MCQ - Applied Maths STD 11 Science Questions - Vidyadip

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18 questions · timed · auto-graded

MCQ 11 Mark

Let $n(A)=m$, and $n(B)=n$. Then the total number of possible relations that can be defined from $A$ to $B$ is

✓
$2^{\mathrm{mn}}$
B
$m^{n}-1$
C
$n^{m}-1$
D
$\mathrm{m}^{\mathrm{n}}$

Answer

Correct option: A.

$2^{\mathrm{mn}}$

(A) $2^{\mathrm{mn}}$
Explanation: Given, $n(A)=m$, and $n(B)=n$
$\therefore \mathrm{n}(\mathrm{A} \times \mathrm{B})=\mathrm{n}(\mathrm{A}) . \mathrm{n}(\mathrm{B})=\mathrm{mn}$
So, the total number of non-empty relations from $A$ to $B: 2^{m n}.$

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MCQ 21 Mark

Out of 5 men and 2 women, a committee of 3 persons is to be formed so as to include at least one woman. The number of ways in which it can be done is

✓
25
B
35
C
45
D
10

Answer

Correct option: A.

25

(A) 25
Explanation: We may have:
i. 1 Woman and 2 men
ii. 2 Women and 1 man
$\therefore$ required number of ways
$=\left({ }^{2} \mathrm{C}_{1} \times{ }^{5} \mathrm{C}_{2}\right)+\left({ }^{2} \mathrm{C}_{2} \times{ }^{5} \mathrm{C}_{1}\right)$
$=\left(2 \times \frac{5 \times 4}{2 \times 1}\right)+(1+5)$
$=(20+5)=25$

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MCQ 31 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 41 Mark

A bag contains 5 red and 3 blue balls. If 3 balls are drawn at random without replacement the probability of getting exactly one red ball is

A
$\frac{45}{196}$
✓
$\frac{15}{56}$
C
$\frac{15}{29}$
D
$\frac{135}{392}$

Answer

Correct option: B.

$\frac{15}{56}$

(B) $\frac{15}{56}$
Explanation: Probability of getting exactly one red (R) ball $=P_{R} \cdot P_{B} \cdot P_{B}+P_{B} \cdot P_{R} \cdot P_{R}+P_{B} \cdot P_{B} \cdot P_{R}$
$=\frac{5}{8} \cdot \frac{3}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{6}+\frac{3}{8} \cdot \frac{2}{7} \cdot \frac{5}{6}$
$=\frac{15}{4 \cdot 7 \cdot 6}+\frac{15}{4 \cdot 7 \cdot 6}+\frac{15}{4 \cdot 7 \cdot6}$
$=\frac{5}{56}+\frac{5}{56}+\frac{5}{56}=\frac{15}{56}$.
Which is the required solution

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MCQ 51 Mark

A man is known to speak truth in 3 out of 4 times. He throws a dice and reports that it is a six. Then the probability that it is actually a six is

A
$\frac{4}{9}$
B
$\frac{2}{7}$
C
$\frac{1}{6}$
✓
$\frac{3}{8}$

Answer

Correct option: D.

$\frac{3}{8}$

(D) $\frac{3}{8}$
Explanation: As Probability $=\frac{\frac{3}{4} \cdot \frac{1}{6}}{\frac{3}{4} \cdot \frac{1}{6}+\frac{1}{4} \cdot \frac{5}{6}}=\frac{3}{8}$

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MCQ 61 Mark

In how many ways can a cricket team be chosen out of a batch of 15 players, if a particular player is always chosen?

A
965
B
1364
✓
1001
D
364

Answer

Correct option: C.

1001

(C) 1001
Explanation: When a particular player is always chosen, then we have to select 10 players out of 14.
$\therefore$ Required number of ways $={ }^{14} \mathrm{C}_{10}={ }^{14} \mathrm{C}_{4}=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}=1001$.

