ML AGGARWAL CLASS 8 Chapter 12 Linear Equations and Inqualities In one Variable Exercise 12.2

 Exercise 12.2

Question 1

Sol :
Let the number be 'x' 

Three more than twice a number is 

= 2x + 3 -------1

Four less than the number = x -4 ----------2

① = ②

2 x+3=x-4

2 x-x=-4-3

x=-7

The number is -7 

Question 2

Sol :

Let the four consecutive integers are $x+1, x+2, x+3$,

x + 4 

Given sum of them =46

$\begin{array}{c}x+1+x+2+x+3+x+y=46 \\4 x+10=46 \\4 x=46-10=36\end{array}$

x=9

The integers are 9+1 , 9+2, 9+3, 9+4

10,11,12,13

Question 3

Sol :

Let the number be 'x '

Manjula subtracts $\frac{7}{3}$ from it $=x-\frac{7}{3}$

The above result is multiplied by 6 i.e $6\left(x-\frac{7}{3}\right)$

Now it is equal 2 less than twice the number 'x'

$6\left(x-\frac{7}{3}\right)=2 x-2$

6 x-14=2 x-2

6 x-2 x=14-2

4 x=12

x=3

Question 4

Sol :

Let the number be x , 7x

15 is added to both number then 

it becomes x + 15 , 7x + 15

Then one new numbers becomes $\frac{5}{2}$ times the other

new number.

$7 x+15=\frac{5}{2}(x+15)$

2(7 x+15)=5(x+15)

14 x+30=5 x+75

14 x-5 x=75-30

9 x=45

x=5

Therefore the number are 5 , 35

Question 5

Sol :

Let the three consecutive even integers are x, x+2 ,x+4

Given Sum = 0

x+x+2+x+4=0 

3 x+6=0

3 x=-6

x=-2

∴ The integers are -2 , -2+2 ,-2+4

= -2, 0 , 2

Question 6

Sol :

Let the two consecutive odd integers are x+ 1 , x + 3 

Given two fifth of smaller exceeds two ninth of greater by 4

$\frac{2}{5}(x+1)+4=\frac{2}{9}(x+3)$

$\frac{2 x+2+20}{5}=\frac{2 x+6}{9}$

Cross Multiplication.

(2 x+22) 9=5(2 x+6)

18 x+198=10 x+30

18 x-10 x=30-198

8 x=-168

x=-21 

x+1=-21+1=-20

x+3=-21+3=-18

The consecutive odd integers are -20,-18

Question 7

Sol :

Given the denominator of a fraction is 1 more than 

twice its numerator

It is written as $\frac{x}{2 x+1}$

Given numerator and denominator are both increased by 5 

it becomes $\frac{3}{5}$

$\frac{x+5}{2 x+1+5}=\frac{3}{5}$

$\frac{x+5}{2 x+6}=\frac{3}{5}$

Cross multiplication.

5(x+5)=3(2 x+6)

5x+25=6 x+18

6 x-5 x=25-18

x=7

$\begin{aligned} \therefore \frac{x}{2 x+1}=\frac{7}{2(7)+1} &=\frac{7}{14+1} \\ &=\frac{7}{15} . \end{aligned}$

Therefore the original fraction is $\frac{7}{15}$

Question 8

Sol :

Let the two number are 2x , 5x 

Given their difference is 15 

5 x-2 x=15

3 x=15

$x=15 / 3=5$

x=5

Therefore the two positive numbers are 10 , 25

Question 9

Sol :

Let the number added to each be x 

on adding "x" to each of the numbers it becomes 

 

12 + x , 22 + x , 42 + x , 72 + x

for the number to be in proportion

$\frac{12+x}{22+x}=\frac{42+x}{72+x}$

Cross Multiplication

 $(x+12)(x+72)=(x+22)(x+42)$

$x^{2}+72 x+12 x+864=x^{2}+22 x+42 x+924$

84 x+864=64 x+924

84 x-64 x=924-864

20 x=60

x=3

So the number that must be added is x = 3

Question 10

Sol :

Let the unit's digit be x

As the difference of both digits is 3 , the ten's digit is x + 3

∴ The number is 10x (x + 3) 

By adding both number , We get 143

10(x+3)+x+10 x+(x+3)=143

10 x+30+x+10 x+x+3=143

22 x+33=143

22 x=143-33

22x=110

x=5

∴ 10(x+3)+x=10(5+3)+5

=80+5 = 85

∴ The two digit number is 85

Question 11

Sol :

Let the unit's digits be ' x ' 

then ,the ten's digits number be 11 - x 

The two digit  number = 10 (11- x ) + x

When we inter charge the digits , the resulting new number 

is greater than original number by 63 

$\begin{aligned} 10(x)+(11-x) &=10(11-x)+x+63 \\ 10 x+11-x &=110-10 x+x+63 \\ 9 x+11 &=173-9 x \\ 9 x+9 x &=173-11 \\ 18 x &=162 \\ x &=9 \end{aligned}$

