Walmart – 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 A train travels a certain distance at a uniform speed. If the train had traveled 10 km/hr faster, it would have taken 2 hours less. If the train had traveled 10 km/hr slower, it would have taken 3 hours more. What is the distance the train travels?

Check Solution

Ans: A

Let the original speed be ‘s’ km/hr and time be ‘t’ hours. Distance = s*t. From the problem, (s+10)(t-2) = s*t => st – 2s + 10t – 20 = st => 2s – 10t = -20 => s – 5t = -10 …(1) Also, (s-10)(t+3) = s*t => st + 3s – 10t – 30 = st => 3s – 10t = 30 …(2) Multiplying equation (1) by 2, we get 2s – 10t = -20. Subtracting this from equation (2), we get s = 50. Substituting s = 50 in equation (1), we have 50 – 5t = -10 => 5t = 60 => t = 12. Distance = s*t = 50*12 = 600 km.

Q.2 A boat travels a certain distance downstream in 3 hours and the same distance upstream in 5 hours. If the speed of the stream is 3 km/hr, what is the speed of the boat in still water?

Check Solution

Ans: A

Let the speed of the boat in still water be ‘x’ km/hr. Downstream speed = (x + 3) km/hr Upstream speed = (x – 3) km/hr Distance = Speed x Time Since the distance is the same: 3(x + 3) = 5(x – 3) 3x + 9 = 5x – 15 2x = 24 x = 12

Q.3 A farmer sold 40% of his mangoes to the first customer. He then sold 60% of the remaining mangoes to the second customer. Finally, he sold 50% of the mangoes left to the third customer. If the farmer was left with 30 mangoes, how many mangoes did he initially have?

Check Solution

Ans: A

Let ‘x’ be the initial number of mangoes. After selling to the first customer, remaining mangoes = x * (1 – 0.40) = 0.6x After selling to the second customer, remaining mangoes = 0.6x * (1 – 0.60) = 0.6x * 0.4 = 0.24x After selling to the third customer, remaining mangoes = 0.24x * (1 – 0.50) = 0.24x * 0.5 = 0.12x We are given that 0.12x = 30 Therefore, x = 30 / 0.12 = 250

Q.4 The ratio of the present ages of a father and his son is 7:2. Five years ago, the ratio of their ages was 6:1. What will be the father’s age after 10 years?

Check Solution

Ans: A

Let the present age of the father be 7x and the present age of the son be 2x. Five years ago, the father’s age was 7x – 5 and the son’s age was 2x – 5. We are given that (7x – 5) / (2x – 5) = 6/1. Therefore, 7x – 5 = 12x – 30. 5x = 25, so x = 5. The father’s present age is 7 * 5 = 35 years. The father’s age after 10 years will be 35 + 10 = 45 years.

Q.5 If sin α = 12/13 and α is acute, then cot α is

Check Solution

Ans: B

Since sin α = 12/13, we can consider a right-angled triangle where the opposite side is 12 and the hypotenuse is 13. Using the Pythagorean theorem, the adjacent side is sqrt(13^2 – 12^2) = sqrt(169 – 144) = sqrt(25) = 5. cot α = adjacent/opposite = 5/12.

Q.6 A tree casts a shadow of 15 meters at a certain time of day. Nearby, a 3-meter tall pole casts a shadow of 2 meters at the same time. What is the height of the tree?

Check Solution

Ans: B

The ratio of the tree’s height to its shadow length is equal to the ratio of the pole’s height to its shadow length. Let h be the height of the tree. Then h/15 = 3/2. Solving for h, we get h = (3/2) * 15 = 22.5 meters.