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Aptitude - Arithmetic Online 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.
Answer : C
Explanation
Here a = 2, d = 7 - 2 = 5, Let there be n term. Using formula Tn = a + (n - 1)d Tn = 2 + (n - 1) x 5 = 87 => 5n - 3 = 87 => n = 18
Q 2 - The divisor is five times the quotient and five times the remainder. If the remainder is 29, the dividend is?
Answer : B
Explanation
Divisor = (5 x 29) = 145 = 5 x Quotient = Divisor => Quotient = 145/5 = 29 Dividend = (Divisor x Quotient) + Remainder Dividend = (145* 29) + 29 = 4234.
Answer : D
Explanation
Sum of n natural numbers is Sn=n(n+1)/2 =99(99+1)/2 =4950
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 3-digits numbers are there which are completely divisible by 6?
Answer : B
Explanation
Here numbers are 102, 108, ..., 996 which is an A.P. Here a = 102, d = 108 - 102 = 6, Using formula Tn = a + (n - 1)d Tn = 102 + (n - 1) x 6 = 996 => 96 - 6n = 996 => n = 900 / 6 = 150
Answer : D
Explanation
Required sum = 101 + 103 + ... + 199 which is an A.P. where a = 101, d = 2, l = 199. Using formula Tn = a + (n - 1)d Tn = 101 + (n-1)2 = 199 => 2n = 199 - 99 = 100 => n = 50 Now Using formula Sn = (n/2)(a + l) ∴ Required sum = (50/2)(101+199) = 50 x 150 = 7500
Q 7 - If first term of a G.P. is 5, common ratio is 2 what is the 8th term?
Answer : C
Explanation
Here a = 5, r = 2, n = 8. Using formula Tn = arn- 1 Tn = 5 x 2(8-1) =5 x 27 =5 x 128 =640
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 Sn = (n/2)[a+l] Sn = (12/2)[1+12] = 6 x 13 = 78 Thus total number of buzzes = 2 x 78 = 156.
Answer : A
Explanation
Here a = 7, d = 11 - 7 = 4, Let there be n term. Using formula Tn = a + (n - 1)d Tn = 7 + (n - 1) x 4 = 151 => 4n + 3 = 151 => n = 37
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