Problems on Numbers Quiz Set 019

Advertisement


Your Score
Correct Answers:
Wrong Answers:
Unattempted:

Question 1

x should be replaced by which minimum number so that 936219x862 is completely divisible by 3?

 A

2.

 B

3.

 C

5.

 D

4.

Soln.
Ans: a

If the above number has to be divisible by 3, the sum of the digits, i.e., 9 + 3 + 6 + 2 + 1 + 9 + x + 8 + 6 + 2, should be divisible by 3. So we can see that $x + 46$ should be divisible by 3. By inspection, x = 2.


Question 2

8 times the middle of three consecutive even numbers is 22 more than 5 times the smallest of the three numbers. What is the middle number?

 A

4.

 B

5.

 C

3.

 D

6.

Soln.
Ans: a

This is the general solution. Let the numbers be 2n - 2, 2n and 2n + 2. We are given 8 × 2n = 22 + 5 × (2n - 2).
⇒ 8 × 2n = 22 + 5 × 2n - 5 × 2.
⇒ 2n × (8 - 5) = 22 - 5 × 2, so 2n = ${22 - 5 × 2}/{8 - 5} = 4$


Question 3

x should be replaced by which minimum number so that 275198x123 is completely divisible by 3?

 A

1.

 B

2.

 C

4.

 D

3.

Soln.
Ans: a

If the above number has to be divisible by 3, the sum of the digits, i.e., 2 + 7 + 5 + 1 + 9 + 8 + x + 1 + 2 + 3, should be divisible by 3. So we can see that $x + 38$ should be divisible by 3. By inspection, x = 1.


Question 4

x should be replaced by which minimum number so that 99938x3687 is completely divisible by 9?

 A

1.

 B

2.

 C

4.

 D

3.

Soln.
Ans: a

If the above number has to be divisible by 9, the sum of the digits, i.e., 9 + 9 + 9 + 3 + 8 + x + 3 + 6 + 8 + 7, should be divisible by 9. So we can see that $x + 62$ should be divisible by 9. By inspection, x = 1.


Question 5

What is 0.99999999...?

 A

$1/1$.

 B

$2/1$.

 C

$1/2$.

 D

$1/3$.

Soln.
Ans: a

0.99999999... is same as $0.\ov 99$. Let y = $0.\ov 99$. Multiply by 100, 100y = 99 + $0.\ov 99$ which is same as 99 + y. So 99y = 99, ⇒ y = $99/99$. So answer = 1.


buy aptitude video tutorials


Creative Commons License
This Blog Post/Article "Problems on Numbers Quiz Set 019" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2017-06-24.

Posted by Parveen(Hoven),
Aptitude Trainer


Advertisement