Question 13 Marks
A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Geetapaid $\text{Rs}. 27$ for a book kept for seven days, whileMohitpaid $\text{Rs.}21$ for the book he kept for five days. Find the fixed charges and the charge for each extra day.
AnswerLet the fixed charges be $\text{Rs.} x$ and the charge for each extra day be $\text{Rs.} y.$
According to the question,
$x + 4y = 27 ...(1)$
and $x + 2y = 21 ....(2)$Solving the equations, we get
$2y = 6$
$\therefore y = 3$
and $x = 21 - 2y = 21 - 2(3) = 15$
Hence, the fixed charges is $\text{Rs.} 15$ and the charge for each extra day is $\text{Rs}. 3.$
View full question & answer→Question 23 Marks
The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of $10 \ km,$ the charge paid is $\text{Rs.} 315$ and for a journey of $15 \ km,$ the charge paid is $\text{Rs.} 465.$ What are the fixed charges and the charge per kilometer ? How much does a person have to pay for travelling a distance of $32 \ km$ ?
AnswerLet the fixed charge be $\text{Rs.} x$ and the charge per kilometer be $\text{Rs.} y.$
The charges for $10 \ km =\text{ Rs.} 10y$
The charges for $15 \ km = \text{Rs.} 15y$
According to the question,
$x+ 10y = 315 ...(1)$
$x+ 15y = 465 ....(2)$
Solving the equations, we get
$- 5y = - 150$
$\Rightarrow y = 30$
and $x = 315 - 10y = 315 - 10(30) = 15$
So, the fixed charges is $\text{Rs.} 15$ and the charges per kilometer is $\text{Rs.} 30.$
To travel $32 \ km,$ a personal has to pay
$=\text{ Rs.} 15 + \text{Rs.} 30( 32 )$
$= \text{Rs.} 15 +\text{ Rs.} 960$
$= \text{Rs.}975.$
View full question & answer→Question 33 Marks
$1250$ persons went to sea a circus$-$show. Each adult paid $\text{Rs.} 75$ and each child paid $\text{Rs.} 25$ for the admission ticket. Find the number of adults and number of children, if the total collection from them amounts to $\text{Rs.} 61,250.$
AnswerLet the number of adults $= x$
and the number of children $= y$
According to the question,
$x + y = 1250 ...(1)$
$75x + 25y = 61250 ...(2)$
Subtracting equation $(2)$ from $(1)$
$3x + y = 2450$
$- x + y = 1250- - - $
$2x = 1200$
$x = 600$
From$ (1)$
$600 + y = 1250$
$y = 650$
Thus, number of adults $= 600$ and the number of children $= 650.$
View full question & answer→Question 43 Marks
The age of a man is twice the sum of the ages of his two children. After $20$ years, his age will be equal to the sum of the ages of his children at that time. Find the present age of the man.
AnswerLet the present age of the man be $x $ years.
and let the sum of the ages of his two children be $y $ years.
According to the question,
$x = 2y ...(1)$
and $ x + 20 = y + 40 ....(2) [$ Since he has two children $]$Solving $(1)$ and $(2),$ We get,
$2y + 20 = y + 40$
$⇒ y = 20$
So, $x = 2y$
$⇒ x = 40.$
Hence, the present age of the man is $40$ years.
View full question & answer→Question 53 Marks
The sum of two digit number and the number obtained by interchanging the digits of the number is $121.$ If the digits of the number differ by $3,$ find the number.
