A can do \(1 \over 3\) of a piece of work in 32 days, B can do \(37{1 \over 2}\)% of the same work in 24 days, while C can do 60% of the same work in 48 days. B and C together started and worked for x days. After x days, B left the work and A joined C and both completed the remaining work in (x + 8) days. If the ratio of the work done by (B + C) together to the work done by (A + C) together is 9 ∶ 11, then what fraction of the same work can be completed by C alone in 3.5x days?
Understanding the Work and Time Problem
This question involves calculating the work rates of three individuals, A, B, and C, and then determining the fraction of work C can complete alone based on conditions involving their combined work and a specific time duration.
First, let's find the time each person takes to complete the entire work alone and their respective daily work rates.
Calculating Individual Work Rates (Daily Work)
- For A:
A can do \( \frac{1}{3} \) of the work in 32 days.
Time taken by A to complete the whole work = \( 32 \text{ days} \div \frac{1}{3} = 32 \times 3 = 96 \text{ days} \).
A's daily work rate = \( \frac{1}{\text{Time taken by A}} = \frac{1}{96} \) of the work per day.
- For B:
B can do \( 37\frac{1}{2}\)% of the work in 24 days.
\( 37\frac{1}{2}\)% as a fraction is \( \frac{37.5}{100} = \frac{75/2}{100} = \frac{75}{200} = \frac{3}{8} \).
So, B can do \( \frac{3}{8} \) of the work in 24 days.
Time taken by B to complete the whole work = \( 24 \text{ days} \div \frac{3}{8} = 24 \times \frac{8}{3} = 8 \times 8 = 64 \text{ days} \).
B's daily work rate = \( \frac{1}{\text{Time taken by B}} = \frac{1}{64} \) of the work per day.
- For C:
C can do 60% of the work in 48 days.
60% as a fraction is \( \frac{60}{100} = \frac{3}{5} \).
So, C can do \( \frac{3}{5} \) of the work in 48 days.
Time taken by C to complete the whole work = \( 48 \text{ days} \div \frac{3}{5} = 48 \times \frac{5}{3} = 16 \times 5 = 80 \text{ days} \).
C's daily work rate = \( \frac{1}{\text{Time taken by C}} = \frac{1}{80} \) of the work per day.
| Person | Time to complete work alone | Daily Work Rate |
|---|---|---|
| A | 96 days | \( \frac{1}{96} \) |
| B | 64 days | \( \frac{1}{64} \) |
| C | 80 days | \( \frac{1}{80} \) |
Analyzing the Two Phases of Work Completion
The work is completed in two phases:
- Phase 1: B and C work together for x days.
- Phase 2: B leaves, and A joins C. A and C work together for (x + 8) days to complete the remaining work.
Combined Work Rates
- Combined daily work rate of B and C:
\( \text{Rate}_{B+C} = \text{Rate}_B + \text{Rate}_C = \frac{1}{64} + \frac{1}{80} \)
To add these fractions, find the Least Common Multiple (LCM) of 64 and 80. LCM(64, 80) = 320.
\( \text{Rate}_{B+C} = \frac{1 \times 5}{64 \times 5} + \frac{1 \times 4}{80 \times 4} = \frac{5}{320} + \frac{4}{320} = \frac{5+4}{320} = \frac{9}{320} \) of the work per day.
- Combined daily work rate of A and C:
\( \text{Rate}_{A+C} = \text{Rate}_A + \text{Rate}_C = \frac{1}{96} + \frac{1}{80} \)
To add these fractions, find the LCM of 96 and 80. LCM(96, 80) = 480.
\( \text{Rate}_{A+C} = \frac{1 \times 5}{96 \times 5} + \frac{1 \times 6}{80 \times 6} = \frac{5}{480} + \frac{6}{480} = \frac{5+6}{480} = \frac{11}{480} \) of the work per day.
Work Done in Each Phase and Finding x
- Work done by (B+C) in x days (Phase 1):
\( W_{B+C} = \text{Rate}_{B+C} \times x = \frac{9}{320} \times x = \frac{9x}{320} \)
- Work done by (A+C) in (x+8) days (Phase 2):
\( W_{A+C} = \text{Rate}_{A+C} \times (x+8) = \frac{11}{480} \times (x+8) \)
The question states that the ratio of the work done by (B + C) together to the work done by (A + C) together is 9 : 11. Based on the problem structure, this ratio likely refers to the work done during the specific periods mentioned for each group (Phase 1 vs Phase 2).
So, \( W_{B+C} : W_{A+C} = 9 : 11 \).
