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Maths 3 Marks Question

P-1 Linear Equations in Two Variables · Maharashtra Board STD 10 (MSBSHSE - English Medium). 109 questions.

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Question 13 Marks
Solve the following simultaneous equation graphically.
5x – 6y + 30 = 0; 5x + 4y – 20 = 0
Answer
Coming soon
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Question 43 Marks
Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Answer
Let the speed of Joseph = x km/hLet the speed of Hamid be = y km/h
When approaching each other, combined speed = (x+y) Km/h
Time taken to meet = $\frac{30}{x+y} = \frac{1}{3}(20\ mins)$
$\therefore x + y = 90 \dots(I)$
When moving away from each other, combined speed = (x-y) Km/h
Time taken for Hamid to catch up = $\frac{30}{x-y} = 3$
$\therefore x - y = 10 \dots(II)$
Equating I and II,
$x+y=90 $
$x-y=10 $
$2 x=100 $
$x=\frac{100}{2}=50$
Substituting $x=50$ in eq. I
$50+y=90 $
$y=90-50 $
$y=40$
Hamid's speed 50 km/hr.
Joseph's speed 40 km/hr.
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Question 53 Marks
In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Answer
Ratio of skilled and unskilled worker’s salary = 5:3Let it be 5x and 3x
Total of one day’s salary = ₹720
So, $5 x +3 x =720$
$8 x=720 $
$x=\frac{720}{8}$
$x=90$
killed worker's wages $=5 x =5 \times 90=₹ 450$.
unskilled worker's wages $3 x =3 \times 90=₹ 270$
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Question 63 Marks
Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Answer
Let Manish’s present age be xLet Savita’s present age be y
According to $1^{st}$ situation,
$x+y=31 \ldots . .( I )$
According to second situation,
$x-3=4(y-3)$
$x-3=4 y-12 $
$x-4 y=-12+3 $
$x-4 y=-9 \ldots . (II)$
Subtracting Eq. II from I
$x+y=31$
$-x+4 y=9 $
$5 y=40 $
$y=\frac{40}{5} $
$y=8$
Substitute $y =8$ in eq. I
$x+8=31 $
$x=31-8 $
$x=23$
Manisha's age 23 years
Savita's age 8 years.
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Question 73 Marks
To find number of notes that Anushka had, complete the following activity
Answer
According to $1^{\text {st }}$ situation
$100 x+50 y=2500\dots(I)$
According to $2^{\text {nd }}$ situation,
$50 x+100 y=2000\dots(II)$
Adding I and II,
$150 x+150 y=4500$
$x+y=30 \ldots(III)$
Subtracting I from II
$50 x-50 y=-500$
$x-y=-10 \ldots (IV)$
Equating Eq. III with Eq. IV
$x+y=30 $
$x-y=-1 $
$2 x=20 $
$x=\frac{20}{2}=10$
File succe
Substituting $x=10$ in Eq. III
$10+y=30$
$y=20 $
$₹ 100 \text { notes }=10$
$₹ 50 \text { notes }=20$
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Question 83 Marks
Kantabai bought $1\frac{1}{2}$ kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare for rickshaw. Total expense was Rs 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs 880 for that. Find the rate of sugar and tea per kg.
Answer
Let x be the cost of tea and y be the cost of sugarAs she paid ₹50 as return fare
$₹ 700-₹ 50=₹ 650 $
$\therefore \frac{3}{2} x+5 y=650 \Rightarrow 3 x+10 y=1300 \ldots \ldots$
According to second situation,
$2 x+7 y=880 \ldots \ldots( II )$
Multiplying Eq. I by 2 and Eq. II by 3
$6 x+20 y=2600 $
$6 x+21 y=2640 \ldots $
$\text { Subtracting Eq. III from IV } $
$6 x+21 y=2640 $
$-6 x-20 y=-2600 $
$y=40$
Substituting $y=40$ in Eq. I
$3 x+10 \times 40=1300 $
$3 x+400=1300 $
$3 x=1300-400 $
$3 x=900 $
$x=\frac{900}{3} $
$x=300$
Tea; ₹ 300 per kg.
Sugar ; ₹ 40 per kg.
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Question 93 Marks
Solve the following simultaneous equations.
$\frac{1}{2(3 x+4 y)}+\frac{1}{5(2 x-3 y)}=\frac{1}{4} ; \frac{5}{(3 x+4 y)}-\frac{2}{(2 x-3 y)}=-\frac{3}{2}$
Answer
$\text { Let } \frac{1}{3 x+4 y}=m \text { and } \frac{1}{2 x-3 y}=n $
$\frac{1}{2} m+\frac{1}{5} n=\frac{1}{4} \Rightarrow 5 m+2 n=\frac{10}{4} \Rightarrow 20 m+8 n=10 \Rightarrow 10 m+4 n=5 \dots(I)$
$5 m-2 n=-\frac{3}{2} \Rightarrow 10 m-4 n=-3 \ldots \text { (II) }$
Equating Eq. I and II
$10 m+4 n=5$
$\underline {10 m-4 n=-3} $
$20 m=2 $
$m=\frac{2}{20} $
$m=\frac{1}{10}$
Substituting $m =\frac{1}{10}$ in Eq. I
$10 \times \frac{1}{10}+4 n =5 $
$1+4 n =5 $
$4 n =5-1$
$4 n =4 $
$n =\frac{4}{4} $
$n =1 $
$\therefore \frac{1}{3 x+4 y}= m \Rightarrow \frac{1}{3 x+4 y}=\frac{1}{10} \Rightarrow 3 x +4 y =10 \ldots (III)$
$\therefore \frac{1}{2 x -3 y }= n \Rightarrow \frac{1}{2 x -3 y }=1 \Rightarrow 2 x -3 y =1 \ldots {IV}$
Multiply Eq. III by 3 and Eq. IV by 4 and Equate
$9 x+12 y=30 \ldots(V) $
$\underline{8 x-12 y=4 \ldots(V I)} $
$17 x=34 $
$x=\frac{34}{17}$
$x=2$
Substituting $x =2$ in Eq. $V$
$9 \times 2+12 y =30 $
$18+12 y =30 $
$12 y =30-18 $
$12 y =12$
$y =\frac{12}{12}$
$y =1$
Hence, $(x, y)=(2,1)$
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Question 103 Marks
Solve the following simultaneous equations.
$\frac{7 x-2 y}{x y}=5 ; \frac{8 x+7 y}{x y}=15$
Answer
$\frac{7 x-2 y}{x y}=5 \Rightarrow \frac{7 x}{x y}-\frac{2 y}{x y}=5 \Rightarrow \frac{7}{y}-\frac{2}{x}=5 \ldots \text { (I) } $
$\frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15 \Rightarrow \frac{8}{y}+\frac{7}{x}=15 \ldots (II)$
$\frac{1}{x}=m \text { and } \frac{1}{y}=n $
$7 n-2 m=5 \ldots \text { (III) }$
$8 n+7 m=15 \ldots(\text { IV) }$
Multiply Eq. 1 by 7 and Eq.Il by 2
$49 n -14 m=35 \ldots( V ) 16 n +14 m=30 \ldots( VI ) $
$65 n =65 $
$n =\frac{65}{65} $
$n =1$
Substituting value in Eq.VI
$16 \times 1+14 m=30$
$14 m=30-16 $
$14 m=14$
$m=\frac{14}{14}$
$m=1$
$\therefore \frac{1}{x}= m \Rightarrow \frac{1}{x}=1 \Rightarrow x =1 $
$\frac{1}{ y }= n \Rightarrow \frac{1}{y}=1 \Rightarrow y =1$
Hence, $(x, y)=(1,1)$
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Question 113 Marks
Solve the following simultaneous equations.
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} ; \frac{231}{x}+\frac{148}{y}=\frac{610}{x y}$
Answer
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} \Rightarrow \frac{148 y+231 x}{x y}=\frac{527}{x y} \Rightarrow 231 x+148 y=527 \ldots(I) $
$\frac{231}{x}+\frac{148}{y}=\frac{610}{x y} \Rightarrow \frac{231 y+148 x}{x y}=\frac{610}{x y} \Rightarrow 148 x+231 y=610 \ldots(II)$
Adding Eq. I and II
$379 x+379 y=1137$
$x+y=3\dots(III)$
Subtracting Eq. I and II
$83 x-83 y=-83 $
$x-y=-1 \ldots( IV )$
Equating I and II
$x+y=3 $
$x-y=-1 $
$2 x=2 $
$x=\frac{2}{2} $
$x=1$
Substituting $x=1$ in Eq. I
$1+y=3 $
$y=3-1 $
$y=2$
Hence, $(x, y)=(1,2)$
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Question 123 Marks
Solve the following simultaneous equations.
$\frac{7}{2 x+1}+\frac{13}{y+2}=27 ; \frac{13}{2 x+1}+\frac{7}{y+2}=33$
Answer
Let $\frac{1}{2 x+1}=m$ and $\frac{1}{y+2}=n 7 m+13 n=27 \ldots(I)$
$13 m+7 n=33\dots(II)$
Adding Eq. I and II
$20 m+20 n=60 \Rightarrow m+n=3 \ldots \text { (III) }$
Subtract Eq. I and II
$-6 m+6 n=-6 \Rightarrow-m+n=-1 \ldots(IV)$
Equating Eq. III and IV
$m + n =3 $
$\underline{ - m + n =-1} $
$2 n =2 $
$n =1$
Substituting $n=1$ in Eq. III
$m +1=3 $
$m=3-1 $
$m=2 $
$\therefore \frac{1}{2 x +1}= m \Rightarrow \frac{1}{2 x +1}=2 \Rightarrow 2(2 x +1)=1 \Rightarrow 4 x +2=1 \Rightarrow 4 x =1-2 $
$\Rightarrow 4 x =-1 \Rightarrow x =-\frac{1}{4} $
$\therefore \frac{1}{ y +2}= n \Rightarrow \frac{1}{ y +2}=1 \Rightarrow y +2=1 \Rightarrow y =1-2 \Rightarrow y =-1 $
$\text { Hence, }( x , y )=\left(-\frac{1}{4},-1\right)$
 
