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

10th Grade > Mathematics

REAL NUMBERS MCQs

Total Questions : 58 | Page 3 of 6 pages

Question 21. Which of the following is not an irrational number?

5−√3
5+√3
4+√2
5+√9

Discuss Question

Answer: Option D. -> 5+√9
:
D
If p is a prime number, then

\sqrt{p}

is an irrational number.
3 is a prime number.

\Rightarrow

\sqrt{3}

is an irrational number.

\Rightarrow

5 - \sqrt{3}

is an irrational number.
Similarly,

5 + \sqrt{3}

is an irrational number.
2 is a prime number.

\Rightarrow

\sqrt{2}

is an irrational number.

\Rightarrow

4 + \sqrt{2}

is an irrational number.
9 is not a prime number.

\sqrt{9} = 3

\Rightarrow

5 + \sqrt{9} = 5 + 3 = 8

which is a rational number.

\Rightarrow

5 + \sqrt{9}

is not anirrational number.

Question 22. HCF of two numbers is __ given that their LCM is 210 and the numbers are 42 and 70.

Discuss Question

:
Product of two numbers = HCF

\times

LCM

\Rightarrow

\times

70 = HCF

\times

210

\Rightarrow

HCF =

\frac{2940}{210}

= 14

Question 23. If

132300 = 2^{2} \times 3^{3} \times 5^{2} \times a^{b},

then which of the following is true ?

a=2b
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 24. 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 of15, 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 25. Which of the following represents the correct prime factorisation of 272?

22×32×13
24×17
33×13
2×3×7×11

Discuss Question

Answer: Option B. -> 24×17
:
B
272 can be factorised as following:

Which Of The Following Represents The Correct Prime Factoris...

Thus 272 =

2^{4} \times 17

Question 26. 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)

Discuss Question

Answer: Option B. -> 2
:
B
Given: a =

2^{3} \times 3

b =

2 \times 3 \times 5

c =

3^{n} \times 5

LCM (a, b, c) =

2^{3} \times 3^{2} \times 5

... (1)
Since, to find LCM we need to take the prime factors with their highest degree:
LCM will be

2^{3} \times 3^{n} \times 5

... (2)

(n \geq 1)

On comparing we get,
n = 2

Question 27. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. The sum of the digits in N is:

Discuss Question

Answer: Option A. -> 4
:
A
Since the remainder is same in each case, hence we will use the following formula
H.C.F (x, y, z) = H.C.F of (x -y), (y- z), ( z-x)
N = H.C.F. of (6905 - 4665), (4665 - 1305), and (6905 - 1305)
= H.C.F. of 2240, 3360and 5600
HCF of 5600 and 3360:

5600 = 3360 \times 1 + 2240

3360 = 2240 \times 1 + 1120

2240 = 1120 \times 2 + 0

∴

HCF of5600 and3360= 1120
Now, HCF of 2240 and 1120 = 1120
So, the HCF of (3360, 2240 and 5600) = N = 1120
Sum of digits in N = (1 + 1 + 2 + 0) = 4

Question 28. A number when divided by 6 leaves a remainder 3. When the square of the number is divided by 6, the remainder is 3.

True
False

Discuss Question

Answer: Option A. -> True
:
A
Let the number be

x

Given:

x = 6 q + 3

where q is a whole number.
Squaring both sides,

x^{2} = (6 q + 3)^{2}

\Rightarrow

x^{2} = 36 q^{2} + 36 q + 9

\Rightarrow

x^{2} = 6 (6 q^{2} + 6 q + 1) + 3

∴

When

x^{2}

is divided by 6, then the remainder is 3.

Question 29.

If L.C.M. (150, 100) = 300, then find H.C.F. (150, 100).

Discuss Question

Answer: Option B. -> 50
:
B

Given: L.C.M. (100,150) = 300
We know that product of two numbers is equal to the product of their HCF and LCM.
$\Rightarrow$ Product of numbers 100 and 150 = HCF $\times$ LCM
$\Rightarrow$ $100 \times 150$ = HCF $\times$ 300
$\Rightarrow$ HCF = $\frac{100 \times 150}{300}$
$\Rightarrow$ HCF = $\frac{15000}{300}$
$\Rightarrow$ HCF = 50
$∴$ HCF of 100 and 150 is 50.

Question 30.

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Discuss Question

Answer: Option D. -> 8
:
D

Maximum number of columns = HCF of 616 and 32
$616 = 2^{3} \times 7 \times 11$
$32 = 2^{5}$
$∴$ HCF of 616 and 32 = $2^{3}$ =8