# Free Real Numbers 03 Practice Test - 10th Grade

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

A.

True

B.

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.

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:

A.

4

B.

5

C.

6

D.

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 5600

HCF of 5600 and 3360:

5600=3360×1+2240
3360=2240×1+1120
2240=1120×2+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

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

A.

9000

B.

9400

C.

9600

D.

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

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

A.

74

B.

94

C.

184

D.

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).

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

A.

480

B.

240

C.

840

D.

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.

Which of the following are irrational numbers?

A.

0.2

B.

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

C.

3

D.

0.22

#### SOLUTION

Solution : B and C

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

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

A.

a=2b

B.

a+b=8

C.

LCM of a and b is 14.

D.

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.

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

A.

22×32×13

B.

24×17

C.

33×13

D.

2×3×7×11

#### SOLUTION

Solution : B

272 can be factorised as following: Thus 272 = 24×17

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)

A. 1
B. 2
C. 3
D. 4

#### 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) (n1)
On comparing we get,
n = 2

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.
(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.