Ex. No. 1: Ratio, Proportion & Percentage Problems
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1. The ratio of number of boys and girls in a school is $3:2$. If 20% of the boys and 30% of the girls are scholarship holders, find the percentage of students who are not scholarship holders.
Solution:
Let the number of boys be $3x$ and the number of girls be $2x$.
Total number of students = $3x + 2x = 5x$.Number of boys who are scholarship holders = $20\%$ of $3x = \frac{20}{100} \times 3x = 0.2 \times 3x = 0.6x$.
Number of girls who are scholarship holders = $30\%$ of $2x = \frac{30}{100} \times 2x = 0.3 \times 2x = 0.6x$.Total number of scholarship holders = $0.6x + 0.6x = 1.2x$.
Number of students who are not scholarship holders = Total students - Total scholarship holders
$= 5x - 1.2x = 3.8x$.Percentage of students who are not scholarship holders = $\left(\frac{\text{Number of non-scholarship holders}}{\text{Total students}}\right) \times 100$
$= \left(\frac{3.8x}{5x}\right) \times 100 = \frac{3.8}{5} \times 100 = 0.76 \times 100 = 76\%$.Therefore, 76% of the students are not scholarship holders.
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2. If the numerator of a fraction is increased by 20% and its denominator be diminished by 10%, the value of the fraction is $\frac{16}{21}$. Find the original fraction.
Solution:
Let the original fraction be $\frac{N}{D}$.
The numerator is increased by 20%: New numerator = $N + 0.20N = 1.2N$.
The denominator is diminished by 10%: New denominator = $D - 0.10D = 0.9D$.The new fraction is $\frac{1.2N}{0.9D}$, and its value is $\frac{16}{21}$.
So, $\frac{1.2N}{0.9D} = \frac{16}{21}$.$\frac{12N}{9D} = \frac{16}{21}$ (multiplying numerator and denominator by 10)
$\frac{4N}{3D} = \frac{16}{21}$ (simplifying $\frac{12}{9}$)To find $\frac{N}{D}$, we can write:
$\frac{N}{D} = \frac{16}{21} \times \frac{3}{4}$
$\frac{N}{D} = \frac{16 \times 3}{21 \times 4} = \frac{4 \times 1}{7 \times 1} = \frac{4}{7}$.The original fraction is $\frac{4}{7}$.
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3. The ratio of incomes of Salim and Jawed was $20:11$. Three years later income of Salim has increased by 20% and income of Jawed was increased by Rs. 500. Now ratio of their incomes becomes $3:2$. Find original incomes of Salim and Jawed.
Solution:
Let the original income of Salim be $20x$ and Jawed be $11x$.
Salim's new income after 20% increase: $20x + 0.20(20x) = 20x + 4x = 24x$.
Jawed's new income after Rs. 500 increase: $11x + 500$.The new ratio of their incomes is $3:2$.
So, $\frac{24x}{11x + 500} = \frac{3}{2}$.Cross-multiplying:
$2 \times 24x = 3 \times (11x + 500)$
$48x = 33x + 1500$
$48x - 33x = 1500$
$15x = 1500$
$x = \frac{1500}{15} = 100$.Original income of Salim = $20x = 20 \times 100 = \text{Rs. } 2000$.
Original income of Jawed = $11x = 11 \times 100 = \text{Rs. } 1100$.Salim's original income was Rs. 2000 and Jawed's original income was Rs. 1100.
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4. In a class, 60% students are boys and 40% are girls. By admitting 16 boys and 8 girls the ratio of boys to girls becomes $8:5$. What must be the number of boys and number of girls originally in the class?
Solution:
Let the total number of students originally be $T$.
Original number of boys = $60\%$ of $T = 0.6T$.
Original number of girls = $40\%$ of $T = 0.4T$.Alternatively, the ratio of boys to girls is $60:40$, which simplifies to $3:2$. Let the original number of boys be $3k$ and girls be $2k$.
After admitting 16 boys and 8 girls:
New number of boys = $3k + 16$.
New number of girls = $2k + 8$.The new ratio of boys to girls is $8:5$.
So, $\frac{3k + 16}{2k + 8} = \frac{8}{5}$.Cross-multiplying:
$5(3k + 16) = 8(2k + 8)$
$15k + 80 = 16k + 64$
$80 - 64 = 16k - 15k$
$16 = k$.Original number of boys = $3k = 3 \times 16 = 48$.
Original number of girls = $2k = 2 \times 16 = 32$.Originally, there were 48 boys and 32 girls in the class.
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5. Incomes of Mr. Shah, Mr. Patel and Mr. Mehta are in the ratio $1:2:3$; while their expenditure are in the ratio $2:3:4$. If Mr. Shah saves 20% of his income, find the ratio of their savings.
Solution:
Let the incomes of Mr. Shah, Mr. Patel, and Mr. Mehta be $I_S, I_P, I_M$.
$I_S : I_P : I_M = 1:2:3$. So, let $I_S = k, I_P = 2k, I_M = 3k$ for some constant $k$.Let their expenditures be $E_S, E_P, E_M$.
