Aptitude - Questions & Answers

Aptitude - Arithmetic Online Quiz

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 105^th term of A.P. 4, 4.5, 5, 5.5 ... ?

Answer : A

Explanation

Here a = 4, d = 4.5 - 4 = 0.5, n = 105 
Using formula T_n = a + (n - 1)d 
T₁₀₅ = 4 + (105 - 1) x 0.5 
= 56

Q 2 - How many 3 digits numbers are there which are divisible by 6?

Answer : B

Explanation

  
Here series is 102, 108, 114, ... 996.  
Here a = 102,  d = 108 - 102 = 6     
Using formula T_n = a + (n - 1)d     
T_n = 102 + (n - 1) x 6 = 996     
=> 102 + 6n - 6 = 996    
=> 6n = 900  
=> n = 150

Q 3 - If 3^X-1 +3^X+1 = 270, then X=?

Answer : D

Explanation

  
 3^{X-1^{+3^X+1 = 270  
 => 3^X(3^-1+3¹) = 270  
 => 3^X x (10/3) = 270  
 => 3^X-1 = 270/10 
 = 27 
 =3³  
 => X-1=3  
 => X=4}}

Q 4 - In a three digit number the digit in the unit's place is twice the digit in the ten's place and 1.5 times the digit in the hundred's place. If the sum of all the three digits of the number is 13, what is the number?

Answer : B

Explanation

 
 Let the ten's digit be y. 
 ∴Unit's digit = 2y And Hundred's digits = 2y/1.5 
 According to the question,       
 y + 2y + 2y/1.5 = 13 or, 
 (1.5y + 3y + 2y) /1.5 = 13 or, 
 6.5y = 13 x 1.5 or, 
 y = (13 x 1.5)/6.5 or, 
 y = 3 
 ∴Unit's digit = 6 and hundred's digit = 6/1.5 = 4 
 ∴Number = 436

Q 5 - The number obtained by interchanging the digits of a two digit number is less than the original number by 18. If sum of the digits is 6, what was the original two digit number?

Answer : C

Explanation

  
 Let the original number be 10x + y.  
 Number obtained by interchanging the digits = 10y + x  
 ∴ (10x + y) - (10y + x) = 18  
 Or, x - y = 2 ... (i)  
 Also, x + y = 6 ... (ii)  
 From equations (i) and (ii), 
 x = 4 and y = 2.  
 ∴ Original number = (10 x 4) + 2 = 42

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

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 S_n = (n/2)[2a + (n-1)d]  
 ∴ Required sum = (20/2)[2+(20-1)x2]  = 10 x 40  = 400

Q 7 - If a ≠ b then which of the following statement is correct?

Answer : C

Explanation

For any two unequal number a and b, arithmetic mean is always greater than their geometric mean. ∴ c is the correct answer.

Q 8 - One has to pay 3600 in 40 installments which are in A.P. After 30th installment being paid, amount left will be one third. What will be the 8^th installment?

Answer : C

Explanation

   
 Installments = 40, Total debt = 3600  Installments = 30, 
 Total debt = (2/3) x 3600 = 2400  
 Let the installments be a, a + d, a + 2d, ...  
 Using formula S_n = (n/2)[2a+(n-1)d]   
 S₃₀ = (30/2)[2a+(30-1)d] = 3600  
 => 2a + 29d = 160 ... (i)  
 S₄₀ = (40/2)[2a+(40-1)d] = 2400  
 => 2a + 39d = 180 ... (ii)  
 Subtracting (i) from (ii)  
 => 10d = 20  
 => d = 2  
 Using (i)  2a = 160 - 29d = 160 - 58 = 102  
 => a = 51  
 Using formula T_n = a + (n-1)d  
 ∴ T₈ = 51 + 7 x 2 = 51 + 14  = 65

Q 9 - What is the 115^th term of A.P. 4, 4.5, 5, 5.5 ... ?

Answer : C

Explanation

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

Q 10 - 14² + 15² ... + 50² = ?

Answer : D

Explanation

  
 Using formula  (1² + 3² ... +  n²) = [n(n+1)(2n+1)]/6  
 14² + 15² ... + 50² = (1² + 2² ... + 50²) - (1² + 2² ... + 13²)  
 = (50 x 51 x 101)/ 6 - (13 x 14 x 27) / 6  
 = 6739225 - 819  
 = 6738406

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