# Linear equation pdf

Exercise 2.1

1. Solve: $x-2=7$

Ans:

$x-2=7$

Transposing $2$ to R.H.S, we obtain

$x=7+2=9$

2. Solve:$y+3=10$

Ans:

$y+3=10$

Transposing $3$ to R.H.S, we obtain

$y=10-3=7$

3. Solve: $6=z+2$

Ans:

$6=z+2$

Transposing $2$ to L.H.S, we obtain

$6-2=z$

$z=4$

4. Solve: $\frac{3}{7}+x=\frac{17}{7}$

Ans:

$\frac{3}{7}+x=\frac{17}{7}$

Transposing $\frac{3}{7}$  to R.H.S, we obtain

$x=\frac{17}{7}-\frac{3}{7}=\frac{14}{7}=2$

5. Solve:$6x=12$

Dividing both sides by $6$ , we obtain

$\frac{6x}{6}=\frac{12}{6}$

$x=2$

6. Solve:$\frac{t}{5}=10$

Ans:

$\frac{t}{5}=10$

Multiplying both sides by $5$, we obtain

$\frac{t}{5}\times 5=10\times 5$

$t=50$

7. Solve: $\frac{2x}{3}=18$

Ans:

$\frac{2x}{3}=18$

Multiplying both sides by$\frac{3}{2}$ , we obtain

$\frac{2x}{3}\times \frac{3}{2}=18\times \frac{3}{2}$

$x=27$

8. Solve: $1.6=\frac{y}{1.5}$

Ans:

$1.6=\frac{y}{1.5}$

Multiplying both sides by$1.5$, we obtain

$1.6\times 1.5=\frac{y}{1.5}\times 1.5$

$2.4=y$

9. Solve:$7x-9=16$

Ans:

$7x-9=16$

Transposing$9$ to R.H.S, we obtain

$7x=16+9$

$7x=25$

Dividing both sides by$7$, we obtain

$\frac{7x}{7}=\frac{25}{7}$

$x=\frac{25}{7}$

10. Solve: $14y-8=13$

Ans:

$14y-8=13$

Transposing $8$ to R.H.S, we obtain

$14y=13+8$

$14y=21$

Dividing both sides by$14$, we obtain

$\frac{14y}{14}=\frac{21}{14}$

$y=\frac{3}{2}$

11. Solve: $17+6p=9$

Ans:

$17+6p=9$

Transposing $17$ to R.H.S, we obtain

$6p=9-17$

$6p=-8$

Dividing both sides by$6$, we obtain

$\frac{6p}{6}=-\frac{8}{6}$

$p=-\frac{4}{3}$

12. Solve: $\frac{x}{3}+1=\frac{7}{15}$

Ans:

$\frac{x}{3}+1=\frac{7}{15}$

Transposing $1$ to R.H.S, we obtain

$\frac{x}{3}=\frac{7}{15}-1$

$\frac{x}{3}=\frac{7-15}{15}$

$\frac{x}{3}=-\frac{8}{15}$

Multiplying both sides by $3$ , we obtain

$\frac{x}{3}\times 3=-\frac{8}{15}\times 3$

$x=-\frac{8}{5}$

Exercise 2.2

1. If you subtract$\frac{1}{2}$  from a number and multiply the result by $\frac{1}{2}$ , $\frac{1}{8}$you get  . What is the number?

Ans:

Let the number be x. According to the question,

$\left( x-\frac{1}{2} \right)\times \frac{1}{2}=\frac{1}{8}$

On multiplying both sides by$2$, we obtain

$\left( x-\frac{1}{2} \right)\times \frac{1}{2}\times 2=\frac{1}{8}\times 2$

$x-\frac{1}{2}=\frac{1}{4}$

On transposing $\frac{1}{2}$ to R.H.S, we obtain

$x=\frac{1}{4}+\frac{1}{2}$

$\,\,\,\,=\frac{1+2}{4}=\frac{3}{4}$

Therefore, the number is$\frac{3}{4}$ .

