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 y if 2y, y+10, 3y+2 are in an A.P.?

Answer : C

Explanation

    
As numbers are in A.P.  
Thus (y + 10) - 2y = (3y + 2) - (y + 10)  
=> 10 - y = 2y - 8  
=> -3y = -18  
=> y = 6

Q 2 - If X and Y are two different prime numbers, then find the number of divisors of X²Y²?

Answer : B

Explanation

 
X²Y² = X² x Y²  
Number of divisors = (2+1)(2+1)= 9

Q 3 - (312x312+389x389+2x312x389) =?

Answer : C

Explanation

  
 By using a²+b²+2ab = (a + b)²  
 (312+389)²  =701²  
 =491401

Q 4 - Two numbers are less than the third number by 50% and 54% respectively. By how much percent is the second number less than the first number?

Answer : D

Explanation

 
 Let the third number be 100. 
 ∴First Number = 50 and Second Number = 46 
 Decrease = 50 - 46 = 4 
 ∴Required Percentage = (4/50)x100 = 8%

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

Answer : A

Explanation

   
 Using formula T_n = ar^{n- 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 - Suman purchases N.S.C. every year whose value exceed s previous year's N.S.C by 400 Rs. In 8 years, she has bought N.S.Cs of 48000 Rs. What was the value of N.S.C. she bought in first year?

Answer : D

Explanation

   
 Let the required amount is a. 
 Also, d = 400, n = 8, S₈ = 48000  
 Using formula S₈ = (n/2)[2a + (n-1)d  
 => (8/2)[2a + (8-1)400] = 48000  
 => 4(2a + 7 x 400) = 48000  
 => 2a + 3800 = 12000  
 => a = 9200 / 2 = 4600

Q 9 - Which term of 2, 7, 12, 17... is 92?

Answer : D

Explanation

  
 Here a = 2,  d = 7 - 2 = 5,  
 Let there be n term.  
 Using formula T_n = a + (n - 1)d  
 T_n = 2 + (n - 1) x 5 = 92  
 => 5n - 3 = 92  
 => n = 19

Q 10 - (1³ + 2³ ... + 20³) - (1 + 2 + ... + 20)= ?

Answer : C

Explanation

  
 Using formula  (1³ + 2³ ... +  n³) = [(1/2)n(n+1]²  
 (1³ + 2³ ... + 20³) = [(20 x 21)/2]²  = 210²  = 44100  
 Using formula  (1 + 2 + ... n) = [(1/2)n(n+1]  
 ∴ (1³ + 2³ ... + 15³) - (1 + 2 + ... + 15) = 44100 - (1/2) x 15 x 16 
 = 44100 - 120  
 = 43980

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