Question 15 Marks
A part of monthly expenses of a family is constants and the remainingvarywith the number of members in the family. For a family of $4$ person, the total monthly expenses are $Rs. 10,400$ whereas for a family of $7$ persons, the total monthly expenses are $Rs. 15,800$. Find the constant expenses per month and the monthly expenses of each member of a family.
AnswerLet $x$ be the constant expense per month of the family,
and $y$ be the expense per month for a single member of the family,For a family of $4$ people,
the total monthly expense is $Rs. 10,400.$
$x + 4y = 10,400 \dots...(1)$
For a family of $7$ people,
the total monthly expense is $Rs. 15,800.$
$x + 7y = 15,800 \dots...(2)$
Subtracting equation $(1)$ from equation $(2)$,We get :
$x + 7y = 15800$
$- x + 4y = 10400$
$- - - $
$3y = 5400$
$y = 1800$
Substituting $y = 1800$ in equation $(1)$, We get
$x + 4( 1800 ) = 10,400$
$\Rightarrow x = 3200.$
$\therefore $ The constant expense is $Rs. 3,200$ per month and the monthly expense of each member of a family is $Rs.1,800.$
View full question & answer→Question 25 Marks
The class $XI$ students of school wanted to give a farewell party to the outgoing students of class $XII$. They decided to purchase two kinds of sweets, one costing $Rs. 250$ per $kg$ and other costing $Rs. 350$ per $kg$. They estimated that $40 \ kg$ of sweets were needed. If the total budget for the sweets was $Rs. 11,800$; find how much sweets of each kind were bought ?
AnswerAssume $x \ kg$ of the first kind costing $Rs. 250$ per $kg$ and $y \ kg$ of the second kind costing $Rs. 350$ per$ \ kg$ sweets were bought.It is estimated that $40 \ kg$ of sweets were needed.
$\Rightarrow x + y = 40\dots....(1)$
The total budget for the sweets was $Rs. 11,800.$
$\Rightarrow 250x + 350y = 11,800 ....(2)$
Multiply equation $(1)$ by $250$, We get :
$250x + 250y = 10,000 \dots.....(3)$
Subtracting equation $(2)$ from $(3),$
$250x + 250y = 10,000$
$- 250x + 350y = 11,800$
$- - - $
$- 100y = - 1800$
$y = 18$
Substituting $y = 18$ in equation $(1)$, We get
$x + 18 = 40$
$\Rightarrow x = 22$
$\therefore 22 \ kgs$ of the first kind costing $Rs. 250$ per $kg$ and $18 \ kgs$ of the second kind costing $Rs. 350$ per $kg$ sweets were bought.
View full question & answer→Question 35 Marks
$90\%$ acid solution $(90\%$ pure acid and $10\%$ water$)$ and $97\%$ acid solution are mixed to obtain $21$ litres of $95\%$ acid solution. How manylitresof each solution are mixed.
AnswerLet the quantity of $90\%$ acid solution be $x$ litres and The quantity of $97\%$ acid solution be ylitres
According to the question,
$x + y = 21 \dots...(1)$
and $90\%$ of $x + 97\%$ of $y = 95\%$ of $21$
$90x + 97y = 1995\dots ...(2)$
Multiplying equation no. $(1)$ by $90$, we get,
$90x + 90y = 1890 \dots....(3)$
Subtracting equation $(2)$ from $(3)$
$90x + 90y = 1890$
$- 90x + 97y = 1995$
$- - -$
$- 7y = - 105$
$y = 15$
From$ (1)$
$x + 15 = 21$
$x = 6$
Hence,$ 90\%$ acid solution is $6$ litresand $97\%$ acid solution is $15$ litres.
View full question & answer→Question 45 Marks
The sum of digit of a two digit number is $11$. If the digit at ten's place is increased by $5$ and the digit at unit place is decreased by $5$, the digits of the number are found to be reversed. Find the original number.
AnswerLet $x$ be the number at the ten's place.
and $y$ be the number at the unit's place.
So, the number is $10x + y.$
The sum of digit of a two digit number is $11.$
$\Rightarrow x + y = 11 \dots...(1)$
lf the digit at ten's place is ineased by $5$ and the digit at unit place is decreased by $5,$
the digits of the number are found to be reversed.
$\Rightarrow 10( x + 5 ) + ( y - 5 ) = 10y + x$
$\Rightarrow 9x - 9y = -45$
$\Rightarrow x - y = -5 \dots...(2)$
Subtracting equation $(1)$ from equation $(2), $we get :
$x - y = - 5$
$- x + y = 11$
$- - - $
$- 2y = - 16$
$\Rightarrow y = 8$
Substituting $y = 8$ in equation $(1),$ we get
$x + 8 = 11$
$\Rightarrow x = 3$
$\therefore $ The number is $10x + y = 10(3) + 8 = 38.$
View full question & answer→Question 55 Marks
From Delhi station, if we buy $2$ tickets for station $A$ and $3$ tickets for station $B,$ the total cost is $Rs. 77$. But if we buy $3$ tickets for station $A$ and $5$ tickets for station $B$, the total cost is $Rs. 124$. What are the fares from Delhi to station $A$ and to station $B$ ?
AnswerLet, the fare of ticket for station $A$ be $Rs. x$ and the fare of ticket for station $B$ be $Rs. y$
According, to the question
$2x + 3y = 77\dots ...(1)$ and
$3x + 5y = 124\dots ...(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $2.$
$6x + 9y = 231 \dots...(3)$
$6x + 10y = 248 \dots....(4)$
Subtracting equation $(4)$ from $(3)$
$6x + 9y = 231$
$- 6x + 10y = 248$
$- - - $
$- y = - 17$
$y = 17$
$From (1)$
$2x + 3 (17) = 77$
$2x = 77 - 51$
$2x = 26$
$x = 13$
Thus, fare for station $A = Rs. 13$ and, fare for station $B = Rs. 17.$
View full question & answer→Question 65 Marks
Rohit says to Ajay, $“\ $Give mehundred, I shall then become twice as rich as you.$”\ $Ajay replies, $“\ $if you give me ten, I shall be six times as rich as you.$”\ $ How much does each have originally ?
AnswerLet Rohit has $Rs. x$ and Ajay has$ Rs. y$
When Ajay gives $Rs. 100$ to Rohit
$x + 100 = 2(y - 100)$
$x - 2y = -300 \dots...(1)$
WhenRohitgives $Rs. 10$ to Ajay
$6(x-10) = y + 10$
$6x - y = 70\dots ...(2)$
Multiplying equation no. $(2)$ By $2.$
$12x - 2y = 140\dots ...(3)$
Subtracting equation $(1)$ and $(3)$
$12x - 2y = 140$
$- x - 2y = - 300$
$- + + $
$11x = 440$
$x = 40$
From $(1)$
$40 - 2y = - 300$
$\Rightarrow - 2y = - 340$
$\Rightarrow y = 170$
Thus, Rohit has $Rs. 40$ and Ajay has $Rs. 170.$
View full question & answer→Question 75 Marks
It takes $12$ hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for $4$ hours and the pipe of smaller diameter is used for $9$ hours, only half of the pool is filled. How long would each pipe take to fill the swimming pool ?
AnswerLet the pipe with larger diameter and smaller diameter be pipes $A$ and $B$ respectively.
Also, let pipe $A$ work at a rate of $x$ hours / unit and pipe $B$ work at a rate of $Y$ hours / unit.
