Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 7 are designed as per the revised syllabus.

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Maximum Marks : 80

Time Allowed : 3 hours

General Instructions :

  1. All the questions are compulsory.
  2. The question paper consists of 38 questions divided into 5 sections A, B, C, D and E.
  3. Section A comprises of 16 questions of 1 mark each. Section B comprises of 10 questions of 2 marks each. Section C comprises of 7 questions of 3 marks each. Section D comprises of 3 questions of 5 marks each. Section E comprises of 2 questions of 4 marks each.
  4. There is no overall choice. However, an internal choice has been provided in five questions of 1 mark each, three questions of 2 marks each, two questions of 3 marks each, and two question of 5 marks each. You have to attempt only one of the alternatives in all such questions.
  5. Use of calculators is not permitted.

Section – A

All questions are compulsory. In case of internal choices attempt any one.

Question 1.
Write the base values of binary, octal and hexadecimal number system. [1]
Answer:
(a) Base of binary number system is 2.
(b) Base of octal number system is 8.
(c) Base of hexa decimal number system is 16.

Question 2.
Simplify:\(\left\{\left(x^n\right)^{n-\frac{1}{n}}\right\}^{\frac{1}{n+1}}\)
OR
Find the value of log2 log2 log216. [1]
Answer:
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 3

Commonly Made Error
Some students make mistakes while doing simplification.

Answering Tip
Students should write the index values clearly to avoid simplification errors.

OR

we have
log2 log2 16
= log2 log2 (log2 24)
= log2 log2 (4l0g2 2)
[Applying rule loga mn = n\(\log _a^m\)
= log2 log24 [Applying rule logamn = \(n \log _a{ }^m\)]
= log2 log2 (22)
= l0g2 2 = 1

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 3.
For a group of 8 students, the sum of squares of differences in the ranks for mathematics and statistics marks was found to be 50. What is the value of rank correlation coefficient ? [1]
Answer:
As given, n = 8 and \(d_i^2\) = 50. Hence, the rank correlation coefficient between marks in Mathematics and Statistics is given by
rs = 1 – \(\frac{6 \times 50}{8\left(8^2-1\right)}\) = 1 – \(\frac{300}{8 \times 63}\)
= \(\frac{504-300}{504}\) = 0.40

Question 4.
Find the equation of the circle which touches both the axes in the first quadrant and whose radius is a. [1]
Answer:
As the radius of circle is a units and it touches both the axes in the first quadrant, its centre is C(a, a).
Then equation of the circle is (x – a)2 + (y – a)2 = a2
⇒ x2 + y2 – 2ax – 2ay + a2 = 0

Question 5.
If sum of first n term of an A.P. is 2n2 + 7n, write its nth term.
OR
The first terms of G.P. is 2 and sum to infinity is 6, find common ratio.  [1]
Answer:
Given, Sn = 2n2 + 7n
We know that, an = Sn – Sn-1
an = 2n2 + 7n – {2(n – 1)2 + 7(n – 1)}
= 2n2+ 7n – {2(n2 – 2n + 1) + 7n – 7}
= 2n2 + 7n – (2n2 – 4n + 2 + 7n – 7)
= 2n2 + 7n – (2n2 + 3n – 5)
= 2n2 + 7n – 2n2 – 3n + 5
= 4n + 5

OR

Given, a = 2 and S = 6
∴ S = \(\frac{a}{1-r}\)
∴ 6 = \(\frac{2}{1-r}\)
⇒ r = \(\frac{4}{6}\) = \(\frac{2}{3}\)

Question 6.
If the following function f(x) is continuous at x = 0, then write the value of k.
f(x) = \(\left\{\begin{array}{cc}
\frac{\sin \frac{3 x}{2}}{x}, & x \neq 0 \\
k, & x=0
\end{array}\right.\)  [1]
Answer:
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 4

Question 7.
Find the 10th common term between the series 3 + 7 + 11 + …. and 1 + 6 + 11 + …….  [1]
Answer:
The first common term is 11. Now, the next common term is obtained by adding LCM of the common difference of 4 and 5, i.e,. 20. Therefore, 10th common term = T10 of the A.E whose first term, a = 11 and common difference, d = 20
T10 = a + 9d = 11 + 9(20) = 191

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 8.
Two statements followed by two conclusions are given below. You have to decide which of the given conclusions logically follows from the given two statements disregarding commonly known facts.
Statements : (A) Some vegetables are fruits.
Statements : (B) No fruit is black.

