CBSE Class 9 Maths Ganita Manjari Part 2 Ch 10 How Quantities Combine Understanding Data Short Answer Questions with Answers

Short Answer Questions

Question 1. Section A has 24 students with a mean score of 65. Section B has 16 students. Taken together, all 40 students have a mean score of 69.
(i) Find the mean score of Section B.
(ii) The combined mean 69 is 4 marks above Section A's mean but further than that below Section B's mean. Explain why.
Answer:
(i) Total = 40 × 69 = 2760; Section A = 24 × 65 = 1560; Section B = 2760 − 1560 = 1200.
Mean of B = 1200 ÷ 16 = 75.
(ii) Section A is larger (24 vs 16), so it pulls the combined mean closer to 65. To balance, B's mean must be further away.
Check: 24 × 4 = 96 = 16 × 6, so the two sides pull equally.
In simple words: The bigger section keeps the combined mean close to its own mean, so the smaller section's mean sits further away.

Teacher's Note:
1. Balance idea: size × distance is equal on both sides.
2. Work with totals for part (i).

Question 2. 500 mL of a solution containing 6% acid is mixed with 300 mL of a solution containing 14% acid. Find the percentage of acid in the mixture, and state with a reason whether it is above or below 10%.
Answer:
Acid: 500 × 0.06 = 30 mL and 300 × 0.14 = 42 mL, total 72 mL in 800 mL.
Concentration = 72 ÷ 800 = 9%.
It is below 10% (the midpoint) because more of the mixture comes from the weaker 6% solution.
In simple words: More weak solution than strong solution drags the mixture below halfway.

Teacher's Note:
1. Predict the side of the midpoint before calculating.
2. Then verify with the numbers.

Question 3. Batch A is 150 kg of brass containing 80% copper, and Batch B is 100 kg of brass containing 55% copper. The two batches are melted together. Find the percentage of copper in the combined alloy.
Answer:
Copper: 150 × 0.80 = 120 kg and 100 × 0.55 = 55 kg, total 175 kg in 250 kg.
Percentage = 175 ÷ 250 = 70%.
It is above the midpoint 67.5% because the heavier batch is the richer one.
In simple words: The alloy is 70% copper, nearer the heavier, richer batch.

Teacher's Note:
1. Alloys work exactly like liquid mixtures.
2. Mass acts as the weight.

Question 4. Rehmat scores 55% in the internal tests, 70% in the project and 80% in the final exam. The three are combined in the ratio 2 : 3 : 5. Find Rehmat's annual percentage score.
Answer:
\( \frac{55 \times 2 + 70 \times 3 + 80 \times 5}{10} = \frac{110 + 210 + 400}{10} = \frac{720}{10} = \) 72%.
The plain average would be about 68.3%; weighting helps because the final exam is her best score.
In simple words: The final exam carries half the weight, and it is Rehmat's best result, so her annual score is 72%.

Teacher's Note:
1. Scores are already percentages, so no conversion is needed.
2. Compare with the plain mean to explain.

Question 5. Two collections of data have the same number of values, n. Their averages are a and b. Prove that the average of the combined collection is exactly \( \frac{a + b}{2} \).
Answer:
Sum of first collection = an; sum of second = bn.
Combined: 2n values with total an + bn.
Average = \( \frac{an + bn}{2n} = \frac{(a + b)n}{2n} = \frac{a + b}{2} \). The n cancels.
In simple words: With equal sizes, the size cancels out and the simple average is exact.

Teacher's Note:
1. The cancelling n is the heart of the proof.
2. With unequal sizes, nothing cancels.

Question 6. A team's run rate after 12 overs is 5.5. Find how many runs the team must score in the 13th over so that the run rate after 13 overs becomes exactly 6.
Answer:
Runs after 12 overs = 12 × 5.5 = 66.
Needed after 13 overs = 13 × 6 = 78.
Runs in the 13th over = 78 − 66 = 12.
In simple words: Work out the target total, then subtract what has already been scored.

Teacher's Note:
1. To raise a mean to 6, the new value must beat 6 by enough.
2. Target total − current total = required value.

Question 7. A family's monthly expenditure, in rupees, is: Housing 6000, Food 5000, Education 4000, Transport 3000, Other 2000. Find the percentage share of each category, and state which category would have the longest segment in a 100% stacked bar chart of this data.
Answer:
Total = ₹20 000.
Housing 30%, Food 25%, Education 20%, Transport 15%, Other 10% (total 100%).
Housing would have the longest segment (30%).
In simple words: Divide each amount by ₹20 000; housing takes the biggest share.

Teacher's Note:
1. Check that shares add up to 100%.
2. Largest amount ⇒ largest share within a bar.

