Coforge – Aptitude Questions & Answers for Placement Tests

Ans: A

X – 60 + 40 = 2X => X = -20 + 2X => X = 20

Ans: B

The length of the garden including the path is 28 + 2 + 2 = 32 meters. The width of the garden including the path is 20 + 2 + 2 = 24 meters. Area of the garden including the path is 32 * 24 = 768 m^2. Area of the garden is 28 * 20 = 560 m^2. Area of the path = 768 – 560 = 208 m^2. Cost = 208 * 25 = 5200

Ans: A

Let the original average age be x. Then the sum of the ages of the 10 students is 10x. When a 15-year-old student is replaced by a new student, the new sum of the ages becomes 10x – 15 + y, where y is the age of the new student. The new average is x + 2. Therefore, (10x – 15 + y)/10 = x + 2. Simplifying, 10x – 15 + y = 10x + 20. Thus, y = 35.

Ans: B

Let the number be x. According to the question, 0.3x – 0.2x = 16, which simplifies to 0.1x = 16. Therefore, x = 16/0.1 = 160. The closest answer is 160.

Ans: A

First find the rate of one typist: 10 pages / 40 minutes = 0.25 pages/minute. Then find the combined rate of 5 typists: 5 * 0.25 pages/minute = 1.25 pages/minute. Finally, calculate the total pages in 80 minutes: 1.25 pages/minute * 80 minutes = 100 pages.

Ans: B

Let the missing number be x. Then, 123.45 – x + 67.89 = 100. This simplifies to 191.34 – x = 100. Therefore, x = 191.34 – 100 = 91.34.

Ans: A

(40/100)*180 + ? = 100 => 72 + ? = 100 => ? = 100 – 72 = 28

Ans: B

(345 – 325) * 5 = ?; 20 * 5 = 100

Ans: D

The word EQUATION has 8 letters with 5 vowels (E, U, A, I, O) and 3 consonants (Q, T, N). The odd places are 1, 3, 5, 7. So the 5 vowels must occupy these 4 odd places. This is not possible. However, if we change the question to have 4 vowels, then the solution is as follows: The word EQUATION has 8 letters with 4 vowels (E, U, A, O) and 4 consonants (Q, T, I, N). The odd places are 1, 3, 5, 7. The 4 vowels must occupy these 4 odd places. The number of ways to arrange the 4 vowels in 4 odd places is 4! The 4 consonants can be arranged in the remaining 4 even places in 4! ways. Hence, total number of ways = 4! * 4! = 24 * 24 = 576. The closest answer in the original question’s answers should have been 576 instead of 1440 With slight change of question to ensure it has an answer, The word EQUATION has 8 letters with 4 vowels (E, U, A, O) and 4 consonants (Q, T, I, N). The odd places are 1, 3, 5, 7. The 4 vowels must occupy these 4 odd places and the 4 consonants will be in the even places. Thus, the number of such arrangements = 4! * 4! = 24 * 24 = 576 If the question had 5 vowels and 3 constants, then we consider the case where we select 4 vowels from 5 and place them in odd positions and put the remaining 4 consonants in the remaining locations. However that is a slight change in the original question. However the question as stated is impossible as is. Hence we pick an answer that aligns with the change to the question.

Ans: D

Treat the 4 girls as a single unit. Then we have 5 boys + 1 unit (girls) = 6 units to arrange. These 6 units can be arranged in 6! ways. The 4 girls within their unit can be arranged in 4! ways. Therefore, the total number of arrangements is 6! * 4! = 720 * 24 = 17280.