Conclusions: (A) Some fruits are vegetables.
(B) No vegetable is black.  [1]
Answer:
Conclusion I is immediate inference and follows from statement (i). Conclusion II does not follow as vegetable is distributed in conclusion, but not in the statement, Hence, only conclusion I follows.

Question 9.
Find the domain of the function/is given by
f(x) = \(\frac{x^2+2 x+1}{x^2-8 x+12}\)  [1]
Answer:
∴ Domain off = R – {x : x2 – 8x + 12 = 0}
i.e., for function to be defined
x2 – 8x + 12 ≠ 0
⇒ x2 – 6x – 2x + 12 ≠ 0
⇒ (x – 6) – 2(x – 6) ≠ 0
⇒ (x – 2) (x – 6) ≠ 0
⇒ x ≠ 2 and x ≠ 6
∴ Domain = R – {2, 6}

Question 10.
How many different words can be formed by using all the letters of word ‘SCHOOL’ ?
OR
In how many ways can a student choose a program of 5 courses, if 9 courses are available and 2 specific courses are compulsory for every student ?  [1]
Answer:
Since, ‘SCHOOL has 6 letters, out of these 6 letters there are 2 O’s.
Hence number of permutations
\(\frac{6 !}{2 !}\) = \(\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2}\)
= 360.

OR

Total number of available courses = 9
Out of these 5 courses have to be chosen. But it is given that 2 courses are compulsory for every student i.e., you have to choose only 3 courses instead of 5, out of 7 instead of 9.
It can be done in 7C3 ways = \(\frac{7 \times 6 \times 5}{6}\) = 35 ways.

Question 11.
A coin is tossed. If it shows head, we draw a ball from a bag consisting of 2 red and 3 black balls. If it shows tail, coin is tossed again. Write the sample space.  [1]
Answer:
Let, R1, R2 are the red balls and B1, B2, B3 are the black balls in the bag.
∴ The sample space associated with the experiment is
S = (HR1, HR2, HB1, HB2, HB3, TH, TT}

Question 12.
If R = {(x, y): x, y ∈ N, x + 2y = 21}, then find the range of R.  [1]
Answer:
Given, x + 2y = 21 ⇒ x = 21 – 2y
When y = 1, x = 21 – 2 × 1 = 19
When y = 2, x = 21 – 2 × 2 = 17
When y = 3, x = 21 – 2 × 3 = 15
When y = 10, x = 21 – 2 × 10 = 1
For other values of y e N, we do not get x e N.
Range of R = {1, 2, 3,…, 10}

Question 13.
Find the average of the two digit numbers, which remain the same, when the digits interchange their positions.  [1]
OR
In 8th Feb., 2005 it was Tuesday. What was the day of the week on 8th Feb, 2004 7.
Answer:
The two digit numbers, which remain the same, when the digits interchange their positions are :
11, 22, 33, 44, 55, 66, 77, 88, 99
∴ Average
= \(\frac{11+22+33+44+55+66+77+88+99}{9}\)
= \(\left[\frac{(11+99)+(22+88)+(33+77)+(66+44)+55}{9}\right]\)
= \(\left[\frac{(4 \times 110)+55}{9}\right]\) = \(\frac{495}{9}\) = 55

OR

The year 2004 is a leap year. It has 2 odd days.
The day on 8th Feb, 2004 is 2 days before the day on 8th Feb, 2005.
Hence, this day is Sunday.

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 14.
An interest rate is 5%, the number of periods are 3, and the present value is ₹ 100, then find the future value. [1]
Answer:
Since, F.V. = C.F. (1 + i)n
Here, C.F. is present value i.e., ₹ 100
i = 5% = \(\frac{5}{100}\) = 0.05
n = 3
∴ F.V. = 100 × (1 + 0.05)3
= 100 × (1.05)3
= 100 × 1.1576
= ₹ 115.76
Thus, future value is ₹ 115.76.

