Question 1:Â A number consists of two digits. When it is divided by the sum of its digits, the quotient is 6 with no remainder. When the number is diminished by 9, the digits are reversed. Find the number.

Solution:

Let the tenâ€™s digit and the unitâ€™s digit of required number be x and y respectively.

We know,

Dividend = (divisor) (quotient) + remainder

According to the given question:

10x + y = 6(x + y) + 0

10x – 6x + y – 6y = 0

4x – 5y = 0 â€¦â€¦..(1)

Number obtained by reversing the digits is 10y + x

10x + y – 9 = 10y + x

9x – 9y = 9

9(x – y) = 9

(x – y) = 1 â€¦â€¦..(2)

Multiplying (1) by 1 and (2) by 5, we get

4x – 5y = 0 â€¦â€¦..(3)

5x – 5y = 5 â€¦â€¦..(4)

Subtracting (3) from (4), we get

x = 5

Putting x = 5 in (1),we get

(4)(5) – 5y = 0

-5y = -20

y =4

Hence, the required number = 54.

Question 2:Â A two-digit number is such that the product of its digits is 35. If 18 is added to the number, the digits interchange their places. Find the number.

Solution:

Let the tenâ€™s digit and the unitâ€™s digit of required number be x and y respectively.

Then,

xy = 35

Required number = 10x + y

Also,

10x + y + 18 = 10y + x

9x – 9y = -18

y – x = 2 â€¦â€¦..(1)

Now,

(y + x)2-(y – x)2= 4xy

\( Rightarrow y + x = \sqrt{(y – x)^{2} + 4xy} \)

\( = sqrt{144} \)

= 12

y + x = 12 â€¦â€¦.. (2)

Adding (1) and (2), we get

2y = 12 + 2 = 14

y = 7

Putting y = 7 in (1),

7 – x = 2

x = 5

Hence, the required number = 5(10) + 7 = 57

Question 3:Â A two-digit number is such that the product of its digits is 14. When 45 is added from the number, the digits interchange their places. Find the number.

Solution:

Let the tenâ€™s digit and the unitâ€™s digit of required number be x and y respectively.

Then,

xy = 14

Required number = 10x + y

Number obtained on reversing = 10y + x

Also,

(10x + y) + 45 = 10y + x

9(y – x) = 45

y – x = 5 â€¦â€¦..(1)

Now,

(y + x)2 – (y – x)2 = 4xy

\( \Rightarrow y + x = \sqrt{(y – x)^{2} + 4xy} \)

\( = \sqrt{25 + 4 \times 14} \)

\( = \sqrt{81} \)

y + x = 9 â€¦â€¦.. (2) [digits cannot be negative, hence -9 is not possible]

On adding (1) and (2), we get

2y = 14

y = 7

Putting y = 7 in (2), we get

7 + x = 9

x = (9 -7) = 2

x = 2 and y = 7

Hence, the required number is (2) (10) + 7 = 27.

Question 4:Â A two-digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.

Solution:

Let the tenâ€™s digit and the unitâ€™s digit of required number be x and y respectively.

Then,

xy = 18

Required number = 10x + y

Number obtained on reversing = 10y + x

Also,

(10x + y) – 63 = 10y + x

9(y – x) = 63

y – x = 7 â€¦â€¦..(1)

Now,

(y + x)2 – (y – x)2 = 4xy

\( \Rightarrow y + x = \sqrt{(y – x)^{2} + 4xy} \)

\( = \sqrt{7^{2} + 4 \times 14} \)

\( = \sqrt{121} \)

y + x = 11 â€¦â€¦.. (2) [Digits cannot be negative, hence -11 is not possible]

On adding (1) and (2), we get

2x = 18

x = 9

Putting x = 9 in (1), we get

9 – y = 7

y = (9 -7) = 2

x = 9 and y = 2

Hence, the required number is (9) (10) + 2 = 92.

Question 5:Â The sum of the numerator and denominator of a fraction is 8. If 3 is added to both of the numerator and the denominator, the fraction becomes 3/4. Find the number.

Solution:

Let the numerator and denominator of the fraction be x and y respectively.

According to the question:

x + y = 8 â€¦â€¦.. (1)

And

\( \frac{x + 3}{y + 3} = \frac{3}{4} \)

\( \Rightarrow 4x + 12 = 3y + 9 \)

\( \Rightarrow 4x – 3y = -3 \)

Multiplying (1) by 3 and (2) by 1

3x + 3y = 24 â€¦â€¦.. (3)

4x – 3y = -3 â€¦â€¦.. (4)

Adding (3) and (4), we get

7x = 21

x = 3

Putting x = 3 in (!), we get

3 + y = 8

y = 5

x = 3, y = 5

Hence, the fraction is >\( \frac{x}{y} = \frac{3}{5} \)

Question 6:Â Find a fraction which becomes (1/2) when 1 is subtracted from the numerator and 2 is added to the denominator.

Solution:

Let the numerator and denominator of fraction be x and y respectively.

Then, the fraction is (x/y)

Therefore, >\( \frac{x – 1}{y + 2} = \frac{1}{2} \Rightarrow 2x – 2 = y + 2 \Rightarrow 2x – y = 4 \)

Subtracting (1) from (2), we get

(2)(15) – y = 4

30 – y = 4

y = 26

x = 15 and y = 26

Hence, the given fraction is >\( \frac{x}{y} = \frac{15}{26}\)

Question 7:Â The denominator of a fraction is greater than its numerator by 11. If 8 is added to both its numerator and denominator, it becomes 3/4. Find the fraction.

