ML Aggarwal Solution Class 9 Chapter 5 Simultaneous Linear Equations Exercise 5.2

 Exercise 5.2

Question 1

(i)

⇒3x+4y=10..(1)

⇒2x-2y=2...(2)

⇒$x=\frac{2+2y}{2}$

⇒$3\left(\frac{2+2y}{2}\right)+4y=10$

⇒$\frac{6+6y}{2}+4y=10$

⇒6+6y+8y=20

⇒6+14y=20

⇒14y=20-6

⇒14y=14

⇒y=1

⇒3x+4y=10

⇒3x=10-4

⇒3x=6

⇒$x=\frac{6}{3}$

⇒x=2

∴x=2 ; y=1

(ii)

⇒2x=5y+4

⇒3x-2y+16=0

⇒$x=\frac{5y+4}{2}$

⇒$3\left(\frac{5y+4}{2}\right)-2y+16=0$

⇒$\frac{15y+12}{2}-2y+16=0$

⇒15y+12-4y+32=0

⇒11y+44=0

⇒11y=-44

⇒$y=\frac{-44}{11}=-4$

⇒y=-4

⇒2x=5(-4)+4

⇒2x=-20+4

⇒2x=-16

⇒$x=\frac{-16}{2}$

⇒x=-8

∴x=-8 ; y=-4

Question 2

(i) $\frac{3}{4}x-\frac{2}{3}y=1$

⇒$\frac{3}{8}x-\frac{1}{6}y=1$

⇒$\frac{3}{8}x=1+\frac{1}{6}y$

⇒$x=\left(1+\frac{1}{6}y\right)\frac{8}{3}$

∴$\frac{3}{4}\left(1+\frac{1}{6}y\right)\frac{8}{3}-\frac{2}{3}y=1$

⇒$2+\frac{2}{6}y-\frac{2}{3}y=1$

⇒$2-1=\frac{2}{3}y-\frac{1}{3}y$

⇒$1=\frac{2-1}{3}.y$

⇒$1=\frac{1}{3}y$

∴y=3

∴$\frac{3}{4}x-\frac{2}{3}(3)=1$

⇒$\frac{3}{4}x=1+2$

⇒$\frac{3}{4}x=3$

⇒$x=\frac{3\times 4}{3}$

⇒x=4

(ii) 2x-3y-3=0

⇒$\frac{2}{3}x+4y+\frac{1}{2}=0$

⇒2x-3y-3=0

⇒2x=3y+3

⇒$x=\frac{3y+3}{2}$

∴$\frac{2}{3}\left(\frac{3y+3}{2}\right)+4y+\frac{1}{2}=0$

⇒$\frac{2}{3}\times \frac{3}{2}(y+1)+4y+\frac{1}{2}=0$

⇒y+1+4y$+\frac{1}{2}$=0

⇒$5y+\frac{3}{2}=0$

⇒$5y=\frac{-3}{2}$

⇒$y=\frac{-3}{10}$

∴2x-3y-3=0

⇒$2x-3\left(\frac{-3}{10}\right)-3=0$

⇒$2x+\frac{9}{10}-3=0$

⇒$2x=3-\frac{9}{10}$

⇒$2x=\frac{30-9}{10}$

⇒$x=\frac{21}{20}$

Question 3

(i) 15x-14y=117

⇒14x-15y=115

⇒14x=115+15y

⇒$x=\frac{115+15y}{14}$

∴$15\left(\frac{115+15y}{14}\right)-14y=117$

⇒1725+225y-196y=117×14

⇒1725+225y-196y=1638

⇒29y=1638-1725

⇒29y=-87

⇒$y=\frac{-87}{29}$

⇒y=-3

∴15x-14(-3)=117

⇒15x+42=117

⇒15x=117-42

⇒15x=75

⇒$x=\frac{75}{15}$

⇒x=5

(ii) 41x+53y=135

⇒53x+41y=147

⇒53x=147-41y

⇒$x=\frac{147-41y}{53}$

⇒41x+53y=135

⇒$41\left(\frac{147-41y}{53}\right)+53y=135$

⇒6027-1681y+2809y=135×53

