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

Very Short Answer Questions

Question 1. Write the formula for the weighted mean of the values x₁, x₂, …, x_n having weights w₁, w₂, …, w_n respectively, and state what the denominator represents.
Answer:
\( \bar{x} = \frac{w_1x_1 + w_2x_2 + \cdots + w_nx_n}{w_1 + w_2 + \cdots + w_n} \)
The denominator is the sum of all the weights; it plays the role that 'number of values' plays in an ordinary mean.

Teacher's Note:
1. When all weights are 1, this becomes the ordinary mean.
2. Never divide by n in a weighted mean.

Question 2. In a class, 15 boys have a mean height of 150 cm and 25 girls have a mean height of 146 cm. Find the mean height of the whole class.
Answer:
Boys: 15 × 150 = 2250 cm; girls: 25 × 146 = 3650 cm; total 5900 cm.
Mean = 5900 ÷ 40 = 147.5 cm.

Teacher's Note:
1. Not 148 (the midpoint): there are more girls.
2. The answer leans towards the larger group's mean.

Question 3. Three readings 5, 9 and 14 are combined so that the first carries weight 2, the second carries weight 3 and the third carries weight w. Their weighted mean comes to 10.7. Find w.
Answer:
\( \frac{10 + 27 + 14w}{5 + w} = 10.7 \Rightarrow 37 + 14w = 53.5 + 10.7w \)
\( 3.3w = 16.5 \Rightarrow \) w = 5.
Check: (10 + 27 + 70) ÷ 10 = 10.7 ✓

Teacher's Note:
1. Clear the fraction first, then collect w terms.
2. Always verify by substituting back.

Question 4. 400 mL of a syrup containing 12% sugar is mixed with 600 mL of a syrup containing 7% sugar. Find the percentage of sugar in the mixture.
Answer:
Sugar: 400 × 0.12 = 48 mL and 600 × 0.07 = 42 mL, total 90 mL.
Concentration = 90 ÷ 1000 = 9%.

Teacher's Note:
1. More of the weaker syrup ⇒ below the midpoint 9.5%.
2. Work in amounts of sugar, not percentages.

Question 5. Explain briefly why the simple average of two group averages can give a wrong value for the average of the combined group.
Answer:
Averaging two averages gives both groups equal importance.
If one group is larger, its values make up more of the combined group and should count for more. Otherwise the smaller group pulls the answer too far towards its own mean.

Teacher's Note:
1. The fix: weight each average by its group size.
2. Correct only when sizes are equal.

Question 6. A cricket team's run rate after 15 overs is 6. Find the total number of runs the team has scored.
Answer:
Runs = run rate × overs = 6 × 15 = 90 runs.

Teacher's Note:
1. Run rate is a mean: runs per over.
2. Any two of runs, overs and rate give the third.

Question 7. A team's run rate after 15 overs is 6. In the 16th over the team scores 22 runs. Find the run rate after 16 overs.
Answer:
Runs after 15 overs = 90; after 16 overs = 90 + 22 = 112.
Run rate = 112 ÷ 16 = 7.

Teacher's Note:
1. Convert rate → total → new rate.
2. 22 runs in one over lifts the rate by 1.

Question 8. The mean of 18 observations is 25. Find the sum of the observations.
Answer:
Sum = mean × number = 18 × 25 = 450.

Teacher's Note:
1. This 'mean × count' step starts most combining problems.
2. Keep it as a habit.

Question 9. The mean of 18 observations is 25. A new observation, 44, is added to the data. Find the new mean.
Answer:
Sum = 450; new sum = 450 + 44 = 494 over 19 observations.
New mean = 494 ÷ 19 = 26.

Teacher's Note:
1. The count increases by 1 as well.
2. 44 is above 25, so the mean rises.

Question 10. Two equal quantities of paint, one containing 30% white and the other 50% white, are mixed together. Find the percentage of white in the mixture.
Answer:
Equal quantities give the midpoint: (30 + 50) ÷ 2 = 40%.

