SChand Composite Mathematics Class 7 Chapter 11 Triangle and it's Properties Exercise 11C

 Exercise 11 C 

Question 1 

Find the lettered angle in each of the following : 

(i) (IMAGE TO BE ADDED)

Question 2

  (IMAGE TO BE ADDED)

Sol: $\angle M=20^{\circ}$

$\angle L+\angle M+\angle N=180^{\circ}$

$b=180^{\circ}-(20+20)$

$b=180^{\circ}-40=140^{\circ}$

Angle opposite equal sides of an isosceles triangle are equal 

Question 3  

(IMAGE TO BE ADDED)

Sol: $m=50^{\circ}$

$\begin{aligned} x &=180^{\circ}-100 \\ x &=30^{\circ} \end{aligned}$

Question 4

(IMAGE TO BE ADDED)

Sol: $\angle Q=59^{\circ}$

$x=180-\left(59^{\circ}+59^{\circ}\right)$

$x=180-118^{\circ}$

$x=62^{\circ}$ Ans

$x=32^{\circ}+y \quad=y=30^{\circ}$

$62-32^{\circ}=y$

Question 5

(IMAGE TO BE ADDED)

Sol:

 $\begin{aligned} \angle D &=63^{\circ} \\ \angle D A E &=180^{\circ}-\left(63^{\circ}+63^{\circ}\right) \\ &=180^{\circ}-126^{\circ} \\ &=180^{\circ}-126^{\circ}=54^{\circ} \\ \angle B+\angle C+54^{\circ}=180^{\circ} \\ x+x &+54^{\circ}=180^{\circ} \end{aligned}$

$2 x=180^{\circ}-54^{\circ}$

$x=\frac{126}{2}$ $x=63^{\circ}$

S Chand Class 10 CHAPTER 11 Coordinate Geometry Exercise 11C

   Exercise 11C

Question 1

Ans: (i) $\begin{aligned}& 3 x+4 y=9 \\\Rightarrow & 4 y=-3 x+9 \\\Rightarrow & y=-\frac{3}{4} x+\frac{9}{4}\end{aligned}$

which is in the form of $y=m x+c$

So slope $(m)=-\frac{3}{4}$

(ii) $3 x+2 y=4$

$\Rightarrow 2 y=-3 x+4$

$\Rightarrow y=\frac{-3}{2} x+\frac{4}{2}$

$\Rightarrow y=-\frac{3}{2} x+2$

so slope(m) $=\frac{-3}{2}$

Question 2

Ans: (i) 

$\begin{aligned} & 3 x+2 y+4=0 \\ \Rightarrow & 2 y=-3 x-4 \\ \Rightarrow & y=-\frac{3}{2} x-2 \end{aligned}$

which is in the form of $y=m x+c$

(ii) slope (m) = $\frac{-3}{2}$ and y- intersect (c) = -2

(iii) The line has been drawn on the graph paper as given here.

(IMAGE TO BE ADDED)

Question 3

Ans: we know that the equation of a line is $y=m x+c$ where $m$ in the slope and $c$ is the $y$. intercept  

Therefore 

(a) $y$-intercept $(c)=2$

slope $(m)=7$

So equation of the line will be 

$y=7 x+2 \Rightarrow 7 x-y+2=0$

(b) slope $(m)=-4$

and $y$ - intercept $(c)=-3$

So Equation on the line will be

$y=-4 x-3 \text { or } 4 x+y+3=0$

(c) y- intercept (c) = -1

If the line is parallel to the line y = 5x-7

slope(m) = 5

 So the required line will be y = 5x-1

(d) y - intercept (c) = 2

So The equation inclined at $45^{\circ}$ to the $x$-axis So slope $(m)=\tan \theta=\tan 45^{\circ}=1$

So Equation of the required line will be

$\begin{aligned}&y=1 \quad x+2 \Rightarrow y=x+2 \\&x-y+2=0\end{aligned}$

(e) $y$ - intercept (c)=-5$

if The line is equally inclined to the axis

So slope $(m)=\tan \theta=\pm 1$

So Equation of the line will be

$\begin{aligned}&y=1 x-5 \Rightarrow y=x-5 \\&x-y=5 \\&\text { or } y=-1 x-5 \Rightarrow y=-x-5 \\&\text { or } x+y+5=0\end{aligned}$

