Verified step-by-step solutions for your class notes.
A stationery shop owner made ₹8000 selling books (30% profit) and ₹1000 selling book covers (50% profit). What is the percentage of profit on the total sales?
Answer: Total sales = ₹8000 + ₹1000 = ₹9000.
Total profit = (0.30 × 8000) + (0.50 × 1000) = 2400 + 500 = ₹2900.
Percentage profit = $\frac{2900}{9000} \times 100 \approx 32.22\%$
A white stork's average daily distance travelled 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?
Answer: Total distance for 20 days = $44.5 \times 20 = 890$ km.
Total distance for 21 days = 890 + 55 = 945 km.
New average = $\frac{945}{21} = 45$ km/day.
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?
Answer: Total volume = 600 + 300 = 900 mL.
Salt concentration = $\frac{600 \times 5\%}{900} = \frac{3000}{900} = 3.33\%$
Sugar concentration = $\frac{300 \times 8\%}{900} = \frac{2400}{900} = 2.67\%$
Correct Option: (v) Salt: 3.33%, Sugar: 2.67%
At a panipuri stall, the spice concentration is 8%. What quantity of regular water should be mixed into 10 litres of pani so that the spice level is reduced to (3/4)th of the original concentration?
Answer: Target concentration = $\frac{3}{4} \times 8\% = 6\%$.
Let $x$ be the amount of water added. The amount of spice remains constant: $10 \times 0.08 = 0.8$ litres.
$\frac{0.8}{10 + x} = 0.06 \implies 10 + x = \frac{0.8}{0.06} \approx 13.33$ litres.
Water added ($x$) = $13.33 - 10 = 3.33$ litres.
Physical fitness marks for strength, flexibility, and agility are weighted 4:5:6. Rashi scored 60, 65, 70. Keerthi scored 55, 65, 75.
(i) Find out whose total is more without calculating.
(ii) What are their final marks?
Answer:
(i) Keerthi's total is more because she scored higher (75) in agility, which carries the highest weight (6), compensating for her lower score in strength (weight 4).
(ii) Rashi's marks = $\frac{60(4) + 65(5) + 70(6)}{15} = \frac{985}{15} \approx 65.67$
Keerthi's marks = $\frac{55(4) + 65(5) + 75(6)}{15} = \frac{995}{15} \approx 66.33$
A restaurant collected ratings from 10 customers... What is the average rating if the metrics are combined with the weights food : ambience : service = 6 : 5 : 4?
| Metric | 5★ | 4★ | 3★ | 2★ | 1★ |
|---|---|---|---|---|---|
| Food | 5 | 3 | 2 | 0 | 0 |
| Ambience | 0 | 4 | 5 | 1 | 0 |
| Service | 1 | 2 | 2 | 4 | 1 |
Answer: First, find the average rating for each category from the 10 customers:
Food avg = $\frac{(5 \times 5) + (4 \times 0) + (3 \times 3) + (2 \times 2) + (1 \times 0)}{10} = 3.8$
Ambience avg = $\frac{(5 \times 0) + (4 \times 4) + (3 \times 5) + (2 \times 1) + (1 \times 0)}{10} = 3.3$
Service avg = $\frac{(5 \times 4) + (4 \times 1) + (3 \times 2) + (2 \times 2) + (1 \times 1)}{10} = 3.5$
Weighted Overall Average = $\frac{(3.8 \times 6) + (3.3 \times 5) + (3.5 \times 4)}{6 + 5 + 4} = \frac{53.3}{15} \approx 3.55$
Using the table of langur weights (Male avg=16.5, Female avg=13.8, Total avg=14.925, Total count=60):
(i) Which expression(s) describes the scenario?
(ii) Find out how many male langurs are present.
(iii) A female langur weighing 15.2 kg is admitted. New average weight of female langurs?
(iv) Two male langurs (16.9 kg and 16.1 kg) are released. New average weight of male langurs?
(v) One male langur loses 1 kg. New average weight of male langurs?
| Category | Male langurs | Female langurs | All langurs |
|---|---|---|---|
| Average weight | 16.5 kg | 13.8 kg | 14.925 kg |
| Number of langurs | $x$ | $y$ | 60 |
Answer:
(i) Expressions (a) $\frac{16.5x + 13.8y}{x + y} = 14.925$ and (b) $\frac{16.5x + 13.8y}{60} = 14.925$ (ignoring the typo negative sign in the textbook) correctly describe the scenario.
(ii) $16.5x + 13.8(60 - x) = 14.925 \times 60 \implies 2.7x = 67.5 \implies x = 25$. There are 25 male langurs (and 35 female langurs).
(iii) New female avg = $\frac{(13.8 \times 35) + 15.2}{36} = \frac{498.2}{36} \approx 13.84$ kg.
(iv) New male avg = $\frac{(16.5 \times 25) - (16.9 + 16.1)}{23} = \frac{412.5 - 33}{23} = \frac{379.5}{23} = 16.5$ kg.
(v) New male avg = $\frac{379.5 - 1}{23} = \frac{378.5}{23} \approx 16.46$ kg.
Using the salinity table (Dead Sea ~34%, Purified ~0.001%, Ground ~0.01%):
(i) Mix 1L Dead Sea with 2L Purified. Salinity?
(ii) Mix Dead Sea and Purified to get Groundwater? Quantity?
(iii) Mix Dead Sea and Ground to get Purified?
| Water Source | Salinity |
|---|---|
| Dead Sea | $\approx 34\%$ |
| Other seas and oceans | $\approx 3.5\%$ |
| Ground water | $\approx 0.01\%$ |
| Purified drinking water | $\approx 0.001\%$ |
Answer:
(i) Salinity = $\frac{(1 \times 34) + (2 \times 0.001)}{3} \approx 11.334\%$
(ii) Yes, it is possible because the target (0.01%) lies between 34% and 0.001%.
$0.01 = \frac{34(1) + 0.001(x)}{1 + x} \implies 0.01 + 0.01x = 34 + 0.001x \implies 0.009x = 33.99 \implies x = 3776.67$ Litres of purified water needed.
(iii) No, it is impossible because both Dead Sea (34%) and Groundwater (0.01%) have a higher salinity than the target Purified water (0.001%). You cannot lower concentration below your least concentrated ingredient.
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