Exercise: 6-C
Multiple Choice Questions
Q1: What per cent of 7.2 kg is 18 g?
Step 1: Convert 18 g into kg:
\[
18 \text{ g} = \frac{18}{1000} = 0.018 \text{ kg}
\]Step 2: Use the percentage formula:
\[
\text{Percentage} = \left( \frac{\text{part}}{\text{whole}} \right) \times 100 = \left( \frac{0.018}{7.2} \right) \times 100
\]Step 3: Divide:
\[
\frac{0.018}{7.2} = 0.0025 \\
0.0025 \times 100 = 0.25\%
\]Answer: c. 0.25%
Q2: A number decreased by \(27\frac{1}{2}\)% gives 87. The number is
Step 1: Let the original number be \(x\)
It is decreased by \(27\frac{1}{2}\% = \frac{55}{2}\%\)
Step 2: Remaining value after decrease:
\[
\text{Remaining} = \left(100\% – \frac{55}{2}\%\right) = \frac{145}{2}\%
\]
So, \( \frac{145}{200} \times x = 87 \)
Step 3: Solve for \(x\):
\[
\frac{145x}{200} = 87 \\
\Rightarrow x = \frac{87 \times 200}{145} = \frac{17400}{145} = 120
\]Answer: c. 120
Q3: A number increased by \(37\frac{1}{2}\)% gives 33. The number is
Step 1: Let the original number be \(x\)
Increase = \(37\frac{1}{2}\% = \frac{75}{2}\%\)
Step 2: After increase, number becomes:
\[
x + \frac{75}{200}x = \left(1 + \frac{75}{200}\right)x = \frac{275}{200}x \\
\frac{275x}{200} = 33
\]Step 3: Solve for \(x\):
\[
x = \frac{33 \times 200}{275} = \frac{6600}{275} = 24
\]Answer: b. 24
Q4: If \(37\frac{1}{2}\)% of a number is 900, then \(62\frac{1}{2}\)% of the number will be
Step 1: Let the number be \(x\)
\[
37\frac{1}{2}\% = \frac{75}{2}\% \\
\Rightarrow \frac{75}{100} \times \frac{1}{2} = \frac{75}{200} \\
\frac{75x}{200} = 900 \\
\Rightarrow x = \frac{900 \times 200}{75} = \frac{180000}{75} = 2400
\]Step 2: Now find \(62\frac{1}{2}\)% of 2400
\[
62\frac{1}{2}\% = \frac{125}{2}\%
= \frac{125}{100} \times \frac{1}{2} = \frac{125}{200} \\
\frac{125}{200} \times 2400 = \frac{300000}{200} = 1500
\]Answer: d. 1500
Q5: By how much per cent is four-fifths of 70 less than five-sevenths of 112?
Step 1: Find four-fifths of 70:
\[
\frac{4}{5} \times 70 = 56
\]Step 2: Find five-sevenths of 112:
\[
\frac{5}{7} \times 112 = 80
\]Step 3: Find the difference:
\[
80 – 56 = 24
\]Step 4: Find the percentage difference (how much 56 is less than 80):
\[
\frac{24}{80} \times 100 = 30\%
\]Answer: b. 30%
Q6: If 11% of a number exceeds 7% of the same by 18, the number is
Step 1: Let the number be \(x\)
11% of \(x\) = \(\frac{11}{100}x\)
7% of \(x\) = \(\frac{7}{100}x\)
Step 2: Given:
\[
\frac{11x}{100} – \frac{7x}{100} = 18 \\
\Rightarrow \frac{4x}{100} = 18
\]Step 3: Solve for \(x\):
\[
x = \frac{18 \times 100}{4} = \frac{1800}{4} = 450
\]Answer: c. 450
Q7: 96% of the population of a town is 23040. The total population of the town is
Step 1: Let the total population be \(x\)
Given that 96% of the population is 23040:
\[
\frac{96}{100} \times x = 23040 \\
\Rightarrow \frac{24x}{25} = 23040
\]Step 2: Solve for \(x\):
\[
x = \frac{23040 \times 25}{24} = \frac{576000}{24} = 24000
\]Answer: a. 24000
Q8: In an examination, 65% of the total examinees passed. If the number of failures is 168, the number of those who passed is
Step 1: Let the total number of examinees be \(x\)
If 65% passed, then 35% failed.
