Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 8 are designed as per the revised syllabus.
CBSE Sample Papers for Class 11 Applied Mathematics Set 8 with Solutions
Maximum Marks : 80
Time Allowed : 3 hours
General Instructions :
- All the questions are compulsory.
- The question paper consists of 38 questions divided into 5 sections A, B, C, D and E.
- 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.
- 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.
- Use of calculators is not permitted.
Section – A
All questions are compulsory. In case of internal choices attempt any one.
Question 1.
Find the sum of binary numbers 111 + 100. [1]
Answer:
Question 2.
Simplify : ya-b × yb-c × yc-a × ya-b. [1]
OR
Find the value of log\(\log _{\sqrt{2}} 64\)
Answer:
Question 3.
Suppose an investor wants to have 1 lakh to retire 5 years from now. How much would he have to invest today with an annual rate of return equal to 15 %? [1]
Answer:
We know that,
Present Value, PV = \(\frac{A_n}{(1+i)^n}\)
Here, An = 1 lakh = ₹ 100000
i = 15%, \(\frac{15}{100}\) = 0.15 n = 5 years
∴ PV = \(\frac{100000}{(1+0.15)^5}\) = \(\frac{100000}{(1.15)^5}\)
= \(\frac{100000}{2.0113}\)
= ₹ 49,719.08
Thus, he would invest ₹ 49,719.08 today.
Question 4.
A dealer in Bhopal (M.P.). He supplies products and services worth ₹ 5,000 to another dealer in Kanpur (UP). If the rate of GST is 28%, find the tax levied under CGST. [1]
Answer:
Since, the given case in of inter-state transaction as the products/services is supplied from one state to another.
∴ CGST = 0.
Question 5.
If in an A.P., 7th term is 9 and 9th term is 7, then find 16th term.
OR
If a, b and c are in G.P., then the value of \(\frac{a-b}{b-c}\) is equal to……….. [1]
Answer:
Let first term and common difference of A.P is a and d, respectively.
Given, T7 = 9 = a + (7 – 1) d
⇒ 9 = a + 6d
and T9 = 7 = a + (9 – 1)d
⇒ 7 = a + 8d …..(ii)
On solving equations (i) and (ii), we get
a = 15 and d = -1
Now T16 = a + (16 – 1)d
15 + 15(-1) = 0
OR
Since, a, b and c are in G.P
∴ \(\frac{b}{a}\) = \(\frac{c}{b}\) = r (common ratio)
⇒ b = ar and c = br
⇒ c = ar.r = ar2
= \(\frac{1}{r}\) = \(\frac{a}{b}\) = \(\frac{b}{c}\)
Question 6.
Evaluate \(\lim _{n \rightarrow \infty} \frac{1+2+3+\ldots+n}{n^2}\) [1]
Answer:
Commonly Made Error
Some students do not apply correct expression of sum of natural numbers.
Answering Tip
Obtaining the result in the required form needs simplification, multiplication, substitution etc.
Question 7.
Write the set of all positive integers whose cube is odd in the builder form. [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 0f 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}.
Question 8.
Showing a lady in the park, Vineet said, “She is the daughter of my grand father’s only son.” How is Vineet related to that lady ? [1]
Answer:
The only son of Vineet’s grand father is the father of Vineet and Daughter of Vineet’s father is the sister of Vineet. Hence, Vineet is the brother of the lady.
Question 9.
Let f be the subset of Z × Z defined by
f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z ? Justify your answer. [1]
Answer:
Given, f = {(ab, a + b): a, b ∈ Z}
Taking a = b = 1, we have (ab, a + b) = (1, 2) ∈ f
Taking a = b = -1, we have (ab, a + b)
= (1, -2) ∈ f
⇒ f-image of 1 is not unique
Hence, f is not a function.
Question 10.
How many different words (with or without meaning) can be made using all the vowels at a time ?
OR
15C8 + 15C9 – 15C6 – 15C7 = ______ [1]
Answer:
There are 5 vowels in 26 alphabets.
Hence, using all 5 vowels at a time, number of different words (with or without meaning) can be made are = 5! = 5 × 4 × 3 × 2 × 1 = 120.
