MATHS M.C.Q (1 Marks)

2. x² + y² - 2x + 2y = 47

Solution:

Given equation of lines is 2x - 3y = 5 .....

(i) and 3x - 4y = 7 .....

(ii)Solving above equations, we get the point of intersection as (1, -1). Eqns (i) and (ii) are diameters of the circle.

We know that the centre of

circle = point of intersection of diameters

 =(1, -1). Now, it is given that the area of the.

circle=154. 

$=\pi\text{r}^2=154 $

$\Rightarrow\text{r}=7$

Hence, the equation of required circle is

(x - 1)² + (y + 1)2 = 72

x² + y² - 2x + 2y = 47

3. x² + 6y² - 13y = 0

Solution:

We have to find equation of a circle with center on the y-axis.

General equation of such circle is (x - 0)² + (y - k)² = k²

It passes through (2, 3) i.e, 2² + (3 - k)² = k²

$\Rightarrow4 + 9 + \text{k}^2 - 6\text{k} = \text{k}^2$

$\Rightarrow\text{k}=\frac{13}{6}$

$∴ $ Equation of circle

 $=\text{x}^2+\Big(\text{y}-\frac{\text{13}}{6}\Big)^2$

$=\Big(\frac{13}{6}\Big)^2$

$=6\text{x}^2+6\text{y}^2-13\text{y}=0$

1. x² + y² - 2x - 2y - 3 = 0

Solution:

Given centroid of an equilateral triangle is G(1,1).

We know that in an equilateral triangle, centroid, circumcenter and incenter are at the same point.

So, the circumcenter is at G(1,1).

Given one vertex of equilateral triangle at A(-1,2)

So, circumradius

$=\text{AG}=\sqrt{5}$  

= So, equation of circumcircle is

= (x - 1)² + (y - 1)² = 5

⇒ x² + y² - 2x - 2y - 3 = 0

Total number of ways in which 6 boys and 6 girls can sit in a row = 12

Consider 6 girls as one group, then 6 boys and one group can arrange in 7 ways.

Now, 6 girls in the group can arrange among themselves in 6. 

So, the number of ways in which all the girls sit together is 7 × 6

 $\therefore$P(all girls sit together) 

$=\frac{\text{Number of ways in which all girls sit together }}{\text{Total Number of ways in which 6 boys and 6 girls sit in a row}}$

$=\frac{7\times6}{12}=\frac{6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8}=\frac{1}{132}$

Hence, the correct answer is option (c).

1. $\frac{5}{84}$​

Solution:

Three balls can be drawn randomly from nine balls in$ \ ^{9}\text{C}_3 = 84$ ways.

Three balls cannot be red as there are only two red balls.

Three balls of the same colour can be drawn in the following ways :

3 blue out of a total of 3 blue balls.

The probability for which is $\frac{\ ^{3}\text{C}_3}{84}=\frac{1}{84}$

3 black out of a total of 4 black balls.

The probability for which is $\frac{\ ^{4}\text{C}_3}{84}=\frac{4}{84}$

Hence, required probability $=\frac{1}{84}+\frac{4}{84}=\frac{5}{84}$

2. $\frac{1}{12}$

Solution:

When two dices are thrown, there are (6 × 6) = 36 outcomes.

The set of all these outcomes is the sample space, given by

S = (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6)

(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)

(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6)

(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)

(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)

(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)

i.e. n(S) = 36

Let E be the event of getting a total score of 10.

Then E = {(4, 6), (5, 5), (6, 4)}

$\therefore$ n(E) = 3

Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{3}{36}=\frac{1}{12}$

We know that,

$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})-\text{P}(\text{A}\cap\text{B})$ $\big[\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cup\text{B})\big]$

$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0\ ...(1)$

But, 

$\text{P(A)}-\text{P}(\text{A }\cap\text{ B})\geq0$

$\text{P(B)}-\text{P}(\text{A }\cap\text{ B})\geq0$

$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})\geq0\ ...(2)$

From (1) and (2), we have,

$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0$

$\Rightarrow\text{P}(\text{A)}=\text{P}(\text{A}\cap\text{B})\text{ and }\text{P}(\text{B)}=\text{P}(\text{A}\cap\text{B})$

$\Rightarrow\text{P(A)}=\text{P(B)}$

Hence, the correct answer is option (a).

