Exercise: 6-C
Competency Focused Questions
Q1: If x% of y is 1000 and y% of z is 2000, then:
Given:
x% of y = 1000
y% of z = 2000
We are to find the value of x in terms of z.
Step 1: Convert both statements into equations:
\[
\frac{x}{100} \cdot y = 1000 \quad \text{(i)} \\
\frac{y}{100} \cdot z = 2000 \quad \text{(ii)}
\]Step 2: Solve equation (ii) to get value of y:
From (ii):
\[
\frac{y}{100} \cdot z = 2000 \Rightarrow y = \frac{2000 \times 100}{z} = \frac{200000}{z}
\]Step 3: Put value of y into equation (i):
\[
\frac{x}{100} \cdot \frac{200000}{z} = 1000 \\
\Rightarrow \frac{200000x}{100z} = 1000 \\
\Rightarrow \frac{2000x}{z} = 1000
\]Step 4: Solve for x:
\[
2000x = 1000z \Rightarrow x = \frac{1000z}{2000} = \frac{z}{2}
\]Answer: c. \(x = \frac{z}{2}\)
Q2: In an examination, 27% students failed in Science, and 24% students failed in Mathematics. If 20% students failed in both the subjects, then the percentage of students who failed in one or both the subjects is:
Let:
S = Students who failed in Science = 27%
M = Students who failed in Mathematics = 24%
S ∩ M = Students who failed in both = 20%
We need to find:
S ∪ M = Students who failed in either Science or Mathematics or both
Step 1: Use the union formula:
\[
\text{S ∪ M} = S + M – S ∩ M \\
= 27\% + 24\% – 20\% = 31\%
\]Answer: b. 31%
Q3: What is 25% of 25% equal to?
We are asked to calculate:
\[
25\% \text{ of } 25\%
\]Step 1: Convert percentages to decimals:
\[
25\% = \frac{25}{100} = 0.25
\]
So,
\[
25\% \text{ of } 25\% = 0.25 \times 0.25
\]Step 2: Multiply the decimals:
\[
0.25 \times 0.25 = 0.0625
\]Answer: c. 0.0625
Q4: \((x\%\text{ of } y + y\%\text{ of } x)\) is equal to:
We are asked to simplify:
\[
x\% \text{ of } y + y\% \text{ of } x
\]Step 1: Convert percentage terms to fractions:
\[
x\% \text{ of } y = \frac{x}{100} \cdot y,\quad y\% \text{ of } x = \frac{y}{100} \cdot x
\]Step 2: Add both expressions:
\[
\frac{x}{100} \cdot y + \frac{y}{100} \cdot x = \frac{xy}{100} + \frac{xy}{100} = \frac{2xy}{100}
\]Step 3: Final simplified expression:
\[
\frac{2xy}{100} = 2\% \text{ of } (xy)
\]Answer: d. 2% of (xy)
Q5: Subtracting 5% of x from x is equivalent to multiplying x by:
Let’s simplify the expression:
Step 1: Express 5% of x in fractional form:
\[
5\% \text{ of } x = \frac{5}{100} \cdot x = 0.05x
\]Step 2: Subtract 0.05x from x:
\[
x – 0.05x = (1 – 0.05)x = 0.95x
\]Step 3: Therefore, it is equivalent to multiplying x by: 0.95
Answer: b. 0.95
Q6: 75% of a number when added to 75 is equal to the number. The number is:
Let the number be \(x\).
Step 1: Frame the equation based on the given condition:
\[
75\% \text{ of } x + 75 = x \\
\Rightarrow \frac{75}{100}x + 75 = x \\
\Rightarrow 0.75x + 75 = x
\]Step 2: Transpose \(0.75x\) to the right-hand side:
\[
75 = x – 0.75x = 0.25x
\]Step 3: Solve for \(x\):
\[
x = \frac{75}{0.25} = 300
\]Answer: c. 300
Q7: 3 litres of water is added to 15 litres of a mixture of a 20% solution of milk and water. The strength of milk now is:
Given:
Original solution = 15 litres
Milk = 20% of 15 litres = ?
Water added = 3 litresStep 1: Calculate quantity of milk in original mixture:
\[
\text{Milk} = 20\% \text{ of } 15 = \frac{20}{100} \cdot 15 = 3 \text{ litres}
\]Step 2: Total quantity after adding water:
\[
\text{New total} = 15 + 3 = 18 \text{ litres}
\]Step 3: Strength of milk in new mixture:
\[
\text{Milk strength} = \left( \frac{3}{18} \right) \times 100 = \frac{100}{6} = \frac{50}{3}\%
\]Answer: b. \(\frac{50}{3}\)%
Q8: If \(x = \frac{y^2}{z}\) and y and z both are increased in value by 10%, then the value of x is:
Original expression:
\[
x = \frac{y^2}{z}
\]Step 1: Let the original values be:
\(y = y\), \(z = z\)
New values:
\[
y_{\text{new}} = y + 10\% \text{ of } y = 1.1y,\quad z_{\text{new}} = z + 10\% \text{ of } z = 1.1z
\]Step 2: Substitute new values into expression:
\[
x_{\text{new}} = \frac{(1.1y)^2}{1.1z} = \frac{1.21y^2}{1.1z}
\]Step 3: Compare with original \(x = \frac{y^2}{z}\):
\[
x_{\text{new}} = \frac{1.21}{1.1} \cdot \frac{y^2}{z} = \frac{1.21}{1.1}x
\]Step 4: Simplify the fraction:
\[
\frac{1.21}{1.1} = \frac{121}{110} = 1.1
\]So:
\[
x_{\text{new}} = 1.1x
\]Answer: a. increased by 10%
Q9: If 35% of A’s income is equal to 25% of B’s income, then the ratio of their incomes is:
Let A’s income be \(A\) and B’s income be \(B\).
Step 1: Frame the given condition:
\[
35\% \text{ of } A = 25\% \text{ of } B \\
\Rightarrow \frac{35}{100} \cdot A = \frac{25}{100} \cdot B
\]Step 2: Cancel the common factor of 100:
\[
35A = 25B
\]Step 3: Rearranging to find the ratio:
\[
\frac{A}{B} = \frac{25}{35} = \frac{5}{7}
\]Step 4: Therefore, the ratio of A’s income to B’s income is:
\[
A : B = 5 : 7
\]Answer: c. 5 : 7
Q10: Ratio of number of boys to the number of girls in a class is 2 : x. If percentage of girls is 30%, then the value of x is:
Let the number of boys be 2 units and number of girls be x units.
Step 1: Total students = boys + girls = \(2 + x\)
Step 2: Given that girls make up 30% of total students:
\[
\frac{x}{2 + x} \times 100 = 30
\]Step 3: Solve the equation:
\[
\frac{x}{2 + x} = \frac{30}{100} = \frac{3}{10}
\]Step 4: Cross-multiply:
\[
10x = 3(2 + x) = 6 + 3x
\]Step 5: Rearranging terms:
\[
10x – 3x = 6 \Rightarrow 7x = 6 \Rightarrow x = \frac{6}{7}
\]Answer: c. \(\frac{6}{7}\)
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