OR
15C8 + 15C9 – 15C6 – 15C7
= 15C15-8 + 15C15-9 – 15C6 – 15C7 [∵ nCr = nCr-1]
= 15C7 + 15C6 – 15C6 – 15C7
= 0.
Question 11.
One number is chosen at random from the number 1 to 21. What is the probability that it is prime. [1]
Answer:
Sample space n(S) = 21
Prime numbers from 1 to 21 are 2, 3, 5, 7, 11, 13, 17, 19.
If ’E’ be the event of getting a prime number, then n(E) = 8
∴ P(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{8}{21}\)
∴ The probability that the number is prime
= \(\frac{8}{21}\)
Question 12.
Show that for any numbers a and b (S.D.) is given by \(\frac{|a-b|}{2}\). [1]
Answer:
For two numbers a and b, given by \(\bar{x}\) = \(\frac{(a+b)}{2}\)
∴ Standard deviation = \(\frac{|a-b|}{2}\)
Question 13.
A can do a certain job in 12 days, B is 60% more efficient than A. How many days does B alone take to do the same job ?
OR
2.2 cubic dm of lead is to be drawn into a cylindrical wire 0.50 cm in diameter. Find the length of the wire in metres. [1]
Answer:
Ratio of times taken by A and B = 160 : 100 = 8 : 5. Suppose B alone take x days to do the same job. Then, 8 : 5 :: 12 : x
⇒ \(\frac{8}{5}\) = \(\frac{12}{x}\)
⇒ x = \(\frac{12 \times 5}{8}\) = \(\frac{60}{8}\) = 7\(\frac{1}{2}\) days
Hence, B can alone to the job in 7\(\frac{1}{2}\) days.
OR
Let the length of the wire be h meters. Then, πr2h = 2.2 cubic dm
i.e., π\(\left(\frac{0.50}{2 \times 100}\right)^2\) × h = \(\frac{2.2}{1000}\)
and 1 cubic dm = metre]
[d = 0.5 cm ⇒ r = \(\frac{0.5}{2}\) cm or \(\frac{0.5}{2 \times 100}\)m and 1 cubic dm = \(\frac{1}{1000}\) meter}
or, h = \(\left(\frac{2.2}{1000} \times \frac{100 \times 100}{0.25 \times 0.25} \times \frac{7}{22}\right)\)
= 112 m
Hence, length of the wire is 112 metres.
Question 14.
Show that the points (a, 0), (0, b) and (3a, – 2b) are collinear. [1]
Answer:
Let the given points be P(x1, y1) = (a, 0), Q(x2, y2) = (0, b) and R(x3, y3) = (3a, -2b).
∴ Area of ΔPQR
Thus, points (a, 0), (0, b) and (3a, – 2b) are collinear.
Question 15.
Find the centre and the radius of the circle 2x2 + 2y2 + 10x – 6y – 1 = 0. [1]
Answer:
Given equation of circle is 2x2 + 2y2 + 10x – 6y – 1 = 0
One dividing by 2, the equation of circle can be written as
x2 + y2 + 5x – 3y – \(\frac{1}{2}\) = 0
On comparing above equation with
x2 + y2 + 2gx + 2fy + c = 0, we get
g = \(\frac{1}{2}\)(coefficient of x) = \(\frac{1}{2}\)(5) = \(\frac{5}{2}\)
f = \(\frac{1}{2}\)(coefficient of y) = \(\frac{1}{2}\)(-3) = –\(\frac{3}{2}\)
and c = constant term = –\(\frac{1}{2}\)
Hence, the centre of the circle is (-g, -f) i.e., \(\left(-\frac{5}{2}, \frac{3}{2}\right)\) and radius of the circle = \(\sqrt{g^2+f^2-c}\)
Question 16.
If set A = {-2, 2} and B = {x : x OR
Let n{A) = m, and n(B) = n. Then find the total number of possible relations that can be defined from A to B. [1]
Answer:
Given
A = {- 2, 2}
and B = {x : x ∈ I, x2 – 4 = 0},
∴ x2 – 4 = 0
⇒ (x – 2) (x + 2) = 0
⇒ x = -2, 2
or, B = {- 2, 2}
Hence, A = B
Commonly Made Error
Some students commit errors in applying correct sign convention while square root of 4. So, it leads to prove the result unequal.
