Important Formulae:
1. If we add a positive integer (x) to both terms of the ratio, we have the following results:- If a < b then (a+x) : (b+x) > a : b
- If a > b then (a+x) : (b+x) < a : b
- If a = b then (a+x) : (b+x) = a : b
2. If two quantities are in the ratio a : b, then the first quantity will be a / (a+b) times the total of the two quantities and the second quantity will be b / (a+b) times the total of the two quantities.
3. If a/b=c/d, then a,b,c and d are proportion. This is represented as a : b : : c : d
The following relationships are possible:
(a + b) : b = (c + d) : d -----> COMPONENDO
(a - b) : b = (c - d) : d -----> DIVIDENDO
(a + b) : (a - b) = (c + d) : (c - d) -----> COMPONENDO DIVIDENDO
If a/b = c/d = e/f ......., then each of these ratios is equal to (a + c + e + ...) / (b + d + f + ...)
4. Direct Variation: The ratio of the two quantities is a constant. A/B = k
Inverse Variation: The product of the two quantities is a constant. AB = k
Examples
Example 1. The weights of Ashok and Baskar are in the ratio 4 : 5. If their total weight is 45 kg, find the weight of Baskar.
(A) 20 kg (B) 25 kg (C) 15 kg (D) 30 kg
Solution:
Let the weights of Ashok and Baskar be 4x kg and 5x kg respectively.
4x + 5x = 45 ⇒ x = 5
∴ weight of Baskar = 5x or 25 kg. Answer: B
Example 2. A bag has coins in the denominations of 50 p, 25 p and 20 p in the ratio 4 : 2 : 1. If the total value of the coins is Rs. 54, find the number of 25 p coins in the bag.
(A) 30 (B) 50 (C) 20 (D) 40
Solution:
Let the number of 50 p, 25 p and 20 p coins be 4x, 2x and x respectively.
Total value of the coins
= 50(4x) + 25(2x) + 20x = 5400
⇒ 270x = 5400 ⇒ x = 20
∴ the number of 25 p coins = 2x = 40. Answer: D
Example 3. a is 60% of b, b is 25% of c, c is 40% of d. Find a : d.
(A) 3 : 50 (B) 9 : 20 (C) 3 : 25 (D) 9 : 40
Solution:
Given, a = 60b/100, b = 25c/100, c = 40d/100
a = (60/100) (25/100) (40/100)d
= 3d/50 ⇒ a : d = 3 : 50 Answer: A
Example 4. A party is attended by a total of 60 children. If the ratio of boys and girls is 3 : 2, how many more boys must join the party such that the ratio becomes 7 : 3?
(A) 15 (B) 25 (C) 10 (D) 20
Solution:
Let the number of boys who must join the party be b.
Number of girls in the party = (2/(2 + 3)) (60) = 24
After the boys have joined, number of girls is 3 parts. (Boys/Girls = 7/3)
Boys = 7/3 (Girls)
Number of boys in the party would then be = 7/3 (24) = 56
But, the number of boys originally = (3/(2 + 3)) (60) = 36
Hence, 20 boys must join the party. Answer: D
Example 5. The volume of a cube varies directly with the cube of its side. If three cubes of sides 6 cm, 8 cm and 10 cm are taken together and melted to form a fourth cube, find the side of the fourth cube.
(A) 12 cm (B) 16 cm (C) 10 cm (D) 14 cm
Solution:
Let the side of the fourth cube be y cm.
Volume of a cube = K (Side of that cube)³ where K is a constant (K ≠ 0)
Volumes of the cubes of sides 6cm, 8cm and 10cm are K(6)³, K(8)³ and K(10)³ respectively.
Volume of the fourth cube = Ky³ = K(6)³ + K(8)³ + K(10)³ = K(1728)
Hence y = ∛1728 = 12 Answer: A
Example 6. The ratio of earnings to expenditure of P is 5 : 3 and that of Q is 7 : 6. If the savings of P is double that of Q. Then what could be the ratio of total earnings of P and Q together to the total expenditure of P and Q together?
(A) 3 : 5 (B) 2 : 1 (C) 4 : 3 (D) 5 : 3
Solution:
Let the earnings of P be 5x, then the expenditure of P will be 3x.
∴ Savings of P = 5x - 3x = 2x
Let the earnings of Q be 7y, then the expenditure of Q will be 6y.
