# Exercise : 4

^{1}⁄

_{2}days more to complete the job than A and B worked together. What time would they take if both A and B worked together?

(b) 5 days

(c) 4 days

(d) 6 days

(e) None of these

**Ans.d**

Let if both A and B work together, they take x days.

∴ (A + B)’s 1 days’s work = ^{1}⁄_{x} th work

A’s 1 day’s work = ^{1}⁄_{(x + 8)} th work.

B’s 1 day’s work = ^{1}⁄_{(x + 9/2)} th work.

Now, \(\frac{1}{x + 8} + \frac{2}{2x + 9} = \frac{1}{x}\)

⇒ x(2x + 9 + 2x + 16) = (x + 8)(2x + 9)

⇒ 4x^{2} + 25x = 2x^{2} + 25x + 72

⇒ x^{2} = 36 ⇒ x = 6 days

(b) 125

(c) 145

(d) 225

(e) None of these

**Ans.d**

1 man’s 1 day’s work = ^{1}⁄_{100}

(10 men + 15 women)’s 1 day’s work = ^{1}⁄_{6}

15 women’s 1 day’s work

= (^{1}⁄_{6} - ^{10}⁄_{100}) = (^{1}⁄_{6} - ^{1}⁄_{10}) = ^{1}⁄_{15}

∴ 1 woman’s 1 day’s work = ^{1}⁄_{225}

∴ 1 woman alone can complete the work in 225 days.

(b) 6

(c) 13

(d) 9

(e) None of these

**Ans.a**

Let the number of men originally employed be x.

9x = 15(x – 6)

or x = 15

^{1}⁄

_{3}of the work has been done. He employs Rakesh who is 60 % efficient as Anil. How many more days will Anil take to complete the job?

(b) 12 days

(c) 10 days

(d) 8 days

(e) None of these

**Ans.c**

In 8 days, Anil does = ^{1}⁄_{3} rd work

∴ in 1 day, he does = ^{1}⁄_{24} th work.

∴ Rakesh’s one day’s work

= 60% of ^{1}⁄_{24} = ^{1}⁄_{40} th work.

Remaining work = 1 - ^{1}⁄_{3} = ^{2}⁄_{3}

(Anil and Rakesh)’s one day’s work

= ^{1}⁄_{24} + ^{1}⁄_{40} = ^{1}⁄_{15} th work.

Now, ^{1}⁄_{15} th work is done by them in one day.

∴ ^{2}⁄_{3} rd work is done by them in 15 × ^{2}⁄_{3} = 10 days

^{14}⁄

_{3}

(b)

^{16}⁄

_{3}

(c)

^{15}⁄

_{3}

(d)

^{17}⁄

_{3}

(e) None of these

**Ans.b**

A's one day's work = ^{1}⁄_{32}

B's one day's work = ^{1}⁄_{20}

(B + C)'s one day's work = ^{1}⁄_{12}

∴ C's one day's work = ^{1}⁄_{12} - ^{1}⁄_{20} = ^{1}⁄_{30}

D's one day's work = ^{1}⁄_{24}

∴ (A + B + C + D)'s one day's work

\(\frac{1}{32} + \frac{1}{20} + \frac{1}{30} + \frac{1}{24} = \frac{75 + 120 + 80 + 100}{2400} \)= ^{375}⁄_{2400} = ^{15}⁄_{96} = ^{5}⁄_{32}

∴ Out of ^{5}⁄_{32} of work done

^{1}⁄_{30} of the work is done by C.

⇒ Out of 25 paid for the work, C will receive

\(\frac{\frac{1}{30}}{\frac{5}{32}} \times 25\), i.e,^{1}⁄

_{30}×

^{32}⁄

_{5}× 25, i.e,

^{16}⁄

_{3}

(b) 3 days

(c) 1 day

(d) 4 days

(e) None of these

**Ans.b**

A’s one day’s work = ^{1}⁄_{15} th work.

B’s one day’s work = ^{1}⁄_{10} th work.

(A + B)’s one day’s work = ^{1}⁄_{15} + ^{1}⁄_{10} = ^{1}⁄_{6} th work.

Let A left after x days.

∴ (A + B)’s x days’ work = ^{x}⁄_{6} th work.

Remaining work = 1 - ^{x}⁄_{6} = ^{(6 - x)}⁄_{6} th work.

Now, in 5 days, work done by B = ^{(6 - x)}⁄_{6} th work.

∴ in 1 day work done by B = ^{(6 - x)}⁄_{30} th work.

and ^{(6 - x)}⁄_{30} = ^{1}⁄_{10}

∴ x = 3 days

(b) 24 days

(c) 40 days

(d) 42 days

(e) None of these

**Ans.a**

Let Suresh undertakes a tour of x days.

Then, expenses for each day = ^{360}⁄_{x}

now, \(\frac{360}{x + 4} = \frac{360}{x} - 3\)

or \(360\left(\frac{1}{x} - \frac{1}{x + 4} \right) = 3\)

or x^{2} + 4x - 480 = 0 or x = – 24 or x = 20

Since, x ≠ -24 we have x = 20

(b) 2 days

(c) 4

^{1}⁄

_{2}days

(d) 3 days

(e) None of these

**Ans.a**

A’s one day’s work = ^{1}⁄_{3} rd work.

B’s one day’s work = ^{1}⁄_{6} rd work.

(A + B)’s one day’s work = ^{1}⁄_{3} + ^{1}⁄_{6} = ^{1}⁄_{2} nd work.

∴ A and B together can complete the work (knit a pair of socks) in 2 days.

∴ They together knit two pair of socks in 4 days.

(b) 12 days, 15 days

(c) 13 days, 16 days

(d) 14 days, 11 days

(e) None of these

**Ans.b**

Let B can finish the work in x days.

Then A can finish the work in (x – 3) days.

B’s one day’s work = ^{1}⁄_{x}th work.

A’s one day’s work = ^{1}⁄_{(x - 3)}th work.

A’s 4 days’ work = ^{4}⁄_{(x - 3)}th work.

Remaining work = 1 - ^{4}⁄_{(x - 3)} = \(\frac{x - 7}{x - 3}\)

The remaining work done by B in 14 – 4 = 10 days.

Now, in 10 days, work done by B = \(\frac{x - 7}{x - 3}\) th work.

∴ in 1 day, work done by B = \(\frac{1}{10}\left(\frac{x - 7}{x - 3} \right)\) th work.

and \(\frac{1}{10}\left(\frac{x - 7}{x - 3} \right) = \frac{1}{x}\)

⇒ x = 15 days

∴ B → 15 days and A → 12 days

^{1}⁄

_{3}as efficiently as he actually did, the work would have completed in 3 days. Find the time for A to complete the job alone.

^{1}⁄

_{4}days

(b) 5

^{3}⁄

_{4}days

(c) 5 days

(d) 3 days

(e) None of these

**Ans.a**

(A + B)’s one day’s work ^{1}⁄_{5}th work

Let A can do job in x days.

Then,A’s one day’s work = ^{1}⁄_{x} th work.

and B’s one day’s work = ^{1}⁄_{5} - ^{1}⁄_{x} = \(\frac{x - 5}{5x}\)th work.

Now, (2A)'s work + (^{1}⁄_{3}) B's work = ^{1}⁄_{3} rd work.

⇒ \(\frac{2}{x} + \frac{1}{3}\left(\frac{x - 5}{5x} \right) = \frac{1}{3}\) ⇒ x = ^{25}⁄_{4} = 6^{1}⁄_{4} days