Class 9: Simultaneous Linear Equations – Exercise 7.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Problems Based on Articles and their cost

Question 4: $5$ pens and $6$ pencils together cost $Rs. \ 9$ and $3$ pens and $2$ pencils cost $Rs. \ 5$ . Find the cost of one pen and one pencil.

Answer:

Let the cost of a pen $= x \ Rs$ . and the cost of a pencil $= y \ Rs$ .

Therefore given:

$5x+6y = 9$ … … … … … (i)

$3x+2y = 5$ … … … … … (ii)

Now multiplying equation( (i) by $3$ and equation (ii) by $5$ we get

$15x+18y=27$ … … … … … (iii)

$15x+10y = 25$ … … … … … (iv)

Subtracting (iv) from (iii) we get

$15x+18y=27$

$\underline{ (-) \hspace{0.5cm} 15x+10y = 25}$

$8y = 2$

$\Rightarrow y = 0.25 \ Rs.$

Substituting $y = 0.25 Rs.$ in (i) we get

$5x+6(0.25) = 9 \Rightarrow 5x = 7.50 \Rightarrow x = 1.50 \ Rs.$

$\\$

Question 5: A person has pens and pencils with together are $40$ in number.If she had $5$ more pencils and $5$ less pens, then the number of pencils would become $4$ times the number of pens. Find the number of pens and pencils that the person had.

Answer:

Let the number of pens $= x \ Rs$ . and the number of pencil $= y \ Rs$ .

Therefore given:

$x+y = 40$ … … … … … (i)

$4(x-5) = y+5 \Rightarrow 4x-y = 25$ … … … … … (ii)

Now multiplying equation( (i) by $4$ we get

$4x+4y=160$ … … … … … (iii)

Subtracting (i1) from (iii) we get

$4x+4y=160$

$\underline{ (-) \hspace{0.5cm} 4x-y = 25}$

$5y = 135$

$\Rightarrow y = 27$

Substituting $y = 27$ in (i) we get

$x = 40 -(27) = 13$

$\\$

Question 6: $5$ books and $7$ pens together cost $Rs. \ 79$ whereas $7$ books and $5$ pens together cost $Rs. \ 77$ . Find the cost of one books and two pens.

Answer:

Let the cost of a book $= x \ Rs$ . and the cost of a pen $= y \ Rs$ .

Therefore given:

$5x+7y = 79$ … … … … … (i)

$7x+5y = 77$ … … … … … (ii)

Now multiplying equation( (i) by $7$ and equation (ii) by $5$ we get

$35x+49y=553$ … … … … … (iii)

$35x+25y = 385$ … … … … … (iv)

Subtracting (iv) from (iii) we get

$35x+49y=553$

$\underline{ (-) \hspace{0.5cm} 35x+25y = 385}$

$24y = 168$

$\Rightarrow y = 7 \ Rs.$

Substituting $y = 7 Rs.$ in (i) we get

$5x+7(7) = 79 \Rightarrow 5x = 30 \Rightarrow x = 6 \ Rs.$

$\\$

Question 7: On selling a TV at $5\%$ gain and a fridge at $10\%$ gain, a shopkeeper gains $Rs. \ 2000$ . But if he sells the TV at $10\%$ gain and the fridge at $5\%$ loss, he gains $Rs. \ 1500$ on the transaction. Find the actual prices of the TV and the fridge.

Answer:

Let the cost of TV $= x \ Rs$ . and the cost of Fridge $= y \ Rs$ .

Therefore given:

$0.05x+0.10y=2000$ … … … … … (i)

$0.10x-0.05y=1500$ … … … … … (ii)

Now multiplying equation( (i) by $2$ we get

$0.20x-0.10y=3000$ … … … … … (iii)

Adding (iii) from (i) we get

$0.05x+0.10y=2000$

$\underline{ (-) \hspace{0.5cm} 0.20x-0.10y=3000}$

$0.25x=5000$

$\Rightarrow x = 20000 \ Rs.$

Substituting $y = 20000 Rs.$ in (i) we get

$0.05(20000)+0.10y=2000 \Rightarrow y = 10000 \ Rs.$

$\\$

Question 8: A lending library has a fixed charge for the first three days and an additional charge for each additional day. Person A paid $Rs. \ 27$ for a book kept for $7$ days while Person B paid $Rs. \ 21$ for the book kept for $5$ days. Find the fixed charge and the charge for each additional day.

