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Applied Maths MCQ

Model Paper 7 · CBSE STD 11 Science (CBSE - English Medium). 18 questions.

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18 questions · timed · auto-graded

MCQ 11 Mark
If A = {1, 2, 4}, B = {2, 4, 5}, c = {2, 5}, then (A - B) × (B - C) is
  • ✓
    {(1, 4)}
  • B
    (2, 5)
  • C
    {(1, 2), (1, 5), (2, 5)}
  • D
    (1, 4)
Answer
Correct option: A.
{(1, 4)}
(a) {(1, 4)}
Explanation: (A – B) = {1, 2, 4} – {2, 4, 5} = {1}
and (B - C) = {2, 4, 5} – {2, 5}
= {4}
Now, (A−B) × (B − C) = {1} × {4}
= {(1, 4)}
Since, it is a set so it is written in curly braces.
Therefore, option B is correct.
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MCQ 21 Mark
In how many ways can the letters of the word 'PENCIL' be arranged so that N is always next to E?
  • A
    1440
  • ✓
    120
  • C
    240
  • D
    720
Answer
Correct option: B.
120
(b) 120
Explanation: Keeping EN together and considering it as one letter, we have to arrange 5 letters at 5 places.
This can be done in ${ }^5 P_5=5!=120$ ways.
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MCQ 31 Mark
Time value of money supports the comparison of cash flows recorded at different time period by
i. Discounting all cash flows to a common point of time
ii. Compounding all cash flows to a common point of time
iii. Using either Discounting all cash flows to a common point of time or Compounding all cash flows to a common point of time
iv. not Discounting all cash flows to a common point of time nor Compounding all cash flows to a common point of time
  • ✓
    Statement (c) is correct
  • B
    Statement (b) is correct.
  • C
    Statement (a) is correct.
  • D
    Statement (d) is correct.
Answer
Correct option: A.
Statement (c) is correct
(a) Statement (c) is correct.
Explanation: Time value of money supports the comparison of cash flows recorded at different time period by either compounding or discounting all cash flows to common point of time.
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MCQ 41 Mark
If A and B are two events, then $P (\bar{A} \cap B)=$
  • ✓
    $P ( B )- P ( A \cap B )$
  • B
    $P(\bar{A}) P(\bar{B})$
  • C
    1 - P(A) - P(B)
  • D
    $P ( A )+ P ( B )- P ( A \cap B )$
Answer
Correct option: A.
$P ( B )- P ( A \cap B )$
(a) $P ( B )- P ( A \cap B )$
Explanation: $P(B)-P(A \cap B)$
Image
From the diagram, we get $A \cap B$ and $\bar{A} \cap B$ are mutually exclusive events such that $( A \cap B ) \cup(\bar{A} \cap B)= B$ Therefore by addition theorem of probability we have
$P(A \cap B)+P(\bar{A} \cap B)= P ( B )$
$\therefore P(A \cap B)=P(B)-P(A \cap B)$
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MCQ 51 Mark
Four cards are drawn from a well-shuffled pack of 52 cards. The probability of obtaining 3 diamonds and one spade is:
  • ✓
    $\frac{{ }^{13} C _3 \times{ }^{13} C _1}{{ }^{52} C _4}$
  • B
    $\frac{{ }^{13} C _3 \times{ }^{10} C _1}{{ }^{52} C _4}$
  • C
    $\frac{{ }^{26} C_2 \times{ }^{26} C_2}{{ }^{52} C_1}$
  • D
    $\frac{{ }^{26} C _4}{{ }^{52} C _4}$
Answer
Correct option: A.
$\frac{{ }^{13} C _3 \times{ }^{13} C _1}{{ }^{52} C _4}$
(a) $\frac{{ }^{13} C _3 \times{ }^{13} C _1}{{ }^{52} C _4}$
Explanation: Number of ways of drawing 4 cards from 52 cards = ${ }^{52} C _4$
In a deck of 52 cards, there are 13 diamonds and 13 spades.
