MCQ

MCQ 11 Mark

Let $A=\{1,2,3\}$, then the domain of the relation $R=\{(1,1),(2,3),(2,1)\}$ defined on $A$ is

A
$\{1,3\}$
✓
$\{1,2\}$
C
None of these
D
$\{2,3\}$

Answer

Correct option: B.

$\{1,2\}$

(b) $\{1,2\}$
Explanation:
Since the domain is represented by the x - coordinate of the ordered pair ( $\mathrm{x}, \mathrm{y}$ ). Therefore, the domain of the given relation is $\{1,2\}$.

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MCQ 21 Mark

20 students can compete for a race. The number of ways in which they can win the first three places is (given that no two students finish in the same place).

A
1140
B
8000
✓
6840
D
6000

Answer

Correct option: C.

6840

(c) 6840
Explanation:
For first place we have 20 students, for second we have 19 and for the third we have 18
${ }^{20} \mathrm{P}_{3}=20 \times 19 \times 18$

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MCQ 31 Mark

The present value of an annuity of ₹ 3993 payable at the end of each year for 3 years, if money is worth $10 \%$ effective.

A
₹ 9039
B
₹ 9903
✓
₹ 9930
D
₹ 9390

Answer

Correct option: C.

₹ 9930

(c) ₹ 9930
Explanation:
R = ₹ 3993, n = 3, i = 0.1
$\therefore$ Present value = ₹ 3993 $\left[\frac{1-(1.1)^{-3}}{0.1}\right]$
= ₹ 39930[1-0.7513] = ₹ 9930.

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MCQ 41 Mark

If $A$ and $B$ are independent events, then $P(A \cap B)$ is equal to:

A
$\frac{P(A)}{P(B)}$
✓
$\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
C
$P(A)-P(B)$
D
$P(A)+P(B)$

Answer

Correct option: B.

$\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$

(b) $\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})$
Explanation:
Given A and B are independent events, then
$P(A \cap B)=P(A) \cdot P(B)$

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MCQ 51 Mark

There are three coins. One is a two headed coin (having head on both faces), another is a biased coin that comes up heads $75 \%$ of the time and third is an unbiased coin. One of the three coins is chosen at random and tossed, it shows heads, what is the probability that it was the two headed coin?

✓
$\frac{4}{9}$
B
$\frac{5}{9}$
C
$\frac{1}{9}$
D
$\frac{2}{9}$

Answer

Correct option: A.

$\frac{4}{9}$

(a) $\frac{4}{9}$
Explanation: Let
$E_{1}, E_{2}$ and $E_{3}$ and are events of selection of a two headed coin, biased coin and unbiased coin respectively.
$\therefore P\left(E_{1}\right)=P\left(E_{2}\right)=P\left(E_{2}\right)=\frac{1}{3}$.
Let $\mathrm{A}=$ event of getting head.
$P\left(A / E_{1}\right)=1, P\left(A / E_{2}\right)=\frac{3}{4}, P\left(A / E_{3}\right)=\frac{1}{2}$.
$P\left(E_{1} / A\right)=\frac{P\left(A / E_{1}\right) \cdot P\left(E_{1}\right)}{P\left(A / E_{1}\right) \cdot P\left(E_{1}\right)+P\left(A / E_{2}\right) \cdot P\left(E_{2}\right)+P\left(A / E_{3}\right) \cdot P\left(E_{3}\right)}$
$=\frac{\frac{1}{3} \cdot 1}{\frac{1}{3} \cdot 1+\frac{1}{3} \cdot \frac{3}{4}+\frac{1}{3} \frac{1}{2}}=\frac{4}{9}$

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MCQ 61 Mark

The number of ways in which the letters of the word CONSTANT can be arranged without changing the relative positions of the vowels and consonants is

A
444
B
372
✓
360
D
256

Answer

Correct option: C.

360

(c) 360
Explanation:
The word CONSTANT consists of two vowels that are placed at the $2^{\text {nd }}$ and $6^{\text {th }}$ position, and six consonants.
The two vowels can be arranged at their respective places, i.e. $2^{\text {nd }}$ and $6^{\text {th }}$ place, in 2 ! ways.
The remaining 6 consonants can be arranged at their respective places in $\frac{6!}{2!2!}$ ways.
$\therefore$ Total number of arrangements $=360$

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MCQ 71 Mark

Find the simple interest on ₹ 5200 for 2 years at $6 \%$ per annum:

A
₹ 600
✓
₹ 624
C
₹ 524
D
₹ 450

Answer

Correct option: B.

