# Free Real Numbers 02 Practice Test - 10th Grade

### Question 1

If L.C.M. (150, 100) = 300, then find H.C.F. (150, 100).

25

50

45

75

#### SOLUTION

Solution :B

Given: L.C.M. (100,150) = 300

We know that product of two numbers is equal to the product of their HCF and LCM.

⇒ Product of numbers 100 and 150 = HCF × LCM

⇒ 100×150 = HCF × 300

⇒ HCF = 100×150300

⇒ HCF = 15000300

⇒ HCF = 50

∴ HCF of 100 and 150 is 50.

### Question 2

An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

9

6

7

8

#### SOLUTION

Solution :D

Maximum number of columns = HCF of 616 and 32

616=23×7×11

32=25

∴ HCF of 616 and 32 = 23 =8

### Question 3

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting point?

24

36

3

6

#### SOLUTION

Solution :B

Number of minutes = LCM of 18 and 12

Prime Factorisation:

12 = 2 × 2 × 318 = 2 × 3 × 3

L.C.M of the two numbers = 22×32=36

### Question 4

What is the least number that must be added to 1056 so the number is divisible by 23?

2

#### SOLUTION

Solution :B

On dividing 1056 by 23, we get,

45231056 1035 ↓ 21

1056 = 23 × 45 + 21

Here, the remainder is 21 and divisor is 21.

The diffrenece between Divisor and remainder = 23 - 21 = 2.

⇒ If we add 2 to the dividend 1056, we will get a number completely divisible by 23.

∴ The least number to be added = 2

### Question 5

The largest 4 digit number exactly divisible by 88 is:

9944

9988

9966

8888

#### SOLUTION

Solution :A

We have,

Largest 4-digit number = 9999

88) 9999 (113

9944

-----------

55

⇒ When 9999 is divided by 88, remainder is 55.

∴ Required number = (9999 - 55) = 9944

### Question 6

The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and 15 as remainder. What is the smaller number ?

240

270

295

360

#### SOLUTION

Solution :B

Let the smaller number be x.

⇒ Larger number = x+1365

By Euclid's division lemma, the larger number can also be written as 6x+15.

⇒x+1365=6x+15

⇒5x=1350

⇒x=270

∴ The smaller number is 270 and the larger number is 270 + 1365 = 1635.

### Question 7

Find the largest number which can divide both 324 and 144.

21

9

36

18

#### SOLUTION

Solution :C

The largest number that can divide both 324 and 144 is the HCF of both the numbers.

Prime factorising the two numbers we get,

324=2×2×3×3×3×3

144=2×2×2×2×3×3

Now, taking the common factors between them will give us the HCF.

∴ HCF =2×2×3×3=36

### Question 8

Which of the following is not an irrational number?

√2

π

√3

4.8

#### SOLUTION

Solution :D

√2=1.414213562373095..., is non-recurring and non-terminating. So it is an irrational number.

The value of π is 3.14159265358..., the digits neither terminate nor recur and hence it is an irrational number.

√3=1.732142857…, is non-recurring and non-terminating. So it is an irrational number..

4.8 = 4810 satisfies all the properties of a rational number. Therefore, it is a rational number.

### Question 9

HCF of 1848, 3058 and 1331 is

#### SOLUTION

Solution :Two numbers 1848 and 3058, where 3058 > 1848

3058 = 1848 × 1 + 1210

1848 = 1210 × 1 + 638 [Using Euclid's division algorithm to the given number 1848 and 3058]

1210 = 638 × 1 + 572

638 = 572 × 1 + 66

572 = 66 × 8 + 44

66 = 44 × 1 + 22

44 = 22 × 2 + 0

Therefore HCF of 1848 and 3058 is 22.

HCF (1848 and 3058) = 22

Let us find the HCF of the numbers 1331 and 22.

1331 = 22 × 60 + 11

22 = 11 × 2 + 0

HCF of 1331 and 22 is 11

HCF (22, 1331) = 11

Hence the HCF of the three given numbers 1848, 3058 and 1331 is 11.

HCF (1848, 3058, 1331) = 11

### Question 10

Which of the following have a non terminating decimal expansion?

17210

239

1780

3550

#### SOLUTION

Solution :A and B

If pq 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 2n5m, where n and m are non - negative integers.

Consider the fraction 17210.

Here 210 = 2 × 3 × 5 ×7

Hence 210 is not in the form of 2n5m.

Consider the fraction 239.

Here 9 = 3 × 3 = 32

Hence 9 is not in the form of 2n5m.

Consider the fraction 1780.

Here 80 = 2 × 2 × 2 × 2 × 5 = 24×5

Hence 80 is in the form of 2n5m with n = 4 and m = 1.

Consider the fraction 3550=710.

Here 10 = 2 × 5 = 21×51

Hence 10 is in the form of 2n5m with n = 1 and m = 1.

Therefore, 17210 and 239 will have non - terminating decimal expansion.