Episode 8 — Aptitude and Reasoning / 8.5 — Ratio and Proportion
8.5.c Solved Examples -- Ratio and Proportion
Basic Level
Problem 1: Simplifying a Ratio
Simplify the ratio 84 : 126.
Solution:
Find HCF(84, 126).
84 = 2 x 2 x 3 x 7
126 = 2 x 3 x 3 x 7
HCF = 2 x 3 x 7 = 42
84 : 126 = 84/42 : 126/42 = 2 : 3
Answer: 2 : 3
Problem 2: Ratio Involving Fractions
Find the ratio (3/4) : (5/6) : (7/8).
Solution:
LCM of denominators 4, 6, 8 = 24
Multiply each fraction by 24:
(3/4) x 24 = 18
(5/6) x 24 = 20
(7/8) x 24 = 21
Ratio = 18 : 20 : 21
Answer: 18 : 20 : 21
Problem 3: Dividing an Amount in a Given Ratio
Divide Rs 2450 among A, B, and C in the ratio 3 : 5 : 6.
Solution:
Sum of ratio terms = 3 + 5 + 6 = 14
A's share = 2450 x 3/14 = Rs 525
B's share = 2450 x 5/14 = Rs 875
C's share = 2450 x 6/14 = Rs 1050
Verification: 525 + 875 + 1050 = 2450 (correct)
Answer: A = Rs 525, B = Rs 875, C = Rs 1050
Problem 4: Finding a Fourth Proportional
Find the fourth proportional to 5, 8, and 15.
Solution:
Let the fourth proportional be x.
5 : 8 :: 15 : x
By cross product rule:
5 x x = 8 x 15
5x = 120
x = 24
Verification: 5 : 8 :: 15 : 24 => 5 x 24 = 120 = 8 x 15 (correct)
Answer: 24
Problem 5: Finding a Mean Proportional
Find the mean proportional between 8 and 32.
Solution:
Mean proportional = sqrt(8 x 32) = sqrt(256) = 16
Verification: 8 : 16 :: 16 : 32 => 8 x 32 = 256 = 16 x 16 (correct)
Answer: 16
Problem 6: Finding a Third Proportional
Find the third proportional to 9 and 12.
Solution:
Let the third proportional be x.
9 : 12 :: 12 : x
9 x x = 12 x 12
9x = 144
x = 16
Verification: 9 : 12 :: 12 : 16 => 9 x 16 = 144 = 12 x 12 (correct)
Answer: 16
Problem 7: Direct Proportion
If 12 notebooks cost Rs 156, what is the cost of 18 notebooks?
Solution:
More notebooks => More cost (Direct proportion)
12 / 156 = 18 / x
x = (156 x 18) / 12
x = 2808 / 12
x = Rs 234
Answer: Rs 234
Medium Level
Problem 8: Combining Ratios
If A : B = 3 : 4 and B : C = 6 : 7, find A : B : C.
Solution:
B appears in both ratios. Make B the same.
B values: 4 and 6. LCM(4, 6) = 12.
A : B = 3 : 4 => multiply by 3 => 9 : 12
B : C = 6 : 7 => multiply by 2 => 12 : 14
Therefore A : B : C = 9 : 12 : 14
Answer: 9 : 12 : 14
Problem 9: Ratio with Addition
The ratio of two numbers is 5 : 7. If 3 is added to each number, the ratio becomes 3 : 4. Find the numbers.
Solution:
Let the numbers be 5k and 7k.
After adding 3:
(5k + 3) / (7k + 3) = 3/4
Cross multiply:
4(5k + 3) = 3(7k + 3)
20k + 12 = 21k + 9
12 - 9 = 21k - 20k
k = 3
Numbers: 5(3) = 15 and 7(3) = 21
Verification: (15 + 3):(21 + 3) = 18:24 = 3:4 (correct)
Answer: 15 and 21
Problem 10: Inverse Proportion
If 15 men can complete a job in 20 days, how many men are needed to complete it in 12 days?