Teacher's Note:
1. The midpoint shortcut needs equal quantities.
2. Unequal quantities pull towards the larger one.

Question 11. A student scores 60 marks in the internal test and 80 marks in the final exam. The two are combined in the ratio 1 : 3. Find the student's combined score.
Answer:
\( \frac{60 \times 1 + 80 \times 3}{1 + 3} = \frac{300}{4} = \) 75 marks.

Teacher's Note:
1. The ratio gives the weights directly.
2. The final exam counts three times as much.

Question 12. State what a stacked bar chart is used for, and say what the full length of one complete bar represents.
Answer:
A stacked bar chart shows totals and their parts together.
The full length of a bar represents the total of all its categories added together.

Teacher's Note:
1. Segments = parts; whole bar = total.
2. Contrast with 100% charts, where every bar is the same length.

Question 13. State one thing that a 100% stacked bar chart can not tell you, and give a reason.
Answer:
It cannot tell you the actual total (or any actual amount) of a bar.
Every bar is drawn to the same length whatever its true total, so identical-looking bars may stand for very different quantities.

Teacher's Note:
1. Shares yes, amounts no.
2. Totals must be given separately.

Question 14. A house used 2500 units of electricity in a year, of which 600 units were used for lighting. Find the percentage share of lighting.
Answer:
Share = \( \frac{600}{2500} \) × 100 = 24%.

Teacher's Note:
1. Part ÷ whole × 100.
2. Just under a quarter of the bar.

Question 15. In a weighted mean, every weight is multiplied by 5 while the values stay the same. State what happens to the weighted mean.
Answer:
It does not change. Numerator and denominator are both multiplied by 5, which cancels.

Teacher's Note:
1. Only the ratio of weights matters.
2. Contrast with adding to each weight.

Question 16. A pit 24 m long is divided along its length into sections of 6 m, 8 m and 10 m, dug to depths of 5 m, 3 m and 6 m respectively. Find the mean depth of the pit.
Answer:
Weight each depth by its length: \( \frac{5 \times 6 + 3 \times 8 + 6 \times 10}{24} = \frac{114}{24} = \) 4.75 m.

Teacher's Note:
1. This is Brahmagupta's mean depth of a pit.
2. The deepest section is also the longest, so the answer is above 4.5 m.

Question 17. A goldsmith melts 5 units of gold of 14 varna with 5 units of gold of 10 varna. Find the purity of the mixture in varna.
Answer:
Equal quantities, so the midpoint: (14 + 10) ÷ 2 = 12 varna.

Teacher's Note:
1. Equal weights ⇒ ordinary mean.
2. 16 varna = pure gold.

Question 18. A customer rates a snack 5 for taste and 2 for price. If taste and price carry weights in the ratio 3 : 1, find the customer's overall rating.
Answer:
\( \frac{5 \times 3 + 2 \times 1}{4} = \frac{17}{4} = \) 4.25.

Teacher's Note:
1. Taste has three times the weight, so the result is near 5.
2. Check the answer lies between 2 and 5.

Question 19. Is it possible to draw a stacked bar chart if only the corresponding 100% stacked bar chart is given? Give a reason.
Answer:
No. The 100% chart keeps only proportions; the totals were discarded. Many different totals give the same proportions, so the actual amounts cannot be recovered.

Teacher's Note:
1. The conversion works only from amounts to shares.
2. Give an example: 30/70 could be 3/7 or 300/700.

Question 20. The mean weight of 20 boxes is 8 kg. One box weighing 8 kg is removed. Find the mean weight of the remaining boxes.
Answer:
Total = 160 kg; after removal 152 kg over 19 boxes: 152 ÷ 19 = 8 kg.

Teacher's Note:
1. Removing a value equal to the mean leaves the mean unchanged.
2. A neat shortcut worth remembering.

