Real Numbers(10th Grade > Mathematics ) Questions and answers for exam Preparation

10th Grade > Mathematics

REAL NUMBERS MCQs

Total Questions : 58 | Page 6 of 6 pages

Question 51.

If $132300 = 2^{2} \times 3^{3} \times 5^{2} \times a^{b},$ then which of the following is true ?

$a = 2 b$
$a + b = 8$
LCM of $a$ and $b$ is 14.
$a = b$

Discuss Question

Answer: Option C. -> LCM of

a

and

b

is 14.
:
C

$132300 = 2^{2} \times 3^{3} \times 5^{2} \times a^{b}$
By fundamental theorem of arithmetic, 132300 can be written as $2^{2} \times 3^{3} \times 5^{2} \times 7^{2} .$
On comparison, we get $a = 7$ and $b = 2.$
LCM of 7 and 2 is 14.

Question 52.

The least multiple of 7 which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is ___.

Discuss Question

Answer: Option D. -> 364
:
D

We first find the L.C.M of 6, 9, 15 and 18, which is 90.

For getting remainder 4, we need to add 4 to it.
i.e., 90 +4 = 94.

But, 94 is not divisible by 7. So, we proceed with the next multiple of 90, which is 180.

Adding 4 to 180, we get 184, which is also not divisible by 7.

The next multiple of 90 is 270.

Adding 4 to 270, we get 274, again not divisible by 7.

The next multiple of 90 is 360.

Adding 4 to it gives us 364, divisible by 7( as $\frac{364}{7} = 52$ ).

Hence, the answer is 364.

Question 53.

Which of the following are irrational numbers?

0.2
3.1415926535... (non-repeating and non-terminating)
$\sqrt{3}$
${0.2}^{2}$

Discuss Question

Answer: Option B. -> 3.1415926535... (non-repeating and non-terminating)
:
B and C

Non-terminating and non-repeating decimals are irrational. Hence, 3.1415926535... and $\sqrt{3}$ are irrational.

Question 54.

If the HCF of 60 and 168 is 12, what is the LCM?

Discuss Question

Answer: Option C. -> 840
:
C

Given: HCF of 60 and 168 = 12
Product of two given numbers = Product of their HCF and LCM
$\Rightarrow$ 60 $\times$ 168 = 12 $\times$ LCM
$\Rightarrow$ LCM = $\frac{60 \times 168}{12}$
$\Rightarrow$ LCM = 840
$∴$ LCM of 60 and 168 is 840.

Question 55.

The greatest four digits number which is divisible by 15, 25, 40 and 75 is

9000
9400
9600
9800

Discuss Question

Answer: Option C. -> 9600
:
C

L.C.M of 15, 25, 40 and 75 is 600

Largest 4 digits number is 9999.

When 9999 is divided by 600, the remainder will be 399.

So, the required number is (9999 - 399) = 9600

Question 56.

Which of the following represents the correct prime factorisation of 272?

$2^{2} \times 3^{2} \times 13$
$2^{4} \times 17$
$3^{3} \times 13$
$2 \times 3 \times 7 \times 11$

Discuss Question

Answer: Option B. ->

2^{4} \times 17

:
B

272 can be factorised as following:

Thus 272 = $2^{4} \times 17$

Question 57.

If a = $2^{3} \times 3$ , b = $2 \times 3 \times 5$ , c = $3^{n} \times 5$ and LCM (a, b, c) = $2^{3} \times 3^{2} \times 5$ , then n = ?
(Here, n is a natural number)