According to the question,
$x+y=\frac{1}{12}$
$\Rightarrow 12 x+12 y=1\ldots .(1)$
and
$4 x+9 y=\frac{1}{2}$
$\Rightarrow 8 x+18 y=1\ldots(2)$
Multiply (1) by 2 and (2) by 3 , We get
$24 x+24 y=2\ldots(3)$
$24 x+54 y=3\ldots(4)$
Subtracting equation $(4)$ from $(3),$
$24 x+24 y=2$
$- 24 x+54 y=3$
$-30 y=-1$
$y=\frac{1}{30}$
Putting $y=\frac{1}{30}$ in equation $(1)$
$12 x+12 y=1$
$ \therefore 12 x+12 x \frac{1}{30}=1$
$ \therefore 12 x+\frac{2}{5}=1$
$ \therefore 12 x=1-\frac{2}{5}$
$ \therefore x=\frac{3}{5} \times \frac{1}{12}$
$ \therefore x=\frac{1}{20}$
Hence, the pipe with larger diameter will take $20$ hours to fill the swimming pool and the pipe with smaller diameter will take $30$ hours to fill the swimming pool.
View full question & answer→Question 85 Marks
The area of a rectangle gets reduced by $9$ square units, if its length is reduced by $5$ units and breadth is increased by $3$ units. However, if the length of this rectangle increases by $3$ units and the breadth by $2$ units, the area increases by $67$ square units. Find the dimensions of the rectangle.
AnswerLet the length of the rectangle be $x$ units and the breadth of the rectangle be $y$ units.
We know that, area of a rectangle $=$ length $x$ breadth $=x y$ According to the question,
$ x y-9=(x-5)(y+3)$
$ \Rightarrow x y-9=x y+3 x-5 y-15$
$\Rightarrow 3 x-5 y=6\ldots .(1)$
$ x y+67=(x+3)(y+2)$
$ \Rightarrow x y+67=x y+2 x+3 y+6$
$\Rightarrow 2 x+3 y=61\ldots(2)$
Multiply $(1)$ by $2$ and $(2)$ by $3 $, we get
$6 x-10 y=12 \dots...(3)$
$6 x+9 y=183\ldots(4)$
Subtracting equation $(4)$ from $(3),$
$6 x-10 y =12$
$- 6 x+9 y =183$
$- - -$
$-19 y =-171$
$y =9$
Putting $y=9$ in equation $(1)$
$3 x-5 y=6$
$ 3 x-5(9)=6$
$ 3 x=6+45$
$ x=\frac{51}{3}$
$ x=17$
Hence, the length of the rectangle is $17$ units and the breadth of the rectangle is $9$ units.
View full question & answer→Question 95 Marks
Two articles $A$ and $B$ are sold for $Rs. 1,167$ making $5\%$ profit on $A$ and $7\%$ profit on $A$ and $7\%$ profit on $B$. IF the two articles are sold for $Rs. 1,165$, a profit of $7\%$ is made on $A$ and a profit of $5\%$ is made on $B$. Find the cost prices of each article.
AnswerLet the cost price of article $A= Rs. x$
and the cost price of articles $B= Rs. y$
According to the question,
$(x+5 \%$ of $x)+(y+7 \%$ of $y)=1167$
$ {\left[x+\frac{5}{100} x\right]+\left[y+\frac{7}{100} y\right]=1167}$
$ \frac{21 x}{20}+\frac{107 y}{100}=1167$
$105 x+107 y=116700\ldots(1)$
and,
$[107 x] / 100+[105 y] / 100=1165$
$107 x+105 y=116,500\ldots(2)$
Adding $(1)$ and $(2)$
$105 x+107 y=116700$
$+ 107 x+105 y=116500$
$ 212 x+212 y=233200$
Dividing by$ 212 ,$
$x+y=1100\ldots(3)$
subtracting $(2)$ from $(1)$
$105 x+107 y=116700$
$- 107 x+105 y=116500$
$-2 x+2 y=200$
Dividing by $2 ,$
$-x+y=100\ldots(4)$
Adding equation $(4)$ and $(3)$
$-x+y =100$
$+ x+y =1100$
$2 y =1200$
$y =600$
from $(3)$
$x+600=1100$
$ x=500$
Thus, cost price of article $A$ is $Rs. 500$ and that of article $B$ is $Rs. 600.$
View full question & answer→Question 105 Marks
The annual incomes of $A$ and $B$ are in the ratio $3 :4$ and their annual expenditure are in the ratio $5 : 7.$ If each $Rs. 5000;$ find their annual incomes.
AnswerLet $A's$ annual in come $= Rs. x$
and $B's$ annual income $= Rs. y$
According to the question,
$\frac{x}{y}=\frac{3}{4}$
$4 x-3 y=0$
$\ldots(1)$
and,
$\frac{x-5000}{y-5000}=\frac{5}{7}$
$7 x-5 y=10000$
$\ldots(2)$
Multiplying equation no. $(1)$ by $7$ and $(2)$ by $4 .$
$28 x-21 y=0 \dots....(3)$
$28 x-20 y=40,000 \dots.....(4)$
Subtracting equation $(4)$ from $(3)$
$28 \mathrm{x}-21 \mathrm{y}=0$
$ -28 x-20 y=40,000$
$y=-40,000$
$ y=40,000$
From $(1)$
$4 x-3(40,000)=0$
$ x=30,000$
From $(1)$
$4 x-3(40,000)=0$
$ x=30,000$
Thus, $A's$ income in $Rs. 30,000$ and $B's$ income is $Rs. 40,000 .$
View full question & answer→Question 115 Marks
Four years ago, a mother was four times as old as her daughter. Six years later, the mother will be two and a half times as old as her daughter at that time. Find the present ages of the mother and her daughter.
AnswerLet the present age of the mother be $x$ years.
and the present age of the daughter be $y$ year.
$4$ yrs ago age $=(x-4)$
mother's age $=4(y-4)=4 y-16$
According to the equation,
$x-4=4(y-4)$
$ \Rightarrow x-4=4 y-16$
$\Rightarrow x-4 y=-12\dots ...(1)$
And,
$6$ yrs later daughter's age $=x+6$
mother's age $=2 \frac{1}{2}(y+6)$
$x+6=2 \frac{1}{2}(y+6)$
$\Rightarrow \mathrm{x}+6=\frac{5}{2} \mathrm{y}+15$
$\Rightarrow \mathrm{x}-\frac{5}{2} \mathrm{y}=9$
Solving $(1)$ and $(2),$
We get $y=14$ and $x=44$
Hence, the present age of the mother is $44$ years and the present age of the daughter is $14$ years.
View full question & answer→Question 125 Marks
$A$ is $20$ years older than $B. 5$ years ago, $A$ was $3$ times as old as $B$. Find their present ages.
AnswerLet $A’s$ presentage be $x$ years
and $B’s$ present age be $y$ years
According to the question
$x = y + 20$
$x - y = 20\dots ...(1)$
Five years ago,
$x - 5 = 3(y - 5)$
$x - 5 = 3y - 15$
$x - 3y = - 10 \dots...(2)$
Subtracting $(1)$ and $(2),$
$x - 3y = - 10$
$- x - y = 20 $
$- + - $
$- 2y = - 30$
$y = 15$
From$ (1)$
$x = 15 + 20$
$x = 35$
Thus, present ages of $A$ and $B$ are $35$ years and $15 $years.