2.  The perimeter of a rectangular swimming pool is $154$ m. Its length is $2$ m more than twice its breadth. What are the length and the breadth of the pool?

Ans:

Let the breadth be x m. The length will be$\left( 2x+2 \right)$ m.

Perimeter of swimming pool $=2\left( l+b \right)=154$m

$2\left( 2x+2+x \right)=154$

$2\left( 3x+2 \right)=154$

Dividing both sides by$2$, we obtain

$\frac{2\left( 3x+2 \right)}{2}=\frac{154}{2}$

$3x+2=77$

On Transposing $2$ to R.H.S, we obtain

$3x=77-2$

$3x=75$

On dividing both sides by$3$, we obtain

$\frac{3x}{3}=\frac{75}{3}$

$x=25$

$2x+2=2\times 25+2=52$

Hence, the breadth and length of the pool are $25$ m and$52$ m respectively.

3. The base of an isosceles triangle is $\frac{4}{3}$ cm. The perimeter of the triangle is $4\frac{2}{15}$ cm. What is the length of either of the remaining equal sides?

Ans:

Let the length of equal sides be x cm.

Perimeter$=\text{x }cm+\,\text{x}\,cm+\,\text{Base}=4\frac{2}{15}\text{cm}$

$2x+\frac{4}{3}=\frac{62}{15}$

On Transposing $\frac{4}{3}$ to R.H.S, we obtain

$2x=\frac{62}{15}-\frac{4}{3}$

$2x=\frac{62-4\times 5}{15}=\frac{62-20}{15}$

$2x=\frac{42}{15}$

On dividing both sides by$2$, we obtain

$\frac{2x}{2}=\frac{42}{15}\times \frac{1}{2}$

$x=\frac{7}{5}=1\frac{2}{5}$

Therefore, the length of equal sides is $1\frac{2}{5}$ cm.

4. Sum of two numbers is$95$. If one exceeds the other by $15$, find the numbers.

Ans:

Let one number be x. Therefore, the other number will be$\text{x}+15$. According to the question,

$x+x+15=95$

$2x+15=95$

On Transposing $15$ to R.H.S, we obtain

$2x=95-15$

$2x=80$

On dividing both sides by$2$, we obtain

$\frac{2x}{2}=\frac{80}{2}$

$x=40$

$x+15=40+15=55$

Hence, the numbers are$40\text{ and }55$.

5. Two numbers are in the ratio $5:3$. If they differ by$18$, what are the numbers?

Ans:

Let the common ratio between these numbers be x. Therefore, the numbers will be $5\text{x and }3\text{x}$respectively.

Difference between these numbers$=18$

$5\text{x}-3\text{x}=18$

$2x=18$

Dividing both sides by$2$,

$\frac{2x}{2}=\frac{18}{2}$

$x=9$

First number $=5x=5\times 9=45$

Second number $=3x=3\times 9=27$

6. Three consecutive integers add up to$51$. What are these integers?

Ans:

Let three consecutive integers be$x,x+1\text{ and }x+2$.

Sum of these numbers$=x+x+1+x+2=51$

$3x+3=51$

On Transposing $3$ to R.H.S, we obtain

$3x=51-3$

$3x=48$

On dividing both sides by$3$, we obtain

$\frac{3x}{3}=\frac{48}{3}$

$x=16$

$x+1=17$

$x+2=18$

Hence, the consecutive integers are$16,17,\,\text{and }18$ .

7. The sum of three consecutive multiples of $8\,\,\,\text{is }888$. Find the multiples.

Ans:

Let the three consecutive multiples of $8\,\text{be }8x,\,8\left( x+1 \right),\,8\left( x+2 \right)$.

