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Applied Maths MCQ

Model Paper 5 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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18 questions · timed · auto-graded

MCQ 21 Mark
For the post of 5 teachers, there are 23 applicants. 2 posts are reserved for SC candidates and there are 7 SC candidates among the applicants. In how many ways can the selection be made?
  • A
    3920
  • B
    11760
  • C
    4880
  • D
    5880
Answer
(b) 11760
Explanation: We have to select 2 posts out of 7 SC and 3 posts out of 16.
Required number of ways $=\left({ }^{7} C_{2} \times{ }^{16} C_{3}\right)=\left(\frac{7 \times 6}{2} \times \frac{16 \times 15 \times 14}{3 \times 2 \times 1}\right)=11760$.
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MCQ 31 Mark
The difference between simple interest and compound interest on ₹ 15000 for one yea $8 \%$ per annum calculated half-yearly is:
  • A
    ₹ 24
  • B
    ₹ 20
  • C
    ₹ 22
  • D
    ₹ 26
Answer
(a) ₹ 24
Explanation: For C.I.:
C.I. $=15000\left(1+\frac{4}{100}\right)^{2}-15000$
$=15000\left[\frac{26}{25} \times \frac{26}{25}-1\right]=\frac{15000 \times 51}{25 \times 25}=₹ 1224$.
S.I. $=\frac{15000 \times 51}{25 \times 25}=₹ 1200$
$\therefore$ Difference between C.I. and S.I. $=₹ 1224-₹ 1200=₹ 24$
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MCQ 41 Mark
5 boys and 5 girls are sitting in a row randomly. The probability that boys and girls sit alternately is:
  • A
    $\frac{5}{126}$
  • B
    $\frac{3}{126}$
  • C
    $\frac{1}{126}$
  • D
    $\frac{4}{126}$
Answer
(c) $\frac{1}{126}$
Explanation: Total number of ways $=10$!
Total number of ways in which 5 boys and 5 girls are sitting in a row $=2 \times 5!\times 5$!
$\therefore$ Required probability
$=\frac{2 \times 5!\times 5!}{10!}=\frac{2 \times 5!\times 5 \times 4 \times 3 \times 2}{10 \times 9 \times 8 \times 7 \times 6 \times 5!}=\frac{1}{126}$
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MCQ 51 Mark
Two men hit at a target with probabilities $\frac{1}{2}$ and $\frac{1}{3}$, respectively. What is the probability that exactly one of them hits the target?
  • A
    $\frac{1}{6}$
  • B
    $\frac{1}{2}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{2}{3}$
Answer
(b) $\frac{1}{2}$
Explanation: Let A be the event that Mr. A hit the target and B be the event that Mr. B hit the target
$\therefore \mathrm{P}(\mathrm{A})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{B})=\frac{1}{3}$
Now, P (exactly one of them hits the target)
$=\mathrm{P}(\mathrm{A} \cap \bar{B}$ or $\bar{A} \cap \mathrm{~B})$
$=\mathrm{P}(\mathrm{A} \cap \bar{B})+\mathrm{P}(\bar{A} \cap \mathrm{~B})$
$=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\bar{B})+\mathrm{P}(\bar{A}) \cdot \mathrm{P}(\mathrm{B})$
$=\frac{1}{2} \cdot \frac{2}{3}+\frac{1}{2} \cdot \frac{1}{3}=\frac{3}{6}=\frac{1}{2}$
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MCQ 61 Mark
There are 4 bus routes between $A$ and $B$ and 3 bus routes between $B$ and $C$. A man can travel round trip in number of ways by bus from A to C via B. If he does not want to use a bus route more than once, in how many ways can he make the round trip?
  • A
    14
  • B
    142
  • C
    72
  • D
    19
Answer
(c) 72
Explanation: From A to B there are 4 choices and from B to C, there are 3 choices.
And from C to B, only 2 choices are left and from B to A, only 3 choices are left.
So total ways $=4 \times 3 \times 2 \times 3=72$
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MCQ 71 Mark
At what rate percent per annum will a sum of ₹ 12000 become ₹ 13230 in 2 years?
  • A
    $5 \%$
  • B
    $6 \%$
  • C
    $6.5 \%$
  • D
    $5.5 \%$
Answer
(a) $5 \%$
Explanation: Here, $\mathrm{P}=₹ 12000, \mathrm{~A}=₹ 13230, \mathrm{n}=2$. Let rate be $\mathrm{r} \%$.
