Simultaneous Linear Equations — Exercise 5(C)

The sum of two numbers is 53 and their difference is 25. Find the numbers.

Answer

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

Given,

Sum of numbers = 53

⇒ x + y = 53     .......(1)

Difference of numbers = 25

⇒ x - y = 25     ......(2)

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

⇒ (x + y) + (x - y) = 53 + 25

⇒ x + x + y - y = 78

⇒ 2x = 78

⇒ x = $\dfrac{78}{2}$ = 39.

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

⇒ x + y = 53

⇒ 39 + y = 53

⇒ y = 53 - 39 = 14.

Hence, the numbers are 39 and 14.

The sum of two numbers exceeds thrice the smaller by 2. If the difference between them is 19, find the numbers.

Answer

Let the numbers be x and y, such that x < y.

Given,

The sum of two numbers exceeds thrice the smaller by 2.

∴ (x + y) - 3x = 2

⇒ x - 3x + y = 2

⇒ y - 2x = 2     ......(1)

Given,

Difference between the numbers = 19

⇒ y - x = 19     ......(2)

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

⇒ y - x - (y - 2x) = 19 - 2

⇒ y - x - y + 2x = 19 - 2

⇒ x = 17.

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

⇒ y - 2x = 2

⇒ y - 2 × 17 = 2

⇒ y - 34 = 2

⇒ y = 36.

Hence, the numbers are 17 and 36.

The sum of two numbers is 51. If the larger is doubled and the smaller is tripled, the difference is 12. Find the numbers.

Answer

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

Given,

Sum of numbers = 51

⇒ x + y = 51

⇒ x = 51 - y     ......(1)

Given,

Difference of numbers when larger is doubled and the smaller is tripled = 12

⇒ 2x - 3y = 12     ......(2)

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

⇒ 2(51 - y) - 3y = 12

⇒ 102 - 2y - 3y = 12

⇒ -5y = 12 - 102

⇒ -5y = -90

⇒ y = $\dfrac{90}{5} = 18$.

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

⇒ x = 51 - y

⇒ x = 51 - 18

⇒ x = 33.

Hence, the numbers are 33 and 18.

Find two numbers such that the sum of twice the first and thrice the second is 103 and four times the first exceeds seven times the second by 11.

Answer

Let two numbers be x and y.

Given,

Sum of twice the first and thrice the second = 103

⇒ 2x + 3y = 103

⇒ 2x = 103 - 3y

⇒ x = $\Big(\dfrac{103 - 3y}{2}\Big)$     ......(1)

Given,

Four times the first number exceeds seven times the second by 11.

⇒ 4x - 7y = 11

⇒ 4x = 7y + 11     .....(2)

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

⇒ $4\Big(\dfrac{103 - 3y}{2}\Big)$ = 7y + 11

⇒ 2(103 - 3y) = 7y + 11

⇒ 206 - 6y = 7y + 11

⇒ 206 - 11 = 7y + 6y

⇒ 195 = 13y

⇒ y = $\dfrac{195}{13} = 15$.

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

$\Rightarrow x = \Big(\dfrac{103 - 3y}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{103 - 3 \times 15}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{103 - 45}{2}\Big) \\[1em] \Rightarrow x = \Big(\dfrac{58}{2}\Big) = 29.$

Hence, the numbers are 29 and 15.

Find two numbers such that the sum of thrice the first and the second is 142 and four times the first exceeds the second by 138.

Answer

Let two numbers be x and y.

Given,

Sum of thrice the first and the second = 142

⇒ 3x + y = 142

⇒ y = 142 - 3x     .....(1)

Given,

Four times the first exceeds the second by 138.

⇒ 4x - y = 138

⇒ 4x = y + 138     ......(2)

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

⇒ 4x = 142 - 3x + 138

⇒ 4x + 3x = 142 + 138

⇒ 7x = 280

⇒ x = $\dfrac{280}{7} = 40$.

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

⇒ y = 142 - 3x

⇒ y = 142 - 3(40)

⇒ y = 142 - 120

⇒ y = 22.

Hence, the numbers are 40 and 22.

Of the two numbers, 4 times the smaller one is less than 3 times the larger one by 6. Also, the sum of the numbers is larger than 6 times their difference by 5. Find the numbers.

Answer

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

Given,

4 times the smaller one is less than 3 times the larger one by 6.

⇒ 3x - 4y = 6

⇒ 4y = 3x - 6

⇒ y = $\Big(\dfrac{3x - 6}{4}\Big)$     .............(1)

Given,

Sum of the numbers is larger than 6 times their difference by 5.

⇒ x + y - 6(x - y) = 5

⇒ x + y - 6x + 6y = 5

⇒ x - 6x + y + 6y = 5

⇒ 7y - 5x = 5     ...........(2)

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

$\Rightarrow 7\Big(\dfrac{3x - 6}{4}\Big) - 5x = 5 \\[1em] \Rightarrow \Big(\dfrac{21x - 42}{4}\Big) - 5x = 5 \\[1em] \Rightarrow \Big(\dfrac{21x - 42 - 20x}{4}\Big) = 5 \\[1em] \Rightarrow \dfrac{x - 42}{4} = 5 \\[1em] \Rightarrow x - 42 = 20 \\[1em] \Rightarrow x = 20 + 42 \\[1em] \Rightarrow x = 62.$

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

$\Rightarrow y = \Big(\dfrac{3x - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{3 \times 62 - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{186 - 6}{4}\Big) \\[1em] \Rightarrow y = \Big(\dfrac{180}{4}\Big) = 45.$

Hence, the numbers are 62 and 45.

If from twice the greater number of the two numbers, 45 is subtracted, the result is the other number. If from twice the smaller number, 21 is subtracted, the result is the greater number. Find the numbers.

Answer

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

Given,

If from twice the greater number of the two numbers, 45 is subtracted, the result is the other number.

⇒ 2x - 45 = y     .....(1)

Given,

If from twice the smaller number, 21 is subtracted, the result is the greater number.

⇒ 2y - 21 = x     .....(2)

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

⇒ 2y - 21 = x

⇒ 2(2x - 45) - 21 = x

⇒ 4x - 90 - 21 = x

⇒ 4x - x - 111 = 0

⇒ 3x = 111

⇒ x = $\dfrac{111}{3} = 37.$

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

⇒ 2x - 45 = y

⇒ 2(37) - 45 = y

⇒ 74 - 45 = y

⇒ y = 29.