Answering Tip
Solve set B properly, so it results equal to set A.
OR
Given, n(A) = m, and n(B) = n
∴ n(A × B) = n(A). n(B) = mn
So, the total number of non-empty relations from A to B = 2mn
Section – B
All questions are compulsory. In case of internal choices attempt any one.
Question 17.
Divide (1001011)2 by (11)2. [2]
Answer:
Hence, quotient is (11001)2
Question 18.
Simplify : \(\frac{2 a^{\frac{1}{2}} \times a^{\frac{2}{3}} \times 6 a^{-\frac{7}{3}}}{9 a^{-\frac{5}{3}} \times a^{\frac{3}{2}}}\), if a = 4 [2]
Answer:
Question 19.
The calendar for the year 2007 will be same for which year ?
OR
Four men A, B, C and D and four women W, X, Y and Z are sitting round a table facing each other,
(A) No two men or women are sitting together
(B) W is the right to B
(C) Y is facing X and is to the left of A
(D) C is to the right of Z.
Find, who are the two persons sitting adjacent to D ? [2]
Answer:
Count the number of odd days from the year 2007 onwards to get the sum equal to 0 odd day.
Sum = 14 odd days= 0 odd day
∴ Calender for the year 2018 will be same as for the year 2007.
The above figure shows the sitting arrangement of the 8 persons.
OR
It is clear from the figure. W and Y are adjacent to D.
Question 20.
What is represented by the shaded regions in each of the following Venn-diagrams.
(A)
(B)
[2]
Answer:
(i)
∴ (A – B) ∪ (B – A)
(ii)
∴ (A ∩ B) ∪ (A ∩ C)
or A ∩ (B ∪ C)
Question 21.
In how many ways can the letter of the word “PENCIL” be arranged so that I is always next to L. [2]
Answer:
There are 6 letters in the word “PENCIL”. Consider LI as one letter. Now, 5 letters (P, E, N, C, .,. LI) can be arranged in = 5! = 120 ways.
Hence, the total number of ways in which I is always next to L is 120.
Commonly Made Error
Sometimes students take permutation of 5 letters arrangement as 6P5 instead of 5P5 which leads to incorrect result.
Answering Tip
Deeply understand the fundamental principles of counting.
Question 22.
Find the 20th term and the nth term of the sequence \(\frac{5}{2}\), \(\frac{5}{4}\), \(\frac{5}{8}\), …….
OR
The straight lines l1, l2, l3 are parallel and lie in the same plane. A total number of m points are taken on l1; n points on l2; k points on l3. Find the maximum number of triangles formed with vertices at these points. [2]
Answer:
The given sequence \(\frac{5}{2}\), \(\frac{5}{4}\), \(\frac{5}{8}\), ….. is a G.P. with first
term, a = \(\frac{5}{2}\) and common ratio r = \(\frac{1}{2}\)
OR
Here, total number of points are (m + n + k) which must give m + n + kC3 number of triangles but m points on line l1 taking 3 points at a time gives mC3 combinations which produce no triangle. Similarly, nC3 and kC3 number of triangles cannot be formed. Therefore, the required number of triangle is
m + n + kC3 – mC3 – nC3 – kC3
Question 23.
In each of the following question, a few statements, followed by four conclusions numbered I, II, III and IV are given. You have to consider every given statement as true, even if it dose not confirm to the well known facts. Read the conclusions and then decide which of the conclusions can be logically derived.
(a) Statements:
(A) Some toys are pens.
(B) Some pens are papers.
(C) Some papers are black.
Conclusions:
I. Some toys are black.
II. No pen is black.
III. No toy is black.
IV. Some pens are black.
I. (b) Statements:
(A) Some books are copies.
(B) All copies are green.
(C) Some green are yellow.
