Simultaneous (Linear) Equations — Exercise 6(C)

The sum of two positive whole numbers is 15 and their difference is 9; the numbers are :

1. 12 and 3

2. 9 and 0

3. 7 and 8

4. 15 and 0

Answer

Let two whole numbers be x and y, where x > y.

Given,

Sum of numbers = 15

⇒ x + y = 15 .........(1)

Difference of numbers = 9

⇒ x - y = 9 .........(2)

Adding equation (1) and (2), we get :

⇒ (x + y) + (x - y) = 15 + 9

⇒ x + x + y - y = 24

⇒ 2x = 24

⇒ x = $\dfrac{24}{2}$ = 12.

Substituting value of x in equation (1), we get :

⇒ 12 + y = 15

⇒ y = 15 - 12 = 3.

Hence, Option 1 is the correct option.

The sum of numerator and denominator of a fraction is 22. If numerator is 8 less than its denominator; the fraction is :

1. $\dfrac{9}{13}$

2. $\dfrac{7}{15}$

3. $\dfrac{13}{9}$

4. $\dfrac{15}{7}$

Answer

Let numerator be x and denominator be y.

Given,

Sum of numerator and denominator of a fraction is 22.

∴ x + y = 22 ........(1)

Given,

Numerator is 8 less than its denominator.

∴ y - x = 8 ........(2)

Adding equation (1) and (2), we get :

⇒ (x + y) + y - x = 8 + 22

⇒ x - x + y + y = 30

⇒ 2y = 30

⇒ y = $\dfrac{30}{2}$ = 15.

Substituting value of y in equation (1), we get :

⇒ x + 15 = 22

⇒ x = 22 - 15 = 7.

Fraction = $\dfrac{x}{y} = \dfrac{7}{15}$.

Hence, Option 2 is the correct option.

The sum of the digits of a two digit number is 8. The difference between them is 4. If the ten's digit is greater than the unit's digit then the number is :

1. 26

2. 53

3. 71

4. 62

Answer

Let ten's digit be x and unit's digit be y.

Given,

Sum of digits = 8

∴ x + y = 8 .........(1)

Difference between digits = 4

∴ x - y = 4 .........(2)

Adding equation (1) and (2), we get :

⇒ (x + y) + (x - y) = 8 + 4

⇒ x + x + y - y = 12

⇒ 2x = 12

⇒ x = $\dfrac{12}{2}$ = 6.

Substituting value of x in equation (1), we get :

⇒ 6 + y = 8

⇒ y = 8 - 6 = 2.

Number = 10x + y = 10(6) + 2 = 60 + 2 = 62.

Hence, Option 4 is the correct option.

The present ages of two persons are in the ratio 1 : 3. After 5 years, the ratio between their ages will be 2 : 5, their ages are :

1. 10 years and 30 years

2. 15 years and 45 years

3. 20 years and 60 years

4. 12 years and 36 years

Answer

Given,

The present ages of two persons are in the ratio 1 : 3.

Let present age be x and 3x.

Given,

After 5 years, the ratio between their ages will be 2 : 5.

$\Rightarrow \dfrac{x + 5}{3x + 5} = \dfrac{2}{5} \\[1em] \Rightarrow 5(x + 5) = 2(3x + 5) \\[1em] \Rightarrow 5x + 25 = 6x + 10 \\[1em] \Rightarrow 6x - 5x = 25 - 10 \\[1em] \Rightarrow x = 15 \\[1em] \Rightarrow 3x = 3 \times 15 = 45.$

Hence, Option 2 is the correct option.

The ratio of two numbers is 2 : 3. If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio. Find the numbers.

Answer

Let the numbers be x and y.

According to question ratio of numbers = $\dfrac{2}{3}$,

$\Rightarrow \dfrac{x}{y} = \dfrac{2}{3} \\[1em] \Rightarrow x = \dfrac{2y}{3} ........(1)$

Given,

If 2 is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the original ratio.

$\Rightarrow \dfrac{x - 2}{y - 8} = \dfrac{3}{2} \\[1em] \Rightarrow 2(x - 2) = 3(y - 8) \\[1em] \Rightarrow 2x - 4 = 3y - 24 \\[1em]$

Substituting value of x from equation (1) in above equation, we get :

$\Rightarrow 2 \times \dfrac{2y}{3} - 4 = 3y - 24 \\[1em] \Rightarrow \dfrac{4y}{3} - 4 = 3y - 24 \\[1em] \Rightarrow -4 + 24 = 3y - \dfrac{4y}{3} \\[1em] \Rightarrow 20 = \dfrac{9y - 4y}{3} \\[1em] \Rightarrow 20 = \dfrac{5y}{3} \\[1em] \Rightarrow y = \dfrac{20 \times 3}{5} \\[1em] \Rightarrow y = 12.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{2 \times 12}{3} \\[1em] = \dfrac{24}{3} \\[1em] = 8.$

Hence, numbers are 8 and 12.