Hence, the numbers are 37 and 29.

If three times the larger of the two numbers is divided by the smaller, then the quotient is 4 and remainder is 5. If 6 times the smaller is divided by the larger, the quotient is 4 and the remainder is 2. Find the numbers.

Answer

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

Given,

Three times the larger divided by the smaller gives quotient 4.

⇒ 3x = 4y + 5

⇒ x = $\dfrac{4y + 5}{3}$     .....(1)

Given,

Six times the smaller divided by the larger gives quotient 4 and remainder 2.

⇒ 6y = 4x + 2     .....(2)

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

$\Rightarrow 6y = 4 \Big(\dfrac{4y + 5}{3}\Big) + 2 \\[1em] \Rightarrow 6y = \Big(\dfrac{16y + 20}{3}\Big) + 2 \\[1em] \Rightarrow 6y = \Big(\dfrac{16y + 20 + 6}{3}\Big) \\[1em] \Rightarrow 6y = \Big(\dfrac{16y + 26}{3}\Big) \\[1em] \Rightarrow 6y \times 3 = 16y + 26 \\[1em] \Rightarrow 18y = 16y + 26 \\[1em] \Rightarrow 18y - 16y = 26 \\[1em] \Rightarrow 2y = 26 \\[1em] \Rightarrow y = \dfrac{26}{2} \\[1em] \Rightarrow y = 13.$

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

$\Rightarrow x = \dfrac{4y + 5}{3} \\[1em] \Rightarrow x = \dfrac{4 \times 13 + 5}{3} \\[1em] \Rightarrow x = \dfrac{52 + 5}{3} \\[1em] \Rightarrow x = \dfrac{57}{3} = 19.$

Hence, the numbers are 19 and 13.

If 2 is added to each of two given numbers, then their ratio becomes 1 : 2. However, if 4 is subtracted from each of the given numbers, the ratio becomes 5 : 11. Find the numbers.

Answer

Let two numbers be x and y.

Given,

If 2 is added to each, the ratio becomes 1 : 2.

⇒ $\dfrac{x + 2}{y + 2} = \dfrac{1}{2}$

⇒ 2(x + 2) = y + 2

⇒ 2x + 4 = y + 2

⇒ y = 2x + 4 - 2

⇒ y = 2x + 2     ....(1)

Given,

If 4 is subtracted from each number, the ratio becomes 5 : 11.

⇒ $\dfrac{x - 4}{y - 4} = \dfrac{5}{11}$

⇒ 11(x - 4) = 5(y - 4)

⇒ 11x - 44 = 5y - 20

⇒ 11x - 5y = -20 + 44

⇒ 11x - 5y = 24     ....(2)

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

⇒ 11x - 5(2x + 2) = 24

⇒ 11x - 10x - 10 = 24

⇒ x - 10 = 24

⇒ x = 24 + 10

⇒ x = 34.

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

⇒ y = 2x + 2

⇒ y = 2 × 34 + 2

⇒ y = 68 + 2

⇒ y = 70.

Hence, the numbers are 34 and 70.

The difference between two numbers is 12 and the difference between their squares is 456. Find the numbers.

Answer

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

Given,

Difference between two numbers = 12.

⇒ x - y = 12     .........(1)

Given,

Difference between their squares = 456

⇒ x² - y² = 456

By identity,

⇒ x² - y² = (x + y)(x - y)

⇒ (x + y)(x - y) = 456

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

⇒ (x + y)(12) = 456

⇒ (x + y) = $\dfrac{456}{12}$

⇒ x + y = 38     ........(2)

Adding equations (1) and (2),

⇒ x + y + x - y = 38 + 12

⇒ 2x = 50

⇒ x = $\dfrac{50}{2}$

⇒ x = 25.

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

⇒ x - y = 12

⇒ 25 - y = 12

⇒ y = 25 - 12

⇒ y = 13.

Hence, the numbers are 25 and 13.

Find the fraction which becomes $\dfrac{1}{2}$ when its numerator is increased by 6 and is equal to $\dfrac{1}{3}$ when its denominator is increased by 7.

Answer

Let the numerator be x and denominator be y.

Thus, fraction = $\dfrac{x}{y}$

Given,

The fraction becomes $\dfrac{1}{2}$ when its numerator is increased by 6.

⇒ $\dfrac{x + 6}{y} = \dfrac{1}{2}$

⇒ 2(x + 6) = y

⇒ 2x + 12 = y     ..........(1)

Given,

The fraction becomes $\dfrac{1}{3}$ when its denominator is increased by 7.

⇒ $\dfrac{x}{y + 7} = \dfrac{1}{3}$

⇒ 3x = y + 7     .......(2)

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

⇒ 3x = (2x + 12) + 7

⇒ 3x = 2x + 19

⇒ 3x - 2x = 19

⇒ x = 19.

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

⇒ 2 × 19 + 12 = y

⇒ 38 + 12 = y

⇒ y = 50.

Hence, the fraction = $\dfrac{19}{50}$.

A fraction becomes $\dfrac{1}{2}$ when 1 is subtracted from its numerator and 1 is added to its denominator; it becomes $\dfrac{1}{3}$ when 6 is subtracted from its numerator and 1 from its denominator. Find the original fraction.

Answer

Let the numerator be x and denominator be y.

Thus, fraction = $\dfrac{x}{y}$

Given,

The fraction becomes $\dfrac{1}{2}$ when 1 is subtracted from the numerator and 1 is added to the denominator,

⇒ $\dfrac{x - 1}{y + 1} = \dfrac{1}{2}$

⇒ 2(x - 1) = y + 1

⇒ 2x - 2 = y + 1

⇒ y = 2x - 2 - 1

⇒ y = 2x - 3     .........(1)

Given,

The fraction becomes $\dfrac{1}{3}$ when 6 is subtracted from the numerator and 1 from the denominator,

⇒ $\dfrac{x - 6}{y - 1} = \dfrac{1}{3}$

⇒ 3(x - 6) = y - 1

⇒ 3x - 18 = y - 1

⇒ y = 3x - 18 + 1

⇒ y = 3x - 17     .........(2)

From equations (1) in (2), we get :

⇒ 3x - 17 = 2x - 3

⇒ 3x - 2x = -3 + 17

⇒ 3x - 2x = 14

⇒ x = 14.

