Students must start practicing the questions from CBSE Sample Papers for Class 11 Applied Mathematics with Solutions Set 10 are designed as per the revised syllabus.
CBSE Sample Papers for Class 11 Applied Mathematics Set 10 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.
Convert binary 11 11 11 11 00 10 into hexadecimal. [1]
Answer:
Given binary number is
11 11 11 11 00 10
Make a 4-4 pair group from right to left
Hence (1111 1111 0010)2 = (FF2)16
Question 2.
Find value of
{(33)2 × (42)3 × (53)2} ÷ {(32)3 × (43)2 × (52)3}
OR
Find value of log 0.0625 to the base 2. [1]
Answer:
\(\frac{\left(3^3\right)^2 \times\left(4^2\right)^3 \times\left(5^3\right)^2}{\left(3^2\right)^3 \times\left(4^3\right)^2 \times\left(5^2\right)^3}\) = \(\frac{3^6 \times 4^6 \times 5^6}{3^6 \times 4^6 \times 5^6}\)
= 1 [Using (am)n = amn]
OR
log2 0.0625 = log2\(\left(\frac{625}{10000}\right)\)
= log2\(\frac{5^4}{10^4}\)
= log2\(\left(\frac{5}{10}\right)^4\)
= 4log2\(\left(\frac{1}{2}\right)\)
[Applying rule logamn = n loga m]
= 4[log21 – 1og22]
[Applying rule loge\(\left(\frac{m}{n}\right)\) = loga m – loga n
= 4[0 – 1]
[Applying rule loga 1 = 0 and loga a = 1]
= -4
Question 3.
Differentiate sin2(x2) w.r.t. x2. [1]
Answer:
Let x2 = t
So, \(\frac{d\left[\sin ^2(x)^2\right]}{d x^2}\) = \(\frac{d\left(\sin ^2 t\right)}{d t}\)
= 2 sin t × (sin t)
= 2 sin t × cos t
= 2 sin (x2) cos (x2)
Commonly Made Error
Sometimes students forget the chain Rule. They do not differentiate the angle of trigonometric functions.
Answering Tip
A thorough revision is a must.
Question 4.
Mr. Ram owns a house property. He lent it to Laxman at ₹ 10,000 per month Laxman sublet it to Mr. Maruti on monthly rent of ₹ 20,000 p.m. Rental income of Ram is taxable under which income head. [1]
Answer:
Given, Mr. Ram owns a house property and lent it to Laxman at ₹ 10,000 per month Laxman sublet it to Mr. Maruti on monthly rent of ₹ 20,000 per month. Thus, rental income of Ram is taxable under the head Income from House Property.
Question 5.
Justify that the relation t = {(x, 3 | x is a real number} is a function. [1]
Answer:
Here, A = {D, E, H, I, L} and B = {D, L, O}
A ∪ B = (D, E, H, I, L} ∪ {D, L, O}
= {D, E, H, I, L, O}
OR
Given, A = {- 1, 1}, A × A = {- 1, 1} × {- 1, 1} = {(-1, -1), (-1, 1), (1, -1), (1, 1)}
Again, A × A × A = {(-1, -1), (-1, 1), (1, -1), (1, 1)} × {-1, 1}
= {(- 1, – 1, – 1), (- 1, – 1, 1), (-1, 1, -1), (- 1, 1, 1), (1, -1, -1), (1, -1, 1), (1, 1, -1), (1, 1, 1)}
Question 6.
Which term of the sequence 3,10,17…….. is 136 ? [1]
Answer:
Given, t = {(x, 3) | x is a real number}
The given relation is a function because every element in the domain has the image 3.
Question 7.
How many numbers are there between 99 and 1000 having 7 in the unit place ?
OR
In how many ways 7 pictures can be hanged on 9 pegs ? [1]
Answer:
The number between 99 and 1000 are all three digit numbers. If we fix 7 at unit place, then middle digit can be any one of the 10 digits from 0 to 9. The digit in hundred’s place can be any of the 9 digits from 1 to 9. Therefore, by the fundamental principle of counting, there are 10 × 9 = 90 numbers between 99 and 1000 having 7 in the units places.
OR
7 pictures have options of 9 pegs. So, they can be selected in 9C7 ways.
