Amazon Aptitude Paper 3
10 Questions MCQ Test Placement Papers - Technical & HR Questions | Amazon Aptitude Paper 3
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A function f(x) is defined as f(x) = f(x - 2) - x(x + 2) for all the integer values of x and f(1) + f(4) = 0. What is the value of f(1) + f(2) + f(3) + f(4) + f(5) + f(6)?
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.
In ABC, the internal bisectors of ABC and ACB met at 1 and BAC = 50.The measure of BIC is
I is the In-centre
So, BIC = 90 + 1/2 BAC
= 90 + 1/2 50 = 115
The difference between [3/5] of [2/3] a number and [2/5] of[1/4] of the same number is 288. What is the number?
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
A, B, C and D are four consecutive odd numbers and their average is 42. What is the product of B and D?
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
BL and CM are medians of ABC right angled at A and BC =5 cm. If BL =3- 5/2 cm, then the length of CM is
Here, BC = 5cm
BL = 3 5/2 cm
By using the formula
4(BL2 + CM2) = 5BC2
A sum of Rs 731 is divided among A, B and C such that A receive 25% more than B and B receive 25% less than C. What is Cs share in the amount?
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.
In how many different ways can letters of the word "PRAISE" be arranged?
As total number of alphabets in PRAISE are 6, so total no. of ways is 6!=720 So option A is the answer
If the numerator of a fraction is increased by 150% and the denominator of the fraction is increased by 300%, the resultant fraction is [5/18]. What is original fraction?
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]
A car covers the first 30 km of its journey in 45 minutes and the remaining 25 km in 30 minutes. What is the average speed of the car?
Total distance/Total time =(30+25)/[(3/4)+(1/2)]
Average Speed=44kmph
Four examiners can examine a certain number of answer papers in 10 days by working for 5 hours a day. For how many hours a day would 2 examiners have to work in order to examine twice the number of answer papers in 20 days?
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