Question 1

Target profit via cost-price averaging: How many kilograms of ₹ 10/kg rice must be mixed with 25 kg of ₹ 8/kg rice so that, when sold at ₹ 15/kg, the overall profit is 80%?

Accepted Answer

Introduction / Context:We need to choose the mix so that the blended cost price is low enough to secure an 80% profit at the given selling price. This is a weighted-average cost problem solved by equating blend CP to SP / 1.8. Given Data / Assumptions: Component A: ₹ 10/kg, unknown mass x kg. Component B: ₹ 8/kg, mass 25 kg. Selling price of mixture = ₹ 15/kg; target profit = 80%. Concept / Approach:If profit is 80%, mixture CP must be SP / 1.8 = 15 / 1.8 = ₹ 25/3 per kg. Set the weighted-average CP equal to 25/3 and solve for x. Step-by-Step Solution:Weighted CP = (10x + 8*25) / (x + 25).Set (10x + 200) / (x + 25) = 25/3.Cross-multiply: 30x + 600 = 25x + 625 ⇒ 5x = 25 ⇒ x = 5. Verification / Alternative check:Blend mass = 30 kg. Total CP = 10*5 + 8*25 = 50 + 200 = ₹ 250. CP/kg = 250/30 = 25/3. Selling at ₹ 15 gives profit% = (15 − 25/3) / (25/3) = 80%, as required. Why Other Options Are Wrong: 3, 6, and 7 kg do not produce CP = ₹ 25/3 per kg when checked. Common Pitfalls: Using 80% of ₹ 15 as CP (incorrect). The correct CP is SP / 1.8. Final Answer:5 kgs

Question 2

Blending two sugar solutions: Two sugar solutions of 30% and 50% concentrations are mixed to obtain a 45% mixture. In what proportion should the two solutions be mixed (30% : 50%)?

Accepted Answer

Introduction / Context:Alligation quickly gives the ratio in which two concentrations must be combined to reach a target concentration. Here we want 45% from 30% and 50%. Given Data / Assumptions: Solutions: 30% and 50%. Target concentration = 45%. Concept / Approach:Alligation ratio (30% : 50%) = (50 − 45) : (45 − 30) = 5 : 15 = 1 : 3. This gives the required mixing proportion for 30% to 50% solution. Step-by-Step Solution:Difference on high side = 50 − 45 = 5.Difference on low side = 45 − 30 = 15.Therefore, 30% : 50% = 5 : 15 = 1 : 3. Verification / Alternative check:Test with 1 L of 30% and 3 L of 50%. Pure sugar = 0.3 + 1.5 = 1.8 in 4 L → 1.8/4 = 45% exactly. Why Other Options Are Wrong: 3 : 1 would yield 35%, not 45%. 2 : 3 and 3 : 2 give other averages, not 45%. Common Pitfalls: Swapping the alligation differences, which would invert the required ratio. Final Answer:1 : 3

Question 3

Overall 4% loss by splitting sales: A merchant has 160 kg of goods. He sells some at a 10% profit and the rest at a 6% loss. If his overall result is a 4% loss, how many kilograms were sold at a 6% loss?

Accepted Answer

Introduction / Context:We apply weighted-average gain/loss on quantities. By assigning a unit cost for simplicity, the overall revenue condition leads directly to the split between the two sale categories. Given Data / Assumptions: Total quantity = 160 kg. Profit on part = 10%; loss on remainder = 6%. Overall loss = 4%. Concept / Approach:Let x kg be sold at 10% profit and (160 − x) kg at 6% loss. With a notional cost price (CP) of ₹ 1 per kg, compute the total revenue and match it to the overall 4% loss target (i.e., 0.96 of total CP). Solve for x and then obtain the loss quantity (160 − x). Step-by-Step Solution:Revenue = 1.10x + 0.94(160 − x) = 0.16x + 150.4.Overall revenue = 0.96 × 160 = 153.6.0.16x + 150.4 = 153.6 ⇒ 0.16x = 3.2 ⇒ x = 20.Therefore, sold at 6% loss = 160 − 20 = 140 kg. Verification / Alternative check:Check: Profit portion revenue = 1.10 × 20 = 22; loss portion revenue = 0.94 × 140 = 131.6. Total = 153.6 = 0.96 × 160, matching 4% loss overall. Why Other Options Are Wrong: 120, 150, and 160 kg do not simultaneously satisfy the overall 4% loss condition with the given rates. Common Pitfalls: Equating the average of 10% and −6% to −4% (simple averaging is invalid without weighting by quantities). Final Answer:140 kgs

Question 4

Overall 20% profit with partial loss on one-fourth: A man buys sugar for ₹ 8000. He sells one-fourth at a 20% loss. At what profit percentage should he sell the remaining stock so that his overall profit is 20%?

