SChand CLASS 9 Chapter 5 Simultaneous Linear Equations Exercise 5(B)

Exercise 5(B)

Question 1

If one number is thrice the other and their sum is 16, find the numbers.

Sol :

Sum of two number=16

Let first number =x

Second number=y

∴x+y=16...(i)

x=3y

x-3y=0...(ii)

4y=16$=y=\frac{16}{4}=4$

x=3y=3×4=12  [Subtracting (ii) from (i)]

So, number are 4, 12 or 12 , 4


Question 2

The sum of two numbers is 6 and their difference is 4. Find the numbers.

Sol :

Sum of two numbers=16
Difference =4
Let the first number=x
Second number=y
x+y=6...(i)
x-y=11...(ii)

By adding

8x=10=5x$=\frac{10}{2}$

=5

By subtracting

8y=2

$y=\frac{2}{2}=1$

The number are 5 and 1


Question 3

In a two digit number, the sum of digits is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number ?

Sol :

Sum of two digit of a two digit number=13

Let unit digit=x

tens digit=y

Number=x+10y

Number by reversing the digit

=y+10x


According to the conditions

x+y=13...(i)

⇒(y+10x)-(x+10y)=45
⇒y+10x-x-10y=45
⇒3x-9y=115

x-y=5...(ii)
[Dividing by 9]

Adding (i) and (ii)
⇒2x=18
⇒$x=\frac{18}{2}=9$

Subtracting (ii) from (i)
2y=8
$y=\frac{8}{2}$
=4

Number=x+10y
=9+0×4
=9+40
=49

Question 4

A number consists of two digits whose sum is 5. When the digits are reversed, the number becomes greater by 9. Find the number.

Sol :

Sum of two digit of a two digit number=5

Let units digit=x

tens digit=4

Then number=x+10y

By reversing the digit

New number=y+10x

According to the conditions

x+y=9

y+10x-(x+10y)=9

y+10x-x-10y=9

9x-9y=9

x-y=1

Adding (i) and (ii)

2x=6

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

Subtracting (ii) from (i)

2y=4

$y=\frac{4}{2}$

=2

Number=x+10y

=3+10×2

=3+20=23


Question 5

The sum of a 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 numbers.

Sol :

Difference of two digits of a two digit
Number=3

Let unit digit=x
ten digit=y

Number=x+10y
Reversing the digit
=y+10x

According tot he conditions
x-y=3...(i)
x+10y-y+10x=121
11x-11y=121
x-y=11...(ii) [Dividing by 11]
2x=14
$x=\frac{14}{2}$=7

Substituting (ii) from (i)
2y=8
$y=\frac{8}{2}$=4

Number =x+10y
=7+10×4

=7+40=110

or y+70x=11+10×7

=4+70=74


Question 6

Seven times a given two digit number is equal to four times the number obtained by reversing the order of digits. The sum of the digits of the number is 3. Find the number.

Sol :

In two digits number

Sum of digits=3

Let one digit=x

tens digit=y

Numbers=x+10y

Reversing the number

=y+10x


According to the condition

x+y=3...(i)

and 7(x+10y)=4(y+10x)

7x+70y=4y+40x

40x-7x+4y-70y=0

33x-66y=0

x-2y=0...(ii)  [Dividing by 33]

x=2y

Substituting the value of x in (i)

2y+y=3

3y=3

$y=\frac{3}{3}$=1

x=2y=2×1

=2+10=12


Question 7

If three times, the larger of the two numbers is divided by the smaller one. we get 4 as quotient and 3 as remainder. Also, if seven times the smaller number is divided by the larger one, we get 5 as quotient and 1 as remainder. Find the numbers.

Sol :

Let the larger number be x

Smaller number be y

According to the condition given

$\frac{3 x}{y}=4 \times y+3$

3x=4y+3

3x-4y=3...(i)

7y=x×5+1

7y=5x+1

5x-7y=-1...(ii)

Multiplying (i) by 7 and (ii) by 4

$\begin{aligned}21 x-98 y=21\\20 x-28 y=-y\\ -\phantom{20 x}+\phantom{28 y}=+\phantom{y}\\ \hline x\quad\quad=+25\end{aligned}$

From (i)

3×25-4y=3

75-4y=3

4y=75-3=72

$y=\frac{72}{4}$=18

Larger number=25

Smaller number=18


Question 8

Divide 36 into four parts so that if 2 is added to the first part, 2 is subtracted from the second part, the third part is multiplied by 2, and the fourth part is divided by 2, then the resulting number is the same in each case.