⇒6027-1681y+2309y=7155

⇒1128y=1128

⇒$y=\frac{1128}{1128}$

⇒y=1

∴41x+53y=135

⇒41x+53(1)=135

⇒41x=135-53

⇒41x=82

⇒$x=\frac{82}{41}$

⇒x=2

Question 4

(i) $\frac{x}{6}=y-6$

⇒$\frac{3x}{4}=1+y$

⇒$y=\frac{3x}{4}-1$

∴$\frac{x}{6}=\frac{3x}{4}-1-6$

⇒$\frac{x}{6}=\frac{3x}{4}-7$

⇒$\frac{x}{6}=\frac{3x-28}{4}$

⇒$\frac{x}{3}=\frac{3x-28}{2}$

⇒2x=9x-84

⇒9x-2x=84

⇒7x=84

⇒$x=\frac{84}{7}$

⇒x=12

∴$\frac{x}{6}=y-6$

⇒$\frac{12}{6}=y-6$

⇒y=2+6=8

(ii) $x-\frac{2}{3}y=\frac{8}{3}$

⇒$\frac{2x}{5}-y=\frac{7}{5}$

⇒$y=\frac{2x}{5}-\frac{7}{5}$

⇒$y=\frac{2x-7}{5}$

∴$x-\frac{2}{3}\left(\frac{2x-7}{5}\right)=\frac{8}{3}$

⇒$x-\frac{4x-14}{15}=\frac{8}{3}$

⇒$\frac{15x-4x+14}{15}=\frac{8}{3}$

⇒15x-4x+14=8×5

⇒11x+14=40

⇒11x=40-14

⇒11x=26

⇒$x=\frac{26}{11}$

∴$\frac{26}{11}-\frac{2}{3}y=\frac{8}{3}$

⇒$\frac{2}{3}y=\frac{26}{11}-\frac{8}{3}$

⇒$\frac{2}{3}y=\frac{78-88}{33}$

⇒$y=\frac{10}{33}\times \frac{3}{2}=\frac{5}{11}$

Question 5

(i) 9-(x-4)=y+7

⇒2(x+y)=4-3y

⇒9-x+4=y+7

⇒13-x=y+7

⇒y=13-x-7

⇒y=6-x

∴2(x+6-x)=4-3(6-x)

⇒2x+12-2x=4-18+3x

⇒12=-14+3x

⇒3x=12+14

⇒3x=26

⇒$x=\frac{26}{3}$

∴y=6-x

⇒$y=6-\frac{26}{3}$

⇒$y=\frac{-8}{3}$

(ii) $2x+\frac{x-y}{6}=2$

⇒$x-\frac{2x+y}{3}=1$

⇒$\frac{3x-2x-y}{3}=1$

⇒x-y=3

⇒x=3+y

∴2(3+y)$+\frac{3+y-y}{6}=2$

⇒$6+2y+\frac{3}{6}=2$

⇒$6+2y+\frac{1}{2}=2$

⇒$2y=2-\frac{1}{2}-6$

⇒$2y=\frac{4-1-12}{2}$

⇒$2y=\frac{-9}{2}$

∴x=3+y

⇒$x=3-\frac{9}{4}$

⇒$x=\frac{12-9}{4}$

⇒$x=\frac{3}{4}$

Question 6

(i) $4x+\frac{x-y}{8}=17$

⇒$\frac{32x+x-y}{8}=17$

⇒38x-y=136...(i)

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

⇒$\frac{6y+3x-5y-2}{3}=2$

⇒y+3x-2=6

⇒y+3x=6+2

⇒y+3x=8

⇒y=8-3x...(ii)

Equation (ii) in equation (i)

∴33x-8+3x=136

⇒36x=136+8

⇒$x=\frac{144}{36}$

⇒x=4

∴y=8-3(4)

=8-12=-4

(ii) Given : x-3y=3x-1=2x-y

∴x-3y=3x-1

⇒3x-x=1-3y

⇒2x=1-3y

⇒2x+3y=1...(i)

∴3x-1=2x-y

⇒3x-2x=1+y

⇒x=1+y...(ii)

Equation (ii) in equation (i)