$\begin{aligned} \therefore 10(11-x)+x &=10(11-9)+9 \\ &=10(2)+9=20+9=29 . \end{aligned}$

∴ The two digit number is 29

Question 12

Sol :

Let the Raju's present age be ' x' year 

 then , Ritu's present age be 4x years

In four times , Ritu's age will be twice of raju's age 

4 x+4=2(x+4)

4 x+4=2 x+8

4 x-2 x=8-4

2 x=4

x=2

∴ The present ages of Raju's , Ritu are 2 , 8 year

Question 13

Sol :

Let the son's age be ' x ' years

then the father age be 7x years'

Two years ago , father was 13 times as old as his Son'

7 x-2=13(x-2)

7 x-2=13 x-26

13 x-7 x=26-2

6x =24

x=4

Their present age of Son, Father are 4 , 28

Question 14 

Sol :

Let the ages of sona and sonali are 5x , 3x 

five years hence , the ratio of their ages were 10 : 7

$\frac{5 x+5}{3 x+5}=\frac{10}{7}$

Cross Multiplication

$\begin{array}{r}7(5 x+5)=10(3 x+5) \\ 35 x+35=30 x+50 \\ 35 x-30 x=50-35 \\ 5 x=15 \\ x-3\end{array}$

There fire their present ages are 15 ,9

Question 15

Sol :

An employee works on a contract of 30 days 

for that he will receive Rs 200 for each days and 

 he will be find Rs 20 for each day he is absent 

Let the number remain absent be 'x' days

then the no of days he worked are 30 -x days 

∴ 200(30-x)-20 x=3,800

6,000-200 x-20 x=3,800

$220 x=6,000-3,80^{\circ}$

220 x=2,200

x=10

∴The no. of days he remained absent are 10 days 

Question 16

Sol :

Let the no of Rs 5 coins be x 

no of Rs 2 coins be 3x

no of Rs 1 coins be 160 -4x

Total Rs 300 in coins of denominator

∴ 5x(x)+2(3 x)+1(160-4 x)=300

5x+6 x+160-4 x=300

7 x=300-160

7 x=140

x=20

Coins of each denominator are 

Rs Coins = 20

Rs 2 Coins = 60

Rs 1 coins = 80

Question 17

Sol :

Let the no of passenger with Rs 5 tickets be 'X'

The no of passenger with Rs 75 tickets be 40 - x

Total receipts from passenger is Rs 230

$\begin{aligned} 5 x+2.5(40-x) &=230 . \\ 5 x+300-7 \cdot 5 x &=230 \\ 75 x-5 x &=300-230 \\ 2.5 x &=70 \\ x &=\frac{70}{2.5} \\ x &=28 \end{aligned}$

∴ Number of passenger with Rs 5 tickets are 28

Question 18

Sol :

Let the no of students in the group be 'x ' 

 

They paid equally for use a of a full boat and 

pay Rs 10 each i.e 10$\times$x---------(1)

If there are 3 more students in group , each would have 

paid Rs 2 less i.e Rs 8 i.e = 8 $\times$(x +3)

1 = 2

10 x=8(x+3)

10 x=8 x+24

2 x=24

x=12

Question 19

Sol :

Let the number of deer in the herd be ' x ' 

half of a herd of deer are grazing in field i,e = $\frac{1}{2} x$

Three fourth of remaining are playing i,e = $\frac{3}{4}\left(\frac{1}{2} x\right)$

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

The rest 9 are drinking water 

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

$x-\frac{4 x+3 x}{8}=9$

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

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

x=72

Number of deer in the herd are 72

Question 20

Sol :

Let the no. of flower in the beginning be 'x'

At 1st temple she offer = $\frac{1}{2} \times x=\frac{x}{2}$

2nd temple she offer =$\frac{1}{2} \times \frac{x}{2}=\frac{x}{4} .$

3rd temple she offer = $\frac{1}{2} \times \frac{1}{4}=\frac{x}{8}$

Now she is left with 6 flower at end

$x-\left\{\frac{x}{2}+\frac{x}{4}+\frac{x}{8}\right\}=6 .$

$x-\frac{4 x+2 x+x}{8}=6$

$\frac{8 x-7 x}{8}=6$

$\frac{x}{8}=6$

x=48

ஃNo of flower in the beginning are 48

Question 21

Sol :

Let the two supplementary angles be x, 90-x

These angles differ by $50^{\circ}$.

i.e 90-x-x=50

90-2 x=50

2 x=90-50

2 x=40

$x=20^{\circ}$

∴ The two supplementary angles are 20 ,70

Question 22

Sol :