Solution:

Let the numerator and denominator of the fraction be x and y respectively.

Then, the fraction is (x/y).

According to the given equation:

y = x + 11

y – x = 11 â€¦â€¦..(1)

\( \frac{x + 8}{y + 8} = \frac{3}{4} \Rightarrow 4x + 32 = 3y + 24 \Rightarrow 4x – 3y = -8 \)

-3y + 4x = -8 â€¦â€¦.. (2)

Multiplying (1) by 4 and (2) by 1

4y – 4x = 44 â€¦â€¦.. (3)

-3y + 4x = -8 â€¦â€¦.. (4)

Adding (3) and (4), we get

y = 36

Putting y = 36 in (1), we get

y – x = 11

36 – x = 11

x = 25

x = 25, y = 36

Hence, the fraction is >\( \frac{25}{36} \)

Question 8:Â If 2 is added to the numerator of a fraction, it reduces to (1/2) and if 1 is subtracted from the denominator, it reduces to (1/3). Find the fraction.

Solution:

Let the numerator and denominator of fraction be x and y respectively.

Then, the fraction is (x/y).

According to the given equation:

\( \frac{x + 2} {y} = \frac{1}{2} \Rightarrow 2x + 4 = y \Rightarrow 2x – y = -4 \)

And

\( \frac{x}{y – 1} = \frac{1}{3} \Rightarrow 3x = y – 1 \Rightarrow 3x – y = -1 \)

Subtracting (1) from (2), we get

x = 3

Putting x = 3 in (1), we get

y = 10

x = 3, y = 10

Hence, the fraction is >\( \frac{3}{10} \)

Question 9:Â The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2:3. Determine the fraction.

Solution:

Let the fraction be >\( \frac{x}{y} \)

When 2 is added to both the numerator and denominator, the fraction becomes:

\( \frac{x + 2}{y + 2} = \frac{1}{3} \)

3x – y = -4 â€¦â€¦.. (1)

When 3 is added to both the numerator and the denominator, the fraction becomes:

\( \frac{x + 3}{y + 3} = \frac{2}{5} \)

5x – 2y = -9 â€¦â€¦.. (2)

Multiplying (1) by 2 and (2) by 1, we get

6x – 2y = -8 â€¦â€¦.. (3)

Subtracting (4) from (3), we get

x = 1

Putting x = 1 in (1),

(3)(1) – y = 4

y =7

Required fraction = >\( \frac{x}{y} = \frac{1}{7} \)

Question 10:Â The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.

Solution:

Let the two numbers be x and y respectively.

According to the given question:

x + y = 16 â€¦â€¦..(1)

And

\( \frac{1}{x} + \frac{1}{y} = \frac{1}{3} \)

From (2),

\( \frac{x + y} {xy} = \frac{1}{3} \, or \, \frac{16}{xy} = \frac{1}{3} \)

xy = 48

We know,

(x – y)2 = (x + y)2 – 4xy

= 162 – 4(48) = 256 – 192 = 64

x – y = 8 â€¦â€¦..(3)

Adding (1) and (3), we get

2x = 24

x = 12

Putting x = 12 in (1).

y = 16 – x

y = 4

Hence, the required numbers are 12 and 4.

Question 11:Â Two years ago, a man was five times as old as his son. Two years later, his age will be 8 more than three times the age of his son. Find their present ages.

Solutions:

Let the present ages of the man and his son be x years and y rears respectively.

Then,

Two years ago:

(x – 2) = 5(y – 2)

x – 2 = 5y – 10

x – 5y = -8 â€¦â€¦..(1)

Two years later:

(x + 2) = 3(y + 2) + 8

x = 2 = 3y + 6 + 8

x – 3y = 12 â€¦â€¦..(2)

Subtracting (2) from (1), we get

-2y = -20

y =10

Putting y = 10 in (1), we get

x – 5(10) = -8

x = 42

Hence, the present ages of the man and his are 42 years and 10 years respectively.

Question 12:Â Five years hence, a manâ€™s age will be three times the age of his son. Five years ago, the man was seven times as old as his son. Find their present ages.

Solution:

Let the present ages of A and B be x and y respectively.

Five years ago:

(x – 5) = 3(y – 5)

x – 5 = 3y – 15

x – 3y = -10 â€¦â€¦..(1)

Ten years later:

(x + 10) = 2(y + 10)

x + 10 = 2y + 20

x – 2y = 10 â€¦â€¦..(2)

Subtracting (2) from (1), we get

y = 20

Putting y = 20 in (1), we get

x – 3y = -10

x – 3(20) = -10

x = -10 + 60 = 50

x =50, y = 20

Hence, the present ages of A and B are 50 years and 20 years respectively.

Question 13:Â The present age of a woman is 3 years more than three times the age of her daughter. Three years hence, the womanâ€™s age will be 10 years more than twice the age of her daughter. Find their present ages.

Solution:

Let the present ages of woman and daughter be x and y respectively.

Then,

Their present ages:

x = 3y + 3

x – 3y = 3 â€¦â€¦..(1)

Three years later:

(x + 3) = 2(y + 3) + 10

x + 3 = 2y + 6 + 10

x – 2y = 13 â€¦â€¦..(2)

Subtracting (2) from (1), we get

y =10

Putting y = 10 in (1), we get

x – 3(10) = 3

x =33

x =33, y = 10

Hence, the present ages of the mother and her daughter are 33 and 10 years.