Question 4

Ans:  The equation of line is

$y=m x+c \quad-i 1$

so it passes through the points $(3,-4)$ and(-1,2)

Substituting the value of x and y in (i) -4 = 3m+ c................(ii)

and $2=-1 m+c \Rightarrow 2=-m+c$........(iii)

Subtracting we get ,

$-6=4 m$

$\Rightarrow m=\frac{-6}{4}=\frac{-3}{2}$

From (ii)

$\begin{aligned}-4 &=3 \times\left(\frac{-3}{2}\right)+c \\ & \Rightarrow-4=-\frac{9}{2}+c \\ & \Rightarrow c=-4+\frac{9}{2} \\ &=\frac{-8+9}{2}=\frac{1}{2} \end{aligned}$

So $m=-\frac{3}{2} \text { and } c=\frac{1}{2}$

Question 5

Ans: IF The line y = mx+c passes through the points $(1,4)$ and $(-2,-5)$

So substituting the value of x and y , we get 

$4=1 m+c$

$\Rightarrow 4=m+c $............(i)

$-5=-2 m+c$

$\Rightarrow-5=-2 m+c$.........(ii)

Subtracting , we get 

9=3m = $m=\frac{9}{3}=3$

From (i)

$\begin{aligned}&4=3+c \\&\Rightarrow c=4-3=1\end{aligned}$

Hence $m=3$ and $c=1$

Question 6

Ans: (a) The equation of the line will be

$y-y_{1}=m\left(x-x_{1}\right)$

= Here it passes through p (-4, 7) and m = $-\sqrt{3}$

So $y-7=-\sqrt{3}[x-(-4)]$

$\begin{aligned}&\Rightarrow y-7=-\sqrt{3} x-4 \sqrt{3} \\&\Rightarrow \sqrt{3} x+y=7-4 \sqrt{3}\end{aligned}$

(b) So the line passes through the point p (-1,-5)

and has slope $=\frac{-6}{11}$

 

So equation of the line will be 

$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y+5=\frac{-6}{11}(x+1) \\ \Rightarrow & 11 y+55=-6 x-6 \\ \Rightarrow & 6 x+11 y+55+6=0 \\ \Rightarrow & 6 x+11 y+61=0 \end{aligned}$

Question 7

Ans: (a) Slope of the line $x+2 y=4 \Rightarrow 2 y=-x+4$

$\Rightarrow y=-\frac{1}{2} x+2$

$m_{1}=\frac{-1}{2}$ and slope of the line perpendicular to it $\left(-\frac{1}{m}\right)=2$

if The required equation parses through the origin ( 0,0)

So equation of the line will be

$\begin{aligned}& y-y_{1}=m\left(x-x_{1}\right) \\\Rightarrow & y-0=2(x-0) \\\Rightarrow & y=2 x \\\Rightarrow & 2 x-y=0\end{aligned}$

(b) Slope of the given line $3 x+4 y=12$

$\begin{aligned}\Rightarrow 4 y &=-3 x+12 \\\Rightarrow y &=-\frac{3}{4} x+3 \\m &=-\frac{3}{4}\end{aligned}$

 if  it passes through the points $(4,3)$ So equation of the line will be

$\begin{aligned}y-y_{1} &=m\left(x-x_{1}\right) \\\Rightarrow y-3 &=\frac{-3}{4}(x-4)\end{aligned}$

$\Rightarrow 4 y-12=-3 x+12$

$\Rightarrow 3 x+4 y=12+12$

$\Rightarrow 3 x+4 y=24$

$\Rightarrow 3 x+4 y-24=0$

(c) Slope of the given line $3 x-2 y+5=0$

$\begin{gathered}\Rightarrow 2 y=3 x+5 \\\Rightarrow y=\frac{3}{2} x+\frac{5}{2} \\m=\frac{3}{2}\end{gathered}$

(i) Slope as the required line which as parallel to the given line $=m=\frac{3}{2}$

IF it passes through the point (4,5)

So equation as the line will be 

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-5=\frac{3}{2}(x-4) \Rightarrow 2 y-10=3 x-12$