Question 15.
Find the equation of a line, which is equidistant from y = – 3 and y = 13.  [1]
Answer:
The given lines are y = – 3 and y = 13.
So, the resulting line is also of the form y = a,
where a = \(\frac{-3+13}{2}\) = 5
∴ The resulting line is y = 5.

Question 16.
Write the set of all positive integers whose cube is odd in the builders form.
OR
If P = {1, 3}, Q = {2, 3, 5}, find the number of relations from P to Q.  [1]
Answer:
{x : x is an odd positive integer} as we are aware cube of an even positive integer is an even positive integer and cube of an odd integer is always an odd positive integer, therefore, the members of the required set are all positive odd integers. Also, it can be written as {x : x = 2p + 1 and p ∈ W}.
OR
Given, P = {1, 3} and Q = {2, 3, 5}
∴ n(P) = 2 and n(Q) = 3
Number of relations = 2n(P) × n(Q)
= 22 × 3
= 26
= 64

Commonly Made Error
Sometimes students get confused with number of ordered pairs, relations, functions etc.

Answering Tips

  • Learn the formulae of relations of two sets, Number of relations = 2n(p) × n(Q)
  • Practice more such problems based on this formulae.

Section – B

All questions are compulsory. In case of internal choices attempt any one.

Question 17.
Find decimal equivalent of Binary Number (1011.011)2. [2]
Answer:
Binary to Decimal conversion is obtained by multiplying 2 to the power of base index along with the value at that index position.
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 5
= 8 + 0 + 2 + 1 + 0 + 0.25 + 0.125
= 11.375
Hence, (1011.011)2 = (11.375)10

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 18.
A can do a piece of work in 80 days. He works at it for 10 days and then B alone finishes the remaining work in 42 days. In how much time will A and B, working together, finish the work ?  [2]
Answer:
Work done by A in 10 days = \(\frac{1}{80} \times 10\) = \(\frac{1}{8}\)
Remaining work = (1 – \(\frac{1}{8}\)) = \(\frac{7}{8}\)
Now, \(\frac{7}{8}\) work is done by B in 42 days.
Whole work will be done by B in \(\left(42 \times \frac{8}{7}\right)\)
= 48 days
∴ A’s 1 day’s work = \(\frac{1}{80}\) and
B’s 1 days’ work = \(\frac{1}{48}\)
∴ (A + B)’s 1 days’ work = \(\frac{1}{80}\) + \(\frac{1}{48}\) = \(\frac{8}{240}\) = \(\frac{1}{30}\)
Hence, both will finish the work in 30 days.

Commonly Made Error
Some students do not apply the correct approach to solve the problems based on work and time due to the lack of conceptual knowledge.

Answering Tip
In work and time questions, remember that ; men and days are inversely proportional i.e., if the number of men increases, then number of days required to complete the same work decreases and vice-versa.

Question 19.
Are sets A = {1, 2, 3, 4}, B = {x: x ∈ N and 5 < x < 7} disjoint ? Why ?  [2]
Answer:
Yes, sets A and B are disjoint, because A ∩ B = ϕ.
∵ A = {1, 2, 3, 4}
and B = {5, 6, 7}
A ∩ B = {1, 2, 3, 4} ∩ {5, 6, 7} = ϕ

Question 20.
If nP4: nP2 = 12, find n.
OR
Let A = {1, 2, 3, 4}, B = {1, 4, 9,16, 25} and R be a relation defined from A to B as, R = {(x, y): x ∈ A, y ∈ B and y = x2}
(A) Depict this relation using arrow diagram.
(B) Find domain of R.
(C) Find range of R.
(D) Write co-domain of R.  [2]
Answer:
Given,
\(\frac{{ }^n P_4}{{ }^n P_2}\) = 12
nP4 = 12 nP2
⇒ \(\frac{n !}{n-4 !}\) = 12\(\frac{n !}{n-2 !}\)
⇒ \(\frac{n-2 !}{n-4 !}\) = 12
⇒ \(\frac{(n-2)(n-3)(n-4) !}{n-4 !}\) = 12
⇒ n2 – 5n + 6 = 12
⇒ n2 – 6n + n – 6 = 0
⇒ n(n – 6) + 1(n – 6) = 0
⇒ (n – 6)(n + 1) = 0
⇒ n = -1 or n = 6
∵ n cannot be negative
∴ n = 6

