Simplification and Approximation

123
 Question 1
(1015)2 = ?
 A 1040125 B 1030225 C 1050125 D 1025125
Arithmetic Aptitude    Simplification and Approximation
Discuss it

Question 1 Explanation:

10152 = (1000+15)2

= 10002 + 152 +2 x 1000 x 15

=1000000 + 225 + 30000

=1030225

 Question 2
103 x 103 + 97 x 97 = ?
 A 21348 B 20018 C 19648 D 21428
Arithmetic Aptitude    Simplification and Approximation
Discuss it

Question 2 Explanation:
103 x 103 + 97 x 97
= (100+3)2 + (100-3)2
= 2 (1002 + 32 )
[ (X +Y)2 + ( X – Y)2 = 2 ( X2 + Y2 ) ]
= 2 (10000 + 9 )
= 2 x 10009
= 20018
 Question 3
9848 x 125 = ?
 A 1232000 B 1242000 C 1231000 D 1233000
Arithmetic Aptitude    Simplification and Approximation
Discuss it

Question 3 Explanation:
9848 x 125 = 9848 x 53
= 9848 x (10 / 2) 3
= 9848 x (103 / 23)
= 9848 x (1000/8)
= 1231000
 Question 4
512 x 512 + 488 x 488 = ?
 A 512438 B 502568 C 500288 D 514318
Arithmetic Aptitude 2    Simplification and Approximation
Discuss it

Question 4 Explanation:
X2 + Y2 = 1/2[(X + Y)2 + (X – Y)2]
5122 + 4882 = ½ [(512+488)2 + (512-488)2]
= 1/2[10002+242]
= 1/2(1000000+576)
= 500288
 Question 5
What should be the value of x in equation (x / √216) = (√96 / x)
 A 12 B 13 C 9 D 11
Arithmetic Aptitude 2    Simplification and Approximation
Discuss it

Question 5 Explanation:
Let (x / √216) = (√96 / x)

Then x2 = √(216 x 96)
= √(36 x 2 x 3 x 16 x 2 x 3)
= √(62 x 42 x 22 x 32)
= 6 x 4 x 2 x 3
= 144
Or x = 12
 Question 6
7 + 72 + 73...........76 =?
 A 140136 B 142156 C 133256 D 137256
Arithmetic Aptitude 4    Simplification and Approximation
Discuss it

Question 6 Explanation:
Given series is a G. P. with a = 7, r = 7 and n = 6

∴ Sn = a(rn-1) / (r-1)

∴ Sn = 7(76-1) / 6

Sn = = 137256
 Question 7
What could be the maximum value of Y in the following equation given that neither of X, Y, Z is zero? 5X8 + 3Y4 + 2Z1 = 1103
 A 0 B 7 C 8 D 9
Numbers    Simplification and Approximation
Discuss it

Question 7 Explanation:
1 1    <- CARRY

5 X 8
+ 3 Y 4
+ 2 Z 1
--------
11 0 3
--------
Clearly, X + Y + Z + 1 = 10 => X + Y + Z = 9 Now, since neither of X, Y, Z can be zero, the value of Y will be maximum when X = Z = 1. => Max Y = 7
 Question 8
What is the unit's digit in the product (267)153 x (66666)72 ?
 A 7 B 6 C 1 D 2
Numbers    Simplification and Approximation
Discuss it

Question 8 Explanation:
We have to find the unit digit only.
In 267 unit digit is 7 and cyclicity of 7 is 4.
So, (267)153 can be written as (267)Rem(153)/4 = (267)1
Unit digit of (267)1 = 7.
similarly for 66666 unit digit is 6 and cyclicity for 6 is 1.
Unit digit for (66666)72 = 6.

Therefore , Unit digit of the resultant is 7 * 6 = 2
 Question 9
√7321 x 35.999 = ?
 A 3060 B 3204 C 3410 D 2930
Simplification and Approximation
Discuss it

Question 9 Explanation:
85*9*100/25=765*4=3060
 Question 10
5432.91 ÷ 2324.65 × 210.05 =?
 A 471 B 431 C 491 D 501
Simplification and Approximation
Discuss it

Question 10 Explanation:
Approx: 2.3*210 = 483
There are 23 questions to complete.
123

My Personal Notes arrow_drop_up