ITC – Aptitude Questions & Answers for Placement Tests
Reviewing Previous Year Questions is a good start. Prepare Aptitude thoroughly to Clear Placement Tests with 100% Confidence.
Q.1 If √(1 – x/169) = 12/13, then the value of x is
Check Solution
Ans: B
Squaring both sides, 1 – x/169 = 144/169. Therefore, x/169 = 1 – 144/169 = 25/169. Hence, x = 25.
Q.2 If x and y are even numbers, then which of the following MUST also be even?
Check Solution
Ans: D
The sum of two even numbers is even, and the square of an even number is even. Therefore, x^2 + y^2 is even.
Q.3 343 + 1 is divisible by
Check Solution
Ans: D
343 + 1 = 344. Checking divisibility, 344/8 = 43.
Q.4 The single discount equal to two consecutive discounts of 20% and 30% is
Check Solution
Ans: C
Let the marked price be 100. After a discount of 20%, the price becomes 80. After a further discount of 30%, the price becomes 80 * 0.70 = 56. The overall discount is 100-56 = 44. Hence the single discount is 44%.
Q.5 If a : b = 2 : 3 and b : c = 4 : 5, then (a2 : b2 : c2) is
Check Solution
Ans: D
a/b = 2/3 and b/c = 4/5. To find a common ratio, multiply the ratios. First, find a:b:c. Multiply first ratio by 4: a:b = 8:12. Multiply second ratio by 3: b:c = 12:15. So, a:b:c = 8:12:15. Now find the ratio of squares. a2:b2:c2 = 82:122:152 = 64:144:225
Q.6 A mixture contains milk and water in the ratio of 4 : 1. If 5 liters of water is added to the mixture, the ratio of milk to water becomes 3 : 2. What is the original quantity of milk in the mixture?
Check Solution
Ans: A
Let the original quantity of milk be 4x and water be x. After adding 5 liters of water, the ratio becomes (4x)/(x+5) = 3/2. Solving for x: 8x = 3x + 15 => 5x = 15 => x = 3. Original quantity of milk is 4x = 4 * 3 = 12 liters.
Q.7 The ratio of the present ages of a father and his son is 7:2. After 10 years, the ratio of their ages will be 9:4. What is the father’s present age?
Check Solution
Ans: B
Let the father’s present age be 7x and the son’s present age be 2x. After 10 years, the father’s age will be 7x+10 and the son’s age will be 2x+10. Thus, (7x+10)/(2x+10) = 9/4. 4(7x+10) = 9(2x+10). 28x + 40 = 18x + 90. 10x = 50. x = 5. Therefore, the father’s present age is 7*5 = 35 years.
Q.8 The ratio of the ages of two brothers is 3 : 5. Ten years ago, the ratio of their ages was 1 : 2. The difference in their current ages is
Check Solution
Ans: C
Let the current ages of the brothers be 3x and 5x. Ten years ago, their ages were 3x – 10 and 5x – 10. So, (3x – 10) / (5x – 10) = 1 / 2. 2(3x – 10) = 5x – 10 6x – 20 = 5x – 10 x = 10 The current ages are 30 and 50. The difference is 50 – 30 = 20.
Q.9 The ratio of the ages of a father and son is 7:2. After 10 years, the ratio of their ages will be 9:4. What is the present age of the father?
Check Solution
Ans: B
Let the present ages of father and son be 7x and 2x respectively. After 10 years, their ages will be 7x+10 and 2x+10. (7x + 10) / (2x + 10) = 9/4 4(7x + 10) = 9(2x + 10) 28x + 40 = 18x + 90 10x = 50 x = 5 Present age of the father = 7x = 7 * 5 = 35
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