End - OF - Chapter Exercises Chapter 10 Class 9 Maths Ganita Manjari Part 2

 

How Quantities Combine: Understanding Data

End-of-chapter exercises — Questions 1 to 6, with detailed step-by-step solutions. Equations are written in LaTeX and rendered using MathJax.

Question 1: Run rate in cricket

In a T20 match, a team scores 6 runs in the first over. Its run rate is the average number of runs scored per over.

1. In the second over, the team scores 12 runs. What is the run rate now?
2. After 19 overs, its run rate is 6. If the team scores 12 runs in the last over, what is the run rate after 20 overs?

(i) After two overs

Run rate is total runs divided by total overs. The team has scored \(6+12=18\) runs in 2 overs.

\[ \text{Run rate}=\frac{\text{Total runs}}{\text{Total overs}} =\frac{6+12}{2}=\frac{18}{2}=9 \]

Answer: The run rate is 9 runs per over.

(ii) After 20 overs

After 19 overs, the total runs scored are \(19\times 6=114\). In the final over, the team adds 12 runs.

\[ \text{Total runs after 20 overs}=114+12=126 \]

\[ \text{Run rate}=\frac{126}{20}=\frac{63}{10}=6.3 \]

Answer: The run rate after 20 overs is 6.3 runs per over.

Question 2: Average price of shuttlecocks

Five friends collected the following prices. N means nylon and F means feather.

Friend	Prices collected (₹)	Number of prices
Yusuf	40(N), 105(N), 383(F), 108(N), 165(F), 116(F)	6
Srikanth	194(N), 85(N), 93(N), 121(N)	4
Kashvi	49(N), 297(F), 105(N), 275(F), 40(N)	5
Prasanna	124(N), 333(F), 182(N), 258(F)	4
Gracy	220(F), 458(F), 129(F), 183(N)	4

(i) Combine the averages calculated by the five friends

Let the averages calculated by Yusuf, Srikanth, Kashvi, Prasanna and Gracy be \(a_y,a_s,a_k,a_p,a_g\), respectively.

Step 1: Calculate each friend's average.

\[ a_y=\frac{40+105+383+108+165+116}{6} =\frac{917}{6}\approx \text{Rs. }152.83 \]

\[ a_s=\frac{194+85+93+121}{4} =\frac{493}{4}=\text{Rs. }123.25 \]

\[ a_k=\frac{49+297+105+275+40}{5} =\frac{766}{5}=\text{Rs. }153.20 \]

\[ a_p=\frac{124+333+182+258}{4} =\frac{897}{4}=\text{Rs. }224.25 \]

\[ a_g=\frac{220+458+129+183}{4} =\frac{990}{4}=\text{Rs. }247.50 \]

Step 2: Use the combined-average expression. The friends collected 6, 4, 5, 4 and 4 prices respectively, so there are \(6+4+5+4+4=23\) prices altogether.

\[ \text{Combined average} =\frac{6a_y+4a_s+5a_k+4a_p+4a_g}{23} \]

Substitute the values of the averages:

\[ =\frac{ 6\left(\frac{917}{6}\right) +4\left(\frac{493}{4}\right) +5\left(\frac{766}{5}\right) +4\left(\frac{897}{4}\right) +4\left(\frac{990}{4}\right)}{23} \]

\[ =\frac{917+493+766+897+990}{23} =\frac{4063}{23} \approx \text{Rs. }176.65 \]

Answer (i): The combined average price is \(\displaystyle \frac{\text{Rs. }4063}{23}\approx \text{Rs. }176.65\).

(ii) Combine the averages for nylon and feather shuttlecocks

Let \(a_n\) be the average price of nylon shuttlecocks and \(a_f\) the average price of feather shuttlecocks.