OR

Given, A = {1, 2, 3, 4} and B = (1, 4, 9, 16, 25) and R = {(x, y) : x ∈ A, y ∈ B and y = x2}

(i) Relation R = {(1, 1), (2, 4), (3, 9), (4, 16)}
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 6
(ii) Domain of R = {1, 2, 3, 4}
(iii) Range of R = {1, 4, 9, 16}
(iv) Codomain of R = {1, 4, 9, 16, 25}

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 21.
If LINGER is 123456 and FORCE is 56789, then find the code for FIERCE.  [2]
Answer:
L → 1
I → 2
N → 3
G → 4
E → 5
R → 6
F → 5
O → 6
R → 7
C → 8
E → 9

Letters R, E are common to LINGER and FORCE, and numbers 6 and 5 are common to their respective coding. Therefore, 5 and 6 will stand for R an E but not respectively. Now, since word FIERCE has double E, hence either of 5 and 6 will appear twice in its coding.

Besides, coding of FIERCE will have two numbers from coding of FORCE for letters F and C, and one from coding of LINGER for letter I. Hence, coding of FIERCE will be 456678.

Question 22.
(A) The son of M is the father of N and grandfather (Mother’s father) of R. S is the daughter of N and sister of B. On the basis of this information, how is M related to B ?
(B) Shyama says that father of Rajiv’s father is my father. How Shyama is related to Rajiv’s ?
OR
Redefine the function
f(x) = |x – 2| + |2 + x|, -3 ≤ x ≤ 3.  [2]
Answer:
(A)
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 7

From the above chart, it can be concluded that M is the (maternal) great grandfather or (maternal) great grandmother of B, hence the gender of M is not specified.

(B) From the given information, it is clear that Rajiv’s father is the brother of Shyama. Therefore, Shyama is related as father’s sister
(Bua) to Rajiv.

OR

∴ |x – 2| = -(x – 2)x < 2
and x – 2, x ≥ 2
Similarly, |2 + x| = -(2 + x), x < -2
and 2 + x, x ≥ -2
f(x) = |x – 2| + |2 + x|, -3 ≤ x ≤ -3
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 8

Question 23.
Evaluate \(\lim _{x \rightarrow 3}\left(\frac{x^4-81}{2 x^2-5 x-3}\right)\)
If (x2 + y2)2 = xy, find \(\frac{d y}{d x}\)  [2]
Answer:
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 9

OR

Given, (x2 + y2)2 = xy
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 10

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 24.
If R2 = {(x, y) | x and y are integers and x2 + y2 = 64} is a relation, then find the value of R2.  [2]
Answer:
Given, R2 = {(x, y) | x and y are integers and x2 + y2 = 64}
Since, 64 is the sum of squares of 0 and ± 8.
When x = 0,. then y2 = 64 ⇒ y = ±8
When x = 8, then y2 = 64 – 82 ⇒ 64 – 64 = 0
When x = -8, then y2 = 64 – (-8)2 ⇒ 64 – 64 = 0
∴ R2 = {(0,8), (0, -8), (8, 0), (-8, 0)}

Commonly Made Error
Sometimes students forget to take 0 for x.

Answering Tip
Read and understand the question carefully.

Question 25.
Compute Sk if
(A) Mean = 108, Mode = 99, σ = 5
(B) σ = 4, Mean = 20.5, Mode = 22 Also, mention type of skewness.  [2]
Answer:
Sk = \(\frac{\text { Mean-Mode }}{\sigma}\)
(A) Sk = \(\frac{108-99}{5}\) = \(\frac{9}{5}\) = 1.8 > 0
i.e., Sk > 0, therefore, it is positively skewed.

