Exercise 7B
Q1 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations |
myhelper
Question 1
Answer the following.
(a) If x is an integer, what are the next two larger integers?
Sol :m
If x is an integer, the next are larger integers are (x+1), (x+2)
(b) If x is an even integer, what are the next two larger even
integers?
Sol :
If x is an even integer, the next two larger even integers are (x+2), (x+4)
(c) The sum of two integers is -8. If one integer is represented byx,
represent the other integer in terms of x.
Sol :
The sum of two integers is -8
One is x other is (-8-x)
(d) The difference between a number y and another number is 20. Represent
the other number in terms of y.
Sol :
A number =y
Difference =20
Other number=y-20
(e) If Priya is n years old now, how old will she be next year and 5 years
from now? How old was she 10 years ago?
Sol :
Priya's old =x
After 1 year priya's old=x+1
After 5 year priya's old=x+5
Before 1 year priya's old=x-10
(f) Over a period of 4 years, a businessman has tripled his early profit.
If his profit was Rs. x at the beginning of this period, what was his at the
end of this period?
Sol :
Profit=x
Over 4 years it is =3x
(g) The tens digit of a number is 4 less than its ones digit. If the ones
digit is x, then what is the number?
Sol :
The number=10(x-4)+x
(h) What is the number if the digits in (g) part are interchanged?
Sol :
The number=10x+(x-4)
(i) The denominator of a fraction is 3 more than its numerator. If the
numerator is y, then what is the fraction?
Sol :
The fraction$=\frac{y}{y+3}$
(j) What is the angle marked a in the figure?
Sol :
a=360°-x°
Q2 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 2
Forty-six more than a number is 13. What is the number?
Sol :
Let , the number =x
∴x=13-46=-33
Q3 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 3
The sum of three consecutive odd numbers is 33. What are the
numbers?
Sol :
9,11,13
Q4 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 4
The sum of two numbers is 87. If twice the smaller number is added to the
larger number, the result is 120. Find the larger number.
Sol :
Sum of two number=87
Let, small number=x
∴Other number=87-x
ATQ,
2x+(87-x)=120
2x+87-x=120
x=120-87=33
∴Small number=33
Large number=87-33=54
Q5 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 5
Find three consecutive numbers such that the sum of the first and second
is 15 more than the third.
Sol :
16,17,18
Q6 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 6
The sum of three numbers is 57. If twice the first is 6 less than the
second and the second numbers is 10 more than the third, find the
numbers.
Sol :
Let 1st number=x
According to question
x+(2x+6)+(2x+6-10)=57
x+2x+6+2x-4=57
5x+2=57
5x=55
$x=\frac{55}{5}=11$
∴1st number=2x+6=2×11+6=22+6=28
3rd number=(2x+6-10)=2×11+6-10=22-4=18
Q7 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 7
In a 2 digit number, digit in units place is twice the digit in tens
place. If 27 is added to it, digits are reversed. Find the number.
Sol :
Let,unit place digit=x
Ten place digit$=10\times \frac{x}{2}=5x$
∴The number is=5x+x=6x
The reversed number will be$=10x+\frac{x}{2}$
where , unit place digit$=\frac{x}{2}$
Ten place digit=10x
According to question
$6x+27=10x+\frac{x}{2}$
$6x+27=\frac{20x+x}{2}=\frac{21x}{2}$
12x+54=21x
21x-12x=54
9x=54
x=6
∴Unit place digit=6
∴Ten place digit$=10\times \frac{6}{2}=30$
∴The number will be=30+6
=36
Q8 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 8
If a certain number is added both to the numerator and to the denominator
of 9/17, the new fraction reduces to 5/7. Find the number.
Sol :
Let, the number=x
ATQ,
$\frac{9+x}{17+x}=\frac{5}{7}$
63+7x=85+5x
7x-5x=85-63=22
2x=22
x=11
Q9 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 9
Find four consecutive odd numbers such that three times the largest
number added to twice the smallest number is 93. Which of these are prime
numbers?
Sol :
Let consecutive numbers are=2x+1,2x+3,2x+5,2x+7
ATQ,
3(2x+7)+2(2x+1)=93
6x+21+4x+2=93
10x+23=93
10x=93-23
10x=70
x=7
Numbers are
2x+1=2×7+1=15
2x+3=2×7+3=17
2x+5=2×7+5=19
2x+7=2×7+7=21
The prime numbers are 17, 19
Q10 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 10
Sum of the digits of a 2 digit number is 11. The given number is less
than the number obtained by interchanging the digits by 9. Find the
number.
