(7x+5) = 4(5x+5)
⇒21x+15 = 20x+20
⇒21x-20x = 20-15
⇒x = 5
Thus, Age of Tanvi = 5x = 25.
And, Age of Divya = 7x = 35.
Que-29: One year ago, the ratio Amit’s and Arun’s ages was 6:7 respectively. Four years hence, their ages will be in the ratio 7:8. How does is Amit ?
Sol: Age of Tanvi: Age of Divya = 5:7
Therefore, age of Tanvi = 5x
And, age of Divya = 7x
5 years hence, their ages of Tanvi and Divya will be (5x+5) and (7x+5) respectively.
Now, according to the question,
⇒3/4 = 5x+5/7x+5
⇒3(7x+5) = 4(5x+5)
⇒21x+15 = 20x+20
⇒21x-20x = 20-15
⇒x = 5
Thus, Age of Tanvi = 5x = 25.
And, Age of Divya = 7x = 35.
Que-30: Reena reduces her weight in the ratio 5:4. What is her weight now, if originally it was 70 kg ?
Sol: Let us assume that, the new weight of Reena is x kg.
→ 70 : x = 5 : 4
→ 70 * 4 = 5 * x
→ 280 = 5 * x
→ 280 = 5x
→ 280/5 = x
→ 56 = x
→ x = 56
Que-31: 68 kg of a mixture contains milk and water in the ratio 27:7. How much more water is to be added to this mixture to get a new mixture containing milk and water in the ratio 3:1 ?
Sol: 68 litre of mixture contains milk and water in the ratio 27 : 7
i.e 54 litre milk and 14 litre water
X litre of water is added to get new mixture in ratio 3:1
milk/water = 54/14+x = 3/1
54 = 42+3x
X = 4 litre
Que-32: A mixtures contains milk and water in the ratio 5:1. On adding 5 litres of water , the ratio of milk to water becomes 5:2. Find the quantity of milk in the original mixture.
Sol: A mixture contains milk and water in the ratio 5 : 1
On adding 5 liters of water, the ratio of milk and water becomes 5 : 2
Let the quantity of milk and water in the original mixture be 5x and x respectively
According to the qestion,
(5x)/(x + 5) = 5/2
⇒ 10x = 5x + 25
⇒ 5x = 25
⇒ x = 5
∴ Quantity of milk in the original mixture is 5x = 5 × 5 = 25 litres
Que-33: In an examination, the ratio of passes to failures was 4:1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5:1. How many students appeared for the examination ?
Sol: Suppose x candidates passed and y failed; therefore,
x/y = 4/1
x = 4y (1)
In the second case; no. of students appeared = x+y−30
and no. of those who passed = x−20
No of failed
N = x+y−30−(x−20)
N = y−10
from the question
(x−20)/(y−10) = 5/1
x−20 = 5y−50
x−5y = −30
from ( 1)
4y−5y = −30
y = 30
x = 120
So, x+y = 150
Que-34: Find the angles of a triangles which are in the ratio 5:4:3.
Sol: Given that in a △ABC, ∠A : ∠B : ∠C = 5:4:3
Let the angles of the given triangle be 5x°, 4x°, 3x°.
We know that the sum of the angles of a triangle is 180°
5x°+4x°+3x° = 180°
⇒ 12x° = 180°
x° = 110°/12 = 15°
So, the angles of the triangle are 70°, 60°, 45°.
Que-35: The sides of the triangles are in the ratio (1/2):(1/3):(1/4) and its perimeter is 91 cm. Find the lengths of the sides of the triangles.
Sol: LCM of 2,3,4 = 12
therefore ratio = 6:4:3
let the sides be 6x,4x,3x
6x+4x+3x = 91
13x = 91
x = 7
now, sides
6x = 6*7 = 42cm
4x = 4*7 = 28cm
3x = 3*7 = 21cm
Que-36: In a school, the boys and girls are in the ratio 9:5. If there are 425 girls, what is the total number of students in the school ?