Answer:

Let the fixed cost $= x \ Rs$ . and the cost per day $= y \ Rs$ .

Therefore given:

$x+(7-3)y = 27 \Rightarrow x+4y=27$ … … … … … (i)

$x+(5-3)y = 21 \Rightarrow x+2y=21$ … … … … … (ii)

Subtracting (ii) from (i) we get

$x+4y=27$

$\underline{ (-) \hspace{0.5cm} x+2y=21}$

$2y=6$

$\Rightarrow y = 3 \ Rs.$

Substituting $y = 3 Rs.$ in (i) we get

$x+4(3)=27 \Rightarrow x = 15 \ Rs.$

$\\$

Problems Based on Numbers

Question 9: The sum of the digits of a two digit number is $8$ and the difference between the number and that formed by reversing the digits is $18$ . Find the number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given: $x+y = 8$ … … … … … (i)

$10y + x - (10x+y) = 18 \Rightarrow 9y - 9x = 18 \Rightarrow y-x = 2$ … … … … … (ii)

Adding (i) and (ii) we get $y = 5 \Rightarrow x = 8-5 = 3$

Hence $x = 3, y = 5$ and the number is $53$ .

$\\$

Question 10: The sum of two digit number and the number obtained by reversing the order of its digits is $121$ , and the two digits differ by $3$ . Find the number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given:

$10y + x + (10x+y) = 121 \Rightarrow 11x+11y = 121 \Rightarrow x+y = 11$ … … … … … (i)

Also $x-y = \pm 3$ … … … … … (ii)

Consider +ve sign

Solving $x+y = 11$ and $x-y = 3 \Rightarrow x = 7$ and $y =4$ and the number is $74$

Consider -ve sign

Solving $x+y = 11$ and $x-y = -3 \Rightarrow x = 4$ and $y =7$ and the number is $47$

$\\$

Question 11: The sum of two digit number and the number formed by interchanging its digits is $110$ . If $10$ is subtracted from the first number, the new number is $4$ more than $5$ times the sum of the digits in the first number. Find the first number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given:

$10y + x + (10x+y) = 110 \Rightarrow 11x+11y = 110 \Rightarrow x+y = 10$ … … … … … (i)

Also $(10y+x-10) = 4 + 5(x+y) \Rightarrow 5y-4x=14$ … … … … … (ii)

Solving $x+y = 10$ and $5y-4x=14 \Rightarrow x = 4$ and $y =6$ and the number is $64$

$\\$

Question 12: The sum of two numbers is $8$ . If their sum is $4$ times their difference, find the numbers.

Answer:

Let the two numbers be $x$ and $y$

Therefore $x+y = 8$ … … … … … (i)

$x+y = 4(x-y) \Rightarrow 3x - 5y = 0$ … … … … … (ii)

Now multiplying equation (i) by $5$ and adding (i) and (ii) we get

$5x+5y = 40$

$\underline{ (+) \hspace{0.5cm} 3x - 5y = 0}$

$8x=40$

Therefore we get $x = 5$ and $y = (8-5) = 3$

$\\$

Question 13: The sum of the digits of a two digit number is $15$ . The number obtained by reversing the order of the digits of the given number exceeds the given number by $9$ . Find the given number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given:

$10y + x + 9 = 10x + y \Rightarrow 9x-9y = 9 \Rightarrow x-y = 9$ … … … … … (i)

Also $x+y = 15$ … … … … … (ii)

Solving $x+y = 15$ and $x-y = 9$ we get $x =8 \ and \ y = 7$ and the number is $78$

$\\$

Question 14: The sum of two numbers is $1000$ and the difference between their squares is $256000$ . Find the numbers.