$\therefore 13230=12000\left(1+\frac{r}{100}\right)^{2}$
$\Rightarrow\left(1+\frac{r}{100}\right)^{2}=\frac{13230}{12000}=\frac{441}{400}=\left(\frac{21}{20}\right)^{2}$
$\Rightarrow 1+\frac{r}{100}=\frac{21}{20} \Rightarrow \frac{r}{100}=\frac{1}{20} \Rightarrow r=5 \%$.
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MCQ 81 Mark
If $\log _{\frac{1}{3}} 27 \sqrt{3}=x$, then value of x is
  • A
    -7
  • B
    7
  • C
    $-\frac{7}{2}$
  • D
    $\frac{7}{2}$
Answer
(c) $-\frac{7}{2}$
Explanation: $\log _{\frac{1}{3}} 27 \sqrt{3}=\mathrm{x}$
$\log _{\frac{1}{3}} 3^{3} \times \sqrt{3}=x$
$\log _{\frac{1}{2}} 3^{\frac{7}{2}}=x$
$\log _{\frac{1}{3}}\left(\frac{1}{3}\right)^{-\frac{7}{2}}=\mathrm{x}\left[\because \mathrm{a}^{\mathrm{x}}=\left(\frac{1}{\mathrm{a}}\right)^{-\mathrm{x}}\right]$
$\therefore \mathrm{x}=-\frac{7}{2}$
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MCQ 91 Mark
Let $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{\mathrm{n}}$ be n observations. Let $\mathrm{w}_{\mathrm{i}}=1 \mathrm{x}_{\mathrm{i}}+\mathrm{k}$ for $\mathrm{i}=1,2, \ldots, \mathrm{n}$, where l and k are constants. If the mean of $\mathrm{x}_{\mathrm{i}}$'s is 48 and their standard deviation is 12, the mean of $\mathrm{w}_{\mathrm{i}}$'s is 55 and standard deviation of $\mathrm{w}_{\mathrm{i}}$'s is 15, the values of l and k should be.
  • A
    $\mathrm{l}=-1.25, \mathrm{k}=5$
  • B
    $\mathrm{l}= 2.5, \mathrm{k}=-5$
  • C
    $\mathrm{l}=1.25, \mathrm{k}=-5$
  • D
    $\mathrm{l}=2.5, \mathrm{k}=5$
Answer
(c) $\mathrm{l}=1.25, \mathrm{k}=-5$
Explanation: Given $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{\mathrm{n}}$ be n observations
And Mean of these $n$ observations, $\bar{x}=48$
And their standard deviation, $\mathrm{SD}_{\mathrm{x}}=12$.
Another series of $n$ observations is given such that
$\mathrm{w}_{\mathrm{i}}=\mathrm{lx}_{\mathrm{i}}+\mathrm{k}$ for $\mathrm{i}=1,2, \ldots, \mathrm{n}$, where l and k are constants
And mean of these $n$ observations, $\bar{w}=55$
And their standard deviation, $\mathrm{SD}_{\mathrm{w}}=15$
Applying the given condition for mean we get
$\mathrm{w}_{\mathrm{i}}=\mathrm{lx}_{\mathrm{i}}+\mathrm{k}$
Substituting the corresponding given values of means, we get
$55=\mathrm{l}(48)+\mathrm{k} \ldots$ (i)
Now we know
If standard deviation of x series is s, then standard deviation of kx series is ks,
So standard deviation of $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{\mathrm{n}}$ is $\mathrm{SD}_{\mathrm{x}}$,
And hence the standard deviation of $\mathrm{lx}_{1}, \mathrm{~lx}_{2}, \ldots, \mathrm{~lx}_{\mathrm{n}}$ is $\mathrm{lSD}_{\mathrm{x}}$.
Similarly,
If standard deviation of x series is s, then standard deviation of $\mathrm{k}+\mathrm{x}$ series is s,
So standard deviation of $\mathrm{lx}_{1}, \mathrm{~lx}_{2}, \ldots, \mathrm{~lx}_{\mathrm{n}}$ is $\mathrm{lSD}_{\mathrm{x}}$,
And hence the standard deviation of $ \mathrm{Ix}_{1}+\mathrm{k}, \mathrm{lx}_{2}+\mathrm{k}, \ldots, \mathrm{lx}_{\mathrm{n}}+\mathrm{k}$ is $\mathrm{lSD}_{\mathrm{x}}$.