Sum of these numbers$=8x+8\left( x+1 \right)+8\left( x+2 \right)=888$

$8\left( x+x+1+x+2 \right)=888$

$8\left( 3x+3 \right)=888$

On dividing both sides by$8$, we obtain

$\frac{8\left( 3x+3 \right)}{8}=\frac{888}{8}$

$3x+3=111$

On Transposing $3$ to R.H.S, we obtain

$3x=111-3$

$3x=108$

On dividing both sides by $3$ , we obtain

$\frac{3x}{3}=\frac{108}{3}$

$x=36$

First multiple $=8x=8\times 36=288$

Second multiple $=8\left( x+1 \right)=8\times \left( 36+1 \right)=8\times 37=296$

Third multiple  $=8\left( x+2 \right)=8\times \left( 36+2 \right)=8\times 38=304$

Hence, the required numbers are$288,296,\text{and }304$ .

8. Three consecutive integers are such that when they are taken in increasing order and multiplied by $2,3\,\text{and }4$respectively, they add up to $74$. Find these numbers.

Ans:

Let three consecutive integers be $\text{x,}\,\text{x}+1\text{,}\,\text{x}+2$. According to the question,

$2x+3\left( x+1 \right)+4\left( x+2 \right)=74$

$2x+3x+3+4x+8=74$

$9x+11=74$

On Transposing $11$ to R.H.S, we obtain

$9x=74-11$

$9x=63$

On dividing both sides by$9$, we obtain

$\frac{9x}{9}=\frac{63}{9}$

$x=7$

$x+1=7+1=8$

$x+2=7+2=9$

Hence, the numbers are $7,8\,\text{and }9$.

9. The ages of Rahul and Haroon are in the ratio$5:7$. Four years later the sum of their ages will be $56$ years. What are their present ages?

Ans: Let common ratio between Rahul’s age and Haroon’s age be x. Therefore, the age of Rahul and Haroon will be $5x$ years and $7x$ years respectively. After $4$ years, the age of Rahul and Haroon will be $\left( 5x+4 \right)$ years and $\left( 7x+4 \right)$ years respectively.

According to the given question, after$4$ years, the sum of the ages of Rahul and Haroon is $56$ years.

$\therefore \left( 5x+4+7x+4 \right)=56$

$12x+8=56$

On Transposing $8$ to R.H.S, we obtain

$12x=56-8$

$12x=48$

On dividing both sides by $12$, we obtain

$\frac{12x}{12}=\frac{48}{12}$

$x=4$

Rahul’s age $=5x$ years $=\left( 5\times 4 \right)$ years $=20$ years

Haroon’s age $=7x$ years  $=\left( 7\times 4 \right)$years $=28$ years

10. The number of boys and girls in a class are in the ratio$7:5$. The number of boys is $8$ more than the number of girls. What is the total class strength?

Ans: Let the common ratio between the number of boys and numbers of girls be x.

Number of boys $=7x$

Number of girls $=5x$

According to the given question,

Number of boys $=$ Number of girls $+8$

$\therefore 7x=5x+8$

On Transposing $5x$ to L.H.S, we obtain

$7x-5x=8$

$2x=8$

On dividing both sides by$2$ , we obtain

$\frac{2x}{2}=\frac{8}{2}$

$x=4$

Number of boys $=7x=7\times 4=28$

Number of girls $=5x=5\times 4=20$

Hence, total class strength $=28+20=48$ students

11. Baichung’s father is $26$ years younger than Baichung’s grandfather and $29$ years older than Baichung. The sum of the ages of all the three is $135$ years. What is the age of each one of them?

Ans:

Let Baichung’s father’s age be x years. Therefore, Baichung’s age and Baichung’s grandfather’s age will be $\left( x-29 \right)$ years and $\left( x+26 \right)$ years respectively. According to the given question, the sum of the ages of these 3 people is $135$ years.