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Question 133 Marks
Solve the following simultaneous equations.
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
Answer
Let $\frac{1}{ x }= m$ and $\frac{1}{ y }= n _{2 m}+\frac{2}{3} n =\frac{1}{6} \Rightarrow 12 m+\frac{12}{3 n }=1 \Rightarrow 12 m+4 n =1 \ldots (I)$
$3 m+2 n =0 \ldots (II)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots (III)$
Subtract Eq.III from Eq. I
$12 m+4 n=1$
$\underline{-6 m-4 n=0} $
$6 m=1$
$m=\frac{1}{6}$
Substitute $m =1 / 6$ in Eq. I
$12 \times \frac{1}{6}+4 n =1 $
$2+4 n =1$
$4 n =1-2 $
$4 n =-1 $
$n =-\frac{1}{4} $
$\therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 $
$\therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4$
Hence, $(x, y)=(6,-4)$
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Question 143 Marks
Solve the following equations by Cramer’s method.
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
Answer
Let,
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3} $
$\Rightarrow 3 x+3 y-24=2 x+4 y-28$
$\Rightarrow x-y=-4 \ldots(1)$
Also,
Let $\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
$\Rightarrow 4 x+8 y-56=9 x-3 y$
$\Rightarrow 5 x-11 y=-56 \ldots(2)$
Hence the two equations are:
$x-y=-4 \ldots(1) $
$5 x-11 y=-56$
Now,
$D=\left|\begin{array}{cc}1 & -1 \\ 5 & -11\end{array}\right| $
$\Rightarrow D=(-11-(-5))=-6$
Also,
$ D _{ x }=\left|\begin{array}{rr} -4 & -1 \\-56 & -11\end{array}\right| $
$\text { Also, }$
$D _{ x }=44-56=-12$
And,
$D _{ y }=\left|\begin{array}{cc} 1 & -4 \\ 5 & -56 \end{array}\right| $
$\Rightarrow D _{ y }=-56+20=-36$
now,  $ x=\frac{D_x}{D}=\frac{-12}{-6}=2 $
And, $y=\frac{D_y}{D}=\frac{-36}{-6}=6$
Hence, $(2,6)$ is the solution
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Question 153 Marks
Solve the following equations by Cramer’s method.
7x + 3y = 15; 12y – 5x = 39
Answer

$D=\left[\begin{array}{cc}7 & 3 \\ -5 & 12\end{array}\right]=(7 \times 12)-(3 \times-5)=84+15=99$
$D_x=\left[\begin{array}{cc}15 & 3 \\ 39 & 12\end{array}\right]=(15 \times 12)-(3 \times 39)=180-117=63$
$ D_y=\left[\begin{array}{cc}7 & 15 \\ -5 & 39\end{array}\right]=(7 \times 39)-(15 \times-5)=273+75=348 $
$x=\frac{D_x}{D}=\frac{63}{99}=\frac{7}{11} y=\frac{D_y}{D}=\frac{348}{99}=\frac{116}{33} $
$ \therefore(x, y)=\left(\frac{7}{11}, \frac{116}{33}\right)$
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Question 163 Marks
Solve the following equations by Cramer’s method.
$3 x -2 y =\frac{5}{2} ; \frac{1}{3} x +3 y =-\frac{4}{3}$
Answer

$3 x -2 y =\frac{5}{2} \Rightarrow 6 x -4 y =5 $
$ \frac{1}{3} x +3 y =-\frac{4}{3} \Rightarrow \frac{ x +9 y }{3}=-\frac{4}{3} \Rightarrow x +9 y =-4 $
$ D =\left[\begin{array}{cc}6 & -4 \\ 1 & 9\end{array}\right]=(6 \times 9)-(-4 \times 1)=54+4=58$
$ D _{ x }=\left[\begin{array}{cc}5 & -4 \\ -4 & 9\end{array}\right]=(5 \times 9)-(-4 \times-4)=45-16=29$
$ D _{ y }=\left[\begin{array}{cc}6 & 5 \\ 1 & -4\end{array}\right]=(6 \times-4)-(5 \times 1)=-24-5=-29 $
$ x=\frac{D_x}{D}=\frac{1}{2}, y=\frac{D_y}{D}=\frac{(-29)}{58}=\frac{(-1)}{2}$
$ \therefore( x , y )=(1 / 2,-1 / 2)$
 
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Question 173 Marks
Solve the following equations by Cramer’s method.
4m – 2n = –4; 4m + 3n = 16
Answer

$D=\left[\begin{array}{cc}4 & -2 \\ 4 & 3\end{array}\right]=(4 \times 3)-(-2 \times 4)=12+8=20 $
$ D_x=\left[\begin{array}{cc}-4 & -2 \\ 16 & 3\end{array}\right]=(-4 \times 3)-(-2 \times 16)=-12+32=20$
$ D_y=\left[\begin{array}{cc}4 & -4 \\ 4 & 16\end{array}\right]=(4 \times 16)-(-4 \times 4)=64+16=80 $
$ x=\frac{D_x}{D}=\frac{20}{20}=1 y=\frac{D_y}{D}=\frac{80}{20}=4 $
$ \therefore(x, y)=(1,4)$
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Question 183 Marks
Solve the following equations by Cramer’s method.
6x – 3y = –10; 3x + 5y – 8 = 0
Answer
$6x-3y = -10$
$3 x+5 y=8 $
$D=\left[\begin{array}{cc}6 & -3 \\ 3 & 5\end{array}\right]=(6 \times 5)-(-3 \times 3)=30+9=39$
$ D_x=\left[\begin{array}{cc}-10 & -3 \\ 8 & 5\end{array}\right]=(-10 \times 5)-3 \times 8=-50+24=-26 $
$ D_y=\left[\begin{array}{cc}6 & -10 \\ 3 & 8\end{array}\right]=(6 \times 8)-(-10 \times 3)=48+30=78$
$ x=\frac{D_x}{D}=\frac{-26}{39}=\frac{-2}{3} y=\frac{D_y}{D}=\frac{78}{39}=2 $
$ \therefore(x, y)=\left(-\frac{2}{3}, 2\right)$
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Question 193 Marks
Find the values of each of the following determinants.
(1) $\left[\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right]$
(2) $\left[\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right]$
(3) $\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]$
Answer
(1) $D =\left[\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right]=(4 \times 7)-(3 \times 2)=28-6=22$
(2) $D=\left[\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right]=(5 \times 1)-2 \times-3=5-6=-1$
(3) $D=\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]=(3 \times 4)-(-1 \times 1)=12+1=13$
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Question 223 Marks
Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.
Answer
Let the distance travelled by bus = xSpeed of bus = 60 km/hr
As,
$time = \frac{distance}{speed}$
Time taken travelling by bus = $\frac{x}{60}$
Let the distance traveled by plane = y
As, total distance traveled was 1900 km
x + y = 1900
Distance traveled by plane = (1900–x)
Speed of plane = 700 km/hr
Time travelling by plane = $\frac{(1900-x)}{700}$
Given,
Total time = 5 hours
$\frac{x}{60}+\frac{1900-x}{700}=5$
$\Rightarrow \frac{35 x+3(1900-x)}{2100}=5 $
$\Rightarrow \frac{35 x+5700-3 x}{2100}=5$
⇒ 32x + 5700 = 10500
⇒ 32x = 4800
⇒ x = 150 km
and y = 1900 - x
= 1900 - 150 = 1750 km
Vishal travels 150km by bus and 1750 km by plane.
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Question 233 Marks
Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Answer
A – 30kg, B – 55kLet the weight of box ‘A’ = x kg
Let the Weight of box’B’ = y kg
According to question,
150 boxes of type A and 100 boxes of type B are loaded in the truck and it weighs 10tons.
$\therefore 150 x +100 y =10000[\because 1 \text { ton }=1000 kg] $
$\Rightarrow 3 x +2 y =200 \ldots \ldots( I )$
260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, still it weighs 10tons
$\therefore 260 x +40 y =10000[\because 1 \text { ton }=1000 kg] $
$\Rightarrow 13 x+2 y=500$
Solving Equation I and II
$3 x+2 y=200 $
$-13 x-2 y=-500 $
$-10 x=-300 $
$x=\frac{300}{10} $
$x=30$
Putting $x=30$ in Eq. I
$3 \times 30+2 y =200 $
$90+2 y =200 $
$2 y =200-90 $
$2 y =110 $
$y =\frac{110}{2}=55$
Hence, A - 30kg, B - 55kg
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Question 243 Marks
The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Answer
Let the numerator and denominator of the fraction be x and y respectively.
Fraction = $\frac{x}{y}$
Given,
Denominator = 2(Numerator) + 4
$\Rightarrow y = 2x + 4$
⇒ 2x-y=(-4) …I
According to the given condition, we have
$y-6=12(x-6) $
$\Rightarrow y-6=12 x-72 $
$\Rightarrow 12 x-y=66 \ldots$
Equating Eq. I and II,
$2 x-y=-4 $
$-12 x+y=-66 $
$-10 x=-70 $
$x=\frac{70}{10} $
$x=7$
Putting $x=7$ in equation $I$, we get
$\Rightarrow 2 \times 7-y=-4 $
$\Rightarrow 14-y=-4 $
$\Rightarrow y=14+4 $
$\Rightarrow y=18$
Hence, required fraction = $\frac{7}{18}$
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Question 253 Marks
The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Answer
Suppose father’s age(in years) be x and that son’s age be y.
Then,
$x+2 y=70 $
$2 x+y=95$
Multiply Eq.I by 2 and equate
$\begin{array}{c} 2 x+4 y=140 \\ -2 x-y=-95 \\ \hline 3 y=45 \end{array}$
$y=\frac{45}{3}$
$y=15$
Substituting $y=15$ in Eq.II
$2 x+15=95$
$2 x=95-15 $
$2 x=80$
$x=\frac{80}{2} $
$x=40$
$\therefore $ Son’s age is 15 years, father’s age is 40 years.
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Question 263 Marks
Complete the following.
Answer
$\text { Length of rectangle } \Rightarrow 2 x + y +8=4 x - y $
$\Rightarrow x -4 x + y + y =-8$
$\Rightarrow-2 x +2 y =-8$
$\Rightarrow- x + y =-4 \ldots . .(1)$
Breadth of the rectangle $=2 y = x +4$
$\Rightarrow-x+2 y=4 \ldots \ldots .(I I)$
Equating Eq. I and II and change sign of Eq. II
$-x +y = -4$
$x-2y=-4$
$-------$
  $-y=-8$
Substituting $y=8$ in Eq.I
$-x+8=-4$
$-x=-4-8$
$-x=-12$
$x=12$
$\text { Length }=2 \times 12+8+8=40 $
$\text { Breadth }=2 \times 8=16$
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Question 273 Marks
Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Answer
Let the greater no. be x and smaller no. be x–3
As per given situation,
2(x–3) + 3(x) = 19
$\Rightarrow 2 x-6+3 x=19$
$\Rightarrow 5 x-6=19$
$\Rightarrow 5 x=19+6 $
$\Rightarrow 5 x=25$
$\Rightarrow x=\frac{25}{5}=5$
$\therefore$ smaller no is $x -3 \Rightarrow 5-3=2$
Hence, The numbers are 5 and 2 .
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Question 283 Marks
Solve the following simultaneous equation.
$\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} $
$\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}$
Answer