Solution:
More men => Fewer days (Inverse proportion)
15 x 20 = x x 12
300 = 12x
x = 25
Answer: 25 men
Problem 11: Alligation -- Mixing Problem
A shopkeeper mixes two types of wheat, one costing Rs 42/kg and another Rs 30/kg, to sell the mixture at Rs 34/kg. In what ratio must he mix them?
Solution:
Using alligation:
Cheaper (30) Dearer (42)
\ /
\ /
Mean (34)
/ \
/ \
|42-34| |34-30|
= 8 = 4
Ratio of cheaper : dearer = 8 : 4 = 2 : 1
Answer: 2 : 1
Problem 12: Partnership Problem
A starts a business with Rs 25,000. B joins after 4 months with Rs 30,000. At the end of the year, the total profit is Rs 23,000. Find each partner's share.
Solution:
A's capital x time = 25000 x 12 = 300000
B's capital x time = 30000 x 8 = 240000 (B invested for 12-4 = 8 months)
Ratio = 300000 : 240000 = 5 : 4
Total ratio parts = 5 + 4 = 9
A's share = 23000 x 5/9 = Rs 12,777.78 (approx)
B's share = 23000 x 4/9 = Rs 10,222.22 (approx)
For exact answer:
A's share = 23000 x 5/9 = 115000/9 ≈ Rs 12,777.78
B's share = 23000 x 4/9 = 92000/9 ≈ Rs 10,222.22
Answer: A gets Rs 115000/9, B gets Rs 92000/9
(approximately Rs 12,778 and Rs 10,222)
Problem 13: Age-Based Ratio Problem
The present ages of A and B are in the ratio 5 : 3. Four years ago, their ages were in the ratio 3 : 1. Find their present ages.
Solution:
Let present ages = 5k and 3k.
Four years ago:
(5k - 4) / (3k - 4) = 3/1
Cross multiply:
5k - 4 = 3(3k - 4)
5k - 4 = 9k - 12
-4 + 12 = 9k - 5k
8 = 4k
k = 2
Present ages: A = 5(2) = 10 years, B = 3(2) = 6 years.
Verification: 4 years ago = 6:2 = 3:1 (correct)
Answer: A = 10 years, B = 6 years
Problem 14: Income-Expenditure Problem
The incomes of A and B are in the ratio 3 : 2. Their expenditures are in the ratio 5 : 3. If each saves Rs 1000, find their incomes.
Solution:
Let incomes = 3x and 2x.
Let expenditures = 5y and 3y.
3x - 5y = 1000 ... (i)
2x - 3y = 1000 ... (ii)
Multiply (ii) by 5/3: (10/3)x - 5y = 5000/3
Alternatively, solve directly:
From (i): 3x - 5y = 1000
From (ii): 2x - 3y = 1000
Multiply (i) by 3: 9x - 15y = 3000
Multiply (ii) by 5: 10x - 15y = 5000
Subtract: (10x - 15y) - (9x - 15y) = 5000 - 3000
x = 2000
From (ii): 2(2000) - 3y = 1000 => 4000 - 3y = 1000 => 3y = 3000 => y = 1000
Income of A = 3(2000) = Rs 6000
Income of B = 2(2000) = Rs 4000
Verification:
A's expenditure = 5(1000) = Rs 5000, savings = 6000 - 5000 = Rs 1000 (correct)
B's expenditure = 3(1000) = Rs 3000, savings = 4000 - 3000 = Rs 1000 (correct)
Answer: A's income = Rs 6000, B's income = Rs 4000
Problem 15: Compound Ratio
The compound ratio of 3:5, 4:7, and 5:9 is?
Solution:
Compound ratio = product of antecedents : product of consequents
= (3 x 4 x 5) : (5 x 7 x 9)
= 60 : 315
= 4 : 21 (dividing by HCF = 15)
Answer: 4 : 21
Problem 16: Componendo-Dividendo
If (3x + 5y) / (3x - 5y) = 7/3, find x : y.