Now, these pictures can arrange themselves in 7! ways.
∴ Number of ways by which 7 pictures can be hanged on 9 pegs = 9C7 × 7!.
Question 8.
In a G.P. of positive terms, if any term in equal to the sum of the next two terms, then find the common ratio of the G.P. [1]
Answer:
Given, tn = tn+1 + tn+2
⇒ arn-1 = arn + arn+1
⇒ 1 = r + r2
⇒ r2 + r – 1 = 0
∴ r = \(\frac{-1 \pm \sqrt{5}}{2}\)
Since, r > 0 therefore, r = \(\frac{\sqrt{5}-1}{2}\)
Question 9.
Introducing a girl, Vipin said, “Her mother is the only daughter of my mother-in-law.” How is Vipin related to the girl? [1]
Answer:
The only daughter of Vipin’s mother-in-law is the wife of Vipin. Therefore, the statement, in other words implies that “Her mother is my wife”. It is hence clear that Vipin is the father of that girl.
Question 10.
If A = set of letters of the word ‘DELHI’ and B = the set of letters the ‘DOLL’ find A ∪ B.
OR
If A = {- 1, 1}, find A × A × A. [1]
Answer:
Given sequence 3, 10, 17 in A.P
Let nth term be the 136, i.e.,
Tn =a + (n – 1)d
⇒ 136 = 3 + (n – 1)7
[∵ a = 3 and d = 7]
⇒ 7n = 140
⇒ n = 20
Hence, 20th term of the sequence is 136.
Question 11.
Find the domain of function f(x) = [x] + x.
Answer:
Given, f(x) = [x] + x
i.e., f(x) = h(x) + g(x)
where, h(x) = [x] and g(x) = x
The domain of h = R and The domain of g = R
∴ Domain of f = R.
Question 12.
Find n, if \(\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}\) = 80, n ∈ N. [1]
Answer:
Given, \(\lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}\) = 80
⇒ n.2n-1 = 80 [∵ \(\lim _{x \rightarrow a} \frac{x^n-a^n}{x-a}\) = nan-1]
n.2n-1 = 5.25-1
∴ Value of n is 5.
Question 13.
If the mean of a, b,c is M and ab + bc + ca = 0, then find the mean of a2, b2, c2.
OR
Today is Monday, then what is the day after 61 days. [1]
Answer:
We have: \(\left(\frac{a+b+c}{3}\right)\) = M
or a + b + c = 3M
Now, (a + b + c)2 = 9M2
⇒ a2 + b2 +c2 + 2(ab + bc + ac) = 9M2
⇒ a2 + b2 + c2 = 9M2
[∵ Given, ab + bc + ca = 0]
∴ Required mean = \(\left(\frac{a^2+b^2+c^2}{3}\right)\) = \(\frac{9 M^2}{3}\)
= 3M2.
OR
Each day of the week is repeated after 7 days.
So, after 63 days, it will be Monday.
Thus, after 61 days, it will be Saturday.
Question 14.
Find the equation of the circle passing through (0, 0) and which makes intercepts a and b on the coordinate axes. [1]
Answer:
Since, the circle passes through O(0, 0) and cuts off intercepts a and b on the co-ordinate axes, it passes through A(0, a) and B(0, b).
As the axes are perpendicular to each other, Therefore ∠AOB = 90° and hence AB becomes a diameter of this circle.
∴ The equation of the circle is given by
(x – a)(x – 0) + (y – 0)(y – b) = 0
or, x2 + y2 – ax – by = 0.
Question 15.
State two merits of quartile deviation. [1]
Answer:
- Quartile deviation is the best measure of dispersion for open-end classification as quartile deviation does not take into account the first twenty five percent and the last twenty five percent of the observations
- It is independent of hänge of data.
Question 16.
A, B and C can complete a piece of work in 24, 6 and 12 days respectively, working together, they will complete the same work in how many days ? [1]
OR
In a shower, 10 cm of rain falls. What will be the volume of water that falls on 1 hectare area of ground?
Answer:
(A + B + C)’s 1 days work = \(\frac{1}{24}\) + \(\frac{1}{6}\) + \(\frac{1}{12}\) = \(\frac{7}{24}\) th work
Complete work in 1 day = \(\frac{7}{24}\) work
\(\frac{24}{7}\) days = 1 work
So, A, B and C together will complete the job in \(\frac{24}{7}\) days = 3\(\frac{3}{7}\) days.
