Question 13 Marks
Solve the following simultaneous equation graphically.
5x – 6y + 30 = 0; 5x + 4y – 20 = 0
5x – 6y + 30 = 0; 5x + 4y – 20 = 0
Answer
View full question & answer→Coming soon
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Question 23 Marks
Solve the following simultaneous equation graphically.
x – 3y = 1; 3x – 2y + 4 = 0
x – 3y = 1; 3x – 2y + 4 = 0
Answer
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Question 33 Marks
Solve the following simultaneous equation graphically.
2x + 3y = 12; x – y = 1
2x + 3y = 12; x – y = 1
Answer
View full question & answer→Coming soon
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
View full question & 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.
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.
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
View full question & 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 $3 x =3 \times 90=₹ 270$
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 $3 x =3 \times 90=₹ 270$
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
View full question & 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.
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.
Question 73 Marks
To find number of notes that Anushka had, complete the following activity
Answer
View full question & 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 $
$₹ 100 \text { notes }=10$
$₹ 50 \text { notes }=20$
$100 x+50 y=2500\dots(I)$
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 $
$₹ 100 \text { notes }=10$
$₹ 50 \text { notes }=20$
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
View full question & 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.
$₹ 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.
Question 93 Marks
Solve the following simultaneous equations.
$\frac{1}{2(3 x+4 y)}+\frac{1}{5(2 x-3 y)}=\frac{1}{4} ; \frac{5}{(3 x+4 y)}-\frac{2}{(2 x-3 y)}=-\frac{3}{2}$
$\frac{1}{2(3 x+4 y)}+\frac{1}{5(2 x-3 y)}=\frac{1}{4} ; \frac{5}{(3 x+4 y)}-\frac{2}{(2 x-3 y)}=-\frac{3}{2}$
Answer
View full question & 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, $(x, y)=(2,1)$
$\frac{1}{2} m+\frac{1}{5} n=\frac{1}{4} \Rightarrow 5 m+2 n=\frac{10}{4} \Rightarrow 20 m+8 n=10 \Rightarrow 10 m+4 n=5 \dots(I)$
$5 m-2 n=-\frac{3}{2} \Rightarrow 10 m-4 n=-3 \ldots \text { (II) }$
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, $(x, y)=(2,1)$
Question 103 Marks
Solve the following simultaneous equations.
$\frac{7 x-2 y}{x y}=5 ; \frac{8 x+7 y}{x y}=15$
$\frac{7 x-2 y}{x y}=5 ; \frac{8 x+7 y}{x y}=15$
Answer
View full question & 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, $(x, y)=(1,1)$
$\frac{8 x+7 y}{x y}=15 \Rightarrow \frac{8 x}{x y}+\frac{7 y}{x y}=15 \Rightarrow \frac{8}{y}+\frac{7}{x}=15 \ldots (II)$
$\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, $(x, y)=(1,1)$
Question 113 Marks
Solve the following simultaneous equations.
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} ; \frac{231}{x}+\frac{148}{y}=\frac{610}{x y}$
$\frac{148}{x}+\frac{231}{y}=\frac{527}{x y} ; \frac{231}{x}+\frac{148}{y}=\frac{610}{x y}$
Answer
View full question & 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, $(x, y)=(1,2)$
$\frac{231}{x}+\frac{148}{y}=\frac{610}{x y} \Rightarrow \frac{231 y+148 x}{x y}=\frac{610}{x y} \Rightarrow 148 x+231 y=610 \ldots(II)$
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, $(x, y)=(1,2)$
Question 123 Marks
Solve the following simultaneous equations.
