5 Marks Questions

Question 15 Marks

The ratio of incomes of two persons is $11 : 7$ and the ratio of their expenditures is $9 : 5.$ If each of them manages to save $Rs.400$ per month, find their monthly incomes.

Answer

Let the incomes of two persons be $11 x$ and $7 x .$
And the expenditures of two persons be $9 y$ and $5 y$
$\therefore 11 x-9 y=400$
$7 x-5 y=400 \ldots.$
Multiplying $(i)$ by $5$ and $(ii)$ by $9$ and subtracting,
$55 x-45 y=2,000$
$63 x-45 y=3,600$
$-8 x=-1600$
$\therefore-8 x=-1600$
$x=\frac{-1,600}{-8}=200$
Therefore, Their monthly incomes are
$11 \times 200=\text { Rs } 2200$
$7 \times 200=\text { Rs } 1400$

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Question 25 Marks

A hemispherical depression is cut out from one face of a cubical block of side $7 cm$ , such that the diameter of the hemisphere is equal to the edge of the cube. Find the surface area of the remaining solid.

Answer

Edge of the cube, $a=7 \ cm$.
Radius of the hemisphere, $r=\frac{7}{2} \ cm$.
Surface area of remaining solid
$=$ total surface area of the cube $-$ area of the top of hemispherical part $+$ curved surface area of the hemisphere
$=6 a^2-\pi r^2+2 \pi r^2=6 a^2+\pi r^2$
$=\left(6 \times 7 \times 7+\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\right) \ cm^2$
$=(294+38.5) \ cm^2=332.5 \ cm^2$

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Question 35 Marks

In an $A.P.,$ the $n ^{\text {th }}$ term is $\frac{1}{m}$ and the $m ^{\text {th }}$ term is $\frac{1}{n}$. Find $(i) (mn) { }^{\text {th }}$ term, $(ii)$ sum of first $(mn)$ terms.

Answer

$a_n=\frac{1}{m}$
$a+(n-1) d=\frac{1}{m}$
$a_m=\frac{1}{n}$
$a+(m-1) d=\frac{1}{n}$
On solving,
$a=\frac{1}{mn}$
$d=\frac{1}{mn}$
$\text { i. } a_{mn}=\frac{1}{mn}+(mn-1) \times \frac{1}{mn}$
$\quad=\frac{1+mn-1}{mn}=1$
$\text { ii. } S_{mn}=\frac{mn}{2}\left(\frac{1}{mn}+1\right)$
$\quad=\frac{1+mn}{2}$

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Question 45 Marks

In a cylindrical vessel of radius $10 \ cm ,$ containing some water, $9000$ small spherical balls are dropped which are completely immersed in water which raises the water level. If each spherical ball is of radius $0.5 \ cm ,$ then find the rise in the level of water in the vessel.

Answer

Volume of raised water in cylinder $=$ Volume of $9000$ spherical balls
$\pi(10)^2 H=9000 \times \frac{4}{3} \times \pi \times(0.5)^3$
$\therefore H=15 \ cm$

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Question 55 Marks

Find the lengths of the medians of a $\triangle A B C$ having vertices at $A(0,-1), B(2,1)$ and $C(0,3)$.

Answer

Let $A (0,-1), B (2,1)$ and $C (0,3)$ be the given points.
Let $AD , BE$ and CF be the medians
Coordinates of D are $\left(\frac{2+0}{2}, \frac{1+3}{2}\right)=(1,2)$
Coordinates of E are $\left(\frac{0}{2}, \frac{3-1}{2}\right)=(0,1)$
Coordinates of F are $\left(\frac{2+0}{2}, \frac{1-1}{2}\right)=(1,0)$
Length of median $A D=\sqrt{(1-0)^2+(2+1)^2}=\sqrt{10}$ units
Length of median $B E=\sqrt{(2-0)^2+(1-1)^2}=2$ units
Length of median $C F=\sqrt{(1-0)^2+(0-3)^2}=\sqrt{10}$ units

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Question 65 Marks

Abdul travelled $300 \ km$ by train and $200 \ km$ by taxi taking $5$ hours $30$ minutes. But, if he travels $260 \ km$ by train and $240 \ km$ by taxi, he takes $6$ minutes longer. Find the speed of the train and that of the taxi.

Answer

Suppose, speed of the train be $x \ km / hr$ and the speed of taxi be $y \ km / h$. time taken to cover $300 \ km $ by the train $=\frac{300}{x}$ hours time taken to cover $200 \ km$ by the taxi $=\frac{200}{y}$ hours
Total time taken $=5 \frac{30}{60}$ hours $=5 \frac{1}{2}$ hours $=\frac{11}{2}$ hours
$\therefore \frac{300}{x}+\frac{200}{y}=\frac{11}{2}$
$\Rightarrow \frac{600}{x}+\frac{400}{y}=11$
Put $\frac{1}{x}= u$ and $\frac{1}{y}= v$
$\Rightarrow 600 u+400 v=11$
time taken to cover $260 \ km $ by the train $=\frac{260}{x}$ hours time taken to cover $240 \ km$ by the taxi $=\frac{240}{y}$ hours
Total time taken $=5 \frac{36}{60}$ hours $=5 \frac{1}{2}$ hours $=\frac{11}{2}$ hours
$\Rightarrow 1300 u+1200 v=28$
Multiplying $(i)$ by $3$ and subtracting $(ii)$ from it,
$\Rightarrow 500 u=5 $
$\Rightarrow u=\frac{5}{500} $
$\Rightarrow u=\frac{1}{100}$
Substituting $u =\frac{1}{100}$ in $ (i),$
$\Rightarrow v =\frac{1}{80}$
$\therefore u=\frac{1}{100} $
$\Rightarrow \frac{1}{x}=\frac{1}{100}$
$ \Rightarrow x=100$
$v=\frac{1}{80} $
$\Rightarrow \frac{1}{y}=\frac{1}{80} $
$\Rightarrow y=80$
$\therefore$ the speed of the train $=100 \ km / hr$
the speed of the taxi $=80 \ km / hr$

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