$\Rightarrow 3 x-2 y+10-12=0$

$\Rightarrow 3 x-2 y-2=0$

(ii) slope as the line perpendicular to the given line  = $\frac{-1}{m}$

$m_{1}=\frac{-2}{3}$

if it passes through the points (4,5)

so equation of the line will be 

$y-y_{1}=m_{1}\left(x-x_{1}\right)$

$\Rightarrow y-5=\frac{-2}{3}(x-4)$

$\Rightarrow 3 y-15=-2 x+8 .$

$\Rightarrow 2 x+3 y-15-8=0$

$\Rightarrow 2 x+3 y-23=0$

$\Rightarrow 2 x+3 y=23$

Question 8

Ans: The equation of a line which pares through two pointe is $y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

(a) If the required line passes through the points $A(1,1)$ and $B(2,3)$

So Equation of the lines will be

$\begin{aligned}&y-1=\frac{3-1}{2-1}(x-1) \Rightarrow y-1=\frac{2}{1}(x-1) \\&\Rightarrow y-1=2 x-2 \\&\Rightarrow 2 x-y-2+1=0 . \\&\Rightarrow 2 x-y-1=0\end{aligned}$

(b) If the required line passes through the points 

$P(3,3)$ and $Q(7,6)$

So Equation of the line w/11 be

$\begin{aligned}y-y_{1} &=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right) \\\Rightarrow y-3 &=\frac{6-3}{7-3}(x-3)\end{aligned}$

$\begin{aligned} \Rightarrow y-3 &=\frac{3}{4}(x-3) \Rightarrow 4 y-12=3 x-9 \\ 3 x-4 y+12-9 &=0 \\ \Rightarrow 3 x-4 y+3 &=0 \end{aligned}$

 (c) If the required line passes through L(a,b) and M(b,a)

So equation of the line will be 

$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$

$y-b=\frac{a-b}{b-a}(x-a) \Rightarrow y-b=\frac{(a-b)}{(a-b)}(x-a)$

$\Rightarrow y-b=-1(x-a)$

$\Rightarrow y-b=-x+a$

$\Rightarrow x+y=a+b$

Question 9

Ans: Slope of line $3 x+4 y=8$

$\begin{aligned}&\Rightarrow 4 y=-3 x+8 \\&\Rightarrow y=-\frac{3}{4} x+2 \\&\text { So }\left(m_{1}\right)=\frac{-3}{4}\end{aligned}$

  and slope of line Px +2y = 7

$\Rightarrow 2 y=-p x+7$

$y=-\frac{p}{2} x+\frac{7}{2}$

$m_{2}=\frac{-p}{2}$

if the lines are parallel

So Their slope are equal

$\text { so } \begin{aligned}m_{1} &=m_{2} \Rightarrow \frac{-3}{4}=\frac{-p}{2} \\& \Rightarrow p=\frac{3 \times 2}{4}=\frac{3}{2}\end{aligned}$

Question 10

Ans: Co- ordinates of two points are E$(0,4), f(3,7)$

(ii) so slope(m) $=\frac{y_{2}-y_{1}}{n_{2}-x_{1}}=\frac{7-4}{3-0}=\frac{3}{3}=1$

So gradient of EF = 1

(ii) Now equation of EF will be 

$y-y_{1}=m\left(x-x_{1}\right)$

$\Rightarrow y-7=1(x-3)$

$\Rightarrow y-7=x-3$

$\Rightarrow x-y+7-3=0$

$\Rightarrow x-y+4=0$

(iii) Let the line EF intersect the x-axis at p then y- co-ordinates of P will be O 

If P lines on EF 

So it will satisfy the equation of EF 

$x-0+4=0$

$\Rightarrow x+4=0$

$\Rightarrow x=-4$

So co-ordinates of P will be (-4,0)

Question 11

Ans: Co-ordinates of points are $P(-2, L), Q(2,2)$ and R(6,-2)

(i) Gradient of QR(M) = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-2}{6-2}$

$=\frac{-4}{4}=-1$

(ii) Slope of line perpendicular to QR = $\frac{-1}{m}$

$=-(-1)=1$

So if it is passes through P 

So equation of the line will be 

$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-1=1(x+2) \\ \Rightarrow & y-1=x+2 \\ & x-y+2+1=0 \\ \Rightarrow & x-y+3=0 \end{aligned}$

Question 12

Ans: The line $3 x-4 y+12=0$ meets x-axis at P 

So y-coordinate of P will be O

Let the co-ordinates of P be (x,0)

if it lies on the $3 x-4 y+12=0$

So it will satisfy the equation

So $3 x-4 \times 0+12=0$

$\begin{aligned}&\Rightarrow 3 x+12=0 \\&\Rightarrow 3 x=-12\end{aligned}$

$\Rightarrow x=\frac{-12}{3}=-4$

So Co-ordinates of P will be (-4,0)