Accepted Answer

Introduction / Context:This problem blends partial loss and required overall profit. We compute the revenue from the loss-making portion and then determine the necessary gain on the remainder to hit the overall target. Given Data / Assumptions: Total cost = ₹ 8000. One-fourth sold at 20% loss → cost of that portion = ₹ 2000. Overall target profit = 20% → total revenue target = 1.20 × 8000 = ₹ 9600. Concept / Approach:Compute revenue from the first (loss) portion, then find the required revenue from the remainder. Translate this to the needed profit percentage on the remaining cost base and express the answer as a clean percentage. Step-by-Step Solution:Revenue from first quarter = 0.80 × 2000 = ₹ 1600.Required revenue from remaining = 9600 − 1600 = ₹ 8000.Remaining cost = 8000 − 2000 = ₹ 6000.Required profit rate on remaining = (8000 − 6000) / 6000 = 2000 / 6000 = 1/3 = 33.33%. Verification / Alternative check:Check totals: Loss portion gives ₹ 1600; remainder at 33.33% gives ₹ 8000; sum = ₹ 9600 → exactly 20% overall profit on ₹ 8000 cost. Why Other Options Are Wrong: 20% and 30% are too low; 40% is too high. Only 33.33% hits the exact overall target. Common Pitfalls: Averaging percentages instead of using cost-weighted calculations. Final Answer:33.33%

Question 5

Splitting ₹ 5000 between 3% and 8% simple interest: A person invests a total of ₹ 5000, part at 3% p.a. and the remainder at 8% p.a. simple interest. If the total interest in 3 years is ₹ 750, find the amounts invested at each rate.

Accepted Answer

Introduction / Context:We distribute a fixed principal between two simple-interest rates. Using total interest over a known time period allows solving for the split uniquely by linear equations. Given Data / Assumptions: Total principal = ₹ 5000. Time = 3 years. Rates: 3% p.a. and 8% p.a. Total interest (3 years) = ₹ 750. Concept / Approach:Let x be the amount at 3% and (5000 − x) at 8%. Total interest = 3-year interest at 3% plus 3-year interest at 8%. Solve the resulting linear equation for x and then compute the remainder. Step-by-Step Solution:Interest = x*0.03*3 + (5000 − x)*0.08*3.= 0.09x + 0.24(5000 − x) = 0.09x + 1200 − 0.24x.Total = 1200 − 0.15x = 750 ⇒ 0.15x = 450 ⇒ x = ₹ 3000.Remainder = ₹ 2000 at 8%. Verification / Alternative check:3-year interest: 3000*0.03*3 = 270; 2000*0.08*3 = 480; total = 750, which matches the given. Why Other Options Are Wrong: Equal split (₹ 2500 each) gives a different total interest. The other pairs do not satisfy the equation 1200 − 0.15x = 750. Common Pitfalls: Forgetting to multiply by 3 years at both rates. Final Answer:Rs. 3000 and Rs. 2000

Question 6

The average daily wages of staff, consisting of supervisors and labourers, of a company is Rs. 50. The average wages of supervisors is Rs. 150 while that of labourers is Rs. 40 per day. If the number of supervisors is 15, find the number of labourer…

Accepted Answer

As per figure we can calculate the ration as below. Number of supervisors / Number of labourers = (10 / 100) = 1/10 Total number of labourers = Total no. of supervisors × 10 = 15 × 10 = 150.

Question 7

A sum of Rs. 70 is divided among 10 children. Each boy gets Rs. 10 whereas a girl gets Rs. 5. If the number of boys is 4, find the number of girls.