Solution:
Apply componendo-dividendo:
(3x + 5y + 3x - 5y) / (3x + 5y - 3x + 5y) = (7 + 3) / (7 - 3)
6x / 10y = 10/4
3x / 5y = 5/2
x/y = (5 x 5) / (2 x 3) = 25/6
x : y = 25 : 6
Verification:
3(25) + 5(6) = 75 + 30 = 105
3(25) - 5(6) = 75 - 30 = 45
105/45 = 7/3 (correct)
Answer: x : y = 25 : 6
Problem 17: Mixing Solutions (Alligation)
A vessel contains 60 litres of milk. 12 litres are removed and replaced with water. This process is repeated two more times. How much milk remains?
Solution:
Using the replacement formula:
Milk remaining = Initial x (1 - removed/total)^n
n = 3 (process done 3 times)
Milk = 60 x (1 - 12/60)^3
= 60 x (48/60)^3
= 60 x (4/5)^3
= 60 x 64/125
= 3840/125
= 30.72 litres
Answer: 30.72 litres
Advanced Level
Problem 18: Three-Person Partnership with Varying Capital
A starts a business with Rs 40,000. After 3 months, B joins with Rs 50,000. After 2 more months, C joins with Rs 60,000. At the end of 12 months, the profit is Rs 37,500. Find each partner's share.
Solution:
A invests for 12 months.
B invests for (12 - 3) = 9 months.
C invests for (12 - 3 - 2) = 7 months.
A's contribution = 40000 x 12 = 480000
B's contribution = 50000 x 9 = 450000
C's contribution = 60000 x 7 = 420000
Ratio = 480000 : 450000 : 420000
Divide by 30000:
= 16 : 15 : 14
Sum = 16 + 15 + 14 = 45
A's share = 37500 x 16/45 = Rs 13,333.33
B's share = 37500 x 15/45 = Rs 12,500
C's share = 37500 x 14/45 = Rs 11,666.67
Verification: 13333.33 + 12500 + 11666.67 = 37500 (correct)
Answer: A = Rs 13,333.33, B = Rs 12,500, C = Rs 11,666.67
Problem 19: Complex Age Problem
The sum of the present ages of A, B, and C is 90 years. Six years ago, their ages were in the ratio 1 : 2 : 3. Find their present ages.
Solution:
Six years ago, the sum of ages = 90 - 3(6) = 90 - 18 = 72 years.
Let ages 6 years ago = k, 2k, 3k.
k + 2k + 3k = 72
6k = 72
k = 12
Ages 6 years ago: 12, 24, 36.
Present ages: 12+6, 24+6, 36+6 = 18, 30, 42.
Verification: 18 + 30 + 42 = 90, and 12:24:36 = 1:2:3 (correct)
Answer: A = 18 years, B = 30 years, C = 42 years
Problem 20: Ratio of Mixtures
Two containers have milk and water in the ratios 3:2 and 7:3 respectively. In what ratio should the contents be mixed to get a new mixture with milk and water in the ratio 2:1?
Solution:
Container 1: Milk = 3/5 of total, Water = 2/5 of total.
Container 2: Milk = 7/10 of total, Water = 3/10 of total.
Desired: Milk = 2/3 of total.
Apply alligation on the milk fraction:
Milk fraction in Container 1 = 3/5 = 6/10
Milk fraction in Container 2 = 7/10
Desired milk fraction = 2/3
Using alligation:
Container 1 : Container 2 = |7/10 - 2/3| : |2/3 - 6/10|
7/10 - 2/3 = (21 - 20)/30 = 1/30
2/3 - 6/10 = (20 - 18)/30 = 2/30
Ratio = (1/30) : (2/30) = 1 : 2
Answer: 1 : 2
Problem 21: Salary Ratio Problem
The salary of A is 20% more than that of B. By what percentage is B's salary less than A's?
Solution:
Let B's salary = 100.
A's salary = 120.
Difference = 120 - 100 = 20.
B's salary is less than A's by:
(20/120) x 100 = 16.67%
Alternative (using ratio approach):
A : B = 120 : 100 = 6 : 5
Difference as % of A = (1/6) x 100 = 16.67%
Answer: 16.67% (or 16 2/3 %)
Problem 22: Continued Proportion
If a, b, c are in continued proportion and a = 4, c = 16, find b.
Solution:
In continued proportion: a : b = b : c
=> b^2 = a x c
=> b^2 = 4 x 16 = 64
=> b = 8
Verification: 4 : 8 :: 8 : 16 => 4 x 16 = 64 = 8 x 8 (correct)
Answer: b = 8
Problem 23: Multi-Step Mixture Problem
A merchant has 100 kg of sugar, part of which he sells at 7% profit and the rest at 17% profit. He gains 10% on the whole. How much is sold at 7% profit?
Solution:
Using alligation on profit percentages:
7% profit 17% profit
\ /
\ /
10% (overall)
/ \
/ \
|17-10| |10-7|
= 7 = 3
Ratio (7% portion : 17% portion) = 7 : 3
Total = 100 kg
Sold at 7% profit = 100 x 7/10 = 70 kg
Sold at 17% profit = 100 x 3/10 = 30 kg
Verification:
Profit from 70 kg at 7% = 0.07 x 70 = 4.9 (in units of cost)
Profit from 30 kg at 17% = 0.17 x 30 = 5.1
Total profit = 10 on 100 units = 10% (correct)
Answer: 70 kg sold at 7% profit
Problem 24: Ratio with Variation
If x varies directly as y and inversely as the square of z, and x = 8 when y = 4 and z = 2, find x when y = 6 and z = 3.
Solution:
x = k * y / z^2
Using given values:
8 = k * 4 / (2^2)
8 = k * 4 / 4
8 = k
k = 8
When y = 6, z = 3:
x = 8 * 6 / (3^2)
x = 48 / 9
x = 16/3
x = 5.33 (approx)
Answer: x = 16/3
Problem 25: Wages and Ratio
A, B, and C are employed to do a piece of work for Rs 5290. A and B together are supposed to do 19/23 of the work. What does C get?
Solution:
A and B together do 19/23 of the work.
C does = 1 - 19/23 = 4/23 of the work.
Payment is proportional to work done.
C's share = 5290 x 4/23 = 21160/23 = Rs 920
Verification: A+B get 5290 x 19/23 = 4370. Total = 4370 + 920 = 5290 (correct)
Answer: C gets Rs 920
Problem 26: Proportion Applied to Speed
Two trains cover the same distance. The speed of the first train is 60 km/h and the speed of the second is 80 km/h. The first train takes 40 minutes more than the second. Find the distance.
Solution:
Speed ratio = 60 : 80 = 3 : 4
Time ratio = 4 : 3 (inverse of speed for same distance)
Let times = 4t and 3t.
Difference = 4t - 3t = t = 40 minutes = 2/3 hour.
Time of first train = 4(2/3) = 8/3 hours.
Time of second train = 3(2/3) = 2 hours.
Distance = Speed x Time = 60 x 8/3 = 160 km.
(Or: 80 x 2 = 160 km.)
Answer: 160 km
Problem 27: Complex Alligation
A container has 40 litres of pure milk. How many litres of water must be added to make the milk-to-water ratio 3:2?
Solution:
After adding water:
Milk = 40 litres (unchanged)
Water = x litres (to find)
Milk : Water = 3 : 2
40 / x = 3 / 2
3x = 80
x = 80/3 = 26.67 litres
Verification: 40 : 26.67 = 120 : 80 = 3 : 2 (correct)
Answer: 80/3 litres (approximately 26.67 litres)
Problem 28: Multiple Ratios and Unknown Total
In a class, the ratio of boys to girls is 5:3. If 20 more girls join, the ratio becomes 5:5 (i.e., 1:1). Find the original number of students.
Solution:
Let boys = 5k, girls = 3k.
After 20 more girls join:
5k / (3k + 20) = 5/5 = 1
5k = 3k + 20
2k = 20
k = 10
Boys = 50, Girls = 30.
Original total = 80.
Verification: After adding 20 girls: 50:50 = 1:1 (correct)
Answer: 80 students