Free Real Numbers 03 Practice Test - 10th Grade

Question 1

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

SOLUTION

Solution : 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 2

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:

SOLUTION

Solution : 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 3

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

9000

9400

9600

9800

SOLUTION

Solution : 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 4

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

184

364

SOLUTION

Solution : 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 5

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

480

240

840

420

SOLUTION

Solution : 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 6

Which of the following are irrational numbers?

0.2

3.1415926535... (non-repeating and non-terminating)

$\sqrt{3}$

${0.2}^{2}$

SOLUTION

Solution : B and C

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

Question 7

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.

$a = b$

SOLUTION

Solution : 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 8

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

$2^{2} \times 3^{2} \times 13$

$2^{4} \times 17$

$3^{3} \times 13$

$2 \times 3 \times 7 \times 11$

SOLUTION

Solution : B

272 can be factorised as following:

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

Question 9

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)

A. 1

B. 2

C. 3

D. 4

SOLUTION

Solution : 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 10

Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

A. 4663

B. 2244

C. 7893

D. 4416

SOLUTION

Solution : A

Let the required number is $x .$
$\Rightarrow (x + 17)$ is the smallest number divisible by 520 and 468.
$\Rightarrow (x + 17) = LCM (520, 468)$

Prime factorisation of 520 and 468:
$520 = 2^{3} \times 5 \times 13$
$468 = 2^{2} \times 3^{2} \times 13$

$\Rightarrow LCM = 2^{3} \times 3^{2} \times 5 \times 13 = 4680$

Thus, $x + 17 = 4680.$
$\Rightarrow x = 4663$

Therefore, the required number is 4663.

Free Real Numbers 03 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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