Accepted Answer

Number of girls / Number of boys = (3 / 2) Number of girls = (3 / 2) × 4 = 6

Question 8

A vessel contains 80 litres of milk. 16 litres of milk was taken out of the vessel and replaced by water. Then 16 litres of mixture was withdrawn and again replaced by water. The operation was repeated for third time. How much milk is now left in th…

Accepted Answer

Amount of milk left = 80 [1 - (16 / 80)]3 = 80 (4 / 5)3 80 × (64 / 125) = 40.96 liters.

Question 9

In what ratio must a grocer mix two varieties of tea worth Rs. 60 a kg and Rs. 65 a kg so that by selling the mixture at Rs. 68.20 a kg he may gain 10%?

Accepted Answer

S.P. of 1 kg mixture = Rs. 68.20, Gain % = 10 % Hence, C.P. of 1 kg mixture = (100 / 110) x 68 20 . = Rs. 62 By the rule of alligation Hence, required ratio = 3 : 2

Question 10

A shopkeeper has 100 kgs of tea. He sells a part of it at 20% profit and the rest at 5% loss. If his overall profit is 10%, find the quantity for each part.

Accepted Answer

As per figure, we can calculate as below. Quantity sold at 20% profit = 3 / (3 + 2) × 100 = 60 kgs. Quantity sold at 5% loss = (100 – 60) = 40 kgs.

Question 11

A cask full of wine from which 8 litres are drawn and is then filled with water. This operation is performed three more times. The ratio of quantity of wine left in the cask to that of the water is 16: 81. How much wine did the cask hold originally?

Accepted Answer

Let A liters wine cask hold originally, then A[1 - (8 / A)]4 / A = 16 / 81 ⇒ (1 - (8 / A))4 = (2 / 3 )4 ⇒ (1 - (8 / A)) = (2 / 3) ⇒ (8 / A) = (1 - (2 /3)) ⇒ (8 / A) = (1 / 3) ∴ A = 24 liters

Question 12

Allocation of Rs. 675 among 75 students (boys and girls): each boy receives Rs. 20 and each girl receives Rs. 5. Determine the exact counts of boys and girls in the group.

Accepted Answer

Introduction / Context: This is a classic two-variable allocation problem that tests linear equations and weighted sums. We are given a total headcount and a total amount distributed with different rates for boys and girls. The goal is to determine how many boys and how many girls are present when each boy gets Rs. 20 and each girl gets Rs. 5. Given Data / Assumptions: Total students = 75. Amount distributed = Rs. 675. Each boy receives Rs. 20. Each girl receives Rs. 5. Concept / Approach: Let b = number of boys and g = number of girls. Then b + g = 75. The money equation is 20*b + 5*g = 675. Solving this simultaneous system yields the unique pair (b, g) that satisfies both headcount and payment constraints. Step-by-Step Solution: b + g = 75 … (1) 20b + 5g = 675 … (2) Divide (2) by 5 ⇒ 4b + g = 135 … (3) Subtract (1) from (3): (4b + g) − (b + g) = 135 − 75 ⇒ 3b = 60 ⇒ b = 20. Then g = 75 − 20 = 55. Verification / Alternative check: Payment check: 20 boys * Rs. 20 = Rs. 400; 55 girls * Rs. 5 = Rs. 275; total = Rs. 675, which matches the given sum. Why Other Options Are Wrong: 15, 60 → money becomes 15*20 + 60*5 = 300 + 300 = 600 (too low). 25, 50 → 25*20 + 50*5 = 500 + 250 = 750 (too high). 30, 45 → 30*20 + 45*5 = 600 + 225 = 825 (too high). Common Pitfalls: Mixing the rates or forgetting to apply both the headcount and money equations simultaneously. Also, do not round—this is an exact integer solution. Final Answer: 20 , 55

Question 13

Rs. 41 is distributed among 50 students (boys and girls). Each boy receives 90 paise while each girl receives 65 paise. Find the exact number of boys.