Sol :
Let, unit place digit=x
ten place digit=11-x
∴The number=10(11-x)+x
After reversing the digit=10x+(11-x)
ATQ,
{10x+(11-x)}-{10(11-x)+x}=9
10x+11-x-110+10x-x=9
18x=99+9=108
x=6
∴x=6
∴unit digit=6
ten place=11-5=5
∴the number=56
Q11 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 11
In the same innings of a cricket match, four batsman’s scores are
consecutive numbers which are divisible by 5. If their total contribution
to the innings is 70 runs, how many runs, does each batsman score?
Sol :
Let ,Batsmen's score=x, x+5, x+10, x+15
ATQ,
x+(x+5)+(x+10)+(x+15)=70
4x=70-30=40
$x=\frac{40}{4}=10$
Scores are
x=10
x+5=10+5=15
x+10=10+10=20
x+15=10+15=25
Q12 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 12
The numerator of a rational number is 2 less than the denominator.
When is subtracted from both the numerator and denominator, the
number’s simplest form is 1/2. What is the rational number?
Sol :
Let, denominator=x
numerator=x-2
ATQ,
$\frac{x-2-1}{x-1}=\frac{1}{2}$
$\frac{x-3}{x-1}=\frac{1}{2}$
2x-6=x-1
2x-x=-1+6=5
x=5
∴The rational number$=\frac{5-2}{5}=\frac{3}{5}$
Q13 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 13
A boy was p years old r years ago. How old is he now? How old will he be
in (p – r) years time?
Sol :
The recent old is (p+r) years
(p-r) years will be {(p+r)+(p-r)}
=p+r+p-r=2p years
Q14 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 14
Three years ago, Hari was 5 years older than Anushka. If he is now twice
as old as she is, find their present ages.
Sol :
Let , Haris now age=x years
∴Anushka's now age$=\frac{x}{2}$ years
3 years ago Haris age=(x-3) years
3 years ago Anushka's age$=\left(\frac{x}{2}-3\right)$ years
ATQ,
$(x-3)-\left(\frac{x}{2}-3\right)=5$
$x-3-\frac{x}{2}+3=5$
$\frac{x}{2}=5$
x=10
∴Now Hari's age=10 years
Anushka's age$=\frac{10}{2}=5$ years
Q15 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 15
Ravi is now 6 years older than Neeta. Three years ago, he was twice as
old as she was then. Find their present ages.
Sol :
Before 3 years
Ravi's age=x
Neeta's age $=\frac{x}{2}$
Now, Present age of ravi=x+3
Present age of neeta$=\frac{x}{2}+3$
According to question,
$(x+3)-\left(\frac{x}{2}+3\right)=6$
$x+3-\frac{x}{2}-3=6$
x=12
∴Now Ravi's age=12+3=15 years
Neeta's age$=\frac{12}{2}+3$=9 years
Q16 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 16
Deepas mother is 4 years more than 3 times as old as Deepa is now, six
years from now she will 8 years more than twice as old as Deepa will be
then. How old is each of them now?
Sol :
Let , Present age of deepa=x years
Present age of deepa's mom=(3x+4) years
After six years
Deepa's age=(x+6)
Deepa's moms age=(3x+6+4)
3x+10-2x-12=8
x-2=8
x=8+2=10
∴Deepa's present age=10 years
Deepa's mom present age=(3×10+4)
=30+4=34 years
Q17 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 17
The present ages of two children are in the ratio 2:3. Five years ago,
the ratio of their ages was 1:2. Find their present ages.
Sol :
Let, common factor=x
∴1st child age=2x
2nd child's age=3x
Before 5 years
1st child age=2x-5
2nd child age=3x-5
ATQ,
$\frac{2x-5}{3x-5}=\frac{1}{2}$
4x-10=3x-5
4x-3x=10-5=5
x=5
∴1st child age=2×5=10 years
2nd child age=3×5=15 years
Q18 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 18
A playground is 30 m longer than it is wide. If its perimeter is 300 m,
what is its area?
Sol :
Let, Length=x
Wide=x-30
∴Perimeter=2(2+w)
=2{x+(x-30)}
ATQ,
2{x+(x-30)}=300
x+(x-30)=150
2x-30=150
2x=150+30=180
x=90
∴Length=90 m
Wide=90-30=60 m
∴Area=L×Q
=90×60=5400 m2
Q19 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 19
In the figure, all the corners are right angled and the dimensions are
given in centimetres. Find, a if the perimeter of the shaded area is 1
metre.
Sol :
Perimeter of Shaded region =1m [1m=100cm]
AB+BC+CP+TP+TS+SD+DA=100
(3a+2)+3+2a+2+2a+3a+3=100
10a+10=100
10(a+1)=100
$(a+1)=\frac{100}{10}$
a+1=10
a=10-1=9cm
Q20 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 20
The figure shows the the lengths in centimetres of the sides of a
triangle. If the triangle is equilateral, find the value of x and y. Also
find the perimeter of the triangle.
