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Applied Maths MCQ

Model Paper 4 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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18 questions · timed · auto-graded

MCQ 11 Mark
In a city 20 percent of the population travels by car, 50 percent travels by bus and 10 percent travels by both car and bus. Then persons travelling by a car or bus is
  • ✓
    60 percent
  • B
    80 percent
  • C
    70 percent
  • D
    40 percent
Answer
Correct option: A.
60 percent
(a) 60 percent
Explanation: Let A denote the set of persons traveling by car, B denotes the set of persons traveling by bus, then
$n(A)=20, n(B)=50, n(A \cap B)=10$
$\therefore \mathrm{n}(\mathrm{A} \cup \mathrm{B})=\mathrm{n}(\mathrm{A})+\mathrm{n}(\mathrm{B})-\mathrm{n}(\mathrm{A} \cap \mathrm{B})$
$=20+50-10=60$
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MCQ 21 Mark
The number of digits in the binary system are
  • A
    6
  • B
    4
  • C
    10
  • ✓
    2
Answer
Correct option: D.
2
(d) 2
Explanation: 2
In a binary number system, we use only two digits, such as 0 and 1 .
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MCQ 31 Mark
A man borrows ₹ 21000 at $10 \%$ per annum compound interest. How much he has to pay to the end of each year to clear his debt in two years?
  • ✓
    ₹ 12100
  • B
    ₹ 12000
  • C
    ₹ 10500
  • D
    ₹ 11000
Answer
Correct option: A.
₹ 12100
(a) ₹ 12100
Explanation: Let him pay ₹ R at the end of each year.
$\therefore 21000=R\left[\frac{100}{100+10}+\left(\frac{100}{100+10}\right)^{2}\right]$
$\Rightarrow 21000=\mathrm{R}\left[\frac{100}{110}+\left(\frac{100}{110}\right)^{2}\right]=\mathrm{R}\left[\frac{10}{11}+\frac{100}{121}\right]=\mathrm{R}\left[\frac{210}{121}\right]$.
$\therefore \mathrm{R}=\frac{21000 \times 121}{210}=$ ₹ $12100$.
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MCQ 41 Mark
Suppose A and B are two events. Event B has occurred and it is known that $\mathrm{P}(\mathrm{B})<1$. What is $\mathrm{P}\left(\mathrm{A} / \mathrm{B}^{\prime}\right)$ equal to?
  • A
    $\frac{P(A)-P(B)}{1-P(B)}$
  • B
    $\frac{P(A)+P\left(B^{\prime}\right)}{1-P(B)}$
  • ✓
    $\frac{P(A)-P(A B)}{1-P(B)}$
  • D
    $\frac{P(A)+P\left(A B^{\prime}\right)}{1-P(B)}$
Answer
Correct option: C.
$\frac{P(A)-P(A B)}{1-P(B)}$
(c) $\frac{P(A)-P(A B)}{1-P(B)}$
Explanation: $\mathrm{P}\left(\mathrm{A} / \mathrm{B}^{\prime}\right)=\frac{P\left(A \cap B^{\prime}\right)}{P\left(B^{\prime}\right)}$
$=\frac{P(A)-P(A \cap B)}{1-P(B)}=\frac{P(A)-P(A B)}{1-P(B)}$
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MCQ 51 Mark
Simplified from of $\log 12-\log 2-\log 3$ is
  • A
    $\log 4$
  • ✓
    $\log 2$
  • C
    $\log 6$
  • D
    $\log 1$
Answer
Correct option: B.
$\log 2$
(b) $\log 2$
Explanation: $\log 12-\log 2-\log 3$
$=\frac{\log 12}{\log 2+\log 3}$
$=\frac{\log 12}{\log 2 \times 3}=\frac{\log 12}{\log 6}$
$=\log \frac{12}{6}$
$=\log 2$
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MCQ 61 Mark
Deduction of the principal of the home loan is allowed under section _____________.
  • A
    80 D
  • B
    24
  • C
    80 E
  • ✓
    80 C
Answer
Correct option: D.