Two numbers are in the ratio 4 : 7. If thrice the larger be added to twice the smaller, the sum is 59. Find the numbers.

Answer

Let two numbers be x and y, where x is the larger and y is the smaller number.

Given,

Two numbers are in the ratio 4 : 7.

$\therefore \dfrac{y}{x} = \dfrac{4}{7} \\[1em] \Rightarrow y = \dfrac{4x}{7} .......(1)$

Given,

If thrice the larger be added to twice the smaller, the sum is 59.

∴ 3x + 2y = 59

$\Rightarrow 3x + 2 \times \dfrac{4x}{7} = 59 \\[1em] \Rightarrow 3x + \dfrac{8x}{7} = 59 \\[1em] \Rightarrow \dfrac{21x + 8x}{7} = 59 \\[1em] \Rightarrow \dfrac{29x}{7} = 59 \\[1em] \Rightarrow x = \dfrac{59 \times 7}{29} \\[1em] \Rightarrow x = 14\dfrac{7}{29}.$

Substituting value of x in equation (1), we get :

$\Rightarrow y = \dfrac{4}{7} \times 14\dfrac{7}{29} \\[1em] = \dfrac{4}{7} \times \dfrac{413}{29} \\[1em] = \dfrac{4 \times 59}{29} \\[1em] = 8\dfrac{4}{29}.$

Hence, numbers are $8\dfrac{4}{29} \text{ and } 14\dfrac{7}{29}$.

When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is $\dfrac{3}{2}$. When the difference of these numbers is divided by the smaller, the result is $\dfrac{1}{2}$ . Find the numbers.

Answer

Let two numbers be x and y, where x is the larger and y is the smaller number.

Given,

When the greater of the two numbers increased by 1 divides the sum of the numbers, the result is $\dfrac{3}{2}$.

$\therefore \dfrac{x + y}{x + 1} = \dfrac{3}{2} \\[1em] \Rightarrow 2(x + y) = 3(x + 1) \\[1em] \Rightarrow 2x + 2y = 3x + 3 \\[1em] \Rightarrow 3x - 2x = 2y - 3 \\[1em] \Rightarrow x = 2y - 3 ...........(1)$

Given,

When the difference of these numbers is divided by the smaller, the result is $\dfrac{1}{2}$.

$\therefore \dfrac{x - y}{y} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x - y) = y \\[1em] \Rightarrow 2x - 2y = y \\[1em] \Rightarrow 2y + y = 2x \\[1em] \Rightarrow 2x = 3y \\[1em] \Rightarrow x = \dfrac{3y}{2} .......(2)$

From (1) and (2), we get :

$\Rightarrow 2y - 3 = \dfrac{3y}{2} \\[1em] \Rightarrow 2(2y - 3) = 3y \\[1em] \Rightarrow 4y - 6 = 3y \\[1em] \Rightarrow 4y - 3y = 6 \\[1em] \Rightarrow y = 6.$

Substituting value of y in equation (2), we get :

$\Rightarrow x = \dfrac{3 \times 6}{2} \\[1em] = \dfrac{18}{2} \\[1em] = 9.$

Hence, numbers are 6 and 9.

The sum of two positive numbers x and y (x > y) is 50 and the difference of their squares is 720. Find the numbers.

Answer

Given,

Sum of numbers = 50

∴ x + y = 50

⇒ x = 50 - y .........(1)

Difference of squares = 720

∴ x² - y² = 720

⇒ (50 - y)² - y² = 720

⇒ 50² + y² - 2 × 50 × y - y² = 720

⇒ 2500 - 100y = 720

⇒ 100y = 2500 - 720

⇒ 100y = 1780

⇒ y = $\dfrac{1780}{100}$ = 17.80

Substituting value of y in equation (1), we get :

⇒ x = 50 - y = 50 - 17.80 = 32.20

Hence, numbers are 32.20 and 17.80

The sum of two numbers is 8 and the difference of their squares is 32. Find the numbers.

Answer

Let two numbers be x and y, such that x > y.

Given,

Sum of two numbers = 8

⇒ x + y = 8

⇒ x = 8 - y ........(1)

Given,

Difference of squares = 32

⇒ x² - y² = 32

⇒ (8 - y)² - y² = 32

⇒ 8² + y² - 2 × 8 × y - y² = 32

⇒ 64 - 16y = 32

⇒ 16y = 64 - 32

⇒ 16y = 32

⇒ y = $\dfrac{32}{16}$ = 2.

Substituting value of y in equation (1), we get :

⇒ x = 8 - 2 = 6.

Hence, numbers are 2 and 6.

The difference between two positive numbers x and y (x > y) is 4 and the difference between their reciprocal is $\dfrac{4}{21}$. Find the numbers.

Answer

Given,

⇒ x > y

$\Rightarrow \dfrac{1}{x} \lt \dfrac{1}{y}$

Difference between two positive numbers x and y (x > y) is 4.

∴ x - y = 4

⇒ x = y + 4 ........(1)

Difference between their reciprocal is $\dfrac{4}{21}$.