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

⇒ y = 2(14) - 3 = 28 - 3 = 25.

Fraction = $\dfrac{x}{y} = \dfrac{14}{25}$.

Hence, the fraction = $\dfrac{14}{25}$.

The denominator of a fraction is greater than its numerator by 9. If 7 is subtracted from both, its numerator and denominator, the fraction becomes $\dfrac{2}{3}$. Find the original fraction.

Answer

Let the numerator be x.

Given,

The denominator of a fraction is greater than its numerator by 9.

⇒ Denominator = x + 9

Thus, fraction = $\dfrac{x}{x + 9}$

Given,

The fraction becomes $\dfrac{2}{3}$, when 7 is subtracted from both numerator and denominator,

⇒ $\dfrac{x - 7}{(x + 9) - 7} = \dfrac{2}{3}$

⇒ $\dfrac{x - 7}{x + 2} = \dfrac{2}{3}$

⇒ 3(x - 7) = 2(x + 2)

⇒ 3x - 21 = 2x + 4

⇒ 3x - 2x = 4 + 21

⇒ x = 25.

The denominator is (x + 9) = 25 + 9 = 34.

Fraction = $\dfrac{x}{y} = \dfrac{25}{34}$.

Hence, the fraction = $\dfrac{25}{34}$.

A number consists of two digits, the difference of whose digits is 3. If 4 times the number equals 7 times the number obtained by reversing its digits, find the number.

[Hint. Original number is greater than the number obtained by reversing its digits. ∴ In original number, ten's digit greater than unit's digit.]

Answer

Let the tens and unit digits of required number be x and y. x > y (according to hint)

Given,

Difference of digits of the number is 3.

⇒ x - y = 3

⇒ x = y + 3     .....(1)

Original number = 10x + y

Number obtained by reversing the digits = (10y + x)

Given,

4 times the number equals 7 times the number obtained by reversing its digits.

⇒ 4(10x + y) = 7(10y + x)

⇒ 40x + 4y = 70y + 7x

⇒ 40x - 7x + 4y - 70y = 0

⇒ 33x - 66y = 0     .....(2)

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

⇒ 33(y + 3) - 66y = 0

⇒ 33y + 33 × 3 - 66y = 0

⇒ 99 - 33y = 0

⇒ 99 - 33y = 0

⇒ 33y = 99

⇒ y = $\dfrac{99}{33}$

⇒ y = 3.

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

⇒ x = y + 3

⇒ x = 3 + 3

⇒ x = 6.

Original number = 10x + y

= 10 × 6 + 3

= 63.

Hence, the number is 63.

A number consists of two digits, the difference of whose digits is 5. If 8 times the number is equal to 3 times the number obtained by reversing the digits, find the number.

Answer

According to question,

Original number is smaller than the number obtained by reversing its digits.

∴ In original number, unit's digit greater than ten's digit.

Let the ten's and unit's digit of required number be x and y respectively y > x.

Given,

⇒ y - x = 5

⇒ y = x + 5     .....(1)

Number obtained by reversing the digits = (10y + x)

Given,

8 times the number is equal to 3 times the number obtained by reversing the digits.

⇒ 8(10x + y) = 3(10y + x)

⇒ 80x + 8y = 30y + 3x

⇒ 80x - 3x + 8y - 30y = 0

⇒ 77x - 22y = 0     .....(2)

Substituting the value of x from equation (2) in 77x - 22y = 0, we get:

⇒ 77x - 22y = 0

⇒ 77x - 22 × (x + 5) = 0

⇒ 77x - 22x - 110 = 0

⇒ 55x - 110 = 0

⇒ 55x = 110

⇒ x = $\dfrac{110}{55}$

⇒ x = 2.

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

⇒ y = x + 5

⇒ y = 2 + 5

⇒ y = 7.

Number = (10x + y) = 10 × 2 + 7 = 27.

Hence, the number is 27.

The result of dividing a two-digit number by the number with its digits reversed is $\Big(1\dfrac{3}{4}\Big)$. If the sum of the digits is 12, find the number.

Answer

Let the ten's and unit's digits of required number be x and y respectively.

Given,

Sum of the digits of the number is 12.

⇒ x + y = 12

⇒ x = 12 - y     .....(1)

Original number = 10x + y

Number obtained by reversing the digits = 10y + x

Given,

On dividing the number by the number with its digits reversed, the result is $\Big(1\dfrac{3}{4}\Big)$.

$\therefore \Big(\dfrac{10x + y}{10y + x}\Big) = \Big(1\dfrac{3}{4}\Big) \\[1em] \Rightarrow \Big(\dfrac{10x + y}{10y + x}\Big) = \dfrac{7}{4} \\[1em] \Rightarrow (10x + y) \times 4 = (10y + x) \times 7 \\[1em] \Rightarrow 40x + 4y = 70y + 7x \\[1em] \Rightarrow 40x - 7x + 4y - 70y = 0 \\[1em] \Rightarrow 33x - 66y = 0 \text{.....(2)}$

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

⇒ 33(12 - y) - 66y = 0

⇒ 396 - 33y - 66y = 0

⇒ 396 - 99y = 0

⇒ 99y = 396

⇒ y = $\dfrac{396}{99}$

⇒ y = 4.

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

⇒ x = 12 - y

⇒ x = 12 - 4

⇒ x = 8.

Original number = (10x + y)

= 10 × 8 + 4

= 84.

Hence, the number is 84.

When a two-digit number is divided by the sum of its digits, the quotient is 8. On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1. Find the number.

Answer

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

On dividing the number by the sum of its digits, the quotient is 8.

$\Rightarrow \Big(\dfrac{10x + y}{x + y}\Big) =8 \\[1em] \Rightarrow 10x + y = 8(x + y) \\[1em] \Rightarrow 10x + y = 8x + 8y \\[1em] \Rightarrow 10x - 8x + y - 8y = 0 \\[1em] \Rightarrow 2x - 7y = 0 \text{ .....(1)}$

Given,

On diminishing the ten’s digit by three times the unit’s digit, the remainder obtained is 1.

⇒ x - 3y = 1

⇒ x = 3y + 1     .........(2)

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

⇒ 2(3y + 1) - 7y = 0

⇒ 6y + 2 - 7y = 0

⇒ 2 - y = 0

⇒ y = 2.