∴2(1+y)+3y=1

⇒2+2y+3y=1

⇒5y=1-2

⇒5y=-1

⇒$y=\frac{-1}{5}$

∴x=y+1

⇒$x=\frac{-1}{5}+1$

⇒$x=\frac{-1+5}{5}$

⇒$x=\frac{4}{5}$

Question 7

Given : $\frac{3}{x}+4y=7$

⇒$\frac{5}{x}+6y=13$

⇒$\frac{5}{x}=13-6y$

⇒$6y=13-\frac{15}{x}$

⇒$6y=\frac{13x-5}{x}$

⇒$y=\frac{13x-5}{6x}$

∴$\frac{3}{x}+4y=7$

⇒$\frac{3}{x}+4\left(\frac{13x-5}{6x}\right)=7$

⇒$\frac{3}{x}+\frac{52x-20}{6x}=7$

⇒$\frac{18+52x-20}{6x}=7$

⇒52x-2=42x

⇒52x-42x=2

⇒10x=2

⇒$x=\frac{2}{10}$

⇒$x=\frac{1}{5}$

⇒$y=\frac{13\left(\frac{1}{5}\right)-5}{6\left(\frac{1}{5}\right)}$

⇒$y=\frac{\frac{13}{5}-5}{\frac{6}{5}}$

⇒$y=\frac{\frac{13-25}{5}}{\frac{6}{5}}$

⇒$y=\frac{-12}{6}$

⇒y=-2

(ii) $5 x-9=\frac{1}{y}$

⇒$x+\frac{1}{y}=3$

⇒$x+\frac{1}{y}=3$

⇒$x=3-\frac{1}{y}$

⇒$x=3-\frac{1}{y}$

⇒$5\left(\frac{3 y-1}{y}\right)-9=\frac{1}{y}$

⇒$\frac{15 y-5}{y}-9=\frac{1}{y}$

⇒$\frac{15 y-5-9 y}{y}=\frac{1}{y}$

⇒6y-5=1

⇒6y=1+5

⇒6y=6

⇒$y=\frac{6}{6}$

⇒y=1

⇒$x=\frac{3 y-1}{y}$

⇒$x=\frac{3-1}{1}$

⇒x=2

Question 8

(i) px+qy=p-q

⇒qx-py=p+q

⇒qx=p+q+py

⇒$x=\frac{q+p(1+y)}{q}$

⇒$p\left(\frac{q+p(1+y)}{q}\right)+q y=p-q$

⇒$\frac{p q+p^{2} (1+y)}{q}+qy=p-q$

⇒$\frac{pq+p^{2}(1+y)}{q}+q^{2} y=p-q$

⇒pq+p²(1+y)+q²y=pq-q²

⇒p²(1+y)=-q²-q²y

⇒p²(1+y)=-q²(1+y)

⇒p²+p²y=-q²-q²y

⇒p²y+q²y=-q²-p²

⇒y(p²+q²)=-(p²+q²)

⇒y=-1

⇒∴$x=\frac{q+p(-1+y)}{q}$

⇒$x=\frac{q+p(1-1)}{q}$

⇒$x=\frac{q+p(0)}{q}$

⇒$x=\frac{q+0}{q}$

⇒$x=\frac{q}{q}$

⇒x=1

⇒∴x=1

⇒y=-1

(ii) $\frac{x}{a}-\frac{y}{b}=0$

⇒ax+by=a²+b²

⇒$\frac{x}{a}=\frac{y}{b}$

⇒$x=\frac{a}{b} \cdot y$

⇒$a \cdot \frac{a}{b} \cdot y+b y=a^{2}+b^{2}$

⇒$\frac{a^{2}}{b} y+b y=a^{2}+b^{2}$

⇒$\frac{a^{2} y+b^{2} y}{b}=a^{2}+b^{2}$

⇒$\left(a^{2}+b^{2}\right) y=\left(a^{2}+b^{2}\right) b$

⇒y=b

⇒∴$x=\frac{a}{b} \cdot b .$

⇒∴x=a

Question 9

Given 2x+y=23

⇒4x-y=19

⇒y=23-2x

⇒∴4x-23+2x=19

⇒6x=19+23

⇒6x=42

⇒$x=\frac{42}{6}$

⇒x=7

⇒∴y=23-2x

⇒y=23-2(7)

⇒y=23-14

⇒y=9

⇒∴x-3y⇒7-3(9)

=7-27=-20

∴x-3y=-20

⇒5y-2x⇒5(9)-2(7)

⇒45-14

⇒31

∴5y-2x=31

Question 10

Given expression ax+by

(i) ax+by=7

When x=2 ; y=1

∴2a+b=7..(i)

(ii) ax+by=1

When x=-1 ; y=1

⇒-a+b=1..(iii)

∴b=1+a

⇒2a+1+a=7

⇒3a+1=7

⇒3a=7-1

⇒3a=6

⇒$a=\frac{6}{3}$

⇒a=2

∴b=1+a

b=1+2=3

Question 6

Let the number be xy[10x+y]

Reverse of that number yx[10y+x]

Answer : Given that $\frac{x y[10 x+1]}{y x[10 y+x]}=1 \frac{3}{4}$

⇒$\frac{101+y}{10 y+x}=\frac{7}{4}$

⇒40x+4y=70y+71

⇒33x-66y=0

⇒x-2y=0...(i)

Ans also given that x+y=12...(2)

On solving equation (1) and equation (2)

eq(2)-eq(1)

⇒x+y-x+2y=12

⇒3y=2

⇒y=4

Putting y=4 in (1)

