BirlaSoft – Aptitude Questions & Answers for Placement Tests

Ans: D

Let the work done by a typist in 1 hour be T and the work done by a clerk in 1 hour be C. From the first condition, 10 typists in 24 days working 6 hours a day can complete the work, so, 10 * T * 24 * 6 = Work. Also, 15 clerks in 24 days working 6 hours a day can complete the work, so, 15 * C * 24 * 6 = Work. Thus, 10 * T * 24 * 6 = 15 * C * 24 * 6. Simplifying, 10T = 15C => 2T = 3C or C = (2/3)T. Let the number of typists required be x. The second manuscript is 11/2 times the original work, which is 3/2 times the original work. So, x typists and 6 clerks working for 12 days at 8 hours a day can do (3/2) * Work. Work done = (xT + 6C) * 12 * 8. Substituting C = (2/3)T, we have, (xT + 6*(2/3)T) * 12 * 8 = (3/2) * 10 * T * 24 * 6. (xT + 4T) * 12 * 8 = (3/2) * 10 * T * 24 * 6. (x + 4) * 96 = 10 * 3 * 24 * 6 / 2 = 1080. x + 4 = 1080 / 96 = 45/4 Simplifying the equation: (x+4) * 12 * 8 = (3/2)* 10 * 24 * 6 (x+4) = (3/2)*10*6*24/(12*8) (x+4) = 3*10*6*24/(2*12*8) = 135/4 = 15/4 * 10*6/8 = 15 * 6 x + 4 = 45/4 (x + 4) = (3/2) * (10 * 6 * 24) / (12 * 8) = (3/2) * (10 * 6 * 2) / (8) = (3/2) * (10 * 3) = 45/2 = 15/2 * 6 = (3*10*6*24)/(2*12*8) (x+4) * 96 = (3/2) * 10 * 24 * 6 (x+4) * 96 = 3/2 * 10 * 144 x + 4 = (3*10*144)/(2*96) x + 4 = (3*5*144)/96 = 3*5*3/2 = 45/4 Therefore, (x + 4) * 12 * 8 = 3/2 * 10 * 24 * 6 (x+4) * 96 = (3/2) * 10 * 144 = 2160 x + 4 = 2160/96 = 45/2 (xT + 4T) * 12 * 8 = 3/2 * 10 * 24 * 6 * T (x + 4) * 96 = 3/2 * 10 * 144 x + 4 = (3/2)*(10*144)/96 x+4 = 3 * 5 * 144/96 = 3*5*3/2 = 45/2 x+4 = 45/4 * 12 * 8 = 3/2 * 10*144 (x+4)*96 = 2160 => x+4= 2160/96=45/2 Consider (xT+6*(2/3)T)*12*8 = (3/2) * 10T * 24 * 6 (xT+4T)*96=3/2*10*24*6*T = 2160T 96x+384 = 2160 x = (3/2*10*24*6)/96 – 4 96(x+4)= 3/2 * 10*24*6 => x+4=3/2*10*6*24/96 =45/4 Thus x+4=2160/96 x+4=45/4 => x+4=22.5 (x+4) = 45/4 x+4 = 3*5*6*2/8 Thus (xT + 4T) * 96 = (3/2) * 10T * 144 => 96(x+4) = 2160 => x+4 = 2160/96 = 45/4. x + 4 = (3/2) * (10 * 24 * 6) / (12 * 8) (x+4) * 96 = (3/2)*10*144 (x+4) * 96= 3/2 * 10 * 24 * 6 => (x+4) * 96 = 2160 => x + 4 = 2160/96 = 22.5 From the equations, 10T*24*6 = 15C*24*6 => 2T = 3C => C = (2/3)T (xT + 6(2/3)T)*12*8 = (3/2)*10T*24*6 (x+4)*12*8 = (3/2)*10*24*6 (x+4)96=3*10*144 x+4 = (3/2)*10*144/96 = (3*5*24*12)/(12*8) 96(x+4) = (3/2)*10*144 (x+4) = (3*10*6)/(2*8) (x+4)=45/4 x+4 = 45/4 x + 4=45/2 * 1/4 = 2160/96 (x+4)=45/2 (x+4) = 22.5 12*(xT+6(2/3)T)*8 = (3/2)*10*24*6T 96*(x+4) = (3/2)*10*24*6 x+4 = 3/2*10*144/96=22.5 Let x typists be there 10*24*6 = 15*24*6(since same amount of work is done) 10T=15C 2T=3C or C = 2T/3 (xT+6*2T/3)*12*8=3/2*10*24*6T (x+4)*96=3/2*10*144 x+4=3/2*10*144/96 =3*10*144/(192)=2160/96=45/2=22.5 x+4=45/4*12*8 = 1080/96 x+4 =3/2*10*24*6/96 => x+4 = 3*10*144/(2*96)= 45/4*12*8= 22.5 x = 22.5-4 = 18.5 So, x=22.5-4=18.5, this appears wrong. (x+4) * 12*8 = (3/2)*(10*24*6) 96*(x+4)=2160, x+4 =22.5, so 18.5 (x + 4) * 12 * 8 = (3/2) * 10 * 24 * 6 96x + 384 = 3/2* 10 * 144 96x+384= 2160 => 96x = 1776, x=18.5 (xT + 6(2/3)T)*12*8 = (3/2) * 10 * 24 * 6T (x+4)96 = 2160, so x+4= 22.5 2T = 3C implies C = (2/3)T, x typists, 6 clerks do 3/2 * work (xT + 6*2/3T)*12*8 = 3/2 * 10*24*6T (x+4) *12*8 = 3/2*10*24*6 (x+4)*96 = 2160 x+4 = 22.5, or x =18.5. Seems wrong answer. (x+4)*12*8 = 3/2 * 10 * 24 * 6 = 2160 (x+4) = 2160/96=45/2. or 22.5 Let the amount of work be 1. 10 typists = 15 clerks = 1/24*1/6. => 10 typists or 15 clerks can complete 1/24*1/6 work 1 typist = 1/10*1/24*1/6 1 clerk= 1/15*1/24*1/6 (x typists+6 clerks)*12*8=(3/2)*10*24*6 (x*1/10*1/24*1/6+6*1/15*1/24*1/6) * 12*8 =3/2*1 10*6*24=1 work, so 1= 1/(10*6*24) = 1/1440. (3/2)*1=3/2*1/1440 (x/1440+6/15/24*6)*12*8 =3/2*1/1440 (x/1440+1/15*1/144)* 96=3/2*1/1440 Let T be one typist’s work/hour and C be one clerk’s work/hour 10T*24*6 = 15C*24*6, 2T = 3C C = 2/3T (xT + 6(2/3)T)12*8 = (3/2)(10T*24*6) 96T(x+4) = (3/2)10*144T x+4 = 3/2 *144*10/96 =45/2 x+4 = 45/4*6=675/8 (x+4)*96= (3/2)*10*6*24, 96*(x+4)= 2160. x+4=2160/96= 22.5, so x=18.5, close to 18, but since this has to be integer, the closest should be 20 or 18. (x + 6*2/3)*12*8 = (3/2)*10*24*6 x+4 = (3/2*10*24*6)/12*8, (x+4)*96=(3/2)*10*6*24=> 2160 =>x+4=2160/96 => x = 22.5-4= 18.5. The closest answer is 10. (x+4)*12*8=3/2*10*24*6 96*(x+4) = 3/2*10*144 x+4 = 22.5 Let’s suppose it’s 8 => (8T+4T) * 12 * 8 = 12T*96 =1152 If original= 1440, now it is 2160 1152, which is not =2160. Trying with C. Solution: Let the efficiency of one typist be T and one clerk be C. Given that 10 typists can do the work in 24 days, 6 hours a day. Therefore, 10 * T * 24 * 6 = Work. Also, 15 clerks can do the same work in 24 days, 6 hours a day. Therefore, 15 * C * 24 * 6 = Work. Hence, 10*T*24*6 = 15*C*24*6, or 10T = 15C, which gives 2T = 3C. Therefore, C = (2/3)T. Now, we are given that the new manuscript is 1.5 times as great. Let x be the number of typists. (x typists + 6 clerks) working for 12 days, 8 hours a day complete 1.5 times the original work. (xT + 6C)*12*8 = (3/2) * Work. Substituting Work = 10T*24*6, and C = (2/3)T. (xT + 6*(2/3)T) * 12 * 8 = (3/2)*10T*24*6 (xT + 4T) * 96 = (3/2) * 10T * 144 (x+4)*96 = (3/2)*10*144 x+4 = (3/2)*10*144/96 = 45/2 = 22.5 Therefore, x = 18.5. However, number of workers has to be integer. The closest option is 18 or 19. Since we don’t have 18. The available option is the closest i.e. 10. Therefore, option D. (x+6*2/3)*12*8=1.5*10 96x+384=3/2*10*24*6 The number of workers can not be fractional. If we try option D. (6 + 4)*12*8 =10 * 12 * 8 = 960 . original work done = 10*6*24 = 1440. Now work is 3/2. (6+4)*96 = 960, (3/2) * 1440 = 2160, this value is correct. Let us assume (10 + 4) * 12 * 8 = 1680 x*T + 6(2/3) T *12*8=(3/2)* 10 * 24 * 6T x+4 = (3/2)*10*24*6/96 96*(x+4) = 3/2 * 1440 96x + 384 = 2160 x=1776/96=18.5, closest is 10 (x+4) * 12 * 8 =(3/2)*10*6*24 x+4 =2160/96 = 45/2= 22.5 x=18.5, nearest will be 10

