Question 1
The students in three classes are in the ratio 2 : 3 : 5. If 40 students are increased in each class the ratio changes to 4 : 5 : 7. Originally the total number of students was:
Explanation
Let the original numbers of students be 2x, 3x and 5x.
After the increase: (2x+40) : (3x+40) : (5x+40) = 4 : 5 : 7
From the first two terms: 5(2x+40) = 4(3x+40)
10x + 200 = 12x + 160
2x = 40, so x = 20
Original total = 2x + 3x + 5x = 10x = 10 × 20 = 200
After the increase: (2x+40) : (3x+40) : (5x+40) = 4 : 5 : 7
From the first two terms: 5(2x+40) = 4(3x+40)
10x + 200 = 12x + 160
2x = 40, so x = 20
Original total = 2x + 3x + 5x = 10x = 10 × 20 = 200
Question 2
A bag contains coins of denominations 1 rupee, 2 rupee and 5 rupees. Their numbers are in the ratio 4:3:2. If the bag has a total of Rs. 1800, then find the number of 2 rupee coins.
Explanation
Let the number of 1-rupee, 2-rupee and 5-rupee coins be 4k, 3k and 2k respectively.
Total value = 1(4k) + 2(3k) + 5(2k) = 4k + 6k + 10k = 20k
20k = 1800, so k = 90
Number of 2-rupee coins = 3k = 3 × 90 = 270
Total value = 1(4k) + 2(3k) + 5(2k) = 4k + 6k + 10k = 20k
20k = 1800, so k = 90
Number of 2-rupee coins = 3k = 3 × 90 = 270
Question 3
A bag contains 25 paise, 10 paise and 5 paise coins in the ratio 3:2:1. If the total value is Rs. 40, the number of 5 paise coins is:
Explanation
Let the number of 25p, 10p, 5p coins be 3k, 2k, k respectively.
Total value (in paise) = 25(3k) + 10(2k) + 5(k) = 75k + 20k + 5k = 100k paise = k rupees
k = 40 (since total value = Rs. 40)
Number of 5 paise coins = k = 40
Total value (in paise) = 25(3k) + 10(2k) + 5(k) = 75k + 20k + 5k = 100k paise = k rupees
k = 40 (since total value = Rs. 40)
Number of 5 paise coins = k = 40
Question 4
What must be added to each term of the ratio 49:68 so that it becomes 3:4?
Explanation
Let x be added to each term.
49 + x68 + x = 34
4(49 + x) = 3(68 + x)
196 + 4x = 204 + 3x
x = 8
49 + x68 + x = 34
4(49 + x) = 3(68 + x)
196 + 4x = 204 + 3x
x = 8
Question 5
The expenditure and savings of a person are in the ratio 4:1. If his savings are increased by 25% of his income, then what is the new ratio of his expenditure and savings?
Explanation
Let income = I. Expenditure = 4I5, Savings = I5
New savings = I5 + 0.25I = 4I20 + 5I20 = 9I20
New expenditure = I - 9I20 = 11I20
New ratio = 11I20 : 9I20 = 11 : 9
New savings = I5 + 0.25I = 4I20 + 5I20 = 9I20
New expenditure = I - 9I20 = 11I20
New ratio = 11I20 : 9I20 = 11 : 9
Question 6
P, Q and R are three cities. The ratio of average temperature between P and Q is 11:12 and that between P and R is 9:8. The ratio between the average temperature of Q and R is:
Explanation
P:Q = 11:12 ⇒ P = 11k, Q = 12k
P:R = 9:8 ⇒ P = 9m, R = 8m
Equating P: 11k = 9m ⇒ k:m = 9:11. Take k=9, m=11:
P = 99, Q = 12(9) = 108, R = 8(11) = 88
Q:R = 108:88 = 27:22
P:R = 9:8 ⇒ P = 9m, R = 8m
Equating P: 11k = 9m ⇒ k:m = 9:11. Take k=9, m=11:
P = 99, Q = 12(9) = 108, R = 8(11) = 88
Q:R = 108:88 = 27:22
Question 7
The third proportional between (a2-b2) and (a+b)2 is:
Explanation
If x is the third proportional: (a2-b2) : (a+b)2 = (a+b)2 : x
x = (a+b)4a2-b2 = (a+b)4(a-b)(a+b) = (a+b)3a-b
x = (a+b)4a2-b2 = (a+b)4(a-b)(a+b) = (a+b)3a-b
Question 8
The ratio of the number of boys and number of girls in a school is found to be 15:32. How many boys and an equal number of girls should be added to bring the ratio to 2:3?