Conclusions:
I. All copies are yellow.
II. Some yellow are green.
III. Some copies are yellow.
IV. All green are copies. [2]
Answer:
(a) All the conclusions given are mediate inferences. Since, middle term is not distributed in any one pair of the above statements. But conclusion II and IV form a complementary pair from statements 2 and 3 and conclusion I and II form a complementary pair from statements (i), (ii) and (iii). Hence, either I or III and either II or IV follow.
(b) Conclusions I and II are mediate inferences drawn from statements (ii) and (iii) Since, middle term ‘green’ is not distributed, hence both conclusions do not follow from statements.
Conclusion II is immediate inference drawn from statement (iii). Also, conclusions III and IV do not follow. Hence, only conclusion II follows.
Question 24.
If f(x) = y = \(\frac{a x-b}{c x-a}\) then prove that f(y) = x. [2]
Answer:
Given,
∴ f(y) = x
Hence proved.
Question 25.
Show that \(\lim _{x \rightarrow 4} \frac{|x-4|}{x-4}\) does not exist.
OR
Differentiate ex sin x + xn cos x with respect to x. [2]
Answer:
Let x – 4 = n ⇒ As x → 4, n → 0
OR
Let, f(x) = ex sin x + xn cos x
∴ f'(x) = \(\frac{d}{d x}\){ex sin x + xn cos x}
= Sin x ex + ex cos x + cos x. nxn-1 + xn.(-sin x)
= ex(sin x + cos x) + xn-1 [n cos x – x sin x]
Question 26.
The monthly income ₹ 1000 of 8 persons working in a factory. Find P30 income value 17, 21, 14, 36, 10, 25, 15, 29. [2]
Answer:
Arrange the data in the increasing order:
10, 14, 15, 17, 21, 25, 29, 36
Here, n = 8
P30 = \(\left[\frac{30(n+1)}{100}\right] \text { th }\) item
= \(\frac{30 \times 9}{100} \text { th }\) item = 2.7th item
= \(\frac{30 \times 9}{100}\)th item = 2.7th item
= 2nd item + 0.7(3rd item – 2nd item)
= 14 + 0.7(15 – 14)
= 14 + 0.7
= 14.7
Section – C
All questions are compulsory. In case of internal choices attempt any one.
Question 27.
Coefficient of variation of two distributions are 50 and 60 and their arithmetic means are 30 and 25 respectively. Find the difference of their standard deviation.
OR
Find the first four moments about zero for the set of the number 1, 3, 5, 7. [3]
Answer:
Given, C.V.1 = 50, C.V.2 = 60
\(\bar{x}_1\) = 30, \(\bar{x}_2\) = 25
We know:
⇒ σ1 = 15
and \(\bar{x}_2\) = \(\frac{\sigma_2}{C_{.} V_{\cdot 2}} \times 100\)
⇒ 25 = \(\frac{\sigma_2}{60} \times 100\)
⇒ σ2 = \(25 \times \frac{3}{5}\)
σ2 = 15
⇒ Required difference = σ1 – σ2
= 15 – 15
= 0
OR
The first four moment about zero means, first four raw moments about origin
| X | X | X2 | X3 | X4 |
| 1 | 1 | 1 | 1 | 1 |
| 3 | 3 | 9 | 27 | 81 |
| 5 | 5 | 25 | 125 | 625 |
| 7 | 7 | 49 | 343 | 2401 |
| ΣX = 16 | ΣX2 = 84 | Σx3 = 496 | Σx4 = 3108 |
Question 28.
‘A’ earns the following income during the financial year 2018-19 :
| Particulars | Amount (₹) |
| (a) Interest paid by an Indian company but received in London | 2,00,000 |
| (b) Pension from former employer in India, received in USA | 8,000 |
| (c) Profits earned from business in Paris which is controlled in India, half of the profits being received in India | 40,000 |
| (d) Income from agriculture in Bhutan and remitted to India | 10,000 |
| (e) Income from property in England and received there | 8,000 |
| (f) Past foreign untaxed income brought to India | 20,000 |
Determine the total income of ‘A’ for the assessment year 2019-20 if he is
(A) Resident and ordinarily resident,
(B) Not ordinarily resident, and
(C) Non-resident in India. [3]
Answer:
| Resident and ordinarily resident | Not ordinarily resident | Non-resident | |
| (₹) | (₹) | (₹) | |
| (a) | 2,00,000 | 2,00,000 | 2,00,000 |
| (b) | 8,000 | 8,000 | 8,000 |
| (c) | 40,000 | 40,000 | 20,000 |
| (d) | 10,000 | – | – |
| (d) | 8,000 | Nil | Nil |
| (e) | 8,000 | Nil | Nil |
| (f) | 2,66,000 | 2,48,000 | 2,28,000 |
Question 29.