So applying the given condition for standard deviation, we get
$\mathrm{SD}_{\mathrm{w}}=\mathrm{lSD}_{\mathrm{x}}$
Substituting the given values, we get
$15=1(12)$
$\Rightarrow 1=\frac{15}{12}=1.25$
Now substituting the value of l in equation (i), we get
$55=(1.25)(48)+\mathrm{k}$
$55=60+\mathrm{k}$
$\Rightarrow \mathrm{k}=55-60=-5$
Hence the values of $k$ and l are -5 and 1.25 respectively
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MCQ 101 Mark
A man can do a piece of work in 5 days, but with the help of his son, he can do it in 3 days. In what time can the son do it alone?
  • A
    8 days
  • B
    $6 \frac{1}{2}$ days
  • C
    7 days
  • D
    $7 \frac{1}{2}$ days
Answer
(d) $7 \frac{1}{2}$ days
Explanation: Son's 1 day's work $=\left(\frac{1}{3}-\frac{1}{5}\right)=\frac{2}{15}$
$\therefore$ The son alone can do the work in $\frac{15}{2}=7 \frac{1}{2}$ days
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MCQ 111 Mark
The equation of two diameters of a circle are $x-y=5$ and $2 x+y=4$ and the radius of the circle is 5, then the equation of the circle is:
  • A
    $x^2+y^2+6 x+4 y+12=0$
  • B
    $x^{2}+y^{2}-6 x+4 y+12=0$
  • C
    $x^{2}+y^{2}+6 x-4 y-12=0$
  • D
    $x^{2}+y^{2}-6 x+4 y-12=0$
Answer
(d) $x^{2}+y^{2}-6 x+4 y-12=0$
Explanation: Given that the equation of two diameters of a circle are $\mathrm{x}-\mathrm{y}=5$ and $2 \mathrm{x}+\mathrm{y}=4$ and the radius is 5.
We know that the intersection point of the diameters be a centre
So, solving equation $\mathrm{x}-\mathrm{y}=5$ and $2 \mathrm{x}+\mathrm{y}=4$
for x and y
We get $\mathrm{x}=3, \mathrm{y}=-2$
$\therefore$ Centre $=(\mathrm{h}, \mathrm{k})=(3,-2)$
radius $=r=5$
$\therefore$ The required equation of the circle is $(\mathrm{x}-\mathrm{h})^{2}+(\mathrm{y}-\mathrm{k})^{2}=\mathrm{r}^{2}$
$\Rightarrow(\mathrm{x}-3)^{2}+(\mathrm{y}+2)^{2}=5^{2}$
$\Rightarrow x^{2}+y^{2}-6 x+4 y+9+4=25$
$\Rightarrow x^{2}+y^{2}-6 x+4 y-12=0$
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MCQ 121 Mark
If $E_{1}$ and $E_{2}$ are two independent events, then $\mathrm{P}\left(E_{1} \cap E_{2}\right)$ is equal to
  • A
    $P\left(E_{1}\right)+P\left(E_{2}\right)$
  • B
    $P\left(E_{1}\right)+P\left(E_{2}\right)+P\left(E_{1}+\cup E_{2}\right)$
  • C
    $P\left(E_{1}\right) P\left(E_{2}\right)$
  • D
    $P\left(E_{1}\right)-P\left(E_{2}\right)$
Answer
(c) $P\left(E_{1}\right) P\left(E_{2}\right)$
Explanation: We have, $P\left(E_{1} \cap E_{2}\right)=P\left(E_{1}\right) . P\left(\frac{E_{2}}{E_{1}}\right)$
Since $E_{1}$ and $E_{2}$ are independents, therefore
$P=\left(\frac{E_{2}}{P\left(E_{1}\right)=P\left(E_{2}\right)}\right)$
$\therefore P\left(E_{1} \cap E_{2}\right)=P\left(E_{1}\right) . P\left(E_{2}\right)$
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MCQ 131 Mark
If $\frac{1}{\log _{x} 10}=\frac{2}{\log _{a} 10}-2$, then $\mathrm{x}=$
  • A
    $\frac{a^{2}}{10}$
  • B
    $\frac{a}{100}$
  • C
    $\frac{a}{2}$
  • D
    $\frac{a^{2}}{100}$
Answer
(d) $\frac{a^{2}}{100}$
Explanation: $\frac{a^{2}}{100}$
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MCQ 141 Mark
A relation R is defined from $\{2,3,4,5\}$ to $\{3,6,7,10\}$ by: $\mathrm{x} \mathrm{R} y \Leftrightarrow \mathrm{x}$ is relatively prime to y. Then domain of $R$ is
  • A
    $\{2,3,4,5\}$
  • B
    $\{3,5\}$
  • C
    $\{2,3,5\}$
  • D
    $\{2,3,4\}$
Answer
(a) $\{2,3,4,5\}$
Explanation: Relatively prime numbers are those numbers that have only 1 as the common factor.