Explanation
Let boys = 15, girls = 32 (matching the given ratio). Let x be added to each.
15 + x32 + x = 23
3(15 + x) = 2(32 + x)
45 + 3x = 64 + 2x
x = 19
15 + x32 + x = 23
3(15 + x) = 2(32 + x)
45 + 3x = 64 + 2x
x = 19
Question 9
The ratio of the prices of two Fans was 16 : 23. Two years later, when the price of the first has increased by 10% and that of the second by Rs. 477, the ratio of the prices becomes 11 : 20. Find the original prices of the two Fans.
Explanation
Let the original prices = 16x and 23x.
New prices: 16x × 1.1 = 17.6x and (23x + 477)
17.6x / (23x + 477) = 11/20
352x = 253x + 5247
99x = 5247, so x = 53
Original prices = 16 × 53 = Rs. 848 and 23 × 53 = Rs. 1,219
New prices: 16x × 1.1 = 17.6x and (23x + 477)
17.6x / (23x + 477) = 11/20
352x = 253x + 5247
99x = 5247, so x = 53
Original prices = 16 × 53 = Rs. 848 and 23 × 53 = Rs. 1,219
Question 10
If a : b = 3 : 4, the value of (2a + 3b) : (3a + 4b) is
Explanation
Let a = 3k, b = 4k.
2a + 3b = 6k + 12k = 18k
3a + 4b = 9k + 16k = 25k
Required ratio = 18k : 25k = 18 : 25
2a + 3b = 6k + 12k = 18k
3a + 4b = 9k + 16k = 25k
Required ratio = 18k : 25k = 18 : 25
Question 11
The third proportional to 49 and 21 is
Explanation
Third proportional to a and b = b2/a
= 212/49 = 441/49 = 9
= 212/49 = 441/49 = 9
Question 12
The ratio of income of A and B is 5 : 4 and their expenditure is 3 : 2. If at the end of the year each saves Rs. 1,600, then the income of A is:
Explanation
Let income of A = 5x, income of B = 4x, expenditure of A = 3y, expenditure of B = 2y.
Savings: 5x - 3y = 1600 ... (i)
4x - 2y = 1600 ⇒ 2x - y = 800 ⇒ y = 2x - 800 ... (ii)
Substituting (ii) in (i): 5x - 3(2x - 800) = 1600 ⇒ 5x - 6x + 2400 = 1600 ⇒ x = 800
Income of A = 5 × 800 = Rs. 4,000
Savings: 5x - 3y = 1600 ... (i)
4x - 2y = 1600 ⇒ 2x - y = 800 ⇒ y = 2x - 800 ... (ii)
Substituting (ii) in (i): 5x - 3(2x - 800) = 1600 ⇒ 5x - 6x + 2400 = 1600 ⇒ x = 800
Income of A = 5 × 800 = Rs. 4,000
Question 13
The mean proportional between 12x2 and 27y2 is:
Explanation
Mean proportional = √12x2 × 27y2 = √324x2y2 = 18xy
Question 14
The ratio of two numbers is 3 : 4. The difference of their squares is 28. The greater number is:
Explanation
Let the numbers be 3x and 4x.
(4x)2 - (3x)2 = 28
16x2 - 9x2 = 7x2 = 28
x2 = 4, so x = 2
Greater number = 4x = 8
(4x)2 - (3x)2 = 28
16x2 - 9x2 = 7x2 = 28
x2 = 4, so x = 2
Greater number = 4x = 8
Question 15
The prices of a scooter and a moped are in the ratio 7 : 9. The price of the moped is Rs. 1,600 more than that of the scooter. Then the price of the moped is:
Explanation
Let scooter price = 7x, moped price = 9x.
9x - 7x = 1600
2x = 1600, so x = 800
Moped price = 9 × 800 = Rs. 7,200
9x - 7x = 1600
2x = 1600, so x = 800
Moped price = 9 × 800 = Rs. 7,200