Accepted Answer

Introduction / Context: Allocation problems with different per-person amounts are solved using simultaneous linear equations. Here, the total number of students is known, and the total payout is fixed, with boys and girls receiving different amounts (in paise). Given Data / Assumptions: Total students = 50. Total money = Rs. 41 = 4100 paise. Boy gets 90 paise; girl gets 65 paise. Concept / Approach: Let b = number of boys, g = number of girls. Then b + g = 50 and 90b + 65g = 4100 (all in paise). Solve these equations to get b and g as integers satisfying both constraints. Step-by-Step Solution: b + g = 50 … (1) 90b + 65g = 4100 … (2) From (1), g = 50 − b. Substitute in (2): 90b + 65(50 − b) = 4100 ⇒ 90b + 3250 − 65b = 4100. 25b = 850 ⇒ b = 34; then g = 16. Verification / Alternative check: 34 boys * 90 = 3060 paise; 16 girls * 65 = 1040 paise; total = 4100 paise = Rs. 41, so the values fit perfectly. Why Other Options Are Wrong: 16, 14, 36 do not satisfy both equations simultaneously; substituting any of them breaks the total amount constraint or headcount. Common Pitfalls: Forgetting to convert rupees to paise consistently or mixing up which variable stands for boys versus girls. Keep units consistent throughout. Final Answer: 34

Question 14

Out of Rs. 7000, part is lent at 6% p.a. and the remainder at 4% p.a. If the total simple interest in 5 years is Rs. 1600, find the principal amount that was lent at 6% p.a.

Accepted Answer

Introduction / Context: This problem checks understanding of simple interest with different interest rates applied to parts of a single principal. We are to determine how much of the total was placed at the higher rate given the total interest after a fixed time. Given Data / Assumptions: Total principal = Rs. 7000. Rate 1 = 6% p.a. for x rupees. Rate 2 = 4% p.a. for the remainder (7000 − x). Time = 5 years; total interest = Rs. 1600. Concept / Approach: Simple interest formula: SI = P * r * t (with r in decimal per year). Sum interests from both portions and equate to the given total to solve for x, the amount at 6%. Step-by-Step Solution: Interest from 6% part = x * 0.06 * 5 = 0.30x. Interest from 4% part = (7000 − x) * 0.04 * 5 = 0.20(7000 − x) = 1400 − 0.20x. Total SI = 0.30x + (1400 − 0.20x) = 1400 + 0.10x. Set equal to 1600 ⇒ 1400 + 0.10x = 1600 ⇒ 0.10x = 200 ⇒ x = 2000. Verification / Alternative check: Compute SI: 2000 at 6% for 5 years = 600; 5000 at 4% for 5 years = 1000; total = 1600, matching the given sum. Why Other Options Are Wrong: Rs. 5000 and Rs. 3500 produce total SI values different from Rs. 1600; “None of these” is unnecessary because Rs. 2000 satisfies perfectly. Common Pitfalls: Using 6% of total instead of isolating the variable portion or forgetting to multiply by time for simple interest. Keep r as a decimal (e.g., 6% = 0.06). Final Answer: Rs. 2000

Question 15

Sugar at Rs. 15/kg is mixed with sugar at Rs. 20/kg in the ratio 2 : 3. Find the price per kilogram of the resulting mixture (weighted average price).

Accepted Answer

Introduction / Context: Alligation and mixture questions often reduce to computing a weighted average when proportions and individual prices are known. Here, two types of sugar are mixed in a stated ratio; we must compute the average price per kilogram of the mixture. Given Data / Assumptions: Sugar A price = Rs. 15/kg. Sugar B price = Rs. 20/kg. Mixing ratio A : B = 2 : 3. Concept / Approach: Weighted average price = (sum of price * quantity) / total quantity. With a ratio, use the ratio numbers as proxy quantities to keep arithmetic simple and exact. Step-by-Step Solution: Let quantities be 2 kg (at Rs. 15) and 3 kg (at Rs. 20). Total cost = 2*15 + 3*20 = 30 + 60 = Rs. 90. Total quantity = 2 + 3 = 5 kg. Mixture price = 90 / 5 = Rs. 18 per kg. Verification / Alternative check: The result lies between Rs. 15 and Rs. 20 and is closer to 20 (because more of Rs. 20 sugar is used), which matches Rs. 18 being nearer to 20 than 15. Why Other Options Are Wrong: Rs. 16 and Rs. 17 are too low; Rs. 19 is too high relative to the 2 : 3 weighting. Common Pitfalls: Averaging the prices directly (i.e., taking (15 + 20)/2) ignores the ratio and gives a wrong value. Always weight by quantities. Final Answer: Rs. 18

Question 16

Kantilal mixes 80 kg of sugar at Rs. 6.75/kg with 120 kg at Rs. 8/kg. At what selling price per kg should he sell to earn a 20% profit on the mixture?

