12
Q1: A mixture of 160 gallons of wine and water contains 25% water. How much water must be added to the mixture in order to increase the percentage of water to 40% of the new mixture?
Correct Answer: A
Solution:
Let $x$ be the amount of water to be added.
Initially, the mixture contains 160 gallons, with 25% water.
This means there are $160 \times 0.25 = 40$ gallons of water and $160 - 40 = 120$ gallons of wine.
After adding $x$ gallons of water, the total volume becomes $160 + x$ gallons.
The amount of water becomes $40 + x$ gallons.
The new percentage of water is 40%, so we can set up the equation: $$ \frac{40 + x}{160 + x} = 0.40 $$ Solving for $x$: $40 + x = 0.40 \times (160 + x)$ $40 + x = 64 + 0.40x$ $0.60x = 24$ $x = \frac{24}{0.60}$ $x = 40$ Therefore, 40 gallons of water must be added.
Solution:
Let $x$ be the amount of water to be added.
Initially, the mixture contains 160 gallons, with 25% water.
This means there are $160 \times 0.25 = 40$ gallons of water and $160 - 40 = 120$ gallons of wine.
After adding $x$ gallons of water, the total volume becomes $160 + x$ gallons.
The amount of water becomes $40 + x$ gallons.
The new percentage of water is 40%, so we can set up the equation: $$ \frac{40 + x}{160 + x} = 0.40 $$ Solving for $x$: $40 + x = 0.40 \times (160 + x)$ $40 + x = 64 + 0.40x$ $0.60x = 24$ $x = \frac{24}{0.60}$ $x = 40$ Therefore, 40 gallons of water must be added.
Q2: 800 students took the CAT exam in Delhi. 50% of the boys and 90% of the girls cleared the cut off in the examination. If the total percentage of qualifying students is 60%, how many girls appeared in the examination?
Correct Answer: D
Solution:
Let $B$ be the number of boys and $G$ be the number of girls.
$$ \text{Total students} = B + G = 800 $$ Number of boys who cleared = $0.5B$ Number of girls who cleared = $0.9G$ Total students who cleared = $0.5B + 0.9G$ Total percentage of qualifying students is 60%, so the number of students who cleared is $0.6 \times 800 = 480$.
Therefore, $0.5B + 0.9G = 480$ We also know that $B + G = 800$, so $B = 800 - G$.
Substitute $B$ in the first equation: $0.5(800 - G) + 0.9G = 480$ $400 - 0.5G + 0.9G = 480$ $0.4G = 80$ $G = \frac{80}{0.4} = 200$ Therefore, the number of girls who appeared in the examination is 200.
Solution:
Let $B$ be the number of boys and $G$ be the number of girls.
$$ \text{Total students} = B + G = 800 $$ Number of boys who cleared = $0.5B$ Number of girls who cleared = $0.9G$ Total students who cleared = $0.5B + 0.9G$ Total percentage of qualifying students is 60%, so the number of students who cleared is $0.6 \times 800 = 480$.
Therefore, $0.5B + 0.9G = 480$ We also know that $B + G = 800$, so $B = 800 - G$.
Substitute $B$ in the first equation: $0.5(800 - G) + 0.9G = 480$ $400 - 0.5G + 0.9G = 480$ $0.4G = 80$ $G = \frac{80}{0.4} = 200$ Therefore, the number of girls who appeared in the examination is 200.
Q3: If 10 kg of sugar costing Rs. 15/kg and 20 kg of salt costing Rs. 10/kg are mixed, find the average cost of the mixture in Rs. per kilogram
Correct Answer: A
Solution:
This problem involves finding the average cost of a mixture.
We can use the concept of weighted average.
Total cost of sugar = $10 \text{ kg} \times \text{Rs.
} 15/\text{kg} = \text{Rs.
} 150$ Total cost of salt = $20 \text{ kg} \times \text{Rs.
} 10/\text{kg} = \text{Rs.
} 200$ Total weight of the mixture = $10 \text{ kg} + 20 \text{ kg} = 30 \text{ kg}$ Total cost of the mixture = $\text{Rs.
} 150 + \text{Rs.
} 200 = \text{Rs.
} 350$ Average cost of the mixture = $\frac{\text{Total cost}}{\text{Total weight}} = \frac{\text{Rs.
} 350}{30 \text{ kg}} = \frac{\text{Rs.
} 35}{3 \text{ kg}}$ This simplifies to approximately Rs.
$11.67/\text{kg}$.
While the question asks for the average cost, and $\frac{\text{Rs.
} 35}{3}$ is technically correct, we should present it in a more commonly used format with proper rounding.
Solution:
This problem involves finding the average cost of a mixture.
We can use the concept of weighted average.
Total cost of sugar = $10 \text{ kg} \times \text{Rs.
} 15/\text{kg} = \text{Rs.
} 150$ Total cost of salt = $20 \text{ kg} \times \text{Rs.
} 10/\text{kg} = \text{Rs.
} 200$ Total weight of the mixture = $10 \text{ kg} + 20 \text{ kg} = 30 \text{ kg}$ Total cost of the mixture = $\text{Rs.
} 150 + \text{Rs.
} 200 = \text{Rs.
} 350$ Average cost of the mixture = $\frac{\text{Total cost}}{\text{Total weight}} = \frac{\text{Rs.
} 350}{30 \text{ kg}} = \frac{\text{Rs.
} 35}{3 \text{ kg}}$ This simplifies to approximately Rs.
$11.67/\text{kg}$.
While the question asks for the average cost, and $\frac{\text{Rs.
} 35}{3}$ is technically correct, we should present it in a more commonly used format with proper rounding.
Q4: The average salary per head of all workers (Grade A and Grade B) of a company is Rs. 400. The average salary of 100 grade A workers is Rs. 1000. If the average salary per head of the rest of the Grade B workers is Rs. 300, find the total number of workers in the company
Correct Answer: D
Solution:
Let the number of Grade A workers be 100.
Let the number of Grade B workers be $x$.
The total salary of Grade A workers is $100 \times 1000 = 100000$.
The total salary of Grade B workers is $300x$.
The total number of workers is $100 + x$.
The total salary of all workers is $100000 + 300x$.
The average salary per head of all workers is given as Rs.
400.
Therefore, $\frac{100000 + 300x}{100 + x} = 400$ $$100000 + 300x = 40000 + 400x$$ $$100x = 60000$$ $$x = 600$$ The total number of workers is $100 + 600 = 700$.
Solution:
Let the number of Grade A workers be 100.
Let the number of Grade B workers be $x$.
The total salary of Grade A workers is $100 \times 1000 = 100000$.
