# 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=6q+3 where q is a whole number.

Squaring both sides,x2=(6q+3)2

⇒ x2=36q2+36q+9

⇒ x2=6(6q2+6q+1)+3

∴ When x2 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:

4

5

6

7

#### 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 5600HCF of 5600 and 3360:

5600=3360×1+2240

3360=2240×1+1120

2240=1120×2+0

∴ HCF of 5600 and 3360 = 1120Now, HCF of 2240 and 1120 = 1120

So, the HCF of (3360, 2240 and 5600) = N = 1120Sum 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 ___.

74

94

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 3647=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

⇒ 60 × 168 = 12 × LCM

⇒ LCM = 60×16812

⇒ 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)

√3

0.22

#### SOLUTION

Solution :B and C

Non-terminating and non-repeating decimals are irrational. Hence, 3.1415926535... and √3 are irrational.

### Question 7

If 132300=22×33×52×ab, then which of the following is true ?

a=2b

a+b=8

LCM of a and b is 14.

a=b

#### SOLUTION

Solution :C

132300=22×33×52×ab

By fundamental theorem of arithmetic, 132300 can be written as 22×33×52×72.

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?

22×32×13

24×17

33×13

2×3×7×11

#### SOLUTION

Solution :B

272 can be factorised as following:

Thus 272 = 24×17

### Question 9

If a = 23×3, b = 2×3×5, c = 3n×5 and LCM (a, b, c) = 23×32×5, then n = ?

(Here, n is a natural number)

#### SOLUTION

Solution :B

Given: a = 23×3

b = 2×3×5

c = 3n×5

LCM (a, b, c) = 23×32×5 ... (1)

Since, to find LCM we need to take the prime factors with their highest degree:

LCM will be 23×3n×5 ... (2) (n≥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.

#### SOLUTION

Solution :A

Let the required number is x.

⇒(x+17) is the smallest number divisible by 520 and 468.

⇒(x+17)=LCM (520,468)

Prime factorisation of 520 and 468:

520=23×5×13

468=22×32×13

⇒LCM =23×32×5×13=4680

Thus, x+17=4680.

⇒x=4663

Therefore, the required number is 4663.