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MCQ 71 Mark

The simple interest on a certain sum of money for 2 years at $10 \%$ per annum is half the compound interest on ₹ 5000 for 2 years at $10 \%$ per annum. The sum is:

A
₹ 2500
B
₹ 2925
C
₹ 2850
✓
₹ 2625

Answer

Correct option: D.

(D) ₹ 2625
Explanation: Let the sum be P, then
$\frac{P \times 10 \times 2}{100}=\frac{1}{2}\left[5000\left(1+\frac{10}{100}\right)^{2}-5000\right]$
$\Rightarrow \frac{P}{5}=\frac{5000}{2}\left[\frac{11}{10} \times \frac{11}{10}-1\right]$
$\Rightarrow \mathrm{P}=\frac{12500 \times 21}{100}=$ ₹ $2625$.

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MCQ 81 Mark

If $\log (3 x+1)=2$, then the value of $x$ is

A
$\frac{19}{3}$
B
$99$
C
$\frac{1}{3}$
✓
$33$

Answer

Correct option: D.

$33$

(D) $33$
Explanation: $\log (3 x+1)=2 \Rightarrow 10^{2}=3 x+1$
$\Rightarrow 3 x=99 \Rightarrow x=33$

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MCQ 91 Mark

Following are the marks obtained by 9 students in a mathematics test: 50, 69, 20, 33, 53, 39, 40, 65, 59 The mean deviation from the median is:

✓
12.67
B
14.76
C
9
D
10.5

Answer

Correct option: A.

12.67

(A) 12.67
Explanation: Arranging given data in increasing order.
So, we have 20, 33, 39, 40, 50, 53, 59, 65, 69
Median $=\left(\frac{9+1}{2}\right)$ th term $=$ 5th term
$\therefore$ Median $=50$
Mean deviation from median $=\frac{|20-50|+|33-50|+|39-50|+|40-50|+|50-50|+|53-50|+|59-50|+|65-50|+|69-50|}{9}$
$=\frac{30+17+11+10+0+3+9+15+19}{9}=\frac{114}{9}=12.67$

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MCQ 101 Mark

A and B together can do a piece of work in 10 days. C can do the same work alone in 15 days. If A, B and C work together, then number of days to finish the work is:

A
4 days
B
8 days
✓
6 days
D
5 days

Answer

Correct option: C.

6 days

(C) 6 days
Explanation: 6 days

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MCQ 111 Mark

If $(x, 3)$ and $(3,5)$ are the extremities of a diameter of a circle with centre at $(2, y)$, then the values of $x$ and $y$ are

✓
None of these
B
$(3,1)$
C
$x=4, y=1$
D
$x=8, y=2$

Answer

Correct option: A.

None of these

(A) None of these
Explanation: The endpoints of the diameter of a circle are $(x, 3)$ and $(3,5)$.
According to the question, we have:
centre is midpoint of the endpoints of diameters.
$\frac{x+3}{2}=2, \mathrm{y}=\frac{5+3}{2}$
$\Rightarrow \mathrm{x}=1, \mathrm{y}=4$

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MCQ 121 Mark

If $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.8$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3$, then $\mathrm{P}\left(\mathrm{A}^{\prime}\right)+\mathrm{P}\left(\mathrm{B}^{\prime}\right)$ is equal to:

✓
0.9
B
1.1
C
0.7
D
0.5

Answer

Correct option: A.

0.9

(A) 0.9
Explanation: $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow P(A)+P(B)=P(A \cup B)+P(A \cap B)$
$\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.8+0.3$
$\Rightarrow \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=1.1$
$\Rightarrow 1-\mathrm{P}(\bar{A})+1-\mathrm{P}(\mathrm{B})=1.1$
$\Rightarrow \mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})=2-1.1$
$\Rightarrow \mathrm{P}(\bar{A})+\mathrm{P}(\bar{B})=0.9$

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MCQ 131 Mark

The value of $3^{\frac{4}{\log _{4} 9}}+27^{\frac{1}{\log _{36} 9}}+81^{\frac{1}{\log _{5} 3}}$, is:

A
847
B
860
C
890
✓
857

Answer

Correct option: D.

