# 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 : B

### Explanation

Here a = 72, d = 63 - 72 = -9, Using formula T_{n}= a + (n - 1)d T_{n}= 72 + (n - 1) x -9 = 0 => 81 - 9n = 0 => n = 9

### Answer : B

### Explanation

Here series is 102, 108, 114, ... 996. Here a = 102, d = 108 - 102 = 6 Using formula T_{n}= a + (n - 1)d T_{n}= 102 + (n - 1) x 6 = 996 => 102 + 6n - 6 = 996 => 6n = 900 => n = 150

### Answer : C

### Explanation

(3^{n}+3^{n-1})/(3^{n+1}-3^{n}) = (1+3^{-1})/ (3-1) = 2/3

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

Q 5 - The average of four consecutive even numbers is one- fourth of the sum of these numbers. What is the difference between the first and the last number?

### Answer : B

### Explanation

Let the four consecutive even numbers are x, x + 2, x + 4 and x + 6. ∴ Required difference = x + 6 - x = 6

Q 6 - What is the sum of all natural numbers starting from 75 up to 97?

### Answer : D

### Explanation

Here numbers are 75, 76, ..., 97 which is an A.P. Here a = 75, d = 1, l = 97 Using formula T_{n}= a + (n - 1)d T_{n}= 75 + (n - 1) x 1 = 97 => 74 + n = 97 => n = 97 - 74 = 23 Now Using formula S_{n}= (n/2)(a + l) ∴ Required sum = (23/2)(75+97) = 23 x 86 = 1978

### 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 - One has to pay 3600 in 40 installments which are in A.P. After 30th installment being paid, amount left will be one third. What will be the 8^{th} installment?

### Answer : C

### Explanation

Installments = 40, Total debt = 3600 Installments = 30, Total debt = (2/3) x 3600 = 2400 Let the installments be a, a + d, a + 2d, ... Using formula S_{n}= (n/2)[2a+(n-1)d] S_{30}= (30/2)[2a+(30-1)d] = 3600 => 2a + 29d = 160 ... (i) S_{40}= (40/2)[2a+(40-1)d] = 2400 => 2a + 39d = 180 ... (ii) Subtracting (i) from (ii) => 10d = 20 => d = 2 Using (i) 2a = 160 - 29d = 160 - 58 = 102 => a = 51 Using formula T_{n}= a + (n-1)d ∴ T_{8}= 51 + 7 x 2 = 51 + 14 = 65

### Answer : D

### Explanation

Here a = 2, d = 7 - 2 = 5, Let there be n term. Using formula T_{n}= a + (n - 1)d T_{n}= 2 + (n - 1) x 5 = 92 => 5n - 3 = 92 => n = 19

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 T_{17}= -10 + (17 - 1) x 8 = 118