EXL – 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 The simple interest on Rs. 6000 in 4 years at the rate of y% per annum equals the simple interest on Rs. 8000 at the rate of 10% per annum in 3 years. The value of y is
Check Solution
Ans: B
SI = (P * R * T) / 100. For the first case: SI = (6000 * y * 4) / 100 = 240y For the second case: SI = (8000 * 10 * 3) / 100 = 2400 Equating both, 240y = 2400, hence y = 10%
Q.2 If a + 1/a = 4, then the value of a3 + 1/a3 is
Check Solution
Ans: A
Cube both sides of a + 1/a = 4: (a + 1/a)3 = 43 a3 + 3a + 3/a + 1/a3 = 64 a3 + 1/a3 + 3(a + 1/a) = 64 a3 + 1/a3 + 3(4) = 64 a3 + 1/a3 = 64 – 12 a3 + 1/a3 = 52
Q.3 If (5a + 2)/a + (5b + 2)/b + (5c + 2)/c = 10, then the value of a/ (ab+ac+bc) is
Check Solution
Ans: A
(5a+2)/a + (5b+2)/b + (5c+2)/c = 10 => 5 + 2/a + 5 + 2/b + 5 + 2/c = 10 => 2/a + 2/b + 2/c = -5 => 2(bc + ac + ab)/abc = -5 => bc + ac + ab = -5abc/2. Now we want to find a/ (ab+ac+bc). Since (5a+2)/a + (5b+2)/b + (5c+2)/c = 10 , we have 2/a + 2/b + 2/c = -5 => 1/a + 1/b + 1/c = -5/2. Hence, (ab+ac+bc)/abc = -5/2 and a/(ab+ac+bc) = a/(-5abc/2) = -2/(5bc) . However the correct answer is 1/2. The equation implies 2/a + 2/b + 2/c = -5 => 2(1/a + 1/b + 1/c) = -5 => 1/a + 1/b + 1/c = -5/2 => (ab + bc + ac)/abc = -5/2. The prompt provides insufficient info, and some adjustments are needed. Let’s assume the correct equation leads to 2/a+2/b+2/c = -10. Therefore 2/a + 2/b + 2/c = 10 – 15 = -5. This is a mistake in the initial premise of the question.
Q.4 Find the simplest value of 3√27 – 2√12 + √75 (given √3 = 1.732)
Check Solution
Ans: A
3√27 – 2√12 + √75 = 3√(9*3) – 2√(4*3) + √(25*3) = 3*3√3 – 2*2√3 + 5√3 = 9√3 – 4√3 + 5√3 = 10√3 = 10 * 1.732 = 17.32
Q.5 If x2 + y2 + z2 = 2(x – y + z) – 3 and x, y, z are real, then the value of (x + y + z) is
Check Solution
Ans: A
Rearrange the given equation: x2 – 2x + y2 + 2y + z2 – 2z + 3 = 0. Completing the square: (x2 – 2x + 1) + (y2 + 2y + 1) + (z2 – 2z + 1) = 0, which simplifies to (x – 1)2 + (y + 1)2 + (z – 1)2 = 0. Since squares of real numbers are non-negative, each term must be zero. Hence x = 1, y = -1, and z = 1. Therefore, x + y + z = 1 – 1 + 1 = 1.
Q.6 A number z when divided by 420 leaves 27 as a remainder. The same number when divided by 20 leaves w as a remainder. The value of w is
Check Solution
Ans: A
x = 420k + 27. Since we are dividing by 20, we can focus on the remainder part. 27 divided by 20 leaves a remainder of 7.
Q.7 The sum of the digits of a two-digit number is 9. If the digits are reversed, the new number is 27 more than the original number. Find the original number.
Check Solution
Ans: A
Let the tens digit be x and the units digit be y. Then x + y = 9. The original number is 10x + y, and the reversed number is 10y + x. So, 10y + x = 10x + y + 27. Simplifying, 9y – 9x = 27, or y – x = 3. We have two equations: x + y = 9 and y – x = 3. Adding them, 2y = 12, so y = 6. Then x = 3. The original number is 36.
Q.8 If a + b + c = 6, a2 + b2 + c2 = 14 and ab + bc + ca = 11, then the value of a3 + b3 + c3 – 3abc is
Check Solution
Ans: A
We know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca). Substituting the given values: 62 = 14 + 2(11) or 36 = 14 + 22 = 36, this is correct. Now, a3 + b3 + c3 – 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca). So, substituting the given values: (6)(14 – 11) = 6 * 3 = 18.
Q.9 In a right-angled triangle PQR, ∠Q = 90°, and M is a point on PR such that QM is perpendicular to PR. If ∠RPM = 30°, what is the measure of ∠MQR?
Check Solution
Ans: A
Since ∠Q = 90°, and ∠RPM = 30°, then ∠R = 90° – 30° = 60°. In triangle QMR, ∠QMR = 90°. Therefore, ∠MQR = 90° – ∠R = 90° – 60° = 30°.
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