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Applied Maths 5 Marks Questions

Model Paper 9 · CBSE STD 11 Science (CBSE - English Medium). 6 questions.

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Question 15 Marks
A retailer buys a TV from a manufacturer for ₹ 25000 . He marks the price of the TV $20 \%$ above his cost price and sells it to a consumer at $10 \%$ discount on the marked price. If the sales are intra-state and rate of GST is 12%, find:
i. the marked price of the TV.
ii. Consumer's cost price of TV inclusive of tax (under GST).
iii. GST paid by the retailer to the Central and State Governments.
Answer
The cost price of the TV which the retailer pays to the manufacturer $=$ ₹ 25000
i. As the retailer marks the price of TV $20 \%$ above his cost price,
$\therefore$ the marked price of the TV $=₹\left(1+\frac{20}{100}\right) \times 25000$
$=₹\left(\frac{120}{100} \times 25000\right)=₹ 30000$
ii. As the sales are intra-state and the rate of GST is $12 \%$, so GST comprises of CGST at $6 \%$ and SGST at $6 \%$
As the retailer sells the TV to a consumer at $10 \%$ discount on the marked price, selling price of the TV by the retailer $=₹\left(1-\frac{10}{100}\right) \times 30000=₹\left(\frac{9}{10} \times 30000\right)=₹ 27000$
Amount of GST collected by retailer from consumer (or paid by consumer to retailer):
CGST $=6 \%$ of $₹ 27000=₹\left(\frac{6}{10} \times 27000\right)=₹ 1620$
SGST $=6 \%$ of $₹ 27000=₹ 1620$
$\therefore$ Consumer's cost price of TV inclusive of tax (under GST)
= Cost price of TV to consumer + GST paid by consumer
= ₹27000 + CGST paid by consumer + SGST paid by consumer
= ₹ 27000 + ₹ 1620 = ₹ 1620
= ₹30240
iii. Amount of GST collected by manufacturer from retailer:
CGST $=6 \%$ of $₹ 25000=₹\left(\frac{6}{10} \times 25000\right)=₹ 1500$
SGST - $6 \%$ of ₹ $25000=₹ 1500$
Amount of input GST of the retailer:
CGST $=$ ₹ 1500 , SGST $=₹ 1500$
Amount of output GST of the retailer:
CGST = ₹1620, SGST = ₹1620
GST paid by the retailer to the Central Government
= Output CGST - input CGST
= ₹ 1620 - ₹ 1500 = ₹ 120
GST paid by the retailer to the State Government
= Output SGST - input SGST
= ₹ 1620 - ₹ 1500 = ₹ 120
$\therefore$ Total GST paid by the retailer to the Central and State Governments
= ₹ 120 + ₹ 120 = ₹ 240
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Question 25 Marks
Calculate Karl Pearson's coefficient of skewness from the following data:
Profits: (in ₹lacs)Below 20406080100
No. of companies:820506470
Answer
The given frequency distribution is a cumulative frequency distribution. This can be written as a frequency distribution as follows:
Profits: (₹ in lacs)0-2020-4040-6060-8080-100
No. of companies:81230146
We shall now prepare a table for the computation of mean, mode and standard deviation as given below:
Profits (₹ in lacs)No. of companies $f_i$$x_i$$\begin{array}{c} u _{ i }=\frac{x_i-50}{20} \\ A=50, h=20\end{array}$$f _{ i } u _{ i }$$f_i u_i^2$
0-20810-2-1632
20-401230-1-1212
40-603050000
60-80147011414
80-10069021224
  N = 70 $\Sigma f_i u_i=-2$$\Sigma f _{ i } u_i^2=82$
Computation of Mean: We have, $N =70, \Sigma f _{ i } u _{ i }=-2, A=50$, and $h =20$
$\therefore \bar{X}= A + h \left(\frac{\Sigma f_i u_i}{N}\right) \Rightarrow \bar{X}=50+20 \times \frac{-2}{70}=49.42$
Computation of S.D.: We have,
$N =70, \Sigma f _{ i } u _{ i }=-2, \Sigma f _{ i } u_i^2=82, A=50$ and $h =20$
$\therefore \sigma=h \sqrt{\frac{1}{N} \sum f_i u_i^2-\left(\frac{1}{N} \Sigma f_i u_i\right)^2}=20 \sqrt{\frac{82}{70}-\left(\frac{-2}{70}\right)^2}=20 \sqrt{\frac{82 \times 70-4}{70^2}}$
$\Rightarrow \sigma=20 \times \frac{\sqrt{5736}}{70}=\frac{20 \times 75.73}{70}=21.63$
Computation of Mode: Since the maximum frequency '30' is of class 40-60. So, 40-60 is the modal class such that l = 40, h = 20, f = 30, $f _1=12$ and $f _2=14$
Now,