In what ratio the line joining (-1, 1) and (5, 7) is divided by the line x + y = 4 ?
OR
Find the image of the point (3, 8) with respect to the line x + 3y = 7 assuming the line to be a plane mirror. [3]
Answer:
Let the given points be A(x1, y1) = (-1, 1) and B(x2, y2) = (5, 7) and P(x3, y3) be the point which divides AB in the ratio m : n.
∴ Co-ordinates of P are \(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)
= \(\left(\frac{5 m-n}{m+n}, \frac{7 m+n}{m+n}\right)\)
Since, the point P lies on line x + y = 4.
∴ \(\frac{5 m-n}{m+n}\) + \(\frac{7 m+n}{m+n}\) = 4
⇒ \(\frac{12 m}{m+n}\) = 4
⇒ 8m = 4n
⇒ m : n = 1 : 2
Commonly Made Error
Students get confused between internal and external division formulae specially in between signs (+ or -).
Answering Tip
If a point P divides the line segments joining the points A(x1, y1) and Q(x2, y2) internally in the ratio m: n, then its coordinates are
\(\left(\frac{m x_2+n x_1}{m+n}, \frac{m y_2+n y_1}{m+n}\right)\)
OR
Let the image of point P(3, 8) be Q(a, b) with respect to line AB which is given by
x + 3y – 7 = 0
The mid-point of P and Q is \(\left(\frac{3+a}{2}, \frac{8+b}{2}\right)\) which lies on line (i).
From (i), \(\frac{3+a}{2}\) + 3.\(\frac{8+b}{2}\) – 7 = 0
⇒ 3 + a + 24 + 3b – 14 = 0
⇒ a + 3b + 13 = 0
Now, slope of AB = \(-\frac{1}{3}\)
∴ \(-\frac{1}{3} \times \frac{b-8}{a-3}\) = -1
⇒ 3a – b – 1 = 0 ….(iii)
Solving (ii) and (iii), we arrive at a = -1, b = -4
∴ Image of (3, 8) is (-1, -4).
Question 30.
Find the equation of the circle drawn on a diagonal of the rectangle as its diameter whose sides are the lines x = 4, x = – 5, y = 5 and y = – 1. [3]
Answer:
Let ABCD be the rectangle formed by the lines
x = 4
x = -5 ……(ii)
y = 5 …..(iii)
y = -1 …..(iv)
The coordinates of the points A, B, C and D are (-5, -1), (4, -1), (4, 5) and (-5, 5), respectively.
The equation of the circle having the diagonal AC as its diameter is
[(x – (-5)](x – 4) + [y – (-1)(y – 5) = 0
[∵ Equation of circle having ends of diameter
(x1, y1) and (x2, y2) is (x – x1) + (y – y1)(y – y2) = 0]
or, (x + 5)(x – 4) + (y + 1)(y – 5) = 0
or, x2 + y2 + x – 4y – 25 = 0.
Commonly Made Errors
Some students do not plot the figure correctly. Some are confused between diagonals are diameters.
Answering Tap
Since, ∠ABC = 90° = ∠ADC, the above circle passes through both points B and D. Hence, it is also the circle with diagonal BD as a diameter. In fact, the circle circumscribes the rectangle ABCD.
Question 31.
