# 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

As numbers are in A.P. Thus (y + 10) - 2y = (3y + 2) - (y + 10) => 10 - y = 2y - 8 => -3y = -18 => y = 6

Q 2 - Find the number which being increased by 1 will be exactly divisible by 13, 15 and 19?

### Answer : A

### Explanation

LCM of 13, 15 and 19 is 3705 So the desired number is3705-1=3704

Q 3 - The difference between local and the face value of 8 in the numeral is 568012?

### Answer : C

### Explanation

8000 - 8 = 7992

Q 4 - If 12^{3} is subtracted from the square of a number, the answer so obtained is 976. What is the number?

### Answer : D

### Explanation

Let the number be x. According to question: x^{2}- 12^{3}= 976 or, x = 52

Q 5 - How many numbers between 100 and 200 are exactly divisible by 6 and 9?

### Answer : B

### Explanation

L.C.M. of 9 and 6 = 18 ∴ required numbers are 108, 126, 144, 162, 180, 198 which are 6.

### Answer : D

### Explanation

Given sequence = 2(1 + 2 + ... 99) + 100 Using formula S_{n}= (n/2)(a + l) Required sum = 2 x (99/2)(1+99) +100 = 2 x 4950 + 100 = 9900 + 100 = 10000

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 - 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 : B

### Explanation

As numbers are in A.P. Thus (y + 10) - 2y = (3y + 5) - (y + 10) => 10 - y = 2y - 5 => -3y = -15 => y = 5

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