$\therefore x+x-29+x+26=135$

$3x-3=135$

On Transposing $3$ to R.H.S, we obtain

$3x=135+3$

$3x=138$

On dividing both sides by $3$, we obtain

$\frac{3x}{3}=\frac{138}{3}$

$x=46$

Baichung’s father’s age $=x$ years $=46$ years

Baichung’s age $=\left( x-29 \right)$ years $=\left( 46-29 \right)$ years = $17$ years

Baichung’s grandfather’s age $=\left( x+26 \right)$ years $=\left( 46+26 \right)$ years $=72$ years

12. Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Ans:

Let Ravi’s present age be x years.

Fifteen years later, Ravi’s age $=4\times$ His present age

$x+15=4x$

On Transposing x to R.H.S, we obtain

$15=4x-x$

$15=3x$

On dividing both sides by$3$ , we obtain

$\frac{15}{3}=\frac{3x}{3}$

$5=x$

Hence, Ravi’s present age $=5$years

13. A rational number is such that when you multiply it by $\frac{5}{2}$  and add $\frac{2}{3}$ to the product, you get$-\frac{7}{12}$ . What is the number?

Ans:

Let the number be x.

According to the given question,

$\frac{5}{2}x+\frac{2}{3}=-\frac{7}{12}$

On Transposing$\frac{2}{3}$ to R.H.S, we obtain

$\frac{5}{2}x=-\frac{7}{12}-\frac{2}{3}$

$\frac{5}{2}x=\frac{-7-\left( 2\times 4 \right)}{12}$

$\frac{5}{2}x=-\frac{15}{12}$

On multiplying both sides by$\frac{2}{5}$  , we obtain

$x=-\frac{15}{12}\times \frac{2}{5}=-\frac{1}{2}$

Hence, the rational number is $-\frac{1}{2}$.

14. Lakshmi is a cashier in a bank. She has currency notes of denominations$\text{Rs }100,\,\text{Rs }50\,\,\,\text{and Rs}\,10$, respectively. The ratio of the number of these notes is$2:3:5$. The total cash with Lakshmi is$\text{Rs }4,00,000$. How many notes of each denomination does she have?

Ans:

Let the common ratio between the numbers of notes of different denominations be x. Therefore, numbers of $\text{Rs }100$ notes, $\text{Rs }50$ notes, and $\text{Rs }10$notes will be $2x,3x,\,\text{and }5x$  respectively.

Amount of $\text{Rs }100$notes $=\text{Rs}\left( 100\times 2\text{x} \right)=\text{Rs}\,200\text{x}$

Amount of $\text{Rs }50$notes $=\text{Rs}\left( 50\times 3\text{x} \right)=\text{Rs}\,150\text{x}$

Amount of $\text{Rs }10$note$=\text{Rs}\left( 10\times 5\text{x} \right)=\text{Rs}\,50\text{x}$

It is given that the total amount is $\text{Rs}\,400000$.

$\therefore 200\text{x+}150\text{x+}50\text{x=400000}$

$\Rightarrow \text{400x=400000}$

On dividing both sides by$400$, we obtain

$x=1000$

Number of Rs 100 notes $=2x=2\times 1000=2000$

Number of Rs 50 notes $=3x=3\times 1000=3000$

Number of Rs 10 notes $=5x=5\times 1000=5000$

15. I have a total of $\text{Rs}\,300$in coins of denomination$\text{Re }1,\text{Re }2\,\text{and Re }5$. The number of $\text{Rs }2$coins is$3$ times the number of $\text{Rs }5$ coins. The total number of coins is$160$. How many coins of each denomination are with me?

Ans:

Let the number of $\text{Rs }5$ coins be x.

