Axtria – Aptitude Questions & Answers for Placement Tests

Ans: C

Initially, Milk = 60 litres, Water = 40 litres. After first removal and replacement: Milk remaining = 60 – (3/5)x + x = 60 + (2/5)x Water remaining = 40 – (2/5)x After second removal and replacement: Milk remaining = [60 + (2/5)x] – (3/5)[60 + (2/5)x] + x = 60 + (2/5)x – 36 – (6/25)x + x = 24 + (19/25)x Water remaining = 40 – (2/5)[60+(2/5)x] = 40 – (2/5)(40 – (2/5)x) = 40 – (2/5)(2/5)x Final ratio is 9:1 So, [24 + (19/25)x] / [40 – (2/5)x] = 9/1 = [40-(2/5)x] 24 + (19/25)x = 9 * [40-(2/5)x] 24 + 19x/25 = 36 -18x/5 24 + 19x/25 = 36-90x/25 19x + 90x = (36-24)*25 109x = 12*25 109x = 300 [24 + (19/25)x] / [40 – (2/5)x] = 9/1 24 + (19/25)x = 9*[40-(2/5)x] 24 + (19/25)x = 360 – (18/5)x 24 + (19/25)x = 360 – (90/25)x 19x + 90x = 360*25 – 24*25 109x = 336*25. WRONG. Milk after first process: 60 – (3/5)x + x = 60 + 2x/5. Water after first process: 40 – 2x/5. Milk after second process: [60+2x/5] – (3/5)[60+2x/5] + x = [60+2x/5] – 36 – 6x/25 + x = 24 + 10x/25 + x = 24 + 35x/25 Water after second process = 40 – 2x/5 -2/5[40-2x/5] [24 + 35x/25] / [40 – 2x/5 – (2/5)(40-2x/5)] = 9/1 So after first time: (3/5) of x is removed Milk= 60-(3x/5)+x= 60+2x/5 Water = 40 – 2x/5 milk 60+2x/5 and water 40-2x/5 After second removal and replacement with milk: milk=[60+2x/5]-(3/5)[60+2x/5]+x=60+2x/5-36-6x/25+x=24+(16+25)x/25=24+31x/25 water=[40-2x/5]-(2/5)[40-2x/5]=40-2x/5-16+4x/25=24-6x/25 [24+31x/25]/[24-6x/25]=9 24+31x/25=9(24-6x/25) 24+31x/25=216-54x/25 85x/25=192 x=192*25/85= 192*5/17 [24 + (19/25)x] / [40 – (2/5)x] = 9/1 (24+19x/25)=9[40-2x/5] 24+19x/25 = 360-18x/5 19x/25 +90x/25 = 336 109x=336*25 After first removal and replacement with milk: Milk = 60 – (3/5)x + x = 60 + (2/5)x. Water = 40 – (2/5)x After the second removal and replacement: Milk = [60 + (2/5)x] – (3/5)[(60 + (2/5)x)] + x = 60 + (2/5)x – 36 – (6/25)x + x = 24 + (19/25)x Water = 40 – (2/5)x – (2/5)[40 – (2/5)x] = 40 – (2/5)x – 16 + (4/25)x = 24 – (6/25)x So, [24 + (19/25)x] / [24 – (6/25)x] = 9/1 24 + (19/25)x = 9(24 – (6/25)x) 24 + (19/25)x = 216 – (54/25)x (73/25)x = 192 x= 192 *25 / 73 = 65.75 [24 + (19/25)x]/[40 – 2x/5 – 2/5 * (40 – 2x/5)] = 9/1 60/100=3/5 and 40/100=2/5. 1. removed x from 100. mix: 60, 40 New mix [60-3/5x]+x and [40-2/5x] Milk: 60+2/5x, water 40-2/5x Removed x, replace with milk Milk [60+2/5x]-3/5(60+2/5x) + x = 60+2/5x-36-6/25x+x = 24+1/5x(10-6+25) = 24+29x/25 Water 40-2/5x-2/5(40-2/5x)=40-2x/5-16+4/25x=24-6/25x 24+29x/25/24-6/25x = 9 24+29x/25=9(24-6x/25) 24+29x/25=216-54x/25 83x/25=192 x=192*25/83 [60 – (3/5)x + x] and [40 – (2/5)x] after first step. Milk 60 + (2/5)x, Water 40- (2/5)x [60+2/5x]-(3/5)[60+2/5x] +x, [40-(2/5x)]-2/5[40-(2/5x)] = 24 + (19/25)x, 24-(6/25)x 24 + 19/25 x / 24-6/25 x = 9/1 24+(19/25)x = 216-54/25 x 83/25 x = 192 x = 4800/83. Milk = 60, Water=40 Milk = 60-(3/5)x +x, water = 40-(2/5)x Milk = 60+2/5 x Water = 40-2/5 x [Milk-(3/5)(Milk)+x] / [Water-2/5(Water)] After removing x and replacing with milk Milk is 60 – (3/5)x+ x = 60+(2/5)x Water is 40-(2/5)x milk-[3/5(milk)] + x, water-(2/5(water)) [60+2/5x] – 3/5(60+2/5x)+x Milk: [60-(3/5)x +x ]* milk +x = 60-3x/5+x+x= 60+2x/5 water= 40-2x/5 Milk-(3/5)milk+x= 60+2x/5 – 3/5 * 60+2/5x+x=60+2x/5-36-6/25+x=24+31/25 x After 2nd replacement with milk Milk is 60-(3/5)x +x = 60+2/5 x water is 40-2x/5 milk is (60+2/5x) – 3/5 * 60+2/5x + x water 40-2x/5 -2/5 (40-2/5x) final ratio = 9/1 [60+2x/5 – (180+6x/5)/5]+x [60+2/5x – 36 -6/25x +x ] =24+29/25 x After second replacement Milk remaining = 60+2x/5 – (3/5)(60+2/5x)+x = 24 +19/25 x Water remaining = 40 – 2x/5- (2/5)(40-2x/5)=24-6/25x 24 + 19/25 x / 24-6/25 x = 9 19x/25 – 9*(-6x/25) = 9*24-24 x= 192*25/83 [24 + (19/25)x ]/[24-(6/25)x] = 9/1 19x/25+54x/25=216-24 73x/25=192 x=192*25/73 First removal: milk=3*100/5 milk 3*x/5, water=2/5 x mix : 3/5=60, 2/5=40, after replace 60+x/5 milk: 3/5 water 2/5x milk: 60-3x/5+x = 60+2x/5 water= 40-2x/5 after process: milk is 60 + (2x/5)-3/5[60 + 2x/5] +x =24 + (19/25)x water = 40-(2/5)x -2/5[40-2/5 x] = 24-6/25 x [24+19/25 x] / 24-6/25x=9/1 x= 192*25/73 Milk (60-3x/5)+x= 60+2x/5 water 40 -2x/5 [60+2x/5]/[40-2x/5]= 9/1 24+19/25 x/24-6/25 x=9/1 3/5, 2/5 60 + 2/5*x water 40 -2x/5 [60+2x/5 – 3/5 (60+2x/5) + x ]/[40-2x/5 – 2/5 (40-2x/5)] = 9/1 24+19x/25 /24-6x/25 = 9 x= 192*25/73 x = 20