Answer:

Let the two numbers be $x$ and $y$

Therefore $x+y = 1000$ … … … … … (i)

$x^2-y^2 256000 \Rightarrow (x-y)(x+y) = 256000 \Rightarrow x-y = 256$ … … … … … (ii)

Adding (i) and (ii) we get

$x+y = 1000$

$\underline{ (+) \hspace{0.5cm} x-y = 256}$

$2x=1256$

Therefore we get $x = 628$ and $y = (1000-628) = 372$

$\\$

Question 15: A $2$ digit number is four times the sum of its digits. If $18$ is added to the number, the digits are reversed. Find the number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given:

$10y + x + 18 = 10x + y \Rightarrow 9x-9y = 18 \Rightarrow x-y = 2$ … … … … … (i)

Also $10y + x = 4(x+y) \Rightarrow x=2y \Rightarrow x-2y=0$ … … … … … (ii)

Solving $x-2y=0$ and $x-y = 2$ we get $x =4 \ and \ y = 2$ and the number is $24$

$\\$

Question 16: A $2$ digit number is such that the product of its digits is $20$ . If $9$ is added to the number, the digits interchange their places. Find the number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given:

$10y + x +9 = 10x + y \Rightarrow 9x-9y = 9 \Rightarrow x-y = 1$ … … … … … (i)

Also $xy = 20 \Rightarrow y=$ $\frac{20}{x}$ … … … … … (ii)

Substituting (ii) in (i) we get

$x^2 -$ $\frac{20}{x}$ $= 1$

$\Rightarrow x^2 - x - 20 = 0$

$\Rightarrow (x-5)(x+4) = 0 \Rightarrow x = 5 \ or \ x = -4$ (not possible)

For $x = 5$ we get $y =$ $\frac{20}{5}$ $= 4$

We get $x =5 \ and \ y = 4$ and the number is $45$

$\\$

Question 17: $7$ times two digit number is equal to $4$ times the number obtained by reversing the digits. If the difference between the digits is $3$ . Find the number.

Answer:

Let the digit in unit place is $x$ and that in tenth place is $y$ .

Therefore Number $= 10y + x$

Number formed by reversing the digits $= 10x+ y$

Given:

$7(10y + x) = 4(10x + y) \Rightarrow 33x = 66y \Rightarrow x= 2y$ … … … … … (i)

Also $x-y=3$ … … … … … (ii)

Solving $x= 2y$ and $x-y=3$ we get $x =6 \ and \ y = 3$ and the number is $36$

$\\$

Problems Based on Fractions

Question 18: A fraction become $\frac{4}{5}$ if $1$ is added to both numerator and denominator. If, however, $5$ is subtracted from both numerator and denominator, the fraction becomes $\frac{1}{2}$ . What is the fraction?

Answer:

Let the fraction $= \frac{x}{y}$

Therefore based on the given conditions:

$\frac{x+1}{y+1}$ $=$ $\frac{4}{5}$ $\Rightarrow 5x + 5 = 4y + 4 \Rightarrow 5x - 4y = -1$ … … … … … (i)

and $\frac{x-5}{y-5}$ $=$ $\frac{1}{2}$ $\Rightarrow 2x-10 = y - 5 \Rightarrow 2x - y = 5$ … … … … … (ii)

Multiplying (ii) by $4$ and subtracting it from (i) we get

$5x-4y = -1$

$\underline{ (-) \hspace{0.5cm} 8x-4y=20}$

$-3x = -21$

Therefore $x = 7$ .

Substituting $x = 7$ in (i) we get $4y = 5(7)+1 \Rightarrow y = 9$

Therefore the fraction is $\frac{7}{9}$

$\\$

Question 19: A denominator of a fraction is $4$ more than twice the numerator. When both the numerator and denominator are decreased by $6$ , the denominator becomes $12$ times the numerator. Determine the fraction.

Answer:

Let the fraction $= \frac{x}{y}$

Therefore based on the given conditions:

$y = 2x+ 4 \Rightarrow 2x-y = -4$ … … … … … (i)

and $y-6 = 12(x-6) \Rightarrow 12x - y = 66$ … … … … … (ii)

Subtracting (ii) from (i) we get

$2x-y = -4$

$\underline{ (-) \hspace{0.5cm} 12x - y = 66}$

$-10x = -70$

Therefore $x = 7$ .