Let A = {1, 2, 3}, B = {3, 4} and C = {4, 5, 6}. Find
(A) A × (B ∩ C) and
(B) (A × B) ∩ (A × C) [3]
Answer:
Given, A = {1, 2, 3}, B{3, 4}, C = {4, 5, 6},
∴ B ∩ C = {4}
(i) A × (B ∩ C) = {1, 2, 3) × {4} = {(1, 4), (2, 4), (3, 4)}
(ii) (A × B) ∩ (A × C)
({1, 2, 3) × {3, 4}) ∩ ({1, 2, 3) × {4, 5, 6}
= ((1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3,4) ∩ {(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)}
= {(1, 4), (2, 4), (3, 4)}
Commonly Made Error
Many candidates do not know how to do cartesian product in case of three sets.
Answering Tip
Learn the properties of cartesian product of sets and adequate practice should be done on various types of problems.
Question 32.
A polygon has 35 diagonals. Find the number of its sides. [3]
Answer:
Let n be the number of sides of a polygon and D be the number of diagonals of that polygon.
We know that, D = nC2 – n = \(\frac{n(n-3)}{2}\)
∴ 35 = \(\frac{n^2-3 n}{2}\)
⇒ n2 – 3n – 70 = 0
⇒ (n – 10)(n + 7) = 0
⇒ n = 10, -7
Since, sides cannot be negative, therefore n = 10.
Hence, polygon is a decagon.
Commonly Made Error
Students are not able to identify that the given problem is from which topic.
Answering Tip
Remember are relation for number of diagonals of polygon, D = nC2 – n or \(\frac{n(n-3)}{2}\)
Question 33.
In the following table some capital alphabets are written in a row, below them their coding have been given. Now in questions a particular word has been coded in a particular manner using codes as given below the capital letter. You have to understand the pattern of coding and answer the following questions. 
(A) If DASF1 is coded as 2 a 8 4, then what is the code for SMASH ?
(B) If FASHION is coded as aat4cx, then what is the code for POSITION ?
(C) If DEER is coded as vw8 and TOSS is coded as e6e, then what is the code for DOTT ? [3]
Answer:
(i) In the word DASH, letters at the odd places have been moved one place back and letters at the even places have been moved one place ahead regarding the position of codes given in the table. Therefore, SMASH can be coded as 8zqe3.
(ii) The codes for the respective letters of the FASHION as per table have been reversed in the coding. So, POSITION can be coded as z64e47c.
(iii) In both the words DEER and TOSS, codes for each letters have been used just below the letters in the table except for the letter that has been written twice i.e., E and S. For these letters, codes have been used just one position ahead in the table that too only one. Therefore, DOU can be codded as V6r.
Section – D
All questions are compulsory. In case of internal choices attempt any one.
Question 34.
If the A.M. between pth and qth terms of an A.P. be equal to A.M. between rth and sth terms of the A.P., then show that p + q = r + s. [5]
Answer:
Let a be the first term and d be the common difference of the given A.P Then
ap = pth term = a + (p – 1)d;
aq = qth term a + (q – 1)d;
ar = rth term a + (r – 1)d;
and as = 5th term = a + (s – 1)d;
It is given that
A.M. between ap and ap = A.M. between ar and as
⇒ \(\frac{1}{2}\)(ap + aq) = \(\frac{1}{2}\)(ar + as)
ap + aq = ar + as
⇒ a + (p – 1)d + a + (q – 1)d
⇒ a + (r – 1)d + a + (s – 1)d
⇒ (p + q – 2)d = (r + s – 2)d
⇒ p + q = r + s
Question 35.
In a college, 70% students pass in Physics, 75% pass in Mathematics and 10% students fail in both. One student is chosen at random. What is the probability that:
(A) He passes in Physics given that he passes in Mathematics.
(B) He passes in Mathematics given that he passes in Physics.
OR
Bag A contains 2 white, 1 black and 3 red balls. Bag B contains 3 white, 2 black and 4 red balls and Bag C contains 4 white, 3 black and 2 red balls.
