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

10th Grade > Mathematics

REAL NUMBERS MCQs

Total Questions : 58 | Page 5 of 6 pages

Using Euclid's division algorithm, find the HCF of 1650 and 847.

Discuss Question

Answer: Option B. -> 11
:
B

Euclid's division algorithm to find HCF of 1650 and 847:
Step 1: 1650 = 847 $\times$ 1 + 803
Step 2: 847 = 803 $\times$ 1 + 44
Step 3: 803 = 44 $\times$ 18 + 11
Step 4: 44 = 11 $\times$ 4 + 0

Hence, 11 is the HCF of 1650 and 847.

Question 42.

Which of the following is not an irrational number?

$5 - \sqrt{3}$
$5 + \sqrt{3}$
$4 + \sqrt{2}$
$5 + \sqrt{9}$

Discuss Question

Answer: Option D. ->

5 + \sqrt{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 an irrational number.

Question 43.

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

Discuss Question

Answer: Option D. ->

5 + \sqrt{9}

Product of two numbers = HCF $\times$ LCM
$\Rightarrow$ 42 $\times$ 70 = HCF $\times$ 210
$\Rightarrow$ HCF = $\frac{2940}{210}$ = 14

Question 44.

Find the number of prime factors of 5005.

Discuss Question

Answer: Option C. -> 4
:
C

5005 = 5 $\times$ 7 $\times$ 11 $\times$ 13
$∴$ There are four prime factors of 5005.

Question 45.

If $\frac{p}{q}$ is a rational number with terminating decimal expansion where $p$ and $q$ are coprimes, then $q$ can be represented as:
(Here, n and m are non-negative integers.)

$3^{n} 5^{m}$
$\frac{2^{n}}{5^{m}}$
$\frac{7^{n}}{5^{m}}$
$2^{n} 5^{m}$

Discuss Question

Answer: Option D. ->

2^{n} 5^{m}

:
D

If $\frac{p}{q}$ is a rational number with terminating decimal expansion where $p$ and $q$ are coprimes, then the prime factorisation of $q$ will be in the form of $2^{n} 5^{m}$ .
Example:
Consider rational number $\frac{1}{8}$ .
Here, denominator $8 = 2^{3} \times 5^{0}$
Therefore, the rational number $\frac{1}{8}$ will have terminating decimal expansion.
$\frac{1}{8} = 0.125$ which is terminating.
Hence verified.

Question 46.

In a seminar, the number of participants for the subjects Hindi, English and Mathematics are 60, 84 and 108 respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them are for the same subject.

Discuss Question

Answer: Option C. -> 21
:
C
Number of rooms will be minimum if each room accomodates maximum number of participants.
In each room, the same number of participants are to be seated and all of them must be for the same subject.

∴

Number of participants in each room must be the HCF of 60, 84 and 108.
Prime factorisation of 60, 84 and 108:

60 = 2^{2} \times 3 \times 5

84 = 2^{2} \times 3 \times 7

108 = 2^{2} \times 3^{3}

∴

HCF

= 2^{2} \times 3 = 12

In each room, 12 participants can be accommodated.

Number of rooms = \frac{Total number of participants}{12}

= \frac{60 + 84 + 108}{12}

= \frac{252}{12}

= 21

∴

The total number of rooms is 21.

Question 47.

Two tankers contain 850 litres and 680 litres of petrol respectively. Find the maximum capacity of a measuring vessel that can be used to exactly measure the petrol from either tankers with no petrol remaining.

200 litres
170 litres
440 litres
360 litres

Discuss Question

Answer: Option B. -> 170 litres
:
B
Maximum capacity of the measuring vessel = HCF (850, 680)
We use Euclid's division algorithm to find the HCF of 850 and 680.
850 = 680

\times

1 + 170
680 = 170

\times

4 + 0

∴

HCF (850, 680) = 170
So, the maximum capacity of the measuring vessel required is 170 litres.

Question 48.

Find the HCF and LCM of 90 and 144 by prime factorisation method.

18 and 720
720 and 18
360 and 180
180 and 720

Discuss Question

Answer: Option A. -> 18 and 720
:
A
Prime factorisation of 90 and 144:

90 = 2 \times 3 \times 3 \times 5

144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3

\Rightarrow HCF = 2 \times 3^{2} = 18

LCM = 2^{4} \times 3^{2} \times 5 = 720

∴

HCF and LCM of 90 and 144 are 18 and 720 respectively.

Question 49.

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, 3360 and 5600

HCF of 5600 and 3360:

$5600 = 3360 \times 1 + 2240$
$3360 = 2240 \times 1 + 1120$
$2240 = 1120 \times 2 + 0$
$∴$ HCF of 5600 and 3360 = 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 50.

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.