Problems on Numbers Quiz Set 012

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Question 1

What is 0.19191919...?

 A

$19/99$.

 B

$20/99$.

 C

$19/100$.

 D

$19/101$.

Soln.
Ans: a

0.19191919... is same as $0.\ov 19$. Let y = $0.\ov 19$. Multiply by 100, 100y = 19 + $0.\ov 19$ which is same as 19 + y. So 99y = 19, ⇒ y = $19/99$. So answer = ${19/99}$.


Question 2

How many prime factors does 4095000 have?

 A

11.

 B

12.

 C

10.

 D

13.

Soln.
Ans: a

We can see that $4095000 = 2^3 × 3^2 × 5^4 × 7^1 × 13^1$. The number of prime factors is sum of the powers = 11.


Question 3

x should be replaced by which minimum number so that 36318x5944 is completely divisible by 9?

 A

2.

 B

3.

 C

5.

 D

4.

Soln.
Ans: a

If the above number has to be divisible by 9, the sum of the digits, i.e., 3 + 6 + 3 + 1 + 8 + x + 5 + 9 + 4 + 4, should be divisible by 9. So we can see that $x + 43$ should be divisible by 9. By inspection, x = 2.


Question 4

If  $x/6 = y/12 = z/28$, then ${x + y + z}/x$ = ?

 A

$7{2/3}$.

 B

13.

 C

$6{2/3}$.

 D

$6{2/5}$.

Soln.
Ans: a

Let $x/6 = y/12 = z/28$ = k. Then ${x + y + z}/x$ is same as ${6k + 12k + 28k}/{6k}$ which equals $7{2/3}$.

${23/3}$ is same as $7{2/3}$.


Question 5

If ${13/20}$ of a number is 23, what is $1{10/13}$ of that number?

 A

$62{102/169}$.

 B

$63{55/56}$.

 C

$60{151/171}$.

 D

$64{143/171}$.

Soln.
Ans: a

Let the number be N. Then it is given that ${13/20}$ x N = 23. ⇒ N = 23 X ${20/13}$. ⇒ ${23/13}$ x N = ${23/13}$ x 23 X ${20/13}$ = ${10580/169}$.

${23/13}$ is same as $1{10/13}$. ${10580/169}$ is same as $62{102/169}$.


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Creative Commons License
This Blog Post/Article "Problems on Numbers Quiz Set 012" 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


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