Direct and Inverse Variations

Q1: Multiple Choice Type

i. A can do a piece of work in 2 days and B can do the same work in 3 days. If A and B work together, the amount of work done by them in 1 day is:

Step 1: A’s 1 day work = \( \frac{1}{2} \)
Step 2: B’s 1 day work = \( \frac{1}{3} \)
Step 3: Combined work in 1 day = \( \frac{1}{2} + \frac{1}{3} = \frac{3 + 2}{6} = \frac{5}{6} \)
Answer: d. \( \frac{5}{6} \)

ii. If \(1/20\) of a work can be done in 5 days, the amount of work done in one day will be:

Step 1: \( \frac{1}{20} \) work in 5 days ⇒ work in 1 day = \( \frac{1}{20} \div 5 = \frac{1}{100} \)
Answer: a. \( \frac{1}{100} \)

iii. Ritu can knit a sweater in 4 days and Manish can knit the same sweater in 6 days. If they together knit the same sweater, number of days taken by them will be:

Step 1: Ritu’s 1 day work = \( \frac{1}{4} \), Manish’s = \( \frac{1}{6} \)
Step 2: Combined work = \( \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12} \)
Step 3: Time = \( \frac{1}{\frac{5}{12}} = \frac{12}{5} \) days
Answer: c. \( \frac{12}{5} \) days

iv. A and B working together can complete a work in 4 days. If A alone can do the same work in 6 days, then B alone can do the same work in:

Step 1: A + B’s 1 day work = \( \frac{1}{4} \)
Step 2: A’s 1 day work = \( \frac{1}{6} \)
Step 3: B’s 1 day work = \( \frac{1}{4} – \frac{1}{6} = \frac{3 – 2}{12} = \frac{1}{12} \)
Step 4: So, B alone can do the work in 12 days
Answer: a. 12 days

Q2: A can do a piece of work in 10 days and B in 15 days. How long will they take to finish it working together?

Step 1: A’s 1 day work = \( \frac{1}{10} \)
Step 2: B’s 1 day work = \( \frac{1}{15} \)
Step 3: Combined 1 day work = \( \frac{1}{10} + \frac{1}{15} \)
Take LCM of 10 and 15 = 30 \[ \frac{1}{10} = \frac{3}{30},\quad \frac{1}{15} = \frac{2}{30} \\ \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \]Step 4: So, A and B together can finish the work in \( \frac{1}{\frac{1}{6}} = 6 \) days
Answer: A and B will take 6 days working together.

Q3: A and B together can do a piece of work in \(6\frac{2}{3}\) days, but B alone can do it in 10 days. How long will A take to do it alone?

Step 1: Convert \(6\frac{2}{3}\) to improper fraction: \[ 6\frac{2}{3} = \frac{20}{3} \]Step 2: A + B’s 1 day work = \( \frac{1}{\frac{20}{3}} = \frac{3}{20} \)
B’s 1 day work = \( \frac{1}{10} \)
Step 3: A’s 1 day work = One day work of (A + B) – One day work of B \[ = \frac{3}{20} – \frac{1}{10} \] Take LCM = 20 \[ = \frac{3}{20} – \frac{2}{20} = \frac{1}{20} \]Step 4: So, A alone will take \( \frac{1}{\frac{1}{20}} = 20 \) days
Answer: A will take 20 days to do the work alone.

Q4: A can do a work in 15 days and B in 20 days. If they work together on it for 4 days, what fraction of the work will be left?

Step 1: A’s 1 day work = \( \frac{1}{15} \)
B’s 1 day work = \( \frac{1}{20} \)
Step 2: Combined 1 day work = \( \frac{1}{15} + \frac{1}{20} \)
LCM of 15 and 20 = 60 \[ \frac{1}{15} = \frac{4}{60},\quad \frac{1}{20} = \frac{3}{60} \Rightarrow \frac{4 + 3}{60} = \frac{7}{60} \]Step 3: Work done in 4 days = \( 4 \times \frac{7}{60} = \frac{28}{60} = \frac{7}{15} \)
Step 4: Fraction of work left = \( 1 – \frac{7}{15} = \frac{15 – 7}{15} = \frac{8}{15} \)
Answer: Fraction of work left = \( \frac{8}{15} \)

Q5: A, B and C can do a piece of work in 6 days, 12 days and 24 days respectively. In what time will they altogether do it?