$\frac{7}{2 x+1}+\frac{13}{y+2}=27 ; \frac{13}{2 x+1}+\frac{7}{y+2}=33$
$\frac{7}{2 x+1}+\frac{13}{y+2}=27 ; \frac{13}{2 x+1}+\frac{7}{y+2}=33$
Answer
View full question & 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 $
$\text { Hence, }( x , y )=\left(-\frac{1}{4},-1\right)$
$13 m+7 n=33\dots(II)$
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 $
$\text { Hence, }( x , y )=\left(-\frac{1}{4},-1\right)$
Question 133 Marks
Solve the following simultaneous equations.
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
$\frac{2}{x}+\frac{2}{3 y}=\frac{1}{6} ; \frac{3}{x}+\frac{2}{y}=0$
Answer
View full question & 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, $(x, y)=(6,-4)$
$3 m+2 n =0 \ldots (II)$
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, $(x, y)=(6,-4)$
Question 143 Marks
Solve the following equations by Cramer’s method.
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
$\frac{x+y-8}{2}=\frac{x+2 y-14}{3}=\frac{3 x-y}{4}$
Answer
View full question & 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
$\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
Question 153 Marks
Solve the following equations by Cramer’s method.
7x + 3y = 15; 12y – 5x = 39
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)$
View full question & 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)$
Question 163 Marks
Solve the following equations by Cramer’s method.
$3 x -2 y =\frac{5}{2} ; \frac{1}{3} x +3 y =-\frac{4}{3}$
$3 x -2 y =\frac{5}{2} ; \frac{1}{3} x +3 y =-\frac{4}{3}$
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)$
View full question & 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)$
Question 173 Marks
Solve the following equations by Cramer’s method.
4m – 2n = –4; 4m + 3n = 16
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)$
View full question & 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)$
Question 183 Marks
Solve the following equations by Cramer’s method.
6x – 3y = –10; 3x + 5y – 8 = 0
6x – 3y = –10; 3x + 5y – 8 = 0
Answer
View full question & 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)$
$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)$
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) $\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]$
(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
View full question & 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) $D=\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]=(3 \times 4)-(-1 \times 1)=12+1=13$
(2) $D=\left[\begin{array}{cc}5 & -2 \\ -3 & 1\end{array}\right]=(5 \times 1)-2 \times-3=5-6=-1$
(3) $D=\left[\begin{array}{cc}3 & -1 \\ 1 & 4\end{array}\right]=(3 \times 4)-(-1 \times 1)=12+1=13$
Question 203 Marks
Solve the following simultaneous equation graphically.
3x + y = 10; x – y = 2
3x + y = 10; x – y = 2
Answer
View full question & answer→Coming soon
Question 213 Marks
Solve the following simultaneous equation graphically.
3x – y – 2 = 0; 2x + y = 8
3x – y – 2 = 0; 2x + y = 8
Answer
View full question & answer→Coming soon
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
View full question & 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.
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.
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
View full question & 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
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
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
View full question & 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 = $\frac{7}{18}$
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 = $\frac{7}{18}$
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
View full question & 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.
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.
Question 263 Marks
Complete the following.
Answer
View full question & 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 $
$\text { Breadth }=2 \times 8=16$
$\Rightarrow x -4 x + y + y =-8$
$\Rightarrow-2 x +2 y =-8$
$\Rightarrow- x + y =-4 \ldots . .(1)$
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 $
$\text { Breadth }=2 \times 8=16$
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
View full question & 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 .
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 .
Question 283 Marks
Solve the following simultaneous equation.
$\frac{1}{3 x+y}+\frac{1}{3 x-y}=\frac{3}{4} $
$\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{1}{8}$
$\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 $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
View full question & 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 $(x, y)=\left(\frac{1}{2},-\frac{1}{2}\right)$
Question 293 Marks
Solve the following simultaneous equation.
$\frac{27}{x-2}+\frac{31}{y+3}=85 ; \frac{31}{x-2}+\frac{27}{y+3}=89$
$\frac{27}{x-2}+\frac{31}{y+3}=85 ; \frac{31}{x-2}+\frac{27}{y+3}=89$
Answer
View full question & 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 $(x, y)=\left(\frac{5}{2},-2\right)$
$\frac{31}{x-2}+\frac{27}{y+3}=89$
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 $(x, y)=\left(\frac{5}{2},-2\right)$
Question 303 Marks
Solve the following simultaneous equation.
