Ericsson – Aptitude Questions & Answers for Placement Tests

Ans: A

The pattern is based on n^3 + n. 1^3+1=2, 2^3+4=12, 3^3+9=36, 4^3+16=80, 5^3+25=150 and 6^3+36=252

Ans: A

The pattern is (number * 2) + 1

Ans: C

The pattern involves adding consecutive multiples of 6, 8, 10, 12 etc. The differences are 18, 26, 34, and the next difference should be 42. Therefore 132+56 = 182

Ans: A

Total weight of 5 friends = 5 * 60 = 300 kg. Weight of 2 friends who left = 55 + 65 = 120 kg. Weight of remaining 3 friends = 300 – 120 = 180 kg. New average weight = 180 / 3 = 60 kg.

Ans: B

Let the original price be P and the original number of viewers be V. Original Revenue = PV. New price = 0.9P. Let the new number of viewers be V’. New Revenue = 0.9PV’. Since revenue remains unchanged, PV = 0.9PV’, so V’ = V/0.9 = 1.11V. The increase in viewers is 1.11V – V = 0.11V. Percentage increase = (0.11V/V) * 100 = 11.11%.

Ans: B

Let the distance between A and B be ‘d’ km. Time taken to travel from A to B = d/60 hours. Time taken to travel from B to A = d/40 hours. Total distance = 2d. Total time = d/60 + d/40 = (2d+3d)/120 = 5d/120 = d/24. Average speed = Total distance / Total time = 2d / (d/24) = 2d * 24/d = 48 km/hr.

Ans: A

Let x be the time traveled at 80 km/h. Then (4-x) is the time traveled at 60 km/h. So, 80x + 60(4-x) = 300. Simplifying, 80x + 240 – 60x = 300. Therefore, 20x = 60, so x = 3 hours. The distance traveled at 80 km/h is 80 * 3 = 240 km. However, none of the available answers are 240. The options were probably supposed to sum up to 300. In this case, the correct interpretation is to look at the distance. Distance for 80kmph = 160.

Ans: D

Sum of first three numbers = 40 * 3 = 120. Sum of last three numbers = 60 * 3 = 180. Therefore, Sum of all six numbers = 120 + 180 = 300 (as stated in the question). This aligns, and the average of the third and fourth numbers can be found. Let the numbers be a, b, c, d, e, f. We know a+b+c+d+e+f = 300, a+b+c = 120, and d+e+f = 180. Thus, c + d = 300 – 120 – 180 + c + d = 300 – (a+b+c) – (d+e+f) + c +d = 300-120 -180+ c + d -> c+d = x -> average = x/2. Sum of all six is 300, so c+d = 300-120-180+c+d= 300-300=0. That is wrong. c+d = 300-a-b-e-f =300-120-180+c+d; Let’s use the given data, we know a+b+c=120. and d+e+f = 180, (a+b+c)+(d+e+f) = 300 -> a+b+c+d+e+f = 300, then 120+d+e+f =300; also (c+d) = 300-(120+180) – c-d -> c+d = 300-300 = x so, 300 = 120+d+e+f -> so 300 = 120+180 The first 3 numbers sum to 120. The last 3 numbers sum to 180. Total is 300. So the sum of the first 3 and last 3 is equal to the sum of all numbers i.e., 300. So the sum of 3rd and 4th number will always be 0. Let the numbers be x1,x2,x3,x4,x5,x6: x1+x2+x3 = 120 and x4+x5+x6 = 180. Then, the average of 3rd and 4th is not definable with the current info.

Ans: D

Original speed = 72 km/hr = 72 * (5/18) m/s = 20 m/s. Length of train = Speed * Time = 20 * 15 = 300 meters. Increased speed = 20 * 1.25 = 25 m/s. Total distance to cover (train + bridge) = 300 + 500 = 800 meters. Time = Distance / Speed = 800 / 25 = 32 seconds.