Sol :

Total=36
Let the result in each case=x
First number=x-2
Second number =x+2
Third number$=\frac{x}{2}$
Fourth number=2v

According to the condition
$x-2+x+2+\frac{x}{2}+2 x$=36
$4 \frac{1}{2}x=36$
$\frac{9}{2} x=36$

$x=\frac{36 \times 2}{9}$=8

First number=8-2=6

Second number=8+2=10

Third number=$=\frac{8}{2}$=4

Fourth number=2×8=16


Question 9

If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes $\frac{5}{8}$ and if the numerator and the denominator of the same fraction are each increased by 1, the fraction becomes equal to $\frac{1}{2}$. Find the fraction.

Sol :

Let numerator of a fraction=x
Denominator=y
Then function$=\frac{x}{y}$

According to the condition

$\frac{x+2}{y+1}=\frac{5}{8}$

8x+16=5y+5

8x-5y=5-16

8x-5y=-11

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

2x+2=y+1

2x-y=1-2=-1

From (ii) y=2x+1

Substituting the value of y in (i)

8x-5y=-11

8x-5(2x+1)=-11

8x-10x-5=-11

-2x=-11+5=-6

$x=\frac{-6}{-2}$

=3

y=2x+1=2×3+1=6+1=7

Fraction$=\frac{x}{4}=\frac{3}{7}$


Question 10

If 1 is added to the denominator of a fraction, the fraction becomes $\frac{1}{2}$ . . If 1 is added to the numerator of the fraction, the fraction becomes 1. Find the fraction.

Sol :

Let numerator of a fraction=x
Denominator=y
Fraction$=\frac{x}{y}$

According to the condition

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

2x=y+1

2x-y=1

$\frac{x+1}{y}=1$

x+1=y

x-y=-1

Subtracting (ii) from (i)

x=2

From (ii) 2-y=-1

-y=-1-2=-3

y=3

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


Question 11

When die numerator of a fraction is increased by 4, the fraction increases by $\frac{2}{3}$.What is the denominator of the fraction? 

Sol :

Let the numerator of a fraction=x

Denominator=y

Fraction$=\frac{x}{y}$

According to the condition

$=\dfrac{x+4}{y}$

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

$\frac{x+4}{y}-\frac{x}{y}=\frac{2}{3}$

$\frac{x+4-x}{y}=\frac{2}{3}$

$\frac{4}{y}=\frac{2}{3}$

$y=\frac{4\times 3}{2}$=6


Denominator=6



Question 12

Father is six times as old as his son. Four years hence he will be four times as old as his son at that time. Find their present ages.

Sol :

The age of father=x years

age of son= y years

x=6y

4 year hence

Age of father will be x+4

Age of son=y+4

(x+4)=4(y+4)

x+4=4y+16

x=4y+16-4

x=4y+12

From (i) and (ii)

6y=4y+12

6y-4y=12

2y=12

y=6

x=6y=6×6=36

Present age of father=36 years

Age of son=6 years


Question 13

The age of the father is 3 years more than three times the age of the son. Three years hence father’s age will be 10 years more than twice the age of the son. Determine their present ages.

Sol :

The age of son= x years

Age of father= y years

y=3x+3

3 years

So,

Age of son will be (x+3) years

Age of father=(y+3) years

y+3=2(x+3)+10

y+3=2x+6+10

y=2x+6+10-3

y=2x+13

From (i) and (ii)

3x+3=2x+13

3x-2x=13-3

x=10

and y=3x+3

=3×10+3=30+3=33

Present age of son=10 years

Age of father=33 years


Question 14

The age of a man is three times the sum of the ages of his two children and five years hence his age will be double the sum of their ages. Find his present age.

Sol :

Age of the man= x years
Sum of ages of his two children= y years
x=3y
5 years So,
Age of man will be (x+5) years
Sum of ages of two children=(y+10) years
x+5=2(y+10)
x+5=2y+20
x=2y+20-5
x=2y+15
From (i) and (ii)
3y=2y+15
3y-2y=15
y=15
and
x=3y
=3×15=45

Age of the man=45 years

Question 15

Ram’s father is 4 times as old as Ram. Five years ago, his father was 9 times as old as he was then. What are their present ages ?

Sol :

Age of ram= x years

Age of Ram's father= y years

y=4x

5 years ago,

Age of ram=(x-5) years

Age of his father=(y-5) years

y-5=9(x-5)
y-5=9x-45
y=9x-45+5
y=9x-40

From (i) and (ii)
9x-40=4x
9x-4x=40
5x=40
$x=\frac{40}{5}$
=8

y=4x=4×8=32

Present age of father=32 years
Age of Ram=8 years

Question 16

At the time of marriage a man was 6 years older than his wife, but 12 years after his marriage, his age was $\frac{6}{5}$ th of the age of his wife. What were their ages at the time of marriage ?

