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 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 - 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 -2≤X≤3 and 3≤Y≤6, the least possible value of 3Y-2X is

Answer : A

Explanation

  
 For 3Y-2X to be minimum the condition is that Y must be substituted with least value and X must be with large value  
 => 3(3)-2(3) 
 =3.

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 - 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 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 - What is the 8^th term of G.P. 2, 6, 18,...?

Answer : C

Explanation

  
 Here a = 2,  r = 3, n = 8. 
 Using formula T_n = ar^{n- 1} 
 T_n = 2 x 3^(8-1)  
 =2 x 3⁷  
 =2 x 2187  =4374

Q 8 - One has to pay Rs 975 in yearly installments where each installment is less than earlier one by Rs 5. If first installment is of 100 in how many years, the entire amount will be paid?

Answer : C

Explanation

   
 Here a = 100, d = -5 , S_n = 975  
 Using formula S_n = (n/2)[2a + (n-1)d]  
 S_n = (n/2)[200 + (n-1)(-5)] = 975  
 => n(205 - 5n) = 1950  
 => 5n² - 205n + 1950 = 0  
 => n² - 41n + 390 = 0  
 => n² - 26n - 15n + 390 = 0  
 => n(n-26) - 15(n-26)= 0  
 => (n-26)(n-15)  
 => n = 15

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 - 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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