So, according to this definition we get to know that (2, 3), (2, 7), (3, 7), (3, 10), (4, 3), (4, 7), (5, 3), (5, 6), (5, 7) are relatively prime.
So, $R=\{(2,3),(2,7),(3,7),(3,10),(4,3),(4,7),(5,3),(5,6),(5,7)\}$.
Therefore, the Domain of $R$ is the values of $x$ or the first element of the ordered pair.
So, Domain $=\{2,3,4,5\}$
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MCQ 151 Mark
If $\log _{12} 27=\mathrm{a}$, then $\log _{6} 16=$
  • A
    $3\left(\frac{4+a}{4-a}\right)$
  • B
    $\frac{3-a}{3+a}$
  • C
    $3\left(\frac{4-a}{4+a}\right)$
  • D
    $4\left(\frac{3-a}{3+a}\right)$
Answer
(d) $4\left(\frac{3-a}{3+a}\right)$
Explanation: $\log _{12} 27=\mathrm{a}$
$\frac{\log 27}{\log 12}=\mathrm{a}$
$\Rightarrow 3 \log 3=\mathrm{a}[2 \log 2+\log 3]$
$\Rightarrow 3 \log 3-\mathrm{a} \log 3=2 \mathrm{a} \log 2$
$\Rightarrow \log 3=\frac{2 a \log 2}{3-\mathrm{a}} \ldots$ (i)
Now, $\log _{6} 16=\frac{\log 16}{\log 6}$
$\Rightarrow \frac{4 \log 2}{\log 2+\log 3} \ldots$ (ii)
$\Rightarrow \frac{4 \log 2}{\log 2+\frac{2 a \log 2}{3-a}}$ [using (i)]
$\Rightarrow 4\left(\frac{3-a}{3+a}\right)$
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MCQ 161 Mark
The effective rate which is equivalent to nominal rate of $10 \%$ p.a. compounded quarterly is:
  • A
    $10.25 \%$
  • B
    $10.38 \%$
  • C
    $10.53 \%$
  • D
    $10.47 \%$
Answer
(b) $10.38 \%$
Explanation: 10.38%
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MCQ 171 Mark
The quartile coefficient of skewness is negative, if:
  • A
    $\mathrm{Q}_{3}+\mathrm{Q}_{1}>$ Median
  • B
    $\mathrm{Q}_{3}+\mathrm{Q}_{1}<2$ Median
  • C
    $\mathrm{Q}_{3}+\mathrm{Q}_{1}>2$ Median
  • D
    $\mathrm{Q}_{3}+\mathrm{Q}_{1}=2$ Median
Answer
(c) $\mathrm{Q}_{3}+\mathrm{Q}_{1}>2$ Median
Explanation: $\mathrm{Q}_{3}+\mathrm{Q}_{1}>2$ Median
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MCQ 181 Mark
Bag A contains 3 red and 5 black balls and bag B contains 2 red and 4 black balls. A ball is drawn from one of the bags. The probability that ball drawn is red is
  • A
    $\frac{17}{24}$
  • B
    $\frac{17}{48}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{3}{8}$
Answer
(b) $\frac{17}{48}$
Explanation: As $\mathrm{P}(\mathrm{red})=\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}\left(\frac{R}{A}\right)+\mathrm{P}(\mathrm{B}) \cdot \mathrm{P}\left(\frac{R}{B}\right)$
$=\frac{1}{2} \cdot \frac{3}{8}+\frac{1}{2} \cdot \frac{2}{6}=\frac{3}{16}+\frac{1}{6}=\frac{17}{48}$
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MCQ - Applied Maths STD 11 Science Questions - Vidyadip