5 Marks Questions

Question 15 Marks

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Ratio of the radius of the cone to its slant height is $3: 5$. If the volume of the toy is $240 \pi cm^3$, then find the total height of the toy.

Answer

Let the radius and the slant height of the cone be $3 x cm$ and $5 x cm$ respectively
$\therefore$height of the cone $(h)=\sqrt{(5 x)^2-(3 x)^2}=4 x cm$
According to question, volume of toy $=240 \pi$
$
\Rightarrow \frac{2}{3} \pi(3 x)^3+\frac{1}{3} \pi(3 x)^2(4 x)=240 \pi
$
Solving, we get $x=2$
$\therefore$ Total height of toy $=[4(2)+3(2)] cm =14 cm$

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Question 25 Marks

The largest possible hemisphere is drilled out from a wooden cubical block of side $21 \ cm$ such that the base of the hemisphere is on one of the faces of the cube. Find:
$(i)$ the volume of wood left in the block,
$(ii)$ the total surface area of the remaining solid.

Answer

Diameter of hemisphere $=$ side of the cube $=21 \ cm$
$\therefore$ radius of hemisphere $=\frac{21}{2} \ cm (i)$
Volume of the wood left $=$ volume of cube $-$ volume of hemisphere
$=21^3-\frac{2}{3} \times \frac{22}{7} \times\left(\frac{21}{2}\right)^3$
$=6835.5 \ cm^3$
$(ii)$ Total surface area of remaining solid $= TSA$ of cube $-$ base area of hemisphere
$+ CSA$ of hemisphere$=6 \times 21^2-\frac{22}{7} \times\left(\frac{21}{2}\right)^2+2 \times \frac{22}{7} \times\left(\frac{21}{2}\right)^2$
$=2992.5 \ cm^2$

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Question 35 Marks

A shopkeeper buys a number of books for $₹ 1,800$ . If he had bought $15$ more books for the same amount, then each book would have cost him $₹ 20$ less. Find how many books he bought initially.

Answer

Let the number of books bought initially be $x$ According to question,
$\frac{1800}{x}-\frac{1800}{x+15}=20$
$\Rightarrow x^2+15 x-1350=0$
$\Rightarrow(x+45)(x-30)=0$
$x \neq-45$
$\therefore x=30$
So, the number of books bought initially $=30$

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Question 45 Marks

If Nidhi were $7$ years younger than what she actually is, then the square of her age $($in years$)$ would be $1$ more than $5$ times her actual age. What is her present age ?

Answer

Let the present age of Nidhi be $x$ years.
According to question, $(x-7)^2=5 x+1$
$\Rightarrow x^2-19 x+48=0$
$\Rightarrow(x-16)(x-3)=0$
$\Rightarrow x=16,3$
$x \neq 3$
$\therefore x=16$
Hence, the present age of Nidhi $=16$ years

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Question 55 Marks

An age$-$wise list of number of literate people in a block is prepared in the following table. There are total $100$ people and their median age is $41.5$ years. Information about two groups are missing, which are denoted by $x$ and $y.$ Find the value of $x$ and $y.$

Age $($in years$)$	Number of literate people
$10-20$	$15$
$20-30$	$X$
$30-40$	$12$
$40-50$	$20$
$50-60$	$Y$
$60-70$	$8$
$70-80$	$10$

Answer

Age $($in years$)$	Number of literate people $(fi)$	Cumulative frequency
$10-20$	$15$	$15$
$20-30$	$x$	$15 + x$
$30-40$	$12$	$27 + x$
$40-50$	$20$	$47 + x$
$50-60$	$Y$	$47 + x + y$
$60-70$	$8$	$55 + x + y$
$70-80$	$10$	$65 + x + y$

$65+x+y=100$
$\Rightarrow x+y=35 \ldots \text { (i) }$
$\text { Median }=41.5 $
$\therefore 40-50 \text { is the median class. }$
$\Rightarrow 41.5=40+\frac{50-27-x}{20} \times 10$
$\text { Solving, we get } x=20$
$\text { From (i), } y=15$

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Question 65 Marks

In the given figure, $\text{MNOP}$ is a parallelogram and $AB \| MP$. Prove that $QC \| PO.$

Answer

$M P \| A B$
$\Rightarrow \triangle Q M P \sim \Delta Q A B$
$\Rightarrow \frac{M P}{A B}=\frac{Q P}{Q B} \ldots \text { (i) }$
Now, $NO \| M P \| A B$
$\Rightarrow \triangle C N O \sim \triangle C A B$
$\Rightarrow \frac{N O}{A B}=\frac{C O}{C B} \ldots \text { (ii) }$
$\text { As } M P=N O$
From $(i)$ and $(ii), \frac{C O}{C B}=\frac{Q P}{Q B}$
$\frac{C O}{C B}-1=\frac{Q P}{Q B}-1$
$\text { or } \frac{B O}{O C}=\frac{B P}{P Q}$
$\therefore Q C \| P O$

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Maths 5 Marks Questions

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