Aptitude - Arithmetic Online Quiz


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Following quiz provides Multiple Choice Questions (MCQs) related to Basic Arithmetic. 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.

Questions and Answers

Q 1 - If an A.P. have 4th term as 14 and 12th term as 70. What will be its 17th term?

A - 108

B - 107

C - 106

D - 105

Answer : D

Explanation

  
Let's have first term as a, common difference is d then  
a + 3d = 14 ... (i)  
a + 11d = 70 ... (ii)  
Subtracting (i) from (ii)  
=> 8d = 56  => d = 7  
Using (i)  
=> a = 14 - 3d  = -7  
Using formula Tn = a + (n - 1)d    
T17 = -7 + (17 - 1) x 7 = 105  

Q 2 - What digit should be placed in of H in 78423928H9 so as make it divisible by 9?

A - 3

B - 4

C - 7

D - 2

Answer : D

Explanation

  
For the number 78423928H9 to be divisible by 9, the sum of all  its digits should be divisible by 9.  
So                        (7+8+4+2+3+9+2+8+2+9)/9 = (52+H)/9 
= Number divisible by 9  
So H should b replaced by 2  

Q 3 - The least prime number is?

A - 3

B - 2

C - 1

D - 0

Answer : B

Explanation

2 is the least prime number.

Q 4 - The product of two successive positive integers is 462. Which is the smaller integer?

A - 20

B - 22

C - 21

D - 23

Answer : C

Explanation

  
 Suppose the two consecutive integers be y and y + 1 respectively.  According to question,       
 (y) x (y + 1) = 462  
 Or, y2 + y - 462 = 0  
 Or, y2 + 22y - 21y - 462 = 0  
 Or, y(y + 22) - 21(y + 22) = 0  
 Or, (y + 22) (y - 21) = 0  
 Or, y = 21 

Q 5 - What is next in 2, 5, 8,...?

A - 7

B - 9

C - 10

D - 11

Answer : D

Explanation

 
 Here numbers are 2, 5, 8, ... which is an A.P. Here a = 2,  d = 3,    
 So next number will be 8 + 3 = 11. 

Q 6 - What is the sum of first 20 odd numbers?

A - 210

B - 300

C - 400

D - 420

Answer : C

Explanation

 
 Here numbers are 1, 3, ..., upto 20 terms which is an A.P. Here a = 1,  d = 2, n = 20. 
 Now Using formula Sn = (n/2)[2a + (n-1)d]  
 ∴ Required sum = (20/2)[2+(20-1)x2]  = 10 x 40  = 400 

Q 7 - If 2th and 5th terms of a G.P. are 2/3 and 16/81 then what will be the 7th term?

A - 15/524

B - 1/32

C - 32/729

D - 64/729

Answer : D

Explanation

   
 Using formula Tn = arn- 1  
 T2 = ar(2 - 1)  
 => ar = 2/3 ... (i)  
 T5 = ar(5 - 1)  
 => ar4 = 16/81 ... (ii)  
 Dividing (ii) from (i)  
 => r3 = (2/3) x (81/16)  
 = 9/8 = (2/3)3  
 => r = 2/3  
 Using (i)  a = (2/3) / (2/3) = 1  
 => T7 = ar(7 - 1)  
 = (2/3)(6)  
 = 64/729 

Q 8 - If a clock buzzes 1 time at 1 o'clock , 2 times at 2 o'clock and so on then how many times it buzzes in a day?

A - 100

B - 150

C - 156

D - 166

Answer : C

Explanation

  
 Total buzzes = 2(1 + 2 + 3 ... + 12)   
 Here a = 1, d = 1 , l = 12   
 Using formula Sn = (n/2)[a+l]   
 Sn = (12/2)[1+12]  = 6 x 13  = 78 
 Thus total number of buzzes = 2 x 78 = 156. 

Q 9 - Which term of 2, 7, 12, 17... is 92?

A - 16th

B - 17th

C - 18th

D - 19th

Answer : D

Explanation

  
 Here a = 2,  d = 7 - 2 = 5,  
 Let there be n term.  
 Using formula Tn = a + (n - 1)d  
 Tn = 2 + (n - 1) x 5 = 92  
 => 5n - 3 = 92  
 => n = 19 

Q 10 - 12 + 22 ... + x2 = [x(x+1)(2x+1)]/6. What is 12 + 32 +... + 202?

A - 2818

B - 2100

C - 2485

D - 2500

Answer : A

Explanation

  
 (12 + 32 ... + 202)  = (12 + 22 ... + 202) - (22 + 42 ... + 192)  
 Using formula  (12 + 32 ... +  n2) = [n(n+1)(2n+1)]/6  
 [20(20+1)(40+1)]/6 - (1 x 22 +  22 x 22 + 22 x  32 + ... + 22 x  92 + 22 x 102)  = 2870 - 22(12 + 22 + ... + 192)  
 = 2870 -  4(1 x 2 x 39)/6  
 = 2870 - 52  
 = 2818 

aptitude_arithmetic.htm

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