$E_S : E_P : E_M = 2:3:4$. So, let $E_S = 2m, E_P = 3m, E_M = 4m$ for some constant $m$.Mr. Shah saves 20% of his income. This means his expenditure is 80% of his income.
$E_S = 80\%$ of $I_S = 0.8 \times I_S$.
So, $2m = 0.8 \times k \implies 2m = \frac{4}{5}k \implies m = \frac{4}{10}k = \frac{2}{5}k$.Now we can express expenditures in terms of $k$:
$E_S = 2m = 2\left(\frac{2}{5}k\right) = \frac{4}{5}k = 0.8k$.
$E_P = 3m = 3\left(\frac{2}{5}k\right) = \frac{6}{5}k = 1.2k$.
$E_M = 4m = 4\left(\frac{2}{5}k\right) = \frac{8}{5}k = 1.6k$.Savings = Income - Expenditure.
Saving of Mr. Shah ($S_S$) = $I_S - E_S = k - 0.8k = 0.2k$.
Saving of Mr. Patel ($S_P$) = $I_P - E_P = 2k - 1.2k = 0.8k$.
Saving of Mr. Mehta ($S_M$) = $I_M - E_M = 3k - 1.6k = 1.4k$.The ratio of their savings $S_S : S_P : S_M = 0.2k : 0.8k : 1.4k$.
Dividing by $k$: $0.2 : 0.8 : 1.4$.
Multiplying by 10 to remove decimals: $2 : 8 : 14$.
Simplifying by dividing by 2: $1 : 4 : 7$.The ratio of their savings is $1:4:7$.
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6. What must be subtracted from each of the numbers 5, 7 and 10, so that the resulting numbers are in the continued proportion?
Solution:
Let $x$ be the number subtracted from each of the numbers 5, 7, and 10.
The resulting numbers are $(5-x)$, $(7-x)$, and $(10-x)$.For these numbers to be in continued proportion, the ratio of the first to the second must be equal to the ratio of the second to the third.
So, $\frac{5-x}{7-x} = \frac{7-x}{10-x}$.Cross-multiplying:
$(5-x)(10-x) = (7-x)^2$
$5(10-x) - x(10-x) = 7^2 - 2(7)(x) + x^2$
$50 - 5x - 10x + x^2 = 49 - 14x + x^2$
$50 - 15x + x^2 = 49 - 14x + x^2$.Subtract $x^2$ from both sides:
$50 - 15x = 49 - 14x$.Rearrange the terms to solve for $x$:
$50 - 49 = -14x + 15x$
$1 = x$.Let's check: If $x=1$, the numbers are $5-1=4$, $7-1=6$, $10-1=9$.
Is $\frac{4}{6} = \frac{6}{9}$?
$\frac{4}{6} = \frac{2}{3}$ and $\frac{6}{9} = \frac{2}{3}$. Yes, they are equal.The number that must be subtracted is 1.
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7. The employees of a firm have maintained their standard of living in such a manner, that they all have identical percentage of saving from their salaries. Amina and Sabina are two employees of the firm. Amina spends Rs. 12,800 per month from her salary of Rs. 35,000 per month. What would be Sabina's savings per month from her salary of Rs. 48,000 per month?
Solution:
Amina's salary = Rs. 35,000.
Amina's expenditure = Rs. 12,800.Amina's savings = Amina's salary - Amina's expenditure
Amina's savings = Rs. 35,000 - Rs. 12,800 = Rs. 22,200.Amina's percentage of saving = $\left(\frac{\text{Amina's savings}}{\text{Amina's salary}}\right) \times 100$
$= \left(\frac{22200}{35000}\right) \times 100 = \frac{222}{350} \times 100 = \frac{111}{175} \times 100 = \frac{11100}{175}\%$.
$\frac{11100}{175} = \frac{444 \times 25}{7 \times 25} = \frac{444}{7}\%$. So, Amina's percentage saving is $\frac{444}{7}\%$.All employees have an identical percentage of saving.
Sabina's salary = Rs. 48,000.
Sabina's savings will be the same percentage of her salary.Sabina's savings = $\frac{444}{7}\%$ of Rs. 48,000
$= \frac{444/7}{100} \times 48000 = \frac{444}{700} \times 48000$
$= \frac{444 \times 480}{7} = \frac{213120}{7}$.Calculating the value: $213120 \div 7 \approx 30445.714$.
Sabina's savings per month would be Rs. $\frac{213120}{7}$ (or approximately Rs. 30,445.71).
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8. A certain job can be performed by 10 men in 24 days working 8 hours a day. How many days would be needed to perform the same job by 8 men working 12 hours a day?
Solution:
This is a problem of work and time, often solved using the concept of "man-hours" or "man-days-hours".
Let $M_1$ be the number of men in the first case, $D_1$ be the days, and $H_1$ be the hours per day.
Let $M_2, D_2, H_2$ be the corresponding values for the second case.
The total work done is proportional to $M \times D \times H$. Since the job is the same, the total work is constant.$W = M_1 \times D_1 \times H_1 = M_2 \times D_2 \times H_2$.