Step 1: A’s 1 day work = \( \frac{1}{6} \)
B’s 1 day work = \( \frac{1}{12} \)
C’s 1 day work = \( \frac{1}{24} \)
Step 2: Total 1 day work = \( \frac{1}{6} + \frac{1}{12} + \frac{1}{24} \)
LCM of 6, 12, 24 = 24 \[ \frac{1}{6} = \frac{4}{24},\quad \frac{1}{12} = \frac{2}{24},\quad \frac{1}{24} = \frac{1}{24} \\ \Rightarrow \frac{4 + 2 + 1}{24} = \frac{7}{24} \]Step 3: Time to finish work = \( \frac{1}{\frac{7}{24}} = \frac{24}{7} \) days \[ \frac{24}{7} = 3\ \text{days and} \ \frac{3}{7}\ \text{day} \]Answer: They can complete the work together in \( \frac{24}{7} \) days or 3 days \( \frac{3}{7} \) day.

Q6: A and B working together can mow a field in 56 days and with the help of C, they could have mowed it in 42 days. How long would C take to mow the field by himself?

Step 1: A and B together take 56 days ⇒ their 1 day work = \( \frac{1}{56} \)
A, B, and C together take 42 days ⇒ their 1 day work = \( \frac{1}{42} \)
Step 2: C’s 1 day work = One day work of (A + B + C) – One day work of (A + B) \[ = \frac{1}{42} – \frac{1}{56} \]Step 3: Find LCM of 42 and 56 = 168 \[ \frac{1}{42} = \frac{4}{168},\quad \frac{1}{56} = \frac{3}{168} \\ \Rightarrow \frac{4 – 3}{168} = \frac{1}{168} \]Step 4: So, C alone can mow the field in \( \frac{1}{\frac{1}{168}} = 168 \) days
Answer: C will take 168 days to mow the field alone.

Q7: A can do a piece of work in 24 days, A and B can do it in 16 days, and A, B and C in \(10\frac{2}{3}\) days. In how many days can A and C do it working together?

Step 1: Convert \(10\frac{2}{3}\) to improper fraction: \[ 10\frac{2}{3} = \frac{32}{3} \]Step 2: A’s 1 day work = \( \frac{1}{24} \)
(A + B)’s 1 day work = \( \frac{1}{16} \)
(A + B + C)’s 1 day work = \( \frac{1}{\frac{32}{3}} = \frac{3}{32} \)
Step 3: Find C’s 1 day work: One day work of (A + B + C) – One day work of (A + B) \[ \frac{3}{32} – \frac{1}{16} = \frac{3}{32} – \frac{2}{32} = \frac{1}{32} \] So, C’s 1 day work = \( \frac{1}{32} \)
Step 4: (A + C)’s 1 day work = \( \frac{1}{24} + \frac{1}{32} \)
LCM of 24 and 32 = 96 \[ \frac{1}{24} = \frac{4}{96},\quad \frac{1}{32} = \frac{3}{96} \\ \Rightarrow \frac{4 + 3}{96} = \frac{7}{96} \]Step 5: So, A and C can complete the work in \( \frac{1}{\frac{7}{96}} = \frac{96}{7} \) days \[ \frac{96}{7} = 13\ \text{days} \ \frac{5}{7} \]Answer: A and C can do the work together in \( \frac{96}{7} \) days or 13 days \( \frac{5}{7} \).

Q8: A can do a piece of work in 20 days and B in 15 days. They worked together on it for 6 days and then A left. How long will B take to finish the remaining work?

Step 1: A’s 1 day work = \( \frac{1}{20} \)
B’s 1 day work = \( \frac{1}{15} \)
Step 2: Work done by A and B together in 1 day = \[ \frac{1}{20} + \frac{1}{15} = \frac{3 + 4}{60} = \frac{7}{60} \]Step 3: Work done in 6 days = \[ 6 \times \frac{7}{60} = \frac{42}{60} = \frac{7}{10} \]Step 4: Work left = \( 1 – \frac{7}{10} = \frac{3}{10} \)
Step 5: B’s 1 day work = \( \frac{1}{15} \), so time to finish \( \frac{3}{10} \) work: \[ \text{Time} = \frac{\frac{3}{10}}{\frac{1}{15}} = \frac{3}{10} \times \frac{15}{1} = \frac{45}{10} = 4.5 \ \text{days} \]Answer: B will take 4.5 days (or 4 days and 12 hours) to finish the remaining work.