View full question & answer→Question 135 Marks
Five years ago, $A's$ age was four times the age of $B$. Five years hence, $A’s$ age will be twice the age of $B$. Find their preset ages.
AnswerLet present age of $A = x$ years
And present age of $B = y$ years
According to the question,
Five years ago,
$x - 5 = 4(y - 5)$
$x - 4y = -15\dots ...(1)$
Five years later,
$x + 5 = 2(y + 5)$
$x + 5 = 2y + 10$
$x - 2y = 5 \dots....(2)$
Now subtracting $(1)$ from $(2)$
$x - 2y = 5$
$- x - 4y = - 15$
$- + + $
$2y = 20$
$y = 10$
From $(1)$
$x - 4 (10) = -15$
$x = 25$
Present ages of $A$ and $B$ are $25$ years and $10$ years respectively.
View full question & answer→Question 145 Marks
Pooja and Ritu can do a piece of work in $17 \frac{1}{7}$ days. If one day work of Pooja be three fourth of one day work of Ritu' find in how many days each will do the work alone.
AnswerLet Pooja's $1$ day work $=\frac{1}{x}$
and Ritu's $1$ day work $=\frac{1}{y}$
According the question,
$\frac{1}{x}+\frac{1}{y}=\frac{7}{120}\ldots .(1)$
and
$\frac{1}{x}=\frac{3}{4} \cdot \frac{1}{y}$
$y=\frac{3}{4} x\ldots .(2)$
Using the value of $y$ from $(2)$ in $(1)$
$\frac{1}{x}+\frac{4}{3 x}=\frac{7}{120}$
$ \frac{1}{x}\left(\frac{7}{3}\right)=\frac{7}{120}$
$ x=40$
From$ (2)$
$y=\frac{3}{4}(40)$
$ y=30$
Pooja will complete the work in $40$ days and Ritu will complete the work in $30$ days.
View full question & answer→Question 155 Marks
If the numerator of a fraction is increased by $2$ and denominator is decreased by $1$ , it becomes $\frac{2}{3}$. If the numerator is increased by $1$ and denominator is increased by $2 $, it becomes $\frac{1}{3}$.Find the fraction.
AnswerLet the numerator and denominator a fraction be $x$ and $y$ respectively.
According to the question,
$\frac{x+2}{y-1}=\frac{2}{3}$
$ 3 x+6=2 y-2$
$3 x-2 y=-8\dots........(1)$
And,
$\frac{x+1}{y+2}=\frac{1}{3}$
$ 3 x+3=y+2$
$3 x-y=-1\dots .......(2)$
Now,
Subtracting equation $(1)$ from$ (2),$
$3 x-y=-1$
$- 3 x-2 y=-8$
$ - + + $
$ y=7$
From $(1),$
$3 x-2(7)=-8$
$ 3 x=-8+14$
$ x=2$
Required fraction $=\frac{2}{7}$
View full question & answer→Question 165 Marks
Two numbers are in the ratio $4:5$. If $30$ is subtracted from each of the numbers, the ratio becomes $1:2.$ Find the numbers.
AnswerLet the common multiple between the numbers be $x .$
So, the numbers are $x \ y$
$\frac{x}{y}=\frac{4}{5}$
$ \frac{x-30}{y-30}=\frac{1}{2}$
$ 2(x-30)=y-30$
$ 2 x-60=y-30$
$ 2 x-y=30$
$2\left(\frac{4}{5 y}\right)-y=30$
$ \frac{8 y}{5}-y=30$
$ \frac{8 y-5 y}{5}=30$
$ 3 y=150$
$ y=\frac{150}{3}$
$ y=50$
from eq.$(ii) $
From eq. $(i) \ (ii)$
$x=\frac{4}{5}(50)$
$ \mathrm{x}=4 \times 10$
$ \mathrm{x}=40$
View full question & answer→Question 175 Marks
The difference between two positive numbers $x$ and $y(x>y$ ) is $4$ and the difference between their reciprocals is $\frac{4}{21}$. Find the numbers.
AnswerTwo numbers are $x$ and $y$ respectively such that $x>y$.
Then,
$x-y=4\ldots...(i)$
$\Rightarrow x=4+y$
And,
$\frac{1}{y}-\frac{1}{x}=\frac{4}{21}$
$ \Rightarrow \frac{x-y}{x y}=\frac{4}{21}$
$\Rightarrow \frac{4}{x y}=\frac{4}{21} \dots......$[From $(1)]$
$\Rightarrow x y=21$
$ \Rightarrow(4+y) y=21$
$ \Rightarrow 4 y+y 2=21$
$ \Rightarrow y 2+4 y-21=0$
$ \Rightarrow y 2+7 y-3 y-21=0$
$ \Rightarrow y(y+7)-3(y+7)=0$
$ \Rightarrow(y-3)(y+7)=0$
$ \Rightarrow y=3$ and $y=-7$
We reject $y=-7$ since $y$ is positive.
$\Rightarrow y=3$
$ \Rightarrow x=4+y=4+3=7$
Thus, the two numbers are $7$ and $3$ respectively.
View full question & answer→Question 185 Marks
The sum of two numbers is $8$ and the sum of their reciprocals is $\frac{8}{15}$. Find the numbers.
AnswerLet the two numbers be $x$ and $y$ respectively.
Then,
$x+y=8\ldots (i)$
$\Rightarrow x=8-y$
And,
$\frac{1}{x}+\frac{1}{y}=\frac{8}{15}$
$ \Rightarrow \frac{y+x}{x y}=\frac{8}{15}$
$\Rightarrow \frac{8}{x y}=\frac{8}{15}\dots .....[$ From $(1) ]$
$\Rightarrow x y=\frac{8 \times 15}{8}$
$\Rightarrow x y=15 \dots...(iii)$
$\Rightarrow(8-y) y=15$
$ \Rightarrow 8 y-y^2=15$
$ \Rightarrow y^2-8 y+15=0$
$ \Rightarrow y^2-3 y-5 y+15=0$
$ \Rightarrow y(y-3)-5(y-3)=0$
$ \Rightarrow(y-3)(y-5)=0$
$ \Rightarrow y=3 \text { or } y=5$
$ \therefore x=8-y$
$ x=8-3=5$
$ x=8-5=3$
$ \Rightarrow x=5$ or $x=3$
Thus, the two numbers are $3$ and $5$ respectively.
View full question & answer→Question 195 Marks
The sum of two positive numbers $x$ and $y (x > y)$ is $50$ and the difference of their squares is $720$. Find the numbers.
AnswerTwo numbers are $x$ and $y$ such that $x>y$.
$x+y=50 \dots...(i)$
$ x^2-y^2=720$
$ \Rightarrow(x+y)(x-y)=720$
$ \Rightarrow 50(x-y)=720 \ldots[$ Using equation $(i)]$
$\Rightarrow \mathrm{x}-\mathrm{y}=\frac{720}{50}$
$\Rightarrow x-y=14.4\dots ....(ii)$
Adding $(i)$ and $(ii)$, we get
$\begin{gathered}x+y=50,+x-y=14.4, 2 x=64.4\end{gathered}$
$x=\frac{64.4}{2}$
$ \Rightarrow x=32.2$
Substituting the value of $x$ in $(i)$, we have
$x+y=50$
$ 32.2+y=50$
$ \Rightarrow y=50-32.2$
$ \Rightarrow y=17.8$
Thus, the two numbers are $17.8$ and $32.2$ respectively.