(B) Sk = \(\frac{20.5-22}{5}\) = \(\frac{-1.5}{4}\) = -0.375 < 0
Here, Sk < 0, therefore, it is negatively skewed

Question 26.
A machine can be purchased for ₹ 50000. Machine will contribute ₹ 12000 per year for the next five years. Assume borrowing is 10% per annum compounded annually. Determine whether machine should be purchased or not. Given P(5, 0.10) = 3.79079.  [2]
Answer:
The present value of annual contribution
PV = C.F. P(n, i)
Here, C.F.= ₹ 12000, i = 10% = \(\frac{10}{100}\) = 0.10, n = 5
and P(5, 0.10) = 3.79079
∴ PV = 12000 × 3.79079
= ₹ 45489.48
Which is less than the initial.

Section – C

All questions are compulsory. In case of internal choices attempt any one.

Question 27.
The area of a triangle is 5 sq. units and two of its vertices are (2, 1) and (3, – 2). If the third vertex is (x, y), where y = x + 3, then find the co-ordinates of the third vertex. [3]
Answer:
The area of a triangle with vertex (x1, y1), (x2, y2) and (x3, y3) is
Δ = \(\frac{1}{2}\)[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Here x1 = 2, x2 = 3, x3 = x
y1 = 1, x2 = 3, x3 = x + 3
Δ = \(\frac{1}{2}\)[2(-2 – x – 3) + 3(x + 3 – 1) + x(1 + 2)]
5 = \(\frac{1}{2}\)[2(-5 – x) + 3(x + 2) + 3x]
⇒ 10 = |-10 – 2x + 3x + 6 + 3x|
⇒ |4x – 4| = 10
⇒ 4x – 4 = 10 or 4x – 4 = -10
⇒ x = \(\frac{7}{2}\) or x = –\(\frac{3}{2}\)
y = x + 3
⇒ y = \(\frac{13}{2}\) or y = \(\frac{3}{2}\)
y = x + 3
∴ (x, y) = (\(\frac{7}{2}\), \(\frac{13}{2}\)) or (-\(\frac{3}{2}\), \(\frac{3}{2}\))

OR

Given, two diameters of a circle lie along the lines
x – y – 9 = 0 …. (i)
and x – 2y – 7 = 0 …(ii)
So, their point of intersection is the centre of the circle.
Solving eqs. (i) and (ii), simultaneously, we get
x = 11 and y = 2
∴ The centre of the circle is (11, 2).
Let r be the radius of the circle, then
area = πr2 = 385 sq units (given)
⇒ \(\frac{22}{7}\)r2 = \(\frac{77}{2}\)
⇒ r2 = \(\frac{49}{4}\)
⇒ r = \(\frac{7}{2}\)
The equation of the circle is
(x – 11)2 + (y – 2)2 = (\(\frac{7}{2}\))2
⇒ x2 + y2 – 22x – 4y + 125 = \(\frac{49}{4}\)
⇒ x2 + y2 – 22x – 4y + 125 – \(\frac{49}{4}\) = 0
⇒ 4(x2 + y2) – 88x – 16y + 451 = 0.

Commonly Made Error
Sometimes shidents don’t know how to calculate centre using equations of diameter.

Answering Tip
All the diameters of the circle intersect at the centre. So, the intersection point of the given equations is centre.

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 28.
Evaluate:
(A) 00101001 × 00000110
(B) 10000111 ÷ 00000101  [3]
Answer:
(A) 00101001 × 00000110
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 11

(B)
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 12
Thus, 10000111 ÷ 00000101 = 00011011

Question 29.
Out of 25 members in a family, 12 like to take tea, 15 like to take coffee and 7 like to take coffee and tea both. How many like (A) at least one of the two drinks (B) only tea but not coffee (C) only coffee but not tea.  [3]
Answer:
Given that, n(T) = 12
n(C) = 15
n(T ∩ C) = 7

(i) n(T ∪ C) = n(T) + n(C) – n(T ∩ C)
= 12 + 15 – 7
n(T ∪ C) = 20
20 members like at least one of the two drinks.