⇒x-8=0

⇒x=8

The required number is 84

Question 7

Let the number be xy[10x+y] and reverse of number is yx[10y+x]

Given : $\frac{10 x+y}{10 y+x}=\frac{5}{6}$

⇒60x+6y=50y+5x

⇒55x-44y=0

⇒5x-4y=0...(1)

And also given that x-y=1..(2)

Put x=y+1 in equation (1)

⇒5y+5-4y=0

⇒y=-5

⇒Put y-5 in equation (2)

⇒x=-4

∴The required number is 45

Question 8

Let the number be xyz[100x+10y+z]

Given that x=4z...(1)

x+y+z=14...(2)

And reverse of the number is zyx[100z+10y+x]

And given that (100x+10y+z)-(100z+10y+x)=594

⇒99x-99z=594

⇒99[x-z]=594

⇒x-z=6..(3)

On solving eq(1) and eq(3)

⇒4z-z=6

⇒3z=6

⇒z=2

Putting z=2 in eq(3) 

⇒x-2=6

⇒x=8

Putting x=8 and z=2 in eq(2)

⇒8+y+2=14

⇒y+4

∴The required number is 842

Question 9

Let the age of marina and her daughter be 'M' and 'D'

Given that

M-4=3(D-4)..(1)

M+6=2(D+6)..(2)

eq(1)⇒M-4=3D-12

⇒M=3D-8..(3)

eq(2)⇒M+6=2D+12

⇒M=2D+6..(4)

From (3) and (4) 

⇒3D-8=2D+6

⇒D=14

Putting D=14 in eq(4)

⇒M=28+6=34

⇒M=34

∴The present age of marina and her daughter is 34 and 14

eq(1)×12⇒144x+120y=1560000

eq(2)×10⇒100x+120y=1340000

Solving above equations 

⇒44x=220000

⇒x=5000

Putting x=5000 in eq (1)

⇒60000+10y=130000

⇒10y=70000

⇒y=7000

Thus the money inverted at 12% is 5000

10% is 7000

Question 22

Let the cost price of table be "x" and list price of their be "y"

Case (i)

Table is sold at a profit of 8%

∴S.P of table $x+\frac{8x}{100}=\frac{108x}{100}$

Chair is sold at a discount of 10%

∴S.P of chair $=y-\frac{10 y}{100}=\frac{90y}{100}$

And given that $\frac{108x}{100}+\frac{90 y}{100}=1008$

⇒6x+5y=5600..(1)

Case-(ii)

Table is sold at a profit of 10%

∴S.P of table $x+\frac{10 x}{100}=\frac{110y}{100}$

Chair is sold at a discount of 8%

∴S.P of chair $y-\frac{8 y}{100}=\frac{92y}{100}$

Given that 

$\frac{110 x}{100}+\frac{92y}{100}=1028$

⇒110x+92y=102800...(2)

On solving eq-(1)+eq-(2)

x=600 , y=400

∴Cost price of table is 600 and list price of chair is 400

Let the money has by A and B is "x" and "y" respectively

Question 23

Case-(i)

Given that 

⇒x+100=75%(y-100)

⇒$x+100=\frac{375}{100}(y-100)$

⇒4x+400=3y-300

⇒4x-3y=-700..(1)

Case-(ii)

⇒x-100=40%(y+100)

⇒$x-100=\frac{40}{100}(y+100)$

⇒5x-500=2y+200

⇒5x-2y=700...(2)

eq-(1)×2⇒8x-6y=-1400

eq-(2)×3⇒15x-6y=2100

Subtracting above equations , we get

⇒7x=3500

⇒x=500

Putting x=500 in eq-(2) 

⇒2500-2y=700

⇒2y=1800

⇒y=900

∴A and B have 500 and 900 respectively

Question 24

Let the number of rows be 'x'

Let the number of students in one row be 'y'

∴Total number of students=xy

Given (i) If one student is extra in a row , there will be 2 rows less 

⇒(y+1)(x-2)=xy

⇒xy-2y+x-2=xy

⇒x-2y=2..(1)

(ii) If one student is less in a row there will be 3 rows more

⇒(y-1)(x+3)=xy

⇒xy+3y-x-3=xy

⇒x-3y=-3..(2)

Question 25

Let "x" grams of 18 carat gold is added, thus the amount of 12 carat gold added is (120-x) grams

Given purity of gold is 24 carat

⇒$x \cdot \frac{18}{24}+(120-x) \cdot \frac{12}{24}=120 \times \frac{16}{24}$

⇒187+12(120-x)=$120\times \frac{16}{24}$

⇒187+12(120-x)=120×16

⇒189x+12×120-12×x=120×16

⇒6x=120(16-12)

