Ganita Manjari - Grade 9 Mathematics Part II
Exercise Set 10.3 - How Quantities Combine: Understanding Data
Questions 1 to 8 - Complete, Formula-Based Solutions (Blogger Mobile Final)
Important Formulae
\[ \boxed{\text{Percentage}=\frac{\text{Part}}{\text{Whole}}\times100} \] \[ \boxed{\text{Weighted Mean}= \frac{\sum(\text{Value}\times\text{Weight})} {\sum\text{Weights}}} \] \[ \boxed{\text{Amount of a substance}= \text{Total quantity}\times\text{Concentration}} \]Question 1
A stationery shop owner made ₹8000 selling books, of which 30% is the profit amount, and ₹1000 selling book covers, of which 50% is the profit amount. What is the percentage of profit on the total sales?Solution
Step 1: Find the profit from selling books.
\[ \text{Profit from books} = \frac{30}{100}\times 8000 = 2400 \]Step 2: Find the profit from selling book covers.
\[ \text{Profit from book covers} = \frac{50}{100}\times 1000 = 500 \]Step 3: Find the total profit.
\[ \text{Total profit}=2400+500=2900 \]Step 4: Find the total sales.
\[ \text{Total sales}=8000+1000=9000 \]Step 5: Use the percentage formula.
\[ \text{Profit percentage} = \frac{\text{Total profit}}{\text{Total sales}}\times100 \] \[ = \frac{2900}{9000}\times100 = 32.222\ldots\% \] Answer: \[ \boxed{32.22\%\text{ approximately}} \]Question 2
A white stork's migration is tracked. The average daily distance travelled, calculated over 20 days, is 44.5 km. On the 21st day, it flew 55 km. What is the average daily distance travelled over these 21 days? Make a guess before you calculate.Solution
Guess: The new average should be slightly greater than \(44.5\) km because the 21st-day distance, \(55\) km, is greater than the previous average.Step 1: Find the total distance travelled in the first 20 days.
\[ \text{Total distance for 20 days} =44.5\times20 =890\text{ km} \]Step 2: Add the distance travelled on the 21st day.
\[ \text{Total distance for 21 days} =890+55 =945\text{ km} \]Step 3: Find the new average.
\[ \text{New Average} =\frac{945}{21} =45\text{ km} \] Answer: \[ \boxed{45\text{ km}} \]Question 3
A 600 mL solution with 5% salt is mixed with a 300 mL solution with 8% sugar. What are the concentrations of salt and sugar in the mixture?Options:
(i) Salt: 5%, Sugar: 8%
(ii) Salt: 13%, Sugar: 3%
(iii) Salt: 6%, Sugar: 6%
(iv) Salt: 5.55%, Sugar: 8.88%
(v) Salt: 3.33%, Sugar: 2.67%
(vi) Salt: 4.1%, Sugar: 7.08%
Solution
Step 1: Find the amount of salt in the first solution.
\[ \text{Salt}=5\%\text{ of }600 =\frac{5}{100}\times600 =30\text{ mL} \]Step 2: Find the amount of sugar in the second solution.
\[ \text{Sugar}=8\%\text{ of }300 =\frac{8}{100}\times300 =24\text{ mL} \]Step 3: Find the total volume of the mixture.
\[ 600+300=900\text{ mL} \]Step 4: Find the concentration of salt.
\[ \text{Salt concentration} =\frac{30}{900}\times100 =3.333\ldots\% \]Step 5: Find the concentration of sugar.
\[ \text{Sugar concentration} =\frac{24}{900}\times100 =2.666\ldots\% \] Answer: \[ \boxed{\text{Salt}=3.33\%,\qquad\text{Sugar}=2.67\%} \] Therefore, the correct option is \[ \boxed{\text{(v)}} \]Question 4
At a panipuri (golgappa) stall, the concentration of spice in the pani (spiced water) was 8%. Many customers complained that it was too spicy. What quantity of regular water should be mixed into 10 litres of pani so that the spice level is reduced to \(\left(\frac34\right)^{\text{th}}\) of the original concentration?Solution
Step 1: Find the original amount of spice.
\[ \text{Spice}=8\%\text{ of }10 =\frac{8}{100}\times10 =0.8\text{ L} \]Step 2: Find the required new concentration.
\[ \text{New concentration} =\frac34\times8\% =6\% \]Step 3: Let \(x\) litres of regular water be added.
Since regular water contains no spice, the amount of spice remains \(0.8\) L.
\[ \frac{0.8}{10+x}=6\%=\frac{6}{100} \]Step 4: Solve for \(x\).
\[ 0.8=0.06(10+x) \] \[ 0.8=0.6+0.06x \] \[ 0.2=0.06x \] \[ x=\frac{0.2}{0.06} =\frac{10}{3} =3.333\ldots \] Answer: \[ \boxed{3\frac13\text{ litres}} \] of regular water should be added.Question 5
A physical fitness evaluation is being undertaken. The final marks are calculated by combining the marks for strength, flexibility, and agility in the ratio \(4:5:6\). Rashi has scored 60, 65, and 70. Keerthi has scored 55, 65, and 75 respectively.(i) Find out who has scored more without calculating.
