CBSE Class 9 Maths Ganita Manjari Part 2 Ch 10 How Quantities Combine Understanding Data MCQs with Answers Set 01

Answer: (c) 66

Explanation:
1. Convert each mean to a total: 20 × 60 = 1200 and 30 × 70 = 2100.
2. Combined mean = \( \frac{1200 + 2100}{50} \) = \( \frac{3300}{50} \) = 66.
3. 65 is the simple average of 60 and 70; it ignores the fact that Section Q is larger.

Teacher's Note:
1. Always turn means back into totals before combining.
2. The larger group pulls the combined mean towards itself.

Answer: (b) 52

Explanation:
1. Let each class have n students. Combined mean = \( \frac{48n + 56n}{2n} \) = \( \frac{104n}{2n} \) = 52.
2. The n cancels, so when sizes are equal the simple average is correct: (48 + 56) ÷ 2 = 52.

Teacher's Note:
1. Equal sizes are the one case where averaging averages works.
2. 104 forgets to divide by 2.

Answer: (d) 7.6

Explanation:
1. Multiply each value by its weight: 4 × 3 + 7 × 2 + 10 × 5 = 12 + 14 + 50 = 76.
2. Divide by the sum of weights: 76 ÷ 10 = 7.6.
3. 7 is the plain mean (weights ignored).

Teacher's Note:
1. Sum of (value × weight) ÷ sum of weights.
2. Check: the answer must lie between 4 and 10.

Answer: (b) 14%

Explanation:
1. Salt: 300 × 0.10 = 30 mL and 200 × 0.20 = 40 mL, total 70 mL.
2. Mixture volume = 500 mL, so concentration = 70 ÷ 500 = 14%.
3. 15% would be correct only for equal volumes; more of the weaker solution pulls it below 15%.

Teacher's Note:
1. Find the amount of the substance first, then divide by the total.
2. Predict: more of the weaker solution ⇒ below the midpoint.

Answer: (a) the two collections have the same number of values

Explanation:
1. The combined average is \( \frac{an + bm}{n + m} \).
2. When n = m it becomes \( \frac{(a + b)n}{2n} \) = \( \frac{a + b}{2} \), the simple average.
3. For unequal sizes, the larger collection pulls the answer towards its own mean.

Teacher's Note:
1. The key condition is equal sizes.
2. Equal averages give agreement only trivially.

Answer: (c) 6

Explanation:
1. Runs after 10 overs = 10 × 5 = 50.
2. After 11 overs: 50 + 16 = 66 runs, so run rate = 66 ÷ 11 = 6.

Teacher's Note:
1. Run rate is a mean: runs ÷ overs.
2. Convert the rate back to a total first.

Answer: (a) stays exactly the same

Explanation:
1. Both the numerator and the denominator are multiplied by 2.
2. The factor 2 cancels, so the weighted mean does not change.
3. Only the ratio of the weights matters.

Teacher's Note:
1. This is why weights can be given as a ratio like 3 : 2 : 5.
2. Scaling weights never changes the result.

Answer: (c) the proportions of the parts within each bar

Explanation:
1. Every bar is drawn to the same full length, standing for 100% of its own total.
2. So segments show each part's share of its bar.
3. Totals and exact quantities are lost.

Teacher's Note:
1. 100% stacked = shares, not amounts.
2. Equal bar lengths never mean equal totals.

Answer: (d) 83

Explanation:
1. Weighted total = 70 × 2 + 80 × 3 + 90 × 5 = 140 + 240 + 450 = 830.
2. Divide by 2 + 3 + 5 = 10: 830 ÷ 10 = 83.
3. It is above the plain mean 80 because the biggest weight is on the highest mark.

Teacher's Note:
1. A ratio gives the weights directly.
2. Check where the heaviest weight sits to predict the direction.

Answer: (b) 50%

Explanation:
1. Equal masses contribute equally, so the result is exactly midway.
2. Midpoint of 60% and 40% = 50%.

Teacher's Note:
1. Equal quantities ⇒ midpoint.
2. The shortcut fails when quantities differ.

Answer: (c) 16

Explanation:
1. Old total = 20 × 15 = 300. New total = 300 + 36 = 336.
2. New mean = 336 ÷ 21 = 16.
3. The mean rises because 36 is above the old mean.

Teacher's Note:
1. Add to the total, add 1 to the count.
2. A value above the mean raises it.

Answer: (d) A 100% stacked bar chart can be drawn from a stacked bar chart, but not the other way round.

Explanation:
1. A stacked bar chart has the actual amounts; dividing each by its bar total gives the percentages.
2. Going back is impossible: the 100% chart has lost the totals, and many totals give the same shares.

Teacher's Note:
1. Converting is one-way: amounts → shares.
2. Information lost in dividing cannot be recovered.

Answer: (a) all the weights are equal

Explanation:
1. If every weight is w: \( \frac{w x_1 + \cdots + w x_n}{nw} \) = the ordinary mean.
2. Equal values also give agreement, but only for those special values, not 'whatever the values may be'.

Teacher's Note:
1. Equal weights ⇒ ordinary mean.
2. Read 'whatever the values may be' carefully.

Answer: (b) 3.9

Explanation:
1. 4 × 5 + 3 × 3 + 5 × 2 = 20 + 9 + 10 = 39.
2. 39 ÷ 10 = 3.9.
3. It is below the plain mean 4 because the highest rating (ambience) has the smallest weight.

Teacher's Note:
1. Match each rating to its own weight carefully.
2. Compare with the plain mean to check the direction.

Answer: (d) 200 mL

Explanation:
1. Spice stays the same: 600 × 0.08 = 48 mL.
2. New total V must satisfy 48 ÷ V = 0.06, so V = 800 mL.
3. Water added = 800 − 600 = 200 mL. (800 mL is the new total, not the water.)