OR
Since, 1 hectare = 10000 m2
Volume of water = Area of base × height
= 10000 × \(\frac{10}{100}\) m3
= 1000 m3
∴ Volume of water is 1000 m3 = 100 KL
Commonly Made Error
Some students do mistake in conversion of hectare to meter.
Answering Tip
Learn all the basic conversions.
Section – B
All questions are compulsory. In case of internal choices attempt any one.
Question 17.
If a vertex of a triangle is (1, 1) and the mid-points of two sides through this vertex are (-1, 2) and (3, 2). Find the centroid of the triangle.
OR
Show that the triangle whose vertices are (8, 2), (5, – 3) and (0, 0) is an isosceles. [2]
Answer:
Let the triangle be ABC with vertex A(1, 1) and midpoint of AB is E(-1, 2), midpoint of AC is F(3, 2).
Let (x1, y1) and (x2, y2) be the coordinates of B and C respectively.
Since, E is the mid-point of AB.
∴ \(\left(\frac{1+x_1}{2}, \frac{1+y_1}{2}\right)\) = (-1, 2)
and F is the mid-point of AC.
∴ \(\left(\frac{1+x_2}{2}, \frac{1+y_2}{2}\right)\) = (3, 2)
⇒ x2 = 5, y2 = 3
∴ Centroid of the triangle have co-ordinates
\(\left(\frac{1-3+5}{3}, \frac{1+3+3}{3}\right)\) = \(\left(1, \frac{7}{3}\right)\)
Commonly Made Error
Sometimes students consider the mid-points of the sides as the vertices of the triangle and go wrongly.
Answering Tip
If a point R is the mid-point of the line segment joining the points P(x1, y1) and Q(x2, y2) then coordinates of R are
\(\left[\left(\frac{x_1+x_2}{2}\right),\left(\frac{y_1+y_2}{2}\right)\right]\)
OR
Let, the given vertices be A(x1, y1) = (8, 2), B(x2, y2) = (5, -3) and C(x3, y3) = (0, 0)
∴ AB = BC i.e; ∠ACB = ∠BAC
Hence, ΔABC is isosceles
Question 18.
Give two difference between direct and indirect taxes. [2]
OR
A product is sold from Kota (Rajasthan) to Gwalior (M.P.) for ₹ 8,000 and then from Gwalior to Indore (M.P.). If the rate of tax under GST system is 18% and the profit made by the dealer in Gwalior is ₹ 3,000, find net GST payable by the dealer in Gwalior.
Answer:
Cost price in Gwalior = ₹ 8,000
and profit = ₹ 3,000
Selling price in Gwalior = ₹ (8000 + 3000)
= ₹ 11,000 A
CGST = ₹ \(\frac{9}{100}\) × 11,000
= ₹ 990
SGST = ₹ \(\frac{9}{100}\) × 11,000
= ₹ 990
∴ Net GST paid by the dealer in Gwalior
= ₹ (990 + 990)
= ₹ 1980
Question 19.
Sachin deposited ₹ 100000 in his bank for 2 years at simple interest rate of 6%. How much interest would he earn? How much would be the final value of deposit ? [2]
Answer:
Required interest amount is given by
I = P i t
100000 × \(\frac{6}{100}\) × 2
= ₹ 12,000 [1]
Final value of deposit is given by
A = P + I
= ₹ (100000 + 12000)
= ₹ 112000.
Question 20.
The mean and median of 100 observations are 50 and 52 respectively. The value of the largest observation is 100. It was later found that it is 110 not 100. Find the true mean and median.
OR
Find the mean deviation about the median of the following distribution. [2]
| Marks obtained | 10 | 11 | 12 | 14 | 14 |
| No. of students | 2 | 3 | 8 | 3 | 4 |
Answer:
Mean = \(\frac{\Sigma f x}{\Sigma f}\)
⇒ 50 = \(\frac{\Sigma f x}{100}\)
⇒ Σfx = 5000
Correct Σfx = 5000 – 100 + 110 = 5010
∴ Correct mean = \(\frac{5010}{100}\) = 50.1
Median will remain same i.e., median = 52
OR
Question 21.