(ii) Only tea but not coffee
n(T) – n(T ∩ C)
= 12 – 7
= 5

(iii) Only coffee but not tea
= n(C) – n(T ∪ C)
= 15 – 7
= 8

Commonly Made Error

Students get confused between the symbol of union and intersection which leads to calculation errors.

Answering Tip
Adequate practice is necessary on application of set theory, so that confusion between the formulae and symbols gets clear.

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 30.
If \(x \sqrt{1+y}+y \sqrt{1+x}\) = 0 and x ≠ y prove that \(\frac{d y}{d x}\) = \(-\frac{1}{(x+1)^2}\)
OR
If x = a(cos 2θ + 2θ sin 2θ) and y = a (sin 2θ – 20 cos 2θ), find \(\frac{d^2 y}{d x^2}\) at θ = \(\frac{\pi}{8}\).  [3]
Answer:
Given, equation is, \(x \sqrt{1+y}+y \sqrt{1+x}\) = o
\(x \sqrt{1+y}=\) = \(-y \sqrt{1+x}\)
On squaring both sides, we get
x2(1 + y) = y2(1 + x)
x2 + x2y = y2 + y2x
⇒ x2 – y2 = y2x – x2y
⇒ (x + y)(x – y) = -xy(x – y)
Now, consider y + xy + x = 0
or y(1 + x) = -x
or y = \(\frac{-x}{1+x}\) …. (i)
On differentiating both sides w.r.t. x, we get
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 13

x = a(cos 2θ + 20 sin 2θ)
⇒ \(\frac{d x}{d \theta}\) = a(-2 sin 2θ + 2 sin 2θ + 4θ cos 2θ)
⇒ \(\frac{d x}{d \theta}\) = a(4θ cos 2θ)
y = a(sin 2θ – 2θ cos 2θ)
⇒ \(\frac{d x}{d \theta}\) = a(2 cos 2θ + 4θ sin 2θ – 2 cos 2θ)
\(\frac{d x}{d \theta}\) = a(4θ sin 2θ)
Using (1) and (2),
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 14
Differentiating again with respect to x, we get
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 15

Question 31.
Find the value of a and b from the following data:

Marks 0-5 5-10 10-15 15-20 20-25 Total
No. of students 7 A 25 30 B 100

Given that the third decile is 11.  [3]
Answer:
We construct the cumulative frequency table as under:

Marks 0-5 5-10 10-15 15-20 20-25
No. of students 7 a 25 30 b
Cumulative frequency 7 7 + a 32 + a 62 + a 62 + a + b

As the total number of students is 100,
62 + a + b = 100 ⇒ a + b = 38
Third decide, D3 is \(\frac{3}{10}\) of 100th i.e., 30th value. Given third decile is 11, so it lies in the class 10 – 15.
Using formula,
D3 = l + \(\frac{\frac{3}{10} n-c}{f}\) × i, we get
⇒ 11 = 10 + \(\frac{30-(7+a)}{25} \times 5\)
⇒ 1 = \(\frac{23-a}{25}\) ⇒ 5 = 23 – a ⇒ a = 18
From eq. (i) b = 38 – a = 38 – 18 = 20
Hence, a = 18 and b = 20

Question 32.
The marked price of the goods is ₹ 5,000 and rate of GST on it 18%. A shopkeeper buys the goods at a reduced price and sells it at its marked price. If the shopkeeper paid ₹ 144 as CGST to the government, find the amount (inclusive of GST) paid by the shopkeeper.  [3]
Answer:
Let the shopkeeper buys the goods at x.
∴ Cost price = ₹ x
and Selling price = ₹ 5,000
According to the question,
CGST paid = ₹ 144
and SGST paid = ₹ 144
Total GST = ₹ (144 + 144) = ₹ 288
i.e., GST on selling price – GST on cost price = ₹ 288
\(\frac{18}{100}\) × 5000 – \(\frac{18}{100}\) × x = 288
\(\frac{18}{100}\) × x = 900 – 288
x = \(\frac{612 \times 100}{18}\)
x = ₹ 3,400
∴ The shopkeeper buys the goods for ₹ 3400.
Total amount paid by the shopkeeper = Cost price of the goods + GST
= ₹(3400 + \(\frac{18}{100}\) × 3400)
= ₹ 4012