Sol :
The triangle is equilateral
∴5x+3=3x+7
5x-3x=7-3
2x=4
x=2
Again, $3x+7=1+\frac{1}{3}y$
$3\times 2+7=1+\frac{y}{3}$
$6+7-1=\frac{y}{3}$
y=12×3=36
∴Perimeter=3a
=3(5x+3)
=3(5×2+3)
=3(10+3)
=3×13=39 cm2
Q21 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 21
Six paving stones are arranged in a square array as shown in the figure.
If each stone has a length 20 cm greater than its width, find
(a) the dimension of each stone,
(b) the area of the ground space that the stones are occupying.
(c) the perimeter of the same ground space.
Sol :
(1) Since, the array formed is of square shape.
So, all sides must be equal.
We have,
Length of array = (x+20)+(x+20) = 2x+40
Width of array = x+x+x = 3x
Now,
Length = Width
2x+40=3x
x= 40cm
That is,
Length = 2x+40= 120cm
Width = Length = 120cm=Side
(2)
Area of ground space stones occupied = Area of array =
Area of square = Side × Side
= 120×120
=14400
(3) Now, Perimeter of square array = 4×Side = 4×120 = 480cm
Q22 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 22
I travel 25 km in 2 and half hours, walking part of the way at 3 km/h and
covering on scooter the rest at 24 km/h. How far do I walk?
Sol :
Total distance=25km
Total time$=2\frac{1}{2}=\frac{5}{2}$hours
Let, walking part distance x
Walking speed=3km/h
Walking time$=\frac{\text{Distance}}{\text{speed}}=\frac{x}{3}$hour
Covering on scooter distance=(25-x)
Scooter speed=24 km/h
Time taken=distance/speed$=\frac{25-x}{24}$
ATQ,
$\frac{x}{3}+\frac{25-x}{24}=\frac{5}{2}$
$\frac{8x+25-x}{24}=\frac{5}{2}$
$7x+25=\frac{5\times 24}{2}$
7x=60-25=35
x=5
∴walking distance 5km
Q23 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 23
Figure represents a cyclists riding from A to B and back again; the
double journey takes 5 hours. Find x.
Sol :
A to B➝
Distance =x km
speed=8km/hour
Time$=\frac{x}{8}$hour
Total time=5 hour
B to A➝
Distance=x km
Speed=12km/hour
Time$=\frac{x}{12}$hour
According to question
$\frac{x}{8}+\frac{x}{12}=5$
$\frac{3x+2x}{24}=5$
$x=\frac{5\times 24}{5}=24km$
Q24 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 24
A man rows upstream in still water at 3 km/h and back to the same at 5
km/h; he takes 48 minutes altogether. How upstream did he go?
Sol :
Let, the distance between man 1st place to upstream=x km
∴Man rows upstream in speed=3km/hour
∴Man rows upstream in time$=\frac{x}{3}$ hour
Man back to the same place in speed=5km/hour
Man back to the same place in time$=\frac{x}{5}$hour
Total time=48min $=\frac{48}{60}$ hour
According to question:
$\frac{x}{3}+\frac{x}{5}=\frac{48}{60}$
$\frac{5x+3x}{15}=\frac{48}{60}$
$x=\frac{48\times 15}{8\times 60}=\frac{3}{2}=1\frac{1}{2}$km
Q25 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 25
Distance between two places A and B is 210 km. Two cars start
simultaneously from A and B in opposite directions and distance between
them after 3 hours is 54 km. If speed of one car is 8 km/h less than that
of the other, then find the speed of each.
Sol :
For Car1→
Let, speed=x km/h
Time take A to C=3hour
Distance A to C= 3x km
For Car2→
Let, speed=x+8 km/h
Time take B to D=3hour
Distance B to D= 3(x+8)=(3x+24) km
Total distance=210km
ATQ,
3x+3x+24+54=210
6x=210-78=132
$x=\frac{132}{6}=22$km/hr
∴Car1 speed=22 km/hr
Car2 speed=(22+8)=30km/hr
Q26 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 26
A streamer covers the distance between two ports in 3 hours when it goes
downstream and 5 hours when it goes upstream. Find the speed of the
streamer upstream if the stream flows at 3 km/h.
Sol :
A steamer goes downstream and covers the distance between two ports is 3
hours.
It covers the same distance in 5 hours when it goes upstream.
The stream flows at 3 km/hr.
We have to find the speed of the steamer upstream.
