In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1:2 Based on the information, the total number of coins in the collection now becomes
UPSC-2013
This problem involves setting up and solving equations based on given ratios. We will use variables to represent the initial number of gold and non-gold coins, form equations from the given ratio conditions, and then solve for the variables to find the total number of coins.
Define initial quantities: Let the initial number of gold coins be \(G\) and the initial number of non-gold coins be \(N\).
Formulate the first equation from the initial ratio: According to the problem, there is one gold coin for every three non-gold coins. This can be written as a ratio:
\[ \frac{G}{N} = \frac{1}{3} \]From this, we can express \(N\) in terms of \(G\):
\[ N = 3G \quad \text{(Equation 1)} \]Formulate the second equation after adding coins: 10 more gold coins are added. So, the new number of gold coins is \(G + 10\). The number of non-gold coins remains \(N\). The new ratio of gold coins to non-gold coins is 1:2:
\[ \frac{G + 10}{N} = \frac{1}{2} \]From this, we can cross-multiply:
\[ 2(G + 10) = N \] \[ 2G + 20 = N \quad \text{(Equation 2)} \]Solve the system of equations: Now we have two equations:
Substitute Equation 1 into Equation 2:
\[ 3G = 2G + 20 \]Subtract \(2G\) from both sides:
\[ 3G - 2G = 20 \] \[ G = 20 \]Find the initial number of non-gold coins: Using \(G = 20\) in Equation 1:
\[ N = 3G = 3 \times 20 = 60 \]Calculate the total number of coins now: After adding 10 gold coins, the number of gold coins is \(G + 10 = 20 + 10 = 30\). The number of non-gold coins is still \(N = 60\). The total number of coins in the collection now is the sum of the new gold coins and non-gold coins:
\[ \text{Total coins} = (G + 10) + N = 30 + 60 = 90 \]Correct Option: A) 90
Out of 120 applications for a post, 70 are male and 80 have a driver's license. What is the ratio between the minimum to maximum number of males having driver's license?
UPSC-2013
This problem involves finding the minimum and maximum number of males having a driver's license from a given set of applications. We will use the principle of inclusion-exclusion and basic set theory concepts to determine these values.
Identify the given information:
Calculate the number of female applicants:
\[ \text{Female applicants} = N - M = 120 - 70 = 50 \]Calculate the number of applicants without a driver's license:
\[ \text{Applicants without license} = N - D = 120 - 80 = 40 \]Determine the minimum number of males with a driver's license (\(M_{min}\)):
To find the minimum number of males with a driver's license, we assume that as many females as possible have a driver's license first. The number of females is 50. So, 50 females can have a driver's license. The remaining driver's licenses must be held by males.
\[ \text{Males with license (minimum)} = D - \text{Female applicants} \]\[ \text{Males with license (minimum)} = 80 - 50 = 30 \]Alternatively, consider the number of males without a driver's license. The maximum number of people without a driver's license is 40. These 40 people could all be females, or some could be males. To minimize males with licenses, maximize males without licenses. The number of males is 70. If all 40 people without licenses are males, then \(70 - 40 = 30\) males must have licenses. This is the minimum.Determine the maximum number of males with a driver's license (\(M_{max}\)):
To find the maximum number of males with a driver's license, we consider the limiting factors. The number of males is 70, and the total number of driver's licenses is 80. The number of males with a driver's license cannot exceed the total number of males, nor can it exceed the total number of driver's licenses. Therefore, the maximum number of males with a driver's license is the smaller of these two values.
\[ M_{max} = \min(\text{Number of males}, \text{Number of driver's licenses}) \]\[ M_{max} = \min(70, 80) = 70 \]Calculate the ratio:
The ratio between the minimum to maximum number of males having a driver's license is:
\[ \text{Ratio} = \frac{M_{min}}{M_{max}} = \frac{30}{70} = \frac{3}{7} \]C) The ratio is 3 to 7, which matches our calculated value of \(\frac{3}{7}\).
There are two containers X and Y. X contains 100 ml of milk and Y contains 100 ml of water. 20 ml of milk from X is transferred to Y. After mixing well, 20 ml of the mixture in Y is transferred back to X. If m denotes the proportion of milk in X and n denotes the proportion of water in Y, then which one of the following is correct?
UPSC-2022
m=n
m>n
m<n
Cannot be determined due to insufficient data
This problem involves tracking the quantities of milk and water in two containers through a series of transfers. We need to calculate the proportion of milk in container X and the proportion of water in container Y after all transfers are complete.
First, determine the composition of the mixture in Y:
Now, \(20 \text{ ml}\) of this mixture is transferred from Y to X.
We found \(m = \frac{5}{6}\) and \(n = \frac{5}{6}\).
Therefore, \(m = n\).
Correct Option: A) m=n
The monthly incomes of X and Y are in the ratio of 4:3 and their monthly expenses are in the ratio of 3:2. However, each saves ₹ 6, 000 per month. What is their total monthly income?
UPSC-2017
This problem involves ratios and basic algebra. We are given the ratios of incomes and expenses for two individuals, X and Y, and their monthly savings. We need to find their total monthly income. We can set up equations based on the given ratios and savings, then solve for the unknown variables.
Let the monthly income of X be \(I_X\) and that of Y be \(I_Y\). Their ratio is given as 4:3.
\[ \frac{I_X}{I_Y} = \frac{4}{3} \]So, we can write \(I_X = 4k\) and \(I_Y = 3k\) for some constant \(k\).
Let the monthly expense of X be \(E_X\) and that of Y be \(E_Y\). Their ratio is given as 3:2.
\[ \frac{E_X}{E_Y} = \frac{3}{2} \]So, we can write \(E_X = 3m\) and \(E_Y = 2m\) for some constant \(m\).