If A and B are two events such that P\(\left(\frac{A}{B}\right)\) = p, P(A) = p, P(B) = \(\frac{1}{3}\) and P(A ∪ B) = \(\frac{5}{9}\) then find the value of p. [2]
Answer:
Question 22.
If x = a sin pt, y = b cos pt, then find \(\frac{d y}{d x}\) at t = 0. [2]
Answer:
Given x = a sin pt
∴ \(\frac{d x}{d t}\) = ap cos pt [½]
and y = b cos pt
∴ \(\frac{d y}{d t}\) = -bp sin pt [1/2]
∴ \(\left(\frac{d y}{d x}\right)_{t=0}\) = –\(\frac{b}{a}\) tan(pt)
Question 23.
Draw the graph of constant function f : R → R; f(x) = 2 ∀ x ∈ R. Also, find its domain and range. [2]
Answer:
Given, f:R→R; f(x) = 2 ∀ x ∈ R
Domain = R and Range = {2}
Question 24.
If TEMPLE is coded as VHQURL, how would you code CHURCH ? [2]
Answer:
From above diagram it is dear that code for T is V, for E is H, for M is Q. It may be noticed from here that letters of TEMPLE have been replaced by new letters from the alphabets. There is a gap of one letter between T and V, gap of two letters between E and H, gap of three letters between M and Q and so on in the alphabets.
Therefore, coding for CHURCH is
i.e., CHURCH is coded as EKYWIO.
Question 25.
Arrange the following words in a logical and meaningful order. [2]
1. Country
2. Furniture
3. Forest
4. Wood
5. Trees
Answer:
From the given words, we can say that country contains forest, forest has trees, trees have wood that is used to make furniture, Hence, the correct order of the given words is 1, 3, 5, 4, 2.
Question 26.
Determine n, if 2nC3 : nC3 = 11 : 1. [2]
Answer:
Section – C
All questions are compulsory. In case of internal choices attempt any one.
Question 27.
Using properties of sets and their complements prove that:
(i) (A ∪ B ) ∩ (A ∩ B’) = A
(ii) A – (A ∩ B) = A – B
OR
Let A = {2, 3, 4, 5, 6, 7, 8, 9}. Let R be the relation on A defined by {{x, y): x, y ∈A, x is a multiple of y and x ≠ y}.
(i) Find the relation.
(ii) Find the domain of R.
(iii) Find the range of R.
(iv) Find the inverse relation. [3]
Answer:
(i)(A ∪ B) ∩ (A ∩ B’) = A
LH.S. = (A ∪ (B ∩ B’)
(By distributive law)
= A ∪ φ (B ∩ B’ = φ)
= A
= R.H.S. Hence proved.
(ii) A – (A ∩ B) = A – B
LH.S. = A – (A B)
= A ∩ (A ∩ B)’
[∵ A – B = A ∩ B’]
= A ∩ (A’ ∪ B’)
(By De-morgan’s law)
=(A ∩ A’) ∪ (A ∩ B’)
(By distributive law)
= φ ∪ A ∩ B’
= A – B
= R.H.S
Hence Proved
Commonly Made Error
Students apply wrong operations of sets as they get confused while applying operations.
Answering Tip
It is necessary to do sufficient practice on operations of sets. Practice more problems based on this topic.
OR
(i) R = {(4, 2)(6, 2)(8, 2)(6, 3)(9, 3)(8, 4)}
(ii) Domain of R {4, 6, 8, 9}
(iii) Range of R = {2, 3, 4}
(iv) R-1 = ((2, 4) (2, 6) (2, 8) (3, 6) (3, 9) (4, 8))
Question 28.
The frustum of a right circular cone has a diameter of base 10 cm, top of 6 cm and a height of 5 cm, find the area of its whole surface and volume.
OR
Read the information carefully and answer the questions based on it.
Five persons are sitting in a row. One of the two persons at the extreme ends is intelligent and other one is fair. A fat person is sitting to the right of a weak person. A tall person is to left of the fair person and the weak person is sitting between the intelligent and the fat person.
(i) Tall person is at which place counting from right ?
(ii) Person to the left of weak person possesses which characteristic ?