∴ Savings of Q = 7y - 6y = y
Given savings of P is double that of Q
∴ 2x = 2 [y] ⇒ x = y
∴ The ratio of total earnings of P and Q to the total expenditure of P and Q
= (5x + 7y)/(3x + 6y) = (5x + 7x)/(3x + 6x) = 12/9 = 4 : 3 Answer: C
Example 7. A stone is dropped from a height of one km. The distance it falls through varies directly with the square of the time taken to fall through that distance. If it travels 64 m in 4 seconds, find the distance the body covers in the 5th second.
(A) 24 m (B) 44 m (C) 36 m (D) 28 m
Solution:
Let the distance travelled by the stone be d m in t seconds, d 𝛂 t²
hence d = kt² where k is the proportionality constant.
k = d/t² = 64/4² = 4
Distance travelled by the stone in the 5th second = distance travelled by the body in the first five seconds - distance travelled by the body in the first 4 seconds
= k(5)² - k(4)² = 9 k or 36 m Answer: C
Example 8. A string is cut into two parts such that the ratio of the lengths of the complete string and the smaller part is 20 times the ratio of the lengths of the smaller part and the larger part. Find the ratio of the length of the string and the square of the length of the smaller part (taken in cm) if the longer part is 4cm.
(A) 5 : 4 (B) 5 : 1 (C) 5 : 3 (D) 5 : 2
Solution:
Let the length of the smaller part be p cm and the length of the string be x cm
Larger part = 4 cm.
Given that, x/p = 20p/4
⇒ x = 5p²
⇒ x/p² = 5 : 1 Answer: B
Example 9. If a : b : c : d is 3 : 4 : 7 : 6 and a + b = 84, then c + d is
(A) 156 (B) 303 (C) 294 (D) 147
Solution:
Given a : b : c : d is 3 : 4 : 7 : 6
let a = 3x then b = 4x, c = 7x, d = 6x
a + b = 3x + 4x = 7x
Given a + b = 84
⇒ 7x = 84 ⇒ x = 12
c + d = 7x + 6x = 13x = 13 (12) = 156. Answer: A
Example 10. The electricity bill for a month varies as the number of units consumed. The charge per unit is Rs.1.35 upto 50 units used. If the number of units consumed is more than 50 then the cost of the additional units is Rs.2.70. If the consumption in the first month is 97 units, then what should be the consumption in the second month be, such that the average for the two months combined is Rs. 135 per month?
(A) 56 (B) 57 (C) 46 (D) 53
Solution:
Number of units consumed in the first month = 97 = 50 + 47.
The bill for the first month = 50 (1.35) + 47 (2.7) = 67.5 + 126.9 = 194.4
The average bill for two months = Rs.135
Sum of the two bills = 2 (135) = Rs.270
The second month bill is = 270 - 194.4 = Rs. 75.6 which is greater than Rs. 67.5
∴ In the second month too the number of units consumed is more than 50.
∴ The charge for consumed in excess of 50 units = Rs.75.6 - Rs.67.5 = Rs.8.1
Number of additional units = 8.1/2.7 = 3
∴ Total number of units consumed in the second month = 53. Answer: D
(A) 20 kg (B) 25 kg (C) 15 kg (D) 30 kg
Solution:
Let the weights of Ashok and Baskar be 4x kg and 5x kg respectively.
4x + 5x = 45 ⇒ x = 5
∴ weight of Baskar = 5x or 25 kg. Answer: B
(A) 30 (B) 50 (C) 20 (D) 40
Solution:
Let the number of 50 p, 25 p and 20 p coins be 4x, 2x and x respectively.
Total value of the coins
= 50(4x) + 25(2x) + 20x = 5400
⇒ 270x = 5400 ⇒ x = 20
∴ the number of 25 p coins = 2x = 40. Answer: D
Example 3. a is 60% of b, b is 25% of c, c is 40% of d. Find a : d.
(A) 3 : 50 (B) 9 : 20 (C) 3 : 25 (D) 9 : 40
Solution:
Given, a = 60b/100, b = 25c/100, c = 40d/100
a = (60/100) (25/100) (40/100)d
= 3d/50 ⇒ a : d = 3 : 50 Answer: A
Example 4. A party is attended by a total of 60 children. If the ratio of boys and girls is 3 : 2, how many more boys must join the party such that the ratio becomes 7 : 3?
(A) 15 (B) 25 (C) 10 (D) 20
Solution:
Let the number of boys who must join the party be b.