The total salary of Grade B workers is $300x$.
The total number of workers is $100 + x$.
The total salary of all workers is $100000 + 300x$.
The average salary per head of all workers is given as Rs.
400.
Therefore, $\frac{100000 + 300x}{100 + x} = 400$ $$100000 + 300x = 40000 + 400x$$ $$100x = 60000$$ $$x = 600$$ The total number of workers is $100 + 600 = 700$.
Q6: Two types of milk having the rates of Rs. 8/kg and Rs. 10/kg respectively are mixed in order to produce a mixture having the rate of Rs. 9.20/kg. What should be the amount of the second type of milk if the amount of the first type of milk in the mixture is 20 kg?
Correct Answer: B
Solution:
Let $x$ be the amount of milk costing Rs.
$8$/kg and $y$ be the amount of milk costing Rs.
$10$/kg.
The mixture costs Rs.
$9.20$/kg.
We can use alligation to solve this.
The difference between the mixture price and the price of the cheaper milk is $9.20 - 8 = 1.20$ The difference between the price of the dearer milk and the mixture price is $10 - 9.20 = 0.80$ The ratio of the amounts of the two types of milk is given by the inverse ratio of these differences: $x : y = 0.80 : 1.20 = 8 : 12 = 2 : 3$ We are given that $x = 20$ kg.
Therefore, we can set up a proportion: $\frac{2}{3} = \frac{20}{y}$ Solving for $y$: $y = \frac{3 \times 20}{2} = 30$ kg Therefore, the amount of the second type of milk should be $30$ kg.
Solution:
Let $x$ be the amount of milk costing Rs.
$8$/kg and $y$ be the amount of milk costing Rs.
$10$/kg.
The mixture costs Rs.
$9.20$/kg.
We can use alligation to solve this.
The difference between the mixture price and the price of the cheaper milk is $9.20 - 8 = 1.20$ The difference between the price of the dearer milk and the mixture price is $10 - 9.20 = 0.80$ The ratio of the amounts of the two types of milk is given by the inverse ratio of these differences: $x : y = 0.80 : 1.20 = 8 : 12 = 2 : 3$ We are given that $x = 20$ kg.
Therefore, we can set up a proportion: $\frac{2}{3} = \frac{20}{y}$ Solving for $y$: $y = \frac{3 \times 20}{2} = 30$ kg Therefore, the amount of the second type of milk should be $30$ kg.
Q7: How many kilograms of salt worth Rs. 360 per kg should be mixed with 10 kg of salt worth Rs. 420 per kg, such that by selling the mixture at Rs. 480 per kg, there may be a gain of 20%?
Correct Answer: A
Solution:
Let $x$ kg of salt worth Rs.
$360$/kg be mixed with 10 kg of salt worth Rs.
$420$/kg.
The total cost of the mixture is $360x + 420(10) = 360x + 4200$.
The total weight of the mixture is $(x + 10)$ kg.
The selling price of the mixture is Rs.
$480$/kg.
Total selling price = $480(x + 10)$ Gain = 20% Selling Price = Cost Price + 20% of Cost Price = $1.2 \times$ Cost Price $$480(x + 10) = 1.2 (360x + 4200)$$ $$480x + 4800 = 432x + 5040$$ $$48x = 240$$ $$x = 5 \text{ kg}$$
Solution:
Let $x$ kg of salt worth Rs.
$360$/kg be mixed with 10 kg of salt worth Rs.
$420$/kg.
The total cost of the mixture is $360x + 420(10) = 360x + 4200$.
The total weight of the mixture is $(x + 10)$ kg.
The selling price of the mixture is Rs.
$480$/kg.
Total selling price = $480(x + 10)$ Gain = 20% Selling Price = Cost Price + 20% of Cost Price = $1.2 \times$ Cost Price $$480(x + 10) = 1.2 (360x + 4200)$$ $$480x + 4800 = 432x + 5040$$ $$48x = 240$$ $$x = 5 \text{ kg}$$
Q9: A tank contains 500 liters of wine. 50 liters of wine is taken out of it and replaced by water. The process is repeated again. Find the proportion of water and wine in the resulting mixture
Correct Answer: C
Solution:
Step 1: After the first replacement, the amount of wine remaining is $500 \times \frac{450}{500} = 450$ liters.
The amount of water is $50$ liters.
Step 2: In the second replacement, $50$ liters of the mixture is taken out.
The proportion of wine in this $50$ liters is $\frac{450}{500} \times 50 = 45$ liters.
Step 3: After the second removal, the amount of wine remaining is $450 - 45 = 405$ liters.
Step 4: $50$ liters of water is added again.
The total volume remains $500$ liters.
Step 5: The proportion of wine in the final mixture is $\frac{405}{500}$.
Step 6: The proportion of water in the final mixture is $1 - \frac{405}{500} = \frac{95}{500}$.
Step 7: The proportion of water and wine is $95:405$ which simplifies to $19:81$.
Solution:
Step 1: After the first replacement, the amount of wine remaining is $500 \times \frac{450}{500} = 450$ liters.
The amount of water is $50$ liters.
Step 2: In the second replacement, $50$ liters of the mixture is taken out.
The proportion of wine in this $50$ liters is $\frac{450}{500} \times 50 = 45$ liters.
Step 3: After the second removal, the amount of wine remaining is $450 - 45 = 405$ liters.
Step 4: $50$ liters of water is added again.
The total volume remains $500$ liters.
Step 5: The proportion of wine in the final mixture is $\frac{405}{500}$.
Step 6: The proportion of water in the final mixture is $1 - \frac{405}{500} = \frac{95}{500}$.
Step 7: The proportion of water and wine is $95:405$ which simplifies to $19:81$.
Q10: A man purchased a table and a chair for Rs. 2000. He sold the table at a profit of 20% and the chair at a profit of 40%. In this way, his total profit was 25%. Find the cost price (in Rs. ) of the table.
Correct Answer: A
Solution:
Let the cost price of the table be $T$ and the cost price of the chair be $C$.
We are given that $T + C = 2000$.
The table was sold at a 20% profit, so the selling price of the table is $1.2T$.
The chair was sold at a 40% profit, so the selling price of the chair is $1.4C$.
The total selling price is $1.2T + 1.4C$.
The total profit is 25%, so the total selling price is also $1.25(T + C) = 1.25 \times 2000 = 2500$.
Therefore, $1.2T + 1.4C = 2500$.
We have a system of two equations: $$T + C = 2000$$ $$1.2T + 1.4C = 2500$$ From the first equation, we have $C = 2000 - T$.