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
$\frac{10}{x+y}+\frac{2}{x-y}=4 ; \frac{15}{x+y}-\frac{5}{x-y}=-2$
Answer
View full question & 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 $(x, y)=(3,2)$
$\frac{15}{x+y}-\frac{5}{x-y}=-2$
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 $(x, y)=(3,2)$
Question 313 Marks
Solve the following simultaneous equation.
$\frac{2}{x}-\frac{3}{y}=15 ; \frac{8}{x}+\frac{5}{y}=77$
$\frac{2}{x}-\frac{3}{y}=15 ; \frac{8}{x}+\frac{5}{y}=77$
Answer
View full question & 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 $( x , y )=\left(\frac{1}{9}, 1\right)$
$\frac{8}{x}+\frac{5}{y}=77$
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 $( x , y )=\left(\frac{1}{9}, 1\right)$
Question 323 Marks
Solve the following simultaneous equations using Cramer’s rule.
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
$2 x+3 y=2 ; x-\frac{y}{2}=\frac{1}{2}$
Answer
View full question & 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.
$x -\frac{ y }{2}=\frac{1}{2} \Rightarrow 2 x - y =1$
$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.
Question 333 Marks
Solve the following simultaneous equations using Cramer’s rule.
4m + 6n = 54; 3m + 2n = 28
4m + 6n = 54; 3m + 2n = 28
Answer
View full question & 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.
$3 m+2 n =28$
$D =\begin{bmatrix}4 & 6 \\3 & 2 \end{bmatrix}=(4 \times 2)-(6 \times 3)=8-18=10$
$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.
Question 343 Marks
Solve the following simultaneous equations using Cramer’s rule.
6x – 4y = –12; 8x – 3y = –2
6x – 4y = –12; 8x – 3y = –2
Answer
View full question & 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.
$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.
Question 353 Marks
Solve the following simultaneous equations using Cramer’s rule.
$x + 2y = –1; 2x – 3y = 12$
$x + 2y = –1; 2x – 3y = 12$
Answer
View full question & 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.
$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.
Question 363 Marks
Solve the following simultaneous equations using Cramer’s rule.
4x + 3y – 4 = 0; 6x = 8 – 5y
4x + 3y – 4 = 0; 6x = 8 – 5y
Answer
View full question & answer→$4 x+3 y=4 $
$6 x+5 y=8$
$D=\begin{bmatrix}4 & 3 \\6 & 5\end{bmatrix}$$=(4 \times 5)-(3 \times 6)=20-18=2 $
$D_x=\begin{bmatrix} 4 & 3 \\8 & 5 \end{bmatrix}=(4 \times 5)-(3 \times 8)=20-24=-4$
$D_y=\begin{bmatrix} 4 & 4 \\ 6 & 8 \end{bmatrix}=(4 \times 8)-(4 \times 6)=32-24=8$
$x=\frac{D_x}{D}=-\frac{4}{2}=-2 y=\frac{D_y}{D}=\frac{8}{2}=4$
$\therefore(x, y)=(-2,4)$ is the solution.
$6 x+5 y=8$
$D=\begin{bmatrix}4 & 3 \\6 & 5\end{bmatrix}$$=(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.
Question 373 Marks
Solve the following simultaneous equations using Cramer’s rule.