Question 8. The mean of 15 observations is 32. Two of the observations, 24 and 27, are removed from the data. Find the mean of the remaining 13 observations.
Answer:
Total = 15 × 32 = 480; remove 24 + 27 = 51, leaving 429.
New mean = 429 ÷ 13 = 33.
In simple words: Take the two values out of the total and divide by 13.

Teacher's Note:
1. Both removed values are below the mean, so the mean rises.
2. Reduce the count by 2.

Question 9. A stall has 8 litres of pani containing 10% spice. Find the quantity of plain water that must be added so that the spice level falls to 4%.
Answer:
Spice = 8 × 0.10 = 0.8 L.
\( \frac{0.8}{8 + w} = 0.04 \Rightarrow 8 + w = 20 \Rightarrow w = \) 12 litres.
In simple words: The spice stays 0.8 L, so the total must grow to 20 L; that needs 12 L of water.

Teacher's Note:
1. Halving to 5% needs 8 L; going further to 4% needs 12 L.
2. Each further dilution costs more water.

Question 10. The stacked bar chart shows the wickets taken by a bowler in three formats, split between matches played at home and overseas.
(i) Find the total number of wickets taken in Test matches.
(ii) Find the format in which the bowler took the fewest wickets in all.
(iii) State which is easier to read from this chart — the total for each format, or the comparison of home wickets across the three formats — and why.
Answer:
(i) Test: 70 + 130 = 200 wickets.
(ii) Totals: Test 200, ODI 150, T20 100. Fewest: T20.
(iii) The totals are easier: every bar starts at 0, so full bar lengths compare directly. Home wickets also happen to start at 0 here, but overseas segments begin at different points and would need measuring.
In simple words: Bar lengths give the totals at a glance; inner segments are harder to compare.

Teacher's Note:
1. Only segments starting at 0 compare easily across bars.
2. Totals are the strength of stacked bars.

Question 11. Explain, using a numerical example of your own, why a 100% stacked bar chart cannot be used to compare the actual quantities represented by two different bars.
Answer:
Shop P: books ₹3000, stationery ₹1000 (total ₹4000).
Shop Q: books ₹30 000, stationery ₹10 000 (total ₹40 000).
Both give 75% books and 25% stationery, so their 100% bars look identical.
Yet Shop Q sells ten times as much. The chart has replaced every total by 100%, so it cannot show this.
In simple words: Two shops ten times apart in sales can have exactly the same 100% bar.

Teacher's Note:
1. Identical shares never mean identical amounts.
2. Scaling all parts keeps shares unchanged.

Question 12. Shreyas holds 40 shares of a company at an average price of ₹200 per share. He then buys 10 more shares at ₹100 each.
(i) Find his average price per share after the purchase.
(ii) Explain why the new average is not ₹150.
Answer:
(i) Cost = 40 × 200 + 10 × 100 = 8000 + 1000 = ₹9000 for 50 shares. Average = ₹180.
(ii) ₹150 would be right only for equal numbers at each price. 40 shares at ₹200 outweigh 10 at ₹100 (4 : 1), pulling the average up to ₹180.
In simple words: Most shares cost ₹200, so the average stays close to ₹200.

Teacher's Note:
1. Quantities are the weights in share averaging.
2. Simple average of prices ignores quantities.

Question 13. A tank 30 m long is dug in three sections of lengths 6 m, 9 m and 15 m. The first two sections are 8 m and 6 m deep, and the mean depth of the whole tank works out to 7 m. Find the depth of the third section.
Answer:
\( \frac{8 \times 6 + 6 \times 9 + 15d}{30} = 7 \Rightarrow 102 + 15d = 210 \)
\( 15d = 108 \Rightarrow d = \) 7.2 m.
In simple words: The lengths act as weights; solve the mean equation for the missing depth.

Teacher's Note:
1. This is a weighted-mean problem worked backwards.
2. Check: (48 + 54 + 108) ÷ 30 = 7 ✓

Question 14. A 100% stacked bar chart shows that in Fatima's garden 45% of the year's blooms appeared in summer, while in Naveen's garden the summer share was 35%.
(i) State whose garden had the larger share of summer blooms.
(ii) State whether it can be concluded that Fatima's garden had more summer blooms than Naveen's.
(iii) Give a pair of yearly totals for the two gardens for which Naveen's garden has more summer blooms than Fatima's.
Answer:
(i) Fatima's (45% against 35%).
(ii) No. The shares are of different, unknown totals.
(iii) Fatima 200, Naveen 400: Fatima 0.45 × 200 = 90, Naveen 0.35 × 400 = 140.
In simple words: A bigger share of a smaller garden can still be fewer flowers.