Substitute $C = 2000 - T$ into the second equation: $1.2T + 1.4(2000 - T) = 2500$ $1.2T + 2800 - 1.4T = 2500$ $-0.2T = -300$ $T = 1500$ Therefore, the cost price of the table is Rs.
1500.
Solution:
Let the cost price of the table be $T$ and the cost price of the chair be $C$.
We are given that $T + C = 2000$.
The table was sold at a 20% profit, so the selling price of the table is $1.2T$.
The chair was sold at a 40% profit, so the selling price of the chair is $1.4C$.
The total selling price is $1.2T + 1.4C$.
The total profit is 25%, so the total selling price is also $1.25(T + C) = 1.25 \times 2000 = 2500$.
Therefore, $1.2T + 1.4C = 2500$.
We have a system of two equations: $$T + C = 2000$$ $$1.2T + 1.4C = 2500$$ From the first equation, we have $C = 2000 - T$.
Substitute $C = 2000 - T$ into the second equation: $1.2T + 1.4(2000 - T) = 2500$ $1.2T + 2800 - 1.4T = 2500$ $-0.2T = -300$ $T = 1500$ Therefore, the cost price of the table is Rs.
1500.
Q11: A dishonest shopkeeper purchased milk at Rs. 100 per litre and mixed 10 liters of water in it. By selling the mixture at the rate of Rs. 100 per litre he earns a profit of 25%. The quantity of the amount of the mixture that he had was
Correct Answer: A
Solution:
Let the quantity of milk purchased be $x$ liters.
The cost price of $x$ liters of milk is $100x$.
The shopkeeper adds 10 liters of water, so the total quantity of the mixture becomes $(x + 10)$ liters.
The selling price of the mixture is $100(x + 10)$ rupees.
The profit is 25%, so the selling price is 125% of the cost price.
$$100(x + 10) = 1.25 \times 100x$$ $$100x + 1000 = 125x$$ $$25x = 1000$$ $$x = 40$$ Therefore, the quantity of milk purchased is 40 liters.
The total quantity of the mixture is $40 + 10 = 50$ liters.
Solution:
Let the quantity of milk purchased be $x$ liters.
The cost price of $x$ liters of milk is $100x$.
The shopkeeper adds 10 liters of water, so the total quantity of the mixture becomes $(x + 10)$ liters.
The selling price of the mixture is $100(x + 10)$ rupees.
The profit is 25%, so the selling price is 125% of the cost price.
$$100(x + 10) = 1.25 \times 100x$$ $$100x + 1000 = 125x$$ $$25x = 1000$$ $$x = 40$$ Therefore, the quantity of milk purchased is 40 liters.
The total quantity of the mixture is $40 + 10 = 50$ liters.
Q14: In what ratio should two qualities of tea having the rates of Rs. 40 per kg and Rs. 30 per kg be mixed in order to get a mixture that would have a rate ofRs 35per kg?
Correct Answer: B
Solution:
Let the ratio be $x:y$.
We can use the alligation method to solve this.
The cost price of the mixture is Rs 35 per kg.
Using the alligation formula: $$ \frac{\text{Cost of better quality tea} - \text{Cost of mixture}}{\text{Cost of mixture} - \text{Cost of poorer quality tea}} = \frac{x}{y} $$ $$ \frac{40 - 35}{35 - 30} = \frac{x}{y} $$ $$ \frac{5}{5} = \frac{x}{y} $$ $$ \frac{x}{y} = \frac{1}{1} $$ Therefore, the ratio is 1:1.
The two qualities of tea should be mixed in a 1:1 ratio.
Solution:
Let the ratio be $x:y$.
We can use the alligation method to solve this.
The cost price of the mixture is Rs 35 per kg.
Using the alligation formula: $$ \frac{\text{Cost of better quality tea} - \text{Cost of mixture}}{\text{Cost of mixture} - \text{Cost of poorer quality tea}} = \frac{x}{y} $$ $$ \frac{40 - 35}{35 - 30} = \frac{x}{y} $$ $$ \frac{5}{5} = \frac{x}{y} $$ $$ \frac{x}{y} = \frac{1}{1} $$ Therefore, the ratio is 1:1.
The two qualities of tea should be mixed in a 1:1 ratio.
Q15: Raman steals four gallons of liquid soap kept in a train compartment’s bathroom from a container that is full of liquid soap. He then fills it with water to avoid detection. Unable to resist the temptation he steals 4 gallons of the mixture again, and fills it with water. When the liquid soap is checked at a station it is found that the ratio of the liquid soap now left in the container to that of the water in it is 36: 13. What was the initial amount of the liquid soap in the container if it is known that the liquid soap is neither used nor augmented by anybody else during the entire period?
Correct Answer: D
Solution:
Let the initial amount of liquid soap be $x$ gallons.
Raman steals 4 gallons, so the remaining soap is $x - 4$ gallons.
He adds 4 gallons of water.
The mixture now contains $x-4$ gallons of soap and 4 gallons of water.
He steals 4 gallons of the mixture.
The fraction of soap in the mixture is $\frac{x-4}{x-4+4} = \frac{x-4}{x}$.
The amount of soap stolen is $4 \times \frac{x-4}{x} = \frac{4x - 16}{x}$ gallons.
The amount of water stolen is $4 \times \frac{4}{x} = \frac{16}{x}$ gallons.
The remaining soap is $(x-4) - \frac{4x-16}{x} = \frac{x^2 - 4x - 4x + 16}{x} = \frac{x^2 - 8x + 16}{x}$ gallons.
The remaining water is $4 - \frac{16}{x} + 4 = 8 - \frac{16}{x} = \frac{8x - 16}{x}$ gallons.
The ratio of soap to water is 36:13.
Therefore, $\frac{\frac{x^2 - 8x + 16}{x}}{\frac{8x - 16}{x}} = \frac{36}{13}$ $\frac{x^2 - 8x + 16}{8x - 16} = \frac{36}{13}$ $13(x^2 - 8x + 16) = 36(8x - 16)$ $13x^2 - 104x + 208 = 288x - 576$ $13x^2 - 392x + 784 = 0$ $x^2 - 30.15x + 60.3 = 0$ (approximately) Solving the quadratic equation, we get $x \approx 28$ gallons.
(approximately) A precise solution requires solving the quadratic equation, which yields $x = 28$.
Solution:
Let the initial amount of liquid soap be $x$ gallons.
Raman steals 4 gallons, so the remaining soap is $x - 4$ gallons.
He adds 4 gallons of water.
The mixture now contains $x-4$ gallons of soap and 4 gallons of water.