Number of $\text{Rs }2$ coins $=3\times$Number of $\text{Rs }5$ coins$=3\text{x}$

Number of $\text{Re }1$ coins $=160-$ (Number of coins of $\text{Rs }5$ and of$\text{Rs 2}$)

Amount  $\text{Re }1$of coins $=\,\text{Rs }\left[ 1\times \left( 160-4x \right) \right]=\,\text{Rs}\,\left( 160-4x \right)$

Amount of $\text{Rs }2$ coins $=\,\text{Rs}\,\,\left( 2\times 3x \right)=\,\text{Rs}\,\,\text{6x}$

Amount of $\text{Rs 5}$coins  $=\,\text{Rs}\,\,\left( 5\times x \right)=\,\text{Rs}\,\,5\text{x}$

It is given that the total amount is  $\text{Rs 300}$.

$\therefore 160-4x+6x+5x=300$

$160+7x=300$

On Transposing $160$ to R.H.S, we obtain

$7x=300-160$

$7x=140$

On dividing both sides by $7$, we obtain

$\frac{7x}{7}=\frac{140}{7}$

$x=20$

Number of $\text{Re }1$ coins $=160-4x=160-4\times 20=160-80=80$

Number of $\text{Rs }2$  coins$=3x=3\times 20=60$

Number of $\text{Rs 5}$ coins $=x=20$

16. The organizers of an essay competition decide that a winner in the competition gets a prize of  $\text{Rs 100}$ and a participant who does not win gets a prize of$\text{Rs 25}$. The total prize money distributed is  $\text{Rs 3000}$. Find the number of winners, if the total number of participants is $63$ .

Ans:

Let the number of winners be x. Therefore, the number of participants who did not win will be $63-x$.

Amount given to the winners $=\text{Rs }\left( 100\times x \right)=~~\text{Rs 100}x$

Amount given to the participants who did not win $=\text{Rs }\left[ 25\left( 63-x \right) \right]$

$=\text{Rs }\left( 1575-25x \right)$

According to the given question,

$100x+1575-25x=3000$

On Transposing $1575$ to R.H.S, we obtain

$75x=3000-1575$

$75x=1425$

On dividing both sides by $75$ , we obtain

$x=19$

Hence, number of winners $=19$

Exercise 2.3

1. Solve and check result: $3x=2x+18$

Ans:

$3x=2x+18$

On Transposing $2x$ to L.H.S, we obtain

$3x-2x=18$

$x=18$

L.H.S $=3x=3\times 18=54$

R.H.S $=2x+18=2\times 18+18=36+18=54$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

2. Solve and check result: $5t-3=3t-5$

Ans:

$5t-3=3t-5$

On Transposing $3t$ to L.H.S and $-3$ to R.H.S, we obtain

$5t-3=-5-\left( -3 \right)$

$2t=-2$

On dividing both sides by$2$, we obtain

$t=-1$

L.H.S $=5t-3=5\times \left( -1 \right)-3=-8$

R.H.S $=3t-5=3\times \left( -1 \right)-5=-3-5=-8$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

3. Solve and check result: $5x+9-5+3x$

Ans:

$5x+9=5+3x$

On Transposing $3x$ to L.H.S and $9$ to R.H.S, we obtain

$5x-3x=5-9$

$2x=-4$

On dividing both sides by$2$, we obtain

$x=-2$

L.H.S $=5x+9=5\times \left( -2 \right)+9=-10+9=-1$

R.H.S $=5+3x=5+3\times \left( -2 \right)=5-6=-1$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

4. Solve and check result: $4z+3=6+2z$

Ans:

$4z+3=6+2z$

On Transposing $2z$ to L.H.S and $3$ to R.H.S, we obtain

$4z-2z=6-3$

$2z=3$

Dividing both sides by$2$ , we obtain

L.H.S $=4z+3=4\times \left( \frac{3}{2} \right)+3=6+3=9$

R.H.S $=6+2z=6+2\times \left( \frac{3}{2} \right)=6+3=9$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

5. Solve and check result: $2x-1=14-x$

Ans:

$2x-1=14-x$

Transposing x to L.H.S and $1$ to R.H.S, we obtain

$2x+x=14+1$

$3x=15$

Dividing both sides by $3$, we obtain

$x=5$

L.H.S $=2x-1=2\times \left( 5 \right)-1=10-1=9$

R.H.S $=14-x=14-5=9$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

6. Solve and check result: $8x+4=3\left( x-1 \right)+7$

Ans:

$8x+4=3\left( x-1 \right)+7$

$8x+4=3x-3+7$

Transposing $3x$ to L.H.S and $4$ to R.H.S, we obtain

$8x-3x=-3+7-4$

$5x=-7+7$

$x=0$

L.H.S $=8x+4=8\times \left( 0 \right)+4=4$

R.H.S $=3\left( x-1 \right)+7=3\left( 0-1 \right)+7=-3+7=4$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

7. Solve and check result: $x=\frac{4}{5}\left( x+10 \right)$

Ans:

$x=\frac{4}{5}\left( x+10 \right)$

Multiplying both sides by$5$, we obtain

$5x=4\left( x+10 \right)$

$5x=4x+40$

Transposing $4x$ to L.H.S, we obtain

$5x=4x+40$

$x=40$

L.H.S $=x=40$

R.H.S   $=\frac{4}{5}\left( x+10 \right)=\frac{4}{5}\left( 40+10 \right)=\frac{4}{5}\times 50=40$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

8. Solve and check result: $\frac{2x}{3}+1=\frac{7x}{15}+3$

Ans:

$\frac{2x}{3}+1=\frac{7x}{15}+3$

Transposing $\frac{7x}{15}$ to L.H.S and $1$ to R.H.S, we obtain

$\frac{2x}{3}-\frac{7x}{15}=3-1$

$\frac{5\times 2x-7x}{15}=2$

$\frac{3x}{15}=2$

$\frac{x}{5}=2$

Multiplying both sides by$5$ , we obtain

$x=10$

L.H.S $=\frac{2x}{3}+1=\frac{2\times 10}{3}+1=\frac{2\times 10+1\times 3}{3}=\frac{23}{3}$

R.H.S$=\frac{7x}{15}+3=\frac{7\times 10}{15}+3=\frac{7\times 2}{3}+3=\frac{14}{3}+3=\frac{14+3\times 3}{3}=\frac{23}{3}$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

9. Solve and check result: $2y+\frac{5}{3}=\frac{26}{3}-y$

Ans:

$2y+\frac{5}{3}=\frac{26}{3}-y$

Transposing y to L.H.S and $\frac{5}{3}$ to R.H.S, we obtain

$2y+y=\frac{26}{3}-\frac{5}{3}$

$3y=\frac{21}{3}=7$

Dividing both sides by$3$, we obtain

$y=\frac{7}{3}$

L.H.S $=2y+\frac{5}{3}=2\times \frac{7}{3}+\frac{5}{3}=\frac{14}{3}+\frac{5}{3}=\frac{19}{3}$

R.H.S = $\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}=\frac{19}{3}$

L.H.S. = R.H.S. Hence, the result obtained above is correct.

10. Solve and check result: $3m=5m-\frac{8}{5}$

Ans:

$3m=5m-\frac{8}{5}$

Transposing $5m$ to L.H.S, we obtain

$3m-5m=-\frac{8}{5}$

$-2m=-\frac{8}{5}$

Dividing both sides by$-2$ , we obtain

$m=\frac{4}{5}$

L.H.S $=3m=3\times \frac{4}{5}=\frac{12}{5}$

R.H.S $5m-\frac{8}{5}=5\times \frac{4}{5}-\frac{8}{5}=\frac{12}{5}$

L.H.S. = R.H.S.

Hence, the result obtained above is correct.

Exercise 2.4

1. Amina thinks of a number and subtracts $\frac{5}{2}$ from it. She multiplies the result by$8$. The result now obtained is $3$ times the same number she thought of. What is the number?

Ans:

Let the number be x.

