Sandisk – Aptitude Questions & Answers for Placement Tests
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Q.1 The sum of the digits of a three-digit number is 14. The hundreds digit is twice the tens digit. If the digits are reversed, the new number is 396 less than the original number. Find the original number.
Check Solution
Ans: C
Let the three-digit number be 100a + 10b + c. Given: a + b + c = 14 and a = 2b. Also, 100c + 10b + a = 100a + 10b + c – 396. Simplifying the last equation, we get 99a – 99c = 396, so a – c = 4. Substitute a = 2b in a + b + c = 14 to get 2b + b + c = 14 or 3b + c = 14. Since a – c = 4 and a = 2b, we have 2b – c = 4. Subtracting this equation from 3b + c = 14 gives 5b = 10, thus b = 2. Then, a = 2b = 4. And c = a – 4 = 0. So, we have 428.
Q.2 The difference between a two-digit number and the number formed by reversing its digits is 27. The sum of the digits is 9. Find the number.
Check Solution
Ans: A
Let the tens digit be x and the units digit be y. The number is 10x + y. The reversed number is 10y + x. We are given: (10x + y) – (10y + x) = 27 x + y = 9 Simplifying the first equation: 9x – 9y = 27 x – y = 3 Now we have two equations: x + y = 9 x – y = 3 Adding the equations: 2x = 12 x = 6 Substituting x = 6 into x + y = 9: 6 + y = 9 y = 3 The original number is 10(6) + 3 = 63.
Q.3 A contractor hired 100 workers to complete a project in 120 days. After 60 days, he realized that only 1/3rd of the work was done. How many more workers does he need to hire to complete the project on time?
Check Solution
Ans: B
Work remaining = 2/3. Time remaining = 60 days. Workers needed = (100 * (2/3)) / (1/3) = 200 workers. Extra workers needed = 200 – 100 = 100
Q.4 If 12 typists can type a manuscript in 18 days by working 6 hours a day, how many days will it take for 15 typists to type the same manuscript, working 9 hours a day?
Check Solution
Ans: B
Let D be the number of days. Using the formula: M1 * D1 * H1 = M2 * D2 * H2 / Work. So, 12 * 18 * 6 = 15 * D * 9 => D = (12 * 18 * 6) / (15 * 9) = 14.4 days, which can be approximated to 14
Q.5 A train travels a certain distance at 40 km/hr and returns the same distance at 60 km/hr. The average speed for the entire journey is:
Check Solution
Ans: B
Let the distance be d. Time taken to travel the first half (at 40 km/hr) = d/40. Time taken to travel the second half (at 60 km/hr) = d/60. Total distance = 2d. Total time = d/40 + d/60 = (3d + 2d)/120 = 5d/120 = d/24. Average speed = Total distance / Total time = 2d / (d/24) = 48 km/hr
Q.6 A train travels a certain distance at 60 km/hr and then travels the same distance at 40 km/hr. What is the average speed of the train for the entire journey?
Check Solution
Ans: B
Let the distance be ‘d’. Time taken for the first part of the journey = d/60 Time taken for the second part of the journey = d/40 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
Q.7 25% of 120 + 120% of 25 = ?
Check Solution
Ans: B
(25/100)*120 + (120/100)*25 = 30 + 30 = 60
Q.8 Approximate value of Cube root of 1330 + 215 = ?
Check Solution
Ans: B
Cube root of 1330 is approximately 11. Adding 215 results in approximately 226. The closest option is 15.
Q.9 (789 – 259 + ?)/3 = 200
Check Solution
Ans: B
(789 – 259 + ?)/3 = 200 => 530 + ? = 600 => ? = 70
Q.10 The difference between Simple Interest and Compound Interest on a certain sum of money for 2 years at 12% per annum is Rs. 432. Find the sum.
Check Solution
Ans: B
Let the sum be P. SI = P*R*T/100 = P*12*2/100 = 24P/100 CI = P(1 + R/100)^T – P = P(1 + 12/100)^2 – P = P(1.12)^2 – P = 1.2544P – P = 0.2544P CI – SI = 0.2544P – 0.24P = 0.0144P 0.0144P = 432 P = 432/0.0144 = 30000
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