Let the angles of triangle are 5x , 6x , 7x 

Sum of angles of triangle= 180

i .e  5x + 6x + 7 x =180

18x = 180

x = 10

∴ The angles of triangles are $50^{\circ}, 60^{\circ}, 70^{\circ}$

Question 23

Sol :

Two equal sides of an isosceles triangles are 3 x-1,2 x+2

3 x-1=2 x+2

3 x-2 x=2+1

x=3

The third side is 2x = $2 \times 3=6$ Units

The two sides are $3 x-1=3 \times 3-1=9-1=8$ unit

$2 x+2=2 \times 3$+2=6+2=8 unit

∴ Perimeter of triangle = 6+8+8=22 Units

Question 24

Sol :

Let the perimeter of given triangle be x cm 

As each Side is increased by 4cm , So the 

perimeter is increased by $3 \times 4=12 \mathrm{~cm}$

According to given information

$\frac{x+12}{x}=\frac{7}{5}$

Cross Multiplication

7 x=5(x+12)

7 x=5 x+60

7 x-5 x=60

2 x=60

x=30

∴ Perimeter of given triangle = 30cm

Question 25

Sol :

Length of a rectangle is 5cm less than twice its breadth

i .e  $l=2 b-5 \Rightarrow$ 2 b=l+5-------(1)

length is decreased by 3cm i .e (l-3)cm

breadth is increased by 2cm i .e (b+2)cm

Resulting perimeter of rectangle is 72cm

2(1-3+b+2)=72

2 l+2 b-2=72

From eq (1) we get

2l+1+5-2=72

3 x+3=72

3 x=72-3

3 x=69

1=23cm

Length of rectangle =23 cm

From eq (1) 2 b=l+5=23+5

$2 b=28 \Rightarrow b=14 cm$

∴Breadth of rectangle = 14 cm

∴Area of rectangle = l $\times$ b = $23 \times 14=322\mathrm{cm}^{2}$

Question 26

Sol :

Length of rectangle l = 10cm

breadth of rectangle b = 8cm

Each side of rectangle is increased by x cm , its 

perimeter is doubled 

perimeter of rectangle = $\begin{aligned} 2(10+8) &=2 \times 18 \\ &=36 \mathrm{~cm} . \end{aligned}$

i.e $2[(10+x)+(8+x)]=2 \times 36,[\because$ perimeter is doubled]

18+2 x=36-0

2 x=36-18=18

x=18 / 2=9

=9 cm

Area of new rectangle i .e $\begin{aligned} &(10+9) \times(8+9) \\=& 19 \times 17=323 \mathrm{~cm}^{2} \end{aligned}$

Question 27

Sol :

Let the speed of streamer in still water be x km\h 

Given the speed of stream= 5km\h

The speed of streamer down steam = (x+5) km/h

Speed of streamer upward stream= ( x-5)km/h

Both upstream and down stream takes same time 

time taken by streams for down stream is $\frac{90}{x+5}$ hr

Time taken by streamer for upstream is $\frac{60}{x-5} h r$

i.e $\frac{90}{x+5}=\frac{60}{x-5}$

Cross Multiplication

3(x-5)=2(x+5)

3 x-15=2 x+10

3 x-2 x=10+15

x=25

∴ Speed of streamer in still water = 25km/h

Question 28

Sol :

Let the speed of streamer in still water be 'x'

Given the speed of stream = 1km/h

Speed of streamer down stream= (x+1)km/h

Speed of streamer upstream = (x-1)km/h

Distance Covered by streamer , downstream= 5(x+1) km

Distance Covered by streamer upstream = 6(x-1) km

According to given information

5(x+1)=6(x-1)

5 x+5=6 x-6

6 x-5 x=5+6

x=11

∴ Speed of streamer in Still water is 11 km/h

Distance between two parts =$\begin{aligned} 5(x+1) &=5(11+1) \\ &=60 \mathrm{~km} . \end{aligned}$

Question 29

Sol :

Let the speed of faster car be x km/hr 

then the speed of other car be (x - 8)km/hr

Let the faster car starts from place A and the slower

car starts from place B 

Let P and Q be their position after 4 hours

 $A p=4 x \mathrm{~km}, \quad B Q=4(x-8) \mathrm{km}, p Q=62 \mathrm{~km}$

According to the given p b

$\begin{aligned} A P &+P Q+B Q=350 \\ 4 x+62+4(x-8)=350 \end{aligned}$

4 x+62+4 x-32=350

8 x+30=350

8 x=350-30=320

x=320 / 8

x=40 km/hr

∴ Speed of faster car is 40 km/hr and the speed of slower

car is (40-8) i.e 32 km/h