Amazon Aptitude Paper 3 - Quant MCQ

Let S = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) 

As f(1) + f(4) = 0, therefore S = f(2) + f(3) + f(5) + f(6) ------ (1)

f(2) = f(0) - 8

f(3) = f(1) - 15

f(4) = f(2) - 24 = f(0) - 32

f(5) = f(3) - 35 = f(1) - 50

f(6) = f(4) - 48 = f(0) - 80 

Put the above values in equation (1), we get

S = f(0) - 8 + f(1) - 15 + f(1) - 50 + f(0) - 80

S = 2(f(0) + f(1)) - 153 ------ (2) 

As we already know f(1) + f(4) = 0 ⇒f(1) + f(0) - 32 = 0 ⇒f(1) + f(0) = 32

Putting this value in equation 2, we get S = 2(32) - 153 = -89

So, Ans is option C.

Let the current price be Rs. 100.

For getting 12% profit he should sell it at Rs. 112.

Let the label price be x.

Let the no. Be [3/5]*[{2/5}x]-[2/5]*[1/4]x,then = 288 Solving it we will get 960

Work done by A and B in 12 days is

12[112*5/4]=12*5/120=.5

As diff. Is same so average should lie between B and C so B is 41 & C is 43 so D must be 45 as we have to find the product of B and D so it would be 1845

Here, BC = 5cm

BL = 3 5/2 cm 

By using the formula

4(BL2 + CM2) = 5BC2

A + B + C = 731 ..... (i)
A = 1.25B, gives A = 1.25 * 0.75 C = 0.9375 C....(ii)
B = 0.75C....(iii)
Using (ii) and (iii) in (i)
we get 0.9375C + 0.75C + C = 731, gives C = 272. So option E is the answer.

As total number of alphabets in PRAISE are 6, so total no. of ways is 6!=720 So option A is the answer

Let the fraction be [n/d] & after that it becomes [2.5 n / 4 d]=5/18 . So we get the result as [4/9]

Total distance/Total time =(30+25)/[(3/4)+(1/2)]

Average Speed=44kmph