5 Marks Questions

Question 15 Marks

If $(-5)$ is a root of the quadratic equation $2 x ^2+ px +15=0$ and the quadratic equation $p \left( x ^2+ x \right)+ k =0$ has equal roots, then find the values of $p$ and $k .$

Answer

Since $(-5)$ is a root of given quadratic equation $2 x^2+p x+15=0$,then,
$2(-5)^2+p(-5)-15=0$
$50-5 p-15=0$
$5 p=35 \Rightarrow p=7$
Now $p\left(x^2+x\right)+k=0$ has equal roots
$p x^2+p x+k=0$
$\text { So }(b)^2-4 a c=0$
$(p)^2-4 p \times k=0$
$(7)^2-4 \times 7 \times k=0$
$28 k=49$
$k=\frac{49}{28}=\frac{7}{4}$
hence $p =7$ and $k=\frac{7}{4}$

View full question & answer→

Question 25 Marks

A tent is in the shape of a right circular cylinder up to a height of 3 m and then a right circular cone, with a maximum height of 13.5 m above the ground. Calculate the cost of painting the inner side of the tent at the rate of ₹ 2 per square metre, if the radius of the base is 14 m.

Answer

Height of the cylinder $=3 m$.
Total height of the tent above the ground $=13.5 m$ height of the cone $=(13.5-3) m =10.5 m$
Radius of the cylinder $=$ radius of cone $=14 m$
Curved surface area of the cylinder $=2 \pi r h m^2=\left(2 \times \frac{22}{7} \times 14 \times 3\right) m ^2=264 m^2$
$\therefore \quad l=\sqrt{r^2+h^2}=\sqrt{14^2+(10.5)^2}=\sqrt{196+110.25}=\sqrt{306.25}=17.5$
$\therefore$ Cured surface area of the cone $=\pi r l=\left(\frac{22}{7} \times 14 \times 17.5\right) m ^2=770 m^2$
Let S be the total area which is to be painted. Then,
$S=$ Curved surface area of the cylinder + Curved surface area of the cone
$\Rightarrow S=(264+770) m^2=1034 m^2$
Hence, Cost of painting = S × Rate = ₹ (1034 × 2) = ₹ 2068

View full question & answer→

Question 35 Marks

A survey regarding the heights $($in $cm)$ of $50$ girls of $X$ of a school was conducted and the following data was obtained:

Height $($in $cm)$	$120 - 130$	$130 - 140$	$140 - 150$	$150 - 160$	$160 - 170$	Total
Number of girls	$2$	$8$	$12$	$20$	$8$	$50$

Find the mean and mode of the above data

Answer

Height $($in $cm)$	No. of girls	$x _{ i }$	$u _{ i }$	$f _{ i } u _{ i }$
$120 - 130$	$2$	$125$	$- 2$	$- 4$
$130 - 140$	$8$	$135$	$- 1$	$- 8$
$140 - 150$	$12$	$145 = a$	$0$	$0$
$150 - 160$	$20$	$155$	$1$	$20$
$160 - 170$	$8$	$165$	$2$	$16$
Total	$50$			$24$

$\text { Mean }=145+\frac{24}{50} \times 10$
$=149.8$
$\therefore$ mean height is $149.8 \ cm$
Modal class is $150-160$
$\text { Mode }=150+\frac{(20-12)}{(2 \times 20-12-8)} \times 10$
$=154$
$\therefore$ modal height is $154 \ cm$

View full question & answer→

Question 45 Marks

A solid is in the shape of a right-circular cone surmounted on a hemisphere, the radius of each of them being $7 \ cm$ and the height of the cone is equal to its diameter. Find the volume of the solid.

Answer

Given, radius of cone $=$ radius of hemisphere
$=r$
$=7 \ cm$
Height of cone $(h)=2 \times$ radius
$=2 \times 7$
$=14 \ cm$
Volume of solid $=$ Volume of cone $+$ Volume of hemisphere
Volume of solid $( V )=\frac{1}{3} \pi r^2 h+\frac{2}{3} \pi r^3$
$=\frac{1}{3} \pi r^2(2 r)+\frac{2}{3} \pi r^3 \ldots(\because h=2 r)$
$=\frac{2}{3} \pi r^3+\frac{2}{3} \pi r^3$
$=\frac{4}{3} \pi r^3$
$=\frac{4}{3} \times \frac{22}{7} \times 7 \times 7 \times 7$
$=\frac{4312}{3}$
$=1437.33 \ cm^3$

View full question & answer→

Question 55 Marks

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^{\circ},$ which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be $60^{\circ}$. Find the further time taken by the car to reach the foot of the tower from this point.

Answer

In right triangle $\text{ABP},$
$\tan 30^{\circ}=\frac{AB}{BP}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{AB}{BP}$
$BP=AB \sqrt{3} \ldots(i)$
In right triangle $\text{ABQ}$,
$\tan 60^{\circ}=\frac{A B}{B Q}$
$\Rightarrow \sqrt{3}=\frac{A B}{B Q}$
$\Rightarrow{ }_{B Q}=\frac{A B}{\sqrt{3}} \ldots \ldots .(ii)$
$\because P Q=B P-B Q$
$\therefore PQ = AB \sqrt{3}-\frac{A B}{\sqrt{3}}$
$=\frac{3 A B-A B}{\sqrt{3}}$
$=\frac{2 A B}{\sqrt{3}}$
$=2 BQ \ [$From eq. $(ii)]$
$\Rightarrow BQ=\frac{1}{2} PQ$
$\because$ Time taken by the car to travel a distance $P Q=6$ seconds.
$\therefore$ Time taken by the car to travel a distance $BQ,$
i.e. $\frac{1}{2} P Q=\frac{1}{2} \times 6=3$ seconds.
Hence, the further time taken by the car to reach the foot of the tower is $3$ seconds.

View full question & answer→

Question 65 Marks

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that cost of production of each article $($in rupees$)$ was $3$ more than twice the number of articles produced on that day. If, the total cost of production on that day was $₹ 90,$ find the number of articles produced and the cost of each article.

Answer

Let cost of production of each article be $Rs.x$
We are given total cost of production on that particular day $= Rs.90$
Therefore, total number of articles produced that day $=90 / x$
According to the given conditions,
$x=2\left(\frac{90}{x}\right)+3$
$\Rightarrow x=\frac{180}{x}+3$
$\Rightarrow x=\frac{180+3 x}{x}$
$\Rightarrow x^2=180+3 x$
$\Rightarrow x^2-3 x-180=0$
$\Rightarrow x^2-15 x+12 x-180=0$
$\Rightarrow x(x-15)+12(x-15)=0$
$\Rightarrow(x-15)(x+12)=0$
$\Rightarrow x=15,-12$
Cost cannot be in negative, therefore, we discard $x=-12$
Therefore, $x= Rs.15$ which is the cost of production of each article.
Number of articles produced on that particular day $=\frac{90}{15}=6$

View full question & answer→

Maths 5 Marks Questions

Test yourself on this topic