# 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 T_{n}= a + (n - 1)d T_{n}= 2 + (n - 1) x 5 = 87 => 5n - 3 = 87 => n = 18

Q 2 - If X and Y are two different prime numbers, then find the number of divisors of X^{2}Y^{2}?

### Answer : B

### Explanation

X^{2}Y^{2}= X^{2}x Y^{2}Number of divisors = (2+1)(2+1)= 9

### Answer : B

### Explanation

2 is the least prime number.

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 - 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 T_{10}= (a-b) + (10 - 1) x (-b) = 20 => a - 9b = 20 ... (i) T_{20}= (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 - How many odd numbered pages are present in a book of 1089 pages?

### Answer : D

### Explanation

Here pages are 1, 3, ..., 1089 which is an A.P. Here a = 1, d = 2, l = 1089 Using formula T_{n}= a + (n - 1)d T_{n}= 1 + (n - 1) x 2 = 1089 => 2n -1 = 1089 => n = 1090 / 2 = 545

### Answer : C

### Explanation

Here a = 2, r = 3, n = 8. Using formula T_{n}= ar^{n- 1}T_{n}= 2 x 3^{(8-1)}=2 x 3^{7}=2 x 2187 =4374

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

Q 9 - If an A.P. have 4^{th} term as 16 and 12^{th} term as 80. What will be its common difference?

### Answer : C

### Explanation

Let's have first term as a, common difference is d then a + 3d = 15 ... (i) a + 11d = 80 ... (ii) Subtracting (i) from (ii) => 8d = 64 => d = 8

### Answer : A

### Explanation

Here a = 72, d = 63 - 72 = -9, Using formula T_{n}= a + (n - 1)d T_{n}= 72 + (n - 1) x -9 = 9 => 81 - 9n = 9 => n = 8