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

⇒ x = 3 × 2 + 1

⇒ x = 6 + 1

⇒ x = 7.

Number = 10x + y

= 10 × 7 + 2

= 72.

Hence, the number is 72.

A number of two digits exceeds four times the sum of its digits by 6, and the number is increased by 9 on reversing its digits. Find the number.

Answer

Let the ten's and unit's digit of required number be x and y respectively.

Number = 10x + y

Given,

A number of two digits exceeds four times the sum of its digits by 6.

⇒ 10x + y - 4(x + y) = 6

⇒ 10x + y - 4x - 4y = 6

⇒ 6x - 3y = 6     .....(1)

Number obtained by reversing the digits = 10y + x

Given,

Number is increased by 9 on reversing its digits.

⇒ 10y + x = 10x + y + 9

⇒ 10y - y + x - 10x = 9

⇒ 9y - 9x = 9

⇒ y - x = 1

⇒ x = y - 1     .....(2)

Substituting the value of x from equation (2) in equation 1,

⇒ 6(y - 1) - 3y = 6

⇒ 6y - 6 - 3y = 6

⇒ 3y - 6 = 6

⇒ 3y = 6 + 6

⇒ 3y = 12

⇒ y = $\dfrac{12}{3}$

⇒ y = 4.

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

⇒ x = 4 - 1

⇒ x = 3.

Number = 10x + y

= 10 × 3 + 4

= 34.

Hence, the number is 34.

The sum of the digits of a two-digit number is 12. If the digits are reversed, the new number is 12 less than twice the original number. Find the original number.

Answer

Let the ten's and unit's digits of required number be x and y.

Number = 10x + y

Given,

Sum of digits = 12.

⇒ x + y = 12

⇒ x = 12 - y     .....(1)

Given,

Number obtained by reversing the digits = (10y + x)

If the digits are reversed, the new number is 12 less than twice the original number.

⇒ (10y + x) = 2(10x + y) - 12

⇒ 10y + x = 20x + 2y - 12

⇒ 10y - 2y + x - 20x = -12

⇒ 8y - 19x = - 12     .....(2)

Substituting the value of x from equation (1) in equation (2),

⇒ 8y - 19(12 - y) = -12

⇒ 8y - 228 + 19y = -12

⇒ 27y = -12 + 228

⇒ 27y = 216

⇒ y = $\dfrac{216}{27}$

⇒ y = 8.

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

⇒ x = 12 - 8

⇒ x = 4.

The number is,

⇒ (10x + y) = 10 × 4 + 8 = 48.

Hence, the number is 48.

If 11 pens and 19 pencils together cost ₹ 502, while 19 pens and 11 pencils together cost ₹ 758, how much do 3 pens and 6 pencils cost together?

Answer

Let the cost of a pen be ₹ x and ₹ y be the cost of a pencil.

Given,

11 pens and 19 pencils together cost ₹ 502.

⇒ 11x + 19y = 502     ........(1)

Given,

19 pens and 11 pencils together cost ₹ 758.

⇒ 19x + 11y = 758     .........(2)

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

⇒ 19x + 11y - (11x + 19y) = 758 - 502

⇒ 19x - 11x + 11y - 19y = 256

⇒ 8x - 8y = 256

⇒ 8(x - y) = 256

⇒ x - y = $\dfrac{256}{8}$

⇒ x - y = 32

⇒ x = 32 + y .....(3)

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

⇒ 11x + 19y = 502

⇒ 11(32 + y) + 19y = 502

⇒ 352 + 11y + 19y = 502

⇒ 30y = 502 - 352

⇒ 30y = 150

⇒ y = $\dfrac{150}{30}$

⇒ y = ₹ 5.

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

⇒ x = 32 + y

⇒ x = 32 + 5

⇒ x = ₹ 37.

The amount to buy the 3 pens and 6 pencils is,

⇒ 3x + 6y

⇒ 3 × 37 + 6 × 5

⇒ 111 + 30

⇒ ₹ 141.

Hence, 3 pens and 6 pencils costs ₹ 141.

5 kg sugar and 7 kg rice together cost ₹ 258, while 7 kg sugar and 5 kg rice together cost ₹ 246. Find the total cost of 8 kg sugar and 10 kg rice.

Answer

Let cost of sugar be ₹ x/kg and cost of rice be ₹ y/kg.

Given,

5 kg sugar and 7 kg rice together cost ₹ 258,

⇒ 5x + 7y = 258

⇒ 7y = 258 - 5x

⇒ y = $\dfrac{258 - 5x}{7}$     .....(1)

Given,

7 kg sugar and 5 kg rice together cost ₹ 246,

⇒ 7x + 5y = 246     .....(2)

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

$\Rightarrow 7x + 5\Big(\dfrac{258 - 5x}{7}\Big) = 246 \\[1em] \Rightarrow \Big(\dfrac{49x + 1290 - 25x}{7}\Big) = 246 \\[1em] \Rightarrow 24x + 1290 = 246 \times 7 \\[1em] \Rightarrow 24x + 1290 = 1722 \\[1em] \Rightarrow 24x = 1722 - 1290 \\[1em] \Rightarrow x = \dfrac{432}{24} = 18.$

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

$\Rightarrow y = \dfrac{258 - 5 \times 18}{7} \\[1em] \Rightarrow y = \dfrac{258 - 90}{7} \\[1em] \Rightarrow y = \dfrac{168}{7} \\[1em] \Rightarrow y = 24.$

Total cost of 8 kg sugar and 10 kg rice = 8x + 10y

= 8 × 18 + 10 × 24

= 144 + 240 = ₹ 384.

Hence, the total cost of 8 kg sugar and 10 kg rice = ₹ 384.

One year ago a man was four times as old as his son. After 6 years, his age exceeds twice his son’s age by 9 years. Find their present ages.

Answer

Let the present age of man be x years and his son be y years.

Given,

One year ago a man was four times as old as his son,

⇒ x - 1 = 4(y - 1)

⇒ x - 1 = 4y - 4

⇒ x = 4y - 4 + 1

⇒ x = 4y - 3     ....(1)

Given,

After 6 years, the man's age will exceed twice his son's age by 9,

⇒ x + 6 = 2(y + 6) + 9

⇒ x + 6 = 2y + 12 + 9

⇒ x = 2y + 12 + 9 - 6

⇒ x = 2y + 15     .....(2)

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

⇒ 4y - 3 = 2y + 15

⇒ 4y - 2y = 15 + 3

⇒ 2y = 18

⇒ y = $\dfrac{18}{2}$ = 9.