In continuation with precious line random and 2 balls are drawn at random from that bag. If the randomly drawn balls happen to be red and black, what is the probability that both come from Bag B ? [5]
Answer:
Let x% students passes in both Mathematics and Physics
Students who pass in Physics = 70%
∴ P(R) = \(\frac{70}{100}\)
Students who pass in Mathematics = 75%
∴ P(M) = \(\frac{75}{100}\)
Students who fail in both = 10%
Now, 70% – x + x + 75% – x = 90%
⇒ x = 55%
∴ P(M ∩ P) = \(\frac{55}{100}\)
(i) Probability that student passes in Physics given that he passes in Mathematics is =
(ii) Probability that student passes in Mathematics given that he passes in Physics is
Let E1, E2 and E3 be the events of choosing bag A, bag B and bag C, respectively.
∴P(E1) = P(E2) = P(E3) = \(\frac{1}{3}\)
Let E be the event of drawing 1 red and t black ball from the bag, then
\(P\left(\frac{E}{E_1}\right)\) = Probability
(1 red and 1 black balls are drawn from Bag A)
= \(\frac{{ }^3 C_1 \times{ }^1 C_1}{{ }^6 C_2}=\frac{3}{15}=\frac{1}{5}\)
Similarly,
\(P\left(\frac{E}{E_2}\right)\) = Probability
(1 red and 1 black balls are drawn from Bag B)
= \(\frac{{ }^4 C_1 \times{ }^2 C_1}{{ }^9 C_2}=\frac{4 \times 2}{36}=\frac{2}{9}\)
and P\(\left(\frac{E}{E_3}\right)\) = Probability
(1 red and 1 black balls are drawn from Bag C)
= \(\frac{{ }^2 C_1 \times{ }^3 C_1}{{ }^9 C_2}=\frac{2 \times 3}{36}=\frac{1}{6}\)
Required Probability = Probability [2 balls (1 red and 1 black) are drawn from hag B ] is given by P\(\left(\frac{E_2}{E}\right)\)
Question 36.
Leela is an athlete who believes that her playing career will last 3 years.
(a) To prepared for future, she deposits ₹ 24,000 at the end of each year for 3 years in an account paying 6% compounded annually. How much will she have on deposit after 3 years ? Also, find the value of interest earned.
(b) Instead of investing ₹ 24,000 at the end of each year, suppose Leela deposits ₹ 80,000 at the end of each year for 3 years in an account paying 5%. compounded monthly. How much will she have on deposit after 3 years? Also, find the value of interest earned.
OR
Calculate the electricity bill amount for a month of 31 days if the following devices are used as specified :
3 bulbs of 40 watts for 6 hours
4 tube lights of 50 watts for 8 hours
Given rate of electricity is rupees 2.50 per unit. [5]
Answer:
(a) Her yearly payments form an annuity regular with C.F. = ₹ 24,000, Since, the interest is compounded annually, no. of conversion periods, n = 3 and the interest rate per period i = 6% = \(\frac{6}{100}\)
= 0.06.
Using the formula for the future value of an annuity
The interest earned = FV. – 3 × C.F.
= 76406.40 – 3 × 24,000
= 76406.40 – 72,000
= 4,406.40
(b) Interest is compounded semi-annually, so no. of conversion periods n = 3 × 1 = 3 and i = 5%
= \(\frac{5}{100}\) = 0.05, C.F. = ₹ 80,000
= 80,000 × 3.1525
= ₹ 252,200
The interest earned = F.V. – 3 × C.F.
= 252,200 – 3 × 80,000
= 252,200 – 240,000
= ₹ 12,200
OR
Electric energy consumed per day by 3 bulbs
= 40 × 3 = 120 watts
= 0.12kW
Electrical units (or one day by 3 bulbs
= 0.12 × 6 = 0.72 units
Electrical units for 31 days by 3 bulbs
0.72 × 31 = 22.32 units
Electric energy consumed per day by 4 tube lights
= 50 × 4 = 200 watts
= 0.2 kW
Electrical units for one day by tube lights
= 0.2 × 8 = 1.6 units
Electrical units for 31 days by 4 tube lights
= 1.6 × 31 = 49.6 units
Total units= 49.6 + 22.32 = 71.92 units
Electricity bill = 71,92 × 2.50 = ₹ 179.8
Section – E
Both the Case study based questions are compulsory. Each Sub parts carries 1 mark.
Question 37.
Jaspal belongs to a middle class family. He works in a company. His children wanted that their father purchase a car for the family. Jaspal takes a loan from a bank for his car.
