McAfee – Aptitude Questions & Answers for Placement Tests

Ans: B

Total distance = train length + bridge length = 200 + 300 = 500 meters. Time = Distance/Speed = 500/25 = 20 seconds.

Ans: A

The pattern is based on cubes plus or minus a number: 2^3 -1 = 7, 3^3 -1 = 26, 4^3 -1 = 63, 5^3 -1 = 124, 7^3 – 1 = 342. So missing number is 6^3 -1 = 215

Ans: A

Let the work be denoted as W. Carpenter’s 1 day work = W / (12 * 15) = W/180 Mason’s 1 day work = W / (10 * 18) = W/180 Laborer’s 1 day work = W / (15 * 12) = W/180 Combined 1 day work = 6(W/180) + 5(W/180) + 10(W/180) = 21W/180 = 7W/60 Number of days = W / (7W/60) = 60/7, which is not an option. Rethinking the work done by each. Carpenter’s 1 day work = W / (12*15) = W/180 Mason’s 1 day work = W/(10*18) = W/180 Laborer’s 1 day work = W/(15*12) = W/180 6 carpenters work = 6 * W/180 = W/30 5 masons work = 5 * W/180 = W/36 10 laborers work = 10*W/180 = W/18 Combined work = W/30 + W/36 + W/18 = (6W + 5W + 10W)/180 = 21W/180 = 7W/60 Number of days = W/(7W/60) = 60/7. Still not an option. Let’s calculate the work in terms of units. LCM of 15, 18, 12 = 180 units Carpenters rate: 180/15 = 12/12 = 1unit/day Masons rate: 180/18 = 10 units/day Laborers rate: 180/12 = 15 units/day Carpenter daily rate: 180/(12*15) = 1 unit/day Mason daily rate: 180/(10*18) = 1 unit/day Laborer daily rate: 180/(15*12) = 1 unit/day. (12*1) = 12 units/day. 15 days –> Total work 180 units. Carpenters: 12 men in 15 days, work per day = 180/15 = 12. 1 man/day = 12/12 = 1 unit Masons: 10 men in 18 days, work per day = 180/18 = 10. 1 man/day = 10/10 = 1 unit Laborers: 15 men in 12 days, work per day = 180/12 = 15. 1 man/day = 15/15 = 1 unit. 6 carpenters, 5 masons, 10 laborers: 6*1 + 5*(180/180) + 10*(180/180) = 6 + 5/10 + 10/15 = 6+5/10+10/15. 6 carpenters’ work in 1 day = 6 * (180/(12*15)) = 6*1 = 6 units per day. 5 masons’ work in 1 day = 5 * (180/(10*18)) = 5*1 = 5 units per day. 10 laborers work in 1 day = 10 * (180/(15*12)) = 10 units per day. Total work in a day: 6+5+10= 21 units. Total work = 180 units. Total time = 180/21, not an option. 1/15 of the wall/day. 1 carpenter does 1/12*15 part of work/day. Work = 1 1 carpenter’s 1 day work = 1/ (12*15) 1 mason’s 1 day work = 1/ (10*18) 1 laborer’s 1 day work = 1/(15*12) 6 carpenters’ work: 6/(12*15) = 1/30 5 masons’ work: 5/(10*18) = 1/36 10 laborers’ work: 10/(15*12) = 1/18 Combined: 1/30 + 1/36 + 1/18 = (6+5+10)/180 = 21/180 = 7/60 Time = 60/7 Assuming the question originally provided, 12 Carpenters in 15 days. Carpenters : 12C * 15 = 180 work units. 1 man 1 day = 1 unit. Masons 10M * 18 = 180. 1 man 1 day = 1 unit Laborers: 15L * 12 = 180, 1 man, 1 day = 1 unit. 6 C, 5M, 10L –> 6*1 + 5*1+ 10*1 = 21 Days = 180/21 , not right. Work units per day: 12 men: 15. 12C =15 days. Work = 180. Work per day 12 C work per day is 180/15 = 12units/day 1 mason works 18 days. 10 Masons work per day 10, so 18 days = 180. 18 days. Laborers 15 work 12 days 12*15. Work = 180 1 day work: C: 12/15 = 1 unit/day 1 man day = 1 Work / day 6 carpenters + 5 masons + 10 labors: 6*1 + 5*1 +10/12 6+5+10= 21 180/21

Ans: C

A’s investment = 40000 * 12 = 480000. B’s investment = 60000 * 4 + 50000 * 8 = 240000 + 400000 = 640000. Ratio of their profits = 480000:640000 = 3:4. B’s share = (4/7)*14000 = 8000.

Ans: C

(62 + 0.47) * (62 + 0.53) = 62 * 62 + 62 * (0.47 + 0.53) + 0.47 * 0.53 ≈ 62 * 62 + 62 * 1 = 3844 + 62 = 3906. More precisely, it can be seen as (62^2) + (0.47*0.53) + (62*0.47 + 62*0.53) = (62^2) + (0.47*0.53) + 62*1. (62^2) = 3844. 0.47*0.53 is close to 0.25. 3844 + 31.21 + 31.31 = 3906.52

Ans: B

Let the distance be ‘d’. Time taken to travel ‘d’ at 60 km/hr = d/60. Time taken to travel ‘d’ at 40 km/hr = 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.