$\therefore$ Number of ways of drawing 3 diamonds and one spade $={ }^{13} C _3 \times{ }^{13} C _1$
Thus the probability of obtaining 3 diamonds and one spade $=\frac{{ }^{13} C _3 \times{ }^{13} C _1}{{ }^{52} C _4}$
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MCQ 61 Mark
A shopkeeper bought a TV from a distributor at a discount of 25% of the listed price of 7 32000. The shopkeeper sells the TV to a consumer at the listed price. If the sales are intra-state and the rate of GST is 18%, the selling price of the TV including tax (under GST) by the distributor is:
  • A
    ₹ 26160
  • B
    ₹ 32000
  • C
    ₹ 28320
  • ✓
    ₹ 28320
Answer
Correct option: D.
₹ 28320
(d) ₹ 28320
Explanation: ₹ 28320
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MCQ 71 Mark
What does the P stands for in this formula?
Present value $=P\left[\frac{1-(1+i)^n}{i}\right]$
  • A
    The present value
  • B
    The future value
  • C
    The number of payments
  • ✓
    The fixed payment amount
Answer
Correct option: D.
The fixed payment amount
(d) The fixed payment amount
Explanation: Present value = cash flow $\times \frac{\left[(1+i)^n-1\right]}{i(1+i)^n}$
$=$ fixed payment amount $\times\left[\frac{1-(1+i)^n}{i}\right]$
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MCQ 81 Mark
The value of $5^{\sqrt{\log 57}}-7^{\sqrt{\log _7 5}}$, is:
  • ✓
    0
  • B
    1
  • C
    log 2
  • D
    -1
Answer
Correct option: A.
0
(a) 0
Explanation: 0
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MCQ 91 Mark
The variance of first 5 natural numbers is
  • A
    4
  • B
    3
  • C
    1
  • ✓
    2
Answer
Correct option: D.
2
(d) 2
Explanation: The variance of first n natural number is $\frac{n^2-1}{12}$
Here n = 5
$\begin{array}{l}\text { So, Variance }=\frac{5^2-1}{12} \\ =\frac{25-1}{12} \\ =\frac{24}{12} \\ =2\end{array}$
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MCQ 101 Mark
In a certain code CAT in written as SATC and DEAF as SEARD, then SING will be with as
a. TSING
b. TINGS
c. GSING
d. SINGS
  • A
    TSING
  • B
    TINGS
  • ✓
    SINGS
  • D
    GSING
Answer
Correct option: C.
SINGS
(c) SINGS
Explanation: SINGS
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MCQ 111 Mark
The vertex of the parabola $y^2=4 a(x-a)$ is
  • A
    (0, a)
  • ✓
    (a, 0)
  • C
    (0, 0)
  • D
    (a, a)
Answer
Correct option: B.
(a, 0)
(b) (a, 0)
Explanation: $y^2=4 a(x-a)$
Let $Y = y$ and $X = x - a$
For standard parbola $Y^2=4 a X$ the vertex is $(0,0)$
So put $X=0$ and $Y=0$.
i.e. put $x - a =0$ and $y =0$
So vertex is $( a , 0)$
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MCQ 121 Mark
If A and B are any two events having P(A ∪ B) =$\frac{2}{3}$ and $P (\bar{B})=\frac{1}{2}$, then $P ( A \cap \bar{B})$ is:
  • A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{1}{2}$
  • ✓
    $\frac{1}{6}$
Answer
Correct option: D.
$\frac{1}{6}$
(d) $\frac{1}{3}$
Explanation: We know that $P ( A \cap \bar{B})= P ( A - B )= P ( A \cup B )- P ( B )$
$=\frac{2}{3}-\left(1-\frac{1}{2}\right)=\frac{2}{3}-\frac{1}{2}=\frac{1}{6}$.
