5 Marks Questions

Question 15 Marks

The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in ₹)	Frequency
1000-1500	24
1500-2000	40
2000-2500	33
2500-3000	28
3000-3500	30
3500-4000	22
4000-4500	16
4500-5000	7

Answer

We may observe from the given data that maximum class frequency is 40 belonging to 1500 - 2000 interval.
Class size (h) = 500
Mode = $1+\frac{f-f_1}{2 f-f_1-f_2} \times h$
Lower limit $(l)$ of modal class = 1500
Frequency (f) of modal class = 40
Frequency $\left( f _1\right)$ of class preceding modal class = 24
Frequency $\left( f _2\right)$ of class succeeding modal class = 33
mode = $1500+\frac{40-24}{2 \times 40-24-33} \times 500$
$=1500+\frac{16}{80-57} \times 500$
= 1500 + 347.826
= 1847.826 ≈ 1847.83

Expenditure (in ₹.)	Number of families $f _{ i }$	$x _{ i }$	$d_i=x_i-2750$	$u _{ i }$	$u _{ i } f _{ i }$
1000-1500	24	1250	-1500	-3	-72
1500-2000	40	1750	-1000	-2	-80
2000-2500	33	2250	-500	-1	-33
2500-3000	28	2750=a	0	0	0
3000-3500	30	3250	500	1	30
3500-4000	22	3750	1000	2	44
4000-4500	16	4250	1500	3	48
4500-5000	7	4750	2000	4	28
	$\Sigma f_i=200$				$\Sigma f_i d_i=-35$

Mean $\bar{x}=a+\frac{\Sigma f_i d_i}{\Sigma f_i} \times h$
$\bar{x}=2750+\frac{-35}{200} \times 500$
$\bar{x}=2750-87.5$
$\bar{x}=2662.5$

View full question & answer→

Question 25 Marks

A solid toy is in the form of a hemisphere surmounted by a right circular cone. Height of the cone is 2 cm and the diameter of the base is 4 cm. If a right circular cylinder circumscribes the solid. Find how much more space it will cover.

Answer

Let BPC be the hemisphere and ABC be the cone mounted on the base of the hemisphere. Let EFGH be the right circular cylinder circumscribing the given toy.

We have,
Given radius of cone, cylinder and hemisphere (r) = $\frac{4}{2}=2 cm$
Height of cone (l) = 2 cm
Height of cylinder (h) = 4 cm
Now, Volume of the right circular cylinder = $\pi r ^2 h=\pi \times 2^2 \times 4 cm^3=16 \pi cm^3$
Volume of the solid toy = $\left\{\frac{2}{3} \pi \times 2^3+\frac{1}{3} \pi \times 2^2 \times 2\right\} cm ^3=8 \pi cm^3$
$\therefore$ Required space = Volume of the right circular cylinder - Volume of the toy
$=16 \pi cm^3-8 \pi cm^3=8 \pi cm^3$.
Hence, the right circular cylinder covers $8 \pi cm^3$ more space than the solid toy.
So, remaining volume of cylinder when toy is inserted in it = $8 \pi cm^3$

View full question & answer→

Question 35 Marks

A solid is in the shape of a hemisphere surmounted by a cone. If the radius of hemisphere and base radius of cone is 7 cm and height of cone is 3.5 cm, find the volume of the solid. (Take $\pi=\frac{22}{7}$ )

Answer

Volume of solid = $\frac{1}{3} \times \frac{22}{7} \times(7)^2 \times 3.5+\frac{2}{3} \times \frac{22}{7} \times(7)^3$
$=\frac{22}{7} \times(7)^2 \times\left[\frac{3.5}{3}+\frac{2}{3} \times 7\right]$
$=898 \frac{1}{3}$ or $898.33 cm^3$

View full question & answer→

Question 45 Marks

Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that $\triangle A B C \sim \triangle P Q R$.

Answer

Given : In $\triangle A B C$ and $\triangle P Q R$ The AD and PM are their medians,
such that $\frac{A B}{P Q}=\frac{A D}{P M}=\frac{A C}{P R}$
To prove : $\triangle A B C \sim \triangle P Q R$
Construction : Produce AD to E such that AD = DE and produce PM to N such that PM = MN. Join CE and RN.
Proof : In $\triangle A B D$ and $\triangle E D C$
AD = DE
$\angle A D B=\angle E D C$ (vertically opposite angles)
BD = DC(as AD is a median)
$\therefore \quad \triangle A B D \equiv \triangle E D C$ (By SAS congruency)
or, AB = CE (By CPCT)
Similarly, PQ = RN
$\frac{A B}{P Q}=\frac{A D}{P M}=\frac{A C}{P R}$ (Given)
or, $\frac{C E}{R N}=\frac{2 A D}{2 P M}=\frac{A C}{P R}$
or $\frac{C E}{R N}=\frac{A E}{P N}=\frac{A C}{P R}$
So $\triangle A C E \sim \triangle P R N$
$\angle 3=\angle 4$
Similarly $\angle 1=\angle 2$
$\angle 1+\angle 3=\angle 2+\angle 4$
So $\angle A=\angle P$ and
$\frac{A B}{P Q}=\frac{A C}{P R}$ (given)
Hence $\triangle A B C \sim \triangle P Q R$

View full question & answer→

Question 55 Marks

The sum of squares of two consecutive multiples of 7 is 637. Find the multiples.

Answer

According to the question, let the consecutive multiples of 7 be 7x and 7x +7
$(7 x)^2+(7 x+7)^2=637$
or, $49 x^2+49 x^2+49+98 x=637$
or, $98 x^2+98 x-588=0$
or, $x^2+x-6=0$
or, $(x+3)(x-2)=0$
or, $x=-3,2$
Rejecting the value, x=2
Thus,the required multiples are, 14 and 21.

View full question & answer→

Question 65 Marks

Solve for x:$\left(\frac{2 x}{x-5}\right)^2+5\left(\frac{2 x}{x-5}\right)-24=0, x \neq 5$

Answer

We have given,
$\left(\frac{2 x}{x-5}\right)^2+5\left(\frac{2 x}{x-5}\right)-24=0$
Let $\frac{2 x}{(x-5)}$ be $y$
$\therefore \quad u^2+5 u-24=0$
Now factorise,
$y^2+8 y-3 y-24=0$
$y(y+8)-3(y+8)=0$
$(y+8)(y-3)=0$
$y=3,-8$
Putting y=3
$\frac{2 x}{x-5}=3$
2x = -8x + 40
10x = 40
x = 4
Hence, x is 15 , 4

View full question & answer→

Maths 5 Marks Questions

Test yourself on this topic