(iii) Which person is sitting at the centre ? [3]
Answer:
Here, r1 = 5cm, r2 = 3 cm
and h = 5 cm
l = \(\sqrt{h^2+\left(r_1-r_2\right)^2}\)
= \(\sqrt{5^2+(5-3)^2}\) = \(\sqrt{29}\) cm
= 5.383 cm
∴ whole surface of the frustum
= π(r1 + r2) + π\(\pi r_1^2\) + π\(\pi r_2^2\)
= \(\frac{22}{7}\) × 5.383 (5 + 3) + \(\frac{22}{7}\) × 52 + \(\frac{22}{7}\) × 32
= 242.25 sq. cm
and Volume = \(\frac{\pi h}{3}\)\(\left(r_1^2+r_2^2+r_1 r_2\right)\)
= \(\frac{22}{7}\) × \(\frac{5}{3}\)(52 + 32 + 5 × 3) = 256.67 cu.cm
OR
First information given in the question that one of the two persons at extreme ends is intelligent and other one is fair Suggest two figures as shown below in fig (1)and fig(2)
Information that a tall person is sitting to the left of fair person rules out the possibility of fig (1) as no person in fig (1) can sit to the left of fair person. Therefore, only fig (2) shows the correct positions of intelligent and fair persons. No rest of the information regarding the position of other person can easily be inserted. The final ranking of their sitting arrangement is as shown in fìg(3)
It is clear from fig 3,
(i) Tall person is at second place counting from right.
(ii) Person left to weak person possesses intelligence.
(iii) Fat person is sitting at the centre.
Commonly Made Error
Students are confused while doing linear sifting arrangement of objects. Which side should they take left or right.
Answering Tip
Unless given specifically in the question, consider your right the right hand side; and your left, the left hand side while arranging the objects.
Question 29.
Oskie-946 has a decay rate of 13.5%. If the original sample was 50 gm, how long will it take for only 10 gm of the sample to remain ?
[Given, loge 0.2 = -1.60943] [3]
Answer:
We use the formula of exponential decay
i.e., m(t) = m0e-rt
Given, r = 13.5% = \(\frac{13.5}{100}\) = 0.135
m(t) = 10 and m0 = 50
Therefore, 10 = 50 e-0.135t
⇒ 0.2 = e-0.135t
Taking log both sides, we get
log(0.2) = log(e-0.135t)
⇒ log (0.2) = – 0.135t loge
⇒ log (0.2) = -0.135t
⇒ t = –\(\frac{\log (0.2)}{0.135}\)
= \(-\frac{(-1.60943)}{0.135}\)
Hence, sample will take approximately 12 years, for remaining 10 gm.
Question 30.
Do the following conversions : [3]
(i) (514)8 = (?)10
(ii) (789)10 = (?)16
Answer:
(i) (514)8
= 5 × 64 + 1 × 8 + 4 × 1
= 320 + 8 + 4 = 332
Hence, (514)8 = (332)10
(ii) (789)10
To change decimal number into hexadecimal, we divide ven decimal number repeatedly by 16.
Thus, (789)10 = (315)16
Question 31.
Find the equation of circle which passes through the points (0, 2), (3, 0) and (3, 2). Find also the centre and radius of this circle. [3]
Answer:
Let the equation of required circle be
x2 + y2 + 2gx + 2fy + c = 0 …. (i)
This circle will pass through the point (0, 2), (3, 0) and (3, 2), if
4 + 4f + c = 0 …(ii)
9 + 6g + c = 0 …(iii)
Now, on adding eqs. (ii) & (iii), we get
13 + 6g + 4f + 2c = 0 ….. (iv)
and circle will pass through the point (3, 2) we get,
13 + 6g + 4f + c = 0
∴ c = 0 …(v)
Substituting value of c in eqs. (ii) and (iii), we get
4 + 4f = 0 ⇒ f = -1
and 9 + 6g = 0 ⇒ g = \(-\frac{3}{2}\)
Putting these values of g, f and c in eq (i), we get the required equation of circle.
i.e., x2 + y2 – 3x – 2y = 0
Now, centre of the circle, (-g, -f) = \(\left(\frac{3}{2}, 1\right)\)
Question 32.
