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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Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. You can customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. The worksheets suit pre-algebra and algebra 1 courses (grades 6-9).

You can choose from SEVEN basic types of equations, ranging from simple to complex, explained below (such as one-step equations, variable on both sides, or having to use the distributive property). Customize the worksheets using the generator below.

Basic instructions for the worksheets

Each worksheet is randomly generated and thus unique. The answer key is automatically generated and is placed on the second page of the file.

You can generate the worksheets either in html or PDF format — both are easy to print. To get the PDF worksheet, simply push the button titled "Create PDF" or "Make PDF worksheet". To get the worksheet in html format, push the button "View in browser" or "Make html worksheet". This has the advantage that you can save the worksheet directly from your browser (choose File → Save) and then edit it in Word or other word processing program.

Sometimes the generated worksheet is not exactly what you want. Just try again! To get a different worksheet using the same options:

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Worksheets for simplifying expressions

Worksheets for evaluating expressions with variables

Worksheets for writing expressions with variables from verbal expressions

Worksheets for linear inequalities Key to Algebra Workbooks

Key to Algebra offers a unique, proven way to introduce algebra to your students. New concepts are explained in simple language, and examples are easy to follow. Word problems relate algebra to familiar situations, helping students to understand abstract concepts. Students develop understanding by solving equations and inequalities intuitively before formal solutions are introduced. Students begin their study of algebra in Books 1-4 using only integers. Books 5-7 introduce rational numbers and expressions. Books 8-10 extend coverage to the real number system.

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

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