AMCAT Aptitude Question-2
AMCAT Quantitative Ability Previous Papers-2
a)1/2 b)1/3 c)3/5 d)2/5
Ans) A single letter drawn at random from the above given word is 5c1/10 = 1/2
13)Suparna needs to browse through 75pages of a novel before she gives her review to the class. She has 2.5 hrs before the lecture. What should be her reading speed in pages/hour?
a) 16 b)30 c) 20 d) 22
Ans) speed=distance/time
S=75/2.5
S=30
14) The value of log100.1 is :
a) 0 b) -1 c) -10 d) -100
Ans)log100.1 = log1010-1 = -1 log 1010 = -1 (log 1010 =1)
15) A written exam consists of 6 questions with the answer options as yes/no/none. In how many ways can the examinees select the answers?
a) 6 ways b) 6 ways c) 3 .3 .3 .3 .3 d) (3 6
Ans) In 36 ways the examiners select the answers.
16) What is the sum of the two consecutive numbers, If the difference of whose squares is 19?
a. 9 b. 10 c. 18 d. 19
Ans) (n+1)2-n2=19
We get 2n=18, n=917) P is an integer. P>883. If (p-7) is a multiple of 11, then the largest number that will divide (p+4) (p+15) is :
11 121 242 None of the above
Ans) 242
Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)
=(11 a+11)(11 a+22)
=121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242
11 121 242 None of the above
Ans) 242
Given P is an integer>883.
P-7 is a multiple of 11=>there exist a positive integer a such that
P-7=11 a=>P=11 a+7
(P+4)(P+15)=(11 a+7+4)(11 a+7+15)
=(11 a+11)(11 a+22)
=121(a+1)(a+2)
As a is a positive integer therefore (a+1)(a+2) is divisible by 2.Hence (P+4)(P+15) is divisible by 121*2=242
18) Find the least number which when divided by 5, 7 and 13 leaves the same remainder 3 in each case
398 453 458 463
Ans) By trial method we get the answer as 458.Why because if we divide the number with 5,7,13 it leaves the remainder as 3 in all the three cases.
398 453 458 463
Ans) By trial method we get the answer as 458.Why because if we divide the number with 5,7,13 it leaves the remainder as 3 in all the three cases.
19)Which number should be subtracted from 321 so that it becomes prime?
2 4 6 9
Ans) If we apply trial method 4 is the answer.If we subtract 317 from 4 the result will be 317hence this is the prime number
2 4 6 9
Ans) If we apply trial method 4 is the answer.If we subtract 317 from 4 the result will be 317hence this is the prime number
410 210 216 416
28 *22 =28+2 =210
Category: Amcat, Amcat Aptitude
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