A mortgage of ₹ 190,000 is required to purchase a house. The mortgage will be repaid with equal monthly payments over 25 years at 8% compounded monthly.
(a) What is the monthly payment ?
(b) What is the total interest paid over the 25 years ? Given that, (1.006)300 = 7.3402. [3]
Answer:
(a) Here, n = 25 × 12 = 300, i = 8% monthly = \(\frac{8}{100} \times \frac{1}{12}\) = 0.006 yearly
Present value, P.V. = ₹ 190,000
(b) Interest = 1319.80 × 300 – 190,000
= 395, 490 – 190,000
= 205, 940
Question 33.
Compute the coefficient of rank correlation between sales and advertisement expressed in thousands of ? from the following table : [3]
| Sales : | 90 | 85 | 68 | 75 | 82 | 80 | 95 | 70 |
| Advertisement: | 7 | 6 | 2 | 3 | 4 | 5 | 8 | 1 |
Answer:
Let the rank given to sales be denoted by x and rank of advertisement be denoted by y. We note that since the highest sales as in the given data, is 95, it is to be given rank 1, the second highest sales 90 is to be given rank 2 and finally rank S goes to the lower sales, namely 68. We have given rank to the other variable advertisement in a similar manner.
Since n = 8 and Σd2 = 4
rs = 1 – \(\frac{6 \Sigma d^2}{n\left(n^2-1\right)}\)
= 1 – \(\frac{6 \times 4}{8\left(8^2-1\right)}\)
= 1 – \(\frac{24}{504}\)
= \(\frac{480}{504}\) = 0.95
The high positive value of the rank correlation coefficient indicates that there is a very good amount of agreement between sales and advertisement.
Commonly Made Error
Sometimes student just write the value of coefficient rank of correlation, they did not interpret that value.
Answering Tip
It is important to give interpretation of the obtained value of rank correlation coefficient i.e., whether it belong to low or high amount of agreement.
Section – D
All questions are compulsory. In case of internal choices attempt any one.
Question 34.
For a non-domestic connection monthly consumption of water is 150 kilolitres, then find the water bill for a month. Tariff rates can be considered as the table given below: [5]
(NON – DOMESTIC CONNECTIONS-COMMERCIAL/INDUSTRIAL)
| Monthly Consumption (in kilolitre) | Service charge(in ₹) | Volumetric charge(per kl in ₹) |
| 0-06 | 146.41 | 17.57 |
| 06-15 | 292.82 | 26.35 |
| 15-25 | 585.64 | 35.14 |
| 25-50 | 1024.87 | 87.85 |
| 50-100 | 1171.28 | 140.56 |
| >100 | 1317.69 | 175.69 |
| Plus Sewer Maintenance Charge : 60% of water volumetric charge | ||
Answer:
Volumetric Charge for consumption upto 6kl
= ₹ 6 × 17.57 = ₹ 105.42
Volumetric Charge for consumption between 6-15 kl
= ₹ 9 × 26.35 = ₹ 237.15
Volumetric Charge for consumption between 15-25
kl = ₹ 10 × 35.14 = ₹ 351.4
Volumetric Charge for consumption between 25-50
kl = ₹ 25 × 87.85 = ₹ 2196.25
Volumetric Charge for consumption between 50-100
kl = ₹ 50 × 140.56 = ₹ 7028
Volumetric Charge for consumption between 100-150
kl = ₹ 50 × 175.69 = ₹ 8784.5
Total volumetric Charge for consumption of 150
k] = ₹ (105.42 + 237.15 + 351.4 + 2196.25 + 7028 + 8784.5)
= ₹ 18702.72
Service Charge
= ₹ 1317.65
Sewage Charges
= 60% of Volumetric Charges
= 18702.72 × 60% = ₹ 11221.63
Amount of water bill for the given month
= ₹ (18702.72 + 1317.65 + 11221.63) = ₹ 31242
Thus, amount of non- domestic water bill is 31242.
Question 35.
A card from a pack of 52 cards is lost. From the remaining cards of the pack, two cards are drawn at random and are found to be hearts. Find the probability of the missing card to be a heart.
OR
Find the first principle of the derivative \(\frac{x+1}{x-1}\) [5]
Answer:
Let C1, C2, C3 and C4 be the events that the lost card is of heart, spade, diamond or club, respectively.