Step 1: Find the nylon total and average. There are 13 nylon prices.

\[ \begin{aligned} \text{Total nylon price} &=(40+105+108)+(194+85+93+121)\\ &\quad +(49+105+40)+(124+182)+183\\ &=253+493+194+306+183\\ &=\text{Rs. }1429 \end{aligned} \]

\[ a_n=\frac{1429}{13}\approx \text{Rs. }109.92 \]

Step 2: Find the feather total and average. There are 10 feather prices.

\[ \begin{aligned} \text{Total feather price} &=(383+165+116)+(297+275)\\ &\quad +(333+258)+(220+458+129)\\ &=664+572+591+807\\ &=\text{Rs. }2634 \end{aligned} \]

\[ a_f=\frac{2634}{10}=\text{Rs. }263.40 \]

Step 3: Combine the two averages.

\[ \text{Combined average}=\frac{13a_n+10a_f}{13+10} =\frac{13a_n+10a_f}{23} \]

Substitute the exact values of the averages:

\[ =\frac{13\left(\frac{1429}{13}\right) +10\left(\frac{2634}{10}\right)}{23} =\frac{1429+2634}{23} =\frac{4063}{23} \approx \text{Rs. }176.65 \]

Answer (ii): The combined average price is \(\displaystyle \frac{\text{Rs. }4063}{23}\approx \text{Rs. }176.65\). Yes, it is equal to the answer in part (i), because both methods combine exactly the same 23 prices.

Important idea: A combined average is a weighted average. Multiply each group’s average by the number of observations in that group, add these totals, and divide by the total number of observations. Do not simply take the average of the group averages unless all groups have the same number of observations.

Question 3: Average price of shares

Shreyas holds 25 shares at an average price of ₹150 per share. Vaishnavi holds 5 shares of the same company, also at an average price of ₹150 per share.

(i) Shreyas buys 10 more shares at ₹30 each

First calculate the total cost of the shares he already owns and the new shares.

\[ \text{Existing cost}=25\times \text{Rs. }150=\text{Rs. }3750 \]

\[ \text{Cost of new shares}=10\times \text{Rs. }30=\text{Rs. }300 \]

He now owns \(25+10=35\) shares, costing \(\text{Rs. }3750+\text{Rs. }300=\text{Rs. }4050\).

\[ \text{New average price}=\frac{\text{Rs. }4050}{35} =\frac{\text{Rs. }810}{7}\approx \text{Rs. }115.71 \]

Answer (i): The new average price is \(\displaystyle \frac{\text{Rs. }810}{7}\), or approximately ₹115.71 per share.

(ii) Vaishnavi buys shares at ₹30 each; her new average is ₹70

Let \(x\) be the number of shares Vaishnavi buys. Her original 5 shares cost \(5\times \text{Rs. }150=\text{Rs. }750\). The new shares cost \(\text{Rs. }30x\), and the total number of shares becomes \(5+x\).

\[ \frac{750+30x}{5+x}=70 \]

Multiply both sides by \(5+x\):

\[ 750+30x=70(5+x)=350+70x \]

\[ 750-350=70x-30x \quad\Rightarrow\quad 400=40x \quad\Rightarrow\quad x=\frac{400}{40}=10 \]

Answer (ii): Vaishnavi bought 10 shares.

Question 4: Mean depth of a pool

A pool is 30 hastas long. It is divided into five sections of lengths 4, 5, 6, 7 and 8 hastas, with respective depths 9, 7, 7, 3 and 2 hastas. Find the mean depth of the pool.

The sections have different lengths, so the mean depth must be weighted by each section’s length.

Section length (hastas)	Depth (hastas)	Length × depth
4	9	4 × 9 = 36
5	7	5 × 7 = 35
6	7	6 × 7 = 42
7	3	7 × 3 = 21
8	2	8 × 2 = 16
Total = 30		Total = 150

\[ \text{Mean depth} =\frac{4(9)+5(7)+6(7)+7(3)+8(2)}{30} =\frac{36+35+42+21+16}{30} =\frac{150}{30}=5 \]

Answer: The mean depth of the pool is 5 hastas.

Question 5: Average price of gold after a purchase

Suvarna already owns 1 g of gold purchased at ₹15,000 per gram. So her existing gold costs ₹15,000 in total. For each scenario, use:

\[ \text{New average price per gram} =\frac{\text{Existing total cost}+\text{New purchase cost}} {\text{Existing weight}+\text{New weight}} \]

Here, the existing weight is 1 g and the existing cost is ₹15,000.

