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 - How many terms are present in A.P. 7, 11, 15, ..., 139?

Answer : D

Explanation

   
Here a = 7,  d = 11 - 7 = 4,   
Let there be n term.   
Using formula T_n = a + (n - 1)d   
T_n = 7 + (n - 1) x 4 = 139   
=> 4n + 3 = 139  
=> n = 34

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 |x - 5| = 10 and |2y - 12| = 8, what is the minimum possible value of y/x?

Answer : A

Explanation

  
 y =10, 2.  
 x =15,-5.  
 So minimum value would be 10 / -5 = -2

Q 4 - The product of two successive positive integers is 462. Which is the smaller integer?

Answer : C

Explanation

  
 Suppose the two consecutive integers be y and y + 1 respectively.  According to question,       
 (y) x (y + 1) = 462  
 Or, y² + y - 462 = 0  
 Or, y² + 22y - 21y - 462 = 0  
 Or, y(y + 22) - 21(y + 22) = 0  
 Or, (y + 22) (y - 21) = 0  
 Or, y = 21

Q 5 - The average of four consecutive even numbers is one- fourth of the sum of these numbers. What is the difference between the first and the last number?

Answer : B

Explanation

  
 Let the four consecutive even numbers are x, x + 2, x + 4 and x + 6.  
 ∴ Required difference = x + 6 - x = 6

Q 6 - What is the sum of all even numbers between 100 and 200 including both?

Answer : D

Explanation

  
 Required sum = 100 + 102 + ... + 200 which is an A.P. where a = 100, d = 2, l = 200.  
 Using formula T_n = a + (n - 1)d  
 T_n = 100 + (n-1)2 = 200  
 => 2n = 200 - 98 = 102  
 => n = 51  
 Now Using formula S_n = (n/2)(a + l)  
 ∴ Required sum = (51/2)(100+200)  = 51 x 150  = 7550

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 - A piles of logs is arranged in such a way that top layer contains one log and each lower layer has one log more than the layer above. if there are 15 layers in total than how many logs are arranged in that pile?

Answer : B

Explanation

   
 Total logs = 1 + 2 + 3 ... + 15   
 Here a = 1, d = 1 , l = 15   
 Using formula S_n = (n/2)[a+l]   
 S_n = (15/2)[1+15]  = 15 x 8  = 120

Q 9 - If 1² + 2² ... + 10² = 385. What is 2² + 4² +... + 20²?

Answer : C

Explanation

   
 (2² + 4² ... + 20²)  = (1 x 2² + 2² x 2² + 3² x 2² +... +  10² x 2²)  
 = 2²(1² + 2² + .... + 10²) 
 = 4 x  385  
 = 1540

Q 10 - 1² + 2² ... + x² = [x(x+1)(2x+1)]/6. What is 1² + 3² +... + 20²?

Answer : A

Explanation

  
 (1² + 3² ... + 20²)  = (1² + 2² ... + 20²) - (2² + 4² ... + 19²)  
 Using formula  (1² + 3² ... +  n²) = [n(n+1)(2n+1)]/6  
 [20(20+1)(40+1)]/6 - (1 x 2² +  2² x 2² + 2² x  3² + ... + 2² x  9² + 2² x 10²)  = 2870 - 2²(1² + 2² + ... + 19²)  
 = 2870 -  4(1 x 2 x 39)/6  
 = 2870 - 52  
 = 2818

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