Arithmetic: percentage, ratio, averages, time & work

Answer: (a) 360

Q's score is 10% below P's, so the 40-mark gap is 10% of P's score. P scored 400 and Q scored 360.

1. Q scored 10% less than P, so the gap of 40 marks is 10% of P's marks.
2. P = 40 ÷ 0.10 = 400.
3. Q = 400 − 40 = 360.
4. Check: 360 is 90% of 400.

Remember · 'x% less than P' means the difference is x% of P — use it to find P first.

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Answer: (b) ₹ 360

The second condition gives X + 40 = Y = Z − 40, and the first gives Y − 40 = Z/2. Solving, Z = 160, Y = 120 and X = 80, a total of ₹360.

1. Let X, Y and Z have x, y and z rupees.
2. If Z gives X ₹40, all three are equal: x + 40 = y = z − 40.
3. If Y gives X ₹40, Y has half of Z: y − 40 = z/2.
4. Put y = z − 40: z − 80 = z/2, so z = 160, y = 120, x = 80.
5. Total = 80 + 120 + 160 = ₹360.
6. Check: after Y gives ₹40, Y has 80, half of Z's 160; after Z gives ₹40, everyone has 120.

Remember · Turn 'then all will be equal' into a chain of equalities; it reduces everything to one unknown.

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Answer: (a) 9

Full marks would be 90 × 5 = 450, and each wrong answer costs 5 + 2 = 7 marks compared with a right one. The shortfall of 63 marks means 9 wrong answers.

1. If all 90 were correct: 90 × 5 = 450 marks.
2. Each wrong answer loses 5 + 2 = 7 marks compared with a correct one.
3. Shortfall = 450 − 387 = 63, so wrong answers = 63 ÷ 7 = 9.
4. Check: 81 × 5 − 9 × 2 = 405 − 18 = 387.

Remember · Start from full marks and divide the shortfall by (marks for a correct answer + penalty for a wrong one).

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Answer: (a) 1 only

If one part exceeds the other two combined, it certainly exceeds each of them, so A gets the most. But being less than the sum of the other two says nothing about being the smallest, as 1 : 3 : 2 shows.

1. Statement 1: shares are positive, so p > q + r means p > q and p > r — A gets the maximum. Correct.
2. Statement 2: take p : q : r = 1 : 3 : 2. Here r = 2 < p + q = 4, yet A, not C, gets the least. Incorrect.
3. So only Statement 1 is correct.

• ✓ 1. p > q + r forces p above both q and r.
• ✗ 2. r < p + q can hold even when r is not the smallest, e.g. 1 : 3 : 2.

Remember · Test 'if … then' claims with a quick counterexample: 'greater than the sum' implies 'greatest', but 'less than the sum' does not imply 'least'.

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Answer: (b) 2 only

X's age at Y's birth is the age gap X − Y, which equals twice Y's present age; so X = 3Y and Y = 14. Eight years ago they were 34 and 6 (not five times), but in 14 years they will be 56 and 28 (exactly twice).

1. At Y's birth X's age was X − Y, and this equals 2Y, so X = 3Y.
2. X = 42, so Y = 14.
3. Statement 1: 8 years ago X was 34 and Y was 6; 5 × 6 = 30, not 34. Incorrect.
4. Statement 2: after 14 years X will be 56 and Y 28; 56 = 2 × 28. Correct.

• ✗ 1. 34 is not five times 6.
• ✓ 2. 56 is exactly twice 28.

Remember · 'My age at your birth' is simply the age difference, which never changes.

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Answer: (a) 0%

A 20% cut takes ₹100 to ₹80, and 25% of 80 is 20, bringing it back to ₹100. The multipliers 0.8 × 1.25 = 1, so there is no net change.

1. Let the price be ₹100. After a 20% decrease it is ₹80.
2. A 25% increase on ₹80 is ₹20, giving ₹100.
3. Net change = 0%.
4. Check: 0.8 × 1.25 = 1.

Remember · Successive changes multiply: a fall of 1/5 is exactly undone by a rise of 1/4.

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Answer: (d) 15

He does 1/8 of the job every 3 days, so the whole job takes 24 days. When the job grows to 1.5 times its size, 5/8 of the original job remains, which takes 15 more days.

1. 7/8 of the job in 21 days, so the whole job takes 21 ÷ 7/8 = 24 days (1/24 of the job per day).
2. The work is increased by 50%, so the total becomes 1.5 jobs.
3. Work left = 1.5 − 7/8 = 12/8 − 7/8 = 5/8 of the original job.
4. Days needed = 5/8 × 24 = 15.
5. Check: 7/8 + 15/24 = 7/8 + 5/8 = 12/8 = 1.5 jobs.

Remember · Turn work done into a daily rate first; then express the remaining work as a fraction of the original job.

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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 years

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) All the options are equally good for the customer.

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) It remains the same.

