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 - If an A.P. have 4^th term as 14 and 12^th term as 70. What will be its 17^th term?

Answer : D

Explanation

  
Let's have first term as a, common difference is d then  
a + 3d = 14 ... (i)  
a + 11d = 70 ... (ii)  
Subtracting (i) from (ii)  
=> 8d = 56  => d = 7  
Using (i)  
=> a = 14 - 3d  = -7  
Using formula T_n = a + (n - 1)d    
T₁₇ = -7 + (17 - 1) x 7 = 105

Q 2 - Find the number which being increased by 1 will be exactly divisible by 13, 15 and 19?

Answer : A

Explanation

  
LCM of 13, 15 and 19 is 3705  
So the desired number is3705-1=3704

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 - If the sum of four consecutive even numbers is 228, which is the smallest of the numbers?

Answer : B

Explanation

  
 According to the question:       
 x + x + 2 + x + 4 + x + 6 = 228  
 or, 4x + 12 = 228  
 or, x = 54  
 ∴The least even number is 54.

Q 5 - If 10^th term of A.P. a, a-b, a-2b, ... is 20 and 20^th term is 10 then what will be x^th term?

Answer : D

Explanation

  
 Here a = a-b,  d = (a-2b) - (a-b) = -b,    
 Using formula T_n = a + (n - 1)d    
 T₁₀ = (a-b) + (10 - 1) x (-b) = 20    
 => a - 9b = 20 ... (i) 
 T₂₀ = (a-b) + (20 - 1) x (-b) = 10    
 => a - 19b = 10 ... (ii) 
 Subtracting (ii) from (i) 10b = 10 
 => b = 1 
 Using (i) a - 9(1) = 20 
 => a = 29 
 ∴ x^th term = a + (x-1)d 
 = a + (x-1)(-b) = 20 + (x-1)(-1) 
 = 30-x

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 - What is the sum of (1 + 1/2 + 1/4 + ...)?

Answer : A

Explanation

  
 This is an infinite G.P. with a = 1 and r = 1/2.  
 Sum of infinite G.P. = a/(1-r)  = 1/(1-1/2)  = 1/(1/2)  = 2

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 - (1³ + 2³ ... + 15³) - (1 + 2 + ... + 15)= ?

Answer : C

Explanation

  
 Using formula  (1³ + 2³ ... +  n³) = [(1/2)n(n+1]²  
 (1³ + 2³ ... + 15³) = [(15 x 16)/2]²  
 = 120²  = 14400  
 Using formula  (1 + 2 + ... n) = [(1/2)n(n+1]  
 ∴ (1³ + 2³ ... + 15³) - (1 + 2 + ... + 15) 
 = 14400 - (1/2) x 15 x 16 = 14400 - 120  
 = 14280

Q 10 - If an A.P. have 4^th term as 16 and 12^th term as 80. What will be its 17^th term?

Answer : A

Explanation

  
 Let's have first term as a, common difference is d then  
 a + 3d = 16 ... (i)  
 a + 11d = 80 ... (ii)  
 Subtracting (i) from (ii)  
 => 8d = 64  
 => d = 8  
 Using (i)  
 => a = 14 - 3d  = -10 
 Using formula T_n = a + (n - 1)d    
 T₁₇ = -10 + (17 - 1) x 8
 = 118

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