Question 21. 250 mL of a solution containing 20% acid is diluted by adding 250 mL of water. Find the percentage of acid in the diluted solution.
Answer:
Acid = 250 × 0.20 = 50 mL, now in 500 mL.
Concentration = 50 ÷ 500 = 10%.

Teacher's Note:
1. Doubling the volume with water halves the concentration.
2. The acid amount never changes.

Question 22. The mean daily rainfall was 4 mm over the 30 days of April and 9 mm over the 31 days of May. Write an expression that gives the mean daily rainfall over the two months together. Do not evaluate it.
Answer:
\( \frac{30 \times 4 + 31 \times 9}{30 + 31} \) (that is, divided by 61 days).

Teacher's Note:
1. The day counts are the weights.
2. Don't simply average 4 and 9.

Question 23. In a 100% stacked bar chart of a family's monthly budget, the food segment covers 30% of the bar. If the family's total monthly expenditure is ₹24 000, find the amount spent on food.
Answer:
Food = 30% of ₹24 000 = \( \frac{30}{100} \) × 24000 = ₹7200.

Teacher's Note:
1. The share alone needed the total to give an amount.
2. Share × total = amount.

Question 24. Two groups have averages 40 and 60, and their combined average is 45. State which group has more members and give a reason.
Answer:
The group with average 40 is larger.
The combined average lies nearer the larger group's mean: 45 is 5 from 40 but 15 from 60. So the sizes are in the ratio 15 : 5 = 3 : 1.

Teacher's Note:
1. Weights are in the inverse ratio of the distances.
2. A quick 'balance' check.

Question 25. 3 kg of tea costing ₹400 per kg is mixed with 2 kg of tea costing ₹550 per kg. Find the cost per kg of the mixture.
Answer:
Cost = 3 × 400 + 2 × 550 = 1200 + 1100 = ₹2300 for 5 kg.
Cost per kg = 2300 ÷ 5 = ₹460.

Teacher's Note:
1. More of the cheaper tea ⇒ nearer ₹400.
2. Quantities are the weights.

Question 26. The mean of 8 numbers is 14. Each number is increased by 3. Find the new mean.
Answer:
Adding 3 to each number adds 3 to the mean: 14 + 3 = 17.

Teacher's Note:
1. Shifting every value shifts the mean by the same amount.
2. The total rises by 8 × 3 = 24.

Question 27. A single stacked bar shows how the 24 hours of a day are spent. The sleep segment stands for 6 hours. Find the percentage share of sleep in the bar.
Answer:
Share = \( \frac{6}{24} \) × 100 = 25%.

Teacher's Note:
1. With a fixed 24-hour total, hours and shares correspond.
2. One bar can be read both ways.

Question 28. A 100% stacked bar chart is sometimes described as an alternative to the pie chart. State one way in which the two are alike, and one way in which the 100% stacked bar chart is easier to read.
Answer:
Alike: both show how a whole is divided into parts, and neither shows absolute totals.
Easier: comparing lengths along a straight line is easier than comparing angles, and several bars can be lined up for comparison.

Teacher's Note:
1. Our eyes judge lengths better than angles.
2. Bars stack neatly for several groups.

Question 29. In a fitness test, marks for strength, flexibility and agility are combined in the ratio 2 : 3 : 5. Rashi scores 60, 65 and 70 respectively. Find her final mark.
Answer:
\( \frac{60 \times 2 + 65 \times 3 + 70 \times 5}{10} = \frac{665}{10} = \) 66.5.

Teacher's Note:
1. Above the plain mean 65 because the heaviest weight is on her best score.
2. Always match scores to weights in order.

Question 30. Explain briefly why the weighted mean of a set of values always lies between the smallest and the largest of those values.
Answer:
Replacing every value by the smallest makes the numerator smaller, giving exactly the smallest value; replacing every value by the largest gives exactly the largest.
The true numerator lies between these two, so after dividing by the total weight the weighted mean lies between the smallest and the largest value.

Teacher's Note:
1. A quick sanity check for any weighted-mean answer.
2. An answer outside the range must be wrong.