30 g of sugar was mixed in 180 ml water in a vessel A, 40 g of sugar was mixed in 280 ml of water in vessel B and 20 g of sugar was mixed in 100 ml of water in vessel C. The solution in vessel B is:
UPSC-2016
sweeter than that in C
sweeter than that in A
as sweet as that in C
less sweet than that in C
To compare the sweetness of the solutions, we need to determine the concentration of sugar in each vessel. Concentration can be expressed as the amount of sugar per unit volume of water. A higher concentration of sugar will result in a sweeter solution.
Sugar in A = 30 g
Water in A = 180 ml
Concentration in A (\(C_A\)) = \(\frac{\text{Sugar}}{\text{Water}}\)
\[ C_A = \frac{30 \text{ g}}{180 \text{ ml}} = \frac{1}{6} \text{ g/ml} \approx 0.1667 \text{ g/ml} \]Sugar in B = 40 g
Water in B = 280 ml
Concentration in B (\(C_B\)) = \(\frac{\text{Sugar}}{\text{Water}}\)
\[ C_B = \frac{40 \text{ g}}{280 \text{ ml}} = \frac{4}{28} \text{ g/ml} = \frac{1}{7} \text{ g/ml} \approx 0.1429 \text{ g/ml} \]Sugar in C = 20 g
Water in C = 100 ml
Concentration in C (\(C_C\)) = \(\frac{\text{Sugar}}{\text{Water}}\)
\[ C_C = \frac{20 \text{ g}}{100 \text{ ml}} = \frac{1}{5} \text{ g/ml} = 0.2 \text{ g/ml} \]We have the concentrations:
Arranging them in ascending order:
\[ C_B < C_A < C_C \]So, \(0.1429 < 0.1667 < 0.2\).
This means vessel B has the least concentration of sugar, making it the least sweet. Vessel C has the highest concentration, making it the sweetest.
Correct Option: D) less sweet than that in C
From our comparison, \(C_B \approx 0.1429 \text{ g/ml}\) and \(C_C = 0.2 \text{ g/ml}\). Since \(C_B < C_C\), the solution in vessel B is less sweet than that in C.
Rakesh and Rajesh together bought 10 balla and 10 rackets. Rakesh spent ₹ 1300 and Rajesh spent ₹ 1500. If each racket costs three times a ball does, then what is the price of a racket ?
UPSC-2019
The problem involves setting up a system of equations based on the total number of items, their costs, and the relationship between the price of a ball and a racket. We will use variables to represent the prices and quantities, form equations, and solve them simultaneously.
Let \(b\) be the price of one ball and \(r\) be the price of one racket.
According to the problem, each racket costs three times a ball. So, we have the relationship:
\[ r = 3b \quad \text{(Equation 1)} \]Let \(x\) be the number of balls Rakesh bought and \(y\) be the number of rackets Rakesh bought.
Let \(x'\) be the number of balls Rajesh bought and \(y'\) be the number of rackets Rajesh bought.
Together, they bought 10 balls and 10 rackets. So,
\[ x + x' = 10 \quad \text{(Equation 2)} \]\[ y + y' = 10 \quad \text{(Equation 3)} \]Rakesh spent ₹ 1300. The total cost for Rakesh is the sum of the cost of balls and rackets he bought:
\[ x \cdot b + y \cdot r = 1300 \quad \text{(Equation 4)} \]Rajesh spent ₹ 1500. The total cost for Rajesh is the sum of the cost of balls and rackets he bought:
\[ x' \cdot b + y' \cdot r = 1500 \quad \text{(Equation 5)} \]Add Equation 4 and Equation 5:
\[ (x \cdot b + y \cdot r) + (x' \cdot b + y' \cdot r) = 1300 + 1500 \]\[ (x + x')b + (y + y')r = 2800 \]Substitute Equation 2 and Equation 3 into the combined equation:
\[ 10b + 10r = 2800 \]Divide the entire equation by 10:
\[ b + r = 280 \quad \text{(Equation 6)} \]Now we have a system of two equations with two variables \(b\) and \(r\):
Equation 1: \(r = 3b\)
Equation 6: \(b + r = 280\)
Substitute Equation 1 into Equation 6:
\[ b + (3b) = 280 \]\[ 4b = 280 \]\[ b = \frac{280}{4} \]\[ b = 70 \]So, the price of one ball is ₹ 70.
Now, find the price of a racket using Equation 1:
\[ r = 3b \]\[ r = 3 \times 70 \]\[ r = 210 \]The price of a racket is ₹ 210.
C) ₹ 210 is the correct price of a racket.
The ratio of a two-digit natural number to a number formed by reversing its digits is 4:7. The number of such pairs is
UPSC-2019
We need to find the number of two-digit natural numbers such that the ratio of the number to the number formed by reversing its digits is 4:7. We will represent the two-digit number algebraically and set up an equation based on the given ratio. Then we will solve for the possible digits.
Let the two-digit natural number be \(10a + b\), where \(a\) is the tens digit and \(b\) is the units digit.
The number formed by reversing its digits will be \(10b + a\).