Jaspal singh repays his total loan of ₹ 118000 by paying every month starting with the first installment of ₹ 1000. If he increase the installment by ₹ 100 every month. [4]
According to the above information, answer the following questions:
(a) If the given problem is based on AP; then the first term and common difference?
(A) 1000, 100
(B) 100, 1000
(C) 100, 100
(D) 1000, 1000
Answer:
Option (A) is correct.
Explanation:
The number involved in this case form an AP in which a = 1000, d = 100
(b) In how many months the loan will be cleared?
(A) 20
(B) 30
(C) 44
(D) 50
Answer:
Option (C) is correct.
Explanation:
Sn = 118000
⇒ Sn = \(\frac{n}{2}\)[2 × 1000 + (n – 1) × 100]
⇒ 118000 = n[1000 + 50n – 50]
⇒ 118000 = n[950 + 50n]
⇒ 11800 = 5n + 95n
⇒ n2 + 19n – 2360 = 0
⇒ (n + 59)(n – 40) = 0
⇒ n = 40
Thus, the loan will be cleared in 40 months.
(c) The amount paid by him in 30” installment is:
(A) ₹3900
(B) ₹3500
(C) ₹3000
(D) ₹3600
Answer:
Option (A) is correct.
Explanation:
The amount paid by him in 3oth installment is
t30 = a + 29d
= 1000 + 29 × 100
= ₹ 3900
(d) The amount paid by him in 30 installments is:
(A) ₹37000
(B) ₹73500
(C) ₹ 75300
(D) ₹ 53700
Answer:
Option (B) is correct.
Explanation:
The amount paid by him in 30 installments is
⇒ S30 = \(\frac{30}{2}\) [2 × 1000 + (30 – 1) × 100]
= 15[2000 + 2900]
= 15 × 4900
= ₹ 73500
Question 38.
Reenas mother huy a game for her. Reena wanted to play it with her friend Neena. She called Neena and told her to about the new game. Neena came to Reena house to play the game. The game consisted of 8 triangles Out of which 3 were blue and rest were red, and 10 squares out of which 6 were blue and rest were red. While both friends were playing one piece was lost at random. [4]
According to the above information, answer the following questions:
(a) How many triangles and squares are of red colour?
(A) 5, 4
(B) 4, 5
(C) 5, 5
(D) 8, 6
Answer:
Option (A) is correct.
Explanation:
Since, Total no. of triangles = 8
Triangles with blue colour = 3
Triangle with red colour = 8 – 3 = 5
Total no. of squares = 10
Squares with blue colour = 6
Square with red colour = 10 – 6 = 4
(b) Find the probability that last piece is triangle.
(A) 49
(B) 5/9
(C) 1/3
(D) 5/18
Answer:
Option (A) is correct.
Explanation:
Number of favourable outcomes for the event that lost figure is triangle, F(E) = 8
Total figures (square and triangle) = 8 + 10
= 18
i.e., T(E) = 18
∴ Probability (getting a last piece of a triangle) P(E)
= \(\frac{F(E)}{T(E)}\)
= \(\frac{8}{18}\) = \(\frac{4}{9}\)
(c) Find the probability that last piece is square.
(A) 4/9
(B) 5/9
(C) 1/3
(D) 5/18
Answer:
Option (B) is correct.
Explanation:
Number of favourable outcomes for the events that squares is lost i.e,
F(E) = 10
T(E) = 8 + 10 = 18
∴ P(getting a last piece of a square) = P(E1) = \(\frac{10}{18}\)
= \(\frac{5}{9}\)
(d) Find the probability that last piece is square of blue colour.
(A) 4/9
(B) 5/9
(C) 1/3
(D) 5/18
Answer:
Option (C) is correct.
Explanation:
Number of favourable outcomes for the events that lost figure is square of blue colour, i.e.,
F(E) = 6 and T(E) = 18
∴ P(getting a last piece of a blue square)
= P(E2) = \(\frac{F(E)}{T(E)}\)
= \(\frac{6}{18}\) = \(\frac{1}{3}\)