$\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} $

$\frac{1}{[2(3 x+y)]}-\frac{1}{[2(3 x-y)]}=\frac{1}{8} $

$\text { Let } \frac{1}{3 x+y}=m \text { and } \frac{1}{3 x-y}=n$
$m + n =\frac{3}{4} \Rightarrow 4(m+ n )=3 \Rightarrow 4 m+4 n =3 \ldots \text { (I) }$
$\frac{1}{2} m-\frac{1}{2} n =\frac{1}{8} \Rightarrow 8(m- n )=1 \times 2 \Rightarrow 8 m-8 n =2 \ldots \text { (II) }$
Multiply Eq. I by 2
$8 m+8 n =6 \ldots( a )$
$8 m-8 n =2 \ldots \ldots$ (b)Add (a) and (b) to get,
$8 m+8 n+8 m-8 n=8 $
$\Rightarrow 16 m=8$
$m=\frac{8}{16} $
$m=\frac{1}{2} $
$8 \times \frac{1}{2}-8 n=2 $
$\Rightarrow 4-8 n=2 \Rightarrow-8 n=2-4$
$\Rightarrow-8 n=-2 $
$\Rightarrow 8 n =2 $
$\Rightarrow n =\frac{2}{8} $
$\Rightarrow n =\frac{1}{4}$
$\therefore m =\frac{1}{2(3 x + y )} $
$\Rightarrow \frac{1}{2(3 x+y)}=\frac{1}{2}$
$\Rightarrow 2=2(3 x+y)$
$\Rightarrow 2=6 x+2 y \ldots ( I I I)$
$\therefore n=\frac{1}{2(3 x-y)} $
$\Rightarrow \frac{1}{2(3 x-y)}=\frac{1}{4} $
$\Rightarrow 4=2(3 x-y) $
$\Rightarrow 4=6 x-2 y \ldots(I V)$
Add Eq. III and IV
$6 x+2 y=2 $
$\underline{6 x-2 y=4}$
${12 x=6} $
$x=\frac{6}{12} $
$x=\frac{1}{2}$
Substituting $x =\frac{1}{2}$ in Eq. III
$6 \times \frac{1}{2}+2 y=2$
$\Rightarrow 3+2 y=2 $
$\Rightarrow 2 y=2-3 $
$\Rightarrow 2 y=-1 $
$y=-\frac{1}{2}$
Hence $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
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Question 293 Marks
Solve the following simultaneous equation.
$\frac{27}{x-2}+\frac{31}{y+3}=85 ; \frac{31}{x-2}+\frac{27}{y+3}=89$
Answer
$\frac{27}{x-2}+\frac{31}{y+3}=85$
$\frac{31}{x-2}+\frac{27}{y+3}=89$
Let $\frac{1}{x-2}=m \text { and } \frac{1}{y+3}=n$
$27 m+31 n=85 \ldots$
$31 m+27 n=89 \ldots$
Adding both equations
$58 m+58 n=174$
Dividing both sides by $58$
$m + n =3$
Subtracting Eq. $I$ and $II$
$27 m+31 n=85$
$-31 m-27 n=-89$
$-4 m+4 n=-4$
Dividing both sides by $4$
$- m + n =-1 \ldots( IV )$
Equating Eq. $III$ and $IV$
$m + n =3$
$\frac{- m + n =-1}{2 n =2}$
$n =\frac{2}{2}$
$n =1$
Subsituting $n=1$ in Eq. $III$
$m +1=3$
$m=3-1$
$m=2$
$\therefore m =\frac{1}{ x -2} \Rightarrow \frac{1}{ x -2}=2 \Rightarrow 2( x -2)=1 \Rightarrow 2 x -4=1 \Rightarrow 2 x =4+1$
$\Rightarrow 2 x =5 \Rightarrow x =\frac{5}{2}$
$\therefore n =\frac{1}{ y +3} \Rightarrow \frac{1}{ y +3}=1 \Rightarrow y +3=1 \Rightarrow y =1-3 \Rightarrow y =-2$
$y =2$
Hence $(x, y)=\left(\frac{5}{2},-2\right)$
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Question 303 Marks
Solve the following simultaneous equation.
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
Answer
$\frac{10}{x+y}+\frac{2}{x-y}=4$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$
Let $\frac{1}{x+y}=m$ and $\frac{1}{x-y}=n$
$10 m+2 n=4$
$15 m-5 n=-2 \ldots II )$
Multiply Eq. $I$ by $5$ and Eq.$II$ by $2$
$50 m+10 n=20$
$\frac{30 m-10 n=-4}{80 m=16}$
$m=\frac{16}{80}$
$m =\frac{1}{5}$
Substituting $m =\frac{1}{5}$ in Eq. I
$10 \times \frac{1}{5}+2 n =4$
$2+2 n =4$
$2 n =4-2$
$2 n =2$
$n =\frac{2}{2}$
$n =1$
$\therefore m =\frac{1}{ x + y } \Rightarrow \frac{1}{ x + y }=\frac{1}{5} \Rightarrow x + y =5$
$\therefore n =\frac{1}{ x - y } \Rightarrow \frac{1}{ x - y }=1 \Rightarrow x - y =1$
Now, equating Eq. $III$ and $IV$
$x+y=5$
$x-y=1$
$2 x=6$
$x=\frac{6}{2}$
$x=3$
Subsituting value of $x=3$ in Eq. III
$3+y=5$
$y=5-3$
$y=2$
Hence $(x, y)=(3,2)$
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Question 313 Marks
Solve the following simultaneous equation.
$\frac{2}{x}-\frac{3}{y}=15 ; \frac{8}{x}+\frac{5}{y}=77$
Answer
$\frac{2}{x}-\frac{3}{y}=15$
$\frac{8}{x}+\frac{5}{y}=77$
Let $\frac{1}{x}=m$ and $\frac{1}{y}=n$
$2 m-3 n =15 \ldots (I)$
$8 m+5 n =77 \ldots (II)$
Multiply Eq. $1$ by $4$
$8 m-12 n =60 \ldots (III)$
Equating Eq. $II$ and $III$. Change the signs of Eq. $III$
$8 m+5 n=77$
$\frac{-8 m+12 n=-60}{17 n=17}$
$n=\frac{17}{17}$
$n=1$
Substituting $n =1$ in Eq. $II$
$8 m+5 \times 1=77$
$8 m+5=77$
$8 m=77-5$
$8 m=72$
$m=\frac{72}{8}$
$m=9$
$\therefore m =\frac{1}{ x } \Rightarrow \frac{1}{ x }=9 \Rightarrow x =\frac{1}{9}$
$\therefore n =\frac{1}{ y } \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1$
Hence $( x , y )=\left(\frac{1}{9}, 1\right)$
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Question 323 Marks
Solve the following simultaneous equations using Cramer’s rule.
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
Answer
$2 x +3 y =2 $
$x -\frac{ y }{2}=\frac{1}{2} \Rightarrow 2 x - y =1$
$D =\left[\begin{array}{cc}
2 & 3 \\
2 & -1
\end{array}\right]=(2 \times-1)-(3 \times 2)=-2-6=-8 $
$D _{ x }=\left[\begin{array}{cc}
2 & 3 \\
1 & -1
\end{array}\right]=(2 \times-1)-(3 \times 1)=-2-3=-5 $
$D_y=\left[\begin{array}{ll}
2 & 2 \\
2 & 1
\end{array}\right]=(2 \times 1)-(2 \times 2)=2-4=(-2) $
$x =\frac{ D _{ x }}{ D }=\frac{-5}{-8}=\frac{5}{8} y =\frac{ D _{ y }}{ D }=\frac{-2}{-8}=\frac{1}{4}$
$\therefore( x , y )=\left(\frac{5}{8}, \frac{1}{4}\right)$ is solution.
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Question 333 Marks
Solve the following simultaneous equations using Cramer’s rule.
4m + 6n = 54; 3m + 2n = 28
Answer
$4 m+6 n =54 $
$3 m+2 n =28$
$D =\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}=(4 \times 2)-(6 \times 3)=8-18=10$
$D _{ x }=\begin{bmatrix} 54 & 6 \\ 28 & 2\end{bmatrix}=(54 \times 2)-(6 \times 28)=108-168=60 $
$D _{ y }=\begin{bmatrix} 4 & 54 \\ 3 & 28\end{bmatrix}=(4 \times 28)-(54 \times 3)=112-162=50 $
$x =\frac{ D _{ x }}{ D }=\frac{60}{10}=6 y =\frac{ D _{ y }}{ D }=\frac{50}{10}=5$
$\therefore( x , y )=(6,5)$ is solution.
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Question 343 Marks
Solve the following simultaneous equations using Cramer’s rule.
6x – 4y = –12; 8x – 3y = –2
Answer
$6 x-4 y=-12 $
$8 x-3 y=-2 $
$D=\begin{bmatrix} 6 & -4 \\ 8 & -3 \end{bmatrix}=(6 \times-3)-(-4 \times 8)=-18+32=14$
$D_x=\begin{bmatrix} -12 & -4 \\ -2 & -3 \end{bmatrix}=(-12 \times-3)-(-4 \times-2)=36-8=28 $
$D_y=\begin{bmatrix} 6 & -12 \\ 8 & -2 \end{bmatrix}=(6 \times-2)-12 \times 8=12+96=108 $
$x=\frac{D_x}{D}=\frac{28}{14}=2 y=\frac{D_y}{D}=\frac{108}{14}=6$
$\therefore(x, y)=(2,6)$ is solution.
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Question 353 Marks
Solve the following simultaneous equations using Cramer’s rule.
$x + 2y = –1; 2x – 3y = 12$
Answer
$x+2 y=-1 $
$2 x-3 y=12 $
$D=\left[\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right]=(1 \times-3)-(2 \times 2)=-3-4=-7 $
$D_x=\left[\begin{array}{cc} -1 & 2 \\ 12 & -3\end{array}\right]=(-1 \times-3)-(2 \times 12)=3-24=-21$
$D_y=\left[\begin{array}{cc} 1 & -1 \\ 2 & 12 \end{array}\right]=(1 \times 12)-(-1 \times 2)=12+2=14$
$x=\frac{D_x}{D}=-\frac{21}{-7}=3 y=\frac{D_y}{D}=\frac{14}{-7}=-2$
$\therefore( x , y )=(3,-2)$ is solution.
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Question 363 Marks
Solve the following simultaneous equations using Cramer’s rule.
4x + 3y – 4 = 0; 6x = 8 – 5y
Answer
$4 x+3 y=4 $
$6 x+5 y=8$
$D=\begin{bmatrix}4 & 3 \\6 & 5\end{bmatrix}$$=(4 \times 5)-(3 \times 6)=20-18=2 $
$D_x=\begin{bmatrix} 4 & 3 \\8 & 5 \end{bmatrix}=(4 \times 5)-(3 \times 8)=20-24=-4$
$D_y=\begin{bmatrix} 4 & 4 \\ 6 & 8 \end{bmatrix}=(4 \times 8)-(4 \times 6)=32-24=8$
$x=\frac{D_x}{D}=-\frac{4}{2}=-2 y=\frac{D_y}{D}=\frac{8}{2}=4$
$\therefore(x, y)=(-2,4)$ is the solution.
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Question 373 Marks
Solve the following simultaneous equations using Cramer’s rule.
3x – 4y = 10; 4x + 3y = 5
Answer
$\therefore(x, y)=(2,-1)$ is the solution
$3 x-4 y=10 $
$4 x+3 y=5 $
$D=\left|\begin{array}{cc}3 & -4 \\4 & 3 \end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25$
$D_x=\left[\begin{array}{cc} 10 & -4 \\ 5 & 3 \end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50$
$D_y=\left[\begin{array}{cc}3 & 10 \\ 4 & 5\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25$
$x =\frac{ D _{ x }}{ D }=\frac{50}{25}=2 y =\frac{ D _{ y }}{ D }=-\frac{25}{25}=-1$
 