$\Rightarrow \dfrac{1}{y} - \dfrac{1}{x} = \dfrac{4}{21} \\[1em] \Rightarrow \dfrac{x - y}{xy} = \dfrac{4}{21}\\[1em] \Rightarrow 21(x - y) = 4xy \\[1em] \Rightarrow 21x - 21y - 4xy = 0 \text{ ..........(2)}$

Substituting value of x from equation (1) in (2), we get :

⇒ 21(y + 4) - 21y - 4y(y + 4) = 0

⇒ 21y + 84 - 21y - 4y² - 16y = 0

⇒ 4y² + 16y - 84 = 0

⇒ 4(y² + 4y - 21) = 0

⇒ y² + 7y - 3y - 21 = 0

⇒ y(y + 7) - 3(y + 7) = 0

⇒ (y - 3)(y + 7) = 0

⇒ y - 3 = 0 or y + 7 = 0

⇒ y = 3 or y = -7.

Since, numbers are positive.

∴ y = 3.

⇒ x = y + 4 = 3 + 4 = 7.

Hence, numbers are 3 and 7.

Two numbers are in the ratio 4 : 5. If 30 is subtracted from each of the numbers, the ratio becomes 1 : 2. Find the numbers.

Answer

Let two numbers be x and y.

Given,

Numbers are in the ratio 4 : 5.

$\Rightarrow \dfrac{x}{y} = \dfrac{4}{5} \\[1em] \Rightarrow x = \dfrac{4y}{5} .......(1)$

Given,

If 30 is subtracted from each of the numbers, the ratio becomes 1 : 2.

$\Rightarrow \dfrac{x - 30}{y - 30} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x - 30) = y - 30 \\[1em] \Rightarrow 2x - 60 = y - 30 \\[1em] \Rightarrow 2x = y - 30 + 60 \\[1em] \Rightarrow 2x = y + 30 \\[1em] \Rightarrow x = \dfrac{y + 30}{2} .......(2)$

From (1) and (2), we get :

$\Rightarrow \dfrac{4y}{5} = \dfrac{y + 30}{2} \\[1em] \Rightarrow 4y \times 2 = 5(y + 30) \\[1em] \Rightarrow 8y = 5y + 150 \\[1em] \Rightarrow 8y - 5y = 150 \\[1em] \Rightarrow 3y = 150 \\[1em] \Rightarrow y = \dfrac{150}{3} \\[1em] \Rightarrow y = 50.$

Substituting value of y in equation (1), we get :

$\Rightarrow x = \dfrac{4 \times 50}{5} \\[1em] = \dfrac{200}{5} \\[1em] = 40.$

Hence, numbers are 40 and 50.

If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes $\dfrac{2}{3}$. If the numerator is increased by 1 and denominator is increased by 2, it becomes $\dfrac{1}{3}$. Find the fraction.

Answer

Let numerator be x and denominator be y.

Given,

If the numerator of a fraction is increased by 2 and denominator is decreased by 1, it becomes $\dfrac{2}{3}$.

$\therefore \dfrac{x + 2}{y - 1} = \dfrac{2}{3} \\[1em] \Rightarrow 3(x + 2) = 2(y - 1) \\[1em] \Rightarrow 3x + 6 = 2y - 2 \\[1em] \Rightarrow 3x - 2y + 6 + 2 = 0 \\[1em] \Rightarrow 3x - 2y + 8 = 0 \text{ .........(1)}$

Given,

If the numerator is increased by 1 and denominator is increased by 2, it becomes $\dfrac{1}{3}$.

$\Rightarrow \dfrac{x + 1}{y + 2} = \dfrac{1}{3} \\[1em] \Rightarrow 3(x + 1) = 1(y + 2) \\[1em] \Rightarrow 3x + 3 = y + 2 \\[1em] \Rightarrow 3x - y + 3 - 2 = 0 \\[1em] \Rightarrow 3x - y + 1 = 0 \text{ .........(2)}$

Subtracting equation (1) from (2), we get :

⇒ 3x - y + 1 - (3x - 2y + 8) = 0 - 0

⇒ 3x - 3x - y + 2y + 1 - 8 = 0

⇒ y - 7 = 0

⇒ y = 7.

Substituting value of y in equation (1), we get :

⇒ 3x - 2(7) + 8 = 0

⇒ 3x - 14 + 8 = 0

⇒ 3x - 6 = 0

⇒ 3x = 6

⇒ x = $\dfrac{6}{3}$ = 2.

Fraction = $\dfrac{x}{y} = \dfrac{2}{7}$.

Hence, fraction = $\dfrac{2}{7}$.

The sum of the numerator and the denominator of a fraction is equal to 7. Four times the numerator is 8 less than 5 times the denominator. Find the fraction.

Answer

Let numerator be x and denominator be y.

Given,

Sum of the numerator and the denominator of a fraction is equal to 7.

