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 unit digit of 415 x 66?

A - 0

B - 6

C - 8

D - 4

Answer : D

Explanation

  
Unit digit of 415 is 4 and unit digit of  66  is 6.  
So unit digit of 415 x 66  will be 4 x 6 =24 i.e., 4.  

Q 3 - The value of (3n+3n-1)/(3n+1-3n) is?

A - 5/2

B - 3/2

C - 2/3

D - 1/3

Answer : C

Explanation

  
 (3n+3n-1)/(3n+1-3n)  = (1+3-1)/ (3-1)  
 = 2/3  

Q 4 - ((991+771)2-(991-771)2)/(991x771)=?

A - 2

B - 8

C - 4

D - 6

Answer : C

Explanation

 
 ((991+771)2-(991-771)2)/(991x771)=((A+B)2-(A-B)2)/(AxB)  
 (4xAB)/AB=4  

Q 5 - How many 3-digits numbers are there which are completely divisible by 6?

A - 102

B - 150

C - 151

D - 156

Answer : B

Explanation

 
 Here numbers are 102, 108, ..., 996 which is an A.P. 
 Here a = 102,  d = 108 - 102 = 6,    
 Using formula Tn = a + (n - 1)d    
 Tn = 102 + (n - 1) x 6 = 996    
 => 96 - 6n = 996   
 => n = 900 / 6 
 = 150 

Q 6 - How many odd numbered pages are present in a book of 1089 pages?

A - 542

B - 543

C - 544

D - 545

Answer : D

Explanation

 
 Here pages are 1, 3, ..., 1089 which is an A.P. Here a = 1,  d = 2, l = 1089    
 Using formula Tn = a + (n - 1)d 
 Tn = 1 + (n - 1) x 2 = 1089 
 => 2n -1 = 1089 
 => n = 1090 / 2 = 545 

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 - In a club, member's ages are in A.P. with common difference of 3 months. If youngest member is 7 years old and sum of ages of all members is 250 years then how many members are there in the club?

A - 15

B - 20

C - 25

D - 30

Answer : C

Explanation

   
 let the ages be 7 , 7.25, 7.5 and so on  
 Here a = 7, d = 1/4 , Sn = 250   
 Using formula Sn = (n/2)[2a + (n-1)d]   
 => (n/2)[14+(n-1)(1/4)]  = 250  
 => n[14 + (n-1)/4] = 500  
 => n[56 + (n-1)] = 2000  
 => n[n + 55] = 2000  
 => n2 + 55n - 2000 = 0  
 => n2 + 80n -25n - 2000 = 0  
 => n(n-80) -25(n-80) = 0  
 => (n-80)(n-25) = 0  
 => n = 25 

Q 9 - (13 + 23 ... + 153) - (1 + 2 + ... + 15)= ?

A - 12280

B - 13280

C - 14280

D - 14400

Answer : C

Explanation

  
 Using formula  (13 + 23 ... +  n3) = [(1/2)n(n+1]2  
 (13 + 23 ... + 153) = [(15 x 16)/2]2  
 = 1202  = 14400  
 Using formula  (1 + 2 + ... n) = [(1/2)n(n+1]  
 ∴ (13 + 23 ... + 153) - (1 + 2 + ... + 15) 
 = 14400 - (1/2) x 15 x 16 = 14400 - 120  
 = 14280 

Q 10 - If population of a bacteria doubles every 2 minutes. In how much minutes, it will grow from 1000 to 1024000?

A - 20

B - 22

C - 23

D - 19

Answer : A

Explanation

   
 Let the required growth be 1000, 2000, 4000,...1024000.  
 Here, a = 1000, r = 2, Tn = 512000  
 Using formula Tn = arn-1  
 => 1000 x 2n-1 = 1024000  
 => 2n-1 = 1024 = 219  
 => n - 1 = 10  => n = 11 
 ∴ time taken will be 2 x 10 = 20 minutes. 

aptitude_arithmetic.htm

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