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Question 383 Marks
Find the values of following determinants.
(1) $\left|\begin{array}{cc}-1 & 7 \\ 2 & 4\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & 3 \\ -7 & 0\end{array}\right|$
(3)$\left|\begin{array}{cc}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\end{array}\right|$
Answer
we know, determinant of a 2 × 2 matrix
$\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$
is (ad - bc)
(1) $(-1 \times 4)-(7 \times 2)=-4-14=-18$
(2) $(5 \times 0)-(3 \times-7)=0-(-21)=21$
(3) $\frac{7}{3} \times \frac{1}{2}-\frac{5}{3} \times \frac{3}{2}=\frac{7}{6}-\frac{15}{6}$
$=-\frac{8}{6}$
$=-\frac{4}{3}$
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Question 393 Marks
Solve the following simultaneous equation graphically.
3x – 4y = –7; 5x – 2y = 0
Answer
Coming soon
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Question 443 Marks
Complete the following table to draw graph of the equations -
(I) x+y = 3 (II) x – y = 4

Answer
Coming soon
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Question 453 Marks
A certain amount is equally distributed among certain number of students. Each would get ₹ 2 less if 10 students were more and each would get ₹ 6 more if 15 students were less. Find the number of students and the amount distributed.
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Question 493 Marks
Sum of two number is 45 and the greater number is twice the smaller number. Find the numbers.
Answer
30,15
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Question 503 Marks
If 1 is added to the numerator of a certain fraction its value becomes $\frac{1}{2}$ and if 1 is added to its denometer $\frac{1}{3}$. Find the original fraction.
Answer
$\frac{3}{8}$
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Question 513 Marks
Shabana's age 10 years hence, will be twice juhi's present age. 6 years back shabana's age was $\frac{5}{3}$ times Juhi's at that time find their present ages.
Answer
26 years, 18 years
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Question 563 Marks
Places A and B are 30 km apart and they are on a straight road. Hamid travels from A to B on bike. At the same time Joseph starts from B on bike, travels towards A. They meet each other after 20 minutes. If Joseph would have started from B at the same time but in the opposite direction (instead of towards A) Hamid would have caught him after 3 hours. Find the speed of Hamid and Joseph.
Answer
Let the speed of $Joseph = x km / h$
Let the speed of Hamid be $=y km / h$
When approaching each other, combined speed $=(x+y) km / h$
Time taken to meet $=\frac{30}{x+y}=\frac{1}{3}(20 mins )$
$\therefore x + y =90 \ldots I$
When moving away from each other, combined speed $=(x-y) km / h$
Time taken for Hamid to catch up $=\frac{30}{x-y}=3$
$\therefore x - y =10 \ldots II$
Equating I and II,
$\begin{aligned}
& x+y=90 \\
& x-y=10 \\
& 2 x=100 \\
& x=\frac{100}{2}=50
\end{aligned}$
Substituting $x=50$ in eq. I
$50+y=90$
$\begin{aligned}
& y=90-50 \\
& y=40
\end{aligned}$
Hamid's speed $50 km / hr$.
Joseph's speed $40 km / hr$.
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Question 573 Marks
In a factory the ratio of salary of skilled and unskilled workers is 5 : 3. Total salary of one day of both of them is ₹ 720. Find daily wages of skilled and unskilled workers.
Answer
Ratio of skilled and unskilled worker's salary $=5: 3$
Let it be $5 x$ and $3 x$
Total of one day's salary $=₹ 720$
$\begin{aligned}
& \text { So, } 5 x+3 x=720 \\
& 8 x=720 \\
& x=\frac{720}{8} \\
& x=90
\end{aligned}$
Skilled worker's wages $=5 x =5 \times 90=₹ 450$.
unskilled worker's wages $3 x =3 \times 90=₹ 270$
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Question 583 Marks
Kantabai bought $1 \frac{1}{2}$ kg tea and 5 kg sugar from a shop. She paid Rs 50 as return fare for rickshaw. Total expense was Rs 700. Then she realised that by ordering online the goods can be bought with free home delivery at the same price. So next month she placed the order online for 2 kg tea and 7 kg sugar. She paid Rs 880 for that. Find the rate of sugar and tea per kg.
Answer
Let $x$ be the cost of tea and $y$ be the cost of sugar
As she paid $₹ 50$ as return fare
$\begin{aligned}
& ₹ 700-₹ 50=₹ 650 \\
& \therefore \frac{3}{2} x+5 y=650 \Rightarrow 3 x+10 y=1300
\end{aligned}$
According to second situation,
$2 x+7 y=880$
$\qquad$
Multiplying Eq. I by 2 and Eq. II by 3
$\begin{aligned}
& 6 x+20 y=2600 \\
& 6 x+21 y=2640
\end{aligned}$
Subtracting Eq. III from IV
$\begin{aligned}
& 6 x+21 y=2640 \\
& -6 x-20 y=-2600 \\
& y=40
\end{aligned}$
Substituting $y=40$ in Eq. I
$\begin{aligned}
& 3 x+10 \times 40=1300 \\
& 3 x+400=1300 \\
& 3 x=1300-400
\end{aligned}$
3x=900
$\begin{aligned} & x=\frac{900}{3} \\ & x=300\end{aligned}$
Tea; ₹ 300 per kg.
Sugar ; ₹ 40 per kg.
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Question 593 Marks
Solve the following simultaneous equations.
$\frac{1}{2(3 x+4 y)}+\frac{1}{5(2 x-3 y)}=\frac{1}{4} ; \frac{5}{(3 x+4 y)}-\frac{2}{(2 x-3 y)}=-\frac{3}{2}$
Answer
Let $\frac{1}{3 x+4 y}=m$ and $\frac{1}{2 x-3 y}=n$
$\begin{aligned}
& \frac{1}{2} m+\frac{1}{5} n=\frac{1}{4} \Rightarrow 5 m+2 n=\frac{10}{4} \Rightarrow 20 m+8 n=10 \Rightarrow 10 m+4 n=5 \ldots (1) \\
& 5 m-2 n=-\frac{3}{2} \Rightarrow 10 m-4 n=-3 \ldots (2)
\end{aligned}$
Equating Eq. I and II
$\begin{aligned}
& 10 m+4 n=5 \\
& 10 m-4 n=-3 \\
& 20 m=2
\end{aligned}$
$\begin{aligned}
& m =\frac{2}{20} \\
& m =\frac{1}{10}
\end{aligned}$
Substituting $m =\frac{1}{10}$ in Eq. I
$\begin{aligned}
& 10 \times \frac{1}{10}+4 n=5 \\
& 1+4 n=5 \\
& 4 n=5-1
\end{aligned}$
4n=4
$\begin{aligned}
& n =\frac{4}{4} \\
& n =1 \\
& \quad \frac{1}{3 x +4 y }= m \Rightarrow \frac{1}{3 x +4 y }=\frac{1}{10} \Rightarrow 3 x +4 y =10 \ldots (3) \\
& \therefore \frac{1}{2 x -3 y }= n \Rightarrow \frac{1}{2 x -3 y }=1 \Rightarrow 2 x -3 y =1 \ldots (4)
\end{aligned}$
Multiply Eq. III by 3 and Eq. IV by 4 and Equate
$\begin{aligned}
& 9 x+12 y=30 \\
& 8 x-12 y=4 \\
& 17 x=34 \\
& x=\frac{34}{17} \\
& x=2
\end{aligned}$
Substituting $x=2$ in Eq. $V$
$\begin{aligned}
& 9 \times 2+12 y=30 \\
& 18+12 y=30 \\
& 12 y=30-18
\end{aligned}$
$\begin{aligned}
& 12 y=12 \\
& y=\frac{12}{12} \\
& y=1
\end{aligned}$
Hence, $(x, y)=(2,1)$
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Question 603 Marks
Solve the following simultaneous equations.
$\frac{7 x-2 y}{x y}=5 ; \frac{8 x+7 y}{x y}=15$
Answer

$\begin{aligned} & \frac{7 x-2 y}{x y}=5 \Rightarrow \frac{7 x}{x y}-\frac{2 y}{x y}=5 \Rightarrow \frac{7}{y}-\frac{2}{x}=5 \ldots \text { (I) } \\ & \frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15 \Rightarrow \frac{8}{y}+\frac{7}{x}=15 \ldots \text { (II) } \\ & \text { Let } \frac{1}{x}=m \text { and } \frac{1}{y}=n \\ & 7 n-2 m=5 \ldots \text { (III) } \\ & 8 n+7 m=15 \ldots \text { (IV) } \\ & \text { Multiply Eq. } 1 \text { by } 7 \text { and Eq.II by } 2 \\ & 49 n-14 m=35 \ldots \text { (V) } \\ & 16 n+14 m=30 \ldots \text { (VI) } \\ & 65 n=65 \\ & n=\frac{65}{65} \\ & n=1 \text { Substituting value in Eq.VI } \\ & 16 \times 1+14 m=30 \\ & 14 m=30-16 \\ & 14 m=14\end{aligned}$
$\begin{aligned}
& m =\frac{14}{14} \\
& m =1 \\
& \therefore \frac{1}{ x }= m \Rightarrow \frac{1}{ x }=1 \Rightarrow x =1 \\
& \quad \frac{1}{ y }= n \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1
\end{aligned}$
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Question 613 Marks
Solve the following simultaneous equations.
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} ; \frac{231}{x}+\frac{148}{y}=\frac{610}{x y}$
Answer