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 33.
Suppose that certain amount P is invested at an annual rate of 6.5%, compounded annually. How long will it take for the amount to triple ?  [3]
[Use, loge 3 = 1.0986]
Answer:
We use the tormul&
A = Pert
Given, A = 3P, r = 6.5% = \(\frac{6.5}{100}\) = 0.065
Therefore 3P = Pe0.065t
⇒ 3 = e0.065t
Taking log both sides, we get
loge3 = logee0.065t
⇒ loge3 = 0.065 t [∵ loge = 1]
⇒ t = \(\frac{\log _e 3}{0.065}\)
⇒ t = \(\frac{1.0986}{0.065}\) = 16.9015 ≅ 16.9
Therefore, it would take 16.9 years for any initial investment P to triple.

Section – D

All question are compulsory. In case of internal choices attempt any one.

Question 34.
Draw the graph of following function and find range (Rf) of f(x) = |x – 2| + |2 – x| ∀ -3 < x ≤ 3. [5]
Answer:
Given, f(x) = |x – 2| + |2 – x| ∀ -3 ≤ x ≤ 3.
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 16
Range of f = {4, 6}

Question 35.
Three persons A, B and C apply for a job of Manager in Private company. Chance of their selection (A, B and C) are in the ratio 1: 2: 4. The probability that A, B and C can introduce changes to improve profits of company are 0.8, 0.5 and 0.3 respectively, if the changes does not take place, find the probability that it is due to the appointment of C.
OR
Bag I contains 1 white, 2 black and 3 red balls; Bag II contains 2 white, 1 black and 1 red balls; Bag III contains 4 white, 3 black and 2 red balls. A bag is chosen at random and two balls are drawn from it with replacement. They happen to be one white and one red. What is probability that they came from Bag III.  [5]
Answer:
Let the events be described as below :
A : No change takes place
E1: Person A gets appointed
E2: Person B gets appointed
E3 : Person C gets appointed
The chances of selection of A, B and C are in the ratio 1 : 2 : 4.
Hence,
P(E1) = \(\frac{1}{7}\) ;P(E2) = \(\frac{2}{7}\); P(E3) = \(\frac{4}{7}\)
Probability of A, B and C introducing changes to improve profits of company are 0.8, 0.5 and 0.3 respectively.
Hence, probability of no changes on appointment of A, B and C are 0.2, 0.5 and 0.7 respectively.
Hence,
P(A/E1) = 0.2 = \(\frac{2}{10}\); P(A/E2) = 0.5 = \(\frac{5}{10}\);
P(A/E3) = 0.7 = \(\frac{7}{10}\)
Therefore, required probability i.e., (E3/A) is P(E3/A)
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 17
∴ If no change takes place, the probability that it is due to appointment of C is \(\frac{7}{10}\)

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

Question 36.
(a) For an industrial connection monthly consumption of electricity units are 550, calculate the electricity bill. Tariff rates can be considered as the table given below: [5]

Unit-slab Rate per unit(in ₹) Fixed Charges
1-300 7.50 330
301-500 8.40 390
501 and above 8.75 450

(b) For a domestic connection monthly consumption of electricity units are 275, calculate the electricity bill. Tariff rates can be considered as the table given below:

Unit slab Rate per unit (in ₹) Fixed charge
1-150 5.50 110
151-300 6.00 125
301-500 6.50 187
501 and above 7.00 221

Electricity duty is considered as 5%.
OR
(a) There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of ₹ 12,000 after 3 years at the same rate ?
Answer:
(a)
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 18
Thus, final bill amount is ₹ 5058.38

(b)
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 19
Thus, final bill amount is ₹ 1785

OR

(a) Let, P 100
Then, simple interest, I = 100 and t = 6 years
∴ simple interest, I = P × i × t
⇒ i = \(\frac{I}{P \times t}\) = \(\frac{100 \times 60}{100 \times 6}\)
= 10% p.a.
Now, P = ₹ 12,000
t = 3 years and i = 10% p.a.
Compound interest = P(1 + i)n – P
= 12000\(\left[\left(1+\frac{10}{100}\right)^3-1\right]\)
= 12000 × \(\frac{331}{1000}\)
= 3972

(b) Compound interest, when interest compound yearly
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 20
Compound interest, when interest compounded half yearly
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 21
Difference = ₹ (5306,04 – 5304)
= ₹ 2.04

Section – E

Both the Case study based questions are compulsory. Each Sub parts carries 1 mark.