Q9: A can finish a piece of work in 15 days and B can do it in 10 days. They worked together for 2 days and then B goes away. In how many days will A finish the remaining work?

Step 1: A’s 1 day work = \( \frac{1}{15} \)
B’s 1 day work = \( \frac{1}{10} \)
Step 2: Combined 1 day work = \( \frac{1}{15} + \frac{1}{10} \)
LCM of 15 and 10 = 30 \[ \frac{1}{15} = \frac{2}{30},\quad \frac{1}{10} = \frac{3}{30},\quad \text{Total} = \frac{5}{30} = \frac{1}{6} \]Step 3: Work done in 2 days together = \( 2 \times \frac{1}{6} = \frac{2}{6} = \frac{1}{3} \)
Step 4: Remaining work = \( 1 – \frac{1}{3} = \frac{2}{3} \)
Step 5: A’s 1 day work = \( \frac{1}{15} \), so time to finish \( \frac{2}{3} \) work: \[ \text{Time} = \frac{\frac{2}{3}}{\frac{1}{15}} = \frac{2}{3} \times 15 = 10\ \text{days} \]Answer: A will finish the remaining work in 10 days.

Q10: A can do a piece of work in 10 days, B in 18 days, and A, B and C together in 4 days. In what time would C do it alone?

Step 1: A’s 1 day work = \( \frac{1}{10} \)
B’s 1 day work = \( \frac{1}{18} \)
(A + B + C)’s 1 day work = \( \frac{1}{4} \)
Step 2: Add A’s and B’s 1 day work: \[ \frac{1}{10} + \frac{1}{18} = \frac{9 + 5}{90} = \frac{14}{90} = \frac{7}{45} \]Step 3: C’s 1 day work = One day work of (A + B + C) − One day work of (A + B) \[ = \frac{1}{4} – \frac{7}{45} \] LCM of 4 and 45 = 180 \[ \frac{1}{4} = \frac{45}{180},\quad \frac{7}{45} = \frac{28}{180} \\ \Rightarrow \frac{45 – 28}{180} = \frac{17}{180} \]Step 4: So, time taken by C alone = \( \frac{1}{\frac{17}{180}} = \frac{180}{17} \) days \[ \frac{180}{17} = 10\ \text{days and} \ \frac{10}{17}\ \text{day} \]Answer: C will complete the work alone in \( \frac{180}{17} \) days or 10 days \( \frac{10}{17} \) day.

Q11: A can do \(\frac{1}{4}\) of a work in 5 days and B can do \(\frac{1}{3}\) of the same work in 10 days. Find the number of days in which both working together will complete the work.

Step 1: A does \( \frac{1}{4} \) work in 5 days.
A can complete work in \(4 \times 5 = 20\) days.
So, A’s 1 day work = \( \frac{1}{4 \times 5} = \frac{1}{20} \)
Step 2: B does \( \frac{1}{3} \) work in 10 days.
B can complete work in \(10 \times 3 = 30\) days.
So, B’s 1 day work = \( \frac{1}{3 \times 10} = \frac{1}{30} \)
Step 3: Combined 1 day work = \( \frac{1}{20} + \frac{1}{30} \)
LCM of 20 and 30 = 60 \[ \frac{1}{20} = \frac{3}{60},\quad \frac{1}{30} = \frac{2}{60} \Rightarrow \frac{3 + 2}{60} = \frac{5}{60} = \frac{1}{12} \]Step 4: Time taken to complete the whole work = \( \frac{1}{\frac{1}{12}} = 12 \) days
Answer: A and B together will complete the work in 12 days.

Q12: One tap can fill a cistern in 3 hours and the waste pipe can empty the full cistern in 5 hours. In what time will the empty cistern be full, if the tap and the waste pipe are kept open together?

Step 1: Tap’s 1 hour work (filling) = \( \frac{1}{3} \)
Waste pipe’s 1 hour work (emptying) = \( \frac{1}{5} \)
Step 2: Net 1 hour work = \( \frac{1}{3} – \frac{1}{5} \)
LCM of 3 and 5 = 15 \[ \frac{1}{3} = \frac{5}{15},\quad \frac{1}{5} = \frac{3}{15} \\ \Rightarrow \frac{5 – 3}{15} = \frac{2}{15} \]Step 3: So, the cistern will be filled in = \( \frac{1}{\frac{2}{15}} = \frac{15}{2} = 7.5 \) hours
Answer: The cistern will be full in 7.5 hours (or 7 hours 30 minutes).