Now slope of the line $3 x+5 y-15=0$

$\Rightarrow 3 y=-3 x+15$

$\Rightarrow y=-\frac{3}{5} x+3$

$(m)=\frac{-3}{5}$

So slope of the line perpendicular to this line 

$=-\frac{1}{m}=-\left(\frac{-5}{3}\right)=\frac{5}{3}$

So Equation of the perpendicular line through P will be

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-0=\frac{5}{3}(x+4)$

$\Rightarrow y=\frac{5}{3}(x+4)$

$\Rightarrow 3 y=5 x+20$

$\Rightarrow 5 x-3 y+20=0$

Question 13

Ans: Point A divides the line joining the point p(5,2) and Q(9,-6) in the ratio 3:1

So $m_{1}: m_{2}=3: 1$

Now co-ordinates of $A$ will be

$\left(\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}, \frac{m_{1} y_{2}+m_{2} y_{1}{m_{1}+m_{2}}\right)$

$=\left(\frac{3 \times 9 + 1 \times 5}{3+1}, \frac{3 \times 6 \times +1 \times(-21)}{3 \times 1}\right)$

$=\left(\frac{27+5}{4}+\frac{18-2}{4}\right)$ or (8,4)

Slope of the line x =$3 y+4=0$

$\Rightarrow 3 y=x+4$

$\Rightarrow y=\frac{1}{3} x+\frac{4}{3}$  will be $(m)\frac{-1}{3}$

So slope of the line perpendicular to it = $\frac{-1}{m}$

$=-\left(\frac{9}{3}\right)=-3$

So equation of the line through A will be 

$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-4=-3(x-8) \\ \Rightarrow & y-4=-3 x+2+\\ \Rightarrow & 3 x+y=2++4 \\ \Rightarrow & 3 x+y=28 \\ \Rightarrow & y+3 x-28=0 \end{aligned}$

Question 14

Ans: The slope of line x- 2y+ 8=0

$\Rightarrow 2 y=x+8$

$\Rightarrow y=\frac{1}{2} x+4$

$(m)=\frac{1}{2}$

So slope of the line parallel to it = $\frac{1}{2}$

IF it passes through the point (1,2)

So equation of the line will be  

$y-y_{1}=m\left(x-x_{1}\right) \Rightarrow y-2=\frac{1}{2}(x-1)$

$\Rightarrow 2 y-4=x-1$

$\Rightarrow x-2 y+4-1=0$

$\Rightarrow x-2 y+3=0$

Question 15

Ans: The equation of the line is given 

$\begin{aligned} & \frac{x}{3}+\frac{y}{4}=1 \\ \Rightarrow & 4 x+3 y=12 \\ \Rightarrow & 3 y=-4 x+12 \\ \Rightarrow & y=-\frac{4}{3} x+4 \end{aligned}$

 So Gradient (m) $=\frac{-4}{3}$

and y- intercept (c) = 4

Question 16

Ans: (i) $\quad 2 x-b y+5=0$

$\Rightarrow b y=2 x+5$

$\Rightarrow y=\frac{2}{b} x+\frac{5}{b}$

So slope $\left(m_{1}\right)=\frac{2}{b}$

$a x+3 y=2$

$\Rightarrow 3 y=-a x+2$

$\Rightarrow y=\frac{-a}{3}+\frac{2}{3}$

So Slope $\left(m_{2}\right)=\frac{-a}{3}$

IF these lines are parallel

So 

$\begin{aligned} & m_{1}=m_{2} \\ \Rightarrow & \frac{2}{b}=\frac{-a}{3} \\ \Rightarrow &-a b=6 \\ \Rightarrow & a b=-6 \end{aligned}$

(ii) We know that equation of a line us $y=m x+c$ where m is slope and C is y - intercept 

If it passes through (1,3) and has y - intercept= 5

So it will also pass through  (0,5)

So slope (m) $=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{5-3}{0-1}=\frac{2}{-1}=-2$

So equation of the line

$\begin{aligned}y &=m x+c \\& \Rightarrow y=-2 x+5 \\& \Rightarrow 2 x+y=5\end{aligned}$