Each person saves ₹ 6,000 per month. Savings are calculated as Income - Expense.
Now we have a system of two linear equations with two variables (\(k\) and \(m\)). We can solve for \(k\) and \(m\).
Multiply Equation 1 by 2 and Equation 2 by 3 to eliminate \(m\):
Subtract Equation 3 from Equation 4:
\[ (9k - 6m) - (8k - 6m) = 18000 - 12000 \]\[ 9k - 8k - 6m + 6m = 6000 \]\[ k = 6000 \]Now that we have the value of \(k\), we can find the monthly incomes of X and Y.
The question asks for their total monthly income.
\[ \text{Total Income} = I_X + I_Y = 24000 + 18000 = \text{₹ } 42000 \]Correct Option: B) ₹ 42, 000
Two glasses of equal volume are respectively half and three-fourths filled with milk. They are then filled to the brim by adding water. Their contents are then poured into another vessel. What will be the ratio of milk to water in this vessel?
UPSC-2012
To solve this problem, we need to calculate the total amount of milk and total amount of water from both glasses when their contents are combined into a new vessel. Since the glasses are of equal volume, we can assume a convenient volume, say 1 unit, for each glass. Then, we calculate the milk and water content in each glass and sum them up to find the total milk and total water in the final vessel. Finally, we determine the ratio of total milk to total water.
Assume a common volume: Let the volume of each glass be \(V\) units. For simplicity, let's assume \(V = 1\) unit.
Calculate contents of the first glass: The first glass is half-filled with milk. So, milk in the first glass is \(\frac{1}{2}\) of its volume.
\[ \text{Milk}_1 = \frac{1}{2}V = \frac{1}{2} \times 1 = \frac{1}{2} \text{ unit} \] It is then filled to the brim with water. So, water in the first glass is the remaining volume. \[ \text{Water}_1 = V - \text{Milk}_1 = 1 - \frac{1}{2} = \frac{1}{2} \text{ unit} \]Calculate contents of the second glass: The second glass is three-fourths filled with milk. So, milk in the second glass is \(\frac{3}{4}\) of its volume.
\[ \text{Milk}_2 = \frac{3}{4}V = \frac{3}{4} \times 1 = \frac{3}{4} \text{ unit} \] It is then filled to the brim with water. So, water in the second glass is the remaining volume. \[ \text{Water}_2 = V - \text{Milk}_2 = 1 - \frac{3}{4} = \frac{1}{4} \text{ unit} \]Calculate total milk in the new vessel: Total milk is the sum of milk from both glasses.
\[ \text{Total Milk} = \text{Milk}_1 + \text{Milk}_2 = \frac{1}{2} + \frac{3}{4} \] To add these fractions, find a common denominator, which is 4. \[ \text{Total Milk} = \frac{2}{4} + \frac{3}{4} = \frac{2+3}{4} = \frac{5}{4} \text{ units} \]Calculate total water in the new vessel: Total water is the sum of water from both glasses.
\[ \text{Total Water} = \text{Water}_1 + \text{Water}_2 = \frac{1}{2} + \frac{1}{4} \] To add these fractions, find a common denominator, which is 4. \[ \text{Total Water} = \frac{2}{4} + \frac{1}{4} = \frac{2+1}{4} = \frac{3}{4} \text{ units} \]Determine the ratio of milk to water: The ratio of milk to water in the new vessel is Total Milk : Total Water.
\[ \text{Ratio} = \text{Total Milk} : \text{Total Water} = \frac{5}{4} : \frac{3}{4} \] To simplify the ratio, multiply both sides by 4. \[ \text{Ratio} = 5 : 3 \]Correct Option: D) 5:3
There is a milk sample with 50% water in it. If 1/3rd of this milk is added to equal amount of pure milk, then water in the new mixture will fall down to
UPSC-2017
This problem involves calculating the percentage of water in a new mixture formed by combining a portion of an initial milk-water mixture with pure milk. We will determine the initial quantities of milk and water, then calculate the amounts of each transferred to the new mixture, and finally find the water percentage in the resulting mixture.
Initial Mixture Composition:
Let the total volume of the initial milk sample be \(V\) units.
Given that the milk sample has 50% water, the amount of water in the initial sample is:
The amount of milk (pure milk component) in the initial sample is:
\[ \text{Milk}_{\text{initial}} = 0.50 \times V \]Portion taken from Initial Mixture:
1/3rd of this milk sample is taken. So, the volume of the portion taken is \(\frac{1}{3}V\).
In this portion, the amount of water will be:
And the amount of milk in this portion will be:
\[ \text{Milk}_{\text{portion}} = \frac{1}{3} \times (0.50 \times V) = \frac{1}{6}V \]Adding Pure Milk:
This portion is added to an equal amount of pure milk. The volume of the portion is \(\frac{1}{3}V\). So, the amount of pure milk added is also \(\frac{1}{3}V\).
This pure milk contains 100% milk and 0% water.
New Mixture Composition:
The total volume of the new mixture will be the volume of the portion plus the volume of pure milk added:
The total amount of water in the new mixture comes only from the portion taken:
\[ \text{Water}_{\text{new}} = \text{Water}_{\text{portion}} = \frac{1}{6}V \]The total amount of milk in the new mixture is the milk from the portion plus the pure milk added:
\[ \text{Milk}_{\text{new}} = \text{Milk}_{\text{portion}} + \text{Pure Milk Added} = \frac{1}{6}V + \frac{1}{3}V = \frac{1}{6}V + \frac{2}{6}V = \frac{3}{6}V = \frac{1}{2}V \]Percentage of Water in New Mixture:
The percentage of water in the new mixture is calculated as:
A) 25% is the correct answer. The calculations show that the water content in the new mixture is 25%.