$\begin{aligned}
& \frac{148}{x}+\frac{231}{y}=\frac{527}{x y} \Rightarrow \frac{148 y+231 x}{x y}=\frac{527}{x y} \Rightarrow 231 x+148 y=527 \ldots(1) \\
& \frac{231}{x}+\frac{148}{y}=\frac{610}{x y} \Rightarrow \frac{231 y+148 x}{x y}=\frac{610}{x y} \Rightarrow 148 x+231 y=610 \ldots(2)
\end{aligned}$
Adding Eq. I and II
$\begin{aligned}
& 379 x+379 y=1137 \\
& x+y=3 \ldots \text { (III) }
\end{aligned}$
Subtracting Eq. I and II
$\begin{aligned}
& 83 x-83 y=-83 \\
& x-y=-1 \ldots \text { (IV) }
\end{aligned}$
Equating I and II
$\begin{aligned}
& x+y=3 \\
& x-y=-1 \\
& 2 x=2 \\
& x=\frac{2}{2} \\
& x=1
\end{aligned}$
Substituting $x =1$ in Eq. I
$\begin{aligned}
& 1+y=3 \\
& y=3-1 \\
& y=2
\end{aligned}$
Hence, $(x, y)=(1,2)$
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Question 623 Marks
Solve the following simultaneous equations.
$\frac{7}{2 x +1}+\frac{13}{ y +2}=27 ; \frac{13}{2 x +1}+\frac{7}{ y +2}=33$
Answer

$\begin{aligned}
& \text { Let } \frac{1}{2 x+1}=m \text { and } \frac{1}{y+2}=n \\
& 7 m+13 n=27 \ldots \text { (I) } \\
& 13 m+7 n=33 \ldots \text { (II) }
\end{aligned}$Adding Eq. I and II
$20 m +20 n =60 \Rightarrow m + n =3 \ldots(III)$
Subtract Eq. I and II
$-6 m+6 n=-6 \Rightarrow-m+n=-1 \ldots(IV)$
Equating Eq. III and IV
$\begin{aligned}
& m+n=3 \\
& -m+n=-1 \\
& \hline 2 n=2 \\
& n=1
\end{aligned}$
Substituting $n=1$ in Eq. III
$\begin{aligned}
& m+1=3 \\
& m=3-1 \\
& m=2
\end{aligned}$
$\begin{aligned}
& \therefore \frac{1}{2 x +1}= m \Rightarrow \frac{1}{2 x +1}=2 \Rightarrow 2(2 x +1)=1 \Rightarrow 4 x +2=1 \Rightarrow 4 x =1-2 \\
& \Rightarrow 4 x =-1 \Rightarrow x =-\frac{1}{4} \\
& \therefore \frac{1}{ y +2}= n \Rightarrow \frac{1}{ y +2}=1 \Rightarrow y +2=1 \Rightarrow y =1-2 \Rightarrow y =-1
\end{aligned}$
Hence, $(x, y)=\left(-\frac{1}{4},-1\right)$
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Question 633 Marks
Solve the following simultaneous equations.
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
Answer

Let $\frac{1}{x}=m$ and $\frac{1}{y}=n$
$2 m+\frac{2}{3} n=\frac{1}{6} \Rightarrow 12 m+\frac{12}{3 n}=1 \Rightarrow 12 m+4 n=1 \ldots(1)$
$3 m+2 n=0 \ldots(2)$
Multiply Eq. II by 2
$6 n+4 n=0 \ldots(3)$
Subtract Eq.III from Eq. I
$\begin{aligned}
& 12 m+4 n=1 \\
& -6 m-4 n=0 \\
& \hline 6 m=1 \\
& m=\frac{1}{6}
\end{aligned}$
Substitute $m=1 / 6$ in Eq. I
$\begin{aligned}
& 12 \times \frac{1}{6}+4 n=1 \\
& 2+4 n=1 \\
& 4 n=1-2
\end{aligned}$
$\begin{aligned}
& 4 n =-1 \\
& n =-\frac{1}{4} \\
& \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{6}=\frac{1}{ x } \Rightarrow x =6 \\
& \therefore n =\frac{1}{ y } \Rightarrow-\frac{1}{4}=\frac{1}{ y } \Rightarrow y =-4
\end{aligned}$
Hence, $(x, y)=(6,-4)$
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Question 643 Marks
Solve the following equations by Cramer’s method.
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
Answer
Let,
$\begin{aligned}
& \frac{x+y-8}{2}=\frac{x+2 y-14}{3} \\
& \Rightarrow 3 x+3 y-24=2 x+4 y-28 \\
& \Rightarrow x-y=-4 \ldots \text { (1) }
\end{aligned}$
Also,
$\begin{aligned}
& \text { Let } \frac{x+2 y-14}{3}=\frac{3 x-y}{4} \\
& \Rightarrow 4 x+8 y-56=9 x-3 y \\
& \Rightarrow 5 x-11 y=-56 \ldots \text { (2) }
\end{aligned}$
Hence the two equations are:
$\begin{aligned}
& x-y=-4 \ldots(1) \\
& 5 x-11 y=-56
\end{aligned}$
Now,
$\begin{aligned}
& D=\left|\begin{array}{cc}
1 & -1 \\
5 & -11
\end{array}\right| \\
& \Rightarrow D=(-11-(-5))=-6
\end{aligned}$
Also, $D _{ x }=\left|\begin{array}{cc}-4 & -1 \\ -56 & -11\end{array}\right|$
$D_x=44-56=-12$
And,
$\begin{aligned}
& D_y=\left|\begin{array}{cc}
1 & -4 \\
5 & -56
\end{array}\right| \\
& \Rightarrow D_y=-56+20=-36
\end{aligned}$
Now,
$x=\frac{D_x}{D}=\frac{-12}{-6}=2$
And,
$y=\frac{D_y}{D}=\frac{-36}{-6}=6$
Hence, $(2,6)$ is the solution
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Question 653 Marks
Solve the following equations by Cramer’s method.
7x + 3y = 15; 12y – 5x = 39
Answer

$\begin{aligned} & D =\left[\begin{array}{cc}7 & 3 \\ -5 & 12\end{array}\right]=(7 \times 12)-(3 \times-5)=84+15=99 \\ & D _{ x }=\left[\begin{array}{cc}15 & 3 \\ 39 & 12\end{array}\right]=(15 \times 12)-(3 \times 39)=180-117=63 \\ & D _{ y }=\left[\begin{array}{cc}7 & 15 \\ -5 & 39\end{array}\right]=(7 \times 39)-(15 \times-5)=273+75=348 \\ & x =\frac{ D _{ x }}{ D }=\frac{63}{99}=\frac{7}{11} y =\frac{ D _{ y }}{ D }=\frac{348}{99}=\frac{116}{33} \\ & \therefore( x , y )=\left(\frac{7}{11}, \frac{116}{33}\right)\end{aligned}$
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Question 663 Marks
Solve the following equations by Cramer’s method.
$3 x-2 y=\frac{5}{2} ; \frac{1}{3} x+3 y=-\frac{4}{3}$
Answer

$\begin{aligned} & 3 x -2 y =\frac{5}{2} \Rightarrow 6 x -4 y =5 \\ & \frac{1}{3} x +3 y =-\frac{4}{3} \Rightarrow \frac{ x +9 y }{3}=-\frac{4}{3} \Rightarrow x +9 y =-4 \\ & D =\left[\begin{array}{cc}6 & -4 \\ 1 & 9\end{array}\right]=(6 \times 9)-(-4 \times 1)=54+4=58 \\ & D _{ x }=\left[\begin{array}{cc}5 & -4 \\ -4 & 9\end{array}\right]=(5 \times 9)-(-4 \times-4)=45-16=29 \\ & D _{ y }=\left[\begin{array}{cc}6 & 5 \\ 1 & -4\end{array}\right]=(6 \times-4)-(5 \times 1)=-24-5=-29 \\ & x=\frac{D_x}{D}=\frac{1}{2}, y=\frac{D_y}{D}=\frac{(-29)}{58}=\frac{(-1)}{2} \\ & \therefore( x , y )=(1 / 2,-1 / 2)\end{aligned}$
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Question 673 Marks
Solve the following equations by Cramer’s method.
4m – 2n = –4; 4m + 3n = 16
Answer

$\begin{aligned} & D =\left[\begin{array}{cc}4 & -2 \\ 4 & 3\end{array}\right]=(4 \times 3)-(-2 \times 4)=12+8=20 \\ & D_x=\left[\begin{array}{cc}-4 & -2 \\ 16 & 3\end{array}\right]=(-4 \times 3)-(-2 \times 16)=-12+32=20 \\ & D_y=\left[\begin{array}{cc}4 & -4 \\ 4 & 16\end{array}\right]=(4 \times 16)-(-4 \times 4)=64+16=80 \\ & x=\frac{D_x}{D}=\frac{20}{20}=1 y=\frac{D_y}{D}=\frac{80}{20}=4 \\ & \therefore(x, y)=(1,4)\end{aligned}$
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Question 683 Marks
Solve the following equations by Cramer’s method.
6x – 3y = –10; 3x + 5y – 8 = 0
Answer

$\begin{aligned} & 6 x-3 y=-10 \\ & 3 x+5 y=8 \\ & D=\left[\begin{array}{cc}6 & -3 \\ 3 & 5\end{array}\right]=(6 \times 5)-(-3 \times 3)=30+9=39 \\ & D_x=\left[\begin{array}{cc}-10 & -3 \\ 8 & 5\end{array}\right]=(-10 \times 5)-3 \times 8=-50+24=-26 \\ & D_y=\left[\begin{array}{cc}6 & -10 \\ 3 & 8\end{array}\right]=(6 \times 8)-(-10 \times 3)=48+30=78 \\ & x=\frac{D_x}{D}=\frac{-26}{39}=\frac{-2}{3} y=\frac{D_y}{D}=\frac{78}{39}=2 \\ & \therefore(x, y)=\left(-\frac{2}{3}, 2\right)\end{aligned}$
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Question 693 Marks
Find the values of each of the following determinants.
(1) $\left|\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right|$
(3) $\left|\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right|$
Answer
(1) $D =\left[\begin{array}{ll}4 & 3 \\ 2 & 7\end{array}\right]=(4 \times 7)-(3 \times 2)=28-6=22$
(2) $D=\left[\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right]=(5 \times 1)-2 \times-3=5-6=-1$
(3) $D =\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]=(3 \times 4)-(-1 \times 1)=12+1=13$
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Question 723 Marks
Solve the following simultaneous equation graphically.
5x – 6y + 30 = 0; 5x + 4y – 20 = 0
Answer
The given simultaneous equations are
$5 x-6 y+30=0$
$\therefore \quad 6 y=5 x+30$
$\therefore \quad y=\frac{5 x+30}{6}$
x0-663
y50107.5
(x, y)(0, 5)(-6, 0)(6, 10)(3, 7.5)