Ans: B

Let the number of days X and Y worked together be ‘d’. In ‘d’ days, they complete d/30 + d/40 of the work. Y works for the total number of days = d + 15. Total work done by Y = (d+15)/40. Total work = 1, So, d/30 + (d+15)/40 = 1 => 4d + 3d + 45 = 120 => 7d = 75 => d = 75/7. But Y worked for 15 days. So (d+15)/40 = (30-d)/30.=> 30d+450 = 1200 – 40d => 70d = 750 => d=75/7. d/30 + 15/40 = 1 4d + 45 = 120 4d=75 d = 75/4 Let the number of days X and Y worked together be ‘d’. The amount of work done by Y in 15 days is 15/40. The remaining work = 1 – 15/40 = 25/40 = 5/8. X and Y together completed 5/8 work. In one day X and Y together do = 1/30 + 1/40 = 7/120 work So they worked together for d days where, (7/120)*d = 5/8 d = (5/8)*(120/7) = 75/7 Work done by Y in 15 days = 15/40 = 3/8. Remaining work = 1-3/8 = 5/8. X and Y together can complete in, x(1/30+1/40) = 5/8. x = 75/7. X and Y together done work in x days, where x/30 + x/40 + 15/40 = 1. 4x +3x + 45 =120. 7x = 75. x is not correct. Let them work ‘d’ days, after ‘d’ days x leave. Y work in 15 days. d(1/30+1/40) + 15/40= 1 => 7d/120 + 15/40 = 1 => 7d/120 = 25/40 => d= 120*25/40*7 = 75/7. Work done by Y in 15 days=15/40=3/8. Thus, A and B worked together to complete 1-3/8=5/8 of work. Combined efficiency of X and Y is 1/30+1/40=7/120. Let the number of days they worked together be d. Then d(7/120)=5/8; d=(5/8)(120/7)=75/7 Let them work ‘d’ days => d/30 + (d+15)/40 = 1. 4d+3d+45=120, 7d=75. No. So work together, then X quit, Y continued. Y worked 15 days, so work done by Y in 15 days = 15/40=3/8. Remaining work 1-3/8 = 5/8. X and Y worked together. Let the worked together be x days. x(1/30 + 1/40) = 5/8. x*7/120=5/8, x= 75/7. No. So, if the X and Y worked for x days together then work completed by X in x days =x/30, and by Y = x/40. Remaining work done only by Y = 15/40. Total work=1. x/30 + x/40 + 15/40 =1. (4x+3x+45)/120 = 1. 7x+45=120. 7x=75. => x=75/7. No Let x is the common days work together. So x/30 + (x+15)/40 = 1 => 4x+3x+45 = 120=> 7x=75. x is not correct. Y works for 15 days after that. So let x days they work together. So work done = x/30 + x/40 + 15/40 = 1. (4x+3x+45)/120=1. => 7x + 45 = 120 => 7x = 75. Amount of work Y did in 15 days = 15/40= 3/8. Remaining work = 1-3/8=5/8. X and Y did remaining work in x days. => x(1/30 + 1/40) = 5/8 => x(7/120)=5/8, x=5*120/8*7= 75/7. Let days they worked together is d. Then, (d/30) + ((d+15)/40) = 1. => 4d + 3d + 45 = 120. 7d = 75. d = 75/7. Work done by Y in 15 days = 15/40 = 3/8. Remaining work = 5/8. (1/30 + 1/40) = 7/120. So the days together = 5/8 / 7/120 = 5/8 * 120/7 = 75/7. Not an option. Work done by Y alone 15/40 = 3/8. so 5/8 done together. d(1/30+1/40)+15/40=1. 7d/120 + 3/8=1. 7d/120 = 5/8. d=5/8*120/7=75/7 days together. Let the days they worked together be ‘x’. Then, x/30 + (x+15)/40 = 1 => (4x + 3x + 45)/120 = 1 => 7x + 45 = 120 => 7x = 75 => x = 75/7 Y work 15 days. 15/40. remaining 5/8. let they done d days. (d/30) + (d+15)/40=1. Then we have 75/7 days. No option. 15/40 = 3/8. X and Y worked for x days. x(1/30 + 1/40)= 1-3/8=5/8. 7x/120=5/8. x=75/7 days. So they worked together and then X left. The correct answer will be between 0 to 30. Work done by Y alone= 15/40. Remaining work 1-15/40 = 25/40=5/8. Both did remaining work, so d(1/30+1/40)=5/8. d= 75/7 They worked together d. X completed x/30 and Y= (d+15)/40. x/30+ (d+15)/40=1. 4d+3d+45=120. 7d =75. d=75/7, not an option. X and Y, so x/30 + y/40=1. Y work 15 days. So. x/30 + (x+15)/40=1. 7x/120+15/40=1. 7x=75, so not an option. let them work together be x days so, x/30 + x/40+ 15/40 =1, so 4x+3x+45=120. so 7x=75. If they worked x days together and Y worked 15 days, then (x/30)+((x+15)/40) = 1. 7x/120 + 3/8 = 1; 7x=75 Total work is 1. Y’s work in 15 days = 15/40 = 3/8. Remaining work = 5/8. (1/30 + 1/40) = 7/120. Days worked together = x. (7/120)*x=5/8, x=75/7 days.

Ans: B

Let the number be x. 0.65x – 0.40x = 82.50 0.25x = 82.50 x = 82.50 / 0.25 x = 330 20% of 330 = 0.20 * 330 = 66

Ans: D

Cost of 1 pen = 96/12 = Rs. 8. Cost of 1 pencil = 30/15 = Rs. 2. Cost of 1 eraser = 36/18 = Rs. 2. Total cost = (15 * 8) + (20 * 2) + (25 * 2) = 120 + 40 + 50 = 210.

Ans: C

Let the maximum marks be x. Ravi failed by 10 marks, which means he needed 200 + 10 = 210 marks to pass. This passing mark is 35% of the total marks. So, 0.35x = 210. Therefore, x = 210 / 0.35 = 600.