Aptitude Test Preparation
Aptitude - Home
Aptitude - Overview
Quantitative Aptitude

Aptitude - Arithmetic Online Quiz

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.

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

Answer : B

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

Q 2 - Which of the following is equal to 205x15?

Answer : B

Explanation

  
205x15= (200+5) x15
=200x15+5x15

Q 3 - 5#2 is a three digit number with # as a missing number. If the number is divisible by 6, the missing number is?

Answer : A

Explanation

If the number is divisible by 6, the sum of the number must be divisible by 6.

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

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 multiples of 3 are available between 15 and 105 including both?

Answer : B

Explanation

 
 Here numbers are 15, 18, ..., 105 which is an A.P. 
 Here a = 15,  d = 3,    
 Using formula T_n = a + (n - 1)d    
 T₁₁ = 15 + (n - 1) x 3 = 105 
 => 12 + 3n = 105 
 => n = 93 / 3 = 31

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

Answer : D

Explanation

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

Q 7 - Sum of three numbers in G.P. is 28 and there product is 512. What are the numbers?

Answer : C

Explanation

   
 let the numbers are a/r, a, ar  
 Then a/r x a x ar = 512  
 => a³ = 8³  
 gt; a = 8  
 Now a/r + a + ar = 28  
 => 8/r + 8 + 8r = 28  
 => 8/r + 8r = 20  
 => 2/r + 2r = 5  
 => 2r² + -5r + 2 = 0  
 => 2r² + -4r -r + 2 = 0  
 => 2r(r-2) - (r-2)=0  
 => (r-2)(2r-1) = 0  
 => r = 2 or r = 1/2  
 ∴ numbers are 4, 8, 16.

Q 8 - 14² + 15² ... + 30² = ?

Answer : D

Explanation

  
 Using formula  (1² + 3² ... +  n²) = [n(n+1)(2n+1)]/6  
 14² + 15² ... + 30² = (1² + 2² ... + 30²) - (1² + 2² ... + 13²)  
 = (30 x 31 x 61)/ 6 - (13 x 14 x 27) / 6  
 = 9455 - 819  = 8636

Q 9 - What is the 15^th term of A.P. 14, 9, -1, -6 ... ?

Answer : C

Explanation

 Here a = 14, d = 9 - 14 = -5, n = 15 
 Using formula T_n = a + (n - 1)d 
 T₁₅ = 14 + (15 - 1) x -5 
 = -56

Q 10 - If n^th term of the series 72, 63, 54, ... is 9. What is n?

Answer : A

Explanation

   
 Here a = 72,  d = 63 - 72 = -9,    
 Using formula T_n = a + (n - 1)d    
 T_n = 72 + (n - 1) x -9 = 9    
 => 81 - 9n = 9   
 => n = 8

aptitude_arithmetic.htm

Print Page