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 nth term of the series 72, 63, 54, ... is 0. What is n?

A - 8

B - 9

C - 10

D - 11

Answer : B

Explanation

     
Here a = 72,  d = 63 - 72 = -9,      
Using formula Tn = a + (n - 1)d      
Tn = 72 + (n - 1) x -9 = 0      
=> 81 - 9n = 0     
=> n = 9    

Q 2 - Find the number which being increased by 1 will be exactly divisible by 13, 15 and 19?

A - 3704

B - 3706

C - 3705

D - 3715

Answer : A

Explanation

  
LCM of 13, 15 and 19 is 3705  
So the desired number is3705-1=3704  

Q 3 - The sum of first 99 natural numbers is?

A - 9801

B - 9800

C - 4851

D - 4950

Answer : D

Explanation

  
 Sum of n natural numbers is  Sn=n(n+1)/2  =99(99+1)/2  
 =4950  

Q 4 - The difference between 58% and 39% of a number is 247. What is 62% of that number?

A - 1300

B - 806

C - 754

D - 1170

Answer : B

Explanation

  
 Let the number be y.  
 According to question,       (58 - 39)% of y = 247  
 Or, y x (19/100) = 247  
 Or, y = 1300  
 ∴62% of 1300 = (62/100) x 1300 = 806 

Q 5 - How many numbers between 10 and 200 are exactly divisible by 7?

A - 21

B - 23

C - 25

D - 27

Answer : D

Explanation

 
 Here numbers are 14, 21, ..., 196 which is an A.P. 
 Here a = 14,  d = 21 - 14 = 7,    
 Using formula Tn = a + (n - 1)d    
 Tn = 14 + (n - 1) x 7 = 196    
 => 7 - 7n = 196   
 => n = 189 / 7 = 27 

Q 6 - What is the sum of all odd numbers between 100 and 200?

A - 3750

B - 6200

C - 6500

D - 7500

Answer : D

Explanation

  
 Required sum = 101 + 103 + ... + 199 which is an A.P. where a = 101, d = 2, l = 199.  
 Using formula Tn = a + (n - 1)d  
 Tn = 101 + (n-1)2 = 199  
 => 2n = 199 - 99 = 100  
 => n = 50  
 Now Using formula Sn = (n/2)(a + l)  
 ∴ Required sum = (50/2)(101+199)  = 50 x 150  = 7500 

Q 7 - If 4th and 9th terms of a G.P. are 54 and 13122 then what will be the 2nd term?

A - 6

B - 12

C - 18

D - 9

Answer : A

Explanation

   
 Using formula Tn = arn- 1  
 T4 = ar(4 - 1)  
 => ar3 = 54 ... (i)  
 T9 = ar(9 - 1)  
 => ar8 = 13122 ... (ii)  
 Dividing (ii) from (i)  
 => r5 = 243 = 35 
 => r = 3  
 Using (i)  a = 54 / 9 = 6 

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 - What is the 115th term of A.P. 4, 4.5, 5, 5.5 ... ?

A - 56

B - 55.5

C - 59

D - 55

Answer : C

Explanation

 Here a = 4, d = 4.5 - 4 = 0.5, n = 115 
 Using formula Tn = a + (n - 1)d 
 T115 = 4 + (115 - 1) x 0.5 
 = 59 

Q 10 - If 12 + 22 ... + 202 = 2870. What is 22 + 42 +... + 402?

A - 770

B - 1155

C - 1540

D - 3852

Answer : C

Explanation

   
 (22 + 42 ... + 402)  = (1 x 22 + 22 x 22 + 32 x 22 +... +  202 x 22)  
 = 22(12 + 22 + .... + 202) 
 = 4 x  385  
 = 1540 

aptitude_arithmetic.htm

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