Aptitude - Questions & Answers

Simplification - Online Quiz

Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Simplification. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.

Q 1 - [(125)² / 50 * 20] / 25 = ?

Answer : D

Explanation

Give Exp = [ 125 * 125/ 50 *20] / 25 = 125 * 50 /  25 = 250

Q 2 - 108 / 36 of 1/3 + 2/5 * 15/4 = ?

Answer : D

Explanation

given exp. = 108 /36 of 1/3+ 2/5 *15/4
= 108/ 12+ 2/5 *15/4 = 9+ 3/2 = 21/2

Q 3 - 5/8 of 24 = 15/7 * ?

Answer : B

Explanation

let 5/8 of 24 = 15/7 of x.  Then , 15x/7 =15  or x= 15*7 /15 = 7

Q 4 - { 15/2+1/2 / 1/2 of 1/4 -2/5 *7/3 /15/8 of (7/5-4/3)}=?

Answer : C

Explanation

Give Exp. = 15/2 +1/2 /1/2 of 1/4 -2/5 *7/3/15/8 of  (7/5 -4/3)
= 15/2 +1/2 /1/2 of 1/4 -2/5 *7/3 /15/8 of 1/15
= 15/2+1/2 /1/8 -2/5*7/3 / 1/8
= 15/2 +1/2 *8/1- 2/5*7/3 *8/1
= 15/2 +4- 112/5 = 23/2- 112/15
=(345-224)/30
= 121/30

Q 5 - 1+ 1/1+ 1/1 + 1/9 = ?

Answer : C

Explanation

given exp.  =  1 + 1/(1+ 1/(9/10))
=1 + 1/(19/10)
=1 + 10/19
= 29/19

Q 6 - [ {1+ 1/ (10+1/10)} * {1+ 1/ (10+1/10)}- {1- 1/ (10+1/10)}* {1- 1/ (10+1/10)}]/[ {1+ 1/ (10+1/10)}+ {1+ 1/( 10+1/10)} simplifies to :

Answer : A

Explanation

given exp.  = (a²-b²)/(a+b)= a-b
= [1 + 1/(10+1/10)] - [1 - 1/(10+1/10)]
= ( 1+ 10/ 101) - (1- 10/101)   = 10/101 +10/101 = 20/101

Q 7 - { 1/2 +1/6 +1 /12 + 1/20 + 1/30 + ... 1/ n (n+1) } =?

Answer : C

Explanation

given exp.  = (1-1/2)+ (1/2 -1/3) + (1/3 -1/4) +(1/4 -1/5) + ...+ [ 1/n - 1/(n+1)]
= (1- 1/n+1) = (n+1-1)/ (n+1)  = n/ (n+1)

Q 8 - if (a-b)=3 and (a² +b²) =29 , then ab =?

Answer : D

Explanation

2ab = (a²+b²) - (a-b)² = 29-(3)² = (29- 9) = 20 ⇒ ab = 10.

Q 9 - x² - (y-z)²/ (x+z)²-y²+ y² -(x-z)²/ (x+y)²-z²+ z²- (x-y)²/ (y+z)²- x² = ?

Answer : C

Explanation

given exp. = [ x+(y-z)][x-(y-z)]/(x+z+y)(x+z-y)
 + [y+(x-z)][y-(x-z)]/(x+y+z) (x+y-z)
 + [z+(x-y)][z-(x-y)]/(y+z+x) (y+z-x)
 = (x+y-z)(x+z-y) / (x+y+z) (x+z-y)
 + (x+y-z) (y+z-x) / (x+y+z)(x+y-z)
 + (z+x-y) (y+z-x) / (y+z+x) (y+z-x)
 = (x+y+z)/(x+y+z) + (y+z-x)/(x+y+z) + (z+x-y)/(x+y+z)
 = (x+y+z)/(x+y+z)
 = 1

Q 10 - If in a examination condition of marks is that for the right answer student gets 3 marks but reduce 2 marks for every wrong answer . if 30 questions attempted and 40 marks gets by the student. Find out the number of question correctly done by the student.

Answer : B

Explanation

Let correctly attempted questions be x. Then ,
3x-2 (30-x) = 40 ⇒  5x = 100  ⇒ x = 20.
Hence, 20 question  correctly done by  the  student.

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