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) lies between v and 2v km/hr

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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Answer: (d) 37.5%

B weighs 60% of A, but the fall is measured against the larger total A + B. Removing 60 from 160 is a 37.5% reduction, not 60%.

1. Let A weigh 100. B on top raises the weight by 60%, so B = 60.
2. Total weight of A and B = 160.
3. Removing B takes away 60 out of 160.
4. Reduction = 60/160 × 100% = 37.5%.

Remember · A rise of r% is undone by a fall of r/(100 + r) × 100% — divide by the new, larger base.

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Answer: (b) 15

12 km is four-fifths of the race, so the full distance is 12 × 5/4 = 15 km.

1. 12 km = 80% of the race.
2. Race = 12 × 100/80 = 15 km.
3. Check: 80% of 15 = 12.

Remember · If x is p% of the whole, the whole is x × 100/p.

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Answer: (a) Raju still does not have enough amount to buy the handset.

The handset costs ₹12,000 because ₹9,000 is only three-quarters of the price. With the ₹2,000 loan Raju has ₹11,000, still ₹1,000 short.

1. ₹9000 is 75% of the price, so the price = 9000 × 100/75 = ₹12000.
2. After borrowing he has 9000 + 2000 = ₹11000.
3. 11000 < 12000: he is still ₹1000 short.

Remember · Find the base first (amount ÷ percentage), then compare; don't mix percentages with rupee amounts.

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Answer: (b) 1994

Writing Meera's 2002 age as three times Meenu's and applying the 'half' relation eight years later gives Meenu's 2002 age as 8, so she was born in 1994.

1. Let Meenu be m years old in 2002; Meera is then 3m.
2. In 2010 both are 8 years older: m + 8 = (3m + 8)/2.
3. 2m + 16 = 3m + 8, so m = 8.
4. Meenu was 8 in 2002, so she was born in 2002 − 8 = 1994.
5. Check: 2002 — 8 and 24 (one-third); 2010 — 16 and 32 (half).

Remember · Both ages move by the same number of years; write them with one variable and shift both together.

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Answer: (c) ₹ 210

Only the joint spending matters: ₹2,800 buys 10 balls and 10 rackets, which is the cost of 40 balls. So a ball is ₹70 and a racket ₹210.

1. Total spent = 1300 + 1500 = ₹2800 for 10 balls and 10 rackets.
2. Let a ball cost b; a racket costs 3b.
3. 10b + 10 × 3b = 40b = 2800, so b = 70.
4. Racket = 3 × 70 = ₹210.
5. Check: 10 × 70 + 10 × 210 = 700 + 2100 = 2800.

Remember · When the split between buyers is irrelevant, add the spends and solve for the combined purchase.

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Answer: (c) Both 1 and 2

Only 30 participants are not Indian. Even if all of them were vegetarian, at least 60 − 30 = 30 vegetarians are Indian; and even if all 30 were non-vegetarian, at least 40 − 30 = 10 non-vegetarians are Indian.

1. Non-Indians = 100 − 70 = 30; non-vegetarians = 100 − 60 = 40.
2. Fewest Indian vegetarians: make all 30 non-Indians vegetarian → Indian vegetarians = 60 − 30 = 30.
3. Fewest Indian non-vegetarians: make all 30 non-Indians non-vegetarian → Indian non-vegetarians = 40 − 30 = 10.
4. Both minimums hold, so both statements are correct.
5. Check: 70 + 60 − 100 = 30 and 70 + 40 − 100 = 10.

• ✓ 1. At most 30 vegetarians can be non-Indian, so at least 30 of the 60 vegetarians are Indian.
• ✓ 2. At most 30 non-vegetarians can be non-Indian, so at least 10 of the 40 non-vegetarians are Indian.

Remember · For a guaranteed overlap, push the other group to the extreme: minimum overlap = A + B − total.

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Answer: (a) 1 only

Since 10% only rested, 90% went shopping, sightseeing or both. Shopping and sightseeing together add up to 130%, so 40% did both, and 80 − 40 = 40% did only shopping — not 20%.

1. Members who went out = 100% − 10% = 90%.
2. Both = 80% + 50% − 90% = 40%.
3. Only shopping = 80% − 40% = 40%.
4. Check: only shopping 40 + only sightseeing 10 + both 40 + rest 10 = 100.

• ✓ 1. 80 + 50 − 90 = 40% did both.
• ✗ 2. Only shopping is 80 − 40 = 40%, not 20%.

Remember · Both = A + B − (at least one); get 'at least one' by removing the 'neither' group first.

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Answer: (b) 380

B's shortfall of 5% is measured on A's marks and equals the 20-mark gap, so A has 400 and B 380.

1. B = A − 20, and B = 95% of A.
2. So 5% of A = 20, giving A = 400.
3. B = 400 − 20 = 380.
4. Check: 95% of 400 = 380.

Remember · When a percentage gap equals a known difference, that difference is the percentage of the base.

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