5 Marks Questions

Question 15 Marks

An air conditioner manufacturer allows a discount of $10 \%$ on marked price to dealer and the dealer sells the air conditioner to a consumer at a discount of $4\%$ on marked price. If the marked price is ₹ $50,000$ and the sales are intra-state sales with GST at $18\%.$ Find:
i. the total amount paid by the consumer to the dealer.
ii. the GST paid by the consumer to the dealer.
iii. the GST paid by the dealer to the central and state governments.
iv. the GST paid by the manufacturer to the central and state governments.

Answer

The sales are intra-state sales with GST at $18 \%$. So, there are two components of GST (i) SGST at $9 \%$ and (ii) CGST at $9 \%$
i. The manufacturer sells the air conditioner to the dealer at discount of $10 \%$ on the marked price of ₹ 50,000 .
$\therefore$ Discount $=10 \%$ of ₹ $50,000=$ ₹$\left(\frac{10}{100} \times 50,000\right)=$ ₹ $5000$
Selling price of the air conditioner (S.P.) $=$ ₹ $50,000-$ ₹ $5000=$ ₹ $45000$.
CGST paid by the dealer to the manufacturer $=9 \%$ of ₹ $45000.$
$=$ ₹$ \left(\frac{9}{100} \times 45000\right)=$ ₹ $4050$
SGST paid by the dealer to the manufacturer $=9 \%$ of ₹ $45000$
$=$ ₹ $\left(\frac{9}{100} \times 45000\right)=$ ₹ $4050$
Total GST paid by the dealer to the manufacturer = ₹ $4050+$ ₹ $4050$
= ₹ 8100 .
i.e. Input GST of the dealer $=$ ₹ $8100$.
The dealer sells the air conditioner at a discount of $4 \%$ on marked price.
$\therefore$ Discount given by the dealer $=4 \%$ of ₹ $50,000=$ ₹ $\left(\frac{4}{100} \times 50,000\right)=$ ₹ $2000$
Selling price of the dealer $=$ ₹ $50,000-$ ₹ $2000= $ ₹ $48000$
$\therefore$ CGST paid by the consumer $=9 \%$ of ₹ $48000=$ ₹ $\left(\frac{9}{100} \times 48000\right)=$ ₹ $4320$
SGST paid by the consumer $=9 \%$ of ₹ $48000=$ ₹ $\left(\frac{9}{100} \times 48000\right)=$ ₹ $4320$
Total GST paid by the consumer (output GST of dealer) = CGST + SGST
$=$ ₹ $4320 ~+$ ₹ $4320 =$ ₹ $8640$
Total amount paid by the consumer to the dealer = ₹ $48000~+$ ₹ $8640=$ ₹ $56640$
ii. GST paid by the consumer to the dealer $=$ ₹ $8640$.
iii. GST paid by the dealer to the central and state governments
= Output GST of dealer - Input GST of dealer
= ₹ $8640~-$ ₹ $8100=$ ₹ 540
iv. GST paid by the manufacturer to the central and state governments
= Output GST of the manufacturer - Input of the manufacturer
= ₹ $8100 ~-$ ₹ $0 =$ ₹ $8100$

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Question 25 Marks

The weights of sons and fathers (in kilograms) are given below:

Weight of father	65	66	67	67	68	69	70	72
Weight of son	67	68	65	68	72	72	69	71

Answer

x	$u - x -A$	$u^2$	y	$v-y-B$	$v^2$	uv
65	-2	4	67	-1	1	2
66	-1	1	68	0	0	0
67	0	0	65	-3	9	0
67	0	0	68	0	0	0
68	1	1	72	4	16	4
69	2	4	72	4	16	4
70	3	9	69	1	1	3
72	5	25	71	3	9	15
	8	44		8	52	32

Here, $\mathrm{N}=8$
$\therefore \mathrm{r}=\frac{\Sigma u v-\frac{1}{\mathrm{~N}} \Sigma u \Sigma v}{\sqrt{\Sigma u^{2}-\frac{(\Sigma u)^{2}}{\mathrm{~N}}} \sqrt{\Sigma v^{2}-\frac{(\Sigma v)^{2}}{\mathrm{~N}}}}=\frac{32-\frac{1}{8} \times 8 \times 8}{\sqrt{44-\frac{8^{2}}{8}} \sqrt{52-\frac{8^{2}}{8}}}$
$=\frac{24}{\sqrt{36} \sqrt{44}}=0.603$

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Question 35 Marks

Find the mean, variance and standard deviation for the following frequency distribution:

Classes	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	15	16	6

Answer

Class	Frequency $f_i$	Mid-point $X _i$	$\frac{y_i\left(x_i-25\right)}{10}$	$y_i^2$	$f _{ i } y _{ i }$	$f_j y_i^2$
0-10	5	5	-2	4	-10	20
10-20	8	15	-1	1	-8	8
20-30	15	25	0	0	0	0
30-40	16	35	1	1	16	16
40-50	6	45	2	4	12	24
	50				10	68