Given: $M_1 = 10$ men
$D_1 = 24$ days
$H_1 = 8$ hours/day$M_2 = 8$ men
$H_2 = 12$ hours/day
$D_2 = ?$ (what we need to find)So, $10 \times 24 \times 8 = 8 \times D_2 \times 12$.
Calculate total work units for the first case:
Work = $10 \times 24 \times 8 = 1920$ man-hours.Now, set this equal to the work units for the second case:
$1920 = 8 \times D_2 \times 12$
$1920 = 96 \times D_2$.$D_2 = \frac{1920}{96}$.
$D_2 = \frac{1920 \div 96}{96 \div 96} = \frac{20}{1} = 20$.It would take 20 days for 8 men working 12 hours a day to perform the same job.
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9. Two metals X and Y are to be used for making two different alloys. If the ratio by weight of X:Y in the first alloy is $6:5$ and that in the second is $7:13$. How many kg of X metal must be melted along with 11 kg of the first alloy and 20 kg of second alloy so as to produce a new alloy containing 40% of metal Y?
Solution:
First Alloy (11 kg):
Ratio X:Y = $6:5$. Total parts = $6+5=11$.
Amount of X in first alloy ($X_1$) = $\frac{6}{11} \times 11 \text{ kg} = 6 \text{ kg}$.
Amount of Y in first alloy ($Y_1$) = $\frac{5}{11} \times 11 \text{ kg} = 5 \text{ kg}$.Second Alloy (20 kg):
Ratio X:Y = $7:13$. Total parts = $7+13=20$.
Amount of X in second alloy ($X_2$) = $\frac{7}{20} \times 20 \text{ kg} = 7 \text{ kg}$.
Amount of Y in second alloy ($Y_2$) = $\frac{13}{20} \times 20 \text{ kg} = 13 \text{ kg}$.Let $m$ kg be the amount of pure metal X added.
New Alloy Composition:
Total amount of metal X = $X_1 + X_2 + m = 6 \text{ kg} + 7 \text{ kg} + m \text{ kg} = (13+m) \text{ kg}$.
Total amount of metal Y = $Y_1 + Y_2 = 5 \text{ kg} + 13 \text{ kg} = 18 \text{ kg}$.Total weight of the new alloy = (Total X) + (Total Y) = $(13+m) + 18 = (31+m) \text{ kg}$.
The new alloy contains 40% of metal Y. This means:
$\frac{\text{Total amount of Y}}{\text{Total weight of new alloy}} = 40\% = \frac{40}{100} = 0.4$.So, $\frac{18}{31+m} = 0.4$.
$18 = 0.4 \times (31+m)$
$18 = 0.4 \times 31 + 0.4m$
$18 = 12.4 + 0.4m$
$18 - 12.4 = 0.4m$
$5.6 = 0.4m$.$m = \frac{5.6}{0.4} = \frac{56}{4} = 14$.
14 kg of metal X must be melted.
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10. Three persons A, B and C whose salaries together amount to Rs. 21,000. Their savings are 20%, 30% and 40% of their salaries respectively. If their expenditures are in the ratio $8:14:3$, find their respective salaries.
Solution:
Let the salaries of A, B, and C be $S_A, S_B, S_C$ respectively.
Given: $S_A + S_B + S_C = 21000$.Savings:
Saving of A = $20\%$ of $S_A = 0.2 S_A$.
Saving of B = $30\%$ of $S_B = 0.3 S_B$.
Saving of C = $40\%$ of $S_C = 0.4 S_C$.Expenditures:
Expenditure = Salary - Saving.
Expenditure of A ($E_A$) = $S_A - 0.2 S_A = 0.8 S_A$.
Expenditure of B ($E_B$) = $S_B - 0.3 S_B = 0.7 S_B$.
Expenditure of C ($E_C$) = $S_C - 0.4 S_C = 0.6 S_C$.The ratio of their expenditures is $E_A : E_B : E_C = 8:14:3$.
Let the common ratio be $k$. So, $E_A = 8k, E_B = 14k, E_C = 3k$.Now we can relate salaries to $k$:
$0.8 S_A = 8k \implies S_A = \frac{8k}{0.8} = 10k$.
$0.7 S_B = 14k \implies S_B = \frac{14k}{0.7} = \frac{140k}{7} = 20k$.
$0.6 S_C = 3k \implies S_C = \frac{3k}{0.6} = \frac{30k}{6} = 5k$.Substitute these into the total salary equation:
$S_A + S_B + S_C = 21000$
$10k + 20k + 5k = 21000$
$35k = 21000$.$k = \frac{21000}{35} = \frac{21000 \div 7}{35 \div 7} = \frac{3000}{5} = 600$.
Now find the respective salaries:
Salary of A ($S_A$) = $10k = 10 \times 600 = \text{Rs. } 6000$.
Salary of B ($S_B$) = $20k = 20 \times 600 = \text{Rs. } 12000$.
Salary of C ($S_C$) = $5k = 5 \times 600 = \text{Rs. } 3000$.The salaries are: A = Rs. 6000, B = Rs. 12000, C = Rs. 3000.