Mode $=1+\frac{f-f_i}{2 f-f_1-f_2} \times h$
$\Rightarrow$ Mode $=40+\frac{30-12}{60-12-14} \times 20=40+\frac{18}{34} \times 20=40+10.58=50.58$
Compilation of Karl Pearson's coefficient of skewness: We have,
$S_{k_p}=\frac{\text { Mean-Mode }}{\text { S.D. }}=\frac{49.42-5058}{21.63}=-0.0536$
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Question 35 Marks
Find the coefficient of variation for the following data:
CI60-7050-6040-5030-4020-3010-20
Frequencies0361012156
Answer
Table for mean and standard deviation
Here A = 35 (say), $\mathrm{h}=10$
CI$x_i$$f _{ i }$$d_i=\frac{x_i-65}{10}$$f _{ i } d _{ i }$$f_i d_i^2$
60-706533927
50-6055621224
40-50451011010
30-403512000
20-302515-1-1515
10-20156-2-1224
  $\sum_{i=1}^5 f_i=52$ $\sum_{i=1}^6 f_i d_i=4$$\sum_{i=1}^6 f i d_i^2=100$
Mean $=\mathrm{A}+\frac{\sum_{i=1}^{6} f_{i} d_{i}}{\sum_{i=1}^{6} f_{i}} \times h=35+\frac{4}{52} \times 10=35+0.77=35.77$
$\sigma=10 \sqrt{\frac{100}{52}-\left(\frac{4}{52}\right)^{2}}=10 \sqrt{1.92-0.0059}=10 \sqrt{1.9141}=10 \times 1.384=13.84$
$\therefore$ Coefficient of variation $=\frac{\sigma}{\text { Mean }} \times 100=\frac{13.84}{35.77} \times 100=38.69$
OR
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Question 45 Marks
Find the domain of the function $\mathrm{f}(\mathrm{x})=\frac{1}{\log (1-x)}+\sqrt{x+3}$
Answer
Let $\mathrm{f}=\mathrm{g}+\mathrm{h}$, then $\mathrm{g}(\mathrm{x})=\frac{1}{\log (1-x)}$ and $\mathrm{h}(\mathrm{x})=\sqrt{x+3}$.
For $\mathrm{D}_{\mathrm{g}}, \mathrm{g}(\mathrm{x})$ must be a real number
$\Rightarrow \frac{1}{\log (1-x)}$ must be a real number
$\Rightarrow 1-\mathrm{x}>0$ and $1-\mathrm{x} \neq 1 \Rightarrow \mathrm{x}<1$ and $\mathrm{x} \neq 0$
$\Rightarrow \mathrm{D}_{\mathrm{g}}=(-\infty, 0) \cup(0,1)$.
For $\mathrm{D}_{\mathrm{h}}$, $\mathrm{h}(\mathrm{x})$ must be a real number
$\Rightarrow \sqrt{x+3}$ must be a real number $\Rightarrow \mathrm{x}+3 \geq 0$
$\Rightarrow \mathrm{x} \geq-3$
$\Rightarrow \mathrm{D}_{\mathrm{h}}=[-3, \infty)$
Image
as $f=g+h$, so $D_{f}=D_{g} \cap D_{h}=[-3,0) \cup(0,1)$.
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Question 55 Marks
The AM between two positive numbers a and $\mathrm{b}(\mathrm{a}>\mathrm{b})$ is twice their GM. Prove that $\mathrm{a}: \mathrm{b}=(2+\sqrt{3}):(2-\sqrt{3})$.
Answer
Here,it is given: Arithmetic mean is twice of geometric mean.
By using Formula:
Arithmetic mean between a and b = $\frac{a+b}{2}$
Geometric mean between a and $b =\sqrt{a b}$
AM = 2(GM)
$\begin{array}{l}\frac{a+b}{2}=2 \sqrt{a b} \\ a + b =4 \sqrt{a b}\end{array}$
Squaring both side ,we get
$\Rightarrow$ $(a+b)^2=16 a b \ldots \ldots$ (i)
We know that $(a-b)^2=(a+b)^2-4 a b$
From eqn.(i)
$\Rightarrow(a-b)^2=16 a b-4 a b$
$\Rightarrow( a - b )^2=12 ab \ldots . .( ii )$
Dividing eq.. (i) and (ii) ,we get
$\frac{(a+b)^2}{(a-b)^2}=\frac{16 a b}{12 a b}$
Taking square root both side
$\Rightarrow \frac{a+b}{a-b}=\frac{2}{\sqrt{3}}$
Applying componendo and dividendo"
$\frac{a+b+a-b}{a+b-a+b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
$\frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Hence proved.
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Question 65 Marks
Find the sum of 30 terms of the sequence 7, 7.7, 7.77, 7.777,...
Answer
7 + 7.7 + 7.77 + 7.777 + ... + 30 terms
=7 [1 + 1.1 + 1.11 + 1.111 + ... 30 terms]
$\begin{array}{l}=\frac{7}{9}[9+9.9+9.99+9.999+\ldots 30 \text { terms }] \\ =\frac{7}{9}[(10-1)+(10-0.1)+(10-0.01)+(10-0.001)+\ldots 30 \text { terms }] \\ =\frac{7}{9}[10 \times 30-\{1+0.1+0.01+0.001+\ldots 30 \text { terms }\}] \\ =\frac{7}{9}\left[300-\frac{1\left\{1-(0.1)^{30}\right\}}{1-0.1}\right] \\ =\frac{7}{9}\left[300-\frac{10}{9}\left\{1-(0.1)^{30}\right\}\right]\end{array}$
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5 Marks Questions - Applied Maths STD 11 Science Questions - Vidyadip