⇒$x=\frac{120 \times 4}{6}-80$

∴80 grams of 18 carat gold added with 120-80=40 grams of 12 carat gold

Question 26

Given A and B both can do work in 15 days

(A+B) one day work $=\frac{1}{15}$

⇒$\frac{A\text {'s 1 days's work}}{B\text {'s 1 days's work}}=\frac{\frac{3}{2}}{1}=\frac{3}{2}$

Let A's 1 day's work be 3x and B's 1 day work is 2x

Then $3 x+2 x=\frac{1}{15}$

⇒$5 x=\frac{1}{15}$

⇒$x=\frac{1}{75}$

∴A's 1 day work $=3 \times \frac{1}{75}=\frac{1}{25}$

B's 1 day work $=2 \times \frac{1}{75}=\frac{1}{37.5}$

∴A and B can do that work in 25 and 37.5 days respectively

Question 27

Let a man's rate be "m"

Let a man's rate be "w"

⇒Given $2m+5w=\frac{1}{4}$

⇒$m+w=\frac{1}{12}$

⇒8m+20w=1..(1)

⇒12m+12w=1...(2)

eq-(1)×3⇒24m+60w=3

eq-(2)×2⇒24m+24w=2

Solving above equation 

⇒36w=1

⇒$w=\frac{1}{36}$

⇒$w=\frac{1}{36}$ in (2)

⇒$m=\frac{1}{18}$

1 Man would take 18 days to complete the work

Question 28

Let the due speed of train be "x" km/h and scheduled time be y km/h

Therefore , Length of the journey=xy km

Given 

(i) (x+30)(y-2)=xy

⇒xy-2x+30y-60=xy

⇒2x-30y+60=0..(1)

(ii) (x-16)(y+2)=xy

⇒xy+2x-15y-30=xy

⇒2x-5y-30=0...(2)

eq-(2)-eq(1)

⇒15y-90=0

⇒15y=90

⇒y=6 hours

Putting y=6 in (1) 

⇒2x-180+60=0

⇒2x=120

⇒x=60 km/h

∴The length of the journey is 60×6=360 km

Question 29

Let speed of boat in still water be "x" km/h

Speed of current be "y" km/h

Time to go with the current is 2 hours

$Time=\frac{\text{Distance}}{\text{Speed}}$

⇒$\frac{40}{x+y}=2$

⇒x+y=20..(1)

Time to go against the current is 4 hours 

⇒$\frac{40}{x-y}=4$

⇒x-y=10..(2)

eq-(1)+eq-(2)

⇒2x=30

x=15

Putting x=15 in (1)

⇒15+y=20

⇒y=5

⇒Speed of boat in still water and speed of current is 15 km/h and 5 km/h respectively

Question 30

Let the speed of boat in still water be "x" km/h

Speed of current be "y" km/h

Time to go with the current is 4 hours

⇒$\frac{44}{x+y}=4$

⇒x+y=11..(1)

Time to go against the current is 4 hours 48 mins

=288mins

$=\frac{24}{5}$ hours

⇒$\frac{44}{x-y}=\frac{24}{5}$

⇒6x-6y=55..(2)

eq-(2)+eq-(1)×6

⇒12x=121

⇒$x=\frac{121}{12}$ km/h [Speed of boat in still water]

Putting $x=\frac{121}{12}$ in (1) 

⇒$y=11-\frac{121}{12}$

⇒$y=\frac{11}{12}$ km/h [Speed of current]

Question 31

Let the plane air speed be "x" km/h and wind speed be "y" km/h

And given that with a head wind it took 3.5 hours

⇒$\frac{1650}{x-y}=3 \cdot 5= \frac{7}{2}$

⇒x-y=480..(1)

On return it took 3 hours

⇒$\frac{1650}{x+y}=3$

⇒x+y=560..(2)

eq-(1)+eq-(2)

⇒2x=1040

⇒x=520

Putting x=520 in (2)

⇒520+y=560

⇒y=40

∴Plane air speed is 520 km/h and wind speed 40 km/h

Question 32

Let the fixed charges be "x" and cost of food per day be "y"

Given that Bhawana paid 2600 for 20 days

⇒x+20y=2600..(1)

and Divya paid 3020 for 26 days

⇒x+26y=3020..(2)

eq-(2)-eq-(1)

⇒6y=420

⇒y=700

Putting y=70 in (1)

⇒x+1400=2600

⇒x=1200

∴Fixed charges 1200

⇒Cost of food per day 70