Sol: Let the total no. Of boys be 9x
No. Of girls be 5x
According to question
5x = 425
X = 425/5
X= 85
Now,
Boys = 9x
= 9*85
= 765
Now,
Girls = 425
Boys = 765
Total = 425 + 765
= 1190.
Que-37: Compare the following ratios :
(i) (7:9) and (11:16) (ii) (19:25) and (17:20) (iii) (1/2:1/5) and (5:2)
Sol: (i) 7/9 = 11/16
cross multiplication
7×16 = 11×9
112 = 99
therefore 7:9 is greater than 11:16
(ii) 17/20 and 19/25
LCM of 20 and 25 = 100
17/20 × 5/5 = 85/100
19/25 × 4/4 = 76/100
By comparing 85/100 and 76/100, we got 85/100 > 76/100
Answer : 17/20 is the greater fraction.
(iii) (1/2 : 1/5) = 5 : 2
1/2 = 1/5
5 : 2 = 5 : 2.
Que-38: Arrange the following ratios in descending order of magnitudes:
(i) (5:6), (8:9), (13:18) and (19:24)
(ii) (6:7), (13:14), (19:21) and (23:28)
(iii) (7:12), (9:16), (13:20) and (5:8).
Sol: (i) 5/6, 8/9, 13/18, 19/24
LCM of denominator is 72 :
{(5×12)/(6×12)}, {(8×8)/(9×8)}, {(13×4)/(18×4)}, {(19×3)/(24×3)}
60/72, 64/72, 52/72, 57/72
64/72 > 60/72 > 57/72 > 52/72
(8:9) > (5:6) > (19:24) > (13:18)
(ii) 6/7, 13/14, 19/21, 23/28
LCM of denominator is 84
{(6×12),(7×12)}, {(13×6)/(14×6)}, {(19×4)/(21×4)}, {(23×3)/(28×3)}
72/84, 78/84, 76/84, 69/84
78/84 > 76/84 > 72/84 > 69/84
(13:14) > (19:21) > (6:7) > (23:28)
(iii) 7/12, 9/16, 13/20, 5/8
LCM of denominator is 240.
{(7×20)/(12×20)}, {(9×15)/(16×15)}, {(13×12)/(20×12)}, {(5×30)/(8×30)}
140/240, 135/240, 156/240, 150/240
156/240 > 150/240 > 140/240 > 135/240.
(13:20) > (5:8) > (7:12) > (9:16).
Que-39: Arrange the following ratios in ascending order of magnitudes :
(i) (4:9), (6:11), (7:13) and (27:50)
(ii) (2:3), (8:15), (11:12) and (7:16)
(iii) (3:5), (4:9), (5:11) and (10:17)
Sol: (i) 4/9, 6/11, 7/13, 27/50
LCM of denominator is 64,350.
{(4×7150)/(9×7150)}, {(6×5850)/(11×5850)}, {(7×4950)/(13×4950)}, {(27×1287)/(50×1287)}
28600/64350, 35100/64350, 34650/64350, 34668/64350
28600/64350 < 34650/64350 < 34668/64350 < 35100/64350
4:9 < 7:13 < 27:50 < 6:11.
(ii) 2/3, 8/15, 11/12, 7/16
0.66, 0.53, 0.92, 0.44
0.44 < 0.53 < 0.66 < 0.92
7:16 < 8:15 < 2/3 < 11/12
(iii) 3/5, 4/9, 5/11, 10/17
0.6, 0.44, 0.45, 0.56
0.44 < 0.45 < 0.56 < 0.6
4:9 < 5:11 < 10:17 < 3:5.
Que-40: If (3a+2b) : (5a+3b) = 18:29, find (a:b)
Sol: (3a+2b) : (5a+3b)
⇒ 29(3a + 2b) = 18(5a + 3b)
⇒ 87a + 58b = 90a + 54b
⇒ 58b – 54b = 90a – 87a
⇒ 4b = 3a
a : b = 4 : 3.
— : End of Ratio and Proportion Class 10 RS Aggarwal Exe-7A Goyal Brothers ICSE Maths Solutions :-
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