$5 x+4 y-20=0$
$\therefore \quad 4 y=20-5 x$
$\therefore \quad y=\frac{20-5 x}{4}$
x04-42
y50102.5
(x, y)(0, 5)(4, 0)(-4, 10)(2, 2.5)

Image

The two lines intersect at point (0, 5).
∴ x = 0 and y = 5 is the solution of the simultaneous equations 5x – 6y + 30 = 0 and 5x + 4y – 20 = 0.
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Question 733 Marks
Solve the following simultaneous equation graphically.
x – 3y = 1; 3x – 2y + 4 = 0
Answer
The given simultaneous equations are
$x-3 y=1$
$\therefore \quad 3 y=x-1$
$\therefore \quad y=\frac{x-1}{3}$
x4-2-51
y1-1-20
(x, y)(4, 1)(-2, -1)(-5, -2)(1, 0)
$3 x-2 y+4=0$
$\therefore \quad 2 y=3 x+4$
$\therefore \quad y=\frac{3 x+4}{2}$
x0-22-4
y2-15-4
(x, y)(0, 2)(-2, -1)(2, 5)(-4, -4)
Image
The two lines intersect at point (-2, -1).
∴ x = -2 and y = -1 is the solution of the simultaneous equations x – 3y = 1 and 3x – 2p + 4 = 0.
 
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Question 753 Marks
Sum of the present ages of Manish and Savita is 31. Manish’s age 3 years ago was 4 times the age of Savita. Find their present ages.
Answer
Let Manish's present age be $x$
Let Savita's present age be y
According to $1^{\text {st }}$ situation,
$x+y=31 \dots (1)$
According to second situation,
$\begin{aligned}
& x-3=4(y-3) \\
& x-3=4 y-12 \\
& x-4 y=-12+3 \\
& x-4 y=-9 \ldots (2)
\end{aligned}$
Subtracting Eq. II from I
$\begin{aligned}
& x+y=31 \\
& -x+4 y=9 \\
& 5 y=40 \\
& y=\frac{40}{5} \\
& y=8
\end{aligned}$
Substitute $y=8$ in eq. I
$\begin{aligned}
& x+8=31 \\
& x=31-8 \\
& x=23
\end{aligned}$
Manisha's age 23 years
Savita's age 8 years.
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Question 763 Marks
Out of 1900 km, Vishal travelled some distance by bus and some by aeroplane. Bus travels with average speed 60 km/hr and the average speed of aeroplane is 700 km/hr. It takes 5 hours to complete the journey. Find the distance, Vishal travelled by bus.
Answer
Let the distance travelled by bus $=x$ speed of bus $=60 km / hr$ 
As,$\text { time }=\frac{\text { distance }}{\text { speed }}$
Time taken travelling by bus $=\frac{x}{60}$
Let the distance traveled by plane $= y$
As, total distance traveled was $1900 km$
$x+y=1900$
Distance traveled by plane $=(1900-x)$
Speed of plane $=700 km / hr$
Time travelling by plane $=\frac{(1900- x )}{700}$
Given,
Total time $=5$ hours
$\begin{aligned}
& \frac{x}{60}+\frac{1900-x}{700}=5 \\
& \Rightarrow \frac{35 x+3(1900-x)}{2100}=5 \\
& \Rightarrow \frac{35 x+5700-3 x}{2100}=5
\end{aligned}$
⇒ 32x + 5700 = 10500
⇒ 32x = 4800
⇒ x = 150 km
and y = 1900 - x
= 1900 - 150 = 1750 km
Vishal travels 150km by bus and 1750 km by plane.
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Question 773 Marks
Two types of boxes A, B are to be placed in a truck having capacity of 10 tons. When 150 boxes of type A and 100 boxes of type B are loaded in the truck, it weighes 10 tons. But when 260 boxes of type A are loaded in the truck, it can still accommodate 40 boxes of type B, so that it is fully loaded. Find the weight of each type of box.
Answer

$A-30 kg , B -55 k$
Let the weight of box ' $A$ ' $=x$ kg
Let the Weight of box'B' $=y$ kg
According to question,
150 boxes of type $A$ and 100 boxes of type $B$ are loaded in the truck and it weighs 10 tons.
$\begin{aligned}
& \therefore 150 x +100 y =10000[\because 1 \text { ton }=1000 kg ] \\
& \Rightarrow 3 x +2 y =200 \ldots \ldots \text { (I) }
\end{aligned}$
260 boxes of type $A$ are loaded in the truck, it can still accommodate 40 boxes of type $B$, still it weighs 10 tons
$\begin{aligned}
& \therefore 260 x +40 y =10000[\because 1 \text { ton }=1000 kg ] \\
& \Rightarrow 13 x +2 y =500 \ldots \ldots \text { (II) }
\end{aligned}$
Solving Equation I and II
$\begin{aligned}
& 3 x+2 y=200 \\
& -13 x-2 y=-500 \\
& -10 x=-300 \\
& x=\frac{300}{10} \\
& x=30
\end{aligned}$
Putting $x=30$ in Eq. I
$\begin{aligned}
& 3 \times 30+2 y=200 \\
& 90+2 y=200 \\
& 2 y=200-90 \\
& 2 y=110 \\
& y=\frac{110}{2}=55
\end{aligned}$
Hence, A - $30 kg , B -55 kg$
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Question 783 Marks
The denominator of a fraction is 4 more than twice its numerator. Denominator becomes 12 times the numerator, if both the numerator and the denominator are reduced by 6. Find the fraction.
Answer
Let the numerator and denominator of the fraction be $x$ and $y$ respectively.
Fraction $=\frac{\bar{x}}{ y }$
Given,
Denominator $=2$ (Numerator $)+4$
$\begin{aligned}
& \Rightarrow y =2 x +4 \\
& \Rightarrow 2 x - y =(-4) \ldots I
\end{aligned}$
According to the given condition, we have
$\begin{aligned}
& y-6=12(x-6) \\
& \Rightarrow y-6=12 x-72 \\
& \Rightarrow 12 x-y=66 \ldots II
\end{aligned}$
Equating Eq. I and II,
$\begin{aligned}
& 2 x-y=-4 \\
& -12 x+y=-66 \\
& -10 x=-70 \\
& x=\frac{70}{10}
\end{aligned}$
$x=7$
Putting $x=7$ in equation $I$, we get
$\begin{aligned}
& \Rightarrow 2 \times 7-y=-4 \\
& \Rightarrow 14-y=-4 \\
& \Rightarrow y=14+4 \\
& \Rightarrow y=18
\end{aligned}$
Hence, required fraction $=\frac{7}{18}$
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Question 793 Marks
The sum of father’s age and twice the age of his son is 70. If we double the age of the father and add it to the age of his son the sum is 95. Find their present ages.
Answer
Suppose father's age(in years) be $x$ and that son's age be $y$.
Then,
$\begin{aligned}
& x+2 y=70 \dots(1) \\
& 2 x+y=95 \dots(2)
\end{aligned}$
Multiply Eq.I by 2 and equate
$\begin{gathered}
2 x+4 y=140 \\
-2 x-y=-95 \\
\hline 3 y=45
\end{gathered}$
$\begin{aligned}
& y=\frac{45}{3} \\
& y=15
\end{aligned}$
Substituting $y=15$ in Eq.II
$\begin{aligned}
& 2 x+15=95 \\
& 2 x=95-15 \\
& 2 x=80 \\
& x=\frac{80}{2} \\
& x=40
\end{aligned}$
∴ Son’s age is 15 years, father’s age is 40 years.
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Question 803 Marks
Answer

$\begin{aligned} & \text { Length of rectangle } \Rightarrow 2 x+y+8=4 x-y \\
& \Rightarrow 2 x-4 x+y+y=-8 \\
& \Rightarrow-2 x+2 y=-8 \\
& \Rightarrow-x+y=-4 \ldots \ldots(I)
\end{aligned}$
$\begin{aligned}
& \text { Breadth of the rectangle }=2 y=x+4 \\
& \Rightarrow-x+2 y=4 \ldots . . \text { (II) }
\end{aligned}$
Equating Eq. I and II and change sign of Eq. II
$\begin{gathered}
-x+y=-4 \\
x-2 y=-4 \\
\hline -y=-8 \\
y=8
\end{gathered}$
Substituting $y=8$ in Eq.I
$\begin{aligned}
& -x+8=-4 \\
& -x=-4-8 \\
& -x=-12 \\
& x=12
\end{aligned}$
$\begin{aligned} & \text { Length }=2 \times 12+8+8=40 \\ & \text { Breadth }=2 \times 8=16 \\ & \text { Area }=\text { Length } \times \text { breadth }=40 \times 16=640 \text { sq. unit } \\ & \text { Perimeter }=2 \text { (Length }+ \text { Breadth) }=2(40+16)=2(56)=112 \text { unit. }\end{aligned}$
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Question 813 Marks
Two numbers differ by 3. The sum of twice the smaller number and thrice the greater number is 19. Find the numbers.
Answer
Let the greater no. be $x$ and smaller no. be $x-3$
As per given situation,
$\begin{aligned}
& 2(x-3)+3(x)=19 \\
& \Rightarrow 2 x-6+3 x=19 \\
& \Rightarrow 5 x-6=19 \\
& \Rightarrow 5 x=19+6 \\
& \Rightarrow 5 x=25 \\
& \Rightarrow x=\frac{25}{5}=5
\end{aligned}$
$\therefore$ smaller no is $x-3 \Rightarrow 5-3=2$
Hence, The numbers are 5 and 2.
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Question 823 Marks
Solve the following simultaneous equation.
$\begin{aligned} & \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \\ & \frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}\end{aligned}$
Answer

