123

Simplification and Approximation

Question 1
(1015)2 = ?
A
1040125
B
1030225
C
1050125
D
1025125
Arithmetic Aptitude    Simplification and Approximation    
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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    
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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    
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Question 3 Explanation: 
9848 x 125 = 9848 x 53
           = 9848 x (10 / 2) 3                              
           = 9848 x (103 / 23) 
           = 9848 x (10008/8)
           = 1231000
Question 4
512 x 512 + 488 x 488 = ?
A
512438
B
502568
C
500288
D
514318
Arithmetic Aptitude 2    Simplification and Approximation    
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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    
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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    
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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    
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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    
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Question 8 Explanation: 
The unit's digit in (267)153 x (66666)72 is same as unit's digit in (7)153 x (6)72. Now, unit's digit in 74 is 1. => Unit's digit in 7z is 1, where 'z' is a multiple of 4. 152 is the nearest number to 153, which is a multiple of 4. => Unit's digit in 7152 is 1 => Unit's digit in 7153 is 1 x 7 = 7   Also, unit's digit when 6 is raised to any power remains 6, i.e., for all 'n', 6n will have 6 in the unit's place. Therefore, unit's digit in (267)153 x (66666)72 = unit's digit in (7)153 x (6)72 = unit's digit in 7 x 6 = 2 Hence, D (2) is the correct answer.
Question 9
√7321 x 35.999 = ?
A
3060
B
3204
C
3410
D
2930
Simplification and Approximation    
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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    
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Question 10 Explanation: 
Approx: 2.3*210 = 483
There are 23 questions to complete.
123

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