Question 17

Ans: (i) Equation of the line is given  $\frac{y}{2}=x-p$

If it passes through the point (-4,4)

So $\frac{4}{2}=-4-p$

$\Rightarrow 2=-4-p$

$\Rightarrow p=-4-2$

$\Rightarrow p=-6$

So p= -6

(ii) In equation 

$\frac{y}{2}=x-p$

$\Rightarrow y=2 x-2 p$

 

Slope $\left(m_{1}\right)=2$

and in equation $a x+5=3 y$

$\begin{aligned} \Rightarrow & y=\frac{a}{3} x+\frac{5}{3} \\ & \text { slope }\left(m_{2}\right)=\frac{a}{3} \end{aligned}$

If these two lines are parallel

So $m_{1}=m_{2}$

$\Rightarrow 2=\frac{a}{3}$

$\Rightarrow 6=a$

So $a=6$

Question 18

Ans: The line passes through a point (-2,0) and has its y-intercept = 3

So it will pass through $(0,3)$.

so slope of the line $(m)=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$=\frac{3-0}{0-(-2)}=\frac{3}{2}$

So Equation of the line will be

$\begin{gathered}y=m x+c \\y=\frac{3}{2} x+3 \Rightarrow 2 y=3 x+6\end{gathered}$

Question 19

Ans: (a) In the figure 

(i) Co - ordinates of A are (2, 3) of B are (-1,2) and of C are (3,0)

(ii) Equation of a line through A and parallel to BC 

Slope of BC (m) $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{0-2}{3+1}=\frac{-2}{4}$

 $=\frac{-1}{2}$

So slope of the required line $=\frac{-1}{2}$

So $y-y_{1}=m\left(x-x_{1}\right)$

$\begin{aligned}&\Rightarrow y-3=-\frac{1}{2}(x-2) \\&\Rightarrow 2 y-6=-x+2 \\&\Rightarrow x+2 y=2+6 \Rightarrow x+2 y=8 \\&\Rightarrow x+2 y-8=0\end{aligned}$

(b) Let ratio be $m_{1}: m_{2}$ Which divides the join of 

$A(0,3)$ and $B(4,1)$ by the $x$-axis at $p$

if $p$ lies on $x$-axis

So co-ordinates of $p$ be $(x, 0)$

so $\begin{aligned} & 0=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}} \Rightarrow \frac{m_{1} \times 1+m_{2} \times 3}{m_{1}+m_{2}}=0 \\ & \frac{m_{1}+3 m_{2}}{m_{1}+m_{2}}=0 \Rightarrow m_{1}+3 m_{2}=0 \\ & \Rightarrow m_{1}=-3 m_{2} \end{aligned}$

$\Rightarrow \frac{m_{1}}{m_{2}}=\frac{-3}{1}$

So ratio $-3: 1$

So and $x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{-3 \times 4+1 \times 0}{-3+1}$

$=\frac{-12+0}{-2}=\frac{-12}{-2}=6$

So co- ordinates of P will be (6,0)

Question 20

Ans: In the given Line 2y = 3x+5

$\Rightarrow y=\frac{3}{2} x+\frac{5}{2}$

slope $\left(m_{1}\right)=\frac{3}{2}$

So slope of the line AB perpendicular to the given line $\left(m_{2}\right)=\frac{-1}{m_{1}}=\frac{-2}{3}$

If it passes through (3,2)

So equation of the line will be 

$\begin{aligned} & y-y_{1}=m\left(x-x_{1}\right) \\ \Rightarrow & y-2=-\frac{2}{3}(x-3) \\ \Rightarrow & 3 y-6=-2 x+6 \\ \Rightarrow & 2 x+3 y=6+6 \\ \Rightarrow & 2 x+3 y=12 \\ \Rightarrow & 2 x+3 y-12=0 \end{aligned}$

If this lines meets x-axis at A and y-axis at B

So substituting the value of y = 0 and x =0

If y= 0 ,then 

$2 x+0 \times y=12$

$\Rightarrow 2 x=12$

$\Rightarrow x=\frac{12}{2}=6$

If x =0, then 

$2 \times 0+3 y=12$

$\Rightarrow 0+3 y=12$

$\Rightarrow 3 y=12$

$\Rightarrow y=\frac{12}{3}=4$

So co-ordinates of A and B will be A (6,0) and B (0,4)

