Free Real Numbers 01 Practice Test - 10th Grade

Question 1

For two numbers 96 and 404, the HCF is 4. What is the LCM of the two numbers?

9696

9698

C. 9600

D. 9700

SOLUTION

Solution : A

Product of two numbers = Product of their HCF and LCM.

$\Rightarrow$ 96 $\times$ 404 = 4 $\times$ LCM
$\Rightarrow$ LCM = $\frac{96 \times 404}{4}$
$\Rightarrow$ LCM = 9696

$∴$ HCF and LCM is 4 and 9696 respectively.

Question 2

The decimal expansion of $\frac{141}{120}$ will terminate after how many places?

Will not terminate

SOLUTION

Solution : A

Given rational number $\frac{141}{120}$

Here, $120 = 2^{3} \times 3 \times 5$

$141 = 3 \times 47$

$\Rightarrow$ $\frac{141}{120}$ = $\frac{3 \times 47}{2^{3} \times 3 \times 5}$

= $\frac{47}{2^{3} \times 5}$

Multiply and divide by $5^{2}$ .

= $\frac{47 \times 5^{2}}{2^{3} \times 5 \times 5^{2}}$

= $\frac{47 \times 25}{(2 \times 5)^{3}}$

= $\frac{1175}{1000}$

= 1.175

Therefore, $\frac{141}{120}$ will terminate after three decimal places.

Question 3

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

SOLUTION

Solution : 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 4

Which of the following is not an irrational number?

$5 - \sqrt{3}$

$5 + \sqrt{3}$

$4 + \sqrt{2}$

$5 + \sqrt{9}$

SOLUTION

Solution : 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 5

Find the number of prime factors of 5005.

SOLUTION

Solution : C

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

Question 6

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

SOLUTION

Solution :
Product of two numbers = HCF $\times$ LCM

$\Rightarrow$ 42 $\times$ 70 = HCF $\times$ 210
$\Rightarrow$ HCF = $\frac{2940}{210}$ = 14

Question 7

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.)

$3^{n} 5^{m}$

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

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

$2^{n} 5^{m}$

SOLUTION

Solution : 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 8

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.

A. 200 litres

B. 170 litres

C. 440 litres

D. 360 litres

SOLUTION

Solution : 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 9

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.

A. 7

B. 14

C. 21

D. 28

SOLUTION

Solution : 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 10

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

A. 18 and 720

B. 720 and 18

C. 360 and 180

D. 180 and 720

SOLUTION

Solution : 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.

Free Real Numbers 01 Practice Test - 10th Grade

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

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