Clearly, P(C1) = P(C2) = P(C3)
= P(C4) = \(\frac{13}{52}\) = \(\frac{1}{4}\)
Let S be the event of drawing two cards of heart from the remaining 51 cards. Here, we wish to find
\(P\left(\frac{C_1}{S}\right)\)
Now, \(P\left(\frac{S}{C_1}\right)\) is the probability of drawing two heart cards from 51 cards given that one heart card is lost
= \(\frac{{ }^{12} C_2}{{ }^{51} C_2}\) = \(\frac{12 \times 11}{1 \times 2}\) × \(\frac{1 \times 2}{51 \times 50}\)
= \(\frac{22}{425}\)
Similarly,
By Baye’s Theorem,
Section – E
Question 36.
Find the sum of first ‘n’ terms of the series 0.7 + 0.77 + 0.777 + ……………
OR
Calculate the income tax payable by following individuals for the assessment year 2019-20: [4]
| Sridevi | Tanuja | Asha | |
| Date of Birth | 14.9.1972 | 10.1.1993 | 15.8.1942 |
| Total income which includes: | ₹ 2,42,000 | ₹ 2,80,000 | ₹ 3,95,000 |
| (a) Long-term capital gains | ₹ 17,000 | ₹ 17,000 | ₹ 17,000 |
| (b) Net-agricultural income | ₹ 20,000 | ₹ 20,000 | ₹ 20,000 |
Answer:
We have, 0.7 + 0.77 + 0.777 + … to n terms
= 7 × 0.1 + 7 × 0.11 + 7 × 0.111 + …. to n terms
= 7{0.1 + 0.11 + 0.111 + … to n terms}
= \(\frac{7}{9}\){0.9 + 0.99 + 0.9999 + …… to n terms}
OR
Income tax payable by following person:
| Sridevi | Tanuja | Asha | |
| Total income (non-agricultural) | 2,42,000 | 2,80,000 | 3,25,000 |
| Tax payable | Nil | 2,000 | 9,200 |
| Less: Rebate u/s 87A | — | 2,000 | 5,000 |
| Add: Education cess & SHEC (S 3% | — | — | 126 |
| Nil | Nil | 4,330 |
(1) Sridevi’s income exclusive of agricultura] income is ₹ 2,42,000 whiçh is less than, exemption limit of ₹ 2,50,000. Hence, there will be no integration of agricultural income with non- agricultural income. Thus tax on ₹ 2,42,000 shall be Nil.
(2) Total income exclusive of agricultural income is ₹ 2,60,000. But it includes ₹ 17,000 on account of LTCG which is taxable at a flat rate of 20%. Since, non-agricultural income exclusive of LTCG (Long Term Capital Gain) is less than maximum exemption limit there cannot be any partial integration.
The tax shall be calculated as under:
| ₹ | |
| Tax on LTCG of ₹ 10,000 (17,000 – 7,000 shifted to other income) @ 20% | 2,000 |
| Tax on other income ₹ 2,43,000 + 7,000 shifted from LTCG | Nil |
| 3,400 | |
| Less: Rebate u/s 87A | 3,400 |
| Nil |
(3) Tax on non-agricultural income of Asha, aged more than 60y cars
| ₹ | |
| Tax on ₹ 3,75,000 + 20,000 = ₹ 3,93,000 | |
| Tax on LTCG of ₹ 17,000 @ 20% | 3,400 |
| Other income ₹ 3,78,000 | 3,900 |
| Less: Tax on ₹ 3,00,000 + 20,000 = ₹ 3,20,000 | 1,000 |
| 6,300 | |
| Less: Rebate u/s 87A (Nil as the total income exceeds ₹ 3,50,000) | Nil |
| 6,300 | |
| Add: Health and education cess @ 4% | 252 |
| Tax | 6,552 |
| Tax (rounded off) | 6,550 |
Section – E
Both the Case study based questions are compulsory. Each Sub parts carries 1 mark.
Question 37.
Shamshad All buys a scooter for ₹ 22000. He pays ₹ 4000 cash and agree to pay the balance in annual installments of ₹ 1000 plus 10% interest on the unpaid amount. [4]
On the basis of the above information answer the following questions:
(a) Interest paid by All on 1 installment ?
