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 - Find the unit digit of 4¹⁵ x 6⁶?

Answer : D

Explanation

  
Unit digit of 4¹⁵ is 4 and unit digit of  6⁶  is 6.  
So unit digit of 4¹⁵ x 6⁶  will be 4 x 6 =24 i.e., 4.

Q 3 - How many digits are required to write numbers between 1 to 50?

Answer : B

Explanation

  
 1 to 9       ------ 9 digits  
 10 to 50   ------ 82 digits

Q 4 - What is the greater of two numbers whose product is 1092 and the sum of the two numbers exceeds their difference by 42?

Answer : C

Explanation

  
 Let the numbers be x and y respectively.  
 According to question,       (x + y) - (x - y) = 42  
 or, y = 21  
 ∴x = 1092/21 = 52  
 Greater number is 52.

Q 5 - How many numbers between 200 and 600 are exactly divisible by 4, 5 and 6?

Answer : B

Explanation

 
 L.C.M. of 4, 5 and 6 (2 x 2) x (5) x (3 x 2) = 4 x 5 x 3 = 60 
 ∴ required numbers are 240, 300, 360, 420, 480, 540 which are 6.

Q 6 - What is the sum of all natural numbers which are multiples of 3 and lies between 100 and 200.

Answer : B

Explanation

 
 Here numbers are 102, 105, ..., 198 which is an A.P. Here a = 102,  d = 23, l = 198. 
 Using formula T_n = a + (n - 1)d 
 T_n = 102 + (n - 1) x 3 = 198 
 => 99 + 3n = 198 
 => n = 99 / 3 = 33 
 Now Using formula S_n = (n/2)(a + l)  
 ∴ Required sum = (33/2)(102+198)  
 = 33 x 150  = 4950

Q 7 - If third term of a G.P. is 4. What will be the product of first 5 terms?

Answer : C

Explanation

  
 Using formula T_n = ar^{n- 1} 
 T₃ = a x r^(3-1) = 4  
 => ar² = 4  
 Thus product of first five terms  = a x ar x ar² x ar³ x ar³  
 = a⁵r¹⁰  
 = (ar²)⁵  
 = 4⁵

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 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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