∴ x + y = 7

⇒ x = 7 - y ..........(1)

Given,

Four times the numerator is 8 less than 5 times the denominator.

⇒ 5y - 4x = 8 .........(2)

Substituting value of x from equation (1) in equation (2), we get :

⇒ 5y - 4(7 - y) = 8

⇒ 5y - 28 + 4y = 8

⇒ 9y = 8 + 28

⇒ 9y = 36

⇒ y = $\dfrac{36}{9}$ = 4

Substituting value of y in equation (1), we get :

⇒ x = 7 - 4 = 3.

Fraction = $\dfrac{x}{y} = \dfrac{3}{4}$.

Hence, fraction = $\dfrac{3}{4}$.

If the numerator of a fraction is multiplied by 2 and its denominator is increased by 1, it becomes 1. However, if the numerator is increased by 4 and denominator is multiplied by 2, the fraction becomes $\dfrac{1}{2}$. Find the fraction.

Answer

Let numerator be x and denominator be y.

Given,

If the numerator of a fraction is multiplied by 2 and its denominator is increased by 1, it becomes 1.

$\therefore \dfrac{2 \times x}{y + 1} = 1 \\[1em] \Rightarrow 2x = 1(y + 1) \\[1em] \Rightarrow 2x = y + 1 \\[1em] \Rightarrow 2x - y - 1 = 0 .........(1)$

Given,

If the numerator is increased by 4 and denominator is multiplied by 2, the fraction becomes $\dfrac{1}{2}$.

$\therefore \dfrac{x + 4}{2 \times y} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x + 4) = 2y \times 1 \\[1em] \Rightarrow 2x + 8 = 2y \\[1em] \Rightarrow 2x - 2y + 8 = 0 ........(2)$

Subtracting equation (2) from (1), we get :

⇒ 2x - y - 1 - (2x - 2y + 8) = 0

⇒ 2x - 2x - y + 2y - 1 - 8 = 0

⇒ y - 9 = 0

⇒ y = 9.

Substituting value of y in equation (1), we get :

⇒ 2x - 9 - 1 = 0

⇒ 2x - 10 = 0

⇒ 2x = 10

⇒ x = $\dfrac{10}{2}$ = 5.

Fraction = $\dfrac{x}{y} = \dfrac{5}{9}$.

Hence, fraction = $\dfrac{5}{9}$.

A fraction becomes $\dfrac{1}{2}$ if 5 is subtracted from its numerator and 3 is subtracted from its denominator. If the denominator of this fraction is 5 more than its numerator, find the fraction.

Answer

Let numerator be x and denominator be y.

Given,

Denominator is 5 more than numerator.

∴ y = x + 5 ......(1)

Given,

A fraction becomes $\dfrac{1}{2}$ if 5 is subtracted from its numerator and 3 is subtracted from its denominator.

$\therefore \dfrac{x - 5}{y - 3} = \dfrac{1}{2} \\[1em] \Rightarrow 2(x - 5) = y - 3 \\[1em] \Rightarrow 2x - 10 = y - 3 \\[1em] \Rightarrow 2x - y = -3 + 10 \\[1em] \Rightarrow 2x - y = 7 \text{ .......(2)}$

Substituting value of y from equation (1) in (2), we get :

⇒ 2x - (x + 5) = 7

⇒ 2x - x - 5 = 7

⇒ x = 7 + 5

⇒ x = 12.

Substituting value of x in equation (1), we get :

⇒ y = 12 + 5 = 17.

Fraction = $\dfrac{x}{y} = \dfrac{12}{17}$.

Hence, fraction = $\dfrac{12}{17}$.

The sum of the digits of a two digit number is 5. If the digits are reversed, the number is reduced by 27. Find the number.

Answer

Let digit at unit's place be x and ten's place be y.

Number = 10 × y + x = 10y + x

Given,

Sum of the digits of a two digit number is 5.

∴ x + y = 5

⇒ x = 5 - y ........(1)

If the digits are reversed, then number = 10x + y.

Given,

If the digits are reversed, the number is reduced by 27.

∴ 10x + y = 10y + x - 27

⇒ 10x - x = 10y - y - 27

⇒ 9x = 9y - 27

⇒ 9x = 9(y - 3)

⇒ x = y - 3 ........(2)

From (1) and (2), we get :

⇒ y - 3 = 5 - y

⇒ y + y = 5 + 3

⇒ 2y = 8

⇒ y = $\dfrac{8}{2}$ = 4.

Substituting value of y in equation (2), we get :

⇒ x = 4 - 3 = 1.

Number = 10y + x = 10(4) + 1 = 40 + 1 = 41.

Hence, the number = 41.

The sum of the digits of a two digit number is 7. If the digits are reversed, the new number decreased by 2, equals twice the original number. Find the number.

Answer

Let digit at unit's place be x and ten's place be y.

Number = 10 × y + x = 10y + x

Given,

Sum of digits = 7

∴ x + y = 7

⇒ x = 7 - y ........(1)

On reversing digits,

New number = 10 × x + y = 10x + y

Given,

The new number decreased by 2, equals twice the original number.