Accepted Answer

Introduction / Context: First compute the cost price (CP) per kg of the mixture using weighted average. Then apply the desired profit percentage to find the required selling price (SP) per kg that yields a 20% gain overall. Given Data / Assumptions: Quantity 1 = 80 kg at Rs. 6.75/kg. Quantity 2 = 120 kg at Rs. 8.00/kg. Desired profit = 20% on mixture CP. Concept / Approach: Mixture CP per kg = (total cost) / (total kg). Then SP per kg = CP per kg * (1 + profit%). Profit% = 20% ⇒ multiplier 1.20. Step-by-Step Solution: Cost1 = 80 * 6.75 = Rs. 540. Cost2 = 120 * 8 = Rs. 960. Total cost = 540 + 960 = Rs. 1500; total quantity = 80 + 120 = 200 kg. Mixture CP/kg = 1500 / 200 = Rs. 7.50. Required SP/kg = 7.50 * 1.20 = Rs. 9.00. Verification / Alternative check: Selling 200 kg at Rs. 9 earns revenue Rs. 1800. Profit = 1800 − 1500 = Rs. 300, which is 20% of 1500, confirming correctness. Why Other Options Are Wrong: Rs. 7.50 is the CP, not SP. Rs. 8.20 and Rs. 8.85 yield profits lower than 20%. Common Pitfalls: Forgetting to compute weighted average CP or applying 20% to the wrong base (quantity instead of cost). Always apply percentage to CP amount. Final Answer: Rs. 9

Question 17

A grocer buys rice at Rs. 1.80/kg and Rs. 1.20/kg. In what proportion should they be mixed so that selling the mixture at Rs. 1.75/kg gives a 25% gain?

Accepted Answer

Introduction / Context: We must determine the cost price (CP) target of the mixture given a selling price and desired profit. Then we use alligation to find the mixing ratio between cheaper and dearer varieties that achieves that CP. Given Data / Assumptions: Prices: Rs. 1.80/kg (dearer) and Rs. 1.20/kg (cheaper). Selling price (SP) of mixture = Rs. 1.75/kg. Desired gain = 25% ⇒ CP of mixture = SP / 1.25. Concept / Approach: First compute mixture CP: CP_mix = 1.75 / 1.25 = Rs. 1.40. Then apply alligation: For prices a (dearer), c (cheaper), and mean m, the ratio (quantity at a) : (quantity at c) = (m − c) : (a − m). Step-by-Step Solution: CP_mix = 1.75 / 1.25 = 1.40. Using alligation with a = 1.80, c = 1.20, m = 1.40: Ratio (dearer : cheaper) = (1.40 − 1.20) : (1.80 − 1.40) = 0.20 : 0.40 = 1 : 2. Therefore, rice at Rs. 1.80 : Rs. 1.20 should be mixed in 1 : 2. Verification / Alternative check: Weighted average with 1 part at 1.80 and 2 parts at 1.20 gives CP = (1.80*1 + 1.20*2)/3 = (1.80 + 2.40)/3 = 4.20/3 = 1.40, as required. Why Other Options Are Wrong: 2 : 1, 3 : 2, and 3 : 4 do not average to a CP of Rs. 1.40, hence they will not yield 25% gain at Rs. 1.75/kg. Common Pitfalls: Using SP in the alligation instead of CP_mix, or reversing the ratio formula. Derive CP first, then apply alligation properly. Final Answer: 1 : 2

Question 18

A 15 L mixture has 20% alcohol and 80% water. After adding 3 L of water, what is the new percentage of alcohol in the mixture?