Mean, $\bar{x}=A+\frac{\sum_{i=1}^{5} f_{i} y_{i}}{N} \times \mathrm{h}=25+\frac{10}{50} \times 10=25+2=27$
Variance $\left(\sigma^{2}\right)=\frac{h^{2}}{N^{2}}\left[N \sum_{i=1}^{5} f_{i} y_{i}^{2}-\left(\sum_{i=1}^{5} f_{i} y_{i}\right)^{2}\right]$
$=\frac{(10)^{2}}{(50)^{2}}\left[50 \times 68-(10)^{2}\right]$
$=\frac{1}{25}[3400-100]=\frac{3300}{25}$
$=132$

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Question 45 Marks

Consider the real function $\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}: \mathrm{f}(\mathrm{x})=\mathrm{x}+5$ for all $\mathrm{x} \in \mathrm{R}$. Find its domain and range. Draw the graph of this function.

Answer

Here we have, $f(x)=x+5 \forall x \in R$
We need to find the Domain and Range of $f(x)$.
The domain of the given function is all real numbers except where the expression is undefined. In this case, there is no real number which makes the expression undefined.
As $f(x)$ is a polynomial function, we can have any value of $x$
Therefore,
Domain (f) $=(-\infty, \infty)\{x \mid x \in R\}$
Now,
Let $\mathrm{y}=\mathrm{f}(\mathrm{x})$
$y=x+5$
$x=y-5$
The range is set of all valid values of $y$.
Therefore,
Range(f) $=(-\infty, \infty)\{y \mid y \in R\}$
Graph:

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Question 55 Marks

Divide 32 into four parts which are in AP such that the product of extremes is to the product of means as $7: 15$.

Answer

Let four parts be ( $a-3 d$ ), $(a-d),(a+d)$ and $(a+3 d)$.
Then, Sum of four parts $=32$
$\Rightarrow(\mathrm{a}-3 \mathrm{~d})+(\mathrm{a}-\mathrm{d})+(\mathrm{a}+\mathrm{d})+(\mathrm{a}+3 \mathrm{~d})=32$
$\Rightarrow 4 \mathrm{a}=32$
$\Rightarrow \mathrm{a}=8$
and $\frac{(a-3 d)(a+3 d)}{(a-d)(a+d)}=\frac{7}{15}$
$\Rightarrow \frac{a^{2}-9 d^{2}}{a^{2}-d^{2}}=\frac{7}{15}$
$\Rightarrow \frac{64-9 d^{2}}{64-d^{2}}=\frac{7}{15} \quad[$ put $\mathrm{a}=8]$
$\Rightarrow 960-135 \mathrm{~d}^{2}=448-7 \mathrm{~d}^{2}$
$\Rightarrow 128 \mathrm{~d}^{2}=512$
$\Rightarrow d^{2}=4$
$\therefore \mathrm{d}= \pm 2$
Hence, the required parts are
$8-3 \times 2,8-2,8+2,8+3 \times 2$ or $8-3(-2), 8-(-2), 8-2,8+3(-2)$ i.e., $2,6,10,14$

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Question 65 Marks

Find four numbers forming a geometric progression in which the third term is greater than the first term by 9, and the second term is greater than the $4^{\text{th }}$ by 18.

Answer

Let the four numbers in G.P. be a, $\mathrm{ar},\mathrm{ar}^{2}, \mathrm{ar}^{3}$
$\therefore \mathrm{ar}^{2}=\mathrm{a}+9$ and $\mathrm{ar}=\mathrm{ar}^{3}+18$
Now, $a r^{2}-\mathrm{a}=9$
$\Rightarrow \mathrm{a}\left(\mathrm{r}^{2}-1\right)=9\ldots\text{(i)}$
And ar $-\mathrm{ar}^{3}=18$
$\Rightarrow \operatorname{ar}\left(1-\mathrm{r}^{2}\right)=18+$
$\Rightarrow-\operatorname{ar}\left(r^2-1\right)=18 \ldots\text{(ii)}$
Dividing eq. (ii) by eq. (i), we have
$\frac{-a r\left(r^{2}-1\right)}{a\left(r^{2}-1\right)}=\frac{18}{9}$
$\Rightarrow \mathrm{r}=-2$
Putting value of $r$ in eq. (i), we get
$a(4-1)=9$
$\Rightarrow \mathrm{a}=3$
$\therefore$ ar $=3 \times(-2)=-6$
$\operatorname{ar}^{2}=3 \times(-2)^{2}=12 \operatorname{ar} \wedge\{3\}$
$=3 \times(-2)^{3}=-24$
Therefore, the required numbers are $3,-6,12,-24$

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