IF 0 is the origin 

So area of $\triangle O A B=\frac{1}{2} \times O A \times O B$

$=\frac{1}{2} \times 6 \times 4=12$ Sq.units

Question 21

Ans: (i) Let co-ordinates of point p be (x,y) 

If p divides the line joining A $(-4,1)$ and $B(17.10)$

in the ratio $1: 2$ i.e. $m_{1}: m_{2}=1: 2$

$x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}=\frac{1 \times 17+2 \times(-4)}{1+2}$

$=\frac{17-8}{3}=\frac{9}{3}=3$

and $y=\frac{m_{1} y_{2}+m_{2} y_{1}}{m_{1}+m_{2}}=\frac{1 \times 10+2 \times 1}{1+2}$

$=\frac{10+2}{3}=\frac{12}{3}=4$

So co-ordinates of p arc (3,4)

(ii) If O is the origin (0,0)

$\begin{aligned} O P &=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \\ &=\sqrt{(3-0)^{2}+(4-0)^{2}} \\ &=\sqrt{(3)^{2}+(412} \\ &=\sqrt{9+16}=\sqrt{25}=5 \end{aligned}$

(ii) Let the ratio be $m_{1}: m_{2}$ In which y-axis divides the line segment AB. Let R be the point on y- axis which divides AB in the ratio  $m_{1}: m_{2}$

co-ordinales of $R$ be $(0, y)$

$\text { so } x=\frac{m_{1} x_{2}+m_{2} x_{1}}{m_{1}+m_{2}}$

$\begin{aligned} o &=\frac{m_{1} \times 17+m_{2} \times(-4)}{m_{1}+m_{2}} \\ & \Rightarrow 17 m_{1}-4 m_{2}=0 \\ & \Rightarrow 17 m_{1}=4 m_{2} \\ & \Rightarrow \frac{m_{1}}{m_{2}}=\frac{4}{17} \end{aligned}$

So Ratio = 4:17

RS Aggarwal solution class 8 chapter 11 Compound Interest Exercise 11C

Exercise 11C

Page-154

Q1 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 1:

Find the amount and the compound interest on Rs 8000 for 1 year at 10% per annum, compounded half-yearly.

Answer 1:

Q2 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 2:

Find the amount and the compound interest on Rs 31250 for $1\frac{1}{2}$ years at 8% per annum, compounded half-yearly.

Sol :

Principal, P = Rs. 31250

Annual rate of interest, R = 8% 

Time, (n) $=1\frac{1}{2}=\frac{3}{2}$ years 

Then the amount with the compound interest is given by

$A=P\left(\frac{1}{1}+\frac{R}{2 \times 100}\right)^{2 n}$

$=31250\left(\frac{1}{1}+\frac{8}{2 \times 100}\right)^{2 \times \frac{3}{2}}$

$=31250\left(\frac{100+4}{100}\right)^3$

$=31250 \times \frac{104}{100}\times \frac{104}{100} \times \frac{104}{100}$

$=31250 \times \frac{26}{25}\times \frac{26}{25}\times \frac{26}{25}$

A=35152

Therefore, compound interest = amount - principal=Rs (35152- 31250) = Rs 3902

Q3 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 3:

Find the amount and the compound interest on Rs 12800 for 1 year at $7\frac{1}{2}\%$ per annum, compounded semi-annually.

Answer 3:

Q4 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 4:

Find the amount and the compound interest on Rs 160000 for 2 years at 10% per annum, compounded half-yearly.

Answer 4:

Q5 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 5:

Swati borrowed Rs 40960 from a bank to buy a piece of land. If the bank charges $12\frac{1}{2}\%$ per annum, compounded half-yearly, what amount will she have to pay after $1\frac{1}{2}$ years? Also, find the interest paid by her.

Answer 5:

Q6 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 6:

Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.

Answer 6:

Q7 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 7:

Sheela deposited Rs 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year?

Answer 7:

Q8 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 8:

Neeraj lent Rs 65536 for 2 years at $12\frac{1}{2}\%$ per annum, compounded annually. How much more could he earn if the interest were compounded half-yearly?

Answer 8:

Q9 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 9:

Sudershan deposited Rs 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months?

Answer 9:

Q10 | Ex-11C | Class 8 | RS AGGARWAL | chapter 11 | Compound Interest | myhelper

OPEN IN YOUTUBE

Question 10:

Arun took a loan of Rs 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year?

Answer 10: