Arithmetic: percentage, ratio, averages, time & work

Answer: (a) 7·8

The overall English average of 8·8 lies four times nearer the girls' 9 than the boys' 8, so girls outnumber boys 4 : 1. Using the same weights on Hindi gives (4 × 8 + 1 × 7)/5 = 7·8.

1. Let there be g girls and b boys. English: 9g + 8b = 8·8(g + b).
2. So 0·2g = 0·8b, giving g : b = 4 : 1.
3. Hindi: x = (4 × 8 + 1 × 7)/5 = 39/5 = 7·8.
4. Check: English with the same weights = (4 × 9 + 8)/5 = 44/5 = 8·8.

Remember · Use the known overall average to find the group ratio, then apply that ratio to the other subject.

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Answer: (c) 35 वर्ष

The four ages total 80. If each student is s years old, the teacher is s + 20, so 4s + 20 = 80 gives s = 15 and the teacher is 35.

1. Total of the four ages = 4 × 20 = 80.
2. Let each student be s years; the teacher is s + 20.
3. 3s + (s + 20) = 80, so 4s = 60 and s = 15.
4. Teacher = 15 + 20 = 35 years.
5. Check: (35 + 15 + 15 + 15)/4 = 80/4 = 20.

Remember · Turn an average into a total first; then express every person in one unknown.

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Answer: (d) ₹3,75,000

A 20% loss means the selling price is 80% of the cost. So the cost is 3,00,000 ÷ 0·8 = ₹ 3,75,000.

1. Selling price = 80% of cost price.
2. Cost price = 3,00,000 ÷ 0·8 = ₹ 3,75,000.
3. Check: 20% of 3,75,000 = 75,000, and 3,75,000 − 75,000 = 3,00,000.

Remember · Loss and profit percentages are on cost price: CP = SP ÷ (1 − loss%).

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Answer: (d) ग्राहक के लिए सभी विकल्प समान रूप से अच्छे हैं।

Each option multiplies the price by the same three factors — 0·9, 0·8 and 1·1 — only in a different order. Multiplication does not depend on order, so the customer pays the same in every case.

1. Take the price as ₹ 100.
2. Option 1: 100 × 0·9 × 0·8 × 1·1 = ₹ 79·2.
3. Option 2: 100 × 0·8 × 0·9 × 1·1 = ₹ 79·2.
4. Option 3: 100 × 1·1 × 0·8 × 0·9 = ₹ 79·2.
5. All three cost the same, so all options are equally good.

Remember · Successive percentage changes are multiplying factors; their order never changes the final result.

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Answer: (c) 34 2/7%

Every married man has exactly one wife in the city, so married men equal married women: 40% of men = 30% of women, which makes men : women = 3 : 4. Then 1·2 + 1·2 = 2·4 of every 7 adults are married, i.e. 34 2/7%.

1. Married men = married women, so 0·4M = 0·3W, giving M : W = 3 : 4.
2. Take 3 men and 4 women: married men = 0·4 × 3 = 1·2, married women = 0·3 × 4 = 1·2.
3. Married adults = 2·4 out of 7 = 24/70 = 12/35.
4. 12/35 × 100 = 240/7 = 34 2/7%.

Remember · In marriage-count problems, equate married men and married women first; that fixes the ratio of the two groups.

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Answer: (c) ₹ 400

Clearing the fractions by multiplying by 12 turns the ratio into 6 : 9 : 10, a total of 25 parts worth ₹ 100 each. The largest share (10 parts) exceeds the smallest (6 parts) by 4 parts, or ₹ 400.

1. Multiply 1/2 : 3/4 : 5/6 by 12 to get 6 : 9 : 10.
2. Total parts = 25; one part = 2,500 ÷ 25 = ₹ 100.
3. Maximum − minimum = (10 − 6) × 100 = ₹ 400.
4. Check: shares 600 + 900 + 1,000 = 2,500.

Remember · For a ratio of fractions, multiply every term by the LCM of the denominators to get whole-number parts.

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Answer: (c) यह वही रहेगा।

Shifting students between groups inside the class changes the group averages but not the class as a whole: the same students with the same total weight remain. So the class average stays the same.

1. Class average = total weight of all students ÷ number of students in the class.
2. Moving students from A and B to C keeps every student in the class.
3. Both the total weight and the head count are unchanged, so the class average is unchanged.

Remember · Ask which total is being averaged; moving members within that total leaves its average unchanged.

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Answer: (c) 16 : 9

Each removal takes away 4/20 = 1/5 of whatever A is present, leaving 4/5 of it. After two rounds A = 20 × (4/5)² = 12·8 litres and B = 7·2 litres, a ratio of 16 : 9.

1. After the first replacement: A = 20 − 4 = 16 litres.
2. Second removal takes 1/5 of the mixture, so A falls to 16 × 4/5 = 12·8 litres.
3. B = 20 − 12·8 = 7·2 litres.
4. A : B = 12·8 : 7·2 = 128 : 72 = 16 : 9.

Remember · Repeated replace-and-remove: remaining original = initial × (1 − removed/total)ⁿ.

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Answer: (c) 62

Turn both averages into totals: 50 × 46·4 = 2320 and 60 × 49 = 2940. The last ten innings added 620 runs, an average of 62.

1. Runs after 50 innings = 50 × 46·4 = 2320.
2. New average = 46·4 + 2·6 = 49; runs after 60 innings = 60 × 49 = 2940.
3. Runs in the last 10 innings = 2940 − 2320 = 620.
4. Average of the last 10 = 620 ÷ 10 = 62.

Remember · Averages over different counts: convert each to a total, subtract, and divide by the new count.

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Answer: (b) ₹ 40

A 25% price rise means the same money buys 1/1·25 = 4/5 of the earlier quantity, i.e. 1/5 less. That lost fifth is 6 kg, so ₹ 1,200 used to buy 30 kg, at ₹ 40 a kg.

1. New price = 5/4 of old, so quantity for ₹ 1,200 becomes 4/5 of the old quantity — a fall of 1/5.
2. 1/5 of old quantity = 6 kg, so old quantity = 30 kg.
3. Original price = 1,200 ÷ 30 = ₹ 40 per kg.
4. Check: new price ₹ 50 buys 1,200 ÷ 50 = 24 kg, which is 6 kg less.

Remember · Price up by 1/n means quantity down by 1/(n + 1) for the same spending; here 25% = 1/4, so quantity falls by 1/5.

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Answer: (a) 6

X would finish the whole work in 40 days and Y in 24 days, so together they do 1/40 + 1/24 = 1/15 of it each day. 40% of the work therefore takes 0·4 × 15 = 6 days.

1. X: 20% in 8 days, so 100% in 40 days. Y: 25% in 6 days, so 100% in 24 days.
2. Together per day: 1/40 + 1/24 = 3/120 + 5/120 = 8/120 = 1/15.
3. Whole work together = 15 days; 40% of it = 0·4 × 15 = 6 days.
4. Check: in 6 days X does 6/40 = 15% and Y does 6/24 = 25%; total 40%.

Remember · Convert partial work to full-job time first, add daily rates, then scale to the fraction of work asked.

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Answer: (b) v और 2v km/hr के बीच रहती है

All four legs cover the same distance, so the slow legs take the most time and pull the average down. Total time = (d/v)(1 + 1/2 + 1/3 + 1/4) = 25d/12v for a distance of 4d, giving an average of 48v/25 = 1·92v.

1. Let XY = d km. Total distance = 4d.
2. Total time = d/v + d/2v + d/3v + d/4v = (d/v)(12 + 6 + 4 + 3)/12 = 25d/12v.
3. Average speed = 4d ÷ (25d/12v) = 48v/25 = 1·92v.
4. 1·92v lies between v and 2v.

Remember · For equal distances at different speeds, average speed is the harmonic mean — always closer to the slower speeds.

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Answer: (d) 3 : 1

Half the time over the same distance means the downstream speed is twice the upstream speed. So b + c = 2(b − c), which gives b = 3c: still-water speed : current = 3 : 1.

1. Let the speed in still water be b and the current c.
2. Downstream time is half the upstream time, so downstream speed = 2 × upstream speed.
3. b + c = 2(b − c), so b = 3c.
4. b : c = 3 : 1.
5. Check: b = 3, c = 1 gives speeds 4 and 2 — half the time downstream.

Remember · Same distance: time ratio is the inverse of speed ratio. Downstream = b + c, upstream = b − c.

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Answer: (a) 10 kg

Going from full to one-third full removes two-thirds of the water and 20 kg of weight, so the full water weighs 30 kg. The empty vessel is 40 − 30 = 10 kg.

1. Full: vessel + water = 40 kg. One-third full: vessel + water/3 = 20 kg.
2. Subtracting: (2/3) × water = 20 kg, so water = 30 kg.
3. Empty vessel = 40 − 30 = 10 kg.
4. Check: 10 + 30/3 = 20 kg.

Remember · Subtract the two readings: the difference belongs to the water alone, which fixes the water's full weight.

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Answer: (b) 29

Each jump-and-slide gains a net 15 cm, but the final jump does not slide back. The frog needs to be within 30 cm of the top, i.e. at 420 cm, before its last jump; 28 jumps with slides get it there, and the 29th jump takes it out.

1. Well depth = 450 cm; net gain per jump-and-slide = 30 − 15 = 15 cm.
2. The last jump of 30 cm must start from at least 450 − 30 = 420 cm.
3. 420 ÷ 15 = 28 jumps (with slides) to reach 420 cm.
4. The 29th jump lifts it from 420 to 450 cm and out — 29 jumps.
5. Check: after 27 jumps it is at 405 cm; the 28th reaches only 435 cm and slides back to 420 cm.

Remember · Climb-and-slip problems: subtract the last full jump from the height, count net gains to reach that point, then add one jump.

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