View full question & answer→Question 205 Marks
When the greater of the two numbers increased by $1$ divides the sum of the numbers, the result is $\frac{3}{2}$. When the difference of these numbers is divided by the smaller, the result $\frac{1}{2}$. Find the numbers.
AnswerLet the two numbers be $a$ and $b$ respectively such that $b>a$.
According to given condition,
$\frac{a+b}{b+1}=\frac{3}{2}$
$\Rightarrow 2 a+2 b=3 b+3$
$\Rightarrow 2 a-b=3\ldots . . .(1)$
Also,
$\frac{b-a}{a}=\frac{1}{2}$
$ \Rightarrow 2 \mathrm{\sim b}-2 \mathrm{a}=\mathrm{a}$
$\Rightarrow 2 b-3 a=0\ldots . . . .(2)$
Multiplying $(1)$ by $2,$ We get
$4 a-2 b=6 \dots.......(3)$
Adding $(2)$ and $(3),$ We get
$\begin{gathered}4 a-2 b=6,+\quad-3 a+2 b=0, a=6\end{gathered}$
Substituting $a=6$ in equation $(1),$ We get
$2(6)-b=3$
$ \Rightarrow 12-b=3$
$ \Rightarrow b=9$
Thus, the two numbers are $6$ and $9$ respectively.
View full question & answer→Question 215 Marks
Two numbers are in the ratio $4 :7.$ If thrice the larger be added to twice the smaller, the sum is $59$. Find the numbers.
AnswerLet the smaller number be $x$ and the larger number be $y$.
According to the question,
$ \frac{x}{y}=\frac{4}{7}$
$ 7 x=4 y$
$7 x-4 y=0\ldots . . .(1)$
and, $3 y+2 x=59\ldots . . .(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $4 .$
$21 x-12 y=0\ldots . . .(3)$
$8 x+12 y=236\dots ........(4)$
Adding equation $(3)$ and $(4),$
$\begin{gathered}21 x-12 y=0,+ 8 x+12 y=236,29 x=236x=\frac{236}{29}\end{gathered}$
From $(1)$
$7\left(\frac{236}{29}\right)=4 y$
$y=7\left(\frac{59}{29}\right)$
$ y=\left(\frac{413}{29}\right)$
Hence, the numbers are $\frac{236}{29}$ and $\frac{413}{29}$.
View full question & answer→Question 225 Marks
The ratio of two numbers is $\frac{2}{3}$. If $2$ is subtracted from the first and 8 from the second, the ratio becomes the reciprocal of the oriainal ratio. Find the numbers.
AnswerLet the two numbers be $x$ and $y$.
According to the question,
$\frac{x}{y}=\frac{2}{3}$
$3 x-2 y=0$
$\ldots . .(1)$
Also, $\frac{x-2}{y-8}=\frac{3}{2}$
$2 x-3 y=-20\ldots . . .(2)$
Multiplying equations no. $(1)$ by $2$ and $(2)$ by $3$
$6 x-4 y=0\ldots . .(3)$
$6 x-9 y=-60\ldots . . .(4)$
Subtracting equation $(4)$ from $(3)$
$6 x-4 y =0$
$- 6 x-9 y =-60$
$ - + + $
$5 y =60$
$y =12$
From $(1)$, we get
$ 3 x-2(12)=0$
$ x=\frac{24}{3}$
$ x=8$
Thus, the numbers are $8$ and $12 .$
View full question & answer→Question 235 Marks
Four times a certain two digit number is seven times the number obtained on interchanging its digits. If the difference between the digits is $4;$ find the number.
AnswerLet $x$ be the number at the ten's place.
and $y$ be the number at the unit's place.
So, the number is $10x + y$.
Four times a certain two$-$digit number is seven times
the number obtained on interchanging its digits.
$\Rightarrow 4( 10x + y ) = 7( 10y + x )$
$\Rightarrow 40x + 4y = 70y + 7x$
$\Rightarrow 33x - 66y = 0$
$\Rightarrow x - 2y = 0 ....(1)$
If the difference between the digits is $4$, then
$\Rightarrow x - y = 4 ...(2)$
Subtracting equation $(1)$ from equation $(2),$ we get :
$x - y = 4$
$- x - 2y = 0$
$- + - $
$y = 4$
Subtracting $y = 4$ in equation $(1),$ We get
$x - 2(4) = 0$
$\Rightarrow x = 8$
$\therefore $ The number is $10x + y = 10(8) + 4 = 84.$
View full question & answer→Question 245 Marks
A two$-$digit number is such that the ten’s digit exceeds twice the unit’s digit by $2$ and the number obtained by inter$-$changing the digits is $5$ more than thethesum of the digits. Find the two digit number.
AnswerLet the digit a unit’s place be $x$ and the digit at ten’s place be $y.$
Required no. $= 10y + x.$
According to the question,
$y - 2x = 2$
$-2x + y = 2 \dots...(1)$
and,
$(10x + y) -3 (y + x) = 5$
$7x - 2y = 5 \dots...(2)$
Multiplying equation no. $(1)$ by $2.$
$- 4x + 2y = 4 \dots...(3)$
Now adding $(2)$ and $(3),$
$- 4x + 2y = 4$
$+ 7x - 2y = 5$
$3x = 9$
$x = 3$
From $(1)$ ,we get
$-2(3) + y = 2$
$y = 8$
Required number is $10(8) + 3 = 83.$
View full question & answer→Question 255 Marks
The ten’s digit of a two digit number is three times the unit digit. The sum of the number and the unit digit is $32$. Find the number.
AnswerLet the digit at unit’s place be $x$ and the digit at ten’s place be $y.$
Required no. $= 10y + x$
According to the question,
$y = 3x$
$\Rightarrow 3x - y = 0 \dots...(1)$
and,
$10y + x + x = 32$
$10y + 2x = 32 \dots...(2)$
Multiplying equation no. $(1)$ by $10.$
$30x - 10y = 0\dots ....(3)$
Now,
Adding equation $(3)$ and $(2)$
$30x - 10y = 0$
$+ 2x + 10y = 32$
$32x = 32$
$x = 1$
From $(1)$, we get
$y = 3(1) = 3$
Required no is $=10(3) + 1 = 31.$
View full question & answer→Question 265 Marks
The sum of the digits of a two digit number is $7$. If the digits are reversed, the new number decreased by $2,$ equals twice the original number. Find the number.
AnswerLet the digit at unit’s place be $x$ and the digit at ten’s place be $y.$
Required no. $= 10y + x$
If the digit’s are reversed
Reversed no. $= 10x + y$
According to the question,
$x + y = 7\dots ...(1)$
and,
$10x + y - 2 = 2(10y + x).$
$8x - 19y = 2\dots ...(2)$
Multiplying equation no. $(1)$ by $19.$
$19x + 19y = 133\dots ...(3)$
Now adding equation $(2)$ and $(3)$
$19x + 19y = 133$
$+ 8x - 19y = 2 $
$27x = 135$
$x = 5$
From $(1)$
$5 + y = 7$
$y = 2$
Required number is $= 10(2) + 5 = 25.$
View full question & answer→Question 275 Marks
The sum of the digits of the digits of two digit number is $5$. If the digits are reversed, the number is reduced by $27$. Find the number.