He steals 4 gallons of the mixture.
The fraction of soap in the mixture is $\frac{x-4}{x-4+4} = \frac{x-4}{x}$.
The amount of soap stolen is $4 \times \frac{x-4}{x} = \frac{4x - 16}{x}$ gallons.
The amount of water stolen is $4 \times \frac{4}{x} = \frac{16}{x}$ gallons.
The remaining soap is $(x-4) - \frac{4x-16}{x} = \frac{x^2 - 4x - 4x + 16}{x} = \frac{x^2 - 8x + 16}{x}$ gallons.
The remaining water is $4 - \frac{16}{x} + 4 = 8 - \frac{16}{x} = \frac{8x - 16}{x}$ gallons.
The ratio of soap to water is 36:13.
Therefore, $\frac{\frac{x^2 - 8x + 16}{x}}{\frac{8x - 16}{x}} = \frac{36}{13}$ $\frac{x^2 - 8x + 16}{8x - 16} = \frac{36}{13}$ $13(x^2 - 8x + 16) = 36(8x - 16)$ $13x^2 - 104x + 208 = 288x - 576$ $13x^2 - 392x + 784 = 0$ $x^2 - 30.15x + 60.3 = 0$ (approximately) Solving the quadratic equation, we get $x \approx 28$ gallons.
(approximately) A precise solution requires solving the quadratic equation, which yields $x = 28$.
Q16: In what ratio should water be mixed with soda costing Rs. 12 per litre so as to make a profit of 50% by selling the diluted liquid at Rs. 15 per litre?
Correct Answer: C
Solution:
Let the cost price of soda be Rs.
$12$ per litre.
Let $x$ litres of water be mixed with $1$ litre of soda.
The total volume of the mixture is $(1+x)$ litres.
The selling price of the mixture is Rs.
$15$ per litre.
The total selling price of $(1+x)$ litres is $15(1+x)$ Rs.
The cost price of $1$ litre of soda is Rs.
$12$.
The profit percentage is $50\%$.
Profit = Selling Price - Cost Price Profit = $15(1+x) - 12$ Profit percentage = $\frac{\text{Profit}}{\text{Cost Price}} \times 100$ $50 = \frac{15(1+x) - 12}{12} \times 100$ $\frac{50}{100} = \frac{15 + 15x - 12}{12}$ $\frac{1}{2} = \frac{3 + 15x}{12}$ $6 = 3 + 15x$ $15x = 3$ $x = \frac{3}{15} = \frac{1}{5}$ Therefore, the ratio of water to soda is $x:1 = \frac{1}{5} : 1 = 1:5$.
Water should be mixed with soda in the ratio $1:5$.
Solution:
Let the cost price of soda be Rs.
$12$ per litre.
Let $x$ litres of water be mixed with $1$ litre of soda.
The total volume of the mixture is $(1+x)$ litres.
The selling price of the mixture is Rs.
$15$ per litre.
The total selling price of $(1+x)$ litres is $15(1+x)$ Rs.
The cost price of $1$ litre of soda is Rs.
$12$.
The profit percentage is $50\%$.
Profit = Selling Price - Cost Price Profit = $15(1+x) - 12$ Profit percentage = $\frac{\text{Profit}}{\text{Cost Price}} \times 100$ $50 = \frac{15(1+x) - 12}{12} \times 100$ $\frac{50}{100} = \frac{15 + 15x - 12}{12}$ $\frac{1}{2} = \frac{3 + 15x}{12}$ $6 = 3 + 15x$ $15x = 3$ $x = \frac{3}{15} = \frac{1}{5}$ Therefore, the ratio of water to soda is $x:1 = \frac{1}{5} : 1 = 1:5$.
Water should be mixed with soda in the ratio $1:5$.
Q17: A sum of Rs. 4 is made up of 20 coins that are either 10 paise coins or 60 paise coins. Find out how many 20 paise coins are there in the total amount.
Correct Answer: C
Solution:
Let $x$ be the number of 10 paise coins and $y$ be the number of 60 paise coins.
We know that: $$x + y = 20 \quad \text{(Equation 1: Total number of coins)}$$ $$0.10x + 0.60y = 4 \quad \text{(Equation 2: Total value in Rupees)}$$ From Equation 1, we get $x = 20 - y$.
Substitute this into Equation 2: $$0.10(20 - y) + 0.60y = 4$$ $$2 - 0.10y + 0.60y = 4$$ $$0.50y = 2$$ $$y = \frac{2}{0.50}$$ $$y = 4$$ Now substitute $y = 4$ back into Equation 1: $$x + 4 = 20$$ $$x = 16$$ Therefore, there are 16 coins of 10 paise and 4 coins of 60 paise.
The question asks for the number of 20 paise coins, which is not present in the given information.
There seems to be a mistake in the question.
Assuming the question meant to ask for the number of 10 paise coins or 60 paise coins, the answer would be 16 and 4 respectively.
Solution:
Let $x$ be the number of 10 paise coins and $y$ be the number of 60 paise coins.
We know that: $$x + y = 20 \quad \text{(Equation 1: Total number of coins)}$$ $$0.10x + 0.60y = 4 \quad \text{(Equation 2: Total value in Rupees)}$$ From Equation 1, we get $x = 20 - y$.
Substitute this into Equation 2: $$0.10(20 - y) + 0.60y = 4$$ $$2 - 0.10y + 0.60y = 4$$ $$0.50y = 2$$ $$y = \frac{2}{0.50}$$ $$y = 4$$ Now substitute $y = 4$ back into Equation 1: $$x + 4 = 20$$ $$x = 16$$ Therefore, there are 16 coins of 10 paise and 4 coins of 60 paise.
The question asks for the number of 20 paise coins, which is not present in the given information.
There seems to be a mistake in the question.
Assuming the question meant to ask for the number of 10 paise coins or 60 paise coins, the answer would be 16 and 4 respectively.
Q18: Pinku a dishonest grocer professes to sell pure butter at cost price, but he mixes it with adulterated fat and thereby gains 25%. Find the percentage of adulterated fat in the mixture assuming that adulterated fat is freely available
Correct Answer: A
Solution:
Let the cost price of pure butter be $CP$.
Pinku sells the mixture at $CP$, but gains 25\%.
This means the selling price of the mixture is $1.25CP$.
The gain is achieved by mixing cheaper adulterated fat.
Let $x$ be the amount of pure butter and $y$ be the amount of adulterated fat.
The cost price of the mixture is $x \cdot CP + y \cdot 0$ (assuming adulterated fat is free).
The total quantity of mixture is $x+y$.