(ii) What are their final marks?
Solution
(i) Find who has scored more without calculating
The weights are:
\[ \text{Strength}:\text{Flexibility}:\text{Agility}=4:5:6 \]Compare Keerthi with Rashi:
| Component | Rashi | Keerthi | Difference | Weight |
|---|---|---|---|---|
| Strength | 60 | 55 | -5 | 4 |
| Flexibility | 65 | 65 | 0 | 5 |
| Agility | 70 | 75 | +5 | 6 |
Keerthi loses \(5\) marks in strength, whose weight is \(4\), but gains \(5\) marks in agility, whose weight is \(6\). Agility has the larger weight.
\[ (-5)(4)+(0)(5)+(5)(6) =-20+0+30 =+10 \]Thus, Keerthi's weighted total is higher.
Answer to (i): \[ \boxed{\text{Keerthi has scored more.}} \](ii) Find their final marks
Rashi's final marks:
\[ \frac{60(4)+65(5)+70(6)}{4+5+6} \] \[ =\frac{240+325+420}{15} =\frac{985}{15} \] \[ =65.666\ldots \] \[ \boxed{\text{Rashi's final marks}=65.67\text{ approximately}} \]Keerthi's final marks:
\[ \frac{55(4)+65(5)+75(6)}{4+5+6} \] \[ =\frac{220+325+450}{15} =\frac{995}{15} \] \[ =66.333\ldots \] \[ \boxed{\text{Keerthi's final marks}=66.33\text{ approximately}} \]Question 6
A restaurant collected ratings from 10 customers on a scale of 1 to 5. The resulting data is shown below.| 5★ | 4★ | 3★ | 2★ | 1★ | |
|---|---|---|---|---|---|
| Food | 5 | 3 | 2 | 0 | 0 |
| Ambience | 0 | 4 | 5 | 1 | 0 |
| Service | 1 | 2 | 2 | 4 | 1 |
Solution
Step 1: Find the average rating for Food.
\[ \text{Food average} = \frac{5(5)+4(3)+3(2)+2(0)+1(0)}{10} \] \[ =\frac{25+12+6}{10} =\frac{43}{10} =4.3 \]Step 2: Find the average rating for Ambience.
\[ \text{Ambience average} = \frac{5(0)+4(4)+3(5)+2(1)+1(0)}{10} \] \[ =\frac{16+15+2}{10} =\frac{33}{10} =3.3 \]Step 3: Find the average rating for Service.
\[ \text{Service average} = \frac{5(1)+4(2)+3(2)+2(4)+1(1)}{10} \] \[ =\frac{5+8+6+8+1}{10} =\frac{28}{10} =2.8 \]Step 4: Use the weights \(6:5:4\).
\[ \text{Combined average} = \frac{4.3(6)+3.3(5)+2.8(4)}{6+5+4} \] \[ = \frac{25.8+16.5+11.2}{15} \] \[ =\frac{53.5}{15} =3.5666\ldots \] Answer: \[ \boxed{3.57\text{ approximately}} \]Question 7
The following table shows the weight data of langurs in an animal facility. Without doing any computations, can you tell whether there are more male langurs or more female langurs? Or are they equal in number?| Male langurs | Female langurs | All langurs | |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | \(x\) | \(y\) | 60 |
(a) \(\displaystyle\frac{16.5x+13.8y}{x+y}=14.925\)
(b) \(\displaystyle\frac{16.5x+13.8y}{60}=14.925\)
(c) \(\displaystyle\frac{16.5x+13.8y}{2}=14.925\)
(d) \(\displaystyle\frac{16.5x+13.8y}{16.5+13.8}=14.925\)
(ii) Find out how many male langurs are present.
(iii) A female langur weighing 15.2 kg is admitted to the facility. What is the average weight of the female langurs after this?
(iv) Two male langurs weighing 16.9 kg and 16.1 kg are released from the facility. What is the average weight of the male langurs after this?
(v) Now, suppose one of the male langurs lost 1 kg of weight. What is the average weight of all the male langurs after this?
Solution
(i) Identify the correct expression
The combined average is:
\[ \text{Overall Average} = \frac{\text{Total weight of all langurs}}{\text{Total number of langurs}} \]Total weight of male langurs is \(16.5x\), and total weight of female langurs is \(13.8y\).
Also, \(x+y=60\). Therefore: \[ \frac{16.5x+13.8y}{x+y}=14.925 \] \[ \boxed{\text{Correct expression: (a)}} \] Since \(x+y=60\), option (b) is also mathematically equivalent to (a): \[ \boxed{\frac{16.5x+13.8y}{60}=14.925} \] Thus, if the question asks which expressions describe the scenario, (a) and (b) are equivalent and correct.