Accepted Answer

Introduction / Context: This question examines percentage composition when a component (water) is added. The amount of alcohol remains constant while the total volume increases, changing the alcohol percentage. Given Data / Assumptions: Total initial volume = 15 L. Alcohol = 20% of 15 L = 3 L; water = 12 L. Add 3 L of water; no alcohol is removed or added. Concept / Approach: After addition, alcohol quantity is unchanged; only the total volume increases. New alcohol percentage = (alcohol volume / new total volume) * 100. Step-by-Step Solution: Initial alcohol = 0.20 * 15 = 3 L. Add 3 L water ⇒ new total volume = 15 + 3 = 18 L. New alcohol % = (3 / 18) * 100 = (1/6) * 100 = 16 2/3%. Verification / Alternative check: Because we added only water, the alcohol fraction must decrease; 16 2/3% is lower than 20%, which is consistent. Why Other Options Are Wrong: 17 and 18 1/2 do not match the exact fraction 3/18. 15% would require an even larger dilution. Common Pitfalls: Recalculating alcohol amount as if it changes. Only water is added, so alcohol volume stays 3 L; only the denominator changes. Final Answer: 16 2 3

Question 19

A milkman claims to sell at cost price but mixes water with milk and thus gains 25%. What percentage of water is present in the mixture?

Accepted Answer

Introduction / Context: When a trader adds a free component (water) to a product sold at the original cost price of the valuable component (milk), the effective gain equals the fraction of free component relative to the paid component. This is a standard result used in mixture-profit problems. Given Data / Assumptions: Selling price equals cost price of pure milk (CP). Water is free; profit arises solely from water content. Stated gain = 25%. Concept / Approach: If water is added to milk and the mixture is sold at the CP of milk, then Profit% = (quantity of water / quantity of milk) * 100. Set this equal to 25% and solve for the water-to-milk ratio, then convert to mixture percentage. Step-by-Step Solution: Let milk = 4 units; to get 25% gain, water must be 1 unit (since 1/4 = 0.25). Total mixture = 4 + 1 = 5 units. Water% in mixture = 1 / 5 * 100 = 20%. Verification / Alternative check: Selling 5 units (4 milk + 1 water) at CP of 4 units yields revenue equivalent to CP of milk; effectively the extra 1 unit water is pure profit, i.e., 25% gain on milk cost. Why Other Options Are Wrong: 25% refers to profit, not water percentage. 4% is too small. “None of these” is not needed since 20% is valid. Common Pitfalls: Confusing profit% with water%. Always relate profit to the paid component (milk), not to the total mixture directly. Final Answer: 20%

Question 20

In what ratio should sugar at 72 paise/kg be mixed with sugar at 48 paise/kg so that selling the mixture at 63 paise/kg yields a profit of 1/6 of the outlay?

Accepted Answer

Introduction / Context: We seek a cost price (CP) for the mixture that will produce a profit of 1/6 when sold at 63 paise/kg. After finding CP_mix, apply alligation between 72p and 48p to determine the mixing ratio (in the order listed in the question). Given Data / Assumptions: Prices: 72p/kg (dearer) and 48p/kg (cheaper). Selling price (SP) = 63p/kg. Profit = 1/6 of outlay ⇒ SP = CP_mix * (1 + 1/6) = CP_mix * (7/6). Concept / Approach: First compute CP_mix = SP / (7/6) = 63 * 6/7 = 54p. Then apply alligation: ratio (quantity at 72p) : (quantity at 48p) = (CP_mix − 48) : (72 − CP_mix). Step-by-Step Solution: CP_mix = 63 / (7/6) = 63 * 6/7 = 54p. Alligation differences: 72 − 54 = 18; 54 − 48 = 6. Therefore, quantities (72p : 48p) = (54 − 48) : (72 − 54) = 6 : 18 = 1 : 3. Verification / Alternative check: Weighted average with 1 part at 72p and 3 parts at 48p gives CP = (72 + 3*48)/4 = (72 + 144)/4 = 216/4 = 54p; selling at 63p yields profit = 9/54 = 1/6, as required. Why Other Options Are Wrong: 3 : 1 inverts the order; 2 : 3 and 3 : 2 do not give CP 54p. Common Pitfalls: Using SP in alligation or reversing the ratio. Always compute CP_mix first, then use alligation with correct order. Final Answer: 1 : 3

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