According to the problem, the ratio of the original number to the reversed number is 4:7. So, we can write the equation:
\[ \frac{10a + b}{10b + a} = \frac{4}{7} \]Cross-multiply to solve for \(a\) and \(b\):
\[ 7(10a + b) = 4(10b + a) \] \[ 70a + 7b = 40b + 4a \]Rearrange the terms to group \(a\) and \(b\):
\[ 70a - 4a = 40b - 7b \] \[ 66a = 33b \]Simplify the equation by dividing both sides by 33:
\[ \frac{66a}{33} = \frac{33b}{33} \] \[ 2a = b \]Now we need to find pairs of digits \((a, b)\) that satisfy \(2a = b\), keeping in mind the constraints for \(a\) and \(b\):
Let's list the possible values for \(a\) and the corresponding \(b\):
Therefore, there are 4 such pairs of numbers: 12, 24, 36, and 48.
Correct Option: B) 4
In a group of persons, 70% of the persons are male and 30% of the persons are married. If two-sevenths of the males are married, what fraction of the female is single?
UPSC-2011
We need to find the fraction of females who are single. We are given percentages of males and married persons, and the fraction of married males. We can assume a total number of persons (e.g., 100) to simplify calculations and then determine the number of males, females, married persons, married males, and finally single females.
Assume Total Persons:
Let the total number of persons in the group be 100.
Calculate Number of Males and Females:
Given that 70% of persons are male:
Number of females:
\[ \text{Number of females} = 100 - 70 = 30 \]Calculate Total Number of Married Persons:
Given that 30% of persons are married:
Calculate Number of Married Males:
Given that two-sevenths of the males are married:
Calculate Number of Married Females:
The total number of married persons is 30. Since 20 of them are males, the remaining married persons must be females.
Calculate Number of Single Females:
The total number of females is 30. Out of these, 10 are married. So, the single females are:
Calculate the Fraction of Single Females:
The fraction of single females is the number of single females divided by the total number of females.
D) 2/3 — As calculated, the fraction of single females is 2/3.
The monthly incomes of Peter and Paul are in the ratio of 4 : 3. Their expenses are in the ratio of 3:2. If each saves ₹ 6,000 at the end of the month, their monthly incomes respectively are (in ₹)
UPSC-2015
24, 000 and 18, 000
28, 000 and 21, 000
32, 000 and 24, 000
34, 000 and 26, 000
This problem involves ratios and basic algebraic equations. We are given the ratio of incomes, the ratio of expenses, and the savings of both individuals. We can set up equations based on these relationships and solve for their actual incomes.
Define variables for incomes:
Let the monthly income of Peter be \(4x\) and the monthly income of Paul be \(3x\), since their incomes are in the ratio of 4:3.
Define variables for expenses:
Let the monthly expense of Peter be \(3y\) and the monthly expense of Paul be \(2y\), since their expenses are in the ratio of 3:2. Note that we use a different variable \(y\) because the ratio of expenses is independent of the ratio of incomes.
Formulate equations using savings:
We know that \(\text{Income} - \text{Expense} = \text{Savings}\). Both Peter and Paul save ₹6,000 each month.
Solve the system of linear equations:
We have two equations with two variables \(x\) and \(y\). We can solve this system using elimination or substitution.
Multiply Equation 1 by 2 and Equation 2 by 3 to eliminate \(y\):
\[ 2 \times (4x - 3y) = 2 \times 6000 \\ 8x - 6y = 12000 \quad (\text{Equation 3}) \] \[ 3 \times (3x - 2y) = 3 \times 6000 \\ 9x - 6y = 18000 \quad (\text{Equation 4}) \]Subtract Equation 3 from Equation 4:
\[ (9x - 6y) - (8x - 6y) = 18000 - 12000 \\ 9x - 6y - 8x + 6y = 6000 \\ x = 6000 \]Calculate their monthly incomes:
Now that we have the value of \(x\), we can find their incomes:
Correct Option: A) 24, 000 and 18, 000
Two equal glasses of same type are respectively 1/3 and 1/4 full of milk. They are then filled up with water and the contents are mixed in a pot. What is the ratio of milk and water in the pot?
UPSC-2015
To find the ratio of milk and water in the pot, we need to calculate the total amount of milk and the total amount of water from both glasses. We will assume the capacity of each glass to be 'V' units. Then, we calculate the milk and water content in each glass and sum them up.
Assume the capacity of each glass:
Let the capacity of each glass be \(V\) units.
Calculate milk and water in the first glass:
First glass is \(1/3\) full of milk.
Calculate milk and water in the second glass:
Second glass is \(1/4\) full of milk.
Calculate total milk in the pot:
Total milk \( = \) Milk from glass 1 \( + \) Milk from glass 2
Calculate total water in the pot:
Total water \( = \) Water from glass 1 \( + \) Water from glass 2
Find the ratio of milk to water:
\[ \text{Ratio (Milk : Water)} = \frac{\text{Total Milk}}{\text{Total Water}} = \frac{\frac{7}{12}V}{\frac{17}{12}V} = \frac{7}{17} \]So, the ratio of milk and water in the pot is \(7:17\).
Correct Option: A) 7:17