Question 37.
In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line (see figure below).
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 1
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. [4]

According to the above information, answer the following questions:
(a) What is the distance travelled by competitor to pick 1st potato ?
(A) 10 m
(B) 16 m
(C) 22 m
(D) 48 m
Answer:
Option (A) is correct.
Explanation:
The distance travelled by the competitor to pick 1st potato = 2 × 5 = 10

(b) What is the distance travelled by competitor to pick 2nd potato ?
(A) 10 m
(B) 16 m
(C) 22m
(D) 48 m
Answer:
Option (B) is correct
Explanation:
The distance travelled by the competitor to pick 2nd potato = 2 × (5 + 3) = 16m

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

(c) What is the distance travelled by competitor to pick 3rd potato ?
(A) 10 m
(B) 16 m
(C) 22 m
(D) 48 m
Answer:
Option (C) is correct.
Explanation:
The distance travelled by the competitor to pick 3rd potato = 2 × (5 + 6) = 22 m

(d) The given problem is based on which concept ?
(A) AP
(B) GP
(C) HP
(D) None of these
Answer:
Option (A) is correct.
Explanation:
The numbers involved in this case is in AP in which a = 10, d = 6, n = 10

Question 38.
Shyam and Ekta live in the same complex. There is a shopping mall in the complex. Both Shyam and Ekta visit a particular grocery shop in the same week (Tuesday to Saturday), in the mall. Each is equally likely to visit the shop on any day as on another day. [4]
CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions 2

According to the above information, answer the following questions:

(a) What are the total numbers of favourable outcomes ?
(A) 5
(B) 10
(C) 25
(D) 20
Answer:
Option (C) is correct.
Explanation:
Since, total favourable outcomes associated to the random experiment of visiting a particular shop in the same week (Tuesday to Saturday) by two customers Shyam and Etka are:
(T, T) (T W) (T, Th) (T, F) (T, S)
(W, T) (W, W) (W, Th) (W, F) (W, S)
(Th, T) (Th, W) (Th, Th) (Th, F) (Th, S)
(F, T) (F, W) (F, Th) (F, F) (F, S)
(S, T) (S, W) (S, Th) (S, F) (S, S)
Where, T = Tuesday, W = Wednesday, Th = Thursday, F = Friday, S = Saturday.
Total number of favourable outcomes 5 × 5 = 25

(b) What is the probability that both will visit the shop on same day ?
(A) 1/5
(B) 4/5
(C) 2/5
(D) 8/25
Answer:
Option (A) is correct
Explanation:
The favourable outcomes of visiting on the same day are (T, T), (W, W), (TH, TH), (F, F) and (S, S).
Number of favourable outcomes = 5
Hence, required probability = \(\frac{5}{25}\) = \(\frac{1}{5}\)

(c) What are the total number of favourable outcomes if both will visit the shop on consecutive day?
(A) 5
(B) 10
(C) 8
(D) 25
Answer:
Option (C) is correct
Explanation:
The favourable outcomes of visiting on
consecutive days arc (T, W), (W, T), (W, Th), (Th, W), (Th, F), (F, Th), (S, F) and (F, S).
Number of favourable outcomes = 8

CBSE Sample Papers for Class 11 Applied Mathematics Set 7 with Solutions

(d) What is the probability that both will visit the shop on consecutive day ?
(A) 1/5
(B) 1/10
(C) 2/5
(D) 8/25
Answer:
Option (D) is correct.
Explanation:
Number of favourable outcomes = 8
Hence, required probability = \(\frac{8}{25}\)