(i) Purchase 1 g at ₹30,000 per gram

\[ \frac{15000+(1\times30000)}{1+1} =\frac{45000}{2}=\text{Rs. }22500 \]

(ii) Purchase 2 g at ₹30,000 per gram

\[ \frac{15000+(2\times30000)}{1+2} =\frac{75000}{3}=\text{Rs. }25000 \]

(iii) Purchase 1 g at ₹10,000 per gram

\[ \frac{15000+(1\times10000)}{1+1} =\frac{25000}{2}=\text{Rs. }12500 \]

(iv) Purchase 10 g at ₹10,000 per gram

\[ \frac{15000+(10\times10000)}{1+10} =\frac{115000}{11} \approx \text{Rs. }10454.55 \]

(v) Purchase 0.5 g at ₹30,000 per gram

\[ \frac{15000+(0.5\times30000)}{1+0.5} =\frac{30000}{1.5}=\text{Rs. }20000 \]

(vi) Purchase 0.5 g at ₹15,000 per gram

\[ \frac{15000+(0.5\times15000)}{1+0.5} =\frac{22500}{1.5}=\text{Rs. }15000 \]

Answers:

1. ₹22,500 per gram
2. ₹25,000 per gram
3. ₹12,500 per gram
4. \(\displaystyle \frac{\text{Rs. }115000}{11}\approx \text{Rs. }10,454.55\) per gram
5. ₹20,000 per gram
6. ₹15,000 per gram

Number-line diagrams for scenarios (i)–(vi)

In each number line, the hollow point O = 15 marks the original average price. The red point marks the new average after the purchase. The scale is in thousands of rupees per gram.

(i) New average = ₹22.5 thousand

0 5 10 15 20 25 30 O = 15 22.5

(ii) New average = ₹25 thousand

0 5 10 15 20 25 30 O = 15 25

(iii) New average = ₹12.5 thousand

0 5 10 15 20 25 30 O = 15 12.5

(iv) New average = ₹10.45 thousand

0 5 10 15 20 25 30 O = 15 10.45

(v) New average = ₹20 thousand

0 5 10 15 20 25 30 O = 15 20

(vi) New average = ₹15 thousand

0 5 10 15 20 25 30 O = 15

Question 6: What happens if all weights are doubled?

Suppose the data values are \(x_1,x_2,\ldots,x_n\), and their corresponding weights are \(w_1,w_2,\ldots,w_n\).

Step 1: Write the original weighted average

\[ A=\frac{w_1x_1+w_2x_2+\cdots+w_nx_n} {w_1+w_2+\cdots+w_n} \]

Step 2: Double every weight

The new weights are \(2w_1,2w_2,\ldots,2w_n\). The new weighted average is

\[ A'=\frac{(2w_1)x_1+(2w_2)x_2+\cdots+(2w_n)x_n} {2w_1+2w_2+\cdots+2w_n} \]

Step 3: Factor out 2 and simplify

\[ A'=\frac{2(w_1x_1+w_2x_2+\cdots+w_nx_n)} {2(w_1+w_2+\cdots+w_n)} =\frac{w_1x_1+w_2x_2+\cdots+w_nx_n} {w_1+w_2+\cdots+w_n}=A \]

Answer: The weighted average does not change when every weight is doubled. The factor of 2 appears in both the numerator and denominator and cancels out. More generally, multiplying all weights by the same positive constant leaves the weighted average unchanged.

Quick example

Suppose values are 10 and 20, with weights 1 and 3.

\[ A=\frac{1(10)+3(20)}{1+3} =\frac{10+60}{4} =\frac{70}{4}=17.5 \]

Now double both weights: 1 becomes 2 and 3 becomes 6.

\[ A'=\frac{2(10)+6(20)}{2+6} =\frac{20+120}{8} =\frac{140}{8}=17.5 \]

Therefore, the weighted average stays 17.5 in this example too. Doubling every weight changes the total weighted sum and the total weight by the same factor, so their ratio remains unchanged.

End of Questions 1–6. Check that every average is calculated as total weighted value divided by total weight or total number of observations.