Since he sells at cost price, $$ \frac{x \cdot CP + y \cdot 0}{x+y} \cdot 1.25 = CP $$ This simplifies to $\frac{x}{x+y} = 0.8$.
Solving for $\frac{y}{x}$: $\frac{y}{x} = \frac{0.2}{0.8} = \frac{1}{4}$.
To express this as percentage, we multiply by 100\%: $\frac{y}{x+y} \cdot 100\% = \frac{1}{5} \cdot 100\% = 20\%$.
Therefore, 20\% of the mixture is adulterated fat.
Solution:
Let the cost price of pure butter be $CP$.
Pinku sells the mixture at $CP$, but gains 25\%.
This means the selling price of the mixture is $1.25CP$.
The gain is achieved by mixing cheaper adulterated fat.
Let $x$ be the amount of pure butter and $y$ be the amount of adulterated fat.
The cost price of the mixture is $x \cdot CP + y \cdot 0$ (assuming adulterated fat is free).
The total quantity of mixture is $x+y$.
Since he sells at cost price, $$ \frac{x \cdot CP + y \cdot 0}{x+y} \cdot 1.25 = CP $$ This simplifies to $\frac{x}{x+y} = 0.8$.
Solving for $\frac{y}{x}$: $\frac{y}{x} = \frac{0.2}{0.8} = \frac{1}{4}$.
To express this as percentage, we multiply by 100\%: $\frac{y}{x+y} \cdot 100\% = \frac{1}{5} \cdot 100\% = 20\%$.
Therefore, 20\% of the mixture is adulterated fat.
Q19: A mixture of 75 liters of alcohol and water contains 20% of water. How much water must be added to the above mixture to make the water 25% of the resulting mixture?
Correct Answer: A
Solution:
Let the initial mixture be 75 liters.
Water in the initial mixture = 20\% of 75 liters = $\frac{20}{100} \times 75 = 15$ liters Alcohol in the initial mixture = $75 - 15 = 60$ liters Let $x$ liters of water be added to the mixture.
The total volume of the new mixture will be $75 + x$ liters.
The amount of water in the new mixture will be $15 + x$ liters.
The percentage of water in the new mixture is given as 25\%.
Therefore, we can set up the equation: $$ \frac{15 + x}{75 + x} = \frac{25}{100} $$ $$ 15 + x = 0.25 \times (75 + x) $$ $$ 15 + x = 18.75 + 0.25x $$ $$ x - 0.25x = 18.75 - 15 $$ $$ 0.75x = 3.75 $$ $$ x = \frac{3.75}{0.75} $$ $$ x = 5 $$ Therefore, 5 liters of water must be added to the mixture.
Solution:
Let the initial mixture be 75 liters.
Water in the initial mixture = 20\% of 75 liters = $\frac{20}{100} \times 75 = 15$ liters Alcohol in the initial mixture = $75 - 15 = 60$ liters Let $x$ liters of water be added to the mixture.
The total volume of the new mixture will be $75 + x$ liters.
The amount of water in the new mixture will be $15 + x$ liters.
The percentage of water in the new mixture is given as 25\%.
Therefore, we can set up the equation: $$ \frac{15 + x}{75 + x} = \frac{25}{100} $$ $$ 15 + x = 0.25 \times (75 + x) $$ $$ 15 + x = 18.75 + 0.25x $$ $$ x - 0.25x = 18.75 - 15 $$ $$ 0.75x = 3.75 $$ $$ x = \frac{3.75}{0.75} $$ $$ x = 5 $$ Therefore, 5 liters of water must be added to the mixture.
Q20: A mixture of 40 liters of milk and water contains 10% water. How much water should be added to it to increase the percentage of water to 25%?
Correct Answer: D
Solution:
Let the amount of milk in the mixture be $M$ and the amount of water be $W$.
Initially, the mixture contains 40 liters, with 10% water.
Therefore: $W = 0.10 \times 40 = 4$ liters of water $M = 40 - 4 = 36$ liters of milk Let $x$ liters of water be added to the mixture.
The new amount of water will be $4 + x$ liters.
The total volume of the new mixture will be $40 + x$ liters.
The new percentage of water is 25%, so: $$ \frac{4 + x}{40 + x} = 0.25 $$ Multiply both sides by $(40 + x)$: $4 + x = 0.25 \times (40 + x)$ $4 + x = 10 + 0.25x$ $x - 0.25x = 10 - 4$ $0.75x = 6$ $x = \frac{6}{0.75}$ $x = 8$ Therefore, 8 liters of water should be added.
Solution:
Let the amount of milk in the mixture be $M$ and the amount of water be $W$.
Initially, the mixture contains 40 liters, with 10% water.
Therefore: $W = 0.10 \times 40 = 4$ liters of water $M = 40 - 4 = 36$ liters of milk Let $x$ liters of water be added to the mixture.
The new amount of water will be $4 + x$ liters.
The total volume of the new mixture will be $40 + x$ liters.
The new percentage of water is 25%, so: $$ \frac{4 + x}{40 + x} = 0.25 $$ Multiply both sides by $(40 + x)$: $4 + x = 0.25 \times (40 + x)$ $4 + x = 10 + 0.25x$ $x - 0.25x = 10 - 4$ $0.75x = 6$ $x = \frac{6}{0.75}$ $x = 8$ Therefore, 8 liters of water should be added.
Q21: Two vessels contain a mixture of spirit and water. In the first vessel the ratio of spirit to water is 8 : 3 and in the second vessel the ratio is 5 : 1. A 35 litre cask is filled from these vessels so as to contain a mixture of spirit and water in the ratio of 4 :1. How many liters are taken from the first vessel?
Correct Answer: A
Solution:
Let $x$ liters be taken from the first vessel and $(35-x)$ liters from the second vessel.
In the first vessel, the ratio of spirit to water is 8:3.
The fraction of spirit is $\frac{8}{8+3} = \frac{8}{11}$.
In the second vessel, the ratio of spirit to water is 5:1.
The fraction of spirit is $\frac{5}{5+1} = \frac{5}{6}$.
In the mixture, the ratio of spirit to water is 4:1.
The fraction of spirit is $\frac{4}{4+1} = \frac{4}{5}$.
Using alligation: $$ \frac{\frac{5}{6} - \frac{4}{5}}{\frac{4}{5} - \frac{8}{11}} = \frac{\frac{25-24}{30}}{\frac{44-40}{55}} = \frac{\frac{1}{30}}{\frac{4}{55}} = \frac{1}{30} \times \frac{55}{4} = \frac{11}{24} $$ Therefore, the ratio of the amount taken from the first vessel to the amount taken from the second vessel is 11:13.