Teacher's Note:
1. When diluting, hold the dissolved amount fixed.
2. Subtract the original volume at the end.

Answer: (a) a stacked bar chart

Explanation:
1. A stacked bar keeps actual amounts: full bar length = total, segments = categories.
2. A 100% chart loses the totals, and pie charts make totals hard to compare.

Teacher's Note:
1. Totals + parts ⇒ stacked bar.
2. Shares only ⇒ 100% stacked bar.

Answer: (c) in Garden 1, more blooms appeared in summer than in winter

Explanation:
1. Within one bar, segments are shares of the same total, so a longer segment means more blooms: 55% > 20%.
2. (a) and (d) compare different bars, whose totals are unknown.
3. (b) is wrong: equal bar lengths in a 100% chart say nothing about totals.

Teacher's Note:
1. Within a bar: comparisons are safe.
2. Across bars: not without the totals.

Answer: (b) correct only when the two sections have the same number of students

Explanation:
1. Halving the sum treats both sections as equally important.
2. That is right only when they have the same size; otherwise the smaller section gets too much say.

Teacher's Note:
1. Link to the 'average of averages' trap.
2. Weights = group sizes.

Answer: (a) 25% of the bar

Explanation:
1. Share = \( \frac{200}{800} \) × 100 = 25%.
2. 20% comes from wrongly dividing by 1000.

Teacher's Note:
1. Share = part ÷ whole × 100.
2. A quarter of the bar.

Answer: (c) a larger weight than 8

Explanation:
1. A weighted mean lies nearer the value with the larger weight.
2. 11 is 1 away from 12 but 3 away from 8, so 12 has more weight.
3. Check: weights 1 and 3 give \( \frac{8 + 36}{4} \) = 11.

Teacher's Note:
1. Distances are in the inverse ratio of the weights.
2. Here 1 : 3 distances ⇒ 3 : 1 weights.

Answer: (d) may change, and generally does

Explanation:
1. Adding 1 changes the ratio of the weights, so nothing cancels.
2. Example: values 6, 10, 15 with weights 2, 3, 5 give 11.7; weights 3, 4, 6 give 148 ÷ 13 ≈ 11.38.

Teacher's Note:
1. Multiplying weights: no change. Adding to weights: usually a change.
2. Adding pulls the result towards the ordinary mean.

Answer: (b) ₹120

Explanation:
1. Cost of old shares = 25 × 150 = ₹3750; new shares = 15 × 70 = ₹1050.
2. Average = (3750 + 1050) ÷ 40 = 4800 ÷ 40 = ₹120.
3. ₹110 is the simple average of 150 and 70; it ignores the numbers of shares.

Teacher's Note:
1. Share-averaging is a weighted mean with quantities as weights.
2. More shares at ₹150 pull the average up.

Answer: (a) 4 litres

Explanation:
1. Salt stays at 1 × 0.36 = 0.36 L.
2. 0.36 ÷ V = 0.09, so V = 4 L.
3. 3 litres is the water added, not the total volume.

Teacher's Note:
1. Read whether the question asks for water added or total volume.
2. Quartering the concentration needs four times the volume.

Answer: (d) The total number of items represented by a bar

Explanation:
1. Every bar is drawn the same length whatever its total, so totals are exactly what is lost.
2. Largest share, order and 'more than half' are all visible.

Teacher's Note:
1. Totals are the missing information in 100% charts.
2. Shares are visible; amounts are not.

Answer: (c) 32

Explanation:
1. Total = 12 × 30 = 360. Remove 8: 352.
2. New mean = 352 ÷ 11 = 32.
3. Removing a value below the mean raises the mean.

Teacher's Note:
1. Subtract from the total and from the count.
2. 22 comes from wrongly subtracting 8 from the mean.

Answer: (b) a little more than 10%

Explanation:
1. The mixture's concentration lies between 10% and 20%.
2. It is nearer the solution present in the larger quantity, the 10% one.
3. So it is just above 10%.

Teacher's Note:
1. Estimate before calculating: which side is heavier?
2. Result can never go below 10% here.

Answer: (c) \( \frac{31 \times 3 + 30 \times 10}{61} \)

Explanation:
1. Total rain = 31 × 3 + 30 × 10 mm over 31 + 30 = 61 days.
2. The day counts act as weights.

Teacher's Note:
1. Mean = total ÷ total count.
2. Divide by 61 days, not by 2 months.

Answer: (d) 100

Explanation:
1. Copper: 140 + 0.4x = 0.6(200 + x) = 120 + 0.6x.
2. 20 = 0.2x, so x = 100.
3. Check: 180 kg copper in 300 kg = 60%.

Teacher's Note:
1. 60% is 10 from 70% and 20 from 40%, so weights are 2 : 1.
2. 200 : 100 = 2 : 1 confirms the answer.

Answer: (a) 10 varna

Explanation:
1. Weighted mean with quantities as weights: \( \frac{12 \times 6 + 7 \times 4}{6 + 4} \) = \( \frac{100}{10} \) = 10 varna.
2. It lies nearer 12 because more of the purer gold was used.

Teacher's Note:
1. Śrīdharācārya used exactly this weighted mean for gold.
2. 19 is the undivided numerator ÷ something wrong.

Answer: (b) can be read both as a stacked bar chart and as a 100% stacked bar chart

Explanation:
1. Each segment shows actual hours, so it is a stacked bar.
2. The whole bar is always 24 hours, so segment lengths are also shares of the day: a 100% stacked bar too.

Teacher's Note:
1. When all totals are equal, both readings coincide.
2. A single bar always has this double reading.