⇒ (10x + y) - 2 = 2(10y + x)

⇒ 10x + y - 2 = 20y + 2x

⇒ 10x - 2x = 20y - y + 2

⇒ 8x = 19y + 2

⇒ 8(7 - y) = 19y + 2 ........[From (1)]

⇒ 56 - 8y = 19y + 2

⇒ 56 - 2 = 19y + 8y

⇒ 54 = 27y

⇒ y = $\dfrac{54}{27}$ = 2.

Substituting value of y in (1), we get :

⇒ x = 7 - 2 = 5.

Number = 10y + x = 10 × 2 + 5 = 25.

Hence, number = 25.

The ten's digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is 32. Find the number.

Answer

Let ten's digit be x and unit's digit be y.

Given,

The ten's digit of a two digit number is three times the unit digit.

∴ x = 3y ...........(1)

Given,

The sum of the number and the unit digit is 32.

Number = 10x + y, unit's digit = y

∴ 10x + y + y = 32

⇒ 10x + 2y = 32

⇒ 2(5x + y) = 32

⇒ 5x + y = 16

⇒ 5(3y) + y = 16 ........[From (1)]

⇒ 15y + y = 16

⇒ 16y = 16

⇒ y = 1.

⇒ x = 3y = 3(1) = 3.

Number = 10x + y = 10(3) + 1 = 31.

Hence, number = 31.

A two-digit number is such that the ten's digit exceeds twice the unit's digit by 2 and the number obtained by inter-changing the digits is 5 more than three times the sum of the digits. Find the two digit number.

Answer

Let x be the digit at ten's place and y be the digit at unit's place.

Given,

Ten's digit exceeds twice the unit's digit by 2.

∴ x - 2y = 2

⇒ x = 2y + 2 ........(1)

Given,

Number obtained by inter-changing the digits is 5 more than three times the sum of the digits.

∴ 10y + x = 3(x + y) + 5

⇒ 10y + x = 3x + 3y + 5

⇒ 10y - 3y + x - 3x = 5

⇒ 7y - 2x = 5 ........(2)

Substituting value of x from equation (1) in equation (2), we get :

⇒ 7y - 2(2y + 2) = 5

⇒ 7y - 4y - 4 = 5

⇒ 3y - 4 = 5

⇒ 3y = 9

⇒ y = $\dfrac{9}{3}$ = 3.

Substituting value of y in equation (1), we get :

⇒ x = 2y + 2 = 2(3) + 2 = 6 + 2 = 8.

Original number = 10x + y = 10(8) + 3 = 80 + 3 = 83.

Hence, original number = 83.

Five years ago, A's age was four times the age of B. Five years hence, A's age will be twice the age of B. Find their present ages.

Answer

Let present age of A be x years and B be y years.

Five years's ago their age will be :

A = (x - 5) years

B = (y - 5) years

Given,

Five years ago, A's age was four times the age of B.

⇒ (x - 5) = 4(y - 5)

⇒ x - 5 = 4y - 20

⇒ x = 4y - 20 + 5

⇒ x = 4y - 15 .........(1)

Five years's later their age will be :

A = (x + 5) years

B = (y + 5) years

Given,

Five years hence, A's age will be twice the age of B.

⇒ (x + 5) = 2(y + 5)

⇒ x + 5 = 2y + 10

⇒ x = 2y + 10 - 5

⇒ x = 2y + 5 .........(2)

From (1) and (2), we get :

⇒ 4y - 15 = 2y + 5

⇒ 4y - 2y = 5 + 15

⇒ 2y = 20

⇒ y = $\dfrac{20}{2}$ = 10 years.

Substituting value of y in equation (2), we get :

⇒ x = 2(10) + 5 = 20 + 5 = 25 years.

Hence, present age of A = 25 years and B = 10 years.

A is 20 years older than B. 5 years ago, A was 3 times as old as B. Find their present ages.

Answer

Let present age of B be x years and so, age of A = (x + 20) years.

Given,

5 years ago, A was 3 times as old as B.

∴ (x + 20) - 5 = 3(x - 5)

⇒ x + 15 = 3x - 15

⇒ 3x - x = 15 + 15

⇒ 2x = 30

⇒ x = $\dfrac{30}{2}$ = 15 years.

Age of A = (x + 20) = 15 + 20 = 35 years.

Hence, present age of A = 35 years and B = 15 years.

Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of mother and her daughter.

Answer

Let present ages of mother and her daughter be x and y years respectively.

Given,

Four years ago, a mother was four times as old as her daughter.

⇒ (x - 4) = 4(y - 4)

⇒ x - 4 = 4y - 16

⇒ x = 4y - 16 + 4

⇒ x = 4y - 12 .......(1)

Given,

Six years later, the mother will be two and a half times as old as her daughter at that time.