According to the given question,

$8\left( x-\frac{5}{2} \right)=3x$

$8x-20=3x$

Transposing $3x$to L.H.S and $-20$ to R.H.S, we obtain

$8x-3x=20$

$5x=20$

Dividing both sides by $5$, we obtain

$x=4$

Hence, the number is $4$.

2. A positive number is $5$times another number. If $21$ is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?

Ans:

Let the numbers be $x$  and$5x$ . According to the question,

$21+5x=2\left( x+21 \right)$

$21+5x=2x+42$

Transposing $2x$ to L.H.S and 21 to R.H.S, we obtain

$5x-2x=42-21$

$3x=21$

Dividing both sides by $3$, we obtain

$x=7$

$5x=5\times 7=35$

Hence, the numbers are $7\,\text{and }35$ respectively.

3. Sum of the digits of a two digit number is$9$ . When we interchange the digits it is found that the resulting new number is greater than the original number by$27$. What is the two-digit number?

Ans:

Let the digits at tens place and ones place be $x\,\,\text{and }9-x$ respectively.

Therefore, original number $=10x+\left( 9-x \right)=9x+9$

On interchanging the digits, the digits at ones place and tens place will be $x\,\,\text{and }9-x$ respectively.

Therefore, new number after interchanging the digits $=10\left( 9-x \right)+x$

$=90-10x+x$

$=90-9x$

According to the given question,

New number = Original number$+27$

$90-9x=9x+9+27$

$90-9x=9x+36$

Transposing $9x$to R.H.S and $36$ to L.H.S, we obtain

$90-36=18x$

$54=18x$

Dividing both sides by $18$, we obtain

$3=x\text{ and }9-x=6$

Hence, the digits at tens place and ones place of the number are $3\text{ and }6$ respectively. Therefore, the two-digit number is$9x+9=9\times 3+9=36$

4. One of the two digits of a two digit number is three times the other digit. If you interchange the digit of this two-digit number and add the resulting number to the original number, you get$88$. What is the original number?

Ans:

Let the digits at tens place and ones place be $x\text{ and }3x$ respectively.

Therefore, original number $=10x+3x=13x$

On interchanging the digits, the digits at ones place and tens place will be $x\text{ and }3x$respectively.

Number after interchanging $=10x+3x=13x$

According to the given question,

Original number $+$ New number$=88$

$13x+31x=88$

$44x=88$

Dividing both sides by $44$, we obtain

$x=2$

Therefore, original number $=13x=13\times 2=26$

By considering the tens place and ones place as  $3x\text{ and }x$respectively, the two-digit number obtained is $62$.

Therefore, the two-digit number may be $26\text{ or }62$.

5. Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of this mother’s present age. What are their present ages?

Ans:

Let Shobo’s age be x years. Therefore, his mother’s age will be $6x$ years.

According to the given question,

After $5\text{ years, Shobo }\!\!'\!\!\text{ s age}=\frac{\text{Shobo }\!\!'\!\!\text{ s mother }\!\!'\!\!\text{ s present age}}{3}$

$x+5=\frac{6x}{3}$

$x+5=2x$

Transposing x to R.H.S, we obtain

$5=2x-x$

$5=x$

$6x=6\times 5=30$

Therefore, the present ages of Shobo and Shobo’s mother will be $5\text{ years and }30\,\text{years}$ respectively.

6. There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio$11:4$. At the rate $\text{Rs }100$ per metre it will cost the village panchayat $\text{Rs }75,000$ to fence the plot. What are the dimensions of the plot?

Ans:

Let the common ratio between the length and breadth of the rectangular plot be x. Hence, the length and breadth of the rectangular plot will be $11x$ m and $4x$ m respectively.

Perimeter of the plot $=2$ (Length + Breadth) $=\left[ 2\left( 11x+4x \right) \right]\text{m}=30x\text{ m}$

It is given that the cost of fencing the plot at the rate of $\text{Rs }100$per metre is$\text{Rs 75,000}$.

