4
Q:

# In a Plane there are 37 straight lines, of which 13 pass through the point A and 11 pass through the point B. Besides, no three lines pass through one point, no lines passes through both points A and B , and no two are parallel. Find the number of points of intersection of the straight lines.

 A) 525 B) 535 C) 545 D) 555

Explanation:

The number of points of intersection of 37 lines is $\inline ^{37}\textrm{C}_{2}$. But 13 straight lines out of the given 37 straight lines pass through the same point A. Therefore instead of getting $\inline ^{13}\textrm{C}_{2}$ points, we get only one point A. Similarly 11 straight lines out of the given 37 straight lines intersect at point B. Therefore instead of getting $\inline ^{11}\textrm{C}_{2}$ points, we get only one point B.

Hence the number of intersection points of the lines is $\inline ^{37}\textrm{C}_{2}-^{13}\textrm{C}_{2}-^{11}\textrm{C}_{2}+2$ = 535

Q:

Out of seven consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed ?

 A) 25200 B) 25000 C) 25225 D) 24752

Explanation:

Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)

=( $\inline \fn_jvn \small 7_{}^{C}\textrm{3}$ $\fn_jvn \small \times 4_{}^{C}\textrm{2}$ ) = ($\inline \fn_jvn \small \frac{7\times 6\times 5}{3\times 2\times 1}$ $\inline \fn_jvn \small \times \frac{4\times 3}{2\times 1}$) = 210.

Number of groups, each having 3 consonants and 2 vowels =210

Each group contains 5 letters.

Number of ways of arranging 5 letters among themselves = 5! = (5 $\fn_jvn&space;\small&space;\times$ 4 $\fn_jvn&space;\small&space;\times$ 3 $\fn_jvn&space;\small&space;\times$ 2 $\fn_jvn&space;\small&space;\times$ 1)=120.

Required number of words = (210 $\fn_jvn&space;\small&space;\times$ 120) = 25200.

8 54
Q:

How many 3-digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9 which are divisible by 5 and none of the digits is repeated ?

 A) 15 B) 20 C) 5 D) 10

Explanation:

Since each number to be divisible by 5, we must have 5 0r 0 at the units place. But in given digits we have only 5.
So, there is one way of doing it.
Tens place can be filled by any of the remaining 5 numbers.So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 $\fn_jvn&space;\small&space;\times$ 5 $\fn_jvn&space;\small&space;\times$ 4) = 20.

8 75
Q:

In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there ?

 A) 205 B) 194 C) 209 D) 159

Explanation:

We may have (1 boy and 3 girls)or(2boys and 2 girls)or(3 boys and 1 girl)or(4 boys).
Required number of ways = ($\inline \fn_jvn \small 6_{}^{C}\textrm{1}$$\inline \fn_jvn \small \times 4_{}^{C}\textrm{3}$) + ($\inline \fn_jvn \small 6_{}^{C}\textrm{2}$$\inline \fn_jvn \small \times 4_{}^{C}\textrm{2}$) + ($\inline \fn_jvn \small 6_{}^{C}\textrm{3}$$\inline \fn_jvn \small \times 4_{}^{C}\textrm{1}$) + ($\inline \fn_jvn \small 6_{}^{C}\textrm{4}$)

= ($\inline \fn_jvn \small 6\times 4$) + $\inline \fn_jvn \small \left ( \frac{6\times 5}{2\times 1}\times \frac{4\times 3}{2\times 1} \right )$ + $\inline \fn_jvn \small \left ( \frac{6\times 5\times 4}{3\times 2\times 1} \times 4\right )$ + $\inline \fn_jvn \small \left ( \frac{6\times 5}{2\times 1} \right )$

= (24+90+80+15)

= 209.

7 61
Q:

In a bag, there are 8 red, 7 blue and 6 green flowers. One of the flower is picked up randomly. What is the probability that it is neither red nor green ?

$\inline \fn_jvn \small A) \frac{1}{3}$  $\inline \fn_jvn \small B) \frac{8}{21}$ $\inline \fn_jvn \small C) \frac{6}{21}$ $\inline \fn_jvn \small D) \frac{20}{21}$

 A) Option A B) Option B C) Option C D) Option D

Explanation:

Total number of flowers = (8+7+6) = 21.
Let E = event that the flower drawn is neither red nor green.
= event taht the flower drawn is blue.
$\fn_jvn&space;\small&space;\therefore$ n(E)= 7
$\fn_jvn&space;\small&space;\therefore$ P(E)=  $\inline \fn_jvn \small \frac{7}{21} = \frac{1}{3}$

7 53
Q:

A box contains 5 green, 4 yellow and 5 white pearls. Four pearls are drawn at random. What is the probability that they are not of the same colour ?

A) $\inline \fn_jvn \small \frac{11}{91}$ B) $\inline \fn_jvn \small \frac{4}{11}$ C) $\inline \fn_jvn \small \frac{1}{11}$ D) $\inline \fn_jvn \small \frac{90}{91}$

 A) Option A B) Option B C) Option C D) Option D

Explanation:

Let S be the sample space. Then,
n(s) = number of ways of drawing 4 pearls out of 14
= $\inline \fn_jvn \small 14_{}^{C}\textrm{4}$ ways = $\inline \fn_jvn \small \frac{(14\times 13\times 12\times 11)}{(4\times 3\times 2\times 1)}$ = 1001
Let E be the event of drawing 4 pearls of the same colour.
Then, E = event of drawing (4 pearls out of 5) or (4 pearls out of 4) or (4 pearls out of 5)

$\inline \fn_jvn \small \Rightarrow 5_{}^{C}\textrm{1}$ + $\inline \fn_jvn \small 4_{}^{C}\textrm{4}$ + $\inline \fn_jvn \small 5_{}^{C}\textrm{1}$ = 5+1+5 =11

$\fn_jvn \small \Rightarrow$ P(E) = $\inline \fn_jvn \small \frac{n(E)}{n(S)} = \frac{11}{1001} = \frac{1}{91}$

$\inline \fn_jvn \small \therefore$ Required probability = $\inline \fn_jvn \small 1-$ $\inline \fn_jvn \small \frac{1}{91}=\frac{90}{91}$