Sol :

At the time of marriage

Age of man=x years

Age of his wife= y years

x=y+6

12 years after,

Age of a man=(x+12) years

Age of his wife=(y+12) years

$x+12=\frac{6}{5}(y+12)$

5x+60=6y+72 (By cross multiplication)

5x-6y=72-60=12....(ii)

From (i) x=y+6
5(y+6)-6y=12
5y+30-6y=12
-y=12-38=-18
and 
x=y+6
=18+6=24

At the time of their marriage
Age of man=24 years
Age of his wife=18 years


Question 17

The present ages of Ram and Shyam are in the ratio 5 : 6. Five years ago, the ratio was 4:5. Find their present ages.
Sol :
Ratio in the present age of Ram and Shyam=5 : 6
Age of Ram=x years
Age of Shyam= y years
$\frac{x}{y}=\frac{5}{6}$
6x=5y

$x=\frac{5}{6} y$
5 years ago,

Age of Ram=(x-5) years
Age of Shyam=(y-5) years
$\frac{x-5}{y-5}=\frac{4}{5}$
5x-25=4y-20
5x-4y=-20+20=5
5x-4y=5
From (i)
$5 \times \frac{5}{6} y-4 y=5$
$\frac{25}{6} y-4 y=5$
25y-24y=30
y=30

$\begin{aligned} x &=\frac{5}{6} y \\ &=\frac{5}{6} \times 30 \end{aligned}$
=25

Present age of Ram=25 years
Age of Shyam=30 years

Question 18

The cost of 5 pencils and 6 erasers is Rs. 1.80 whereas that of 3 pencils and 2 erasers is 92 paise. Find the cost of each of a pencil and an eraser.
Sol :

Cost of one pencil=x paise

Cost of one eraser= y paise
According to the conditions,
5x+6y=180
3x+2y=92

Multiplying (i) by 1 and (ii) by 3, then subtracting
$\begin{aligned}5x+6y&=180\\9x+6y&=276 \\ \hline -4x&=-96 \end{aligned}$

$x=\frac{-96}{-4}$=24

From (i)
5×24+6y=180
120+6y=180
6y=180-120=60

$y=\frac{60}{6}$=10
Cost of one pencil=24 paise
Cost of one eraser=10 paise

Question 19

9 chairs and 5 tables cost Rs. 90 while 5 chairs and 4 tables cost Rs. 61. Find the price of 6 chairs and 3 tables.
Sol :
Cost of one chair=x
Cost of one Table =y

According to the conditions
9x+5x=90
5x+4y=61

Multiplying (i) by 4 and (ii) by 5, then subtracting we get
$\begin{aligned} 36 x+20 y &=360 \\ 25 x+20 y &=305 \\\hline11x&=55 \end{aligned}$

$x=\frac{55}{11}$=5

From (i)
9×5+5y=90
45+5y=90
90-45=5y
$y=\frac{45}{5}$=9

Cost of 1 chair=5
Cost of 1 table=9
Cost of 6 chair and 3 table
=6×5+9×3=30+37=57

Question 20

A horse and two cows together cost Rs. 680. If a horse cost Rs. 80 more than a cow, find the cost of each.
Sol :
Cost of one horse=x
Cost of one cow=y

According to the condition
x-2y=680
x=y+80

From (ii)
y+80+2y=680
3y=680-80=600
$y=\frac{600}{3}$=200
and
x=y+80=200+80=280

Cost of one horse=280
Cost of one cow=200

Question 21

If one kg of sweets and 3 kg of apples cost Rs. 33 and 2 kg of sweets and one kg of apples cost Rs. 31, what is the cost per kg of each ?

Sol :

Cost of 1 kg sweet=x
Cost of 1 kg apples=y

According to the condition,
x+3y=33...(i)
2x+y=31...(ii)

Multiplying (i) by 1 and (ii) by 3

$\begin{aligned}x+3y=33\\6x+3y=93\\ \hline -5x=-60 \end{aligned}$

$x=\frac{-60}{-5}$=12

From (i)

21+3y=33
3y=33-12=21
$y=\frac{21}{3}$=7

Price of 1 kg sweets=12
Price of 1 kg apples=7

Question 22

“A pen costs Rs. 3.50 more than a pencil and the cost of 3 pens and 2 pencils is Rs. 13”. Taking x and y as the cost (in Rs) of a pen and a pencil respectively, write two simultaneous equations in x and y which satisfy the statement given above. (Do not solve the equations). 
Sol :
Cost of one pen=x
Cost of one pencil=y
x=y+3.5
x-y=3.5
and 3x+2y=13
3(y+3.5)+2y=13
3y+10.5+2y=13
5y=13-10.5=2.5
$y=\frac{2.5}{5}$
=0.5=50 paise