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MCQ 131 Mark
log (1 × 2 × 3) is equal to
  • A
    log 3
  • B
    log 1
  • C
    log 2
  • ✓
    log 1 + log 2 + log 3
Answer
Correct option: D.
log 1 + log 2 + log 3
(d) log 1 + log 2 + log 3
Explanation: Since, $\log _a(m n)=\log _a m+\log _a n$
One extending this rule for three variables, we get
$\log _a(m n p)=\log _a m+\log _a n+\log _a p$
Therefore, log(1 × 2 × 3) = log 1 + log 2 + log 3
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MCQ 141 Mark
For two non empty sets A and B, if n(A) = m and n(B) = n then the total number of relations from set A to set B is
  • A
    $2^m+2^n$
  • ✓
    $2^{ mn }$
  • C
    mn
  • D
    m + n
Answer
Correct option: B.
$2^{ mn }$
(b) $2^{m n}$
Explanation: as n(A) = m, n(B) = n
$\Rightarrow$n(A × B) = mn
So, number of relations = $2^{m n}$
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MCQ 151 Mark
$2^4=16$ in logarithmic form is
  • A
    4 log 2 = log 16
  • B
    $\log _4 16=2$
  • C
    $\log _4 2=16$
  • ✓
    $\log _2 16=4$
Answer
Correct option: D.
$\log _2 16=4$
(d) $\log _2 16=4$
Explanation: $2^4=16$ in logarithmic form.
As we know that
if $a^y=x$
then $\log _a x=y$
$\therefore \log _2 16=4$
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MCQ 161 Mark
A shopkeeper bought a TV from a distributor at a discount of 25% of the listed price of ₹ 32000. The shopkeeper sells the TV to a consumer at the listed price. If the sales are intra-state and the rate of CST is 18%, the selling price of the TV including tax (under GST) by the distributor is
  • A
    ₹32000
  • ✓
    ₹28320
  • C
    ₹24000
  • D
    ₹26160
Answer
Correct option: B.
₹28320
(b) ₹ 28320
Explanation: Listed price = ₹ 32000, discount = 25% of ₹ 32000 = ₹ 8000
$\therefore$ S.P. of distributor = ₹ 32000 - ₹ 8000 = ₹ 24000
CGST = 9% of ₹ 24000 = ₹ 2160
SGST = 9% of ₹ 24000 = ₹ 2160
$\therefore$ S.P. including tax = ₹ 24000 + ₹ 2160 + ₹ 2160 = ₹ 28320
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MCQ 171 Mark
The mean and standard deviation of 100 items are 50, 5 and that of 150 items are 40, 6 respectively. What is the combined mean of all 250 items?
  • A
    43
  • B
    46
  • ✓
    44
  • D
    45
Answer
Correct option: C.
44
(c) 44
Explanation: Given that, mean of 100 items, $\bar{x}_{100}=50$
Mean of 150 items, $\bar{x}_{100}=50$ and standard deviation of 100 items, $\sigma_{100}=5$
Standard deviation of 150 items, $\sigma_{100}=6$
Here, $n _1=100, \bar{x}_{100}=50$ and $n _2=150, \bar{x}_{150}=40$
$\therefore$ Combined mean of all 250 items, $\bar{x}_{250}=\frac{n_1 \cdot \bar{x}_{100}+n_2 \cdot \bar{x}_{150}}{n_1+n_2}$
$=\frac{100 \times 50+150 \times 40}{100+150}=\frac{5000+6000}{250}=\frac{11000}{250}=44$
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MCQ 181 Mark
A bag contains 5 black balls, 4 white balls and 3 red balls. If a ball is selected random wise, the probability that it is black or red ball is
  • ✓
    $\frac{2}{3}$
  • B
    $\frac{5}{12}$
  • C
    $\frac{1}{3}$
  • D
    $\frac{1}{4}$
Answer
Correct option: A.
$\frac{2}{3}$
(a) $\frac{2}{3}$
Explanation: $\frac{2}{3}$
We know that the bag contains 5 (black), 4W (white) and 3R (red) balls.
Now,
$\begin{array}{l}P(B)=\frac{5}{12} \\ P(R)=\frac{3}{12} \\ P ( B \text { or } R )= P ( B )+ P ( R ) \\ =\frac{5}{12}+\frac{3}{12} \\ =\frac{8}{12}=\frac{2}{3}\end{array}$
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MCQ - Applied Maths STD 11 Science Questions - Vidyadip