Number of girls in the party = (2/(2 + 3)) (60) = 24
After the boys have joined, number of girls is 3 parts. (Boys/Girls = 7/3)
Boys = 7/3 (Girls)
Number of boys in the party would then be = 7/3 (24) = 56
But, the number of boys originally = (3/(2 + 3)) (60) = 36
Hence, 20 boys must join the party. Answer: D
Example 5. The volume of a cube varies directly with the cube of its side. If three cubes of sides 6 cm, 8 cm and 10 cm are taken together and melted to form a fourth cube, find the side of the fourth cube.
(A) 12 cm (B) 16 cm (C) 10 cm (D) 14 cm
Solution:
Let the side of the fourth cube be y cm.
Volume of a cube = K (Side of that cube)³ where K is a constant (K ≠ 0)
Volumes of the cubes of sides 6cm, 8cm and 10cm are K(6)³, K(8)³ and K(10)³ respectively.
Volume of the fourth cube = Ky³ = K(6)³ + K(8)³ + K(10)³ = K(1728)
Hence y = ∛1728 = 12 Answer: A
Example 6. The ratio of earnings to expenditure of P is 5 : 3 and that of Q is 7 : 6. If the savings of P is double that of Q. Then what could be the ratio of total earnings of P and Q together to the total expenditure of P and Q together?
(A) 3 : 5 (B) 2 : 1 (C) 4 : 3 (D) 5 : 3
Solution:
Let the earnings of P be 5x, then the expenditure of P will be 3x.
∴ Savings of P = 5x - 3x = 2x
Let the earnings of Q be 7y, then the expenditure of Q will be 6y.
∴ Savings of Q = 7y - 6y = y
Given savings of P is double that of Q
∴ 2x = 2 [y] ⇒ x = y
∴ The ratio of total earnings of P and Q to the total expenditure of P and Q
= (5x + 7y)/(3x + 6y) = (5x + 7x)/(3x + 6x) = 12/9 = 4 : 3 Answer: C
Example 7. A stone is dropped from a height of one km. The distance it falls through varies directly with the square of the time taken to fall through that distance. If it travels 64 m in 4 seconds, find the distance the body covers in the 5th second.
(A) 24 m (B) 44 m (C) 36 m (D) 28 m
Solution:
Let the distance travelled by the stone be d m in t seconds, d 𝛂 t²
hence d = kt² where k is the proportionality constant.
k = d/t² = 64/4² = 4
Distance travelled by the stone in the 5th second = distance travelled by the body in the first five seconds - distance travelled by the body in the first 4 seconds
= k(5)² - k(4)² = 9 k or 36 m Answer: C
Example 8. A string is cut into two parts such that the ratio of the lengths of the complete string and the smaller part is 20 times the ratio of the lengths of the smaller part and the larger part. Find the ratio of the length of the string and the square of the length of the smaller part (taken in cm) if the longer part is 4cm.
(A) 5 : 4 (B) 5 : 1 (C) 5 : 3 (D) 5 : 2
Solution:
Let the length of the smaller part be p cm and the length of the string be x cm
Larger part = 4 cm.
Given that, x/p = 20p/4
⇒ x = 5p²
⇒ x/p² = 5 : 1 Answer: B
Example 9. If a : b : c : d is 3 : 4 : 7 : 6 and a + b = 84, then c + d is
(A) 156 (B) 303 (C) 294 (D) 147
Solution:
Given a : b : c : d is 3 : 4 : 7 : 6
let a = 3x then b = 4x, c = 7x, d = 6x
a + b = 3x + 4x = 7x
Given a + b = 84
⇒ 7x = 84 ⇒ x = 12
c + d = 7x + 6x = 13x = 13 (12) = 156. Answer: A
Example 10. The electricity bill for a month varies as the number of units consumed. The charge per unit is Rs.1.35 upto 50 units used. If the number of units consumed is more than 50 then the cost of the additional units is Rs.2.70. If the consumption in the first month is 97 units, then what should be the consumption in the second month be, such that the average for the two months combined is Rs. 135 per month?
(A) 56 (B) 57 (C) 46 (D) 53
Solution:
Number of units consumed in the first month = 97 = 50 + 47.
The bill for the first month = 50 (1.35) + 47 (2.7) = 67.5 + 126.9 = 194.4
The average bill for two months = Rs.135
Sum of the two bills = 2 (135) = Rs.270
The second month bill is = 270 - 194.4 = Rs. 75.6 which is greater than Rs. 67.5
∴ In the second month too the number of units consumed is more than 50.
∴ The charge for consumed in excess of 50 units = Rs.75.6 - Rs.67.5 = Rs.8.1
Number of additional units = 8.1/2.7 = 3
∴ Total number of units consumed in the second month = 53. Answer: D
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