Substituting $x = 7$ in (i) we get $y = 2(7) + 4 = 18$

Therefore the fraction is $\frac{7}{18}$

$\\$

Question 20: A fraction becomes $\frac{1}{3}$ is $1$ is subtracted from both its numerator and denominator. If $1$ is added to both numerator and denominator, it becomes $\frac{1}{2}$ . Find the fraction.

Answer:

Let the fraction $= \frac{x}{y}$

Therefore based on the given conditions:

$\frac{x-1}{y-1}$ $=$ $\frac{1}{3}$ $\Rightarrow 3x-3 = y-1 \Rightarrow 3x-y=2$ … … … … … (i)

and $\frac{x+1}{y+1}$ $=$ $\frac{1}{2}$ $\Rightarrow 2x+2 = y +1 \Rightarrow 2x - y = -1$ … … … … … (ii)

Subtracting (ii) from (i) we get

$3x-y=2$

$\underline{ (-) \hspace{0.5cm} 2x - y = -1}$

$x = 3$

Therefore $x = 3$ .

Substituting $x = 3$ in (i) we get $y = 3(3) -2 = 7$

Therefore the fraction is $\frac{3}{7}$

$\\$

Question 21: If the numerator of the fraction is multiplied by $2$ and the denominator is reduced by $5$ the fraction becomes $\frac{6}{5}$ . And if the denominator is doubled and the numerator is increased by $8$ , the fraction becomes $\frac{2}{5}$ . Determine the fraction.

Answer:

Let the fraction $= \frac{x}{y}$

Therefore based on the given conditions:

$\frac{2x}{y-5}$ $=$ $\frac{6}{5}$ $\Rightarrow 10x = 6y - 30 \Rightarrow 5x-3y=-15$ … … … … … (i)

and $\frac{x+8}{2y}$ $=$ $\frac{2}{5}$ $\Rightarrow 5x+40=4y \Rightarrow 5x-4y=-40$ … … … … … (ii)

Subtracting (ii) from (i) we get

$5x-3y=-15$

$\underline{ (-) \hspace{0.5cm} 5x-4y=-40}$

$y=25$

Therefore $y = 25$ .

Substituting $y=25$ in (i) we get $5x = 3(25) - 15 = 60 \Rightarrow x = 12$

Therefore the fraction is $\frac{12}{25}$

$\\$

Question 22: The sum of the numerator and denominator of a fraction is $18$ . If the denominator is increased by $2$ , the fraction reduces to $\frac{1}{3}$ . Find the fraction.

Answer:

Let the fraction $= \frac{x}{y}$

Therefore based on the given conditions:

$x+y = 18$ … … … … … (i)

and $\frac{x}{y+2}$ $=$ $\frac{1}{2}$ $\Rightarrow 3x=y-2 \Rightarrow 3x-y = 2$ … … … … … (ii)

Adding (i) and (i) we get

$x+y = 18$

$\underline{ (-) \hspace{0.5cm} 3x-y = 2}$

$4x = 20$

Therefore $x = 5$ .

Substituting $x = 5$ in (i) we get $y = 18 - 5 = 13$

Therefore the fraction is $\frac{5}{18}$

$\\$

Question 23: The sum of the numerator and denominator of a fraction is $3$ less than twice the denominator. If the numerator and denominator are decreased by $1$ , the numerator becomes half the denominator. Determine the fraction.

Answer:

Let the fraction $= \frac{x}{y}$

Therefore based on the given conditions:

$x+y + 3= 2y \Rightarrow x - y = -3$ … … … … … (i)

and $x-1 = \frac{1}{2} (y-1) \Rightarrow 2x - 2 = y - 1 \Rightarrow 2x - y = 1$ … … … … … (ii)

Subtracting (ii) from (i) we get

$x - y = -3$

$\underline{ (-) \hspace{0.5cm} 2x - y = 1}$

$-x = -4$

Therefore $x = 4$ .

Substituting $x = 4$ in (i) we get $y = 4+3 = 7$

Therefore the fraction is $\frac{4}{7}$

$\\$

Exercise 7.2 continued …

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Class 9: Simultaneous Linear Equations – Exercise 7.2

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