AnswerLet the digit at unit’s place be $x$ and the digit at ten’s place $y.$
Required no.$ = 10y + x$
If the digit’s are reversed,
Reversed no.$ = 10x + y$
According to the question,
$x + y = 5 \dots...(1)$
and,
$(10y + x) - (10x + y) = 27$
$9y - 9x = 27$
$y - x = 3\dots ...(2)$
Now,
Adding equation $(1)$ and $(2),$
$y - x = 3 \dots....(2)$
$+ y + x = 5\dots ....(1)$
$2y = 8$
$y = 4$
From $(1)$
$x + 4 = 5$
$x = 1$
Require no is
$10 (4) + 1 = 41$
View full question & answer→Question 285 Marks
Solve :$\frac{2 x y}{x+y}=\frac{3}{2}; \frac{x y}{2 x-y}=-\frac{3}{10}; x+y \neq 0$ and $2 x-y \neq 0$
Answer$ \frac{2 x y}{x+y}=\frac{3}{2}$
$ \Rightarrow \frac{x+y}{x y}=\frac{4}{3}$
$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{4}{3}\ldots . .(1)$
$ \frac{x y}{2 x-y}=-\frac{3}{10}$
$ \Rightarrow \frac{2 x-y}{x y}=-\frac{10}{3}$
$\Rightarrow-\frac{1}{x}+\frac{2}{y}=-\frac{10}{3}\ldots . . .(2)$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
Then, equations $(1)$ and $(2)$ become
$u+v=\frac{4}{3}$ and $ -u+2 v=-\frac{10}{3}$
$ \Rightarrow 3 u+3 v=4$ and $-3 u+6 v=-10$
Adding, We have
$9 \mathrm{v}=-6$
$ \Rightarrow \mathrm{v}=-\frac{6}{9}=-\frac{2}{3}$
$ \Rightarrow \frac{1}{y}=-\frac{2}{3}$
$ \Rightarrow \mathrm{y}=-\frac{3}{2}$
Substituting $y=-\frac{3}{2}$ in $(1),$ We have
$\frac{1}{x}-\frac{2}{3}=\frac{4}{3}$
$\Rightarrow \frac{1}{x}=\frac{6}{3}=2$
$\Rightarrow \mathrm{x}=\frac{1}{2}$
Hence, $x=\frac{1}{2} \quad$ and $y=-\frac{3}{2}$
View full question & answer→Question 295 Marks
Solve:$\frac{a}{x}-\frac{b}{y}=0, \frac{a b^2}{x}+\frac{a^2 b}{y}=a^2+b^2$
AnswerGiven equation are $\frac{a}{x}-\frac{b}{y}=0$ and $\frac{a b^2}{x}+\frac{a^2 b}{y}=a^2+b^2$
Taking $\frac{1}{x}=u$ and $\frac{1}{y}=v$, the above system of equations become
$a u-b v+0=0$
$ a b^2 u+a^2 b v-\left(a^2+b^2\right)=0$
By cross$-$multiplication, we have
$\frac{u}{-b \times\left[-\left(a^2+b^2\right)\right]-a^2 b \times 0}=\frac{-v}{a \times\left[-\left(a^2+b^2\right)\right]-a b^2 \times 0}=\frac{1}{a \times a^2 b-a b^2 \times(-b)}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{-v}{-a\left(a^2+b^2\right)}=\frac{1}{a^3 b+a b^3}$
$ \Rightarrow \frac{u}{b\left(a^2+b^2\right)}=\frac{v}{a\left(a^2+b^2\right)}=\frac{1}{a b\left(a^2+b^2\right)}$
$ \Rightarrow u=\frac{b\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$ and $v=\frac{a\left(a^2+b^2\right)}{a b\left(a^2+b^2\right)}$
$ \Rightarrow \mathrm{u}=\frac{1}{a}$ and $\mathrm{v}=\frac{1}{b}$
$ \Rightarrow \frac{1}{x}=\frac{1}{a}$ and $\frac{1}{y}=\frac{1}{b}$
$ \Rightarrow \mathrm{x}=\mathrm{a}$ and $\mathrm{y}=\mathrm{b}$
View full question & answer→Question 305 Marks
Solve :$\frac{34}{3 x+4 y}+\frac{15}{3 x-2 y}=5;\frac{25}{3 x-2 y}-\frac{8.50}{3 x+4 y}=4.5$
AnswerLet $a=3 x+4 y$ and $b=3 x-2 y$
$\therefore \frac{34}{3 x+4 y}+\frac{15}{3 x-2 y}=5$
$\Rightarrow \frac{34}{a}+\frac{15}{b}=5\ldots . .(1)$
$\frac{25}{3 x-2 y}-\frac{8.50}{3 x+4 y}=4.5$
$\Rightarrow-\frac{8.50}{a}+\frac{25}{b}=4.5\ldots . . .(2)$
Multiply equation $(2)$ by $4$ , We get:
$-\frac{34}{a}+\frac{100}{b}=18\ldots . . .(3)$
Adding equation $(1)$ and $(3)$
$-\frac{34}{a}+\frac{100}{b}=18$
$+\frac{34}{a}+\frac{15}{b}=5$
$\frac{115}{b}=23$
$b=5$
$3 x-2 y=5\ldots . . .(4)$
Substituting $b=5$ in equation $(1)$, We get
$\frac{34}{a}+\frac{15}{b}=5$
$ \frac{34}{a}+\frac{15}{5}=5$
$ \frac{34}{a}=2$
$ 2 a=34$
$ a=17$
$3 x+4 y=17\ldots . . .(5)$
Subtracting equation $(5)$ from $(4)$, We get:
$3 x-2 y =5$
$-3 x+4 y =17$
$- -$
$-6 y =-12$
$y =2$
Substituting $y=2$ in equation $(4)$, We get
$3 x-2(2)=5$
$ 3 x=9$
$ x=3$
$\therefore$ Solution is $\mathrm{x}=3$ and $\mathrm{y}=2$.