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

⇒ x = 4y - 3

⇒ x = 4(9) - 3

⇒ x = 36 - 3 = 33.

Hence, man's present age = 33 years and son's present age = 9 years.

5 years ago, A was thrice as old as B and 10 years later A shall be twice as old as B. What are the present ages of A and B?

Answer

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

Given,

5 years ago, A was thrice as old as B.

⇒ x - 5 = 3(y - 5)

⇒ x - 5 = 3y - 15

⇒ x = 3y - 15 + 5

⇒ x = 3y - 10     ...,,,,.(1)

Given,

10 years later, A shall be twice as old as B.

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x = 2y + 20 - 10

⇒ x = 2y + 10     .........(2)

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

⇒ 3y - 10 = 2y + 10

⇒ 3y - 2y = 10 + 10

⇒ y = 20.

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

⇒ x = 3y - 10

⇒ x = 3(20) - 10

⇒ x = 60 - 10 = 50.

Hence, A's present age = 50 years and B's present age = 20 years.

The monthly incomes of A and B are in the ratio 7 : 5 and their expenditures are in the ratio 3 : 2. If each saves ₹ 1500 per month, find their monthly incomes.

Answer

Let monthly income of A be 7x and B be 5x.

Let monthly expenditure of A be 3y and B be 2y.

Given,

Both A and B save ₹ 1500 per month.

For A :

⇒ 7x - 3y = 1500

⇒ 7x = 1500 + 3y

⇒ x = $\dfrac{1500 + 3y}{7}$     .....(1)

For B :

⇒ 5x - 2y = 1500     ......(2)

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

$\Rightarrow 5x - 2y = 1500 \\[1em] \Rightarrow 5\Big(\dfrac{1500 + 3y}{7}\Big) - 2y = 1500 \\[1em] \Rightarrow \Big(\dfrac{7500 + 15y}{7}\Big) - 2y = 1500 \\[1em] \Rightarrow \Big(\dfrac{7500 + 15y - 14y}{7}\Big) = 1500 \\[1em] \Rightarrow 7500 + y = 10500 \\[1em] \Rightarrow y = 10500 - 7500 \\[1em] \Rightarrow y = 3000.$

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

$\Rightarrow x = \dfrac{1500 + 3y}{7} \\[1em] \Rightarrow x = \dfrac{1500 + 3 \times 3000}{7} \\[1em] \Rightarrow x = \dfrac{1500 + 9000}{7} \\[1em] \Rightarrow x = \dfrac{10500}{7} = 1500.$

A's income = ₹ 7x = ₹ 7 × 1500 = ₹ 10,500,

B's income = ₹ 5x = ₹ 5 × 1500 = ₹ 7,500.

Hence, A's income = ₹ 10,500, B's income = ₹ 7,500.

A 90% acid solution is mixed with a 97% acid solution to obtain 21 litres of a 95% solution. Find the quantity of each the solutions to get the resultant mixture.

Answer

Let x litres be the quantity of the 90% acid solution and y litres be the quantity of the 97% acid solution.

Given,

Total volume = 21 litres

⇒ x + y = 21

⇒ x = 21 - y     ....(1)

Als,

⇒ 90% of x + 97% of y = 95% of 21

⇒ 0.90x + 0.97y = 0.95 × 21

⇒ 0.90x + 0.97y = 19.95     ....(2)

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

⇒ 0.90(21 - y) + 0.97y = 19.95

⇒ 18.9 - 0.90y + 0.97y = 19.95

⇒ 18.9 + 0.07y = 19.95

⇒ 0.07y = 19.95 - 18.9

⇒ 0.07y = 1.05

⇒ y = $\dfrac{1.05}{0.07} = 15$.

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

⇒ x = 21 - y

⇒ x = 21 - 15

⇒ x = 6.

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

There are two examination halls A and B. If 12 pupils are sent from A to B, the number of pupils in each room becomes the same. If 11 pupils are sent from B to A, then the number of pupils in A is double their number in B. Find the number of pupils in each room.

Answer

Let x and y be the initial number of pupils in examination hall A and B respectively.

Given,

If 12 pupils are sent from A to B, the number of pupils in each room becomes the same.

⇒ x - 12 = y + 12

⇒ x = y + 12 + 12

⇒ x = 24 + y     .....(1)

Given,

If 11 pupils are sent from B to A, then the number of pupils in A is double their number in B.

⇒ x + 11 = 2(y - 11)

⇒ x + 11 = 2y - 22

⇒ x - 2y = -22 - 11

⇒ x - 2y = -33     ........(2)

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

⇒ 24 + y - 2y = -33

⇒ 24 - y = -33

⇒ y = 24 + 33

⇒ y = 57.

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

⇒ x = 24 + y

⇒ x = 24 + 57

⇒ x = 81.

Hence, no. of pupils in examination hall A = 81 and examination hall B = 57.

A and B each have a certain number of marbles. A says to B, “If you give 30 to me, I will have twice as many as left with you.” B replies, “If you give me 10, I will have thrice as many as left with you.” How many marbles does each have?

Answer

Let the number of marbles A has be x, and the number of marbles B has be y.

After B gives 30 marbles to A,

⇒ x + 30 = 2(y - 30)

⇒ x + 30 = 2y - 60

⇒ x = 2y - 60 - 30

⇒ x = 2y - 90     .......(1)

After A gives 10 marbles to B,

⇒ y + 10 = 3(x - 10)

⇒ y + 10 = 3x - 30

⇒ y = 3x - 30 - 10

⇒ y = 3x - 40     ........(2)

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

⇒ y = 3(2y - 90) - 40

⇒ y = 6y - 270 - 40

⇒ y = 6y - 310

⇒ y - 6y = -310

⇒ -5y = -310

⇒ y = $\dfrac{-310}{-5} = 62$.

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

⇒ x = 2y - 90

⇒ x = 2(62) - 90

⇒ x = 124 - 90

⇒ x = 34.

Hence, A has 34 marbles and B has 62 marbles.