80 C
(d) 80 C
Explanation: The Principal portion of the EMI paid for the year is allowed as a deduction under Section 80C. The maximum amount that can be claimed is up to Rs 1.5 lakh. But to claim this deduction, the house property should not be sold within 5 years of possession.
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MCQ 71 Mark
In what time will a sum of ₹ 1562.50 produce ₹ $195.10$ at $4 \%$ per annum compound interest?
  • A
    $1 \frac{1}{2}$ years
  • ✓
    3 years
  • C
    $2 \frac{1}{2}$ years
  • D
    2 years
Answer
Correct option: B.
3 years
(b) 3 years
Explanation: Let the time required be n, then
$1562.50\left(1+\frac{4}{100}\right)^{n}-1562.50=195.10$
$\Rightarrow\left(\frac{26}{25}\right)^{n}=\frac{1562.50+195.10}{1562.50}=\frac{1757.60}{1562.50}=\left(\frac{26}{25}\right)^{3}$
$\Rightarrow \mathrm{n}=3$.
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MCQ 81 Mark
The value of $\log _{2} \log _{2} \log _{4} 256+2 \log _{\sqrt{2}} 2$ is:
  • A
    3
  • B
    2
  • C
    7
  • ✓
    5
Answer
Correct option: D.
5
(d) 5
Explanation: Use the properties:
$\log _{a}(m)^{n}=n \log _{a} m$ and $\log _{a} a=1$
Consider, $\log _{2} \log _{2} \log _{4} 256+2 \log _{\sqrt{2}} 2$
$=\log _{2} \log _{2} \log _{4}(4)^{4}+2 \log _{\sqrt{2}}(\sqrt{2})^{2}$
$=\log _{2} \log _{2} 4+2(2)$
$=\log _{2} \log _{2}(2)^{2}+4$
$=\log _{2} 2+4$
$=1+4=5$
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MCQ 91 Mark
The mean deviation of the numbers $3,4,5,6,7$ from the mean is
  • A
    25
  • B
    0
  • ✓
    1.2
  • D
    5
Answer
Correct option: C.
1.2
(c) 1.2
Explanation: Mean $(X)=\frac{3+4+5+6+7}{5}$
$=\frac{25}{5}=5$
Taking the absolute value of deviation of each term from the mean, we get:
$\mathrm{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$
$=\frac{2+1+0+1+2}{5}$
$=\frac{6}{5}$
$=1.2$
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MCQ 101 Mark
$\quad A$ is a brother of $F$ and $F$ is daughter of $D$. $P$ is brother of $D$. $X$ is sister of $P$ and $M$ is father of $X$ then $P$ is related to A as
  • A
    niece
  • B
    father
  • C
    uncle
  • D
    brother
Answer
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MCQ 111 Mark
The equations of the lines through $(-1,-1)$ and making angles of $45^{\circ}$ with the line $x+y=0$ are
  • A
    $x-1=0, y-1=0$
  • ✓
    $x+1=0, y+1=0$
  • C
    $x-1=0, y-x=0$
  • D
    $x+y=0, y+1=0$
Answer
Correct option: B.
$x+1=0, y+1=0$
(b) $\mathrm{x}+1=0, \mathrm{y}+1=0$
Explanation: The lines $\mathrm{x}+1=0$ and $\mathrm{y}+1=0$ are perpendicular to each other.
The slope of the line $\mathrm{x}+\mathrm{y}=0$ is -1
Hence the angle made by this line with respect to X -axis is $45^{\circ}$
In other words, the angle made by this line with $\mathrm{x}+1=0$ is $45^{\circ}$
Clearly the other line with which it can make $45^{\circ}$ is $\mathrm{y}+1=0$
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MCQ 121 Mark
Let $A=\{1,2,3\}, B=\{1,3,5\}$. If relation $R$ from $A$ to $B$ is given by $R=\{(1,3),(2,5),(3,3)\}$. Then, $R^{-1}$ is
  • A
    $\{(1,3),(5,2)\}$
  • ✓
    $\{(3,3),(3,1),(5,2)\}$
  • C
    $\{(1,3),(2,5),(3,3)\}$
  • D
    $\{(5,2)\}$
Answer
Correct option: B.