View full question & answer→Question 315 Marks
Solve :$\frac{20}{x+y}+\frac{3}{x-y}=7; \frac{8}{x-y}-\frac{15}{x+y}=5$
Answer$\frac{20}{x+y}+\frac{3}{x-y}=7\ldots . .(1)$
$\frac{8}{x-y}-\frac{15}{x+y}=5\ldots . . .(2)$
Multiplying equation no $(1)$ by $8$ and $(2)$ by $3 .$
$\frac{160}{x+y}+\frac{24}{x-y}=56\ldots . . .(3)$
$-\frac{45}{x+y}+\frac{24}{x-y}=15\ldots . . .(4)$
Subtracting equation $(4)$ from $(3)$
$\frac{160}{x+y}+\frac{24}{x-y}=56$
$-\frac{45}{x+y}+\frac{24}{x-y}=15$
$\frac{205}{x+y}= 41$
$\frac{205}{x+y}=41$
$x+y=5\ldots . . . .(5)$
From $(1)$
$ \frac{20}{5}+\frac{3}{x-y}=7$
$ \frac{3}{x-y}=3$
$x-y=1\ldots . . .(6)$
Adding equation $(5)$ and $(6)$
$x+y=5$
$+ x-y=1$
$ 2 x=6$
$x=3$
From $(5)$
$3+y=5$
$ y=5-3$
$ y=2$
View full question & answer→Question 325 Marks
Solve :$\frac{3}{x}-\frac{2}{y}=0$ and $\frac{2}{x}+\frac{5}{y}=19$ Hence, find $'a\ '$ if $\mathrm{y}=\mathrm{ax}+ 3.$
Answer$\frac{3}{x}-\frac{2}{y}=0\ldots . . .(1)$
$\frac{2}{x}+\frac{5}{y}=19\ldots . . .(2)$
Multiplying equation no. $(1)$ by $5$ and $(2$) by $2 .$
$\frac{15}{x}-\frac{10}{y}=0\dots........(3)$
$\frac{4}{x}+\frac{10}{y}=38\dots .........(4)$
Adding $(3)$ and $(4),$
$ \frac{15}{x}-\frac{10}{y}=0$
$ +\frac{4}{x}+\frac{10}{y}=38$
$\frac{19}{x}=38$
$ x=\frac{1}{2}$
From $(1)$
$ 3\left(\frac{1}{2}\right)-\frac{2}{y}=0$
$ y=\frac{1}{3}$
$ \therefore y=a x+3$
$ \frac{1}{3}=a\left(\frac{1}{2}\right)+3$
$ \frac{a}{2}=-\frac{8}{3}$
$ a=-\frac{16}{3}$
View full question & answer→Question 335 Marks
Solve :$4 x+\frac{6}{y}=15$ and $3 x-\frac{4}{y}=7$. Hence, find $a$ if $y=a x-2$
Answer$4 x+\frac{6}{y}=15\ldots . .(1)$
$3 x-\frac{4}{y}=7\ldots . .(2)$
Multiplying $(1)$ by $4$ and $(2)$ by $6$
$16 x+\frac{24}{y}=60\ldots . .(3)$
$18 x-\frac{24}{y}=42\ldots .(4)$
Adding $(3)$ and $(4),$ We get
$\begin{gathered}16 x+\frac{24}{y}=60+ 18, x-\frac{24}{y}=42,34 x=102,x=3\end{gathered}$
Substituting $x=3$ in $(1),$ We get
$4(3)+\frac{6}{y}=15$
$\Rightarrow \frac{6}{y}=15-12$
$\Rightarrow y=\frac{6}{3}=2$
Now,
$y=a x-2$
$ 2=a(3)-2$
$ 3 a=4$
$ a=\frac{4}{3}=1 \frac{1}{3}$
View full question & answer→Question 345 Marks
Solve :$5 x+\frac{8}{y}=19, 3 x-\frac{4}{y}=7$
Answer$5 x+\frac{8}{y}=19\ldots .(1)$
$3 x-\frac{4}{y}=7\ldots . .(2)$
Multiplying equation $(2)$ by $2$ , We get,
$6 x-\frac{8}{y}=14 \dots......(3)$
Adding $(1)$ and $(3),$ We get,
$\begin{gathered}5 x+\frac{8}{y}=19+6 x-\frac{8}{y}=14,11 x=33,x=3\end{gathered}$
Substituting $x=3$ in equation $(1),$ We get
$5(3)+\frac{8}{y}=19$
$ 15+\frac{8}{y}=19$
$ \frac{8}{y}=19-15$
$ y=\frac{8}{4}$
$ y=2$
View full question & answer→Question 355 Marks
Solve the pairs of equations:$\frac{3}{x}+\frac{2}{y}=10 ;\frac{9}{x}-\frac{7}{y}=10.5$
Answer$\frac{3}{x}+\frac{2}{y}=10\ldots . .(1)$
$\frac{9}{x}-\frac{7}{y}=10.5\ldots . .(2)$
Multiplying equation $(1)$ by $3$ , We get
$\frac{9}{x}+\frac{6}{y}=30\ldots . . .(3)$
Subtracting $(2)$ from $(3),$ We get
$\frac{9}{x}+\frac{6}{y}=30$
$ \frac{9}{x}-\frac{7}{y}=10.5$
$ \frac{ - +-} {\frac{13}{y}=19.5}$
$ y=\frac{13}{19.5}=\frac{2}{3}$
From $(1),$
$\frac{3}{x}+\frac{2 \times 3}{2}=10$
$ \Rightarrow \frac{3}{x}+3=10$
$ \Rightarrow \frac{3}{x}=7$
$ \Rightarrow x=\frac{3}{7}$
View full question & answer→Question 365 Marks
Solve :$\frac{9}{x}-\frac{4}{y}=8; \frac{13}{x}+\frac{7}{y}=101$
Answer$\frac{9}{x}-\frac{4}{y}=8\ldots .(1)$
$\frac{13}{x}+\frac{7}{y}=101\ldots . .(2)$
Multiplying equation no. $(1)$ by $7$ and $(2)$ by $4 .$
$\frac{63}{x}-\frac{28}{y}=56\ldots . .(3)$
$\frac{52}{x}+\frac{28}{y}=404\ldots . . .(4)$
Adding equation $(3)$ and $(4)$
$\frac{63}{x}-\frac{28}{y}=56$
$+\frac{52}{x}+\frac{28}{y}=404$
$\frac{115}{x}=460$
$x=\frac{115}{460}=x=\frac{1}{4}$
From $(1)$
$9 \times\left(\frac{4}{1}\right)-\frac{4}{y}=8$
$ -\frac{4}{y}=-28$
$ y=\frac{1}{7}$
View full question & answer→Question 375 Marks
Solve :$\frac{3}{2 x}+\frac{2}{3 y}=-\frac{1}{3}; \frac{3}{4 x}+\frac{1}{2 y}=-\frac{1}{8}$
AnswerGiven equations are $\frac{3}{2 x}+\frac{2}{3 y}=-\frac{1}{3}$ and $\frac{3}{4 x}+\frac{1}{2 y}=-\frac{1}{8}$
Let $\frac{1}{x}=u$ and $\frac{1}{y}=v$
Then, the system of equations become
$\frac{3}{2} u+\frac{2}{3} v=-\frac{1}{3}$ and $\frac{3}{4} u+\frac{1}{2} v=-\frac{1}{8}$
$ \Rightarrow \frac{9 u+4 v}{6}=-\frac{1}{3}$ and $\frac{3 u+2 v}{4}=-\frac{1}{8}$
$\Rightarrow 27 u+12 v=-6$ and $24 u+16 v=-4$
$ \Rightarrow 27 u+12 v+6=0$ and $24 u+16 v+4=0$
$\Rightarrow$
$ \frac{u}{12 \times 4-16 \times 6}=\frac{-v}{27 \times 4-24 \times 6}=\frac{1}{27 \times 16-24 \times 12}$