⇒ (x + 6) = 2.5(y + 6)

⇒ x + 6 = 2.5y + 15

⇒ x = 2.5y + 15 - 6

⇒ x = 2.5y + 9 ........(2)

From (1) and (2), we get :

⇒ 4y - 12 = 2.5y + 9

⇒ 4y - 2.5y = 9 + 12

⇒ 1.5y = 21

⇒ y = $\dfrac{21}{1.5}$ = 14.

Substituting value of y in equation (1), we get :

⇒ x = 4 × 14 - 12 = 56 - 12 = 44.

Hence, age of mother is 44 years and age of daughter is 14 years.

The age of a man is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children at that time. Find the present age of the man.

Answer

Let present age of man be x years and sum of the ages of his two children be y years.

Given,

The age of a man is twice the sum of the ages of his two children.

∴ x = 2y ........(1)

Given,

After 20 years, his age will be equal to the sum of the ages of his children at that time.

After 20 years,

Age of man = (x + 20) years

Sum of ages of his children = (y + 40), as each child's age will increase by 20 years.

⇒ x + 20 = y + 40

⇒ 2y + 20 = y + 40 .........[From (1)]

⇒ 2y - y = 40 - 20

⇒ y = 20.

⇒ x = 2y = 2(20) = 40.

Hence, present age of man = 40 years.

The annual incomes of A and B are in the ratio 3 : 4 and their annual expenditure are in the ratio 5 : 7. If each saves ₹ 5,000; find their annual incomes.

Answer

Given,

Annual incomes of A and B are in the ratio 3 : 4.

Let annual income of A be 3x and B be 4x.

Annual expenditure are in the ratio 5 : 7.

Let annual expenditure of A be 5y and B be 7y.

Given,

Each save ₹ 5,000.

For A,

⇒ 3x - 5y = 5000 ...........(1)

For B,

⇒ 4x - 7y = 5000 ..........(2)

Multiplying equation (1) by 4, we get :

⇒ 4(3x - 5y) = 4 × 5000

⇒ 12x - 20y = 20000 .........(3)

Multiplying equation (2) by 3, we get :

⇒ 3(4x - 7y) = 3 × 5000

⇒ 12x - 21y = 15000 .........(4)

Subtracting equation (4) from (3), we get :

⇒ 12x - 20y - (12x - 21y) = 20000 - 15000

⇒ 12x - 12x - 20y + 21y = 5000

⇒ y = 5000.

Substituting value of y in (1), we get :

⇒ 3x - 5(5000) = 5000

⇒ 3x - 25000 = 5000

⇒ 3x = 25000 + 5000

⇒ 3x = 30000

⇒ x = $\dfrac{30000}{3}$ = 10000.

⇒ 3x = 3(10000) = 30000 and 4x = 4(10000) = 40000.

Hence, annual income of A = ₹ 30,000 and B = ₹ 40,000.

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1. Find the number of students who appeared for the examination.

Answer

Given,

Ratio of passes to failures = 4 : 1

Let no. of students passed be 4x and failed be x.

So, total students who appeared for examination (originally) = 5x (4x + x)

Given,

Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1.

So, now students appeared = 5x - 30 and no. of students passed = 4x - 20

No. of students failed = (5x - 30) - (4x - 20) = 5x - 4x - 30 + 20 = x - 10.

Now ratio = 5 : 1.

$\Rightarrow \dfrac{4x - 20}{x - 10} = \dfrac{5}{1} \\[1em] \Rightarrow 4x - 20 = 5(x - 10) \\[1em] \Rightarrow 4x - 20 = 5x - 50 \\[1em] \Rightarrow 5x - 4x = 50 - 20 \\[1em] \Rightarrow x = 30$

No. of students who appeared for examination originally = 5x = 5 x 30 = 150.

Hence, no. of students who appeared for examination are 150.

A and B both have some pencils. If A gives 10 pencils to B, then B will have twice as many pencils as are left with A. And if B gives 10 pencils to A, then they will have the same number of pencils. How many pencils does each have ?

Answer

Let A have x pencils and B have y pencils.

Given,

If A gives 10 pencils to B, then B will have twice as many as A.

∴ 2(x - 10) = y + 10

⇒ 2x - 20 = y + 10

⇒ 2x - y = 10 + 20

⇒ 2x - y = 30 .........(1)

Given,

If B gives 10 pencils to A, then they will have the same number of pencils.

∴ x + 10 = y - 10

⇒ x - y = -10 - 10

⇒ x - y = -20 ..........(2)

Subtracting equation (2) from (1), we get :

⇒ 2x - y - (x - y) = 30 - (-20)

⇒ 2x - x - y + y = 30 + 20

⇒ x = 50

Substituting value of x in equation (1), we get :

⇒ 2(50) - y = 30

⇒ 100 - y = 30

⇒ y = 100 - 30 = 70.

Hence, A has 50 pencils and B has 70 pencils.

1250 persons went to see a circus-show. Each adult paid ₹ 75 and each child paid ₹ 25 for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to ₹ 61,250.

Answer

Let there be x adults and y children.