And
x=y+3.5
=0.50+3.50=4.0

Cost of one pen=4
Cost of one pencil=50 paise

Question 23

A man buys postage stamps of denominations 3 paise and 5 paise, for Re 1.00. He buys 22 stamps in all. Find the number of 3 paise stamps bought by him.
Sol :
Total stamps=22
Number of stamps of 3 paise=x
Number of stamps of 5 paise=y

According to the condition given
x+y=22...(i)
and 
$\frac{x \times 3}{100}+\frac{y \times 5}{100}=1$
3x+5y=100

From (i)
y=22-x
Substituting the value of y in (ii)
3x+5(22-x)=100
3x+110-5x=100
-2x=100-110=-10

$x=\frac{-10}{-2}$=5

Number of stamps of 3 paise=5

Question 24

There were 2500 persons who bought tickets to see a village fair. The adults paid 75 paise each for their admission tickets but the children paid 25 paise each. If the total receipts amounted to Rs. 1503, using an equation method, find how many children saw the fair?
Sol :
Total person=2500
Cost of adult ticket=75 paise
Cost of child ticket=25 paise

Total amount of sold tickets=1503 ₹ or 150300 paise

Number of children=x and number of adults=y

According to the condition
25x+75y=150300...(i)

x+y=2500
y=2500-x...(ii)

Substituting the value of y in (i)
25x+75(2500-x)=150300
25x+187500-75x=150300
-50x=150300-187500
-50x=-37200
$x=\frac{-37200}{-50}$
=744

Number of children=744
Number of adults=2500-744=1756

Question 25

In a triangle, the sum of the two angles is equal to the third. If the difference between them is 50°, determine the angles.
Sol :
Sum of three angles of a triangle=180°
Sum of two angles= third angle

Difference between them=50
First angle=x
Second angle=y
Third angle=x+y

x+y+x=180
2x+2y=180
x+y=50

2x=140 [adding we get]
$x=\frac{140}{2}$
=70

Subtracting ,
2y=40
$y=\frac{40}{2}$
=20

First angle=70°
Second angle=20°
Third angle=70°+20°=90°

Question 26

In a parallelogram, one angle is $\frac{2}{5}$ th of the adjacent angles. Determine the angles of this parallelogram.
Sol :
In a parallelogram, one angle$=\frac{2}{5}$ adjacent angle
First angle=x
Second angle=y

According to the conditions
x+y=180....(i)  [co-interior angles]

$x=\frac{2}{5} y$...(ii)

$\frac{2}{5} y+y=180$
2y+5y=900
7y=900
$y=\frac{900}{7}$

$x=\frac{2}{5} \times \frac{900}{7}$
=360

Opposite angles of the parallelogram equal
Angles of parallelogram are
$\frac{900^{\circ}}{7}, \frac{360^{\circ}}{7}, \frac{900^{\circ}}{7}, \frac{360^{\circ}}{7}$

Question 27

If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.
Sol :
Let length of rectangle (l)=x
and breadth (b)=y

Area=l×b
=xy m2

First case,
Length=x+1
and Breadth=y+1

(x+1)(y+1)
=xy+21
=xy+x+y+1
=xy+21
=x+y
=21-1=20...(i)

Second case,
Length=x+1
and Breadth=y-1

Area=(x+1)(y-1)
=xy-5
=xy-x+y-1
=xy-5


=-x+y
=-5+1=-4
=x-y=4....(ii)

Now, adding (i) and (ii)
2x=24
$=x=\frac{24}{2}$
=12...(iii)

Subtracting (iii) from (i)
2y=16
$=y=\frac{16}{2}$
=8

Length=12m
and Breadth=8m

Perimeter=2(l+b)
=2(12+8)=2×20=40m

Question 28

A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.
Sol :
Time taken downstream=40 min or $=\frac{40}{60}$ hour

Distance=8m
and time taken upstream=60 min or 1 hour

Let speed of sailor in still water=x km\hr
and Speed of the current=y km\hr

According to the condition,
$\frac{8}{x+y}=\frac{40}{60}=\frac{2}{3}$

$=x+y=\frac{8 \times 3}{2}$

=12...(i)

$\frac{8}{x-y}=1$

x-y=8...(ii)