$ \Rightarrow \frac{u}{48-96}=\frac{-v}{108-144}=\frac{1}{432-288}$
$ \Rightarrow \frac{u}{-48}=\frac{-v}{-36}=\frac{1}{144}$
$ \Rightarrow \frac{u}{-48}=\frac{v}{36}=\frac{1}{144}$
$ \Rightarrow u=\frac{-48}{144}=\frac{1}{3}$ and $v=\frac{36}{144}=\frac{1}{4}$
$ \Rightarrow \frac{1}{x}=-\frac{1}{3}$ and $\frac{1}{y}=\frac{1}{4}$
$ \Rightarrow \mathrm{x}=-3$ and $\mathrm{y}=4 .$
View full question & answer→Question 385 Marks
Solve, using cross-multiplication :$0.4 x-1.5 y=6.5; 0.3 x+0.2 y=0.9$
AnswerGiven equation are $0.4 x-1.5 y=6.5$ and $0.3 x+0.2 y=0.9$
Comparing with $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$, We have
$a_1=0.4, b_1=-1.5, c_1=-6.5$ and $a_2=0.3, b_2=0.2, c_2=0.9$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x=\frac{(-1.5) \times(-0.9)-(0.2) \times(-6.5)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$ and $y=\frac{(-6.5) \times(0.3)-(-0.9) \times(0.4)}{0.4 \times(0.2)-(0.3) \times(-1.5)}$
$ \Rightarrow x=\frac{1.35+1.3}{0.08+0.45}$ and $y=\frac{-1.95+0.36}{0.08+0.45}$
$ \Rightarrow x=\frac{2.65}{0.53}$ and $y=\frac{-1.59}{0.53}$
$ \Rightarrow x=5$ and $y=-3$
View full question & answer→Question 395 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$13x+ 11y = 70;11x + 13y = 74$
Answer$13x + 11y = 70 \dots...(1)$
$11x + 13y = 74\dots ...(2)$
Adding $(1)$ and $(2)$
$13x + 11y = 70$
$+ 11x + 13y = 74$
$24x + 24y = 144$
Dividing by $24,$
$x + y = 6\dots ....(3)$
Subtracting $(2)$ from $(1)$
$13x + 11y = 70$
$- 11x + 13y = 74$
$- - - $
$2x - 2y = - 4$
Dividing by $2$
$x - y = - 2 \dots....(4)$
Adding equation $(3)$ and $(4)$
$x - y = - 2$
$+ x + y = 6$
$2x = 4$
$x = 2$
From $(3)$
$2 + y = 6$
$y = 4$
View full question & answer→Question 405 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$2 x-3 y-3=0; \frac{2 x}{3}+4 y+\frac{1}{2}=0$
Answer$2 x-3 y-3=0$
$\Rightarrow 2 x-3 y=3\dots .....(1)$
$\frac{2 x}{3}+4 y+\frac{1}{2}=0$
Multiply by $6 ,$
$6 \times \frac{2 x}{3}+6 \times 4 y+\frac{1}{2} \times 6=0 \times 6$
$4 x+24 y=-3\ldots . . .(2)$
Multiplying equation no. $(1)$ by $8$
$16 x-24 y=24\dots .....(3)$
Adding equation $(3)$ and $(2)$
$16 x-24 y =24$
$+4 x+24 y =-3$
$20 x =21$
$x =\frac{21}{20}$
From $(1)$
$\therefore 2\left[\frac{21}{20}\right]-3 y=3$
$ \therefore-3 y=3-\frac{21}{20}$
$ \therefore y=-\frac{3}{10}$
View full question & answer→Question 415 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{x-y}{6}=2(4-x);2 x+y=3(x-4)$
AnswerThe given pair of linear equations are
$ \frac{x-y}{6}=2(4-x)$
$ \Rightarrow x-y=12(4-x)$
$ \Rightarrow x-y=48-12 x$
$\Rightarrow 13 x-y=48\dots....(1) [$ On simplifying$]$
$ 2 x+y=3(x-4)$
$ \Rightarrow 2 x+y=3 x-12$
$\Rightarrow x-y=12\dots.....(2) [$ On simplifying $]$
Multiply equation $(2)$ by $13$ , We get,
$13 x-13 y=156\dots .....(3)$
Multiply equation $(2)$ by $13 ,$ We get,
$13 x-13 y=156$
Subtracting equation $(1)$ from $(3)$
$13 x-13 y=156$
$- 13 x-y=48$
$-+y$
$-12 y=108$
$y=-9$
Substituting $y=-9$ in equation $(1)$, we get
$13 x-(-9)=48$
$ \Rightarrow 13 x=39$
$ \Rightarrow x=3$
$\therefore$ Solution is $x=3$ and $y=-9$.
View full question & answer→Question 425 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{1}{5}(x-2)=\frac{1}{4}(1-y); 26 x+3 y+4=0$
Answer$ \frac{1}{5}(x-2)=\frac{1}{4}(1-y)$
$ \Rightarrow 4(x-2)=5(1-y)$
$ \Rightarrow 4 x-8=5-5 y$
$\Rightarrow 4 x+5 y=13\dots.....(1)$
$26 x+3 y=-4\dots.....(2)$
Multiplying equation no. $(1)$ by $3$ and $(2)$ by $5 .$
$12 x+15 y=39\dots.....(3)$
$130 x+15 y=-20\dots.....(4)$
Subtracting equation $(4)$ from $(3)$
$12 x+15 y =39$
$-130 x+15 y =-20$
$- +$
$-118 x =59$
$x =-\frac{59}{118}$
$x =-\frac{1}{2}$
From $(1)$
$4\left(-\frac{1}{2}\right)+5 y=13$
$ 5 y=13+2$
$ y=3$
View full question & answer→Question 435 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$\frac{5 y}{2}-\frac{x}{3}=8; \frac{y}{2}+\frac{5 x}{3}=12$
AnswerThe given pair of linear equations are
$\frac{5 y}{2}-\frac{x}{3}=8$
$\Rightarrow-\frac{x}{3}+\frac{5 y}{2}=8\dots.....(1) [$ On Similifying $]$
$\frac{y}{2}+\frac{5 x}{3}=12$
$\Rightarrow \frac{5 x}{3}+\frac{y}{2}=12\dots.....(2) [$ On Similifying $]$
Multiply equation $(1)$ by $5 ,$ we get
$-\frac{5 x}{3}+\frac{25 y}{2}=40\dots......(3)$
Adding equation $(3)$ and $(2)$
$-\frac{5 x}{3}+\frac{25 y}{2}=40$
$ + \frac{5 x}{3}+\frac{y}{2}=12$
$\frac{26 y}{2}=52$
$ \Rightarrow 13 y=52$
$ \Rightarrow \mathrm{y}=4$
Substituting $y=4$ in equation $(1)$, We get
$-\frac{x}{3}+\frac{5(4)}{2}=8$
$ \Rightarrow-\frac{x}{3}=8-10$
$ \Rightarrow \mathrm{x}=6$
$\therefore$ Solution is $\mathrm{x}=6$ and $\mathrm{y}=4$.
