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.
Q 1 - If an A.P. have 4th term as 14 and 12th term as 70. What will be its common difference?
Answer : C
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
Q 2 - If X and Y are two different prime numbers, then find the number of divisors of X2Y2?
Answer : B
Explanation
X2Y2 = X2 x Y2 Number of divisors = (2+1)(2+1)= 9
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 - 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 - 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 : C
Explanation
Here numbers are 1, 3, ..., upto 20 terms which is an A.P. Here a = 1, d = 2, n = 20. Now Using formula Sn = (n/2)[2a + (n-1)d] ∴ Required sum = (20/2)[2+(20-1)x2] = 10 x 40 = 400
Answer : A
Explanation
Here a = 3, r = 3 , Sn = 363 Using formula Sn = a(rn - 1) / (r-1) Sn = 3 x (3n - 1) / (3 - 1) = 363 => 3n - 1 = (363 x 2) / 3 => 3n = 243 = 35 => n = 5
Answer : D
Explanation
Using formula (13 + 23 ... + n3) = [(1/2)n(n+1]2 (13 + 23 ... + 153) = [(15 x 16)/2]2 = 1202 = 14400
Answer : D
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 = 92 => 5n - 3 = 92 => n = 19
Q 10 - Suman purchases N.S.C. every year whose value exceed s previous year's N.S.C by 500 Rs. In 10 years, she has bought N.S.Cs of 50000 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 = 500, n = 10, S10 = 50000 Using formula S10 = (n/2)[2a + (n-1)d => (10/2)[2a + (10-1)500] = 50000 => 5(2a + 9 x 500) = 50000 => 2a + 4500 = 10000 => a = 5500 / 2 = 2750
