Ratio and Proportion
Key Concepts
Solved Examples: 66
Practice Questions: 7
Solved Examples
Question 1: The ratio of two numbers is 7 : 9. If their sum is 160, find the two numbers.
Let the numbers be 7x and 9x.
7x + 9x = 160
16x = 160
x = 10
Therefore,
First number = 7 × 10 = 70
Second number = 9 × 10 = 90
7x + 9x = 160
16x = 160
x = 10
Therefore,
First number = 7 × 10 = 70
Second number = 9 × 10 = 90
Shortcut method:
Total parts = 7 + 9 = 16
One part = 160 ÷ 16 = 10
Numbers = 7 × 10 = 70, 9 × 10 = 90
Total parts = 7 + 9 = 16
One part = 160 ÷ 16 = 10
Numbers = 7 × 10 = 70, 9 × 10 = 90
Question 2 : The ratio of two numbers is 11 : 14. Their difference is 39. Find the numbers.
Difference in ratio = 14 − 11 = 3 parts
3 parts = 39
1 part = 13
Numbers
= 11 × 13 = 143
= 14 × 13 = 182
3 parts = 39
1 part = 13
Numbers
= 11 × 13 = 143
= 14 × 13 = 182
Shortcut method:
Difference in ratio = 3
One part = 39 ÷ 3 = 13
Numbers = 143 and 182
Difference in ratio = 3
One part = 39 ÷ 3 = 13
Numbers = 143 and 182
Question 3: Three numbers are in the ratio 4 : 7 : 9. If their sum is 400, find the numbers.
Total parts = 4 + 7 + 9 = 20
One part
= 400 ÷ 20
= 20
Numbers
4 × 20 = 80
7 × 20 = 140
9 × 20 = 180
One part
= 400 ÷ 20
= 20
Numbers
4 × 20 = 80
7 × 20 = 140
9 × 20 = 180
Question 4: The ratio of the incomes of A and B is 5 : 7. If each of their incomes is increased by ₹4,000, the new ratio becomes 3 : 4. Find their original incomes.
Let the original incomes of A and B be:
A = 5x
B = 7x
After increasing both incomes by ₹4,000:
(5x + 4000) : (7x + 4000) = 3 : 4
Cross multiplication:
4(5x + 4000) = 3(7x + 4000)
20x + 16000 = 21x + 12000
x = 4000
Therefore,
A’s income = 5 × 4000 = ₹20,000
B’s income = 7 × 4000 = ₹28,000
A = 5x
B = 7x
After increasing both incomes by ₹4,000:
(5x + 4000) : (7x + 4000) = 3 : 4
Cross multiplication:
4(5x + 4000) = 3(7x + 4000)
20x + 16000 = 21x + 12000
x = 4000
Therefore,
A’s income = 5 × 4000 = ₹20,000
B’s income = 7 × 4000 = ₹28,000
Question 5: Two numbers are in the ratio 5 : 8. If 20 is subtracted from each number, the ratio becomes 1 : 2. Find the numbers.
Let the numbers be 5x and 8x
After subtraction, (5x − 20)/(8x − 20) = 1/2
Cross multiplying, 2(5x − 20) = 8x − 20
10x − 40 = 8x − 20
x = 10
Numbers = 50 and 80
After subtraction, (5x − 20)/(8x − 20) = 1/2
Cross multiplying, 2(5x − 20) = 8x − 20
10x − 40 = 8x − 20
x = 10
Numbers = 50 and 80
Question 6: If A : B = 5 : 3 and B : C = 4 : 7, find A : B : C.
Given: A : B = 5 : 3 and B : C = 4 : 7
The common term is B.
The values of B are 3 and 4.
LCM of 3 and 4 = 12
Multiply the first ratio by 4: A : B = 20 : 12
Multiply the second ratio by 3: B : C = 12 : 21
Therefore, A : B : C = 20 : 12 : 21
The common term is B.
The values of B are 3 and 4.
LCM of 3 and 4 = 12
Multiply the first ratio by 4: A : B = 20 : 12
Multiply the second ratio by 3: B : C = 12 : 21
Therefore, A : B : C = 20 : 12 : 21
Shortcut method:
Make the common term equal by multiplying the ratios with the opposite value.
A : B = 5 : 3 → ×4 → 20 : 12
B : C = 4 : 7 → ×3 → 12 : 21
Combined ratio:
A : B : C = 20 : 12 : 21
Make the common term equal by multiplying the ratios with the opposite value.
A : B = 5 : 3 → ×4 → 20 : 12
B : C = 4 : 7 → ×3 → 12 : 21
Combined ratio:
A : B : C = 20 : 12 : 21
Question 7: If A : B = 3 : 5, B : C = 2 : 7 and C : D = 4 : 9, find A : B : C : D.
Given: A : B = 3 : 5, B : C = 2 : 7 and C : D = 4 : 9
First combine A : B and B : C.
The common term is B.
B values are 5 and 2.
LCM of 5 and 2 = 10
A : B = 6 : 10
B : C = 10 : 35
Therefore,
A : B : C = 6 : 10 : 35
Now combine C : D = 4 : 9.
C values are 35 and 4.
LCM of 35 and 4 = 140
Multiply:
A : B : C = 24 : 40 : 140
C : D = 140 : 315
Therefore,
A : B : C : D = 24 : 40 : 140 : 315
First combine A : B and B : C.
The common term is B.
B values are 5 and 2.
LCM of 5 and 2 = 10
A : B = 6 : 10
B : C = 10 : 35
Therefore,
A : B : C = 6 : 10 : 35
Now combine C : D = 4 : 9.
C values are 35 and 4.
LCM of 35 and 4 = 140
Multiply:
A : B : C = 24 : 40 : 140
C : D = 140 : 315
Therefore,
A : B : C : D = 24 : 40 : 140 : 315
Question 8: The ratio of the number of boys to girls in Class A is 4 : 5. The ratio of girls to teachers is 10 : 1. Find the ratio of boys to girls to teachers.
Given:
Boys : Girls = 4 : 5
Girls : Teachers = 10 : 1
The common term is Girls.
Girls values are 5 and 10.
LCM of 5 and 10 = 10
Multiply the first ratio by 2:
Boys : Girls = 8 : 10
Girls : Teachers = 10 : 1
Therefore, Boys : Girls : Teachers = 8 : 10 : 1
Boys : Girls = 4 : 5
Girls : Teachers = 10 : 1
The common term is Girls.
Girls values are 5 and 10.
LCM of 5 and 10 = 10
Multiply the first ratio by 2:
Boys : Girls = 8 : 10
Girls : Teachers = 10 : 1
Therefore, Boys : Girls : Teachers = 8 : 10 : 1
Question 9: The ratio of the incomes of A and B is 7 : 9. The ratio of the incomes of B and C is 3 : 5. If the total income of A, B and C is ₹31,000, find the income of each person.
A : B = 7 : 9
B : C = 3 : 5
Convert B:
B : C = 9 : 15
Therefore,
A : B : C = 7 : 9 : 15
Total parts:
7 + 9 + 15 = 31
One part:
31000 ÷ 31 = 1000
A’s income = 7 × 1000 = ₹7,000
B’s income = 9 × 1000 = ₹9,000
C’s income = 15 × 1000 = ₹15,000
B : C = 3 : 5
Convert B:
B : C = 9 : 15
Therefore,
A : B : C = 7 : 9 : 15
Total parts:
7 + 9 + 15 = 31
One part:
31000 ÷ 31 = 1000
A’s income = 7 × 1000 = ₹7,000
B’s income = 9 × 1000 = ₹9,000
C’s income = 15 × 1000 = ₹15,000
Question 10: Divide ₹1,200 between A and B in the ratio 5 : 3.
Ratio of A and B: A : B = 5 : 3
Total parts: 5 + 3 = 8
Value of one part: 1200 ÷ 8 = 150
A’s share: 5 × 150 = ₹750
B’s share: 3 × 150 = ₹450
Total parts: 5 + 3 = 8
Value of one part: 1200 ÷ 8 = 150
A’s share: 5 × 150 = ₹750
B’s share: 3 × 150 = ₹450
Question 11: A sum of ₹4,800 is divided among A, B and C in the ratio 3 : 5 : 8. Find the share of each person.