\( \frac{9x}{320} : \frac{11(x+8)}{480} = 9 : 11 \)
\( \frac{\frac{9x}{320}}{\frac{11(x+8)}{480}} = \frac{9}{11} \)
\( \frac{9x}{320} \times \frac{480}{11(x+8)} = \frac{9}{11} \)
Cancel out 9 from both numerators and 11 from both denominators:
\( \frac{x}{320} \times \frac{480}{x+8} = 1 \)
\( \frac{480x}{320(x+8)} = 1 \)
Simplify the fraction \( \frac{480}{320} = \frac{48}{32} = \frac{3}{2} \).
\( \frac{3x}{2(x+8)} = 1 \)
\( 3x = 2(x+8) \)
\( 3x = 2x + 16 \)
\( 3x - 2x = 16 \)
\( x = 16 \)
Verifying the value of x
If \( x = 16 \) days:
- Work done by (B+C) in Phase 1 (\( x \) days) = \( \frac{9}{320} \times 16 = \frac{9 \times 16}{320} = \frac{144}{320} \). Divide numerator and denominator by 16: \( \frac{9}{20} \).
- Work done by (A+C) in Phase 2 (\( x+8 \) days) = \( \frac{11}{480} \times (16+8) = \frac{11}{480} \times 24 = \frac{11 \times 24}{480} \). Divide numerator and denominator by 24: \( \frac{11}{20} \).
Total work done = Work in Phase 1 + Work in Phase 2 = \( \frac{9}{20} + \frac{11}{20} = \frac{20}{20} = 1 \). This confirms that \( x=16 \) is correct as the total work is completed.
The ratio of work done in Phase 1 to Phase 2 is \( \frac{9}{20} : \frac{11}{20} = 9 : 11 \), which matches the condition given in the question.
Calculating Work Done by C Alone
We need to find the fraction of work C can complete alone in \( 3.5x \) days.
- Value of \( x = 16 \) days.
- Time for C = \( 3.5x = 3.5 \times 16 \) days.
- \( 3.5 \times 16 = \frac{7}{2} \times 16 = 7 \times 8 = 56 \) days.
- C's daily work rate = \( \frac{1}{80} \).
- Work done by C alone in 56 days = C's daily work rate \( \times \) Time
- Work done by C = \( \frac{1}{80} \times 56 = \frac{56}{80} \)
Simplify the fraction \( \frac{56}{80} \). Both numerator and denominator are divisible by 8.
\( \frac{56 \div 8}{80 \div 8} = \frac{7}{10} \).
So, C alone can complete \( \frac{7}{10} \) of the work in \( 3.5x \) days.
Final Answer Fraction of Work
The fraction of the same work that can be completed by C alone in \( 3.5x \) days is \( \frac{7}{10} \).
Revision Table: Key Calculations
| Description | Calculation | Result |
|---|---|---|
| A's time to complete work | \( 32 / (1/3) \) | 96 days |
| B's time to complete work | \( 24 / (3/8) \) | 64 days |
| C's time to complete work | \( 48 / (3/5) \) | 80 days |
| B + C daily work rate | \( 1/64 + 1/80 \) | \( 9/320 \) |
| A + C daily work rate | \( 1/96 + 1/80 \) | \( 11/480 \) |
| Equation from ratio | \( \frac{9x/320}{11(x+8)/480} = \frac{9}{11} \) | \( x = 16 \) |
| Time for C alone | \( 3.5 \times x \) | 56 days |
| Work by C in 56 days | \( 1/80 \times 56 \) | \( 7/10 \) |
Additional Information: Time and Work Concepts
Time and work problems often involve calculating individual efficiency and then combining or comparing the work done by multiple individuals over specific periods.
- Daily Work Rate: If a person completes a work in 'N' days, their daily work rate is \( 1/N \) of the total work. This is the amount of work done per day.
- Total Work: The total work is typically considered as '1 unit' or '100%'.
- Combined Work Rate: If multiple people work together, their individual daily work rates are added to find the combined daily work rate. For example, if A's rate is \( R_A \) and B's rate is \( R_B \), their combined rate is \( R_A + R_B \).
- Work Done = Rate \( \times \) Time: If a person or group works at a rate 'R' for 'T' days, the total work done is \( R \times T \).
- Completing Work: When the work is completed, the total work done equals 1.
- Efficiency: Work rate is directly related to efficiency. A higher work rate means higher efficiency (completing more work in less time).
Understanding these basic principles is key to solving complex time and work problems like the one discussed here.
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