Let the speed of the steamer in still water be x km/hr
The speed of the stream is 3 km/hr
Speed of the steamer downstream = (x + 3) km/hr
Speed of the steamer upstream = (x - 3) km/hr
According to the question,
Distance covered by steamer downstream in 3 hours = distance covered by
steamer upstream in 5 hours.
3(x + 3) = 5(x - 3)
3(x) + 3(3) = 5(x) - 5(3)
3x + 9 = 5x - 15
By transposing,
5x - 3x = 9 + 15
2x = 24
$x = \frac{24}{2}$
x = 12
Now, speed of the steamer upstream = 12 - 3
= 9 km/hr
Therefore, the required speed of the steamer upstream is 9 km/hr
Q27 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 27
Find an angle such that the measure of its supplement is 10 degrees less
than 3 times the measure of its complement.
Sol :
Let , angle be x
Supplement of angle=180°-x
Complement of angle=90°-x
ATQ→
180°-x-3(90°-x)=10°
180°-x-270°+3x=10°
2x=270°+10°-180°
2x=270°-170°=100°
x=50°
Q28 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 28
The figure shows in degrees the angles at which a line is cut by two
parallal lines. Find the value of x.
Sol :
Both lines are parallel
∴ATQ,
180-{4(x+7)}=5(x-20)
180-4x-28=5x-100
5x+4x=180-28+100=252
9x=252
x=28
Q29 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 29
In the figure, ∠ACB = 4∠ABC, and CP bisects ∠ACB, Find
(a) ∠ABC
(b) ∠BPC
Sol :
∵∠ACB=4∠ABC
Let, ∠ABC=x°
∴∠ACB=4x
∠BAC=90°
We know that
90°+x+4x=180°
5x=180°-90°=90°
x=18°
∴∠ABC=18°
∠ACB=4×18°=72°
Though, CP bisects ∠ACB
∴∠BCP$=\frac{72}{2}$=36°
∴∠BPC=180°-(36+18)=180-54=126°
∠BPC=126°
Q30 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 30
An apple has 29 more calories than a peach and 13 fewer calories than a
banana. If 3 apples have 43 fewer calories than 2 bananas and 2 peaches,
how many calories does an apple have?
Sol :
Let ,A apples's calories=x
A peach calories =x-29
A banana calories=x+13
∴3 Apple's calories=3x
2 peach calories=2(x-2a)
2 banana calories=2(x+13)
ATQ,
{2(x-29)+2(x+13)}-3x=43
2x-54+2x+26-3x=43
4x-3x-32=43
x=43+32=75
∴apple's calories 75
Q31 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 31
Rishi’s mother gave him Rs. 245 with which to buy New Year cards. If he
got some Rs. 10 cards, 2/3 as many Rs. 5 cards, and 1/5 as many Rs. 15
cards, how many of each kind did he get?
Sol :
Let, He get x card in 10
∴Number of card of 10 =10x
Number of card of 5$=\frac{2n}{3}\times 5=\frac{10n}{3}$
Number of card of 15$=\frac{x}{5}\times 15=3x$
Total money=245
ATQ,
$10x+\frac{10x}{3}+3x=245$
$\frac{30x+10x+9x}{3}=245$
49x=245×3
x$=\frac{245\times 3}{49}=15$
Number of card of 10=15
Number of card of 5$=\frac{2}{3}\times 15$=10
Number of card of 15$=\frac{1}{5}\times 15$=3
Q32 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 32
One-fifth of an estate is left to a sister, one fourth to a son,
one-sixth to a brother and the remainder, which is Rs. 5,75,000 to the
widow. What is the value of the estate?
Sol :
Let, Total estate=x
Sister get$=\frac{x}{5}$
Son get$=\frac{x}{4}$
Widow get=5,75,000
ATQ,
$x-\left(\frac{x}{5}+\frac{h}{4}+\frac{h}{6}\right)=5,75,000$
$x-\left(\frac{12x+15x+10x}{60}\right)=5,75,000$
$x-\frac{37x}{60}=5,75,000$
$\frac{60x-37x}{60}=5,75,000$
23x=5,75,000
$x=\frac{575000\times 6}{23}$
x=1500000
Q33 | Ex-7B | Class 8 | SChand New Learning | Chapter 7 | Linear Equations | myhelper
Question 33
An employee works in a company on a contract of 30 days on the condition
that he will receive Rs. 120 for each day he works and he will be fined
Rs. 10 for each day he is absent. If he receives Rs. 2300 in all, for how
many days did he remain absent?
Sol :
In present he will get 120 per day
In 30 days=120×30
=3600
ATQ
120(30-x)-10x=2300
3600-120x-10x=2300
-130x=2300-3600
$x=\frac{1300}{130}=10$