Given,

There are total 1250 persons.

∴ x + y = 1250

⇒ x = 1250 - y .......(1)

Given,

Total collection is of ₹ 61,250.

∴ 75x + 25y = 61250 .......(2)

Substituting value of x from equation (1) in equation (2), we get :

⇒ 75(1250 - y) + 25y = 61250

⇒ 93750 - 75y + 25y = 61250

⇒ 93750 - 50y = 61250

⇒ 50y = 93750 - 61250

⇒ 50y = 32500

⇒ y = $\dfrac{32500}{50}$ = 650.

Substituting value of y in equation (1), we get :

⇒ x = 1250 - 650 = 600.

Hence, number of adults = 600 and number of children = 650.

Rohit says to Ajay, "Give me a hundred, I shall then become twice as rich as you." Ajay replies, "if you give me ten, I shall be six times as rich as you." How much does each have originally?

Answer

Let Rohit have ₹ x and Ajay have ₹ y.

According to first part of question :

⇒ x + 100 = 2(y - 100)

⇒ x + 100 = 2y - 200

⇒ 2y - x = 100 + 200

⇒ 2y - x = 300 ...........(1)

According to second part of question :

⇒ y + 10 = 6(x - 10)

⇒ y + 10 = 6x - 60

⇒ y - 6x = -60 - 10

⇒ y - 6x = -70

Multiplying both sides of the above equation by 2, we get :

⇒ 2(y - 6x) = 2 × -70

⇒ 2y - 12x = -140 ...........(2)

Subtracting equation (2) from (1), we get :

⇒ 2y - x - (2y - 12x) = 300 - (-140)

⇒ 2y - x - 2y + 12x = 300 + 140

⇒ 11x = 440

⇒ x = $\dfrac{440}{11}$ = ₹ 40.

Substituting value of x from equation (1), we get :

⇒ 2y - 40 = 300

⇒ 2y = 300 + 40

⇒ 2y = 340

⇒ y = $\dfrac{340}{2}$ = ₹ 170.

Hence, originally Rohit has ₹ 40 and Ajay has ₹ 170.

The sum of a two digit number and the number obtained by reversing the order of the digits is 99. Find the number, if the digits differ by 3.

Answer

Let digit at ten's place be x and unit's place be y.

Number = 10(x) + y = 10x + y

On reversing the digits,

Reversed number = 10(y) + x = 10y + x

Given,

Sum of a two digit number and the number obtained by reversing the order of the digits is 99.

∴ (10x + y) + (10y + x) = 99

⇒ 11x + 11y = 99

⇒ 11(x + y) = 99

⇒ x + y = 9

⇒ x = 9 - y ............(1)

Given,

Digits differ by 3.

∴ x - y = 3 or y - x = 3

Considering x - y = 3

⇒ x = 3 + y ..........(2)

From (1) and (2), we get :

⇒ 9 - y = 3 + y

⇒ 9 - 3 = 2y

⇒ 2y = 6

⇒ y = 3.

Substituting value of y = 3 in equation (1), we get :

⇒ x = 9 - 3 = 6.

Number = 10x + y = 10(6) + 3 = 60 + 3 = 63.

Considering y - x = 3

⇒ x = y - 3 ..........(3)

From (1) and (3), we get :

⇒ 9 - y = y - 3

⇒ 9 + 3 = 2y

⇒ 2y = 12

⇒ y = 6.

Substituting value of y = 6 in equation (1), we get :

⇒ x = 9 - 6 = 3.

Number = 10x + y = 10(3) + 6 = 30 + 6 = 36.

Hence, number = 36 or 63.

Seven times a two digit number is equal to four times the number obtained by reversing the digits. If the difference between the digits is 3, find the number.

Answer

Let digit at ten's place be x and digit at unit's place be y.

Number = 10(x) + y = 10x + y

On reversing the digits,

Reversed number = 10(y) + x = 10y + x

Given,

Seven times a two digit number is equal to four times the number obtained by reversing the digits.

⇒ 7(10x + y) = 4(10y + x)

⇒ 70x + 7y = 40y + 4x

⇒ 40y - 7y = 70x - 4x

⇒ 33y = 66x

⇒ y = $\dfrac{66x}{33}$

⇒ y = 2x ..........(1)

Given,

Difference between the digits is 3.

Let x > y

⇒ x - y = 3 ..........(2)

Let y > x

⇒ y - x = 3 ..........(3)

Substituting value of y from equation (1) in equation (2), we get :

⇒ x - 2x = 3

⇒ -x = 3

⇒ x = -3

This is not possible as digits cannot be negative.

Substituting value of y from equation (1) in equation (3), we get :

⇒ 2x - x = 3

⇒ x = 3

Substituting value of x in equation (1), we get :

⇒ y = 2x = 2(3) = 6.

Number = 10x + y = 10(3) + 6 = 30 + 6 = 36.

Hence, number = 36.

From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is ₹ 77. But if we buy 3 tickets for station A and 5 tickets for station B, the total cost is ₹ 124. What are the fares from Delhi to station A and to station B ?

