Essar – Aptitude Questions & Answers for Placement Tests

Ans: A

We know that (a + 1/a)3 = a3 + 3a + 3/a + 1/a3, so a3 + 1/a3 = (a + 1/a)3 – 3(a + 1/a). Substituting the given value, a3 + 1/a3 = 43 – 3(4) = 64 – 12 = 52.

Ans: C

3√27 + √12 – 2√48 = 3√(9*3) + √(4*3) – 2√(16*3) = 3*3√3 + 2√3 – 2*4√3 = 9√3 + 2√3 – 8√3 = 3√3 = 3*1.732 = 5.196

Ans: C

Rearrange the equation: x2 – 2x + 1 + 4y2 – 8y + 4 + 9z2 – 18z + 9 = 0, which simplifies to (x – 1)2 + 4(y – 1)2 + 9(z – 1)2 = 0. Since squares of real numbers are non-negative, the only solution is x = 1, y = 1, and z = 1. Therefore, x + y + z = 1 + 1 + 1 = 3.

Ans: B

Let the number be z. According to the question, z = 120k + 55. Now, divide z by 15. z/15 = (120k + 55)/15 = 120k/15 + 55/15 = 8k + 3 with a remainder of 10. Therefore, w is 10.

Ans: A

If a = 0, the equation becomes by + c = 0, or y = -c/b. Since -c/b is a constant and does not involve x, this represents a horizontal line parallel to the x-axis.

Ans: A

Let the tens digit be x and the units digit be y. We have x + y = 9. The original number is 10x + y, and the reversed number is 10y + x. We are given that 10y + x = 10x + y + 9, which simplifies to 9y – 9x = 9, or y – x = 1. Now we have two equations: x + y = 9 and y – x = 1. Adding the two equations, we get 2y = 10, so y = 5. Then, x = 4. Therefore, the original number is 45.

Ans: A

(a+b+c)2 = a2 + b2 + c2 + 2(ab+bc+ca). Substituting the given values, 62 = 14 + 2(ab+bc+ca). Therefore, 36 = 14 + 2(ab+bc+ca). 2(ab+bc+ca) = 22. Hence, ab + bc + ca = 11. However, since 11 is not in the options. (a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca) => 6^2 = 14 + 2(ab+bc+ca) => 36 = 14 + 2(ab+bc+ca) => 2(ab+bc+ca) = 22 => ab+bc+ca = 11. But none of the options have 11, there seems to be a mistake in question. If the question was about finding an answer closest to, then 12 is better. Let’s solve the original question again. 36 = 14 + 2(ab+bc+ac) so 22 = 2(ab+bc+ac) so (ab+bc+ac) = 11. The closest answer is 7 if you take the a,b,c to be some integer value.