The present age of a man is 3 years more than three times the age of his son. Three years hence, the man’s age will be 10 years more than twice the age of his son. Determine their present ages.

Answer

Let x be present age of the man and y be the present age of son.

Given,

The present age of a man is 3 years more than three times the age of his son,

⇒ x = 3y + 3     ........(1)

Given,

Three years hence, the man’s age will be 10 years more than twice the age of his son.

⇒ x + 3 = 2(y + 3) + 10

⇒ x + 3 = 2y + 6 + 10

⇒ x = 2y + 16 - 3

⇒ x = 2y + 13     ........(2)

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

⇒ 3y + 3 = 2y + 13

⇒ 3y - 2y = 13 - 3

⇒ y = 10.

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

⇒ x = 3y + 3

⇒ x = 3(10) + 3

⇒ x = 30 + 3

⇒ x = 33.

Hence, the present age of son = 10 years and that of man = 33 years.

The length of a room exceeds its breadth by 3 meters. If the length is increased by 3 m and breadth is decreased by 2 meters, the area remains the same. Find the length and breadth of the room.

Answer

Let x meters be the length of a room and y meters be the breadth of the room.

Given,

The length of a room exceeds its breadth by 3 m,

⇒ x = y + 3     ....(1)

Given,

If length is increased by 3 and breadth decreased by 2, area remains the same,

⇒ (x + 3)(y - 2) = xy

⇒ (xy - 2x + 3y - 6) = xy

⇒ xy - xy = 2x - 3y + 6

⇒ 2x - 3y + 6 = 0     ....(2)

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

⇒ 2(y + 3) - 3y + 6 = 0

⇒ 2y + 6 - 3y + 6 = 0

⇒ -y + 12 = 0

⇒ y = 12.

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

⇒ x = y + 3

⇒ x = 12 + 3

⇒ x = 15.

Hence, length and breadth of the room are 15 m and 12 m respectively.

The area of a rectangle gets reduced by 8 m², if its length is reduced by 5 m and breadth increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 74 m². Find the length and breadth of the rectangle.

Answer

Let x meters be the length and y meters be the breadth of the rectangle.

Given,

If the length is reduced by 5 m and breadth is increased by 3 m, then the area reduces by 8 m².

⇒ (x - 5)(y + 3) = xy - 8

⇒ (xy + 3x - 5y - 15) = xy - 8

⇒ 3x - 5y - 15 + 8 = xy - xy

⇒ 3x - 5y - 7 = 0

⇒ 3x = 5y + 7

⇒ x = $\dfrac{5y + 7}{3}$     ........(1)

Given,

If the length is increased by 3 m and breadth by 2 m, then the area increases by 74 m².

⇒ (x + 3)(y + 2) = xy + 74

⇒ (xy + 2x + 3y + 6) = xy + 74

⇒ 2x + 3y + 6 - 74 = xy - xy

⇒ 2x + 3y - 68 = 0     .......(2)

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

$\Rightarrow 2 \Big(\dfrac{5y + 7}{3}\Big) + 3y - 68 = 0 \\[1em] \Rightarrow \Big(\dfrac{10y + 14}{3}\Big) + 3y = 68 \\[1em] \Rightarrow \Big(\dfrac{10y + 14 + 9y}{3}\Big) = 68 \\[1em] \Rightarrow 19y + 14 = 68 \times 3 \\[1em] \Rightarrow 19y + 14 = 204 \\[1em] \Rightarrow 19y = 204 - 14 \\[1em] \Rightarrow 19y = 190 \\[1em] \Rightarrow y = \dfrac{190}{19} = 10.$

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

$\Rightarrow x = \dfrac{5y + 7}{3} \\[1em] \Rightarrow x = \dfrac{5 \times 10 + 7}{3} \\[1em] \Rightarrow x = \dfrac{57}{3} = 19.$

Hence, the length and breadth of rectangles are 19 m and 10 m respectively.

A motorboat takes 6 hours to cover 100 km downstream and 30 km upstream. If the motorboat goes 75 km downstream and returns back to its starting point in 8 hours, find the speed of the motorboat in still water and the rate of the stream.

Answer

Let x km/hr be the speed of motorboat in still water and y km/hr be the speed of stream.

Downstream speed = x + y km/h

Upstream speed = x - y km/h

Given,

Time $= \dfrac{\text{Distance}}{\text{Speed}}$

Motorboat takes 6 hours to cover 100 km downstream and 30 km upstream.

$\Rightarrow \dfrac{100}{x + y} + \dfrac{30}{x - y} = 6$     .........(1)

Given,

Motorboat goes 75 km downstream and returns back to its starting point in 8 hours.

$\Rightarrow \dfrac{75}{x + y} + \dfrac{75}{x - y} = 8$     ....(2)

Substituting, u = $\dfrac{1}{x + y}$, v = $\dfrac{1}{x - y}$ in equation (1), we get :

⇒ 100u + 30v = 6

⇒ 10(10u + 3v) = 6

⇒ (10u + 3v) = $\dfrac{6}{10}$     ........(3)

Substituting, u = $\dfrac{1}{x + y}$, v = $\dfrac{1}{x - y}$ in equation (2), we get :

⇒ 75u + 75v = 8

⇒ 75(u + v) = 8

⇒ u + v = $\dfrac{8}{75}$

⇒ u = $\dfrac{8}{75} - v$     ........(4)

Substituting value of u from equation (4) in (3), we get :

$\Rightarrow 10 \Big(\dfrac{8}{75} - v\Big) + 3v = \dfrac{6}{10} \\[1em] \Rightarrow \dfrac{80}{75} - 10v + 3v = \dfrac{6}{10} \\[1em] \Rightarrow \dfrac{80}{75} - 7v = \dfrac{6}{10} \\[1em] \Rightarrow 7v = \dfrac{80}{75} - \dfrac{6}{10} \\[1em] \Rightarrow 7v = \dfrac{160 - 90}{150} \\[1em] \Rightarrow 7v = \dfrac{70}{150} \\[1em] \Rightarrow v = \dfrac{7}{15 \times 7} \\[1em] \Rightarrow v = \dfrac{1}{15}.$

Substituting value of v in equation (4), we get :

$\Rightarrow u = \dfrac{8}{75} - v \\[1em] \Rightarrow u = \dfrac{8}{75} - \dfrac{1}{15} \\[1em] \Rightarrow u = \dfrac{8 - 5}{75} \\[1em] \Rightarrow u = \dfrac{3}{75} \\[1em] \Rightarrow u = \dfrac{1}{25}.$

Since,

$\Rightarrow u = \dfrac{1}{x + y} \\[1em] \Rightarrow \dfrac{1}{25} = \dfrac{1}{x + y}$

⇒ x + y = 25     .......(5)

$\Rightarrow v = \dfrac{1}{x - y} \\[1em] \Rightarrow \dfrac{1}{15} = \dfrac{1}{x - y}$

⇒ x - y = 15     ........(6)

Adding equations (5) and (6),

⇒ x + y + x - y = 25 + 15

⇒ 2x = 40

⇒ x = $\dfrac{40}{2} = 20$.