3x – 4y = 10; 4x + 3y = 5
3x – 4y = 10; 4x + 3y = 5
Answer
View full question & answer→$\therefore(x, y)=(2,-1)$ is the solution
$3 x-4 y=10 $
$4 x+3 y=5 $
$D=\left|\begin{array}{cc}3 & -4 \\4 & 3 \end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25$
$D_x=\left[\begin{array}{cc} 10 & -4 \\ 5 & 3 \end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50$
$D_y=\left[\begin{array}{cc}3 & 10 \\ 4 & 5\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25$
$x =\frac{ D _{ x }}{ D }=\frac{50}{25}=2 y =\frac{ D _{ y }}{ D }=-\frac{25}{25}=-1$
$3 x-4 y=10 $
$4 x+3 y=5 $
$D=\left|\begin{array}{cc}3 & -4 \\4 & 3 \end{array}\right|=(3 \times 3)-(-4 \times 4)=9+16=25$
$D_x=\left[\begin{array}{cc} 10 & -4 \\ 5 & 3 \end{array}\right]=(10 \times 3)-(-4 \times 5)=30+20=50$
$D_y=\left[\begin{array}{cc}3 & 10 \\ 4 & 5\end{array}\right]=(3 \times 5)-(10 \times 4)=15-40=-25$
$x =\frac{ D _{ x }}{ D }=\frac{50}{25}=2 y =\frac{ D _{ y }}{ D }=-\frac{25}{25}=-1$
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)$\left|\begin{array}{cc}\frac{7}{3} & \frac{5}{3} \\ \frac{3}{2} & \frac{1}{2}\end{array}\right|$
(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
View full question & 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) $\frac{7}{3} \times \frac{1}{2}-\frac{5}{3} \times \frac{3}{2}=\frac{7}{6}-\frac{15}{6}$
$=-\frac{8}{6}$
$=-\frac{4}{3}$
$\left|\begin{array}{ll}a & b \\ c & d\end{array}\right|$
is (ad - bc)
(1) $(-1 \times 4)-(7 \times 2)=-4-14=-18$
(2) $(5 \times 0)-(3 \times-7)=0-(-21)=21$
(3) $\frac{7}{3} \times \frac{1}{2}-\frac{5}{3} \times \frac{3}{2}=\frac{7}{6}-\frac{15}{6}$
$=-\frac{8}{6}$
$=-\frac{4}{3}$
Question 393 Marks
Solve the following simultaneous equation graphically.
3x – 4y = –7; 5x – 2y = 0
3x – 4y = –7; 5x – 2y = 0
Answer
View full question & answer→Coming soon
Question 403 Marks
Solve the following simultaneous equation graphically.
3x – y = 2; 2x – y = 3
3x – y = 2; 2x – y = 3
Answer
View full question & answer→Coming soon
Question 413 Marks
Solve the following simultaneous equation graphically.
x + y = 0; 2x – y = 9
x + y = 0; 2x – y = 9
Answer
View full question & answer→Coming soon
Question 423 Marks
Solve the following simultaneous equation graphically.
x + y = 5; x – y = 3
x + y = 5; x – y = 3
Answer
View full question & answer→Coming soon
Question 433 Marks
Solve the following simultaneous equation graphically.
(1) x + y = 6; x – y = 4
(1) x + y = 6; x – y = 4
Answer
View full question & answer→Coming soon
Question 443 Marks
Complete the following table to draw graph of the equations -
(I) x+y = 3 (II) x – y = 4
(I) x+y = 3 (II) x – y = 4
Answer
View full question & answer→Coming soon
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.
Question 463 Marks
Solve : 15x + 17y = 21; 17x + 15y = 11
Question 473 Marks
Solve :
$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
View full question & answer→$\frac{4}{x-y}+\frac{1}{x+y}=3 ; \frac{2}{x-y}-\frac{3}{x+y}=5$
Question 483 Marks
Solve the following simultaneous equations.
$5 x-3 y=8 ; 3 x+y=2$
View full question & answer→$5 x-3 y=8 ; 3 x+y=2$
Question 493 Marks
Sum of two number is 45 and the greater number is twice the smaller number. Find the numbers.
Answer
View full question & answer→30,15
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
View full question & answer→$\frac{3}{8}$