$\begin{aligned}& \frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} \\
& \frac{1}{[2(3 x+y)]}-\frac{1}{[2(3 x-y)]}=\frac{1}{8} \\
& \text { Let } \frac{1}{3 x+y}=m \text { and } \frac{1}{3 x-y}=n \\
& m+n=\frac{3}{4} \Rightarrow 4(m+n)=3 \Rightarrow 4 m+4 n=3 \ldots \text { (I) } \\
& \frac{1}{2} m-\frac{1}{2} n=\frac{1}{8} \Rightarrow 8(m-n)=1 \times 2 \Rightarrow 8 m-8 n=2 \ldots (II)
\end{aligned}$
Multiply Eq. I by 2
$8 m+8 n=6$ $\dots$ (a)
$8 m-8 n=2$ $\dots$ (b)Add (a) and (b) to get,
$8 m+8 n+8 m-8 n=8$
$\Rightarrow 16 m =8$
$m =\frac{8}{16}$
$m =\frac{1}{2}$
Substituting $m =\frac{1}{2}$ in Eq. II
$8 \times \frac{1}{2}-8 n=2$
⇒ 4 - 8n = 2
⇒ - 8n = 2 - 4
⇒ - 8n = -2
⇒ 8n = 2
$\begin{aligned}
\Rightarrow & n=\frac{2}{8} \\
\Rightarrow & n=\frac{1}{4} \\
& \therefore m=\frac{1}{2(3 x+y)} \\
& \frac{1}{2(3 x+y)}=\frac{1}{2} \\
\Rightarrow & 2=2(3 x+y) \\
\Rightarrow & 2=6 x+2 y \ldots \ldots . \text { III } \\
& n=\frac{1}{2(3 x-y)} \\
\Rightarrow & \frac{1}{2(3 x-y)}=\frac{1}{4} \\
\Rightarrow & 4=2(3 x-y) \\
\Rightarrow & 4=6 x-2 y \ldots . I V
\end{aligned}$
Add Eq. III and IV
$\begin{aligned}
& 6 x+2 y=2 \\
& \frac{6 x-2 y=4}{12 x=6} \\
& x=\frac{6}{12} \\
& x=\frac{1}{2}
\end{aligned}$
Substituting $x =\frac{1}{2}$ in Eq. III
$\begin{aligned}
& 6 \times \frac{1}{2}+2 y=2 \\
& \Rightarrow 3+2 y=2 \\
& \Rightarrow 2 y=2-3 \\
& \Rightarrow 2 y=-1 \\
& y=-\frac{1}{2}
\end{aligned}$
Hence $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
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Question 833 Marks
Solve the following simultaneous equation.
$\frac{27}{x-2}+\frac{31}{y+3}=85 ; \frac{31}{x-2}+\frac{27}{y+3}=89$
Answer

$\begin{aligned}& \frac{27}{x-2}+\frac{31}{y+3}=85 \\
& \frac{31}{x-2}+\frac{27}{y+3}=89
\end{aligned}$
Let $\frac{1}{x-2}=m$ and $\frac{1}{y+3}=n$
$27 m+31 n=85 \dots (1)$
$31 m+27 n=89 \dots (2)$
Adding both equations
$58 m+58 n=174$
Dividing both sides by 58
$m + n =3 \dots (3)$
Subtracting Eq. I and II
$\begin{aligned}
& 27 m+31 n=85 \\
& -31 m-27 n=-89 \\
& -4 m+4 n=-4
\end{aligned}$
Dividing both sides by 4
$-m+n=-1 \dots (4)$
Equating Eq. III and IV
$\begin{gathered}
m+n=3 \\
-m+n=-1 \\
\hline 2 n=2
\end{gathered}$
$\begin{aligned}
& n =\frac{2}{2} \\
& n =1
\end{aligned}$
Subsituting $n=1$ in Eq. III
$\begin{aligned}
& m+1=3 \\
& m=3-1 \\
& m=2
\end{aligned}$
$\begin{aligned}
& \therefore m =\frac{1}{ x -2} \\
& \Rightarrow \frac{1}{ x -2}=2 \Rightarrow 2( x -2)=1 \\
& \Rightarrow 2 x -4=1 \Rightarrow 2 x =4+1 \\
& \Rightarrow 2 x =5 \Rightarrow x =\frac{5}{2} \\
& \therefore n =\frac{1}{ y +3} \\
& \Rightarrow \frac{1}{ y +3}=1 \\
& \Rightarrow y +3=1 \\
& \Rightarrow y =1-3 \\
& \Rightarrow y =-2 \\
& y =2
\end{aligned}$
Hence $(x, y)=\left(\frac{5}{2},-2\right)$
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Question 843 Marks
Solve the following simultaneous equation.
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
Answer

$\begin{aligned} & \frac{10}{x+y}+\frac{2}{x-y}=4 \\
& \frac{15}{x+y}-\frac{5}{x-y}=-2 \\
& \text { Let } \frac{1}{x+y}=m \text { and } \frac{1}{x-y}=n \\
& 10 m+2 n=4 \ldots \text { (I) } \\
& 15 m-5 n=-2 \ldots \text { (II) }
\end{aligned}$
Multiply Eq. I by 5 and Eq.II by 2
$\begin{gathered}
50 m+10 n=20 \\
\frac{30 m-10 n=-4}{80 m=16}
\end{gathered}$
$\begin{aligned}
m & =\frac{16}{80} \\
m & =\frac{1}{5}
\end{aligned}$
$\begin{aligned}
& \text { Substituting } m=\frac{1}{5} \text { in Eq. I } \\
& 10 \times \frac{1}{5}+2 n=4 \\
& 2+2 n=4 \\
& 2 n=4-2 \\
& 2 n=2 \\
& n=\frac{2}{2} \\
& n=1
\end{aligned}$
$\begin{aligned}
\therefore & m =\frac{1}{ x + y } \Rightarrow \frac{1}{ x + y }=\frac{1}{5} \Rightarrow x + y =5 \dots (3)\\
& n =\frac{1}{ x - y } \Rightarrow \frac{1}{x-y}=1 \Rightarrow x - y =1 \dots (4)
\end{aligned}$
Now, equating Eq. III and IV
$\begin{gathered}
x+y=5 \\
x-y=1 \\
\hline 2 x=6 \\
x=\frac{6}{2} \\
x=3
\end{gathered}$
Subsituting value of $x=3$ in Eq. III
$\begin{aligned}
& 3+y=5 \\
& y=5-3 \\
& y=2
\end{aligned}$
Hence $(x, y)=(3,2)$
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Question 853 Marks
Solve the following simultaneous equation.
$\frac{2}{x}-\frac{3}{y}=15 ; \frac{8}{x}+\frac{5}{y}=77$
Answer

$\begin{aligned} & \frac{2}{x}-\frac{3}{y}=15 \\
& \frac{8}{x}+\frac{5}{y}=77 \\
& \text { Let } \frac{1}{x}=m \text { and } \frac{1}{y}=n \\
& 2 m-3 n=15 \ldots \text { (I) } \\
& 8 m+5 n=77 \ldots \text { (II) }
\end{aligned}$
Multiply Eq. I by 4
$8 m-12 n=60 \ldots \text { (III) }$
Equating Eq. II and III. Change the signs of Eq. III
$\begin{gathered}
8 m+5 n=77 \\
-8 m+12 n=-60 \\
\hline 17 n=17
\end{gathered}$
$\begin{aligned}
& n =\frac{17}{17} \\
& n =1
\end{aligned}$
Substituting $n =1$ in Eq. II
$\begin{aligned}
& 8 m+5 \times 1=77 \\
& 8 m+5=77 \\
& 8 m=77-5 \\
& 8 m=72 \\
& m=\frac{72}{8} \\
& m=9
\end{aligned}$
$\begin{aligned} & \therefore m =\frac{1}{ x } \Rightarrow \frac{1}{ x }=9 \Rightarrow x =\frac{1}{9} \\ & \therefore n =\frac{1}{ y } \Rightarrow \frac{1}{ y }=1 \Rightarrow y =1 \\ & \text { Hence }( x , y )=\left(\frac{1}{9}, 1\right)\end{aligned}$
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Question 863 Marks
Solve the following simultaneous equations using Cramer’s rule.
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
Answer

$\begin{aligned} & 2 x+3 y=2 \\ & x-\frac{y}{2}=\frac{1}{2} \Rightarrow 2 x-y=1 \\ & D=\left[\begin{array}{cc}2 & 3 \\ 2 & -1\end{array}\right]=(2 \times-1)-(3 \times 2)=-2-6=-8 \\ & D_x=\left[\begin{array}{cc}2 & 3 \\ 1 & -1\end{array}\right]=(2 \times-1)-(3 \times 1)=-2-3=-5 \\ & D_y=\left[\begin{array}{ll}2 & 2 \\ 2 & 1\end{array}\right]=(2 \times 1)-(2 \times 2)=2-4=(-2) \\ & x=\frac{D_x}{D}=\frac{-5}{-8}=\frac{5}{8} y=\frac{D_y}{D}=\frac{-2}{-8}=\frac{1}{4} \\ & \therefore(x, y)=\left(\frac{5}{8}, \frac{1}{4}\right) \text { is solution. }\end{aligned}$
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Question 873 Marks
Solve the following simultaneous equations using Cramer’s rule.
4m + 6n = 54; 3m + 2n = 28
Answer

$\begin{aligned} & 4 m+6 n=54 \\
& 3 m+2 n=28 \\
& D=\left[\begin{array}{ll}
4 & 6 \\
3 & 2
\end{array}\right]=(4 \times 2)-(6 \times 3)=8-18=10 \\
& D_x=\left[\begin{array}{ll}
54 & 6 \\
28 & 2
\end{array}\right]=(54 \times 2)-(6 \times 28)=108-168=60 \\
& D_y=\left[\begin{array}{ll}
4 & 54 \\
3 & 28
\end{array}\right]=(4 \times 28)-(54 \times 3)=112-162=50 \\
& x=\frac{D_x}{D}=\frac{60}{10}=6 y=\frac{D_y}{D}=\frac{50}{10}=5
\end{aligned}$
$\therefore(x, y)=(6,5)$ is solution.
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Question 883 Marks
Solve the following simultaneous equations using Cramer’s rule.
6x – 4y = –12; 8x – 3y = –2
Answer

$\begin{aligned} & 6 x-4 y=-12 \\
& 8 x-3 y=-2 \\
& D=\left[\begin{array}{ll}
6 & -4 \\
8 & -3
\end{array}\right]=(6 \times-3)-(-4 \times 8)=-18+32=14 \\
& D_x=\left[\begin{array}{cc}
-12 & -4 \\
-2 & -3
\end{array}\right]=(-12 \times-3)-(-4 \times-2)=36-8=28 \\
& D_y=\left[\begin{array}{cc}
6 & -12 \\
8 & -2
\end{array}\right]=(6 \times-2)-12 \times 8=12+96=108 \\
& x=\frac{D_x}{D}=\frac{28}{14}=2 y=\frac{D_y}{D}=\frac{108}{14}=6
\end{aligned}$
$\therefore(x, y)=(2,6)$ is solution.
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Question 893 Marks
Solve the following simultaneous equations using Cramer’s rule.
x + 2y = –1; 2x – 3y = 12
Answer