Substituting value of x in equation (6), we get :

⇒ x - y = 15

⇒ 20 - y = 15

⇒ 20 - 15 = y

⇒ y = 5.

Hence, the speed of the motorboat in still water = 20 km/hr and the speed of the stream = 5 km/hr.

A man sold a chair and a table for ₹ 2,178, thereby making a profit of 12% on the chair and 16% on the table. By selling them for ₹ 2,154, he gains 16% on the chair and 12% on the table. Find the cost price of each.

Answer

Let x be the cost of chair and y be the cost of table.

Given,

When sold for ₹ 2,178, he makes profit of 12% on the chair and 16% on the table.

$\therefore x + \dfrac{12}{100}x + y + \dfrac{16}{100}y = 2178 \\[1em] \Rightarrow \dfrac{100x + 12x}{100} + \dfrac{100y + 16y}{100} = 2178 \\[1em] \Rightarrow \dfrac{112x}{100} + \dfrac{116y}{100} = 2178 \\[1em] \Rightarrow 112x + 116y = 2178 \times 100 \\[1em] \Rightarrow 112x + 116y = 217800\\[1em] \Rightarrow 4(28x + 29y) = 217800 \\[1em] \Rightarrow 28x + 29y = \dfrac{217800}{4} \\[1em] \Rightarrow 28x + 29y = 54450 \text{ ....(1) }$

Given,

When sold for ₹ 2,154, he makes profit of 16% on the chair and 12% on the table.

$\therefore x + \dfrac{16}{100}x + y + \dfrac{12}{100}y = 2154 \\[1em] \Rightarrow \dfrac{100x + 16x}{100} + \dfrac{100y + 12}{100} = 2154 \\[1em] \Rightarrow \dfrac{116x}{100} + \dfrac{112y}{100} = 2154 \\[1em] \Rightarrow 116x + 112y = 2154 \times 100 \\[1em] \Rightarrow 116x + 112y = 215400 \\[1em] \Rightarrow 4(29x + 28y) = 215400 \\[1em] \Rightarrow 29x + 28y = \dfrac{215400}{4} \\[1em] \Rightarrow 29x + 28y = 53850 \text{ ........(2) }$

Subtracting equation eqn 1 from 2, we get :

⇒ 29x + 28y - (28x + 29y) = 53850 - 54450

⇒ 29x + 28y - 28x - 29y = -600

⇒ x - y = -600

⇒ x = y - 600     ....(3)

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

⇒ 28x + 29y = 54450

⇒ 28(y - 600) + 29y = 54450

⇒ 28y - 16800 + 29y = 54450

⇒ 57y = 54450 + 16800

⇒ 57y = 71250

⇒ y = $\dfrac{71250}{57}$

⇒ y = ₹ 1,250

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

⇒ 29x + 28 × 1250 = 53850

⇒ 29x + 35000 = 53850

⇒ 29x = 53850 - 35000

⇒ 29x = 18850

⇒ x = $\dfrac{18850}{29}$

⇒ x = ₹ 650

Hence, cost price of table = ₹ 1,250 and cost price of chair = ₹ 650.

A man travels 600 km partly by train and partly by car. If he covers 120 km by train and the rest by car, it takes him 8 hours. But if he travels 200 km by train and the rest by car, he takes 20 minutes longer. Find the speed of the car and that of the train.

Answer

Let x km/hr be the speed of train and y km/hr be the speed of car.

$Time = \dfrac{\text{Distance}}{\text{Speed}}$

Given,

120 km by train, rest (600 - 120 = 480 km) by car takes 8 hours.

$\Rightarrow \dfrac{120}{x} + \dfrac{480}{y} = 8$ ........(1)

Given,

200 km by train, rest (600 - 200 = 400 km) by car takes 8 hours 20 mins = $8 + \dfrac{20}{60} = 8 + \dfrac{1}{3} = \dfrac{25}{3}$ hours.

$\Rightarrow \dfrac{200}{x} + \dfrac{400}{y} = \dfrac{25}{3}$ ..........(2)

Substituting, u = $\dfrac{1}{x}$, v = $\dfrac{1}{y}$ in equation (1), we get :

⇒ 120u + 480v = 8

⇒ 120u + 480v - 8 = 0

⇒ 8(15u + 60v - 1) = 0

⇒ 15u + 60v - 1 = 0

⇒ 15u = 1 - 60v

⇒ u = $\dfrac{1 - 60v}{15}$     ..........(3)

Substituting, u = $\dfrac{1}{x}$, v = $\dfrac{1}{y}$ in equation (2), we get :

⇒ 200u + 400v = $\dfrac{25}{3}$     .........(4)

Substituting value of u from equation (3) in equation (4), we get :

$\Rightarrow 200 \Big(\dfrac{1 - 60v}{15}\Big) + 400v = \dfrac{25}{3} \\[1em] \Rightarrow \Big(\dfrac{200 - 12000v}{15}\Big) + 400v = \dfrac{25}{3} \\[1em] \Rightarrow \Big(\dfrac{200 - 12000v + 6000v}{15}\Big) = \dfrac{25}{3} \\[1em] \Rightarrow 200 - 6000v = \dfrac{25}{3} \times 15 \\[1em] \Rightarrow 200 - 6000v = 25 \times 5 \\[1em] \Rightarrow 200 - 6000v = 125 \\[1em] \Rightarrow 6000v = 200 - 125 \\[1em] \Rightarrow 6000v = 75 \\[1em] \Rightarrow v = \dfrac{75}{6000} \\[1em] \Rightarrow v = \dfrac{1}{80}.$

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

$\Rightarrow u = \dfrac{1 - 60 \times \dfrac{1}{80}}{15} \\[1em] \Rightarrow u = \dfrac{1 - \dfrac{3}{4}}{15} \\[1em] \Rightarrow u = \dfrac{\dfrac{1}{4}}{15} \\[1em] \Rightarrow u = \dfrac{1}{60}.$

Substituting,

⇒ u = $\dfrac{1}{x}$

⇒ $\dfrac{1}{60} = \dfrac{1}{x}$

⇒ x = 60.