$\{(3,3),(3,1),(5,2)\}$
(b) $\{(3,3),(3,1),(5,2)\}$
Explanation: Inverse of a relation is given by interchanging the element's position in each pair.
Ex: Inverse of relation $P=\{(x, y)\}$ is given by $P^{-1}=\{(y, x)\}$.
Therefore, $\mathrm{R}^{-1}=\{(3,1),(5,2),(3,3)\}$.
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MCQ 131 Mark
Characteristic of $\log 0.0003798$ is
  • ✓
    $\overline{4}$
  • B
    $\overline{3}$
  • C
    4
  • D
    3
Answer
Correct option: A.
$\overline{4}$
(a) $\overline{4}$
Explanation: $0.0003798=3.798 \times 10^{-4}$
$\therefore$ Characteristic of $\log 0.0003798=-4$ i.e. $\overline{4}$
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MCQ 141 Mark
Number of relations that can be defined on the set $A=\{a, b, c, d\}$ is
  • A
    24
  • B
    $4^{4}$
  • C
    16
  • ✓
    $2^{16}$
Answer
Correct option: D.
$2^{16}$
(d) $2^{16}$
Explanation: No. of elements in the set $\mathrm{A}=4$. Therefore, the no. of elements in $A \times A=4 \times 4=16$. As, the no. of relations in $A \times A=$ no. of subsets of $A \times A=2^{16}$.
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MCQ 151 Mark
If Karl Pearson's coefficient of skewness of a distribution is 2.5, standard deviation is 8 and mean is 30, then mode of the distribution is:
  • A
    5
  • B
    20
  • ✓
    10
  • D
    25
Answer
Correct option: C.
10
(c) 10
Explanation: 10
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MCQ 161 Mark
Integrated Goods and Services Tax is applicable when:
  • A
    Sold in Union territory
  • B
    Sold within a state
  • C
    Sold from one GST dealer to another GST dealer
  • ✓
    There is interstate supply
Answer
Correct option: D.
There is interstate supply
(d) There is interstate supply
Explanation: Integrated Goods and Services Tax is applicable when there is interstate supply.
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MCQ 171 Mark
The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is:
  • ✓
    38
  • B
    30
  • C
    35
  • D
    28
Answer
Correct option: A.
38
(a) 38
Explanation: L et the numbers are $\mathrm{x}_{1}, \mathrm{x}_{2}, \mathrm{x}_{3}, \mathrm{x}$ and $\mathrm{x}_{5}$. Then,
we have, $\frac{x_{1}+x_{2}+x_{3}+x_{4}+x_{5}}{5}=30$
$\Rightarrow \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}+\mathrm{x}_{4}+\mathrm{x}_{5}=150 \ldots$ (i)
Now, suppose $\mathrm{X}_{1}$ is excluded, then $\frac{x_{2}+x_{3}+x_{4}+x_{5}}{4}=28$ [given]
$\Rightarrow x_{2}+x_{3}+x_{4}+x_{5}=112 \ldots$ (i)
From Eqs. (i) and (ii), we get $x_{1}=150-112=38$
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MCQ 181 Mark
Which of the following binary numbers is equivalent to decimal number 24 ?
  • A
    111111
  • ✓
    11000
  • C
    11001
  • D
    1101111
Answer
Correct option: B.
11000
(b) 11000
Explanation: $1101111=1 \times 2^{6}+1 \times 2^{5}+0+1 \times 2^{3}+1 \times 2^{2}+1 \times 2^{1}+1 \times 2^{0}=111$
$11000=1 \times 2^{4}+1 \times 2^{3}+0+0+0=16+8=24$
$111111=1 \times 2^{5}+1 \times 2^{4}+1 \times 2^{3}+1 \times 2^{2}+1 \times 2^{1}+1 \times 2^{0}=63$
$11001=1 \times 2^{4}+1 \times 2^{3}+0+0+1 \times 2^{0}=16+4+1=21$
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