AnswerLet the tens digit of the number be $x$ and the units digit be $ y.$
So, the number is $10x + y.$
The number obtained by interchanging the digits will be $10y + x.$
According to question, we have
$10x + y + 10y + x = 121$
$\Rightarrow 11x + 11y = 121$
$\Rightarrow 11( x + y ) = 121$
$\Rightarrow x + y = 11 ...(1)$
And,
$x - y = 3 ...(2)$Adding $(1)$ and $(2),$ We get
$\therefore 2x = 14$
$x = 7$
$\therefore y = 11 - x = 11 - 7 = 4$
Hence, the number is $74.$
View full question & answer→Question 63 Marks
Solve $:x+ y = 7xy,2x - 3y = - xy$
Answer$x+y=7 x y\ldots(1)$
$2 x-3=-x y\ldots(2)$
Multiplying equation no.$(1)$ by $3.$
$3 x+3 y=21 x y\ldots .(3)$
Adding equation $(3)$ and $(2)$
gathered
$3 x+3 y=21 x y+ 2 x-3 y=-x y-5 x=20 x y$
$y=\frac{1}{4}$
gathered From $(1)$
$\mathrm{x}+\frac{1}{4}=7 x\left(\frac{1}{4}\right)$
$\frac{1}{4}=\frac{3}{4} x$
$x=\frac{1}{3}$
View full question & answer→Question 73 Marks
Solve :$x + y = 2xy,x - y = 6xy$
Answer$x+y=2 x y\ldots . .(1)$
$x-y=6 x y\ldots . . .(2)$
Adding equation $( 1 )$ and $( 2 )$
$x+y=2 x y + x-y=6 x y- 2 x=8 x y$
$ 2=8 y$
$ y=\frac{1}{4}$
From $(1)$
$x+\frac{1}{4}=2 \times\left(\frac{1}{4}\right)$
$ \frac{1}{2} x=-\frac{1}{4}$
$ x=-\frac{1}{2}$
View full question & answer→Question 83 Marks
Solve, using cross-multiplication :$3x + 4y = 11,2x + 3y = 8$
AnswerGiven equations are $3 x+4 y=11$ and $2 x+3 y=8$
Comparing with $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$,
we have
$a_1=3, b_1=4, c_1=-11$ and $a_2=2, b_2=3, c_2=-8$
Now, $x=\frac{b_1 c_2-b_2 c_1}{a_1 b_2-a_2 b_1}$ and $y=\frac{c_1 a_2-c_2 a_1}{a_1 b_2-a_2 b_1}$
$\Rightarrow x=\frac{4 \times(-8)-3 \times(-11)}{3 \times 3-2 \times 4}$ and $y=\frac{-11 \times 2-(-8) \times 3}{3 \times 3-2 \times 4}$
$\Rightarrow x=\frac{-32+33}{9-8}$ and $y=\frac{-22+24}{9-8}$
$\Rightarrow x=1$ and $y=2$
View full question & answer→Question 93 Marks
For solving pair of equation, in this exercise use the method of elimination by equating coefficients $:13 + 2y = 9x,3y = 7x$
Answer$13+2 y=9 x\ldots(1)$
$3 y=7 x\ldots(2)$
Multiplying equation no.$ (1)$ by $ 3$ and $(2)$ by $2$ , we get,
$39+6 y =27x\dots(1)$
$6 y =14x\dots(2)$
$39 =13x$
$x =3$
From $(2)$
$ 3 y=7 x$
$ \therefore 3 y=7(3)$
$ \therefore y=\frac{21}{3}$
$=7$
View full question & answer→Question 103 Marks
$10%$ of $x + 20%$ of $y = 24,3x - y = 20$
Answer$10\%$ of $x + 20\%$ of $y = 24$
$\Rightarrow 0.1x + 0.2y = 24 .....(1) [$ On Simplyfying$ ]$
$3x - y = 20 .....(2)$ Multiply equation $(2)$ by $0.2,$
We get $:$
$0.6x - 0.2y = 4 ......(3)$
Adding equation $(3)$ and $(1)$
$0.6x - 0.2y = 4$
$+ 01x + 0.2y = 24$
$0.7x = 28$
$x = 40$
Substituting $x = 40$ in equation $(1),$
We get
$0.1(40) + 0.2y = 24$
$\Rightarrow 0.2y = 20$
$\Rightarrow y = 100$
