Mathematics Test

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Mathematics Test - 31

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Weekly Quiz Competition

Question 1
3 / -1

India plays two matches each with West Indies and Australia. In any match, the probabilities of India getting points 0, 1 and 2 are 0.45, 0.05 and 0.50 respectively. Assuming that the outcomes are independent, the probability of India getting atleast 7 points is

A
0.8750

B
0.0875

C
0.0625

D
0.0250

Solution

Probability of getting atleast seven points

= Probability of getting 7 points or 8 points

= Prob. of getting 7 points + Prob. of getting 8 points.

Seven points in four matches can be obtained in the following four different ways :

2, 2, 2, 1;

2, 2, 1, 2;

2, 1, 2, 2;

1, 2, 2, 2

The probability of each of these ways = (0.50)^{3 x} (0.05) (by multiplication theorem for independent events)

= 0.00625

Hence, probability of getting 7 points = ^{4 x} (0.00625) (by add. Theorem)

= 0.0250.

Eight points in four matches can be obtained only in one way i.e. 2, 2, 2, 2.

Hence, Prob. of getting 8 points = (0.50)⁴ = 0.0625

Thus, the required prob. = 0.250 + 0.0625 = 0.0875.
Question 2
3 / -1

Find the area of the region(in square units) bounded by the two parabolas y = x² and y² = x.

A
$\frac{1}{3}$

B
$\frac{1}{2}$

C
$\frac{2}{5}$

D
$\frac{1}{7}$

Solution

The two parabolas given are $y=x^{2}$ and $y^{2}=x$.
The point of intersection of these two parabolas is $\mathrm{O}(0,0)$ and $\mathrm{A}(1,1)$ as shown in the figure given below-

Now,

$y^{2}=x$
$y=\sqrt{x}=f(x)$
$y=x^{2}=g(x),$ where, $f(x) \geq g(x)$ in [0,1]

Area of the shaded region

$$
\begin{array}{l}
=\int_{0}^{1}[f(x)-g(x)] d x \\
=\int_{0}^{1}\left[\sqrt{x}-x^{2}\right] d x \\
=\left[\frac{2}{3} x^{\frac{3}{2}}-\frac{x^{3}}{3}\right]_{0}^{1} \\
=(2 / 3)-(1 / 3) \\
=1 / 3
\end{array}
$$
Question 3
3 / -1

Let f : R be a differentiable function and f(1) = 4. Find the value of

A
16

B
8

C
4

D
2

Solution
Question 4
3 / -1

A
0

B
2

C
1

D
-1

Solution
Question 5
3 / -1

A
π/4

B
π/3

C
π/2

D
none of these

Solution
Question 6
3 / -1

The cartesian equation of a line $\mathrm{AB}$ is $\frac{2 x-1}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$
Find the direction cosines of a line parallel to $\mathrm{AB}$.

A

$\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}$

B
$\frac{\sqrt{4}}{\sqrt{55}}, \frac{4}{\sqrt{56}}, \frac{6}{\sqrt{55}}$

C
$\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{10}{\sqrt{57}}$

D
$\frac{\sqrt{11}}{\sqrt{55}}, \frac{4}{\sqrt{59}}, \frac{6}{\sqrt{55}}$

Solution

The line given is $\frac{2\left(x-\frac{1}{2}\right)}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3}$
$\Rightarrow \quad \frac{x-\frac{1}{2}}{\sqrt{3}}=\frac{y+2}{4}=\frac{z-3}{6}$
Direction ratios of line $A B$ are $\sqrt{3}, 4$ and 6 .
Direction ratios of line parallel to $A B$ are $\sqrt{3}, 4$ and 6.
Dividing by $\sqrt{3+16+36}=\sqrt{55},$ we get direction cosines as $\frac{\sqrt{3}}{\sqrt{55}}, \frac{4}{\sqrt{55}}, \frac{6}{\sqrt{55}}$.
Question 7
3 / -1

Out of 40 consecutive integers, two are chosen at random. The probability that their sum is odd is

A
1/200

B
20/39

C
1/2

D
None of these

Solution
Question 8
3 / -1

A

B

C

D

Solution
Question 9
3 / -1

If three natural numbers between $1$ and $100$ are selected randomly, then the probability that all are divisible by both $2$ and $3$ is

A
$\frac{4}{105}$

B
$\frac{4}{33}$

C
$\frac{4}{35}$

D
$\frac{4}{1155}$

Solution

If a number is divisible by both $2$ and $3$ that means it is divisible by $6$.

So, sample space $S$ is a number divisible by $6$, is

$S=\{6,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96\}$

$\Rightarrow n(S)=16$

So, required probability $\frac{{ }^{16} C_{3}}{{ }^{100} C_{3}}$

$=\frac{16 \times 15 \times 14}{100 \times 99 \times 98}$

$=\frac{4}{1155}$
Question 10
3 / -1

A man of height 2m walks at a uniform speed of 5km/hr away from a lamp post which is 6m high. Find the rate at which the length of his shadow increases.

A
2.5 km/h

B
1.5 km/h

C
3.5 km/h

D
6 km/h

Solution

Let $\mathrm{AB}$ be the lamp post and let $\mathrm{M} \mathrm{N}$ be the man of height $2 \mathrm{~m}$ and let $\mathrm{AM}=$ 1 meter.

and MS is the shadow of the man.

Let length of shadow $\mathrm{MS}=\mathrm{s}$

Given: Man walks at speed of $5 \mathrm{kmph}$

$\therefore \frac{\mathrm{dl}}{\mathrm{dt}}=5 \mathrm{kmph}$

We need to find rate at which the length of his shadow increases $=\frac{\mathrm{ds}}{\mathrm{dt}}$.

In $\triangle \mathrm{ASB}, \tan \theta=\frac{\mathrm{AB}}{\mathrm{A} \mathrm{S}}=\frac{6}{1+\mathrm{s}}$

In $\triangle \mathrm{MSN}, \tan \theta=\frac{\mathrm{M} \mathrm{N}}{\mathrm{MS}}=\frac{2}{\mathrm{~s}}$

From $(1)$ and $(2)$

$\frac{6}{1+s}=\frac{2}{s}$

$\Rightarrow 6 \mathrm{~s}=2 \mathrm{l}+2 \mathrm{~s}$

$\Rightarrow 4 s=21$

or $l=2$ s

$\frac{d l}{d t}=2 \frac{d s}{d t}$

$5=2 \frac{\mathrm{ds}}{\mathrm{dt}}$ since given that $\frac{\mathrm{dl}}{\mathrm{dt}}=5 \mathrm{kmph}$

$\therefore \frac{\mathrm{ds}}{\mathrm{dt}}=\frac{5}{2} \mathrm{kmph}=2.5 \mathrm{kmph}$