# Exercise : 1

^{3}⁄

_{4}of the same job. How many days will 10 men and 8 women together take to complete the same job?

(b) 13

^{1}⁄

_{2}

(c) 12

(d) Data inadequate

(e) None of these

**Ans.b**

12 M × 18 = 12 W × 18 × ^{4}⁄_{3}

\ w = 3/4 M

10M + 8W = 10M + 8 × ^{3}⁄_{4}M = 16 M

\16 men can complete the same work

in \(\frac{12 \times 18}{16} = \frac{27}{2} = 13\tfrac{1}{2} \; days\)

(b) 4

(c) 6

(d) Cannot be determined

(e) None of these

**Ans.e**

M = 2B

∴ 7M + 4B = 14B + 4B = 18B

5M + 4B = 10B + 4B = 14B

∴ 18 boys complete the work in 6 days.

∴ 14 boys complete the work in

\(\frac{6 \times 18}{14} = 7\tfrac{5}{7} \; days\)Note: 7 men and 4 boys complete the work in 6 days. We have to find out the no. of days in which 5 men and 4 boys complete the work. Here, we see that 4 boys are common in both the cases, therefore, 5 men will take more time to complete the work, i.e., more than 6 days, which is not given in any options. Therefore, without calculating we can say that our answer is(e).

^{1}⁄

_{3}days

(b) 3

^{2}⁄

_{3}days

(c) 3 days

(d) 1

^{1}⁄

_{2}days

(e) None of these

**Ans.d**

8W = 6M = 12B

12M + 12W + 12B Þ 12M + 9M + 6M = 27M

\9 men can complete the work by working 1 hour per day in 6 × 6 days

\27 men working 8 hours per day = \(\frac{6 \times 6 \times 9}{27 \times 8} = 1\tfrac{1}{2} \; days\)

^{2}⁄

_{13}

(b) 15

^{5}⁄

_{13}

(c) 8

^{2}⁄

_{13}

(d) 31

^{11}⁄

_{19}

(e) None of these

**Ans.d**

Tank filled in 1 minute = ^{1}⁄_{25} + ^{1}⁄_{40} - ^{1}⁄_{30} part

= \(\frac{24 + 15 - 20}{600} = \frac{19}{600}\) part

∴ tank will be filled complete in minutes

= ^{600}⁄_{19} = 31^{11}⁄_{19}

(b) 1.30 pm

(c) 1.00 pm

(d) 11.30 am

(e) None of these

**Ans.c**

Part of print done by A, B and C in 2 hours = 2(^{1}⁄_{8} + ^{1}⁄_{10} + ^{1}⁄_{12}) = ^{37}⁄_{60}

Remaining = 1 - ^{37}⁄_{60} = ^{23}⁄_{60}

If B and C print together, then they can print in \(\frac{10 \times 12}{10 + 12}\)

Therefore, remaining part can be printed by

B and C in \(\frac{10 \times 12}{22} \times \frac{23}{60}\) ≈ 2 hrs.

Hence, the job will be finished at

9 am + 2 + 2 = 1.00 p.m.

(b) 5

(c) 6

(d) 12

(e) None of these

**Ans.a**

The part of job that Suresh completes in 9 hours

= ^{9}⁄_{15} = ^{3}⁄_{5}

Remaining job = 1 - ^{3}⁄_{5} = ^{2}⁄_{5}

Remaining job can be done by Ashutosh in ^{2}⁄_{5} × 10 = 4 hours

(b) 150 days

(c) 90 days

(d) 225 days

(e) None of these

**Ans.d**

15 women's work of a day = ^{1}⁄_{6} - ^{1}⁄_{10} ⇒ ^{1}⁄_{15} part

∴ for 1 whole part a woman will take

= 15 × 15 = 225 days.

(b) 25 hours

(c) 20 hours

(d) Cannot be determined

(e) None of these

**Ans.a**

Here ratio of efficiencies of pipes A, B and C are as follows:

\(\frac{\begin{matrix} C \;\;\; & B \;\;\; & A\\ 2 \;\;\; & 1 \;\;\; & \\ & 2 \;\;\; & 1 \end{matrix}} {\begin{matrix} 4 & : & 2 & : & 1 \end{matrix}}\)Suppose the efficiencies of pipes C, B and A are 4K, 2K and K.

Since, the tank is filled in 5 hours by the three pipes having combined efficiency equal to 7K, the time required to fill the tank by A alone = \(\frac{7K \times 5}{K} = 35 \; hours\)

(b) 32

(c) 34

(d) 36

(e) None of these

**Ans.b**

m_{1} × d_{1} × t_{1} × w_{2} = m_{2} × d_{2} × t_{2} × w_{1}

24 × 10 × 8 × 1 = m_{2} × 6 × 10 × 1

⇒ m_{2} = \(\frac{24 \times 10 \times 8}{6 \times 10} = 32 \; men\)

(b) 6 minutes to fill

(c) 9 minutes to empty

(d) 9 minutes to fill

(e) None of these

**Ans.a**

∵ Pipe A in 1 minute fills ^{1}⁄_{10} part and Pipe Bin 1 min.empties ^{1}⁄_{6} part

\ Pipe A + B in 1 min = ^{1}⁄_{10} - ^{1}⁄_{6} = ^{-1}⁄_{15}

∵ ^{1}⁄_{15} part gets emptied in 1 min

\ ^{2}⁄_{5} part is emptied in 15 × ^{2}⁄_{5} min = 6 min.