$\therefore $ Solution is $x = 40$ and $y = 100.$
View full question & answer→Question 113 Marks
Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution:$2x + 7y = 39,3x + 5y = 31$
Answer$2 x+7 y=39\ldots(1)$
$3 x+5 y=31\ldots(2)$
$2 x+7 y=39$
$ \therefore x=\frac{39-7 y}{2}$
Putting this value of $x$ in $(2)$
$3\left(\frac{39-7 y}{2}\right)+5 y=31$
$ 117-21 y+10 y=62$
$ -11 y=-55$
$ y=5$
From $(1) x=\frac{39-7(5)}{2}$
$x=\frac{4}{2}$
$ x=2$
View full question & answer→Question 123 Marks
Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution $:y = 4x - 7,16x - 5y = 25$
Answer$y=4 x-7\ldots .(1)$
$16 x-5 y=25\ldots .(2)$
$y=4 x-7$
Putting this value of $y$ in $(2)$
$\therefore 16 x-5(4 x-7)=25$
$ \therefore 16 x-20 x+35=25$
$ \therefore-4 x=-10$
$ \therefore x=\frac{5}{2}$
From $(1)$
$y=4\left(\frac{5}{2}\right)-7$
$ \Rightarrow y=10-7$
$ \Rightarrow y=3$
Solution is $\mathrm{x}=\frac{5}{2}$ and $\mathrm{y}=3$.
View full question & answer→Question 133 Marks
Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution$:6x = 7y + 7,7y - x = 8$
Answer$6 x=7 y+7\ldots(1)$
$7 y-x=8\ldots(2)$
$7 y-x=8$
$ \therefore x=7 y-8$
Putting this value of $x$ in $(1)$
$6(7 y-8)=7 y+7$
$ \therefore 42 y-48=7 y+7$
$ \therefore 35 y=55$
$ \therefore y=\frac{11}{7}$
From $(2)$
$x=7\left(\frac{11}{7}\right)-8$
$ x=3$
$\therefore x=3, y=\frac{11}{7} \text {. }$
View full question & answer→Question 143 Marks
Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution$:2x + 3y = 8,2x = 2 + 3y$
Answer$2 x+3 y=8\ldots(1)$
$2 x=2+3 y\ldots(2)$
$2 x=2+3 y$
Putting this value of $2 x$ in $(1)$
$2+3 y+3 y=8$
$ \therefore 6 y=8-2$
$ \therefore 6 y=6$
$ \therefore y=1$
From$ (2)$
$2 x=2+3(1)$
$x=\frac{5}{2}$
$ x=2.5$
View full question & answer→Question 153 Marks
Solve the pair of linear $($simultaneous$)$ equation by the method of elimination by substitution$ :2x - 3y = 7,5x + y= 9$
Answer$2x - 3y = 7 ...(1)$
$5x + y = 9 ...(2)$
$5x + y = 9$
$\therefore y = 9 - 5x ...(3)$
Putting this value of $y$ in $(1)$
$2x - 3 (9 - 5x) = 7$
$\therefore 2x - 27 + 15x = 7$
$\therefore 2x + 15x = 7 + 27$
$\therefore 17x = 34$
$\therefore x = 2$
From $(2)$
$y = 9 - 5(2)$
$y = -1$
View full question & answer→Question 163 Marks
Solve the following pair of linear $($simultaneous$)$ equation using method of elimination by substitution $:2x - 3y + 6 = 0,2x + 3y - 18 = 0$
Answer$ 2 x-3 y+6=0$
$ \Rightarrow 2 x=3 y-6$
$\Rightarrow x=\frac{3 y-6}{2}\ldots(1)$
And,
$2 x+3 y-18=0$
$ \Rightarrow 2\left(\frac{3 y-6}{2}\right)+3 y=18 \ldots[$ From$(1)] $
$\Rightarrow 3 y-6+3 y=18$
$\Rightarrow 6 y=24$
$ \Rightarrow y=4$
Substituting the value of $y$ in $(1),$
we have
$x=\frac{3 \times 4-6}{2}=\frac{12-6}{2}=\frac{6}{2}=3$
$\therefore$ Solution is $x=3$ and $y=4$.
View full question & answer→