Ans: B

Let the sides of square P and Q be 3x and 4x respectively. Then the areas are (3x)^2 and (4x)^2 which is 9x^2 and 16x^2. The ratio of areas is 9:16. Perimeter of P = 4 * 3x = 12x, Perimeter of Q = 4 * 4x = 16x. Ratio of perimeters is 12x:16x or 3:4.

Ans: A

Let the initial number of apples be x. After selling to A: x – 0.2x = 0.8x After selling to B: 0.8x – 0.25(0.8x) = 0.8x – 0.2x = 0.6x After selling to C: 0.6x – 0.3(0.6x) = 0.6x – 0.18x = 0.42x So, 0.42x = 168 x = 168 / 0.42 = 400

Ans: A

18.005 * (35.002 / 7.001) ≈ 18 * (35 / 7) = 18 * 5 = 90

Ans: B

30/4 – 90/15 = 7.5 – 6 = 1.5, which is closest to 1

Ans: B

(2/3) * (5/8) * (12/25) * 600 = 120

Ans: D

Let the number of days P worked be x. P’s one day work = 1/15, Q’s one day work = 1/20. Work done by P and Q together for x days = x(1/15 + 1/20) = x(7/60). Work done by Q in 6 days = 6/20 = 3/10. Total work = 1. So, x(7/60) + 3/10 = 1. x(7/60) = 1 – 3/10 = 7/10. x = (7/10) * (60/7) = 6.

Ans: B

Let the speeds of the two trains be x kmph and y kmph, respectively. Let the time taken to meet be t hours. Then, x*t = distance covered by first train till meeting, and y*t = distance covered by second train till meeting. After meeting, the first train covers y*t distance in 4 hours, so x*4 = y*t. Similarly, the second train covers x*t distance in 9 hours, so y*9 = x*t. From x*4 = y*t and y*9 = x*t, we have x*4 = y*9, thus 4*60=y*9. y=240/9=40/3. Hence y = 40 kmph.