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Question 15 Marks
A housewife buys the following articles from departmental store sugar 5 kg at rate of ₹ $55 / \mathrm{kg}$ and GST $5 \%$, butter 500 g at rate of ₹ $600 / \mathrm{kg}$, GST $12 \%$, shampoo 200 ml bottle at rate of ₹ $110 / \mathrm{bottle}$, GST $18 \%$, hair colour 2 packets at rate of 160 per packet GST 18%. Prepare the bill for the above purchases.
Answer
Bill for housewife
ArticleWeight/Number of articlesCost/article (in ₹)Amount (in ₹)Total amount (in ₹)
Sugar5 Kg₹ 55 per kg GST 5%275.0013.75288.75
Butter500 g₹ 600/kg GST 12%300.0036.00336.00
Butter1 bottle₹ 110/bottle CST 18%110.0019.80129.80
Hair colour2 packers₹ 160/packers GST 18%320.0057.60377,60
Total bill1132.15
Amount of bill = ₹ 1132.15.
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Question 25 Marks
Compute mean deviation from mean of the following distribution:
Marks10-2020-3030-4040-5050-6060-7070-8080-90
No. of students8101525201095
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Question 35 Marks
Find the Spearman's rank correlation between marks in Mathematics and Statistics obtained by 10 students:
Maths in Mathematics80389530748491606640
Marks in Statistics85509258706588565246
Answer
We construct the following table:
Marks in MathematicsRank R1Marks in StatisticsRank R2Diff. d = R1 - R2d2
80485311
38950900
95192100
3010586416
74570411
843655-24
91288200
60756700
666528-24
4084610-24
30
$\therefore$ Spearman's rank correlation $=\mathrm{r}=1-\frac{6 \Sigma d^{2}}{n\left(n^{2}-1\right)}=1-\frac{180}{10 \times 99}=+\frac{9}{11}$
$\therefore r=+0.82$
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Question 45 Marks
Evaluate $\lim _{x \rightarrow 2} \frac{x^{3}+4 x-16}{2 x^{3}-3 x-10}$
Answer
$\lim _{x \rightarrow 2} \frac{x^{3}+4 x-16}{2 x^{3}-3 x-10}$
$=\lim _{x \rightarrow 2} \frac{(x-2)\left(x^{2}+2 x+8\right)}{(x-2)\left(2 x^{2}+4 x+5\right)}$
$=\lim _{x \rightarrow 2} \frac{x^{2}+2 x+8}{2 x^{2}+4 x+5}=\frac{4+4+8}{8+8+5}=\frac{16}{21}$
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Question 55 Marks
If ${ }^{n} P_{r}=336,{ }^{n} C_{r}=56$. Find $n$ and $r$ and hence find ${ }^{n-1} C_{r-1}$.
Answer
${ }^{n} P_{r}=336,{ }^{n} C_{r}=56$
We know ${ }^{\mathrm{n}} \mathrm{P}_{\mathrm{r}}=\mathrm{r}$! ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \Rightarrow 336=56 \mathrm{r}$!
$\Rightarrow r!=6=3!\Rightarrow r=3$
Consider ${ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}=56 \Rightarrow{ }^{\mathrm{n}} \mathrm{C}_{3}=56$
$\Rightarrow \frac{n(n-1)(n-2)}{3!}=56$
$\Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)=56 \times 6=8 \times 7 \times 6$
$\Rightarrow \mathrm{n}(\mathrm{n}-1)(\mathrm{n}-2)=8(8-1)(8-2) \Rightarrow \mathrm{n}=8$
Hence $n=8, r=3$
$\therefore{ }^{\mathrm{n}-1} \mathrm{C}_{\mathrm{r}-1}={ }^{7} \mathrm{C}_{2}=\frac{7 \times 6}{2}=21$.
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Question 65 Marks
i. How many different words can be formed with the letters of the word HARYANA?
ii. How many of these begin with H and end with N?
iii. In how many of these H and N are together?
Answer
i. There are 7 letters in the word 'HARYANA' of which 3 are A's and remaining all are each of its own kind.
So, total number of words $=\frac{7!}{3!1!1!1!1!}=\frac{7!}{3!}=840$
ii. After fixing H in first place and N in last place, we have 5 letters out of which three are alike i.e. A's and remaining all are each of its own kind.
So, total number of words $=\frac{5!}{3!}=20$
iii. Considering H and N together we have $7-2+1=6$ letters out of which three are alike i.e A's and others are each of its own kind. These six letters can be arranged in $\frac{6!}{3!}$ ways. But H and N can be arranged amongst themselves in $2!$ ways.
Hence, the requisite number of words $=\frac{6!}{3!} \times 2!=120 \times 2=240$
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5 Marks Questions - Applied Maths STD 11 Science Questions - Vidyadip