Step 2: Given 35% of total = 168:
\[
\frac{35}{100} \times x = 168 \\
\Rightarrow \frac{7x}{20} = 168
\]Step 3: Solve for \(x\):
\[
x = \frac{168 \times 20}{7} = \frac{3360}{7} = 480
\]Step 4: Find number of students who passed:
\[
65\% \text{ of } 480 = \frac{65}{100} \times 480 = 312
\]Answer: b. 312
Q9: The price of an item is increased by 20% and then decreased by 20%. The final price as compared to original price is
Step 1: Use successive percentage change formula:
\[
\text{Net change} = a + b + \frac{ab}{100}
\]
where \(a = +20\%\), \(b = -20\%\)
Step 2: Substitute values:
\[
\text{Net change} = 20 – 20 + \frac{(20 \times -20)}{100} = 0 – \frac{400}{100} = -4\%
\]Step 3: Final price is 4% less than original price
Answer: c. 4% less
Q10: A customer asks for the production of x number of goods. The company produces y number of goods daily out of which z% are unfit for sale. The order will be completed in
Step 1: Goods produced per day = \(y\)
Unfit goods = \(z\%\) of \(y = \frac{z}{100} \cdot y\)
So, usable goods per day = \(y – \frac{z}{100} \cdot y = y\left(1 – \frac{z}{100}\right)\)
Step 2: Days required = Total required goods / usable goods per day:
\[
\text{Days} = \frac{x}{y\left(1 – \frac{z}{100}\right)} = \frac{x}{\frac{y(100 – z)}{100}} = \frac{100x}{y(100 – z)}
\]Answer: d. \(\frac{100x}{y(100 – z)}\)
Q11: A man’s wages were decreased by 50%. Again the reduced wages were increased by 50%. He has a loss of
Step 1: Let original wages = ₹100
Step 2: After 50% decrease:
\[
100 – 50\% \text{ of } 100 = 100 – 50 = ₹50
\]Step 3: Then 50% increase on ₹50:
\[
50 + 50\% \text{ of } 50 = 50 + 25 = ₹75
\]Step 4: Compare with original wages:
\[
\text{Loss} = 100 – 75 = ₹25 \\
\Rightarrow \text{Loss \%} = \frac{25}{100} \times 100 = 25\%
\]Answer: d. 25%
Q12: If A’s income is 30% more than B’s, then how much percent is B’s income less than A’s?
Step 1: Let B’s income = ₹100
Then A’s income = ₹100 + 30% of 100 = ₹130
Step 2: Now calculate how much less B’s income is compared to A’s:
\[
\text{Difference} = 130 – 100 = 30 \\
\text{Percentage less} = \frac{30}{130} \times 100 = \frac{3000}{130} = 23.08\%
\]
Convert to mixed fraction:
\[
\frac{3000}{130} = \frac{3000 \div 13}{10} = 230.77 \div 10 = 23\frac{1}{13}\%
\]Answer: a. \(23\frac{1}{13}\)%
Q13: If A’s income is 30% less than B’s, then how much per cent is income more than A’s?
Step 1: Let B’s income = ₹100
Then A’s income = ₹100 − 30% of 100 = ₹70
Step 2: How much more is B’s income than A’s?
\[
\text{Difference} = 100 – 70 = ₹30 \\
\text{Percentage more} = \frac{30}{70} \times 100 = \frac{3000}{70} = 42.8571\%
\]Step 3: Convert to mixed fraction:
\[
42.8571\% = 42\frac{6}{7}\%
\]Answer: c. \(42\frac{6}{7}\)%
Q14: If the numerator of a fraction is increased by 20% and its denominator be diminished by 10%, the value of the fraction is \(\frac{16}{21}\). The original fraction is
Step 1: Let the original fraction be \(\frac{x}{y}\)
Step 2: According to the question:
Numerator increases by 20% → new numerator = \(x + 20\%\text{ of }x = 1.2x\)
Denominator decreases by 10% → new denominator = \(y – 10\%\text{ of }y = 0.9y\)
So,
\[
\frac{1.2x}{0.9y} = \frac{16}{21}
\]Step 3: Cross-multiply and simplify:
\[
\frac{1.2x}{0.9y} = \frac{16}{21} \\
\Rightarrow \frac{4x}{3y} = \frac{16}{21} \\
\Rightarrow \frac{x}{y} = \frac{16 \times 3}{21 \times 4} = \frac{48}{84} = \frac{4}{7}
\]Answer: c. \(\frac{4}{7}\)
Q15: In an examination, it is required to get 36% of maximum marks to pass. A student got 113 marks and was declared failed by 85 marks. The maximum marks are
Step 1: Let the maximum marks be \(x\)
To pass, the student must score 36% of \(x\):
\[
\text{Passing marks} = \frac{36}{100} \times x = \frac{9x}{25}
\]Step 2: Student scored 113 marks and failed by 85 marks:
\[
\text{Passing marks} = 113 + 85 = 198
\]Step 3: Equate and solve:
\[
\frac{9x}{25} = 198 \\
\Rightarrow x = \frac{198 \times 25}{9} = \frac{4950}{9} = 550
\]Answer: b. 550
Q16: In a class, the number of boys is more than the number of girls by 12% of the total strength. The ratio of boys to girls is
Step 1: Let total number of students = 100
Then 12% of total strength = 12
So, number of boys = number of girls + 12
Step 2: Let number of girls = \(x\)
Then boys = \(x + 12\)
But total = 100, so:
\[
x + (x + 12) = 100 \\
\Rightarrow 2x + 12 = 100 \\
\Rightarrow 2x = 88 \Rightarrow x = 44
\]
Girls = 44, Boys = 56
Step 3: Required ratio:
\[
\text{Boys : Girls} = 56 : 44 = \frac{56}{4} : \frac{44}{4} = 14 : 11
\]Answer: b. 14 : 11
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