Answer

Let cost of ticket from Delhi to station A be ₹ x and cost of ticket from Delhi to station B be ₹ y.

Given,

From Delhi station, if we buy 2 tickets for station A and 3 tickets for station B, the total cost is ₹ 77.

⇒ 2x + 3y = 77 .......(1)

Given,

From Delhi station, if we buy 3 tickets for station A and 5 tickets for station B, the total cost is ₹ 124.

⇒ 3x + 5y = 124 .......(2)

Multiplying equation (1) by 3, we get :

⇒ 3(2x + 3y) = 3 × 77

⇒ 6x + 9y = 231 .......(3)

Multiplying equation (2) by 2, we get :

⇒ 2(3x + 5y) = 2 × 124

⇒ 6x + 10y = 248 .......(4)

Subtracting equation (3) from (4), we get :

⇒ (6x + 10y) - (6x + 9y) = 248 - 231

⇒ 6x - 6x + 10y - 9y = 17

⇒ y = ₹ 17.

Substituting value of y in equation (1), we get :

⇒ 2x + 3(17) = 77

⇒ 2x + 51 = 77

⇒ 2x = 77 - 51

⇒ 2x = 26

⇒ x = $\dfrac{26}{2}$ = ₹ 13.

Hence, fares from Delhi to station A = ₹ 13 and to station B = ₹ 17.

The sum of digits of a two digit number is 11. If the digit at ten's place is increased by 5 and the digit at unit's place is decreased by 5, the digits of the number are found to be reversed. Find the original number.

Answer

Let digit at ten's place be x and digit at unit's place be y.

Given,

The sum of digits of a two digit number is 11.

∴ x + y = 11

⇒ x = 11 - y ..........(1)

Given,

If the digit at ten's place is increased by 5 and the digit at unit's place is decreased by 5, the digits of the number are found to be reversed.

∴ 10(x + 5) + (y - 5) = 10y + x

⇒ 10x + 50 + y - 5 = 10y + x

⇒ 10x + y - 10y - x + 45 = 0

⇒ 9x - 9y + 45 = 0

⇒ 9y = 9x + 45

⇒ y = x + 5

⇒ x = y - 5 ...........(2)

From equation (1) and (2), we get :

⇒ 11 - y = y - 5

⇒ y + y = 11 + 5

⇒ 2y = 16

⇒ y = $\dfrac{16}{2}$ = 8.

Substituting value of y in equation (2), we get :

⇒ x = 8 - 5 = 3.

Number = 10x + y = 10(3) + 8 = 30 + 8 = 38.

Hence, number = 38.

90% acid solution (90% pure acid and 10% water) and 97% acid solution are mixed to obtain 21 litres of 95% acid solution. How many litres of each solution are mixed.

Answer

Let x litres of 90% and y litres of 97% be mixed; then

⇒ x + y = 21

⇒ x = 21 - y .......(1)

and

⇒ 90% of x + 97% of y = 95% of 21

$\Rightarrow \dfrac{90}{100}x + \dfrac{97}{100}y = \dfrac{95}{100} \times 21 \\[1em] \Rightarrow 90x + 97y = 95 \times 21 \\[1em] \Rightarrow 90x + 97y = 1995 \text{ ........(2)}$

Substituting value of x from equation (1) in equation (2), we get :

⇒ 90(21 - y) + 97y = 1995

⇒ 1890 - 90y + 97y = 1995

⇒ 7y = 1995 - 1890

⇒ 7y = 105

⇒ y = $\dfrac{105}{7}$ = 15.

Substituting value of y in equation (1), we get :

⇒ x = 21 - 15 = 6.

Hence, 6 litres of 90% acid solution and 15 litres of 97% of acid solution are mixed.

Class XI students of a school wanted to give a farewell party to the outgoing students of class XII. They decided to purchase two kinds of sweets, one costing ₹ 250 per kg and the other costing ₹ 350 per kg. They estimated that 40 kg of sweets were needed. If the total budget for the sweets was ₹ 11800; find how much sweets of each kind were brought ?

Answer

Let x kg of ₹ 250 per kg and y kg of ₹ 350 per kg sweets were purchased.

Given,

40 kg of sweets were needed.

∴ x + y = 40

⇒ x = 40 - y .........(1)

Given,

Total budget for sweets was ₹ 11800.

∴ 250x + 350y = 11800

⇒ 25x + 35y = 1180 ........(2)

Substituting value of x from equation (1) in equation (2), we get :

⇒ 25(40 - y) + 35y = 1180

⇒ 1000 - 25y + 35y = 1180

⇒ 10y = 1180 - 1000

⇒ 10y = 180

⇒ y = $\dfrac{180}{10}$ = 18.

Substituting value of y in equation (1), we get :

⇒ x = 40 - 18 = 22.

Hence, 22 kg of ₹ 250 per kg and 18 kg of ₹ 350 per kg sweets were purchased.