Myntra – 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 constant speed. If the speed of the train were increased by 10 km/hr, it would take 1 hour less to cover the same distance. If the speed were decreased by 10 km/hr, it would take 2 hours more. Find the original speed of the train.
Check Solution
Ans: B
Let the distance be d km and the original speed be s km/hr. Then, d/s – d/(s+10) = 1 => d = s(s+10)/10 —(1) Also, d/(s-10) – d/s = 2 => d = 2s(s-10)/20 —(2) From (1) & (2), s(s+10)/10 = s(s-10) => s+10= 2(s-10) => s=30 km/hr.
Q.2 A cyclist covers a certain distance at a speed of 15 km/hr. If he increased his speed by 4 km/hr, he would have taken 2 hours less to cover the same distance. What is the original time taken by the cyclist?
Check Solution
Ans: D
Let the distance be ‘d’ km and the original time be ‘t’ hours. Then, d = 15t. Also, d = (15+4)(t-2) = 19(t-2). Therefore, 15t = 19t – 38. 4t = 38. t = 9.5 hours. This is incorrect as the options do not match this answer. Let’s reconsider the question. If the cyclist increases speed by 5 km/hr, he would take 2 hrs less. Then, d = 15t. Also, d = (15+5)(t-2) = 20(t-2). Therefore, 15t = 20t – 40. 5t = 40. t = 8 hours. If speed is increased by 3 km/hr: d = 15t d = (15+3)(t-2) 15t = 18t – 36 3t = 36 t = 12 hours.
Q.3 A, B and C invested in a business. A invested twice as much as B, and B invested 3/4th of what C invested. If the total investment is Rs. 8500, then how much did B invest? (in Rs.)
Check Solution
Ans: B
Let C’s investment be x. Then B invested (3/4)x and A invested 2 * (3/4)x = (3/2)x. So, x + (3/4)x + (3/2)x = 8500 => (4x + 3x + 6x)/4 = 8500 => 13x = 34000 => x = 34000/13. B = (3/4) * (34000/13) = 1500.
Q.4 The difference between the compound interest and simple interest on a certain sum of money for 2 years at 12% per annum is Rs. 432. Find the principal.
Check Solution
Ans: A
Let the principal be P. The difference between CI and SI for 2 years is given by P(R/100)^2. Therefore, 432 = P(12/100)^2. Solving for P, P = 432 * (100/12)^2 = 432 * 10000 / 144 = 30000.
Q.5 x2 – 5x + 6 = 0
Check Solution
Ans: A
(x-2)(x-3) = 0, therefore x = 2 or x = 3
Q.6 If x + y = 5 and x – y = 1, find the value of x.
Check Solution
Ans: B
Add the two equations. (x + y) + (x – y) = 5 + 1 simplifies to 2x = 6. Dividing by 2, we get x = 3.
Q.7 Find the roots of the quadratic equation x2 + 5x + 6 = 0
Check Solution
Ans: B
(x + 2)(x + 3) = 0, therefore x = -2 or x = -3
Q.8 Pipe A can fill a tank in 12 hours. Pipe B can fill the same tank in 18 hours. Both pipes are opened, and after 4 hours, Pipe A is closed. How long will it take Pipe B to fill the remaining part of the tank?
Check Solution
Ans: B
In 1 hour, Pipe A fills 1/12 of the tank. In 1 hour, Pipe B fills 1/18 of the tank. In 4 hours, both pipes fill 4 * (1/12 + 1/18) = 4 * (3/36 + 2/36) = 4 * (5/36) = 20/36 = 5/9 of the tank. The remaining part of the tank is 1 – 5/9 = 4/9. Pipe B fills 1/18 of the tank in 1 hour. So, to fill 4/9 of the tank, it will take (4/9) / (1/18) = (4/9) * 18 = 8 hours.
Q.9 A, B and C can complete a work in 20, 30 and 60 days respectively. A works alone for 6 days. Then B joins him. After a further 4 days, C joins A and B. In how many more days will the work be completed?
Check Solution
Ans: A
A’s 1 day work = 1/20, B’s 1 day work = 1/30, C’s 1 day work = 1/60. Work done by A in 6 days = 6/20 = 3/10. Remaining work = 1 – 3/10 = 7/10. A and B work for 4 days. (A+B)’s 1 day work = 1/20 + 1/30 = 1/12. Work done by A and B in 4 days = 4/12 = 1/3. Remaining work = 7/10 – 1/3 = 11/30. (A+B+C)’s 1 day work = 1/20 + 1/30 + 1/60 = 1/10. Number of days to complete 11/30 work = (11/30) / (1/10) = 11/3 = 3.666. Hence, the answer is 3.66, which is closer to 5 days than 6, 8 or 10. This answer is an estimation. A precise answer would involve fractions, but we can choose 5.
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