9
Q:

# 10 men and 15 women together can complete a work in 6 days. It takes 100 days for one man alone to complete the same work. How many days will be required for one woman alone to complete the same work?

 A) 215 days B) 225 days C) 235 days D) 240 days

Explanation:

Given that

(10M + 15W) x 6 days = 1M x 100 days

=> 60M + 90W = 100M

=> 40M = 90W

=> 4M = 9W.

From the given data,

1M can do the work in 100 days

=> 4M can do the same work in 100/4= 25 days.

=> 9W can do the same work in 25 days.

=> 1W can do the same work in 25 x 9 = 225 days.

Hence, 1 woman can do the same work in 225 days.

Q:

Raghu can do a job in 12 days alone and Sam can do the same job in 15 days alone. A third person Aru whose efficiency is two-third of efficiency of both Ram and Shyam together, can do the same job in how many days alone?

 A) 10 B) 12 C) 13 D) 15

Explanation:

One day work of Raghu and Sam together = 1/12 + 1/15 = 9/60 = 3/20

Aru efficiency = 2/3 of (Raghu + Sam)

Number of days required for Aru to do thw work alone = 3/2 x 20/3 = 10 days.

3 351
Q:

12 men complete a work in 14 days. 5 days after they had started working, 3 men join them. How many more days will all of them take to complete the remaining work?

 A) 5 B) 4 C) 3 D) 2

Explanation:

Let p be the required number of days.

From the given data,

12 x 14 = 12 x 5 + (12+3) x p

12 x 14 = 12[5 + 3p]

14 = 5 + 3p

3p = 9

p = 3 days.

Hence, more 3 days all of them take to complete the remaining work.

0 527
Q:

Lasya alone can do a work in 16 days. Srimukhi’s efficiency is 20 % lesser than that of Laya. If Rashmi and Srimukhi together can do the same work in 12 days, then find the efficiency ratio of Rashmi to that of Lasya?

 A) 19 : 7 B) 30 : 19 C) 8 : 15 D) 31 : 17

Explanation:

Given Lasya can do a work in 16 days.

Now, time taken by Srimukhi alone to complete the work = 16 x 100/80

Time taken by Rashmi = n days

=> (12 x 20)/(20 - 12) = (12 x 20)/n

=> n= 30 days.

Required ratio of efficiencies of Rashmi and Lasya = 1/30  ::  1/16  = 8 : 15.

5 715
Q:

5 boys and 3 girls can together cultivate a 23 acre field in 4 days and 3 boys and 2 girls together can cultivate a 7 acre field in 2 days. How many girls will be needed together with 7 boys, if they cultivate 45 acres of field in 6 days?

 A) 4 B) 3 C) 2 D) 1

Explanation:

Let workdone 1 boy in 1 day be b

and that of 1 girl be g

From the given data,

4(5b + 3g) = 23

20b + 12g = 23 .......(a)

2(3b + 2g) = 7

6b + 4g = 7 ........(b)

Solving (a) & (b), we get

b = 1, g = 1/4

Let number og girls required be 'p'

6(7 x 1 + p x 1/4) = 45

=> p = 2.

Hence, number of girls required = 2

5 618
Q:

70000 a year is how much an hour?

 A) 80 B) 8 C) 0.8 D) 0.08

Explanation:

Given for year = 70000

=> 365 days = 70000

=> 365 x 24 hours = 70000

=>   1 hour = ?

70000/365x24 = 7.990 = 8

2 680
Q:

A, B and C can do a piece of work in 72, 48 and 36 days respectively. For first p/2 days, A & B work together and for next ((p+6))/3days all three worked together. Remaining 125/3% of work is completed by D in 10 days. If C & D worked together for p day then, what portion of work will be remained?

 A) 1/5 B) 1/6 C) 1/7 D) 1/8

Explanation:

Total work is given by L.C.M of 72, 48, 36

Total work = 144 units

Efficieny of A = 144/72 = 2 units/day

Efficieny of B = 144/48 = 3 units/day

Efficieny of C = 144/36 = 4 units/day

According to the given data,

2 x p/2 + 3 x p/2 + 2 x (p+6)/3 + 3 x (p+6)/3 + 4 x (p+6)/3 = 144 x (100 - 125/3) x 1/100

3p + 4.5p + 2p + 3p + 4p = 84 x 3 - 54

p = 198/16.5

p = 12 days.

Now, efficency of D = (144 x 125/3 x 1/100)/10 = 6 unit/day

(C+D) in p days = (4 + 6) x 12 = 120 unit

Remained part of work = (144-120)/144 = 1/6.

6 1840
Q:

A,B,C can complete a work in 15,20 and 30 respectively.They all work together for two days then A leave the work,B and C work some days and B leaves 2 days before completion of that work.how many days required to complete the whole work?

Given A,B,C can complete a work in 15,20 and 30 respectively.

The total work is given by the LCM of 15, 20, 30 i.e, 60.

A's 1 day work = 60/15 = 4 units

B's 1 day work = 60/20 = 3 units

C's 1 day work = 60/30 = 2 units

(A + B + C) worked for 2 days = (4 + 3 + 2) 2 = 18 units

Let B + C worked for x days = (3 + 2) x = 5x units

C worked for 2 days = 2 x 2 = 4 units

Then, 18 + 5x + 4 = 60

22 + 5x = 60

5x = 38

x = 7.6

Therefore, total number of days taken to complete the work = 2 + 7.6 + 2 = 11.6 = 11 3/5 days.

1041
Q:

M, N and O can complete the work in 18, 36 and 54 days respectively. M started the work and worked for 8 days, then N and O joined him and they all worked together for some days. M left the job one day before completion of work. For how many days they all worked together?

 A) 4 B) 5 C) 3 D) 6

Explanation:

Let M, N and O worked together for x days.

From the given data,
M alone worked for 8 days
M,N,O worked for x days
N, O worked for 1 day

But given that
M alone can complete the work in 18 days
N alone can complete the work in 36 days
O alone can complete the work in 54 days

The total work can be the LCM of 18, 6, 54 = 108 units

M's 1 day work = 108/18 = 6 units
N's 1 day work = 108/36 = 3 units
O's 1 day work = 108/54 = 2 units

Now, the equation is
8 x 6 + 11x + 5 x 1 = 108
48 + 11x + 5 = 108
11x = 103 - 48
11x = 55
x = 5 days.

Hence, all M,N and O together worked for 5 days.