Let the amount taken from the first vessel be $11k$ and from the second vessel be $13k$.
Total amount = $11k + 13k = 24k = 35$ liters $k = \frac{35}{24}$ Amount taken from the first vessel = $11k = 11 \times \frac{35}{24} = \frac{385}{24} \approx 16.04$ liters (approximately) However, this method leads to a non-integer solution, which is unusual for these types of problems.
Let's use another approach.
Let $x$ litres be taken from the first vessel.
Then $(35-x)$ litres are taken from the second vessel.
Spirit in $x$ litres from first vessel = $\frac{8}{11}x$ Spirit in $(35-x)$ litres from second vessel = $\frac{5}{6}(35-x)$ Total spirit in 35 litres = $\frac{8}{11}x + \frac{5}{6}(35-x)$ Total spirit should be $\frac{4}{5} \times 35 = 28$ litres $\frac{8}{11}x + \frac{5}{6}(35-x) = 28$ Multiplying by 66 (LCM of 11 and 6): $48x + 55(35-x) = 28 \times 66$ $48x + 1925 - 55x = 1848$ $-7x = 1848 - 1925$ $-7x = -77$ $x = 11$
Solution:
Let $x$ liters be taken from the first vessel and $(35-x)$ liters from the second vessel.
In the first vessel, the ratio of spirit to water is 8:3.
The fraction of spirit is $\frac{8}{8+3} = \frac{8}{11}$.
In the second vessel, the ratio of spirit to water is 5:1.
The fraction of spirit is $\frac{5}{5+1} = \frac{5}{6}$.
In the mixture, the ratio of spirit to water is 4:1.
The fraction of spirit is $\frac{4}{4+1} = \frac{4}{5}$.
Using alligation: $$ \frac{\frac{5}{6} - \frac{4}{5}}{\frac{4}{5} - \frac{8}{11}} = \frac{\frac{25-24}{30}}{\frac{44-40}{55}} = \frac{\frac{1}{30}}{\frac{4}{55}} = \frac{1}{30} \times \frac{55}{4} = \frac{11}{24} $$ Therefore, the ratio of the amount taken from the first vessel to the amount taken from the second vessel is 11:13.
Let the amount taken from the first vessel be $11k$ and from the second vessel be $13k$.
Total amount = $11k + 13k = 24k = 35$ liters $k = \frac{35}{24}$ Amount taken from the first vessel = $11k = 11 \times \frac{35}{24} = \frac{385}{24} \approx 16.04$ liters (approximately) However, this method leads to a non-integer solution, which is unusual for these types of problems.
Let's use another approach.
Let $x$ litres be taken from the first vessel.
Then $(35-x)$ litres are taken from the second vessel.
Spirit in $x$ litres from first vessel = $\frac{8}{11}x$ Spirit in $(35-x)$ litres from second vessel = $\frac{5}{6}(35-x)$ Total spirit in 35 litres = $\frac{8}{11}x + \frac{5}{6}(35-x)$ Total spirit should be $\frac{4}{5} \times 35 = 28$ litres $\frac{8}{11}x + \frac{5}{6}(35-x) = 28$ Multiplying by 66 (LCM of 11 and 6): $48x + 55(35-x) = 28 \times 66$ $48x + 1925 - 55x = 1848$ $-7x = 1848 - 1925$ $-7x = -77$ $x = 11$
Q22: There are two mixtures of milk and water, the quantity of milk in them being 20% and 80% of the mixture. If 2 liters of the first are mixed with three liters of the second, what will be the ratio of milk to water in the new mixture?
Correct Answer: D
Solution:
Let's denote the first mixture as $M_1$ and the second mixture as $M_2$.
In $M_1$, milk is 20% and water is 80%.
In 2 liters of $M_1$, there is $0.2 \times 2 = 0.4$ liters of milk and $2 - 0.4 = 1.6$ liters of water.
In $M_2$, milk is 80% and water is 20%.
In 3 liters of $M_2$, there is $0.8 \times 3 = 2.4$ liters of milk and $3 - 2.4 = 0.6$ liters of water.
When mixed, the total quantity of milk is $0.4 + 2.4 = 2.8$ liters.
The total quantity of water is $1.6 + 0.6 = 2.2$ liters.
The ratio of milk to water in the new mixture is $2.8 : 2.2$.
Simplifying the ratio by dividing by 0.2, we get $14:11$.
Solution:
Let's denote the first mixture as $M_1$ and the second mixture as $M_2$.
In $M_1$, milk is 20% and water is 80%.
In 2 liters of $M_1$, there is $0.2 \times 2 = 0.4$ liters of milk and $2 - 0.4 = 1.6$ liters of water.
In $M_2$, milk is 80% and water is 20%.
In 3 liters of $M_2$, there is $0.8 \times 3 = 2.4$ liters of milk and $3 - 2.4 = 0.6$ liters of water.
When mixed, the total quantity of milk is $0.4 + 2.4 = 2.8$ liters.
The total quantity of water is $1.6 + 0.6 = 2.2$ liters.
The ratio of milk to water in the new mixture is $2.8 : 2.2$.
Simplifying the ratio by dividing by 0.2, we get $14:11$.
Q23: There are two kinds of alloys of silver and copper. The first alloy contains silver and copper such that 93.33% of it is silver. In the second alloy there is 86.66% silver. What weight of the first alloy should be mixed with some weight of the second alloy so as to make a 100 kg mass containing 90% of silver?
Correct Answer: B
Solution:
Let $x$ kg be the weight of the first alloy and $y$ kg be the weight of the second alloy.
The first alloy contains 93.33% silver, so the amount of silver in $x$ kg is $0.9333x$ kg.
The second alloy contains 86.66% silver, so the amount of silver in $y$ kg is $0.8666y$ kg.
The total weight of the mixture is 100 kg, so $x + y = 100$.
The mixture contains 90% silver, so the amount of silver in the mixture is $0.9 \times 100 = 90$ kg.
Therefore, $0.9333x + 0.8666y = 90$.
We have a system of two linear equations: $$ x + y = 100 $$ $$ 0.9333x + 0.8666y = 90 $$ From the first equation, $y = 100 - x$.
Substitute this into the second equation: $0.9333x + 0.8666(100 - x) = 90$ $0.9333x + 86.66 - 0.8666x = 90$ $0.0667x = 90 - 86.66$ $0.0667x = 3.34$ $x = \frac{3.34}{0.0667}$ $x \approx 50$ Therefore, $y = 100 - 50 = 50$ So, 50 kg of the first alloy should be mixed with 50 kg of the second alloy.