(A) ₹ 1600
(B) ₹ 1700
(C) ₹ 1800
(D) none of these
Answer:
Option (C) is correct.
Explanation:
Scooter cost = ₹ 22000
Down payment ₹ 4000.
Balance payment = ₹ 18000
Now, interest on I installment
= \(\frac{18000 \times 10 \times 1}{100}\) = ₹ 1800 (∵ I = \(\frac{P \times R \times T}{100}\))
(b) Interest paid by All on 2nd installment ?
(A) ₹ 1600
(B) ₹ 1700
(C) ₹ 1800
(D) none of these
Answer:
Option (B) is correct.
Explanation:
Unpaid amount = 18000 – 1000 = 17000
Interest on II installment
= \(\frac{17000 \times 10 \times 1}{100}\) = ₹ 17001
(c) Interest paid by Ali on 3rd installment?
(A) ₹ 1600
(B) ₹ 1700
(C) ₹ 1800
(D) none of these
Answer:
Option (A) is correct.
Explanation:
Unpaid amount = 17000 – 1000 = 16000
Interest on III Installment
= \(\frac{16000 \times 10 \times 1}{100}\) = ₹16001
(d) Total interest paid by Ali is:
(A) ₹ 17100
(B) ₹ 22000
(C) ₹ 39100
(D) none of these
Answer:
Option (A) is correct.
Explanation:
Total interest paid by him
1800 + 1700 + 1600 + ….. + 18 terms
which is an A.P. with a = 1800, d = 1700 – 1800
= -100
Therefore, total interèst
= \(\frac{18}{2}\)[2 × 1800 + (18 – 1)(-100)]
(∵ Sum of AP = Sn + (n – 1)d])
= 9 (3600 – 1700) = 9 × 1900 = 17100
Question 38.
One of the four persons John, Rita, Aslam or Gurpreet will be promoted next month. Consequently the sample space consists of four elementary outcomes S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted}. You are told that the chances of John’s promotion is same as that of Gurpreet. Rita’s chances of promotion are twice as likely as Johns. Aslam’s chance are four times that of John. [4]
Answer:
Let Event:
J = John promoted
R = Rita promoted
A = Aslam promoted
G = Gurpreet promoted
Given sample space, S = {John promoted, Rita promoted, Aslam promoted, Gurpreet promoted)
i.e., S = (J, R, A, G)
It is given that, chances of John’s promotion is same as that of Gurpreet
P(J) = P(G)
Rita’s chances o promotion are twice as likely as John.
P(R) = 2P(J)
and Aslam’s chances ol promotion are tour times that of John.
P(A) = 2P(J)
On the basis of the above information answer the following questions :
(a) What is the probability that John got promotion ?
(i) \(\frac{1}{8}\)
(ii) \(\frac{1}{4}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{1}{6}\)
Answer:
Option (A) is correct.
Explanation:
Now, P(J) + P(R) + P(A) + P(G) = 1
⇒ P(J) + 2P(J) + 4P(J) + P(J) = 1
⇒ P(J) = P (John Promoted) = \(\frac{1}{8}\)
(b) What is the probability that Rita got promotion ?
(i) \(\frac{1}{8}\)
(ii) \(\frac{1}{4}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{1}{6}\)
Answer:
Option (B) is correct.
Explanation:
P(Rita promoted) = P(R)
= 2P(J) = 2 × \(\frac{1}{8}\) = \(\frac{1}{4}\)
(c) What is the probability that Aslam got promotion ?
(i) \(\frac{1}{8}\)
(ii) \(\frac{1}{4}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{1}{6}\)
Answer:
Option (C) is correct.
Explanation:
P(Aslam promoted) = P(A) = 4P(J)
= 4 × \(\frac{1}{8}\) = \(\frac{1}{2}\)
(d) What is the probability that Gurpreet got promotion ?
(i) \(\frac{1}{8}\)
(ii) \(\frac{1}{4}\)
(iii) \(\frac{1}{2}\)
(iv) \(\frac{1}{6}\)
Answer:
Option (A) is correct.
Explanation:
P(Gurpreet promoted) = P(G) = P(J) = \(\frac{1}{8}\)