$\begin{aligned} & x+2 y=-1 \\ & 2 x-3 y=12 \\ & D=\left[\begin{array}{cc}1 & 2 \\ 2 & -3\end{array}\right]=(1 \times-3)-(2 \times 2)=-3-4=-7 \\ & D_x=\left[\begin{array}{cc}-1 & 2 \\ 12 & -3\end{array}\right]=(-1 \times-3)-(2 \times 12)=3-24=-21 \\ & D_y=\left[\begin{array}{cc}1 & -1 \\ 2 & 12\end{array}\right]=(1 \times 12)-(-1 \times 2)=12+2=14 \\ & x=\frac{D_x}{D}=-\frac{21}{-7}=3 y=\frac{D_y}{D}=\frac{14}{-7}=-2 \\ & \therefore(x, y)=(3,-2) \text { is solution. }\end{aligned}$
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Question 903 Marks
Solve the following simultaneous equations using Cramer’s rule.
4x + 3y – 4 = 0; 6x = 8 – 5y
Answer

$\begin{aligned}
& 4 x+3 y=4 \\
& 6 x+5 y=8 \\
& D=\left[\begin{array}{ll}
4 & 3 \\
6 & 5
\end{array}\right]=(4 \times 5)-(3 \times 6)=20-18=2 \\
& D_x=\left[\begin{array}{ll}
4 & 3 \\
8 & 5
\end{array}\right]=(4 \times 5)-(3 \times 8)=20-24=-4 \\
& D_y=\left[\begin{array}{ll}
4 & 4 \\
6 & 8
\end{array}\right]=(4 \times 8)-(4 \times 6)=32-24=8 \\
& x=\frac{D_x}{D}=-\frac{4}{2}=-2 y=\frac{D_y}{D}=\frac{8}{2}=4
\end{aligned}$
$\therefore( x , y )=(-2,4)$ is the solution.
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Question 913 Marks
Solve the following simultaneous equations using Cramer’s rule.
3x – 4y = 10; 4x + 3y = 5
Answer

$\begin{aligned}
& 3 x-4 y=10 \\
& 4 x+3 y=5 \\
& D=\left|\begin{array}{cc}
3 & -4 \\
4 & 3
\end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25 \\
& D_x=\left[\begin{array}{cc}
10 & -4 \\
5 & 3
\end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50 \\
& D_y=\left[\begin{array}{cc}
3 & 10 \\
4 & 5
\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25 \\
& x=\frac{D_x}{D}=\frac{50}{25}=2 \\
& x=\frac{D_y}{D}=-\frac{25}{25}=-1
\end{aligned}$
$\therefore(x, y)=(2,-1)$ is the solution
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Question 923 Marks
Find the values of following determinants.
(1) $\left|\begin{array}{cc}-1 & 7 \\ 2 & 4\end{array}\right|$
(2) $\left|\begin{array}{cc}5 & 3 \\ -7 & 0\end{array}\right|$
(3) $\left|\begin{array}{ll}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\end{array}\right|$
Answer
we know, determinant of a $2 \times 2$ matrix
$\left|\begin{array}{ll}
a & b \\
c & d
\end{array}\right|$
(1) $(-1 \times 4)-(7 \times 2)=-4-14=-18$
(2) $(5 \times 0)-(3 \times-7)=0-(-21)=21$
$\begin{aligned}
& \text { (3) } \frac{7}{3} \times \frac{1}{2}-\frac{5}{3} \times \frac{3}{2}=\frac{7}{6}-\frac{15}{6} \\
& =-\frac{8}{6} \\
& =-\frac{4}{3}
\end{aligned}$
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Question 933 Marks
Solve the following simultaneous equation graphically.
2x – 3y = 4 ; 3y – x = 4
Answer
The given simultaneous equations are
$2 x-3 y=4$
$\therefore \quad 3 y=2 x-4$
$\therefore \quad y=\frac{2 x-4}{3}$
x2-158
y0-224
(x. y)(2, 0)(-1, -2)(-5, 2)(8, 4)
$3 y-x=4$
$\therefore \quad 3 y=x+4$
$\therefore \quad y=\frac{x+4}{3}$
x2-45-1
y2031
(x. y)(2, 2)(-4, -0)(5, 3)(-1, 1)

Image
The two lines intersect at point (8, 4).
∴ x = 8 and y = 4 is the solution of the simultaneous equations 2x – 3y = 4 and 3y – x = 4.
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Question 943 Marks
Solve the following simultaneous equation graphically.
3x – 4y = –7; 5x – 2y = 0
Answer

The given simultaneous equations are
$3 x-4 y=-7$
$\therefore \quad 4 y=3 x+7$
$\therefore \quad y=\frac{3 x+7}{4}$
x-1-535
y1-27-5. 5
(x, y)(-1, 1)(5, -2)(3, 4)(5, 5.5)
$5 x-2 y=0$
$\therefore \quad 2 y=5 x$
$\therefore \quad y=\frac{5}{2} x$
x02-21
y05-52
(x, y)(0, 0)(2, 5)(-2 -5)(11, 02)

Image
The two lines intersect at point (1, 2.5).
∴ x = 2 and y = 4 is the solution of the simultaneous equations 3x – y – 2 = 0 and 2x + y = 8.
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Question 1003 Marks
A certain amount is equally distributed among certain number of students. Each would get ₹ 2 less if 10 students were more and each would get ₹ 6 more if 15 students were less. Find the number of students and the amount distributed.
Answer
Let the number of students be $x$ and amount given to each stud be ₹ $y$.
$\therefore$ Total amount distributed is $x y$
From the first condition we get,
$\begin{aligned}
& (x+10)(y-2)=x y \\
& \therefore x y-2 x+10 y-20=x y \\
& \therefore-2 x+10 y=20 \\
& \therefore-x+5 y=10 \ldots \text { (I) }
\end{aligned}$
From the $2^{\text {nd }}$ condition we get,
$\begin{aligned}
& (x-15)(y+6)=x y \\
& \therefore x y+6 x-15 y-90=x y \\
& \therefore 6 x-15 y=90 \\
& \therefore 2 x-5 y=30 \ldots \text { (II) }
\end{aligned}$
Adding equations (I) and (II)
$\begin{aligned}
-x+5 y & =10 \\
^+ 2 x-5 y & =30 \\
\hline x & =40
\end{aligned}$
Substitute this value of $x$ in equation (I)
$\begin{gathered}
-x+5 y=10 \\
\therefore-40+5 y=10 \\
\therefore \quad 5 y=50 \\
\therefore \quad y=10
\end{gathered}$
Total amount distributed is $=x y=40 \times 10=₹ 400$.
$\therefore \quad ₹ 400$ distributed equally among 40 students.
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Question 1013 Marks
Solve : 15x + 17y = 21; 17x + 15y = 11
Answer
$\begin{gathered}
15 x+17 y=21 \ldots \text { (I) } \\
17 x+15 y=11 \ldots \text { (II) }
\end{gathered}$
In the two equations above, the coefficients of $x$ and $y$ are interchanged. While solving such equations we get two simple equations by adding and subtracting the given equations. After solving these equations, we can easily find the solution.
Let's add the two given equations.
$\begin{aligned}
15 x+17 y & =21 \\
+17 x+15 y & =11 \\
\hline 32 x+32 y & =32
\end{aligned}$
Dividing both sides of the equation by 32 .
$x+y=1 \ldots \text {(III)}$
Now, let's subtract equation (II) from (I)
$\begin{aligned}
& \quad 15 x+17 y=21 \\
& \begin{array}{r}
+ \_17 x \pm 15 y=\_11 \\
\hline-2 x+2 y=10
\end{array} \\
\end{aligned}$
dividing the equation by 2 .
$-x+y=5 \ldots \text { (IV) }$
Now let's add equations (III) and (V).
$\begin{aligned}
& \quad x+y=1 \\
+ & -x+y=5 \\
& \hline \therefore \quad 2 y=6 \\
& \therefore \quad y=3 \\
\end{aligned}$
Place this value in equation (III).
$\begin{array}{ll} 
x+y=1 \\
\therefore x+3=1 \\
\therefore x=1-3 \quad \therefore x=-2 \\
(x, y)=(-2,3)  \text { is the solution. }
\end{array}$
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Question 1023 Marks
Solve :
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
Answer
$\begin{aligned}
\frac{4}{x-y}+\frac{1}{x+y} & =3 ; \frac{2}{x-y}-\frac{3}{x+y}=5 \\
4\left(\frac{1}{x-y}\right)+1\left(\frac{1}{x+y}\right) & =3 \ldots \text { (I) } \\
2\left(\frac{1}{x-y}\right)-3\left(\frac{1}{x+y}\right) & =5 \ldots \text { (II) }
\end{aligned}$
Replacing $\left(\frac{1}{x-y}\right)$ by $a$ and $\left(\frac{1}{x+y}\right)$ by $b$ we get
$\begin{aligned}
& 4 a+b=3 \ldots \text { (III) } \\
& 2 a-3 b=5 \ldots \text { (IV) }
\end{aligned}$
On solving these equations we get, $a=1 b=-1$
But $a=\left(\frac{1}{x-y}\right), b=\left(\frac{1}{x+y}\right)$
$\begin{aligned}
\therefore\left(\frac{1}{x-y}\right) & =1,\left(\frac{1}{x+y}\right)=-1 \\
\therefore \quad x-y & =1 \ldots( V ) \\
x+y & =-1 \ldots( VI )
\end{aligned}$
Solving equation (V) and (VI) we get $x=0, y=-1$
$\therefore$ Solution of the given equations is $(x, y)=(0,-1)$
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Question 1033 Marks
Sum of two number is 45 and the greater number is twice the smaller number. Find the numbers.
Answer
30,15
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Question 1043 Marks
Shabana's age 10 years hence, will be twice juhi's present age. 6 years back shabana's age was $\frac{5}{3}$ times Juhi's at that time find their present ages.
Answer
26 years, 18 years
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Question 1053 Marks
If 1 is added to the numerator of a certain fraction its value becomes $\frac{1}{2}$ and if 1 is added to its denometer $\frac{1}{3}$. Find the original fraction.
Answer
$\frac{3}{8}$
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