Amazon Aptitude Paper 1

Suppose the boy got x sums right and 2x sums wrong. Then, x + 2x = 48 3x = 48 x = 16.

Do not confuse it with Tuesday.the correct answer is Monday

Assume there are x participants in the race. In a round race, number of participants in front of a person wil be x-1 and that behind him wil be x-1. i.e, 1/3(x-1) + 3/4(x-1) = x solving x = 13

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7 = 1008 + 7 = 1015

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4. So, the numbers 12 and 16. L.C.M. of 12 and 16 = 48.

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

In this alternating repetition series, the random number 21 is interpolated every other number into an otherwise simple addition series that increases by 2, beginning with the number 9.

Let us name the trains as A and B. Then, (As speed) : (Bs speed) = b : a = 16 : 9 = 4 : 3.

Given that 5 women is equal to 8 girls to complete a work so, 10 women=16 girls. Therefore 10women +5girls=16girls+5girls=21girls. 8 girls can do a work in 84 days then 21 girls= (8*84/21)=32days. Therefore 10 women and 5 girls can a   work in 32days

If men is fixed,work is proportional to time. If work is fixed, then time is inversely proportional to men therefore, (M1*T1/W1)=(M2*T2/W2) From the above formula i.e (m1*t1/w1)=(m2*t2/w2) so (9*6*88/1)=(6*8*d/1) on solving, d=99   days.

Usual speed = S Usual time = T Distance = D New Speed is 3/4 S New time is 4/3 T 4/3 T, T = 5/2 T=15/2 = 7 1/2

Weight increased per person is 1 kg. Total increase in weight = 25 kgs Weight of new man is 70 kgs, (Which means his weight is 25 kgs heavier) The weight of the old man was 70 - 25 = 45 kgs

Speed of boat in still water (b) = 4.5 km/hr. Speed of boat with stream (Down Stream), D = b + u Speed of boat against stream (Up stream), U = b – u It is given upstream time is twice to that of down stream. ⇒ Downstream speed is twice   to that of upstream. So b + u = 2(b – u) ⇒ u =b/3 = 1.5 km/hr.

S.I. for 3 years = Rs. (12005 - 9800) = Rs. 2205. S.I. for 5 years = Rs. (2205/3) x 5 = Rs. 3675 Principal = Rs. (9800 - 3675) = Rs. 6125. Hence, rate = (100 x 3675)/( 6125 x 5) % = 12%

lets us consider t1 = time taken on level road. t2 = uphill. t3 = down hill; the distance traveled uphill and down hill same so t2*3 = 6*t3; => t2 = 2t3;---> (1) total time = 2*t1 + t2 + t3 = 6 hours ---> (2) 2t1+3t3 = 6 -->(3) total distance   = 2*(4t1) + 3*t2+ 6*t3 ---> (4) substitute (1) in (4) 8t1 + 12t3 => 4(2t1+3t3) then from (3) the total distance will become 4*6= 24 => one way distance = 12km

Let the number of cows be x and the number of hens be y. Then, 4x + 2y = 2 (x + y) + 14 4x + 2y = 2x + 2y + 14 2x = 14 x = 7.

Clearly, the required number would be such that it leaves a remainder of 1 when divided by 2, 3 or 4 and no remainder when divided by 5.

In same time ,they cover 110 km & 90 km respectively so ratio of their speed =110:90 = 11:9

Consider 9x - 3y = 12.........divide both sides by 3: 3x - y = 4............now multiply both sides by 2: 6x - 2y = 8 So, it`s (d)

Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A. Then, 2x + 3y = 77 ...(i) and 3x + 2y = 73 ...(ii) Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17. Putting y = 17 in (i), we get: x = 13.

Required sum = (80 - 3 x 3) years = (80 - 9) years = 71 years.

Principal = Rs. (100 x 5400)/(12*3) = Rs. 15000.

From the given data A+B = 12+(B+C) A+B-(B+C) = 12 A-C=12 years. C is 12 years younger than A

Let the present age be x years. Then fathers present age is 3x+3 years. Three years hence (3x+3)+3=2(x+3)+10 x=10 Hence fathers present age = 3x+3 =[ (3 x 10) + 3] = 33 years.

Speed in still water A = ½ ( 11+5) km/hr A= 8 kmph

Speed of train = 60 - 5/18 m/s = 50/3 m/s, Time taken to cross the telegraph post = 110/(50/3) = 6.6 seconds

Let the age of three persons be x, 3x and 5x => 9x/3 = 27 = > x = 9

Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. Required number (9999 - 399) = 9600.