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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 - What is the 16^th term of A.P. 3, 5, 7, 9 ...?

Answer : C

Explanation

   
Here a = 3, d = 5 - 3 = 2, n = 16   
Using formula T_n = a + (n - 1)d   
T₁₆ = 3 + (16 - 1) x 2 = 33

Q 2 - The divisor is five times the quotient and five times the remainder. If the remainder is 29, the dividend is?

Answer : B

Explanation

  
Divisor = (5 x 29) = 145 = 5 x Quotient = Divisor 
=> Quotient = 145/5 = 29 
Dividend = (Divisor x Quotient) + Remainder 
Dividend = (145* 29) + 29 = 4234.

Q 3 - The difference between local and the face value of 8 in the numeral is 568012?

Answer : C

Explanation

  
 8000 - 8 
 = 7992

Q 4 - If (28)³ is subtracted from the square of a number, the answer so obtained is 1457. What is the number?

Answer : D

Explanation

 
 Let the number = x 
 According to question, x² - 283 = 1457 
 or, 
 x = 153

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 - if a, a-2 and 3a are in an A.P. what is a?

Answer : B

Explanation

 
 As a, a-2 and 3a are in an A.P. 
 ∴ (a-2) - a = 3a - (a-2) 
 => -2 = 2a - 2 
 => a = -2

Q 7 - Sum of how many terms of series 3, 9, 27 will be 363?

Answer : A

Explanation

   
 Here a = 3, r = 3 , S_n = 363  
 Using formula S_n = a(rⁿ - 1) / (r-1)  
 S_n = 3 x (3ⁿ - 1) / (3 - 1) = 363  
 => 3ⁿ - 1 = (363 x 2) / 3  
 => 3ⁿ = 243 
 = 3⁵  
 => n = 5

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?

Answer : C

Explanation

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

Q 9 - What is the 18^th term of A.P. 3, 5, 7, 9 ...?

Answer : D

Explanation

 
 Here a = 3, d = 5 - 3 = 2, n = 18 
 Using formula T_n = a + (n - 1)d 
 T₁₈ = 3 + (18 - 1) x 2 
 = 37

Q 10 - How many 2 digits numbers are there which are divisible by 6?

Answer : B

Explanation

  
 Here series is 12, 18, 24, ... 96.  
 Here a = 12,  d = 18 - 12 = 6     
 Using formula T_n = a + (n - 1)d     
 T_n = 12 + (n - 1) x 6 = 96     
 => 12 + 6n - 6 = 96    
 => 6n = 90  
 => n = 15

aptitude_arithmetic.htm