2. (-1, 2)

Solution:

We have,

= y² - 4y - x + 3 = 0

⇒(y - 2)² - 4 - x + 3 = 0

⇒ (y - 2)² = (x + 1)

$∴$ Vertex of the parabola = (-1, 2)

3. $\frac{1}{6}$

Solution:

Total cases = 6 × 6

Let A be the event of getting a multiple of 2 on first dice and a multiple of 3 on the second dice.

Hence, A = {(2, 3), (2, 6), (4, 3), (4, 6), (6, 3), (6, 6)} n(A) = 66

$∴\text{P(A)}=\frac{6}{36}=\frac{1}{6}$

1. $\frac{1}{4}$

Solution:

2 coins tossed sample space = (h, h), (h, t), (t, h), (t, t)

have no head  = 1 (when both tail)

probability is  $\frac{1}{4}$

3. (x - 1)² + (y + 1)² = 49

Solution:

Circle area $=\pi\text{r}^2=154$ ⇒ r = 7 sq.units

intersection of diameter  (2x - 3y = 5) × 3

(3x - 4y = 7) × 2

- y = 1⇒ y = -1

12x = 5 - 3 = 2 ⇒ x = 1

eqn of circle  (x -1)² + (y + 1)² = 49

3. $\frac{1}{3}$

Solution:

Total number of sample space, S = {1, 2, 3, 4, 5, 6}

$\therefore\text{n}\text{(S)} = 6$

Let A be the event of getting the number 1 or 6.

A = {1, 6}

i.e. n(A) = 2

Hence, required probability $=\frac{\text{n(A)}}{\text{n(S)}}=\frac{2}{6}=\frac{1}{3}$

4. $\frac{1}{132}$

Solution:

Total number of ways in which six boys and six girls can be seated in a row = (12)

Taking all the six girls as one person, seven persons can be seated in a row in 7 ways.

The six girls can be arranged among themselves in 6 ways.

Then number of ways in which six boys and six girls can be seated in a row so that all

the girls sit together = 7 × 6

$\therefore\text{Required Probability}=\frac{7\times6}{12}$$\frac{720}{12\times11\times10\times9\times8}=\frac{1}{132}$

3. $\frac{1}{10}$

Solution:

Clearly, the sample space is given by

S = {1, 2, 3, 4, 5 ....97, 98, 99, 100}

$\therefore$ n(S) = 100

Let E = event of getting a square.

Then E = {1, 4, 9, 16, 25, 36, 49, 64, 81, 100}

$\therefore$ n(E) = 10

Hence, required probability $=\frac{\text{n}(\text{E})}{\text{n}(\text{S})}=\frac{10}{100}=\frac{1}{10}$

1. $\frac{2}{\sqrt{13}}$

Solution:

We know that, Latus rectum = 2× (distance from focus to directrix)

$=2.\frac{6\ -\ 12\ +\ 5}{\sqrt{4\ +\ 9}}$

$=\frac{2}{\sqrt{13}}$

2. $\frac{3}{11}$

4. 8

Solution:

Distance of vertex (2, -3) from directrix x = 4 is

$=\frac{2\ -\ 4}{\sqrt{1^2\ +\ 0^2}}=2$

So length of latue rectum of above parabola is =4× distance of vertex to directrix = 4 × 2 = 8.

1. $\sqrt5$

Solution:

The equation of circle x² + y² = 9 The point is (2,c) ⟹ 2² + c² = 94 + c² = 9

c² = 9 - 4

c² = 5

$\text{c}=\sqrt{5}$

1. ​​​​​0.39

Solution:

$\text{p(A)}=0.25\text{ and }\text{P(B)=0.50}$

$\text{P}(\text{A}\cap\text{B})=0.14$

$\therefore\text{Required probability}=1-\text{P}(\text{A}\cup\text{B})$

$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})\big]$

$=1-\big[0.25+0.50-0.14\big]$

$=1-0.61=0.39$

3. $\frac{26}{51}$

Solution:

We know that out of 52 playing cards 26 are of red and 26 are of black colour.

$\therefore$ P(both cards of differents colour)

$=\frac{26}{52}\times\frac{26}{51}+\frac{26}{52}\times\frac{26}{51}$

$=2\times\frac{26}{52}\times\frac{26}{51}=\frac{6}{51}$

1. $0.39$

Solution:

$\text{P(A)}=0.25, \text{P(B)=0.50}\text{ and }\text{P(A}\cap\text{B)}=0 .14$

$\therefore\text{Required Probability}=1-\text{P}(\text{A}\cup\text{B})$

$=1-\big[\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B}\big]$

$=1-\big[0.25+0.50-0.14\big]$

$=1-0.61=0.39$

2. $\frac{1}{26}$

Solution:

Total number of cases = 52

Number of favourable cases (getting a black ace) = 2

Thus, Probability (getting a black ace)

$=\frac{5}{52}=\frac{1}{26}$

1. ​​​​​​$\frac{64}{64}$

Solution:

The answer is one, because the article would be either good or defective as per the question.
Hence, the only option is $\frac{64}{64}=1$

2. $\frac{3}{49}$

Solution:

Total number of possibilities Total number of possibilities = 49 (Since, 3 cards of spade are removed)

Number of black face cards = 3 (3 cards of clubs)

Thus, required probability

$=\frac{3}{49}$

3. $\frac{10}{21}$

Solution:

There are nine persons (three men, two women and four children) out of which four persons can be selected in $\ ^{9}\text{C}_4 = 126\ \text{ways}.$

Total number of elementary events = 126

Exactly two children means selecting two children and two other people from three men and two women.

This can be done in $\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 \text{ways}.$

Favourable number of elementary events$=\ ^{4}\text{C}_2\times\ ^{ 5}\text{C}_2 = 60$

So, required probability $=\frac{60}{120} =\frac{10}{21}$

3. $\frac{15}{29}$​

Solution:

The total number of ways in which two integers can be chosen from the given 30 integers is $\ ^{30}\text{C}_2$.

The sum of the selected numbers is odd if exactly one of them is even or odd.

$\therefore$ Favourable number of outcomes $=\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1$

Hence, required probability $=\frac{\ ^{15}\text{C}_1\times\ ^{15}\text{C}_1}{\ ^{30}\text{C}_2}=\frac{15}{29}$

2. $\frac{2}{5}$

Solution:

Let E be the event of getting a prime number.

E = {2, 3, 5, 7, 11, 13, 17, 19} Hence, P(E)

$\frac{8}{20}=\frac{2}{5}.$

4.  $25\pi$

Solution:

The general equation of circle with centre (h, k) and radius rr is given by

(x - h)² + (y - k)² = (r)².It is given that the equation of the circle is

(x + 3)² + (y + 1)² = 25 Therefore,

= -3, k = -1 and r = 5.

The area of the circle with radius r is

$=\pi\text{r}^2$ thus, the area of the circle with radius r = 5 is:

$=\pi\text{r}^2=\pi(5)^2=25\pi$

Hence, the area of the circle is

$=25\pi.$ 

4. $\frac{2}{3}$​

Solution:

Out of 12 balls, one ball can be drawn in $\ ^{12}\text{C}_1$ ways.

Total number of elementary events $= \ ^{12}\text{C}_1 = 12$

Out of fivne black balls, one black ball can be chosen in $\ ^{5}\text{C}_1 = 5$ ways.

Out of three red balls, one red ball can be chosen in $\ ^{3}\text{C}_1 = 3$ ways

Favourable number of events $= 5 + 3 = 8$

Hence, required probability $=\frac{8}{12}=\frac{2}{3}$

1. $\frac{4}{13}$

Solution:

Since, Total cards = 52

No. of kings = 4

No. of cards that are spade = 13

There is one card of king which is of spad

$∴$ no. of cards which are spade or a king = 16

$∴$ Required probability $=\frac{16}{52}=\frac{4}{13}$

4. $\frac{5}{9}$

Solution:

We have digite 0, 2, 3, 5.

Number of divisible by 5 if unit place digit is '0' or '5'

If unit place is '0' then first three places can be filled in 3! ways.

If unit place is '5' then first place can be filled in two ways and second and thried place can be filled in 2! ways.

So, number of numbers ending with digit '5' is 2 × 2! = 4

$\therefore$ Total number of numbers divisible by 5 = 3! + 4 = 110 = n(E)

Also total number of numbers = 3 × 3! = 18

$\therefore\ \text{Required probability}=\frac{10}{18}=\frac{5}{9}$

2. $\frac{1}{36}$

Solution:

Total number of cases = 6³ = 216

The same number can appear on each of the dice in the following ways:

(1, 1, 1), (2, 2, 2), ………….(3, 3, 3)

So, favourable number of cases = 6

Hence, required probability = 6/216 = 1/3

1. $\frac{47}{66}$

Solution:

Out of  12 balls, two balls can be drawn in $^{12}\text{C}_2$ ways.

$\therefore$ Total number of elementary events, $\text{n}(\text{S})=^{12}\text{C}_2=66$

We observe that at least one ball of each colour can be drawn in one of the following mutually exclusive ways:

Red and 1 white

1 red and 1 blue

1 white and 1 blue

Thus, if we define three events A, B and C as follows:

A = drawing 1 red and 1 white

B = drawing 1 red and 1 blue

C = drawing 1 white and 1 blue

then, A, B and C are mutually exclusive events.

$\therefore$ Required probability $=\text{P}(\text{A}\cup\text{B}\cup\text{C})$

$=\text{P}(\text{A})+\text{P}(\text{B})+\text{P}(\text{C})$

$=\frac{^3\text{C}_1\times^4\text{C}_1}{^{12}\text{C}_2}+\frac{^3\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}+\frac{^4\text{C}_1\times^5\text{C}_1}{^{12}\text{C}_2}$

$=\frac{3\times4}{66}+\frac{3\times5}{66}+\frac{4\times5}{66}$

$=\frac{12}{66}+\frac{15}{66}+\frac{20}{56}=\frac{47}{66}$

4. $5\sqrt{2}\text{ units}$

Solution:

The line x - y = 4 and 2x + 3y + 7 = 0 intersect at the point (1, -3) So, the centre of circle lies at (1, -3)

Point (2, 4) lies on the circle. So, radius = distance of (2, 4) from (1, -3)

$=\sqrt{(2-1)^2+(4-(-3)^2}$

$=\sqrt{1^2+7^2}$

$=\sqrt{50}=5\sqrt{2}$

3. $\frac{17}{19}$​

Solution:

Number of ways in which we can choose three distinct integers from 20 integers $\ ^{20}\text{C}_3=1140$

We know that, if we take three odd numbers, there product will always be an odd number.

Out of 20 consecutive integers, 10 are even and 10 are odd integers.

Number of ways in which we can choose three distinct odd integers from 10 odd integers $=\ ^{10}\text{C} _3=120$

P(product is even) = 1 - P(product is odd),

$=1-\frac{120}{1140}=\frac{1140-120}{1140}=\frac{1020}{1140}=\frac{17}{19}$

2. $\frac{1}{3}$

Solution:

Total number of outcomes = 9

Favourable outcomes (the ball is neither white nor red) = 3

Probability $=\frac{3}{9}=\frac{1}{3}$

1. $(2,3)$

Solution:

(x - 2)² + (y - 3)² = 0

⇒ x² + 4 - 4x + y² + 9 - 6y = 0

x² + y² - 4x + 6y + 13 = 0radius = $\sqrt{\text{g}^2+\text{f}^2-\text{cg}}=-2$

$\text{f}=-3 \text{ c}=13\text{r}$ 

$=\sqrt{(-2)^2+(-3)^2-13}=\sqrt{13-13}=$ 0radius of

the circle is zero .Hence diameter =0i.e., it is a point circle with centre at (2 ,3).

1. $\frac{5}{13}$

Soultion:

Favourable outcome (Getting a male fish) = 5

Total number of outcomes (Male and female fish) =13

Probabilit $=\frac{5}{13}$

2. 60

Solution:

The word ′TOTAL′It can be arranged in $\frac {5!}{2!}$​ waysi.e, Total no. of words = 5

$\therefore$ It can be arranged in 5! ways. No. of repeating words can be arranged in 2! ways.

$\therefore$ Probability $=\frac {5!}{2!}=60.$