After adding we get
2x=20
x=10

and subtracting
2y=4
y=2

Hence the speed of the sailor
=10 km/hr

Question 29

The boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream, in $6\frac{1}{2}$ hours. Find the speed of the boat in still water and also the speed of the stream.
Sol :
In 6 hour,
Upstream distance=24 km
Downstream distance=28 km
in $6 \frac{1}{2}$ her

Upstream distance=30 km
Downstream distance=31 km

Let the speed f boat=x km\hr
and speed of stream= y km\hr

According to the conditions
$\frac{24}{x-y}+\frac{28}{x+y}=6$

$\frac{30}{x-y}+\frac{21}{x+y}=\frac{13}{2}$

Let x+y=a
and x-y=b

then $\frac{24}{b}+\frac{28}{a}=6$...(i)

$\frac{30}{b}+\frac{21}{a}=\frac{13}{2}$

Multiplying (i) by 3 and (ii) by 4
$\frac{72}{b}+\frac{84}{a}=18$

$\frac{120}{b}+\frac{84}{a}=26$

After subtracting we get

$\frac{-48}{b}=-8$

=-48=-8b

$=b=\frac{-48}{-8}$
=6

Subtracting the value of b in (i)
$\frac{24}{6}+\frac{28}{a}=6$
$=4+\frac{28}{a}=6$
$=\frac{28}{a}=6-4=2$
$=a=\frac{28}{2}=14$

Hence , 
$\begin{aligned}x+y&=14 \\x-y&=6\\ \hline 2 x&=20 \end{aligned}$

After adding we get
$=x=\frac{20}{2}=10$

and after subtracting we get
2y=8
$=y=\frac{8}{2}=4$

Speed of boat=10 km/hr
Speed of stream=4 km/hr

Question 30

Satish and Ashok start at the same time from two places 21 km apart. If they walk in the same direction Satish overtakes Ashok in the 21 hours but if they walk in opposite directions, they meet in 3 hours. Find the rate at which each of them walks.

Sol :
Distance between two places=21 km













Let speed of A=x km\hr
Speed of B=y km\hr
x>y

First case:
They meet in 21 hours and 

Second case:
They meet in 3 hours

In the first case
The distance travelled by A
=x×21 km

Hence, 2x-21y=21
=x-y=1...(i)

In second case
The distance travelled by A=3x km
and the distance travelled by B=3y km

3x+3y=21
x+y=7...(ii)

After adding (i) and (ii) , we get
2x=8
$=x=\frac{8}{2}$
=4

After subtracting (i) from (ii)
2y=6
$=y=\frac{6}{2}$=3

Now ,Speed of A=4 km\hr
and Speed of B=3 km\hr

Question 31

A boy walks to a picnic spot from his house in 6 hours, but he can travel the same distance on his cycle in 2 hours. If his average cycling speed is 7 kilometers per hour faster than his average walking speed, find his average walking speed and his average cycling speed.
Sol :
Time taken while walking=6 hours
and Time taken while cycling =2 hours

Now, the average speed of cycling is greater than by walking=7 km\hr

Now, let the speed of walking =x km\hr

Speed of cycling=y km\hr

y-x=7
y=x+7

Now, the distance covered by walking =x×6 km
and the distance covered by cycling=y×2 km

6x=2y
$y=\frac{6}{2} x$
=3x....(ii)

From (i) and (ii)
x+7=3x
=3x-x=7

2x=7
$=x=\frac{7}{2}$=3.5

∴y=3x
=3×3.5=10.5

The speed of a man by walking=3.5 km\hr
and the speed of man by cycling=10.5 km\hr

Question 32

A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.
Sol :
Total distance covered=600 km
Let the speed of train=x km\hr
The speed of car=y km\hr

First case:
The distance travelled by train=400 km

The distance travelled by car=600-200=400 km

$=\frac{400}{x}+\frac{200}{y}$
$=6 \frac{1}{2}$ hours

$=\frac{400}{x}+\frac{200}{y}=\frac{13}{2}$..(i)

and 
$\frac{200}{x}+\frac{400}{y}=7$...(ii)

Now , Multiplying (i) by 2, (ii) by 1, we get

$=\frac{800}{x}+\frac{400}{y}=13$
$=\frac{200}{x}+\frac{400}{y}=7$

After subtracting we get

$\frac{600}{x}$
=6
=x$=\frac{600}{6}$
=100

From (i)

$\frac{400}{100}+\frac{200}{y}=\frac{13}{2}$
$=4+\frac{200}{y}=\frac{13}{2}$
$=\frac{200}{y}$
$=\frac{13}{2}-4$
$=\frac{5}{2}$
$=y=\frac{200 \times 2}{5}$
=80

The speed of train=100 km\hr 
and the speed of car=80 km\hr

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