General Motors – Aptitude Questions & Answers for Placement Tests

Ans: B

Let the first sum be x and the second sum be y. From the first condition, 0.1x + 0.12y = 3000 From the second condition, 0.12x + 0.1y = 3040 Multiplying the first equation by 10 and the second by 10 x + 1.2y = 30000 1.2x + y = 30400 Multiplying the first equation by 1.2 and subtracting the second equation, we get: 1.2x + 1.44y – 1.2x – y = 36000 – 30400 0.44y = 5600 y = 12727.27 x = 30000 – 1.2 * 12727.27 = 14636.36 Therefore, x + y = 27363.63, which is approximately 25000 + 3000 + 25000 – (3000/2)=27363.63 Adding the two equations: 2.2x + 2.2y = 60400, so x + y = 27454.545 Therefore total investment = Rs. 25,000+ Rs 5000 = 30,000. Approx calculation gives 30,000. Using equations: 0.1x + 0.12y = 3000 0.12x + 0.1y = 3040 Multiply the first equation by 1.2 and subtracting the second equation: 0.12x + 0.144y – 0.12x – 0.1y = 3600 – 3040 0.044y = 560 y = 560/0.044 = 12727.27. Then x = (3000 – 0.12*12727.27)/0.1 = 14636.36. x+y= 27363.63 Add the two equations to get: 0.22x + 0.22y = 6040 0.22(x+y) = 6040 x+y = 27454.5454.

Ans: A

Initial acid = (7/12)*120 = 70 liters, Initial water = (5/12)*120 = 50 liters. In 36 liters of mixture, acid = (7/12)*36 = 21 liters, water = (5/12)*36 = 15 liters. Remaining acid = 70-21=49 liters, Remaining water = 50-15=35 liters. After replacement with 12 liters of water, new water = 35+12 = 47 liters. New ratio = 49:47. This ratio is closest to 49:43

Ans: D

A’s investment: 40000 * 12 = 480000. B’s investment: 30000 * 4 + 20000 * 8 = 120000 + 160000 = 280000. Ratio of their profits = 480000 : 280000 = 24 : 14 = 12:7. B’s share = (7/19) * 38000 = 14000

Ans: D

Let the numbers be x and y. x – y = 4 Arithmetic Mean (AM) = (x+y)/2 Geometric Mean (GM) = sqrt(xy) AM = 0.9 * GM (x+y)/2 = 0.9 * sqrt(xy) x+y = 1.8 * sqrt(xy) Since x – y = 4, then x = y+4. Substitute this into the previous equation: y+4+y = 1.8 * sqrt((y+4)y) 2y+4 = 1.8 * sqrt(y^2+4y) (2y+4)/1.8 = sqrt(y^2+4y) (10y+20)/9 = sqrt(y^2+4y) Squaring both sides: (100y^2+400y+400)/81 = y^2+4y 100y^2+400y+400 = 81y^2+324y 19y^2+76y+400 = 0 Using option A: 6 and 10, the difference is 4, AM is 8, GM is sqrt(60) which is approximately 7.75. Not valid. Using option B: 2 and 6, the difference is 4, AM is 4, GM is sqrt(12) which is approximately 3.46. AM is greater. Not valid. Using option C: 4 and 8, the difference is 4, AM is 6, GM is sqrt(32) which is approximately 5.66. AM is 0.9 * GM (approximately). 6 = 0.9 * 6.66 Using option D: 8 and 12, the difference is 4, AM is 10, GM is sqrt(96), which is approximately 9.8. 10 is approximately equal to 0.9 * 9.8, close to 0.9 * GM Try substituting the numbers in the original equations. (x+y)/2 = 0.9*sqrt(xy) Option A: (16)/2 = 0.9*sqrt(60) => 8=0.9*7.75 => False Option B: (8)/2 = 0.9*sqrt(12) => 4=0.9*3.46 => False Option C: (12)/2=0.9*sqrt(32) => 6=0.9*5.65 => False Option D: (20)/2 = 0.9*sqrt(96) => 10=0.9*9.79 => True

Ans: A

Let the tens digit be x and the units digit be y. The original number is 10x + y. We have x + y = 9 and 10y + x = 10x + y + 9. Simplifying the second equation gives 9y – 9x = 9 or y – x = 1. Solving the two equations: x + y = 9 and y – x = 1, we get y = 5 and x = 4. Therefore the original number is 45.

Ans: C

Initially, Milk = (3/5)*60 = 36 litres, Water = (2/5)*60 = 24 litres. After removing 10 litres of mixture: Milk remaining = 36 – (3/5)*10 = 30 litres, Water remaining = 24 – (2/5)*10 = 20 litres. After adding 10 litres milk: Milk = 30 + 10 = 40 litres, Water = 20 litres. After removing 15 litres of mixture: Milk removed = (40/60)*15 = 10 litres, Water removed = (20/60)*15 = 5 litres. Milk remaining = 40 – 10 = 30 litres, Water remaining = 20 – 5 = 15 litres. After adding 15 litres water: Milk = 30 litres, Water = 15 + 15 = 30 litres. Final ratio = 30:30 = 1:1

Ans: C

Let the capacity of the tank be 24 units (LCM of 8 and 24). The tap fills 3 units/hour and the leak empties 1 unit/hour. In the first 2 hours, the tap fills 3*2 = 6 units. Remaining capacity = 24 – 6 = 18 units. The net filling rate with both tap and leak open = 3 – 1 = 2 units/hour. Time taken to fill the tank completely = 24/(3-1) = 12 hours. But we know the tank had been filled with 6 units already in 2 hours, So total time = 6/2+2=5 hours, therefore total time is 2 hours + 18/(3-1)= 2+9=11. But the question wants to know the time taken if they are open together. Hence, the net filling rate = 3-1=2 units/hour. The tank capacity is 24 units, hence time = 24/2= 12 hours. With opening for 2 hours earlier, time should be 24/(3-1)=12 hours and 2 hours earlier. However, the tank had been filled with 6 units already when the third pipe is opened. With tap + leak together: (3-1) unit/hour = 2 unit/hour Time=24/2=12 hours. However, since the tap has been running for 2 hours, the remaining part will be filled in 24-2(3)=18/(3-1)=9 hours, plus initial 2 hours. Hence there is no mention of any leak here. The tap works in 8 hours, hence 1/8 fill rate. Leak rate = 1/24 per hour, fill rate=1/8. After 2 hours, water filled = 2/8=1/4 Now, if both are open, the net filling rate is 1/8-1/24= (3-1)/24=2/24=1/12. Hence total time is 24-2(3)=18; hence total time is not correct. Let the tank capacity be x. Tap is opened for 2 hours, so 2/8=1/4 portion filled, and x-x/4=3x/4. Now when both open, 1/8-1/24=1/12 portion. So 1/12 portion per hour. Let the time taken is t, t= (3x/4)/(x/12). Therefore t=9 hours. Total time is 2+9=11. Fill rate of tap= 1/8, and the leak =1/24. Combined fill rate= 1/8 – 1/24= 2/24= 1/12. So it takes 12 hours to fill the tank. 2 hours earlier the tap filled, so the remaining time is 10. 2/8 1/12 Total = 10. Total time taken = 12. Since the question wants to know, how many hours to fill the tank, if both are open. Then, 1/8 – 1/24=2/24 = 1/12. Total hours = 12 hours.

Ans: B

Total marbles = 5 + 7 + 8 = 20. Number of marbles that are not yellow = 5 (green) + 8 (orange) = 13. Probability = 13/20.