Teacher's Note:
1. Invent totals to test cross-bar claims.
2. Share × total = number.

Question 15. Ten customers rate a restaurant on a scale of 1 to 5 for three aspects. The mean ratings work out to 4.3 for food, 3.8 for ambience and 3.3 for service. The three aspects are combined with weights in the ratio 5 : 3 : 2. Find the restaurant's overall rating.
Answer:
\( \frac{4.3 \times 5 + 3.8 \times 3 + 3.3 \times 2}{10} = \frac{21.5 + 11.4 + 6.6}{10} = \frac{39.5}{10} = \) 3.95.
The plain average is 3.8; the weighted result is higher because food, the best aspect, has the most weight.
In simple words: Food counts most and scored best, so the overall rating is 3.95.

Teacher's Note:
1. Explain the direction using where the heaviest weight lies.
2. Show all three products.

Question 16. A shopkeeper earns ₹12 000 from selling books, on which the profit is 25%, and ₹8000 from selling stationery, on which the profit is 40%. Find the percentage profit on the total sales.
Answer:
Profits: 25% of 12 000 = ₹3000 and 40% of 8000 = ₹3200, total ₹6200.
Total sales = ₹20 000, so profit = 6200 ÷ 20 000 × 100 = 31%.
In simple words: The overall profit rate is a weighted mean of 25% and 40%, weighted by sales.

Teacher's Note:
1. Nearer 25% because more sales came from books.
2. Sales amounts are the weights.

Question 17. The values x₁, x₂, …, x_n have weights w₁, w₂, …, w_n. Prove that if every weight is multiplied by the same non-zero number k, the weighted mean does not change.
Answer:
Original: \( S = \frac{w_1x_1 + \cdots + w_nx_n}{w_1 + \cdots + w_n} \).
New: \( \frac{kw_1x_1 + \cdots + kw_nx_n}{kw_1 + \cdots + kw_n} = \frac{k(w_1x_1 + \cdots + w_nx_n)}{k(w_1 + \cdots + w_n)} = S \), since k ≠ 0 cancels.
So the weighted mean is unchanged.
In simple words: The same factor appears on top and bottom, so it cancels.

Teacher's Note:
1. This is why a ratio like 3 : 2 : 5 fully fixes a weighted mean.
2. k ≠ 0 is needed for cancellation.

Question 18. The average daily electricity use of a house, in units, is given for two years. 2010: Lighting 300, Cooling 250, Kitchen 150, Other 300. 2025: Lighting 180, Cooling 700, Kitchen 400, Other 320. Find the percentage share of each category in each year, and state one change that a 100% stacked bar chart of this data would make clear at a glance.
Answer:
2010 total = 1000: Lighting 30%, Cooling 25%, Kitchen 15%, Other 30%.
2025 total = 1600: Lighting 11.25%, Cooling 43.75%, Kitchen 25%, Other 20%.
The chart would show cooling growing from a quarter to nearly half of use, while lighting shrinks sharply.
In simple words: Electricity use has shifted from lighting to cooling.

Teacher's Note:
1. Divide by each year's own total.
2. 100% charts highlight changes in composition.

Question 19. The table shows the time, in minutes, taken by three athletes in a triathlon. P: swimming 60, cycling 300, running 240. Q: 75, 315, 210. R: 90, 360, 300.
(i) Find the total time taken by each athlete.
(ii) Find the percentage of the total time that Athlete P spent cycling.
(iii) State which of the two charts — a stacked bar chart or a 100% stacked bar chart — would let you see at a glance who finished the race first, and why.
Answer:
(i) P: 600 min; Q: 600 min; R: 750 min.
(ii) 300 ÷ 600 × 100 = 50%.
(iii) A stacked bar chart: bar length = total time. A 100% chart makes all bars equal and hides that R took longer.
In simple words: Finishing order depends on totals, which only a stacked bar shows.

Teacher's Note:
1. Choose the chart by the question being asked.
2. Totals ⇒ stacked; shares ⇒ 100% stacked.

Question 20. A group of n students has an average mark of 70 and a group of m students has an average mark of 85. The average mark of the two groups taken together is 76. Find the ratio n : m.
Answer:
\( \frac{70n + 85m}{n + m} = 76 \Rightarrow 70n + 85m = 76n + 76m \)
\( 9m = 6n \Rightarrow \frac{n}{m} = \frac{9}{6} = \frac{3}{2} \). So n : m = 3 : 2.
Check: 76 is 6 from 70 and 9 from 85; weights are in the inverse ratio 9 : 6 = 3 : 2.
In simple words: The group nearer the combined average is the larger one.

Teacher's Note:
1. Inverse-ratio check gives the answer quickly.
2. Solve the equation for full marks.