857

(D) 857
Explanation: 857

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MCQ 141 Mark

Let R be the relation in the set N given by $\mathrm{R}=\{(\mathrm{a}, \mathrm{b}): \mathrm{a}=\mathrm{b}-2, \mathrm{~b}>6\}$.

✓
$(6,8) \in R$
B
$(8,7) \in R$
C
$(2,4) \in R$
D
$(3,8) \in R$

Answer

Correct option: A.

$(6,8) \in R$

(A) $(6,8) \in R$
Explanation: $(6,8) \in \mathrm{R}$
as $b-2=8-2=6$ and $b>6.$

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MCQ 151 Mark

$(64)^{\frac{2}{3}}$ is equal to

A
12
B
4
✓
16
D
8

Answer

Correct option: C.

16

(C) 16
Explanation: as $(64)^{\frac{2}{3}}=4^{\frac{2}{3} \times 3}=4^{2}=16$

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MCQ 161 Mark

The compound interest on ₹ 50000 at $5 \%$ per annum is ₹ 5125 . The time period is:

✓
2 years
B
$2 \frac{1}{2}$ years
C
3 years
D
$1 \frac{1}{2}$ years

Answer

Correct option: A.

2 years

(A) 2 years
Explanation: Let the time period be n.
$\therefore 50000\left(1+\frac{5}{100}\right)^{n}-50000=5125$
$\Rightarrow 50000 \times\left(\frac{21}{20}\right)^{n}=55125$
$\Rightarrow\left(\frac{21}{20}\right)^{n}=\frac{55125}{50000}=\frac{11025}{10000}=\frac{441}{400}=\left(\frac{21}{20}\right)^{2}$
$\Rightarrow \mathrm{n}=2.$

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MCQ 171 Mark

The mean of 100 observations is 50 and their standard deviation is 5 . The sum of all squares of all the observations is

✓
252500
B
50000
C
250000
D
255000

Answer

Correct option: A.

252500

(A) 252500
Explanation: Given, $\overline{\mathbf{x}}=50, \mathrm{n}=100$ and $\sigma=5$
$\sigma=\frac{\sum x_{i}}{N}$
$\sum x_{i}=50 \times 100$
$\sum x_{i}=5000$
Now, $\sigma^{2}=\frac{\sum x_{i}^{2}}{N}-(\bar{x})^{2}$
$25=\frac{\sum x_{i}^{2}}{100}-(50)^{2}$
$\sum x_{i}^{2}=252500$

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MCQ 181 Mark

Which of the formula is related to Bayes' theorem?

A
$P\left(\frac{A}{B}\right)=\frac{P\left(\frac{B}{A}\right)}{P(B)}$
B
$P\left(\frac{A}{B}\right)=\frac{1}{P(B)}$
✓
$P\left(\frac{A}{B}\right)=\frac{P\left(\frac{B}{A}\right) \cdot P(A)}{P(B)}$
D
$P\left(\frac{A}{B}\right)=\frac{P(A)}{P(B)}$

Answer

Correct option: C.

$P\left(\frac{A}{B}\right)=\frac{P\left(\frac{B}{A}\right) \cdot P(A)}{P(B)}$

(C) $P\left(\frac{A}{B}\right)=\frac{P\left(\frac{B}{A}\right) \cdot P(A)}{P(B)}$
Explanation: Since, $P\left(\frac{B}{A}\right)=\frac{P(B \cap A)}{P(A)}$
$\therefore \frac{P\left(\frac{B}{A}\right) \cdot P(A)}{P(B)}=\frac{P(B \cap A)}{P(A)} \times \frac{P(A)}{P(B)}$
$=\frac{P(B \cap A)}{P(B)}$
$=\frac{P(A \cap B)}{P(B)}$
$=P\left(\frac{A}{B}\right)$

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MCQ - Applied Maths STD 11 Science Questions - Vidyadip