₹ 624

(b) ₹ 624
Explanation:
Given, $\mathrm{P}=5200, \mathrm{n}=2, \mathrm{i}=6 \%=\frac{6}{100}=0.06$
I = Pit
$\therefore \mathrm{I}=5200 \times 0.06 \times 2$
or, $\mathrm{I}=624$.

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MCQ 81 Mark

The product $\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}$ equals

A
$\sqrt[12]{2}$
B
$\sqrt[12]{32}$
✓
2
D
$\sqrt{2}$

Answer

Correct option: C.

(c) 2
Explanation:
$\sqrt[3]{2} \cdot \sqrt[4]{2} \cdot \sqrt[12]{32}=2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot\left(2^{5}\right)^{\frac{1}{12}}=2^{\frac{1}{3}} \cdot 2^{\frac{1}{4}} \cdot 2^{\frac{5}{12}}=2^{\frac{1}{3}+\frac{1}{4}+\frac{5}{12}}$
$=2^{\frac{12}{12}}=2^{1}$

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MCQ 91 Mark

The mean deviation of the data $3,10,10,4,7,10,5$ from the mean is

A
3.75
✓
2.57
C
3
D
2

Answer

Correct option: B.

2.57

(b) 2.57
Explanation:
Given data is $3,10,10,4,7,10,5$. They are total 7 .
Here mean, $\bar{x}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
This can be written in table form as,

Data (x_i)	$d _{ i }=\left\| x _{ i }-\bar{x}\right\|$
3	=\|3-7\|=4
10	=\|10-7\|=3
10	=\|10-7\|=3
4	=\|4-7\|=3
7	=\|7-7\|=0
10	=\|10-7\|=3
5	=\|5-7\|=2
Total	$\Sigma d _{ i }=18$

Hence Mean Deviation becomes,
$
\text { M.D }=\frac{\sum d_i}{7}=\frac{18}{7}=2.57
$
Therefore, the mean deviation about the mean of the distribution is 2.57

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MCQ 101 Mark

How many odd days are there in a century?

A
4
B
2
C
3
✓
5

Answer

Correct option: D.

(d) 5
Explanation:
In a century, we have 24 leap years and 76 non-leap years, so number of odd days
$=2 \times 24+1 \times 76=48+76=124$
$=7 \times 17+5=5$ odd days

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MCQ 111 Mark

Equation of the circle with centre lies on the $y$-axis and passing through the origin and the point $(2,3)$ is:

✓
$3 x^{2}+3 y^{2}-13 y=0$
B
$x^{2}+y^{2}+13 y=0$
C
$6 x^{2}+6 y^{2}-13 x=0$
D
$x^{2}+y^{2}+13 x+3=0$

Answer

Correct option: A.

$3 x^{2}+3 y^{2}-13 y=0$

(a) $3 x^{2}+3 y^{2}-13 y=0$
Explanation:
Given that the circle with centre lies on the $y$-axis and passing through the origin.
General equation of the circle is
$(x-0)^{2}+(y-k)^{2}=k^{2}$
It passes through $(2,3)$
i.e. $2^{2}+(3-k)^{2}=k^{2}$
$\Rightarrow 4+9+\mathrm{k}^{2}-6 \mathrm{k}=\mathrm{k}^{2}$
$\Rightarrow k=\frac{13}{6}$
Therefore, the required equation of the circle is
$x^{2}+\left(y-\frac{13}{6}\right)^{2}=\left(\frac{13}{6}\right)^{2}$
$\Rightarrow 3 x^{2}+3 y^{2}-13 y=0$
$3 x^{2}+3 y^{2}-13 y=0$

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MCQ 121 Mark

6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

A
$\frac{12}{431}$
B
$\frac{1}{432}$
C
$\frac{1}{112}$
✓
$\frac{1}{132}$

Answer

Correct option: D.

$\frac{1}{132}$

(d) $\frac{1}{132}$
Explanation:
If all the girls sit together, then consider it as 1 group.

Total number of persons $=6+1=7$ persons
$\therefore$ Total number of arrangements in a row of 7 persons $=7$ !
and the girls interchanges their seats in 6 ! ways.
Required Probability $=\frac{\text { Number of favourable outcome }}{\text { Total number of outcomes }}$
$\therefore$ Required probability $=\frac{6!7!}{12!}$
$=\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 7!}{12 \times 11 \times 10 \times 9 \times 8 \times 7!}$
$=\frac{6 \times 5 \times 4 \times 3 \times 2}{12 \times 11 \times 10 \times 9 \times 8}$
$=\frac{1}{2 \times 11 \times 2 \times 3 \times 2}$
$=\frac{1}{132}$
Hence, the correct option

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MCQ 131 Mark

Standard form of 192.8 is

A
$19.28 \times 10^{1}$
✓
$1.928 \times 10^{2}$
C
1.928
D
$0.1928 \times 10^{-1}$

Answer

Correct option: B.

$1.928 \times 10^{2}$

(b) $1.928 \times 10^{2}$
Explanation:
Standard farm of
192.8
$=1.928 \times 10^{2}$

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MCQ 141 Mark

A relation $\phi$ from $C$ to $R$ is defined by $x \phi \Leftrightarrow|x|=y$. Which one is correct?

A
$(1+i) \phi 2$
B
$3 \phi(-3)$
✓
i $\phi 1$
D
$(2+3 i) \phi 13$

Answer

Correct option: C.

i $\phi 1$

(c) i $\phi 1$
Explanation:
We have $x \phi$ g given by, $|x|=y$
i $\phi 1$
$\mathrm{x}=\mathrm{i}$;
$|x|=\sqrt{1^{2}}$
$=1$
$1=1$
$|x|=y$.

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MCQ 151 Mark

If $\frac{\log 243}{\log 27}=\mathrm{x}$, then x is

✓
$\frac{5}{3}$
B
$\frac{3}{5}$
C
3
D
5

Answer

Correct option: A.

$\frac{5}{3}$

(a) $\frac{5}{3}$
Explanation:
$\frac{\log 243}{\log 27}=\mathrm{x}$
$\frac{\log 3^{5}}{\log 3^{3}}=\mathrm{x}$
$x \log 3^{3}=\log 3^{5}$
$\log \left(3^{3}\right)^{x}=\log 3^{5}$
$\log 3^{3 x}=\log 3^{5}$
$\therefore 3 x=5$
$\mathrm{x}=\frac{5}{3}$

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MCQ 161 Mark

The amount at the compound interest which is calculated yearly on a certain sum of money is ₹ 1250 is one year and ₹ 1375 in two years. The rate of interest per annum is:

A
9 %
✓
10 %
C
8 %
D
11 %

Answer

Correct option: B.

10 %

(b) 10 %
Explanation:
₹ 1250 is the interest of first year and ₹ 1375 is the interest in second year. Here, the difference is of ₹ 125 which is the interest obtained ₹ 1250 .
Let rate be r%
$\therefore \frac{1250 \times r \times 1}{100}=125$
$\Rightarrow r=\frac{125 \times 100}{1250}=10$.

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MCQ 171 Mark

How many types of measures of dispersion are there?

A
3
✓
5
C
2
D
4

Answer

Correct option: B.

(b) 5
Explanation:
There are five most commonly used measures of dispersion. These are range, variance, standard deviation, mean deviation and quartile deviation.

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MCQ 181 Mark

If $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{2}$ and $\mathrm{P}\left(A^{\prime} \cap B^{\prime}\right)=\frac{1}{3}, \mathrm{P}(\mathrm{A})=\mathrm{p}$ and $\mathrm{P}(\mathrm{B})=2 \mathrm{p}$, then the value of p is:

A
$\frac{4}{9}$
B
$\frac{1}{3}$
✓
$\frac{7}{18}$
D
$\frac{1}{9}$

Answer

Correct option: C.

$\frac{7}{18}$

(c) $\frac{7}{18}$
Explanation:
Given $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\frac{1}{2}, \mathrm{P}^{\prime}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\frac{1}{3}, \mathrm{P}(\mathrm{A})=\mathrm{p}, \mathrm{P}\left(\mathrm{B}^{\prime}\right)=2 \mathrm{p}$
Now, $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=(\mathrm{A} \cup \mathrm{B})^{\prime}=1-(\mathrm{A} \cup \mathrm{B})=\frac{1}{3}$
$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=1-\frac{1}{3}=\frac{2}{3}$
$\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{2}{3}$
$\Rightarrow \mathrm{p}+2 \mathrm{p}-\frac{1}{2}=\frac{2}{3}$
$\Rightarrow 3 \mathrm{p}=\frac{2}{3}+\frac{1}{2}=\frac{4+3}{6}=\frac{7}{6} \Rightarrow \mathrm{p}=\frac{7}{18}$

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Data (x_i)	$d _{ i }=\left\| x _{ i }-\bar{x}\right\|$
3	=\|3-7\|=4
10	=\|10-7\|=3
10	=\|10-7\|=3
4	=\|4-7\|=3
7	=\|7-7\|=0
10	=\|10-7\|=3
5	=\|5-7\|=2
Total	$\Sigma d _{ i }=18$

Applied Maths MCQ

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