5 Marks Questions

Question 15 Marks

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder as shown in the figure. If the height of the cylinder is $10 \ cm$ and its base is of radius $3.5 \ cm,$ find the total surface area of the article.

Answer

$\text { TSA of the article }=2 \pi rh +2\left(2 \pi r ^2\right)$
$=2 \pi(3.5)(10)+2\left[2 \pi(3.5)^2\right]$
$=70 \pi+49 \pi$
$=119 \pi$
$=119 \times \frac{22}{7}$
$=374 \ cm^2$

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Question 25 Marks

The product of Tanay's age $($in years$)$ five years ago and his age ten years later is $16.$ Determine Tanay's present age.

Answer

Let the present age of Tanay be $x$ years
By the question,
$(x-5)(x+10)=16$
or, $x^2+5 x-50=16$
or, $x^2+5 x-66=0$
or, $x^2+11 x-6 x-66=-66$
$x(x+1)-16(x-111)=0$
$(x+11)(x-6)=0$
$=-11,6$
Rejecting $x=-11$, as age cannot be negative.
$\therefore$ Present age of Tanay is $6$ years.

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Question 35 Marks

Find the mean of the following frequency distribution:

Class	$5 - 15$	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$
Frequency	$6$	$11$	$21$	$23$	$14$	$5$

Answer

$CI$	$f_i$	$x_i$	$d_i$	$u_i$	$f_iu_i$
$5 - 15$	$6$	$10$	$-20$	$- 2$	$-12$
$15 - 25$	$11$	$20$	$-10$	$- 1$	$-11$
$25 - 35$	$21$	$30$	$0$	$0$	$0$
$35 - 45$	$23$	$40$	$10$	$1$	$23$
$45 - 55$	$14$	$50$	$20$	$2$	$28$
$55 - 65$	$5$	$60$	$30$	$3$	$15$
$\text{Total}$	$80$				$43$

$\text { Mean }= A +\frac{\Sigma f_i u_i}{\Sigma f_i} \times h$
$=30+\frac{43}{80} \times 10$
$=35.375$

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Question 45 Marks

A conical vessel of radius $6 \ cm$ and height $8 \ cm$ is completely filled with water. A sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed as shown in Figure. What fraction of water over flows?

Answer

Radius $(R)$ of conical vessel $= 6 \ cm$
Height $(H)$ of conical vessel $= 8 \ cm$

Volume of conical vessel $\left( V _{ c }\right)=\frac{1}{3} \pi r ^2 h$
$=\frac{1}{3} \times \pi \times 6 \times 6 \times 8$
$=96 \pi \ cm^3$
Let the radius of the sphere be $\text{rcm}$
In right $\Delta PO'R $ by pythagoras theorem We have
$l^2=6^2+8^2$
$l=\sqrt{36+64}=10 \ cm$
In right triangle $\text{MRO}$
$\sin \theta=\frac{O M}{O R}$
$\Rightarrow \frac{3}{5}=\frac{r}{8-r}$
$\Rightarrow 24-3 r=5 r$
$\Rightarrow 8 r=24$
$\Rightarrow r=3 \ cm$
$\therefore V _1=$ Volume of the sphere $=\frac{4}{3} \pi \times 3^3 \ cm^3=36 \pi \ cm^3$
$V _2=$ Volume of the water $=$ Volume of the cone $=\frac{1}{3} \pi \times 6^2 \times 8 \ cm^3=96 \pi \ cm^3$
Clearly, volume of the water that flows out of the cone is same as the volume of the sphere
i.e., $V _1$.
$\therefore$ Fraction of the water that flows out $=V_1: V_2=36 \pi: 96 \pi=3: 8$

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Question 55 Marks

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then prove that the other two sides are divided in the same ratio.

Answer

Given: $\text{ABC}$ is a triangle in which $DE \| BC$.
To prove: $\frac{A D}{B D}=\frac{A E}{C E}$
Construction: Draw $D N \perp A E$ and $E M \perp A D$.,
Join $BE$ and $CD$.
Proof :

In $\triangle A D E$,
Area of $\Delta A D E=\frac{1}{2} \times A E \times D N...(i)$
In $\Delta D E C,$
Area of $\Delta D C E=\frac{1}{2} \times C E \times D N...(ii)$
Dividing equation $(i)$ by equation $(ii),$
$\Rightarrow \frac{\operatorname{area} \ (\Delta\ A D E)}{\text { area }(\Delta\ D E C)}=\frac{\frac{1}{2} \times A E \times D N}{\frac{1}{2} \times C E \times D N}$
$\Rightarrow \frac{\text { area }(\Delta\ A D E)}{\text { area }(\Delta\ D E C)}=\frac{A E}{C E} \ldots(\text { iii) }$
Similarly, In $\Delta\ A D E$,
Area of $\Delta\ A D E=\frac{1}{2} \times A D \times E M \ldots (iv)$
In $\Delta\ D E B$,
Area of $\Delta\ D E B=\frac{1}{2} \times E M \times B D$
Dividing equation $(iv)$ by equation $(v),$
$\Rightarrow \frac{\operatorname{area}(\Delta\ A D E)}{\text { area }(\Delta\ D E B)}=\frac{\frac{1}{2} \times A D \times E M}{\frac{1}{2} \times B D \times E M}$
$\Rightarrow \frac{\text { area }(\Delta\ A D E)}{\text { area }(\Delta\ D E B)}=\frac{A D}{B D} \ldots(vi)$
$\Delta D E B$ and $\Delta D E C$ lie on the same base $DE$ and between two parallel lines $DE$ and $BC$ .
$\therefore$ Area $(\Delta\ D E B)=$ Area $(\Delta\ D E C)$
From equation $(iii),$
$\Rightarrow \frac{\text { area }(\Delta\ A D E)}{\text { area }(\Delta\ D E B)}=\frac{A E}{C E} ...(i)$
From equation $(vi)$ and equation $(vii),$
$\frac{A E}{C E}=\frac{A D}{B D}$
$\therefore$ If a line is drawn parallel to one side of a triangle to intersect the other two sides in two points, then the other two sides are divided in the same ratio.

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Question 65 Marks

A 2-digit number is such that the product of its digits is 24. If 18 is subtracted from the number, the digits interchange their places. Find the number.

Answer

Let the ten's digit be $x$ and the one's digit be $y$.
The number will be $10 x + y$
Given, a product of digits is 24
$\therefore xy=24$
or, $y =\frac{24}{x}$
Given that when 18 is subtracted from the number, the digits interchange their places.
$\therefore 10 x+y-18=10 y+x$
or, $9 x -9 y =18$
Substituting y from equation (i) in equation (ii), we get
$9 x-9\left(\frac{24}{x}\right)=18$
or, $x-\frac{24}{x}=2$
or, $x ^2-24-2 x =0$
or, $x^2-2 x-24=0$
or, $x^2-6 x+4 x-24=0$
or, $x(x-6)+4(x-6)=0$
or, $(x-6)(x+4)=0$
or, $x-6=0$ and $x+4=0$
or, $x=6$ and $x=-4$
Since, the digit cannot be negative, so, $x=6$
Substituting $x=6$ in equation (i), we get
$y=\frac{24}{6}=4$
$\therefore$ The number $=10(6)+4=60+4=64$

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