Ratio: A : B : C = 3 : 5 : 8
Total parts: 3 + 5 + 8 = 16
Value of one part: 4800 ÷ 16 = 300
A’s share: 3 × 300 = ₹900
B’s share: 5 × 300 = ₹1,500
C’s share: 8 × 300 = ₹2,400
Total parts: 3 + 5 + 8 = 16
Value of one part: 4800 ÷ 16 = 300
A’s share: 3 × 300 = ₹900
B’s share: 5 × 300 = ₹1,500
C’s share: 8 × 300 = ₹2,400
Question 12: The ages of A and B are in the ratio 5 : 7. If the sum of their ages is 48 years, find their ages.
Age ratio: A : B = 5 : 7
Total parts: 5 + 7 = 12
Value of one part: 48 ÷ 12 = 4
A’s age: 5 × 4 = 20 years
B’s age: 7 × 4 = 28 years
Total parts: 5 + 7 = 12
Value of one part: 48 ÷ 12 = 4
A’s age: 5 × 4 = 20 years
B’s age: 7 × 4 = 28 years
Question 13: A sum of money is divided between A and B in the ratio 7 : 11. If B receives ₹2,400 more than A, find the total amount.
Ratio: A : B = 7 : 11
Difference in ratio: 11 − 7 = 4 parts
4 parts = ₹2,400
1 part: 2400 ÷ 4 = ₹600
Total parts: 7 + 11 = 18
Total amount: 18 × 600 = ₹10,800
Difference in ratio: 11 − 7 = 4 parts
4 parts = ₹2,400
1 part: 2400 ÷ 4 = ₹600
Total parts: 7 + 11 = 18
Total amount: 18 × 600 = ₹10,800
Question 14: A person distributes his wealth among his three children in the ratio 3 : 4 : 5. If the youngest child receives ₹24,000, find the total wealth.
Ratio: 3 : 4 : 5
Youngest child’s share:
5 parts = ₹24,000
1 part: 24000 ÷ 5 = ₹4,800
Total parts: 3 + 4 + 5 = 12
Total wealth: 12 × 4800 = ₹57,600
Youngest child’s share:
5 parts = ₹24,000
1 part: 24000 ÷ 5 = ₹4,800
Total parts: 3 + 4 + 5 = 12
Total wealth: 12 × 4800 = ₹57,600
Question 15: A sum of money is divided between A and B in the ratio 3 : 5. If ₹2,000 is added to each share, the ratio becomes 5 : 7. Find the original shares.
Let the shares be:
A = 3x
B = 5x
After adding ₹2,000: (3x + 2000) : (5x + 2000) = 5 : 7
Cross multiplication:
7(3x + 2000) = 5(5x + 2000)
21x + 14000 = 25x + 10000
4000 = 4x
x = 1000
A’s share: 3 × 1000 = ₹3,000
B’s share: 5 × 1000 = ₹5,000
A = 3x
B = 5x
After adding ₹2,000: (3x + 2000) : (5x + 2000) = 5 : 7
Cross multiplication:
7(3x + 2000) = 5(5x + 2000)
21x + 14000 = 25x + 10000
4000 = 4x
x = 1000
A’s share: 3 × 1000 = ₹3,000
B’s share: 5 × 1000 = ₹5,000
Question 16: The present ages of a father and son are in the ratio 5 : 2. After 8 years, the ratio of their ages will become 2 : 1. Find their present ages.
Let: Father = 5x and Son = 2x
After 8 years:
(5x + 8) : (2x + 8) = 2 : 1
Cross multiplication:
5x + 8 = 2(2x + 8)
5x + 8 = 4x + 16
x = 8
Father’s age: 5 × 8 = 40 years
Son’s age: 2 × 8 = 16 years
After 8 years:
(5x + 8) : (2x + 8) = 2 : 1
Cross multiplication:
5x + 8 = 2(2x + 8)
5x + 8 = 4x + 16
x = 8
Father’s age: 5 × 8 = 40 years
Son’s age: 2 × 8 = 16 years
Question 17: The ratio of the ages of A and B is 2 : 3. After 12 years, the ratio becomes 4 : 5. Find their present ages.
Let present ages be: A = 2x and B = 3x
After 12 years:
(2x + 12) : (3x + 12) = 4 : 5
Cross multiplication:
5(2x + 12) = 4(3x + 12)
10x + 60 = 12x + 48
2x = 12
x = 6
A’s age: 2 × 6 = 12 years
B’s age: 3 × 6 = 18 years
After 12 years:
(2x + 12) : (3x + 12) = 4 : 5
Cross multiplication:
5(2x + 12) = 4(3x + 12)
10x + 60 = 12x + 48
2x = 12
x = 6
A’s age: 2 × 6 = 12 years
B’s age: 3 × 6 = 18 years
Question 18: A printing press prints 4,500 pages using 90 kg of paper. How much paper will be required to print 7,000 pages?
Paper required is directly proportional to the number of pages.
Paper required per page = 90 ÷ 4500 = 0.02 kg
Paper required for 7,000 pages
= 7000 × 0.02 = 140 kg
Paper required per page = 90 ÷ 4500 = 0.02 kg
Paper required for 7,000 pages
= 7000 × 0.02 = 140 kg
Question 19: If 12 workers can complete a piece of work in 15 days, how many days will 20 workers take to complete the same work?
Since more workers require fewer days, this is an inverse proportion.
Workers × Days = Constant
12 × 15 = 20 × A
180 = 20A
A = 9
Workers × Days = Constant
12 × 15 = 20 × A
180 = 20A
A = 9
Question 20: A water tank can be emptied by 6 identical pumps in 10 hours. How long will it take if 15 such pumps are used?
Number of pumps and time are inversely proportional.
Pumps × Time = Constant
6 × 10 = 15 × A
60 = 15A
A = 4
Pumps × Time = Constant
6 × 10 = 15 × A
60 = 15A
A = 4
Question 21: A printer prints a book in 30 minutes at a speed of 40 pages per minute. If the printing speed is increased to 50 pages per minute, how long will it take to print the same book?
Printing speed and time are inversely proportional.
Speed × Time = Constant
40 × 30 = 50 × A
1200 = 50A
A = 24
Speed × Time = Constant
40 × 30 = 50 × A
1200 = 50A
A = 24
Question 22: If 20% of A is equal to 25% of B, find the ratio A : B.
Given, 20% of A = 25% of B
20A = 25B
Divide both sides by 5.
4A = 5B
Therefore,
A : B = 5 : 4
20A = 25B
Divide both sides by 5.
4A = 5B
Therefore,
A : B = 5 : 4
Question 23 The incomes of A and B are in the ratio 7 : 9. If A saves 20% of his income and B saves 30% of his income, find the ratio of their savings.
Income ratio A : B = 7 : 9
Savings:
A saves 20% of 7 = 1.4
B saves 30% of 9 = 2.7
Required ratio
1.4 : 2.7
Multiply by 10.
14 : 27
Savings:
A saves 20% of 7 = 1.4
B saves 30% of 9 = 2.7
Required ratio
1.4 : 2.7
Multiply by 10.
14 : 27
Shortcut method:
Savings ratio
= (Income Ratio × Saving Percentage)
= (7 × 20) : (9 × 30)
= 140 : 270
= 14 : 27
Savings ratio
= (Income Ratio × Saving Percentage)
= (7 × 20) : (9 × 30)
= 140 : 270
= 14 : 27
Question 24: The salaries of A and B are in the ratio 5 : 8. If each receives a 20% increase in salary, find the new ratio of their salaries.
Original ratio 5 : 8
Both salaries increase by the same percentage.
Therefore, New salaries
= 5 × 1.20
= 6
and
8 × 1.20
= 9.6
Ratio 6 : 9.6
Simplify 5 : 8
Both salaries increase by the same percentage.
Therefore, New salaries
= 5 × 1.20
= 6
and
8 × 1.20
= 9.6
Ratio 6 : 9.6
Simplify 5 : 8
Question 25: The populations of two cities are in the ratio 5 : 6. If the population of the first city increases by 20% and that of the second city increases by 50%, find the new ratio.
Original ratio 5 : 6
New populations
First city = 5 × 1.20 = 6
Second city = 6 × 1.50 = 9
Required ratio
6 : 9
= 2 : 3
New populations
First city = 5 × 1.20 = 6
Second city = 6 × 1.50 = 9
Required ratio
6 : 9
= 2 : 3
Shortcut method:
Multiply each ratio term by its multiplication factor.
5 × 1.20 : 6 × 1.50 = 6 : 9 = 2 : 3
Multiply each ratio term by its multiplication factor.
5 × 1.20 : 6 × 1.50 = 6 : 9 = 2 : 3
Question 26: If x : 18 = 18 : 72, find the value of x.
Since the numbers are in continued proportion,
18² = x × 72
324 = 72x
x = 324 ÷ 72
x = 4.5
18² = x × 72
324 = 72x
x = 324 ÷ 72
x = 4.5
Question 27: Find the mean proportional between 16 and 81.
Mean proportional
= √(16 × 81)
= √1296
= 36
= √(16 × 81)
= √1296
= 36
Question 28: Find the third proportional to 12 and 18.
Let the third proportional be x.
According to the definition,
12 : 18 = 18 : x
Cross multiplying,
12x = 18 × 18
12x = 324
x = 27
According to the definition,
12 : 18 = 18 : x
Cross multiplying,
12x = 18 × 18
12x = 324
x = 27
Question 29: If three numbers are in continued proportion and the first and third numbers are 9 and 81, find the middle number.
Middle number
= √(9 × 81)
= √729
= 27
= √(9 × 81)
= √729
= 27
Question 30: Three numbers are in continued proportion. If their sum is 42 and the first and third numbers are 8 and 18, determine whether they are actually in continued proportion.
If the numbers are in continued proportion, the middle number should be
√(8 × 18)
= √144
= 12
Now, 8 + 12 + 18 = 38
But the given sum is 42.
Therefore, the given numbers cannot be in continued proportion.
√(8 × 18)
= √144
= 12
Now, 8 + 12 + 18 = 38
But the given sum is 42.
Therefore, the given numbers cannot be in continued proportion.
Question 31: Find the compound ratio of 2 : 5 and 3 : 4.
Multiply all antecedents together and all consequents together.
Compound ratio
= (2 × 3) : (5 × 4)
= 6 : 20
= 3 : 10
Compound ratio
= (2 × 3) : (5 × 4)
= 6 : 20
= 3 : 10
Question 32: Find the compound ratio of 4 : 7, 5 : 8 and 9 : 2.
Compound ratio
= (4 × 5 × 9) : (7 × 8 × 2)
= 180 : 112
Divide both terms by 4.
= 45 : 28
= (4 × 5 × 9) : (7 × 8 × 2)
= 180 : 112
Divide both terms by 4.
= 45 : 28
Question 33: The ratio of boys to girls in a class is 5 : 4 and the ratio of girls to teachers is 8 : 1. Find the compound ratio of boys : girls and girls : teachers.
Compound ratio
= (5 × 8) : (4 × 1)
= 40 : 4
= 10 : 1
= (5 × 8) : (4 × 1)
= 40 : 4
= 10 : 1
Question 34: The ratio of the cost prices of two articles is 7 : 9 and the ratio of their selling prices is 10 : 11. Find the compound ratio of the two given ratios.
Compound ratio
= (7 × 10) : (9 × 11)
= 70 : 99
= (7 × 10) : (9 × 11)
= 70 : 99
Question 35: The ratio of boys to girls in a school is 7 : 5. If there are 96 more boys than girls, find the total number of students.
Ratio Boys : Girls = 7 : 5
Difference in ratio
7 − 5 = 2 parts
2 parts = 96
1 part = 48
Number of boys
= 7 × 48
= 336
Number of girls = 5 × 48 = 240
Total students = 336 + 240 = 576
Difference in ratio
7 − 5 = 2 parts
2 parts = 96
1 part = 48
Number of boys
= 7 × 48
= 336
Number of girls = 5 × 48 = 240
Total students = 336 + 240 = 576
Question 36: The ratio of the incomes of A and B is 4 : 5. A saves 25% of his income, while B saves 20% of his income. If A saves ₹4,000, find B’s savings.
A’s savings = 25% of A’s income
A’s income = 4000 ÷ 25 × 100
= ₹16,000
Income ratio 4 : 5
Therefore, B’s income
= 16000 × 5 ÷ 4
= ₹20,000
B’s savings
= 20% of 20,000
= ₹4,000
A’s income = 4000 ÷ 25 × 100
= ₹16,000
Income ratio 4 : 5
Therefore, B’s income
= 16000 × 5 ÷ 4
= ₹20,000
B’s savings
= 20% of 20,000
= ₹4,000
Question 37: The ratio of the ages of a father and son is 5 : 2. After 6 years, the ratio becomes 2 : 1. Find their present ages.
Let
Father = 5x
Son = 2x
After 6 years
(5x + 6) : (2x + 6) = 2 : 1
5x + 6 = 4x + 12
x = 6
Father’s age = 30 years
Son’s age = 12 years
Answer: Father = 30 years, Son = 12 years
Father = 5x
Son = 2x
After 6 years
(5x + 6) : (2x + 6) = 2 : 1
5x + 6 = 4x + 12
x = 6
Father’s age = 30 years
Son’s age = 12 years
Answer: Father = 30 years, Son = 12 years
Question 38: Milk and water are mixed in the ratio 7 : 3. How much water must be added to 40 litres of the mixture so that the ratio becomes 7 : 5?
Milk in 40 litres
= (7/10) × 40
= 28 litres
Water in 40 litres
= 12 litres
Let x litres of water be added.
Required ratio
28 : (12 + x)
= 7 : 5
Cross multiplication
140 = 84 + 7x
56 = 7x
x = 8
= (7/10) × 40
= 28 litres
Water in 40 litres
= 12 litres
Let x litres of water be added.
Required ratio
28 : (12 + x)
= 7 : 5
Cross multiplication
140 = 84 + 7x
56 = 7x
x = 8
Question 39: The ratio of the monthly salaries of A and B is 3 : 4. If each receives a 25% increase in salary, what will be the new ratio?
Both salaries increase by the same percentage.
Therefore, the ratio remains unchanged.
Answer = 3 : 4
Therefore, the ratio remains unchanged.
Answer = 3 : 4
Question 40: A sum of ₹9,000 is divided among A, B and C in the ratio 2 : 3 : 5. If A gives ₹300 to B, what is the new ratio?
Original shares
A = ₹1,800
B = ₹2,700
C = ₹4,500
After transfer
A = ₹1,500
B = ₹3,000
C = ₹4,500
New ratio
1500 : 3000 : 4500
= 1 : 2 : 3
A = ₹1,800
B = ₹2,700
C = ₹4,500
After transfer
A = ₹1,500
B = ₹3,000
C = ₹4,500
New ratio
1500 : 3000 : 4500
= 1 : 2 : 3
Question 41: If (x² + y²) : (x² − y²) = 25 : 7, find (x⁴ + y⁴) : (x⁴ − y⁴).
Using Componendo & Dividendo,
[(x² + y²) + (x² − y²)] : [(x² + y²) − (x² − y²)] = (25 + 7) : (25 − 7)
Therefore, 2x² : 2y² = 32 : 18
x² : y² = 16 : 9
Let: x² = 16k and y² = 9k
Therefore,
x⁴ = 256k² and y⁴ = 81k²
Hence,
(x⁴ + y⁴) : (x⁴ − y⁴)
= (256k² + 81k²) : (256k² − 81k²)
= 337k² : 175k²
= 337 : 175
[(x² + y²) + (x² − y²)] : [(x² + y²) − (x² − y²)] = (25 + 7) : (25 − 7)
Therefore, 2x² : 2y² = 32 : 18
x² : y² = 16 : 9
Let: x² = 16k and y² = 9k
Therefore,
x⁴ = 256k² and y⁴ = 81k²
Hence,
(x⁴ + y⁴) : (x⁴ − y⁴)
= (256k² + 81k²) : (256k² − 81k²)
= 337k² : 175k²
= 337 : 175
Question 42: If three numerical ratios are locked in a continuous proportion such that a / b = c / d = e / f, find the simplified numerical value of the expression (5a² + 7c² + 2e²) / (5b² + 7d² + 2f²) given that the baseline ratio a / b is equal to 3 / 4.
Since a ⁄ b = c ⁄ d = e ⁄ f = 3 ⁄ 4
we have a = 3b ⁄ 4, c = 3d ⁄ 4, e = 3f ⁄ 4
Squaring, a² = 9b² ⁄ 16, c² = 9d² ⁄ 16, e² = 9f² ⁄ 16
∴ 5a² + 7c² + 2e² = 9⁄16 × (5b² + 7d² + 2f²)
or, (5a² + 7c² + 2e²) ⁄ (5b² + 7d² + 2f²) = 9⁄16
Question 43: If the quantities x, y, and z satisfy the continuous multi-variable proportion rule where x / (2y + 2z) = y / (2z + 2x) = z / (2x + 2y), find the unique numerical value that each of these individual ratios must equal, assuming all tracking values are positive integers.
Given x ⁄ (2y + 2z) = y ⁄ (2z + 2x) = z ⁄ (2x + 2y)
Let the common ratio be k.
Then
x = 2k(y + z)
y = 2k(z + x)
z = 2k(x + y)
Adding these three equations,
x + y + z = 2k[(y + z) + (z + x) + (x + y)]
∴ x + y + z = 4k(x + y + z)
Since x, y and z are positive integers, (x + y + z) ≠ 0
So, 1 = 4k
∴ k = 1⁄4
Let the common ratio be k.
Then
x = 2k(y + z)
y = 2k(z + x)
z = 2k(x + y)
Adding these three equations,
x + y + z = 2k[(y + z) + (z + x) + (x + y)]
∴ x + y + z = 4k(x + y + z)
Since x, y and z are positive integers, (x + y + z) ≠ 0
So, 1 = 4k
∴ k = 1⁄4
Question 44: A vessel contains milk and water mixed in the ratio 4 : 1. If 10 litres of this liquid mixture are drawn out and replaced completely with 10 litres of pure water, the new ratio of milk to water becomes 2 : 3. Find the total volume capacity of the vessel in litres.
Let the total capacity of the vessel be A litres.
Initially
Milk = ⁴⁄₅A litres
Water = ¹⁄₅A litres
When 10 litres of the mixture are removed, milk and water are removed in the ratio 4 : 1.
Milk removed = 8 litres
Water removed = 2 litres
After removing 10 litres
Milk = ⁴⁄₅A − 8
Water = ¹⁄₅A − 2
Then 10 litres of pure water are added
Milk = ⁴⁄₅A − 8
Water = ¹⁄₅A + 8
The new ratio of milk to water is 2 : 3.
(⁴⁄₅A − 8) : (¹⁄₅A + 8) = 2 : 3
or, 3(⁴⁄₅A − 8) = 2(¹⁄₅A + 8)
or, ¹²⁄₅A − 24 = ²⁄₅A + 16
or, ¹⁰⁄₅A = 40
or, 2A = 40
or, A = 20
So capacity of the container is 20 litres.
Initially
Milk = ⁴⁄₅A litres
Water = ¹⁄₅A litres
When 10 litres of the mixture are removed, milk and water are removed in the ratio 4 : 1.
Milk removed = 8 litres
Water removed = 2 litres
After removing 10 litres
Milk = ⁴⁄₅A − 8
Water = ¹⁄₅A − 2
Then 10 litres of pure water are added
Milk = ⁴⁄₅A − 8
Water = ¹⁄₅A + 8
The new ratio of milk to water is 2 : 3.
(⁴⁄₅A − 8) : (¹⁄₅A + 8) = 2 : 3
or, 3(⁴⁄₅A − 8) = 2(¹⁄₅A + 8)
or, ¹²⁄₅A − 24 = ²⁄₅A + 16
or, ¹⁰⁄₅A = 40
or, 2A = 40
or, A = 20
So capacity of the container is 20 litres.
Question 45: A canister contains a blend of two chemical juices, Liquid 1 and Liquid 2, in the ratio 7 : 5. When 9 litres of this blended solution are drawn off and the canister is replenished with 9 litres of pure Liquid 2, the ratio of Liquid 1 to Liquid 2 updates to 7 : 9. Find how many litres of Liquid 1 were originally contained in the canister.
Let the original volumes of Liquid 1 and Liquid 2 be 7A litres and 5A litres respectively.
Total initial volume = 12A litres
Step 1: Calculate the quantities removed
Since 9 litres of the mixture are removed in the ratio 7 : 5
Liquid 1 removed = 9 × ⁷⁄₁₂ = ²¹⁄₄ litres
Liquid 2 removed = 9 × ⁵⁄₁₂ = ¹⁵⁄₄ litres
Step 2: Find the quantities after replacement
Liquid 1 remaining = 7A − ²¹⁄₄
Liquid 2 remaining = 5A − ¹⁵⁄₄ + 9
= 5A + ²¹⁄₄
Step 3: Use the new ratio 7 : 9
(7A − ²¹⁄₄) : (5A + ²¹⁄₄) = 7 : 9
or, 9(7A − ²¹⁄₄) = 7(5A + ²¹⁄₄)
or, 63A − ¹⁸⁹⁄₄ = 35A + ¹⁴⁷⁄₄
or, 28A = ³³⁶⁄₄
or, 28A = 84
or, A = 3
Step 4: Find the original quantity of Liquid 1
Original Liquid 1 = 7A
= 7 × 3
= 21 litres
∴ The original volume of Liquid 1 was 21 litres.
Total initial volume = 12A litres
Step 1: Calculate the quantities removed
Since 9 litres of the mixture are removed in the ratio 7 : 5
Liquid 1 removed = 9 × ⁷⁄₁₂ = ²¹⁄₄ litres
Liquid 2 removed = 9 × ⁵⁄₁₂ = ¹⁵⁄₄ litres
Step 2: Find the quantities after replacement
Liquid 1 remaining = 7A − ²¹⁄₄
Liquid 2 remaining = 5A − ¹⁵⁄₄ + 9
= 5A + ²¹⁄₄
Step 3: Use the new ratio 7 : 9
(7A − ²¹⁄₄) : (5A + ²¹⁄₄) = 7 : 9
or, 9(7A − ²¹⁄₄) = 7(5A + ²¹⁄₄)
or, 63A − ¹⁸⁹⁄₄ = 35A + ¹⁴⁷⁄₄
or, 28A = ³³⁶⁄₄
or, 28A = 84
or, A = 3
Step 4: Find the original quantity of Liquid 1
Original Liquid 1 = 7A
= 7 × 3
= 21 litres
∴ The original volume of Liquid 1 was 21 litres.
Question 46: A laboratory barrel is filled with a mixture of active acid and water in the ratio 5 : 3. What fraction of the total mixture must be drawn off and replaced with an equal volume of pure water so that the final mixture contains active acid and water in the ratio 1 : 1?
Let the total volume of the mixture be 1 unit.
Initially Active acid = ⁵⁄₈ unit and Water = ³⁄₈ unit
Let A be the fraction of the total mixture that is drawn off.
Since the mixture is drawn off in the ratio 5 : 3
Active acid removed = ⁵⁄₈A
Therefore, active acid remaining is = ⁵⁄₈ − ⁵⁄₈A
The removed mixture is replaced with an equal volume of pure water, so the total volume remains 1 unit.
For the final ratio of active acid to water to be 1 : 1, the final amount of active acid must be ¹⁄₂ unit.
Therefore,
⁵⁄₈ − ⁵⁄₈A = ¹⁄₂
or, ⁵⁄₈A = ⁵⁄₈ − ⁴⁄₈
or, ⁵⁄₈A = ¹⁄₈
or, A = ¹⁄₅
∴ ¹⁄₅ of the total mixture must be drawn off and replaced with an equal volume of pure water.
Initially Active acid = ⁵⁄₈ unit and Water = ³⁄₈ unit
Let A be the fraction of the total mixture that is drawn off.
Since the mixture is drawn off in the ratio 5 : 3
Active acid removed = ⁵⁄₈A
Therefore, active acid remaining is = ⁵⁄₈ − ⁵⁄₈A
The removed mixture is replaced with an equal volume of pure water, so the total volume remains 1 unit.
For the final ratio of active acid to water to be 1 : 1, the final amount of active acid must be ¹⁄₂ unit.
Therefore,
⁵⁄₈ − ⁵⁄₈A = ¹⁄₂
or, ⁵⁄₈A = ⁵⁄₈ − ⁴⁄₈
or, ⁵⁄₈A = ¹⁄₈
or, A = ¹⁄₅
∴ ¹⁄₅ of the total mixture must be drawn off and replaced with an equal volume of pure water.
Question 47: A barrel contains a cleaning solvent mixture of concentrate and water in the ratio 3 : 1. A technician draws out a certain volume of the mixture and replaces it with an equal volume of pure water. The ratio of concentrate to water in the barrel then becomes 1 : 1. If the concentrate remaining in the barrel is worth ¤15, find the total value of the concentrate in the original mixture.
Initially, the ratio of concentrate to water is 3 : 1.
Therefore, concentrate forms ³⁄₄ of the original mixture.
After some mixture is removed and replaced with pure water, the ratio becomes 1 : 1.
Therefore, concentrate now forms ¹⁄₂ of the total mixture.
Since the total volume of the mixture remains unchanged:
Original concentrate : Remaining concentrate = ³⁄₄ : ¹⁄₂ = 3 : 2
Therefore, the value of the original concentrate is also in the ratio 3 : 2 with the value of the remaining concentrate.
Given value of remaining concentrate = ¤15
Therefore, value of original concentrate = ¤15 × ³⁄₂ = ¤22.50
∴ The total value of the concentrate in the original mixture was ¤22.50.
Therefore, concentrate forms ³⁄₄ of the original mixture.
After some mixture is removed and replaced with pure water, the ratio becomes 1 : 1.
Therefore, concentrate now forms ¹⁄₂ of the total mixture.
Since the total volume of the mixture remains unchanged:
Original concentrate : Remaining concentrate = ³⁄₄ : ¹⁄₂ = 3 : 2
Therefore, the value of the original concentrate is also in the ratio 3 : 2 with the value of the remaining concentrate.
Given value of remaining concentrate = ¤15
Therefore, value of original concentrate = ¤15 × ³⁄₂ = ¤22.50
∴ The total value of the concentrate in the original mixture was ¤22.50.
Question 48: The value of a precious gemstone varies directly as the square of its weight. A gemstone weighing 10 grams breaks accidentally into two pieces whose weights are in the ratio 2 : 3. Calculate the percentage loss in the total value of the gemstone.
Let the weights of the two pieces be 2A grams and 3A grams.
Since their total weight is 10 grams
2A + 3A = 10
or, 5A = 10
or, A = 2
Therefore, the weights of the two pieces are
2A = 4 grams
3A = 6 grams
Since the value varies directly as the square of the weight
Value ∝ Weight²
Let the value of the original 10-gram gemstone be proportional to 10² = 100
After breaking, the combined value of the two pieces is proportional to
4² + 6²
= 16 + 36
= 52
Therefore, the loss in value is proportional to
100 − 52 = 48
Percentage loss = ⁴⁸⁄₁₀₀ × 100 = 48%
∴ The percentage loss in the total value of the gemstone is 48%.
Since their total weight is 10 grams
2A + 3A = 10
or, 5A = 10
or, A = 2
Therefore, the weights of the two pieces are
2A = 4 grams
3A = 6 grams
Since the value varies directly as the square of the weight
Value ∝ Weight²
Let the value of the original 10-gram gemstone be proportional to 10² = 100
After breaking, the combined value of the two pieces is proportional to
4² + 6²
= 16 + 36
= 52
Therefore, the loss in value is proportional to
100 − 52 = 48
Percentage loss = ⁴⁸⁄₁₀₀ × 100 = 48%
∴ The percentage loss in the total value of the gemstone is 48%.
Question 49: The electrical power consumed by a heating element varies jointly as the square of the current passing through it and directly as its internal resistance. If the current is increased by 20% and the resistance is reduced by 10%, find the net percentage change in the power consumed.
Power varies jointly as the square of the current and the resistance.
∴ P ∝ I²R
Let the original power be P.
The current is increased by 20%
New current = 120% of I
= ¹²⁰⁄₁₀₀ I
= ⁶⁄₅I
The resistance is reduced by 10%
New resistance = 90% of R
= ⁹⁄₁₀R
Therefore, new power ∝ (⁶⁄₅I)² × ⁹⁄₁₀R
= ³⁶⁄₂₅ × ⁹⁄₁₀ × I²R
= ³²⁴⁄₂₅₀ × I²R
= ¹⁶²⁄₁₂₅ × I²R
Therefore, new power = ¹⁶²⁄₁₂₅P
= 129.6% of P
Percentage change
= 129.6% − 100%
= 29.6%
∴ The power consumed increases by 29.6%.
∴ P ∝ I²R
Let the original power be P.
The current is increased by 20%
New current = 120% of I
= ¹²⁰⁄₁₀₀ I
= ⁶⁄₅I
The resistance is reduced by 10%
New resistance = 90% of R
= ⁹⁄₁₀R
Therefore, new power ∝ (⁶⁄₅I)² × ⁹⁄₁₀R
= ³⁶⁄₂₅ × ⁹⁄₁₀ × I²R
= ³²⁴⁄₂₅₀ × I²R
= ¹⁶²⁄₁₂₅ × I²R
Therefore, new power = ¹⁶²⁄₁₂₅P
= 129.6% of P
Percentage change
= 129.6% − 100%
= 29.6%
∴ The power consumed increases by 29.6%.
Question 50: A collection cash box contains coins of three different face values: ¤1 coins, ¤0.50 coins, and ¤0.25 coins, and their quantities are in the ratio 5 : 6 : 8. If the total value of all the coins in the box is exactly ¤420, find the total number of ¤0.50 coins.
Let the numbers of ¤1, ¤0.50, and ¤0.25 coins be 5A, 6A, and 8A respectively.
Their total value is
5A × ¤1 + 6A × ¤0.50 + 8A × ¤0.25
= ¤5A + ¤3A + ¤2A
= ¤10A
Given that the total value is ¤420,
10A = 420
or, A = 42
Therefore, the number of ¤0.50 coins is 6A = 6 × 42
= 252
∴ The total number of ¤0.50 coins is 252.
Their total value is
5A × ¤1 + 6A × ¤0.50 + 8A × ¤0.25
= ¤5A + ¤3A + ¤2A
= ¤10A
Given that the total value is ¤420,
10A = 420
or, A = 42
Therefore, the number of ¤0.50 coins is 6A = 6 × 42
= 252
∴ The total number of ¤0.50 coins is 252.
Question 51: A canvas bag contains identical tokens with face values of ¤5, ¤2, and ¤1. The ratio of the total value contributed by the ¤5, ¤2, and ¤1 token groups is 15 : 8 : 3. If the total number of tokens in the bag is 110, find the number of ¤2 tokens.
Let the total values contributed by the ¤5, ¤2, and ¤1 token groups be,
15A, 8A, and 3A respectively.
Therefore, the corresponding numbers of tokens are
Number of ¤5 tokens = ¹⁵A⁄₅ = 3A
Number of ¤2 tokens = ⁸A⁄₂ = 4A
Number of ¤1 tokens = ³A⁄₁ = 3A
The total number of tokens is 110.
∴ 3A + 4A + 3A = 110
or, 10A = 110
or, A = 11
Therefore, the number of ¤2 tokens is
4A = 4 × 11
= 44
∴ The number of ¤2 tokens is 44.
15A, 8A, and 3A respectively.
Therefore, the corresponding numbers of tokens are
Number of ¤5 tokens = ¹⁵A⁄₅ = 3A
Number of ¤2 tokens = ⁸A⁄₂ = 4A
Number of ¤1 tokens = ³A⁄₁ = 3A
The total number of tokens is 110.
∴ 3A + 4A + 3A = 110
or, 10A = 110
or, A = 11
Therefore, the number of ¤2 tokens is
4A = 4 × 11
= 44
∴ The number of ¤2 tokens is 44.
Question 52: A small lockbox holds ¤10 and ¤5 premium coupon slips. The ratio of the number of ¤10 slips to the number of ¤5 slips is 3 : 4. If a clerk adds 10 more ¤10 slips to the box, the ratio of the total value of the ¤10 slips to the total value of the ¤5 slips becomes 2 : 1. Find the initial total monetary value contained in the lockbox.
Let the initial numbers of ¤10 and ¤5 slips be 3A and 4A respectively.
Therefore, their initial values are
Value of ¤10 slips = 3A × ¤10 = ¤30A
Value of ¤5 slips = 4A × ¤5 = ¤20A
When 10 more ¤10 slips are added:
Additional value = 10 × ¤10 = ¤100
New value of ¤10 slips = ¤30A + ¤100
The value of ¤5 slips remains ¤20A.
Given that the new value ratio is 2 : 1
(30A + 100) : 20A = 2 : 1
or, 30A + 100 = 40A
or, 100 = 10A
or, A = 10
Therefore, the initial total monetary value is
= ¤30A + ¤20A
= ¤50A
= ¤50 × 10
= ¤500
∴ The initial total monetary value contained in the lockbox was ¤500.
Therefore, their initial values are
Value of ¤10 slips = 3A × ¤10 = ¤30A
Value of ¤5 slips = 4A × ¤5 = ¤20A
When 10 more ¤10 slips are added:
Additional value = 10 × ¤10 = ¤100
New value of ¤10 slips = ¤30A + ¤100
The value of ¤5 slips remains ¤20A.
Given that the new value ratio is 2 : 1
(30A + 100) : 20A = 2 : 1
or, 30A + 100 = 40A
or, 100 = 10A
or, A = 10
Therefore, the initial total monetary value is
= ¤30A + ¤20A
= ¤50A
= ¤50 × 10
= ¤500
∴ The initial total monetary value contained in the lockbox was ¤500.
Question 53: A storage pouch contains ¤1 coins and ¤0.50 coins. The ratio of the number of ¤1 coins to the number of ¤0.50 coins is 4 : 7. If the total value of the ¤1 coins exceeds the total value of the ¤0.50 coins by exactly ¤18, find the total number of coins in the pouch.
Let the numbers of ¤1 coins and ¤0.50 coins be 4A and 7A respectively.
Therefore
Value of ¤1 coins = 4A × ¤1 = ¤4A
Value of ¤0.50 coins = 7A × ¤0.50 = ¤3.50A
Given that the value of the ¤1 coins exceeds the value of the ¤0.50 coins by ¤18
¤4A − ¤3.50A = ¤18
or, ¤0.50A = ¤18
or, A = 36
Therefore, the total number of coins is
= 4A + 7A
= 11A
= 11 × 36
= 396
∴ The total number of coins in the pouch is 396.
Therefore
Value of ¤1 coins = 4A × ¤1 = ¤4A
Value of ¤0.50 coins = 7A × ¤0.50 = ¤3.50A
Given that the value of the ¤1 coins exceeds the value of the ¤0.50 coins by ¤18
¤4A − ¤3.50A = ¤18
or, ¤0.50A = ¤18
or, A = 36
Therefore, the total number of coins is
= 4A + 7A
= 11A
= 11 × 36
= 396
∴ The total number of coins in the pouch is 396.
Question 54: If a : b = 3 : 5 and b : c = 4 : 7, find the ratio (2a + 3b) : (3b + 2c).
Given a : b = 3 : 5 and b : c = 4 : 7
Make the value of b common in both ratios.
a : b = 12 : 20
b : c = 20 : 35
Therefore, a : b : c = 12 : 20 : 35
Now, 2a + 3b = 2 × 12 + 3 × 20
= 24 + 60
= 84
And,
3b + 2c = 3 × 20 + 2 × 35
= 60 + 70
= 130
Therefore, (2a + 3b) : (3b + 2c)
= 84 : 130
= 42 : 65
∴ The required ratio is 42 : 65.
Make the value of b common in both ratios.
a : b = 12 : 20
b : c = 20 : 35
Therefore, a : b : c = 12 : 20 : 35
Now, 2a + 3b = 2 × 12 + 3 × 20
= 24 + 60
= 84
And,
3b + 2c = 3 × 20 + 2 × 35
= 60 + 70
= 130
Therefore, (2a + 3b) : (3b + 2c)
= 84 : 130
= 42 : 65
∴ The required ratio is 42 : 65.
Question 55: Three positive numbers are in continued proportion. If the sum of the first and third numbers is 26 and their difference is 10, find the middle number.
Let the three numbers be a, b and c.
Since they are in continued proportion
a : b = b : c
∴ b² = ac
We are given a + c = 26 and c − a = 10
Adding these equations
2c = 36
∴ c = 18
Therefore, a = 26 − 18 = 8
Since b² = ac:
b² = 8 × 18
= 144
or, b = 12 Since, the numbers are positive
∴ The middle number is 12.
Since they are in continued proportion
a : b = b : c
∴ b² = ac
We are given a + c = 26 and c − a = 10
Adding these equations
2c = 36
∴ c = 18
Therefore, a = 26 − 18 = 8
Since b² = ac:
b² = 8 × 18
= 144
or, b = 12 Since, the numbers are positive
∴ The middle number is 12.
Question 56: The mean proportional between two positive numbers is 12. If their difference is 7, find the two numbers.
Let the two numbers be a and b, with a > b.
Since 12 is their mean proportional
12² = ab
or, ab = 144
Also, a − b = 7
Therefore, a = b + 7
Substituting b(b + 7) = 144
or, b² + 7b − 144 = 0
Factoring (b + 16)(b − 9) = 0
Since b is positive b = 9
Therefore,
a = 9 + 7
= 16
∴ The two numbers are 16 and 9.
Since 12 is their mean proportional
12² = ab
or, ab = 144
Also, a − b = 7
Therefore, a = b + 7
Substituting b(b + 7) = 144
or, b² + 7b − 144 = 0
Factoring (b + 16)(b − 9) = 0
Since b is positive b = 9
Therefore,
a = 9 + 7
= 16
∴ The two numbers are 16 and 9.
Question 57: A machine takes 18 hours to complete a certain task at its normal efficiency. If its efficiency is increased by 25%, how long will the machine take to complete the same task?
For a fixed task, the time taken varies inversely with efficiency.
An increase of 25% means the new efficiency is
100% + 25% = 125%
= ¹²⁵⁄₁₀₀
= ⁵⁄₄ of the original efficiency.
Therefore, the new time is 18 × ⁴⁄₅ = ⁷²⁄₅ = 14.4 hours
= 14 hours 24 minutes
∴ The machine will take 14 hours 24 minutes to complete the task.
An increase of 25% means the new efficiency is
100% + 25% = 125%
= ¹²⁵⁄₁₀₀
= ⁵⁄₄ of the original efficiency.
Therefore, the new time is 18 × ⁴⁄₅ = ⁷²⁄₅ = 14.4 hours
= 14 hours 24 minutes
∴ The machine will take 14 hours 24 minutes to complete the task.
Question 58: The time required to manufacture a fixed number of components varies directly with the number of components and inversely with the number of identical machines working simultaneously. Eight machines manufacture 1,200 components in 15 hours. How many hours will 12 machines take to manufacture 1,800 components at the same rate?
Let the time be T, the number of components be N, and the number of machines be M.
Since T varies directly with N and inversely with M T ∝ N⁄M
Therefore, T₂⁄T₁ = N₂⁄N₁ × M₁⁄M₂
Substituting
T₂ ⁄ 15 = ¹⁸⁰⁰ ⁄ ¹²⁰⁰ × ⁸⁄₁₂
or, T₂ ⁄ 15 = ³⁄₂ × ²⁄₃
or, T₂ ⁄ 15 = 1
or, T₂ = 15
∴ 12 machines will take 15 hours to manufacture 1,800 components.
Since T varies directly with N and inversely with M T ∝ N⁄M
Therefore, T₂⁄T₁ = N₂⁄N₁ × M₁⁄M₂
Substituting
T₂ ⁄ 15 = ¹⁸⁰⁰ ⁄ ¹²⁰⁰ × ⁸⁄₁₂
or, T₂ ⁄ 15 = ³⁄₂ × ²⁄₃
or, T₂ ⁄ 15 = 1
or, T₂ = 15
∴ 12 machines will take 15 hours to manufacture 1,800 components.
Question 59: A number is divided into two parts in the ratio 7 : 5. If 20% of the larger part is transferred to the smaller part, what will be the new ratio of the two parts?
Let the two parts be 7A and 5A.
20% of the larger part is
20% of 7A
= ¹⁄₅ × 7A
= ⁷⁄₅A
After the transfer
Larger part = 7A − ⁷⁄₅A
= ²⁸⁄₅A
Smaller part = 5A + ⁷⁄₅A
= ³²⁄₅A
Therefore, the new ratio is
²⁸⁄₅A : ³²⁄₅A
= 28 : 32
= 7 : 8
∴ The new ratio is 7 : 8.
20% of the larger part is
20% of 7A
= ¹⁄₅ × 7A
= ⁷⁄₅A
After the transfer
Larger part = 7A − ⁷⁄₅A
= ²⁸⁄₅A
Smaller part = 5A + ⁷⁄₅A
= ³²⁄₅A
Therefore, the new ratio is
²⁸⁄₅A : ³²⁄₅A
= 28 : 32
= 7 : 8
∴ The new ratio is 7 : 8.
Question 60: The time required to complete a journey varies directly with the distance travelled and inversely with the average speed. A journey of 240 km takes 4 hours at an average speed of 60 km/h. How long would a journey of 315 km take at an average speed of 70 km/h?
Let the time be T, distance be D, and average speed be S.
Since T varies directly with D and inversely with S T ∝ D⁄S
Therefore, T₂⁄T₁ = D₂⁄D₁ × S₁⁄S₂
Substituting
T₂⁄4 = ³¹⁵⁄₂₄₀ × ⁶⁰⁄₇₀
or, T₂⁄4 = ²¹⁄₁₆ × ⁶⁄₇
or, T₂⁄4 = ¹⁸⁄₁₆
or, T₂ = 4 × ⁹⁄₈ = ⁹⁄₂
= 4.5 hours
= 4 hours 30 minutes
∴ The journey would take 4 hours 30 minutes.
Since T varies directly with D and inversely with S T ∝ D⁄S
Therefore, T₂⁄T₁ = D₂⁄D₁ × S₁⁄S₂
Substituting
T₂⁄4 = ³¹⁵⁄₂₄₀ × ⁶⁰⁄₇₀
or, T₂⁄4 = ²¹⁄₁₆ × ⁶⁄₇
or, T₂⁄4 = ¹⁸⁄₁₆
or, T₂ = 4 × ⁹⁄₈ = ⁹⁄₂
= 4.5 hours
= 4 hours 30 minutes
∴ The journey would take 4 hours 30 minutes.
Question 61: The corresponding sides of two similar triangles are in the ratio 3 : 5. If the area of the smaller triangle is 54 cm², find the area of the larger triangle.
For similar figures, the ratio of their areas is equal to the square of the ratio of their corresponding sides.
Therefore, area of smaller triangle : Area of larger triangle
= 3² : 5²
= 9 : 25
Let the area of the larger triangle be A cm².
Then, 54 : A = 9 : 25
or, 9A = 54 × 25
or, A = 150
∴ The area of the larger triangle is 150 cm².
Therefore, area of smaller triangle : Area of larger triangle
= 3² : 5²
= 9 : 25
Let the area of the larger triangle be A cm².
Then, 54 : A = 9 : 25
or, 9A = 54 × 25
or, A = 150
∴ The area of the larger triangle is 150 cm².
Question 62: A sum of ¤2,160 is divided among A, B and C in the ratio 3 : 5 : 7. If C receives ¤480 more than A, verify the amount received by B.
Let the shares of A, B and C be 3A, 5A and 7A.
The difference between C’s and A’s shares is ¤480.
or, 7A − 3A = 480
or, 4A = 480
or, A = 120
Therefore, B’s share is 5A = 5 × 120
= ¤600
∴ B receives ¤600.
The difference between C’s and A’s shares is ¤480.
or, 7A − 3A = 480
or, 4A = 480
or, A = 120
Therefore, B’s share is 5A = 5 × 120
= ¤600
∴ B receives ¤600.
Question 63: If a : b = c : d, prove that (a + b) : (a − b) = (c + d) : (c − d), provided b ≠ d and the denominators are non-zero.
Given a : b = c : d
or, a⁄b = c⁄d
Cross-multiplying ad = bc
Now consider
(a + b) : (a − b)
= (a + b)⁄(a − b)
Dividing numerator and denominator by b
= (a⁄b + 1) : (a⁄b − 1)
Since a⁄b = c⁄d
= (c⁄d + 1) : (c⁄d − 1)
= (c + d) : (c − d)
∴ (a + b) : (a − b) = (c + d) : (c − d).
or, a⁄b = c⁄d
Cross-multiplying ad = bc
Now consider
(a + b) : (a − b)
= (a + b)⁄(a − b)
Dividing numerator and denominator by b
= (a⁄b + 1) : (a⁄b − 1)
Since a⁄b = c⁄d
= (c⁄d + 1) : (c⁄d − 1)
= (c + d) : (c − d)
∴ (a + b) : (a − b) = (c + d) : (c − d).
Question 64: The ratio of the number of books in two shelves is 5 : 8. If 15% of the books on the second shelf are transferred to the first shelf, what is the new ratio of the numbers of books on the two shelves?
Let the original numbers of books be 5A and 8A.
15% of the books on the second shelf is
15% of 8A
= ¹⁵⁄₁₀₀ × 8A
= ⁶⁄₅A
After the transfer books on
First shelf
= 5A + ⁶⁄₅A
= ³¹⁄₅A
Books on second shelf
= 8A − ⁶⁄₅A
= ³⁴⁄₅A
Therefore, the new ratio is ³¹⁄₅A : ³⁴⁄₅A
= 31 : 34
∴ The new ratio is 31 : 34.
15% of the books on the second shelf is
15% of 8A
= ¹⁵⁄₁₀₀ × 8A
= ⁶⁄₅A
After the transfer books on
First shelf
= 5A + ⁶⁄₅A
= ³¹⁄₅A
Books on second shelf
= 8A − ⁶⁄₅A
= ³⁴⁄₅A
Therefore, the new ratio is ³¹⁄₅A : ³⁴⁄₅A
= 31 : 34
∴ The new ratio is 31 : 34.
Question 65: The radii of two circular metal discs are in the ratio 3 : 5. The thicknesses of the discs are in the ratio 10 : 9. Find the ratio of their volumes.
The volume of a cylinder is V = πr²h
Therefore, the volume varies directly as the square of the radius and directly as the thickness.
Hence, V₁ : V₂ = r₁²h₁ : r₂²h₂
Given r₁ : r₂ = 3 : 5 and
h₁ : h₂ = 10 : 9
Therefore,
V₁ : V₂
= 3² × 10 : 5² × 9
= 9 × 10 : 25 × 9
= 90 : 225
= 2 : 5
∴ The ratio of their volumes is 2 : 5.
Therefore, the volume varies directly as the square of the radius and directly as the thickness.
Hence, V₁ : V₂ = r₁²h₁ : r₂²h₂
Given r₁ : r₂ = 3 : 5 and
h₁ : h₂ = 10 : 9
Therefore,
V₁ : V₂
= 3² × 10 : 5² × 9
= 9 × 10 : 25 × 9
= 90 : 225
= 2 : 5
∴ The ratio of their volumes is 2 : 5.
Question 66: Three positive numbers satisfy a : b = 2 : 3 and b : c = 4 : 5. Find the ratio (a + b) : (b + c) : (c + a).
Make the value of b common.
Given a : b = 2 : 3 and b : c = 4 : 5
The least common value of b is 12.
Therefore, a : b = 8 : 12 and b : c = 12 : 15
Hence, a : b : c = 8 : 12 : 15.
Now,
a + b = 8 + 12 = 20
b + c = 12 + 15 = 27
c + a = 15 + 8 = 23
∴ (a + b) : (b + c) : (c + a) = 20 : 27 : 23.
Given a : b = 2 : 3 and b : c = 4 : 5
The least common value of b is 12.
Therefore, a : b = 8 : 12 and b : c = 12 : 15
Hence, a : b : c = 8 : 12 : 15.
Now,
a + b = 8 + 12 = 20
b + c = 12 + 15 = 27
c + a = 15 + 8 = 23
∴ (a + b) : (b + c) : (c + a) = 20 : 27 : 23.
What are Ratios and Proportions?
A ratio is a way of comparing two quantities of the same kind by showing how much one quantity is in relation to another. A ratio is generally written in the form a : b, where a and b represent the quantities being compared and b ≠ 0. Before forming a ratio, both quantities should always be converted into the same unit.
For example, if a basket contains 20 apples and 30 oranges, the ratio of apples to oranges is: 20 : 30 = 2 : 3
This means that for every 2 apples, there are 3 oranges.
A proportion is a mathematical statement that two ratios are equal. In other words, when two ratios represent the same relationship, they are said to be in proportion.
For example: 2 : 3 = 4 : 6
Since both ratios represent the same comparison, they are in proportion.
For example, if a basket contains 20 apples and 30 oranges, the ratio of apples to oranges is: 20 : 30 = 2 : 3
This means that for every 2 apples, there are 3 oranges.
A proportion is a mathematical statement that two ratios are equal. In other words, when two ratios represent the same relationship, they are said to be in proportion.
For example: 2 : 3 = 4 : 6
Since both ratios represent the same comparison, they are in proportion.
Cross Multiplication Property of Proportion
If four quantities are in proportion, then the product of the first and fourth terms is equal to the product of the second and third terms.
If a : b = c : d
then: a/b = c/d
Cross multiplying: a × d = b × c
Therefore, in a proportion:
Product of extremes = Product of means
where a and d are called extremes, and b and c are called means.
If a : b = c : d
then: a/b = c/d
Cross multiplying: a × d = b × c
Therefore, in a proportion:
Product of extremes = Product of means
where a and d are called extremes, and b and c are called means.
Direct Proportion
Two quantities are said to be in direct proportion when an increase in one quantity results in a proportional increase in the other quantity, and a decrease in one results in a proportional decrease in the other.
If two quantities x and y are directly proportional: x ∝ y
or, x/y = constant
Therefore, x₁/y₁ = x₂/y₂
Example: If the cost of 5 books is ₹500, then the cost of 10 books will be ₹1,000, assuming the price per book remains constant.
If two quantities x and y are directly proportional: x ∝ y
or, x/y = constant
Therefore, x₁/y₁ = x₂/y₂
Example: If the cost of 5 books is ₹500, then the cost of 10 books will be ₹1,000, assuming the price per book remains constant.