View full question & answer→Question 445 Marks
Solve :$4 x+\frac{x-y}{8}=17;2 y+x-\frac{5 y+2}{3}=2$
Answer$4 x+\frac{x-y}{8}=17 ($Given$)$
$\Rightarrow 32 x+x-y=136$
$\Rightarrow 33 x-y=136 \dots......(1)$
$2 y+x-\frac{5 y+2}{3}=2 ($Given$)$
$\Rightarrow 6 y+3 x-5 y-2=6$
$\Rightarrow 3 x+y=8\dots.......(2)$
Adding equations $(1)$ and $(2),$ we get
$33 x-y =136$
$+ 3 x+y =8$
$36 x =144$
$x =4$
Substituting $x=4$ in equation $(2),$ We get
$3 \times 4+y=8$
$ \Rightarrow 12+y=8$
$ \Rightarrow y=8-12$
$ \Rightarrow y=-4$
$ \therefore$ Solution is $x=4$ and $y=-4$
View full question & answer→Question 455 Marks
Solve :$\frac{7+x}{5}-\frac{2 x-y}{4}=3 y-5;\frac{5 y-7}{2}+\frac{4 x-3}{6}=18-5 x$
Answer$\frac{7+x}{5}-\frac{2 x-y}{4}=3 y-5($Given$)$
$ \Rightarrow 4(7+x)-5(2 x-y)=20(3 y-5)$
$ \Rightarrow 28+4 x-10 x+5 y=60 y-100$
$\Rightarrow-6 x-55 y=-128\ldots . . .(1)$
$\frac{5 y-7}{2}+\frac{4 x-3}{6}=18-5 x $(Given$)$
$ \Rightarrow 3(5 y-7)+4 x-3=6(18-5 x)$
$ \Rightarrow 15 y-21+4 x-3=108-30 x$
$\Rightarrow 34 x+15 y=132 \dots.......(2)$
Multiplying equation $(1)$ by $34$ and equation $(2)$ by $6 $, We get
$-204 x-1870 y=-4352\dots.....(3)$
$204 x+90 y=792\dots......(4)$
Adding equation $(3)$ and $(4)$, We get
$-204 x-1870 y=-4352\dots.....(3)$
$+204 x+90 y =792$
$-1780 y =-3560$
$\Rightarrow y =2$
Substituting $y=2$ in equation $(1),$ We get
$-6 \mathrm{x}-55 \times 2=-128$
$ \Rightarrow-6 \mathrm{x}-110=-128$
$ \Rightarrow-6 \mathrm{x}=-18$
$ \Rightarrow \mathrm{x}=3$
$\therefore$ Solution is $x=3$ and $y=2$.
View full question & answer→Question 465 Marks
Solve :$11(x - 5) + 10(y - 2) + 54 = 0;7(2x - 1) + 9(3y - 1) = 25$
Answer$11( x - 5 ) + 10( y - 2 ) + 54 = 0($given$)$
$\Rightarrow 11x - 55 + 10y - 20 + 54 = 0$
$\Rightarrow 11x + 10y - 21 = 0$
$\Rightarrow 11x + 10y = 21\dots....(1)$
$7( 2x - 1 ) + 9(3y - 1) = 25($given$)$
$\Rightarrow 14x - 7 + 27y - 9 = 25$
$\Rightarrow 14x + 27y - 16 = 25$
$\Rightarrow 14x + 27y = 41\dots.....(2)$ Multiplying equation $(1)$ by $27$ and equation $(2)$ by $10,$ we get,
$297x + 270y = 567\dots....(3)$
$140x + 270y = 410\dots.....(4)$ Subtracting equation $(4)$ from equation $(3)$, we get
$157x = 157$
$\Rightarrow x = 1$ Substituting $x = 1$ in equation $(1),$ we get,
$11 x 1 + 10y = 21$
$\Rightarrow 10y = 10$
$\Rightarrow y = 1$
$\therefore $ Solution set is $x = 1$ and $y = 1.$
View full question & answer→Question 475 Marks
Solve for $\mathrm{x}$ and $\mathrm{y}$ :$4 x=17-\frac{x-y}{8};2 y+x=2+\frac{5 y+2}{3}$
AnswerThe given pair of linear equations are
$4 \mathrm{x}=17-\frac{x-y}{8}$
$\Rightarrow 33 x-y=136\dots ...(1)($ On Simplifying$)$
$2 y+x=2+\frac{5 y+2}{3}$
$\Rightarrow 3 x+y=8\dots...(2)($ On Simplifying $)$
Multiply equation $(2)$ by $11$ , we get,
$33 x+11 y=88\dots...(3)$
Subtracting equation $(1)$ from $(3)$
$\begin{gathered}33 x+11 y=88,-33 x-y=136,-+-,12 y=-48y=-4\end{gathered}$
Substituting $y=-4$ in equation $(1)$, we get :
$33 x-(-4)=136$
$ \Rightarrow 33 x=132$
$ \Rightarrow x=4$
$\therefore$ Solution is $\mathrm{x}=4$ and $\mathrm{y}=-4$.
View full question & answer→Question 485 Marks
Solve for $x$ and $y$ :$\frac{y+7}{5}=\frac{2 y-x}{4}+3 x-5;\frac{7-5 x}{2}+\frac{3-4 y}{6}=5 y-18$
AnswerThe given pair of linear equation are
$\frac{y+7}{5}=\frac{2 y-x}{4}+3 x-5$
$\Rightarrow 55 x+6 y=128\dots....(1)($ On simplifying $)$
$\frac{7-5 x}{2}+\frac{3-4 y}{6}=5 y-18$
$\Rightarrow 15 x+34 y=132\dots....(2)($ On simplifying $)$
Multiply equation $(1)$ by $3$ and equation $(2)$ by $11$ , we get:
$165 x+18 y=384\dots....(3)$
$165 x+374 y=1452\dots.....(4)$
Subtracting $(4)$ from $(3)$
$\begin{gathered}165 x+18 y=384,- 165 x+374 y=1452,- - -,356 y=-1068,y=3\end{gathered}$
Substituting $y=3$ in equation $(1)$, we get
$55 x+6(3)=128$
$ \Rightarrow 55 x=110$
$ \Rightarrow x=2$
$\therefore$ Solution is $\mathrm{x}=2$ and $\mathrm{y}=3$.
View full question & answer→Question 495 Marks
If $10y = 7x - 4$ and $12x + 18y = 1$; find the values of $4x + 6y$ and $8y - x.$
Answer$ 10 y=7 x-4$
$ -7 x+10 y=-4$
$ -(7 x-10 y)=4$
$7 x-10 y=4\ldots(1)$
$12 x+18 y=1\ldots(2)$
Multiplying equation no. $(1)$ by $12$ and $(2)$ by $7$ .
$84 x-120 y=48\ldots . .(3)$
$84 x+126 y=7\ldots .(4)$
Substract equation $(3)$ and $(4)$
$84 x-120 y=48$
$ 84 x+126 y=7 $
$ - - -$
$-246 y=41$
$y=\frac{41}{-246}$
$y=-\frac{1}{6}$
From (1)
$ 7 x-10\left(-\frac{1}{6}\right)=4$
$ 7 x+\frac{5}{3}=4$
$ 7 x=4-\frac{5}{3}$
$ 7 x=\frac{7}{3}$
$ x=\frac{7}{3} \times 7$
$ x=\frac{1}{3}$
$ \therefore 4 x+6 y=4\left(\frac{1}{3}\right)+6\left(-\frac{1}{6}\right)=\frac{4}{3}-\frac{6}{6}=\frac{4}{3}-1=\frac{1}{3}$
$ \therefore 8 y-x=8\left(-\frac{1}{6}\right)-\frac{1}{3}=-\frac{4}{3}-\frac{1}{3}=-\frac{5}{3}$
View full question & answer→Question 505 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients :$41x + 53y = 135;53x + 41y = 147$
Answer$41x + 53y = 135 \dots...(1)$
$53x + 41y = 147 \dots...(2)$
Adding equation $(1)$ and $(2)$
$41x + 53y = 135$
$+ 53x + 41y = 147$
$94x + 94y = 282$
Dividing by $94,$
$x + y = 3\dots ....(3)$
Subtracting equation $(2)$ from $(1)$
$41x + 53y = 135$
$- 53x + 41y = 147 $
$- - - $
$- 12x + 12y = - 12$
Dividing by $12,$
$- x + y = -1\dots....(4)$
Adding $(3)$ and $(4)$
$x + y = 3$
$+ - x + y = -1$
$2y = 2$
$y = 1$
From $(3)$
$x + y = 3$
$x + 1 = 3$
$x = 2$
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