# 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 4^{th} term as 14 and 12^{th} term as 70. What will be its 17^{th} 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 T_{n}= a + (n - 1)d T_{17}= -7 + (17 - 1) x 7 = 105

Q 2 - What digit should be placed in of H in 78423928H9 so as make it divisible by 9?

### Answer : D

### Explanation

For the number 78423928H9 to be divisible by 9, the sum of all its digits should be divisible by 9. So (7+8+4+2+3+9+2+8+2+9)/9 = (52+H)/9 = Number divisible by 9 So H should b replaced by 2

### Answer : B

### Explanation

2 is the least prime number.

Q 4 - The product of two successive positive integers is 462. Which is the smaller integer?

### Answer : C

### Explanation

Suppose the two consecutive integers be y and y + 1 respectively. According to question, (y) x (y + 1) = 462 Or, y^{2}+ y - 462 = 0 Or, y^{2}+ 22y - 21y - 462 = 0 Or, y(y + 22) - 21(y + 22) = 0 Or, (y + 22) (y - 21) = 0 Or, y = 21

### 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.

### 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 S_{n}= (n/2)[2a + (n-1)d] ∴ Required sum = (20/2)[2+(20-1)x2] = 10 x 40 = 400

Q 7 - If 2^{th} and 5^{th} terms of a G.P. are 2/3 and 16/81 then what will be the 7^{th} term?

### Answer : D

### Explanation

Using formula T_{n}= ar^{n- 1}T_{2}= ar^{(2 - 1)}=> ar = 2/3 ... (i) T_{5}= ar^{(5 - 1)}=> ar^{4}= 16/81 ... (ii) Dividing (ii) from (i) => r^{3}= (2/3) x (81/16) = 9/8 = (2/3)^{3}=> r = 2/3 Using (i) a = (2/3) / (2/3) = 1 => T_{7}= ar^{(7 - 1)}= (2/3)^{(6)}= 64/729

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 S_{n}= (n/2)[a+l] S_{n}= (12/2)[1+12] = 6 x 13 = 78 Thus total number of buzzes = 2 x 78 = 156.

### 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 - 1^{2} + 2^{2} ... + x^{2} = [x(x+1)(2x+1)]/6. What is 1^{2} + 3^{2} +... + 20^{2}?

### Answer : A

### Explanation

(1^{2}+ 3^{2}... + 20^{2}) = (1^{2}+ 2^{2}... + 20^{2}) - (2^{2}+ 4^{2}... + 19^{2}) Using formula (1^{2}+ 3^{2}... + n^{2}) = [n(n+1)(2n+1)]/6 [20(20+1)(40+1)]/6 - (1 x 2^{2}+ 2^{2}x 2^{2}+ 2^{2}x 3^{2}+ ... + 2^{2}x 9^{2}+ 2^{2}x 10^{2}) = 2870 - 2^{2}(1^{2}+ 2^{2}+ ... + 19^{2}) = 2870 - 4(1 x 2 x 39)/6 = 2870 - 52 = 2818