Pair of Linear Equations in Two Variables Test - 42

Let no of $$20$$p coins $$=x$$
No of $$25$$p coins $$=y$$
According to the question
$$ x+y=50$$
$$\Rightarrow x=50-y.....eq1$$
And $$20x+25y=1125\Rightarrow 4x+5y=225....eq2$$
Put the value of $$x$$ from $$eq1$$
$$\Rightarrow 4\left ( 50-y \right )+5y=225......eq3$$
$$\Rightarrow 200-4y+5y=225\Rightarrow y=25$$
Put $$y=25$$ in $$eq1$$
$$\Rightarrow x+25=50\Rightarrow x=25  $$

No of $$20$$p coins $$=25$$

No of $$25$$p coins $$=25$$

Let the first number be $$x$$ and second number $$y.$$

According to question
$$x+y=8\quad\quad\quad\dots(i)$$

$$4(x-y)=8$$

$$\Rightarrow x-y=2\quad\quad\quad\dots(ii)$$

Add equations $$(i)$$ and $$(ii),$$
$$\begin{aligned}{}\left( {x + y} \right) + \left( {x - y} \right)& = 8 + 2\\2x &= 10\\x &= 5\end{aligned}$$

Substitute $$x=5$$ in $$(i),$$
$$\begin{aligned}{}\left( 5 \right) + y &= 8\\y&=8-5\\y &= 3\end{aligned}$$

So, the numbers are $$5$$ and $$3.$$

Clearly the quadrilateral is trapezium
Thus it's area $$=\cfrac{1}{2}($$AD$$+$$BC $$)\times$$ AB
                       $$=\cfrac{1}{2}\times (2+6)\times 2=8 $$ sq. units

let A has $$x$$ mangoes and B has $$y$$ mangoes
According to the question
$$\Rightarrow \left (x+30  \right )=2\left ( y-30 \right )$$
$$\Rightarrow x+30=2y-60\Rightarrow  x-2y=-90.....eq1$$
$$\Rightarrow  3\left (x-10  \right )=y+10$$
$$\Rightarrow 3x-30=y+10\Rightarrow 3x-y=40....eq2$$

$$\Rightarrow \dfrac{x+2y+1}{2x-y+1}=2$$
$$\Rightarrow 4x-2y+2=x+2y+1$$
$$\Rightarrow 3x-4y=-1........eq1$$
$$\Rightarrow \dfrac{3x-y+1}{x-y+3}=5$$
$$\Rightarrow 5x-5y+15=3x-y+1$$
$$\Rightarrow 2x-4y=-14........eq2$$
Substract eq1 and eq2
$$\Rightarrow x=13$$
put x=13 in eq1
$$\Rightarrow 3\times13-4y=-1$$
$$\Rightarrow -4y=-40 \Rightarrow y=10$$
Hence x=13 and y=10

$$\Rightarrow \dfrac{3x-2}{3y+7}=\dfrac{5x-1}{5y+16}$$
$$\Rightarrow 15xy-10y+48x-32=15xy+35x-3y-7$$
$$\Rightarrow 13x-7y=25.........eq1$$
$$\Rightarrow \dfrac{3x-15}{x-9}=\dfrac{6y-5}{2y+3}$$
$$\Rightarrow 6xy-30y+9x-45=6xy-54y-5x+45$$
$$\Rightarrow 14x+24y=90.......eq2$$
Multiply eq1 by 24 and eq2 by 7
$$\Rightarrow 312x-168y=600.......eq3$$
$$\Rightarrow 98x+168y=630.......eq4$$
Add eq3 and eq4
$$\Rightarrow 410x=1230$$
$$\Rightarrow x=3$$
Put x=3 in eq 1
$$\Rightarrow 13\times3-7y=25$$
$$\Rightarrow -7y=-14$$
$$\Rightarrow y=2$$
Hence, x=3 and y=2

Suppose the cost of $$1$$ table $$=x$$ and cost of $$1$$ chair $$=y$$
Then according to the question
$$\Rightarrow 4x+3y=2250$$...........(1)
$$\Rightarrow 3x+4y=1950$$.............(2)

Multiply (1) by $$4$$ and (2) by $$3$$ and Subtract both
$$\Rightarrow  (16x+12y=9000)- (9x+12y=5850)$$

$$\Rightarrow 7x=3150\Rightarrow x=450$$

Put $$x=450$$ in $$eq2$$

$$\Rightarrow 450\times4+3y=2250\Rightarrow 3y=2250-1800\Rightarrow 3y=450$$

$$\Rightarrow y=150$$

Cost of $$1$$ table $$=450$$
Cost of $$1$$ chair $$=150$$
$$\therefore$$ Cost of $$1$$ table and $$2$$ chair $$ =450+150\times2=750$$

Let the fixed charge for $$3$$ days $$=x$$ and additional charge $$=y$$
According to the question
$$\Rightarrow x+4y=27....eq1$$
$$\Rightarrow x+2y=21....eq2$$
Subtract $$eq1$$ and $$eq2$$
$$\Rightarrow \left ( 3x+4y=27 \right )-\left ( 3x+2y=21 \right )\Rightarrow 2y=6\Rightarrow y=3$$
put $$y=3$$ in $$eq1$$
$$\Rightarrow x+4\times3=27\Rightarrow x=15$$ 
Fixed charges $$=$$ Rs. $$15$$
The charge for each extra day $$=$$ Rs. $$3$$
 

let one has x Rs. and others has y Rs.
Then according to the question