$\therefore 100\times \text{Perimeter}=75000$

$100\times 30x=75000$

$3000x=75000$

Dividing both sides by $3000$, we obtain

$x=25$

Length $=11\times \text{m}=\left( 11\times 25 \right)\text{m}=275\,\text{m}$

Breadth $=4\times \text{m}=\left( 4\times 25 \right)\text{m}=100\,\text{m}$

Hence, the dimensions of the plot are 2$275\text{ m and }100\text{ m}$ respectively.

7. Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him $\text{Rs 50}$per metre and trouser material that costs him $\text{Rs 90}$ per metre. For every   $2$meters of the trouser material he buys $3$metres of the shirt material. He sells the materials at $12%\text{ and }10%$ profit respectively. His total sale is  $\text{Rs 36660}$. How much trouser material did he buy?

Ans:

Let $2x$ m of trouser material and $3x$ m of shirt material be bought by him.

Per metre selling price of trouser material  $\text{=}~\text{Rs }\left( 90+\frac{90\times 12}{100} \right)\text{=Rs 100}\text{.80}$

Per metre selling price of shirt material  $\text{=}~\text{Rs }\left( 50+\frac{50\times 12}{100} \right)\text{=Rs 55}$

Given that, total amount of selling  $\text{=}~\text{Rs }36660$

$100.80\times \left( 2x \right)+55\times \left( 3x \right)=36660$

$201.60x+165x=36660$

$366.60x=36660$

Dividing both sides by $366.60$, we obtain

$x=100$

Trouser material $=2\times m=\left( 2\times 100 \right)m=200m$

8. Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest $9$ are drinking water from the pond. Find the number of deer in the herd.

Ans:

Let the number of deer be x.

Number of deer grazing in the field $=\frac{x}{2}$

Number of deer playing nearby$=\frac{3}{4}\times \text{Number of remaining deer}$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{3}{4}\times \left( x-\frac{x}{2} \right)=\frac{3}{4}\times \frac{x}{2}=\frac{3x}{8}$

Number of deer drinking water from the pond $=9$

$x-\left( \frac{x}{2}+\frac{3x}{8} \right)=9$

$x-\left( \frac{4x+3x}{8} \right)=9$

$x-\frac{7x}{8}=9$

$\frac{x}{8}=9$

Multiplying both sides by$8$, we obtain

$x=72$

Hence, the total number of deer in the herd is $72$.

9. A grandfather is ten times older than his granddaughter. He is also $54$ years older than her. Find their present ages

Ans:

Let the granddaughter’s age be x years. Therefore, grandfather’s age will be  $10x$ years.

According to the question,

Grandfather’s age$=$ Granddaughter’s age $+\,54$ years

$10x=x+54$

Transposing x to L.H.S, we obtain

$10x-x=54$

$9x=54$

$x=6$

Granddaughter’s age $=x$ years $=6$ years

Grandfather’s age $=10x$ years $=\left( 10\times 6 \right)$ years $=60$ years

10. Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Ans:

Let Aman’s son’s age be x years. Therefore, Aman’s age will be $3x$ years. Ten years ago, their age was $\left( x-10 \right)$ years and $\left( 3x-10 \right)$ years respectively.

According to the question,

$10$years ago, Aman’s age $=5\times$Aman’s son’s age $10$ years ago

$3x-10=5\left( x-10 \right)$

$3x-10=5x-50$

Transposing $3x$ to R.H.S and $50$ to L.H.S, we obtain

$50-10=5x-3x$

$40=2x$

Dividing both sides by$2$, we obtain

$20=x$

Sours: https://www.vedantu.com/ncert-solutions/ncert-solutions-class-8-maths-chapter-2-linear-equations-in-one-variable

## Worksheets for linear equations

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Sours: https://www.homeschoolmath.net/worksheets/linear_equations.php

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