Arithmetic Aptitude 4

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Question 1
How many numbers between 20 and 451 are divisible by 9?
A
44
B
48
C
50
D
52
Arithmetic Aptitude 4    Number Divisibility    
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Question 1 Explanation: 
The required numbers are 27, 36, 45……450.

This is an A.P. with a = 27 and d = 9

Let it has n terms.

Then Tn = 450 = 27 + (n-1) x9

∴ 450 = 27+ 9n - 9

∴ 9n = 432

∴ n = 48
Question 2
How many terms are there in 3,9,27,81........531441?
A
25
B
12
C
13
D
14
Arithmetic Aptitude 4    Numbers    
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Question 2 Explanation: 
3, 9, 27, 81..............531441 form a G.P. 
with a = 3 and r = 9/3 = 3

Let the number of terms be n

Then 3 x 3n-1 = 531441

∴ 3n = 312

∴ n = 12
Question 3
7 + 72 + 73...........76 =?
A
140136
B
142156
C
133256
D
137256
Arithmetic Aptitude 4    Simplification and Approximation    
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Question 3 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 4
Two numbers are in the ratio of 2:9. If their H. C. F. is 19, numbers are:
A
6, 27
B
8, 36
C
38, 171
D
20, 90
Arithmetic Aptitude 4    HCF    Ratio and Proportion    
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Question 4 Explanation: 
Let the numbers be 2X and 9X
Then their H.C.F. is X, so X = 19
∴ Numbers are (2x19 and 9x19) i.e. 38 and 171
Question 5
Subhash has 90 currency notes, some of which are Rs. 1000 denomination and rest of Rs. 500 denomination. The total amount is Rs 71,000. How many notes he has in denomination of Rs. 500?
A
30
B
32
C
34
D
38
Arithmetic Aptitude 4    
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Question 5 Explanation: 
Let the no. of Rs. 500 note be X
Then the no. of Rs. 1000 note = 90 – X
∴ 500X + 1000(90 – X) = 71000
∴ 500X + 90000 – 1000X = 71000
∴ 500X = 19000
∴ X = 38
Question 6
If the average of four consecutive odd numbers is 12, find the smallest of these numbers?
A
5
B
7
C
9
D
11
Arithmetic Aptitude 4    Numbers    
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Question 6 Explanation: 
Let the numbers be x, x+2, x+4 and x+6
Then (x + x + 2 + x + 4 + x + 6)/4  = 12
∴ 4x + 12 = 48
∴ x = 9 
Question 7
If the sum of two numbers is 13 and the sum of their square is 85. Find the numbers?
A
6 & 7
B
5 & 8
C
4 & 9
D
3 & 10
Arithmetic Aptitude 4    Numbers    
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Question 7 Explanation: 
Let the numbers be x and 13-x Then x2 + (13 – x)2 = 85 ∴ x2 + 169 + x2 – 26x = 85 ∴ 2 x2 – 26x + 84 = 0 ∴ x2 – 13x + 42 = 0 ∴ (x-6)(x-7)=0 Hence numbers are 6 & 7
There are 7 questions to complete.
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