1. When we reverse the digits of the number 13, the increases by 18. How many other two digit numbers increases by 18 when their digits are reversed?

a) 5

b) 6

c) 7

d) 8

Explanation: Let the numbers are in the form of (10x + y), so when the digits of the number are reversed the number becomes (10y + x)

According to question,

(10y + x) - (10x + y) = 18

9(y - x) = 18

→ y - x = 2

So, the possible pairs of (x, y) are (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

we need the number other than 13.

There are 6 possible numbers i.e. 24, 35, 46, 57, 68, 79

So, total numbers of possible numbers are 6

2. Find the LCM and HCF of 2.5, 0.5 and 0.175.

a) 17.5

b) 5

c) 7.5

d) 2.5

Explanation:

$$\eqalign{ & 2.5 = \frac{{25}}{{10}}, \cr & 0.5 = \frac{5}{{10}}, \cr & 0.175 = \frac{{175}}{{1000}}, \cr} $$

Now,

LCM of two or more fractions is given by:

$$\eqalign{ & \frac{{{\text{LCM}}\,{\text{of}}\,{\text{Numerators}}}}{{{\text{HCF}}\,{\text{of}}\,{\text{Denominators}}}} \cr & \frac{{{\text{LCM}}\,{\text{of}}\,25,\,5,\,175}}{{{\text{HCF}}\,{\text{of}}\,10,\,10,\,1000}} \cr & = \frac{{175}}{{10}} \cr & = 17.5 \cr} $$

3. If A381 is divisible by 11, find the value of the smallest natural number A?

a) 9

b) 7

c) 6

d) 5

Explanation: A number is divisible by 11 if the difference of the sum of the digits in the odd places and sum of the digits in even place is zero or divisible by 11.

Hence, (A + 8) - (3 + 1) = 0 or multiple of 11.

To get the difference 0 or multiple of 11, we need 7 at the place of A.

So, sum of odd place - sum of even place

= 15 - 4 = 11. And this is divisible by 11.

4. The greatest number which will divides: 4003, 4126 and 4249, leaving the same remainder in each case:

a) 43

b) 41

c) 45

d) None of these

Explanation: Rule- Greatest number with which if we divide P, Q, R and it leaves same remainder in each case. Number is of form = HCF of (P - Q), (P - R)

Therefore, HCF of (4126 - 4003), (4249 - 4003) = HCF of 123, 246 = 41.

5. LCM of two numbers is 936. If their HCF is 4 and one of the numbers is 72, the other is

a) 62

b) 42

c) 52

d) None of these

Explanation: Let two numbers be N

_{1}and N

_{2}.

Now,

HCF of the numbers × LCM of the numbers = Multiplication of the numbers.

936 × 4 = 72 × N

_{1}

N

_{1}= $$\frac{{936 \times 4}}{{72}}$$

N

_{1}= 52

6. Two alarm clocks ring their alarms at regular intervals of 50 seconds and 48 seconds. If they first beep together at 12 noon, at what time will they beep again for first time?

a) 12:10 PM

b) 12:12 PM

c) 12:11 PM

d) 12:20 PM

Explanation: They will ring together after,

LCM of 48 and 50 secs.

48 = 2 × 2 × 2 × 2 × 3;

50 = 2 × 5 × 5

LCM = 2 × 2 × 2 × 2 × 3 × 5 × 5 = 1200 secs

= 20 min.

They will beep together at 12:20

7. The sum of the digits of two-digit number is 10, while when the digits are reversed, the number decrease by 54. Find the changed number.

a) 28

b) 19

c) 37

d) 46

Explanation: Let number be (10x + y)

According to question,

(10x + y) - (10y + x) = 54

10x - 10y + y - x = 54

9x - 9y = 54

x - y = 6 -------(i)

Sum of digits,

(x + y) = 10 ------- (ii)

(i) - (ii)

So, x - y - x - y = 6 - 10

-2y = -4

y = 2 and, x = 8

Then, the required number is

= (10y + x)

= 10 × 2 + 8

= 28

8. Find the HCF of (3^{125}-1) and (3^{35}-1).

a) 3^{5} - 1

b) 3^{12} - 1

c) 3^{4} - 1

d) None of these

Explanation: Rule - The HCF of (a

^{m}- 1) and (a

^{n}- 1) is given by (a

^{HCF of m, n}- 1)

Thus for this question the answer is (3

^{5}- 1)

Since, 5 is the HCF of 35 and 125

9. The HCF of 2472, 1284 and a third number 'N' is 12. If their LCM is 2^{3} × 3^{2} × 5 × 103 × 107, then the number 'N' is:

a) 2^{2} × 3^{2} × 7

b) 2^{2} × 3^{3} × 10

c) 2^{2} × 3^{2} × 5

d) None of these

Explanation: HCF of the numbers × LCM of the numbers = Multiplication of the numbers.

(12) × (2

^{3}× 3

^{2}× 5 × 103 × 107) = 2472 × 1284 × N

Hence,

N = $$\frac{{\{ \left( {12} \right) \times \left( {{2^3} \times {3^2} \times 5 \times 103 \times 107} \right)\} }}{{2472 \times 1284}}$$

N = 3 × 5

10. The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32, 35 is.

a) 1120

b) 5600

c) 4714

d) 5200

Explanation: LCM of 20, 28, 32, 35 will be the greatest number which is divisible by these numbers.

Firstly, we find LCM of 20, 28, 32, 35

20 = 2 × 2 × 5

28 = 2 × 2 × 7

32 = 2 × 2 × 2 × 2 × 2

35 = 5 × 7

LCM = 2 × 2 × 2 × 2 × 2 × 5 × 7 = 1120

Required greatest number which subtract from 5834 are divide by 20, 28, 32 and 35

= 5834 - 1120

= 4714