Inverse Proportion
Two quantities are said to be in inverse proportion when an increase in one quantity results in a proportional decrease in the other quantity, and vice versa.
If two quantities x and y are inversely proportional: x ∝ 1/y
or, xy = constant
Therefore, x₁y₁ = x₂y₂
Example:
If more workers are employed to complete the same work, the time required to complete the work decreases proportionally.
If two quantities x and y are inversely proportional: x ∝ 1/y
or, xy = constant
Therefore, x₁y₁ = x₂y₂
Example:
If more workers are employed to complete the same work, the time required to complete the work decreases proportionally.
Competitive Exam Shortcuts
Convert Percentages into Ratios
Example
25% of A = 50% of B
1/4 A = 1/2 B
A : B = 2 : 1
25% of A = 50% of B
1/4 A = 1/2 B
A : B = 2 : 1
Equality of Percentages
Whenever a question states:
x% of A = y% of B
immediately write
A : B = y : x
The percentages simply interchange.
Example
15% of A = 25% of B
A : B
= 25 : 15
= 5 : 3
x% of A = y% of B
immediately write
A : B = y : x
The percentages simply interchange.
Example
15% of A = 25% of B
A : B
= 25 : 15
= 5 : 3
Difference Represents Remaining Parts
If the ratio is known and the actual difference is given, first determine the value of one part.
Example
A : B = 7 : 5
Difference = 24
Difference in ratio
= 2 parts
One part
= 24 ÷ 2
= 12
A = 84
B = 60
Example
A : B = 7 : 5
Difference = 24
Difference in ratio
= 2 parts
One part
= 24 ÷ 2
= 12
A = 84
B = 60
Total Represents Sum of Parts
If the ratio and total are known:
Value of one part
= Total ÷ Sum of ratio terms
Example
A : B = 3 : 5
Total = 80
One part
= 80 ÷ 8
= 10
A = 30
B = 50
Value of one part
= Total ÷ Sum of ratio terms
Example
A : B = 3 : 5
Total = 80
One part
= 80 ÷ 8
= 10
A = 30
B = 50
Cross Multiplication Saves Time
Instead of finding equivalent ratios step by step, directly use cross multiplication.
If a : b = c : d
then a × d = b × c
If a : b = c : d
then a × d = b × c
Direct Proportion Shortcut
For directly proportional quantities,
New Value / Old Value
= New Quantity / Old Quantity
This avoids writing lengthy proportion equations.
New Value / Old Value
= New Quantity / Old Quantity
This avoids writing lengthy proportion equations.
Inverse Proportion Shortcut
For inversely proportional quantities,
Old Quantity × Old Value
= New Quantity × New Value
This relation is frequently used in questions on work, speed, and efficiency.
Old Quantity × Old Value
= New Quantity × New Value
This relation is frequently used in questions on work, speed, and efficiency.
Continued Proportion
Instead of finding equivalent ratios step by step, directly use cross multiplication.
If a : b = c : d
then a × d = b × c
If a : b = c : d
then a × d = b × c
Finding a Combined Ratio
Make the common term equal first, then combine the ratios.
Example
If a : b = 5 : 2
and b : c = 4 : 5
The common term b has values 2 and 4.
LCM(2, 4) = 4
Multiply the first ratio by 2: a : b = 10 : 4
The second ratio already has: b : c = 4 : 5
Therefore, a : b : c = 10 : 4 : 5
Example
If a : b = 5 : 2
and b : c = 4 : 5
The common term b has values 2 and 4.
LCM(2, 4) = 4
Multiply the first ratio by 2: a : b = 10 : 4
The second ratio already has: b : c = 4 : 5
Therefore, a : b : c = 10 : 4 : 5
Splitting a Combined Ratio
For example:
If A : B : C = 3 : 5 : 7
then immediately write:
A : B = 3 : 5
B : C = 5 : 7
A : C = 3 : 7
If A : B : C = 3 : 5 : 7
then immediately write:
A : B = 3 : 5
B : C = 5 : 7
A : C = 3 : 7
Standard Proportion Transformations
| Transformation | If | Then |
|---|---|---|
| Invertendo | a/b = c/d | b/a = d/c |
| Alternendo | a/b = c/d | a/c = b/d |
| Addendo | a/b = c/d | (a+b)/b = (c+d)/d |
| Subtrahendo | a/b = c/d | (a−b)/b = (c−d)/d |
| Componendo | a/b = c/d | (a+b)/b = (c+d)/d |
| Dividendo | a/b = c/d | (a−b)/b = (c−d)/d |
| Componendo & Dividendo | a/b = c/d | (a+b)/(a−b) = (c+d)/(c−d) |
| Convertendo | a/b = c/d | a/(b−a) = c/(d−c) |
Summary of Ratios and Proportions
| Term | Explanation |
|---|---|
| Ratio (a : b) | Compares two quantities of the same kind. |
| Proportion (a : b = c : d) | States that two ratios are equal. |
| Cross Multiplication | If a/b = c/d, then a × d = b × c. |
| Continued Proportion | If a : b = b : c, then a, b and c are in continued proportion. |
| Mean Proportional | If a : b = b : c, then b = √(ac). |
| Third Proportional | If a : b = b : c, then c is called the third proportional to a and b. |
| Duplicate Ratio | The ratio of the squares of two quantities: a² : b². |
| Direct Proportion | Both quantities increase or decrease together. |
| Inverse Proportion | One quantity increases as the other decreases so that their product remains constant. |
| Compound Ratio | Obtained by multiplying two or more ratios together, e.g., a:b and c:d give ac:bd. |
| Equivalent Ratios | Ratios representing the same relationship, e.g., 2:4 = 1:2. |
Common Mistakes
- Confusing Ratio with Proportion: A ratio compares two quantities (e.g., 3 : 5), whereas a proportion states that two ratios are equal (e.g., 3 : 5 = 6 : 10). Do not use these terms interchangeably.
- Ignoring the Order of Terms: The ratio 3 : 5 is not the same as 5 : 3. Always maintain the order given in the question.
- Adding Ratios Directly: Ratios can only be added after converting them into actual quantities or making their corresponding terms compatible. Never add ratios term by term without justification.
- Using Different Units: Always convert quantities to the same unit before forming a ratio. For example, convert metres to centimetres or hours to minutes before comparison.
- Forgetting to Simplify the Final Ratio: After obtaining a ratio, divide both terms by their HCF to express it in the simplest form.
- Using the Wrong Formula for Combining Ratios: When combining ratios such as a : b and b : c, first make the common term (b) equal before combining. Never combine them directly.
- Applying Direct Proportion to an Inverse Proportion Problem: More workers take fewer days, and higher speed requires less time. Such situations involve inverse proportion, not direct proportion.
- Changing the Ratio After Equal Percentage Increase or Decrease: If all quantities increase or decrease by the same percentage, their ratio remains unchanged.
- Incorrectly Solving Age Ratio Problems: Remember that the difference between two people’s ages always remains constant. Only the ratio changes with time.
- Forgetting Cross Multiplication in Proportion: If a : b = c : d, then a × d = b × c. This is the quickest way to solve most proportion problems.
- Mistakes in Continued Proportion: For numbers in continued proportion, always remember that the square of the middle term equals the product of the extreme terms.
- Not Simplifying Compound Ratios: After multiplying the corresponding terms of the given ratios, always simplify the final compound ratio before writing the answer.
- Not Checking Whether the Answer Is Reasonable: In age problems, verify that the calculated ages are realistic. In money or quantity problems, ensure the values satisfy all the given conditions before finalising the answer.
Practice Questions
Question 1: The ratio of boys to girls in a school is 7 : 5. If there are 120 more boys than girls, find the total number of students in the school.
Question 2: The monthly incomes of A and B are in the ratio 5 : 7. If A earns ₹18,000 less than B, find the monthly income of each person.
Question 3: A sum of ₹24,000 is divided among A, B and C in the ratio 3 : 5 : 7. Find the amount received by each person.
Question 4: The present ages of a father and his son are in the ratio 6 : 1. After 8 years, the ratio of their ages will become 10 : 3. Find their present ages.
Question 5: The ratio of the populations of two towns is 9 : 11. If the larger town has 18,000 more people than the smaller town, find the population of each town.
Question 6: A mixture contains milk and water in the ratio 7 : 3. If the mixture weighs 60 litres, find the quantity of milk and water separately.
Question 7: A school library has books on Science and Mathematics in the ratio 8 : 5. If 90 Mathematics books are added, the ratio becomes 4 : 3. Find the original number of books in each category.
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