(ii) Number of male langurs
Since \(x+y=60\):
\[ y=60-x \]Using option (a):
\[ \frac{16.5x+13.8y}{60}=14.925 \] \[ 16.5x+13.8y=14.925(60) \] \[ 16.5x+13.8y=895.5 \]Substitute \(y=60-x\):
\[ 16.5x+13.8(60-x)=895.5 \] \[ 16.5x+828-13.8x=895.5 \] \[ 2.7x=67.5 \] \[ x=25 \] Therefore: \[ y=60-25=35 \] Answer: \[ \boxed{25\text{ male langurs}} \] and \[ \boxed{35\text{ female langurs}} \] So there are more female langurs.(iii) A 15.2 kg female langur is admitted
Original total weight of female langurs:
\[ 35\times13.8=483\text{ kg} \]After admitting the new langur:
\[ \text{New total weight}=483+15.2=498.2\text{ kg} \] \[ \text{New number}=35+1=36 \] \[ \text{New average}=\frac{498.2}{36}=13.8388\ldots\text{ kg} \] \[ \boxed{13.84\text{ kg approximately}} \](iv) Two male langurs weighing 16.9 kg and 16.1 kg are released
Original total weight of male langurs:
\[ 25\times16.5=412.5\text{ kg} \]Total weight released:
\[ 16.9+16.1=33\text{ kg} \]Remaining total weight and number:
\[ 412.5-33=379.5\text{ kg} \] \[ 25-2=23 \] \[ \text{New average}=\frac{379.5}{23}=16.5\text{ kg} \] \[ \boxed{16.5\text{ kg}} \](v) One male langur loses 1 kg
The number of male langurs remains \(25\), but the total weight decreases by \(1\) kg.
\[ \text{New total weight}=412.5-1=411.5\text{ kg} \] \[ \text{New average}=\frac{411.5}{25}=16.46\text{ kg} \] \[ \boxed{16.46\text{ kg}} \]Question 8
Given: Dead Sea water = 34%, groundwater = 0.01%, purified drinking water = 0.001%.
(i) What is the salinity of the mixture if 1 litre of Dead Sea water is mixed with 2 litres of purified drinking water?
Solution — (i) Salinity of the mixture
Formula:
Mixture salinity = [(salinity of first water × volume of first water) + (salinity of second water × volume of second water)] ÷ total volume
Substitution:
Mixture salinity = [34 × 1 + 0.001 × 2] ÷ (1 + 2) %
= (34 + 0.002) ÷ 3 %
= 34.002 ÷ 3 %
= 11.334%
Answer: 11.334% approximately.
(ii) Dead Sea water + purified drinking water
Question: Is it possible to mix water from the Dead Sea and purified drinking water to get a mixture with the salinity of groundwater? What quantity of purified drinking water should be mixed with 1 litre of Dead Sea water?
Step 1: Check whether the target salinity is possible.
0.001% < 0.01% < 34%
Therefore, the required salinity 0.01% lies between the two original salinities. So, yes, it is possible.
Step 2: Let x litres of purified drinking water be mixed with 1 litre of Dead Sea water.
Step 3: Use the mixture-salinity formula.
(34 × 1 + 0.001 × x) ÷ (1 + x) = 0.01
(34 + 0.001x) ÷ (1 + x) = 0.01
Step 4: Cross-multiply.
34 + 0.001x = 0.01(1 + x)
34 + 0.001x = 0.01 + 0.01x
Step 5: Collect the terms containing x.
34 − 0.01 = 0.01x − 0.001x
33.99 = 0.009x
Step 6: Solve for x.
x = 33.99 ÷ 0.009
x = 3776.666... litres
Answer: x ≈ 3776.67 litres.
Therefore, approximately 3776.67 litres of purified drinking water must be mixed with 1 litre of Dead Sea water to obtain a mixture having salinity 0.01%.
(iii) Dead Sea water + groundwater
Question: Is it possible to mix water from the Dead Sea and groundwater to get a mixture with the salinity of purified drinking water? What quantity of groundwater should be mixed with 1 litre of Dead Sea water?
Step 1: Check whether the target salinity is possible.
Groundwater salinity = 0.01%, whereas purified drinking water salinity = 0.001%.
0.001% < 0.01% < 34%
A mixture of Dead Sea water (34%) and groundwater (0.01%) can have a salinity between 0.01% and 34%. It cannot reach 0.001%. Therefore, it is not possible.
Step 2: Verify the result using the mixture formula.
Let x litres of groundwater be mixed with 1 litre of Dead Sea water.
(34 × 1 + 0.01 × x) ÷ (1 + x) = 0.001
(34 + 0.01x) ÷ (1 + x) = 0.001
Step 3: Cross-multiply.
34 + 0.01x = 0.001(1 + x)
34 + 0.01x = 0.001 + 0.001x
Step 4: Solve for x.
34 − 0.001 = 0.001x − 0.01x
33.999 = −0.009x
x = 33.999 ÷ (−0.009)
x = −3777.666... litres
A quantity of water cannot be negative. Thus, there is no positive value of x. This confirms mathematically that the required mixture cannot be made.
Answer: It is not possible.
No positive quantity of groundwater can be mixed with 1 litre of Dead Sea water to obtain a mixture having salinity 0.001%.
End of Question 8 — Parts (i), (ii) and (iii) are complete.