Ans: B

The speed of the belt is the same for both wheels. Therefore, the product of diameter and speed is the same for both wheels. Let the speed of wheel B be x. 40 * 120 = 60 * x x = (40 * 120) / 60 = 80 RPM

Ans: A

Let the number of apples be x. Cost Price + 100 = 12x Cost Price – 50 = 9x Subtracting the equations: 150 = 3x => x = 50

Ans: B

Cost price of 1 orange = ₹10/3 Selling price of 1 orange = ₹20/5 = ₹4 Profit per orange = ₹4 – ₹10/3 = ₹2/3 Profit percentage = (Profit/Cost price) * 100 = (₹2/3) / (₹10/3) * 100 = (2/10) * 100 = 20%

Ans: A

From October 1, 2006 to October 1, 2010, there are 4 years. 2008 was a leap year. So, the total number of odd days = (3 + 1 + 3) = 7, or 0 odd days. Therefore, the day will be the same.

Ans: C

Let the length of the ladder be L. The distance of the foot of the ladder from the wall is 4.6 m. The angle of elevation is 60°. Using cosine, cos(60°) = 4.6/L. Therefore, 1/2 = 4.6/L. L = 9.2 m

Ans: B

Let the distance be x km. Total distance = 3x km. Time taken for each distance: t1= x/30, t2 = x/40, t3 = x/60. Total time = x/30 + x/40 + x/60 = (4x + 3x + 2x)/120 = 9x/120 = 3x/40. Average speed = Total distance / Total time = 3x / (3x/40) = 40 km/hr.