Solution:
Let $x$ kg be the weight of the first alloy and $y$ kg be the weight of the second alloy.
The first alloy contains 93.33% silver, so the amount of silver in $x$ kg is $0.9333x$ kg.
The second alloy contains 86.66% silver, so the amount of silver in $y$ kg is $0.8666y$ kg.
The total weight of the mixture is 100 kg, so $x + y = 100$.
The mixture contains 90% silver, so the amount of silver in the mixture is $0.9 \times 100 = 90$ kg.
Therefore, $0.9333x + 0.8666y = 90$.
We have a system of two linear equations: $$ x + y = 100 $$ $$ 0.9333x + 0.8666y = 90 $$ From the first equation, $y = 100 - x$.
Substitute this into the second equation: $0.9333x + 0.8666(100 - x) = 90$ $0.9333x + 86.66 - 0.8666x = 90$ $0.0667x = 90 - 86.66$ $0.0667x = 3.34$ $x = \frac{3.34}{0.0667}$ $x \approx 50$ Therefore, $y = 100 - 50 = 50$ So, 50 kg of the first alloy should be mixed with 50 kg of the second alloy.
Q26: Two vessels contain spirit and water mixed respectively in the ratio of 1: 4 and 4: 1 Find the ratio in which these are to be mixed to get a new mixture in which the ratio of spirit to water is 1: 3.
Correct Answer: A
Solution:
Let the ratio in which the two mixtures are mixed be $x:y$.
Mixture 1: Spirit : Water = $1 : 4$ (1 part spirit, 4 parts water) Mixture 2: Spirit : Water = $4 : 1$ (4 parts spirit, 1 part water) New Mixture: Spirit : Water = $1 : 3$ Using alligation: Spirit ratio in mixture 1 = $\frac{1}{5}$ Spirit ratio in mixture 2 = $\frac{4}{5}$ Spirit ratio in new mixture = $\frac{1}{4}$ By alligation: $\left(\frac{4}{5} - \frac{1}{4}\right) : \left(\frac{1}{4} - \frac{1}{5}\right)$ $\frac{(16 - 5)}{20} : \frac{(5 - 4)}{20}$ $\frac{11}{20} : \frac{1}{20}$ $11 : 1$ Therefore, the ratio in which the mixtures are to be mixed is $11:1$.
Solution:
Let the ratio in which the two mixtures are mixed be $x:y$.
Mixture 1: Spirit : Water = $1 : 4$ (1 part spirit, 4 parts water) Mixture 2: Spirit : Water = $4 : 1$ (4 parts spirit, 1 part water) New Mixture: Spirit : Water = $1 : 3$ Using alligation: Spirit ratio in mixture 1 = $\frac{1}{5}$ Spirit ratio in mixture 2 = $\frac{4}{5}$ Spirit ratio in new mixture = $\frac{1}{4}$ By alligation: $\left(\frac{4}{5} - \frac{1}{4}\right) : \left(\frac{1}{4} - \frac{1}{5}\right)$ $\frac{(16 - 5)}{20} : \frac{(5 - 4)}{20}$ $\frac{11}{20} : \frac{1}{20}$ $11 : 1$ Therefore, the ratio in which the mixtures are to be mixed is $11:1$.
Q27: The price of a table and a chair is Rs. 3000. The table was sold at a 20% profit and the chair at a 10% loss. If in the transaction a man gains Rs. 300, how much is cost price (in Rs. ) of the table?
Correct Answer: C
Solution:
Let the cost price of the table be $T$ and the cost price of the chair be $C$.
Given: $T + C = 3000$ (Equation 1) The table was sold at a 20% profit, so the selling price of the table is $1.2T$.
The chair was sold at a 10% loss, so the selling price of the chair is $0.9C$.
Total selling price = $1.2T + 0.9C$ Total profit = 300 Total selling price = Total cost price + Total profit $1.2T + 0.9C = 3000 + 300 = 3300$ (Equation 2) Now we have a system of two equations with two variables: $$T + C = 3000$$ $$1.2T + 0.9C = 3300$$ From Equation 1, $C = 3000 - T$ Substitute this into Equation 2: $1.2T + 0.9(3000 - T) = 3300$ $1.2T + 2700 - 0.9T = 3300$ $0.3T = 600$ $T = \frac{600}{0.3}$ $T = 2000$ Therefore, the cost price of the table is Rs.
2000.
Solution:
Let the cost price of the table be $T$ and the cost price of the chair be $C$.
Given: $T + C = 3000$ (Equation 1) The table was sold at a 20% profit, so the selling price of the table is $1.2T$.
The chair was sold at a 10% loss, so the selling price of the chair is $0.9C$.
Total selling price = $1.2T + 0.9C$ Total profit = 300 Total selling price = Total cost price + Total profit $1.2T + 0.9C = 3000 + 300 = 3300$ (Equation 2) Now we have a system of two equations with two variables: $$T + C = 3000$$ $$1.2T + 0.9C = 3300$$ From Equation 1, $C = 3000 - T$ Substitute this into Equation 2: $1.2T + 0.9(3000 - T) = 3300$ $1.2T + 2700 - 0.9T = 3300$ $0.3T = 600$ $T = \frac{600}{0.3}$ $T = 2000$ Therefore, the cost price of the table is Rs.
2000.
Q28: A person purchased a pen and a pencil for Rs. 15. He sold the pen at a profit of 20% and the pencil at a profit of 30%. If his total profit was 24%, find the cost price of the pen.
Correct Answer: C
Solution:
Let $CP_{pen}$ be the cost price of the pen and $CP_{pencil}$ be the cost price of the pencil.
We know that $CP_{pen} + CP_{pencil} = 15$.
The pen was sold at a 20% profit, so the selling price of the pen is $SP_{pen} = CP_{pen} \times 1.20$.
The pencil was sold at a 30% profit, so the selling price of the pencil is $SP_{pencil} = CP_{pencil} \times 1.30$.
The total selling price is $SP_{pen} + SP_{pencil} = CP_{pen} \times 1.20 + CP_{pencil} \times 1.30$.
The total profit is 24%, so the total selling price is also $15 \times 1.24 = 18.6$.
Therefore, $1.20 \times CP_{pen} + 1.30 \times CP_{pencil} = 18.6$.
We have a system of two equations: 1) $CP_{pen} + CP_{pencil} = 15$ 2) $1.20 \times CP_{pen} + 1.30 \times CP_{pencil} = 18.6$ From equation (1), $CP_{pencil} = 15 - CP_{pen}$.
