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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.
Q 1 - If an A.P. have 4th term as 14 and 12th term as 70. What will be its 17th 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 Tn = a + (n - 1)d T17 = -7 + (17 - 1) x 7 = 105
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 - 5#2 is a three digit number with # as a missing number. If the number is divisible by 6, the missing number is?
Answer : A
Explanation
If the number is divisible by 6, the sum of the number must be divisible by 6.
Q 4 - The difference between 58% and 39% of a number is 247. What is 62% of that number?
Answer : B
Explanation
Let the number be y. According to question, (58 - 39)% of y = 247 Or, y x (19/100) = 247 Or, y = 1300 ∴62% of 1300 = (62/100) x 1300 = 806
Answer : D
Explanation
Here numbers are 2, 5, 8, ... which is an A.P. Here a = 2, d = 3, So next number will be 8 + 3 = 11.
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 Tn = a + (n - 1)d Tn = 102 + (n - 1) x 3 = 198 => 99 + 3n = 198 => n = 99 / 3 = 33 Now Using formula Sn = (n/2)(a + l) ∴ Required sum = (33/2)(102+198) = 33 x 150 = 4950
Q 7 - If a ≠ b then which of the following statement is correct?
Answer : C
Explanation
For any two unequal number a and b, arithmetic mean is always greater than their geometric mean. ∴ c is the correct answer.
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
Answer : C
Explanation
Here a = 4, d = 4.5 - 4 = 0.5, n = 115 Using formula Tn = a + (n - 1)d T115 = 4 + (115 - 1) x 0.5 = 59
Answer : B
Explanation
Here series is 12, 18, 24, ... 96. Here a = 12, d = 18 - 12 = 6 Using formula Tn = a + (n - 1)d Tn = 12 + (n - 1) x 6 = 96 => 12 + 6n - 6 = 96 => 6n = 90 => n = 15