⇒ v = $\dfrac{1}{y}$

⇒ $\dfrac{1}{80} = \dfrac{1}{y}$

⇒ y = 80.

Hence, speed of train = 60 km/h, speed of car = 80 km/h.

6 men and 8 boys can finish a piece of work in 14 days while 8 men and 12 boys can do it in 10 days. Find the time taken by one man alone and by one boy alone to finish the work.

Answer

Lets assume that one man takes x days to do work and y days for one boy.

So, the amount of work done by 1 man in 1 day = $\dfrac{1}{x}$

So, the amount of work done by 1 boy in 1 day = $\dfrac{1}{y}$

Given,

6 men and 8 boys finish the work in 14 days,

$\Rightarrow \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{14} \text{ ....(1) }$

Given,

8 men and 12 boys finish the same work in 10 days,

$\Rightarrow \dfrac{8}{x} + \dfrac{12}{y} = \dfrac{1}{10}$     ....(2)

Multiply equation (1) by 2, we get:

$\Rightarrow 2 \Big(\dfrac{6}{x} + \dfrac{8}{y}\Big) = \dfrac{1}{14} \times 2 \\[1em] \Rightarrow \dfrac{12}{x} + \dfrac{16}{y} = \dfrac{1}{7} \text{ ....(3) }$

Multiply equation (2) by $\dfrac{3}{2}$, we get:

$\Rightarrow \dfrac{3}{2} \Big(\dfrac{8}{x} + \dfrac{12}{y}\Big) = \dfrac{1}{10} \times \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{12}{x} + \dfrac{18}{y}= \dfrac{3}{20} \text{ ....(4) }$

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

$\Rightarrow \Big(\dfrac{12}{x} + \dfrac{18}{y}\Big) - \Big(\dfrac{12}{x} + \dfrac{16}{y}\Big) = \dfrac{3}{20} - \dfrac{1}{7} \\[1em] \Rightarrow \Big(\dfrac{12}{x} + \dfrac{18}{y} - \dfrac{12}{x} - \dfrac{16}{y}\Big) = \dfrac{21}{140} - \dfrac{20}{140} \\[1em] \Rightarrow \dfrac{18}{y} - \dfrac{16}{y} = \dfrac{1}{140} \\[1em] \Rightarrow \dfrac{2}{y} = \dfrac{1}{140} \\[1em] \Rightarrow y = 2 \times 140 = 280.$

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

$\Rightarrow \dfrac{6}{x} + \dfrac{8}{y} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{6}{x} + \dfrac{8}{280} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{6}{x} = \dfrac{1}{14} - \dfrac{8}{280} \\[1em] \Rightarrow \dfrac{6}{x} = \dfrac{20}{280} - \dfrac{8}{280} \\[1em] \Rightarrow \dfrac{6}{x} = \dfrac{12}{280} \\[1em] \Rightarrow \dfrac{6 \times 280}{12} = x \\[1em] \Rightarrow x = 140.$

Hence, one man can finish the work in = 140 days and one boy finish the work in = 280 days .

A lady has 25-P and 50-P coins in her purse. If in all she has 80 coins totalling ₹ 25, how many coins of each kind does she have ?

Answer

Let Number of 25-P coins be x and number of 50-P coins be y.

Given,

Total number of coins = 80,

⇒ x + y = 80

⇒ x = 80 - y     .........(1)

Given,

Total value = ₹ 25 = 2500 paise,

⇒ 25x + 50y = 2500     .......(2)

Substituting value of x from equation (1) in 25x + 50y = 2500, we get :

⇒ 25x + 50y = 2500

⇒ 25(80 - y) + 50y = 2500

⇒ 2000 - 25y + 50y = 2500

⇒ -25y + 50y = 2500 - 2000

⇒ 25y = 500

⇒ y = $\dfrac{500}{25}$

⇒ y = 20.

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

⇒ x = 80 - y

⇒ x = 80 - 20

⇒ x = 60.

Hence, number of 25-P coins = 60 and number of 50-P coins = 20.

A and B together can do a piece of work in 6 days. If A’s one day’s work is $1\dfrac{1}{2}$ times the one day's work of B, find how many days, each alone can finish the work.

Answer

Let A's one day work be x and B's one day work be y.

According to first condition given in the problem,

A works $1\dfrac{1}{2}$ times of B,

$\Rightarrow x = 1\dfrac{1}{2}y$

$\Rightarrow x = \dfrac{3}{2}y$

⇒ 2x = 3y

⇒ 2x - 3y = 0     ......(1)

Also given, A and B together can do a piece of work in 6 days.

$\therefore x + y = \dfrac{1}{6}$

⇒ 6(x + y) = 1

⇒ 6x + 6y = 1     ...(2)

Multiplying (1) by 2 we get,

⇒ 2(2x - 3y) = 2 × 0

⇒ 4x - 6y = 0     ...(3)

Adding equations (2) and (3) we get,

⇒ 6x + 6y + 4x - 6y = 1 + 0

⇒ 10x = 1

⇒ x = $\dfrac{1}{10}$

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

⇒ 2 × $\dfrac{1}{10}$ - 3y = 0

⇒ $\dfrac{1}{5}$ - 3y = 0

⇒ 3y = $\dfrac{1}{5}$

⇒ y = $\dfrac{1}{15}$

Since, A's one day work is x and B's one day work is y, so A can do complete work in $\dfrac{1}{x}$ and B can do work in $\dfrac{1}{y}$ days.

$\dfrac{1}{x} = \dfrac{1}{\dfrac{1}{10}}$ = 10 days

$\dfrac{1}{y} = \dfrac{1}{\dfrac{1}{15}}$ = 15 days

Hence, A can finish the work in 10 days while B can finish the work in 15 days.