Substitute this into equation (2): $1.20 \times CP_{pen} + 1.30 \times (15 - CP_{pen}) = 18.6$ $1.20 \times CP_{pen} + 19.5 - 1.30 \times CP_{pen} = 18.6$ $-0.10 \times CP_{pen} = 18.6 - 19.5$ $-0.10 \times CP_{pen} = -0.9$ $CP_{pen} = 9$ Therefore, the cost price of the pen is Rs.
9.
Solution:
Let $CP_{pen}$ be the cost price of the pen and $CP_{pencil}$ be the cost price of the pencil.
We know that $CP_{pen} + CP_{pencil} = 15$.
The pen was sold at a 20% profit, so the selling price of the pen is $SP_{pen} = CP_{pen} \times 1.20$.
The pencil was sold at a 30% profit, so the selling price of the pencil is $SP_{pencil} = CP_{pencil} \times 1.30$.
The total selling price is $SP_{pen} + SP_{pencil} = CP_{pen} \times 1.20 + CP_{pencil} \times 1.30$.
The total profit is 24%, so the total selling price is also $15 \times 1.24 = 18.6$.
Therefore, $1.20 \times CP_{pen} + 1.30 \times CP_{pencil} = 18.6$.
We have a system of two equations: 1) $CP_{pen} + CP_{pencil} = 15$ 2) $1.20 \times CP_{pen} + 1.30 \times CP_{pencil} = 18.6$ From equation (1), $CP_{pencil} = 15 - CP_{pen}$.
Substitute this into equation (2): $1.20 \times CP_{pen} + 1.30 \times (15 - CP_{pen}) = 18.6$ $1.20 \times CP_{pen} + 19.5 - 1.30 \times CP_{pen} = 18.6$ $-0.10 \times CP_{pen} = 18.6 - 19.5$ $-0.10 \times CP_{pen} = -0.9$ $CP_{pen} = 9$ Therefore, the cost price of the pen is Rs.
9.
Q29: A container is full of a mixture of kerosene and petrol in which there is 18% kerosene. Eight liters are drawn off and then the vessel is filled with petrol. If the kerosene is now 15%, how much does the container hold?
Correct Answer: D
Solution:
Let the capacity of the container be $C$ liters.
Initially, the amount of kerosene is $0.18C$ liters.
8 liters of mixture are drawn off.
The amount of kerosene removed is $0.18 \times 8 = 1.44$ liters.
The remaining kerosene is $0.18C - 1.44$ liters.
8 liters of petrol are added.
The amount of kerosene remains the same.
The new percentage of kerosene is 15%.
Therefore, $\frac{0.18C - 1.44}{C} = 0.15$ $$0.18C - 1.44 = 0.15C$$ $$0.03C = 1.44$$ $$C = \frac{1.44}{0.03}$$ $$C = 48 \text{ liters}$$ Therefore, the container holds 48 liters.
Solution:
Let the capacity of the container be $C$ liters.
Initially, the amount of kerosene is $0.18C$ liters.
8 liters of mixture are drawn off.
The amount of kerosene removed is $0.18 \times 8 = 1.44$ liters.
The remaining kerosene is $0.18C - 1.44$ liters.
8 liters of petrol are added.
The amount of kerosene remains the same.
The new percentage of kerosene is 15%.
Therefore, $\frac{0.18C - 1.44}{C} = 0.15$ $$0.18C - 1.44 = 0.15C$$ $$0.03C = 1.44$$ $$C = \frac{1.44}{0.03}$$ $$C = 48 \text{ liters}$$ Therefore, the container holds 48 liters.
Q30: Two solutions of 80% and 87% purity are mixed resulting in 35 liters of mixture of 84% purity. How much is the quantity of the first solution in the resulting mixture?
Correct Answer: A
Solution:
Let $x$ liters of 80% solution and $(35-x)$ liters of 87% solution be mixed.
The equation representing the mixture is: $$0.80x + 0.87(35-x) = 0.84 \times 35$$ $$0.80x + 30.45 - 0.87x = 29.4$$ $$-0.07x = 29.4 - 30.45$$ $$-0.07x = -1.05$$ $$x = \frac{-1.05}{-0.07}$$ $$x = 15$$ Therefore, the quantity of the first solution (80% purity) in the resulting mixture is 15 liters.
Solution:
Let $x$ liters of 80% solution and $(35-x)$ liters of 87% solution be mixed.
The equation representing the mixture is: $$0.80x + 0.87(35-x) = 0.84 \times 35$$ $$0.80x + 30.45 - 0.87x = 29.4$$ $$-0.07x = 29.4 - 30.45$$ $$-0.07x = -1.05$$ $$x = \frac{-1.05}{-0.07}$$ $$x = 15$$ Therefore, the quantity of the first solution (80% purity) in the resulting mixture is 15 liters.
Q31: In the Delhi zoo, there are lions and there are hens. If the heads are counted, there are 180, while the legs are 448. What will be the number of lions in the zoo?
Correct Answer: C
Solution:
Let's denote the number of lions as $l$ and the number of hens as $h$.
Each lion has 1 head and 4 legs, while each hen has 1 head and 2 legs.
We can set up a system of two equations based on the given information: Equation 1 (Heads): $l + h = 180$ Equation 2 (Legs): $4l + 2h = 448$ We can solve this system using substitution or elimination.
Let's use elimination.
Multiply Equation 1 by -2: $-2l - 2h = -360$ Now add this modified Equation 1 to Equation 2: $(-2l - 2h) + (4l + 2h) = -360 + 448$ $2l = 88$ $l = 44$ Therefore, there are 44 lions in the zoo.
To find the number of hens, substitute $l = 44$ into Equation 1: $44 + h = 180$ $h = 180 - 44$ $h = 136$ There are 44 lions and 136 hens.
Solution:
Let's denote the number of lions as $l$ and the number of hens as $h$.
Each lion has 1 head and 4 legs, while each hen has 1 head and 2 legs.
We can set up a system of two equations based on the given information: Equation 1 (Heads): $l + h = 180$ Equation 2 (Legs): $4l + 2h = 448$ We can solve this system using substitution or elimination.
Let's use elimination.
Multiply Equation 1 by -2: $-2l - 2h = -360$ Now add this modified Equation 1 to Equation 2: $(-2l - 2h) + (4l + 2h) = -360 + 448$ $2l = 88$ $l = 44$ Therefore, there are 44 lions in the zoo.
To find the number of hens, substitute $l = 44$ into Equation 1: $44 + h = 180$ $h = 180 - 44$ $h = 136$ There are 44 lions and 136 hens.
12
No comments:
Post a Comment