14
Q:

There are three boats B1, B2 and B3 working together they carry 60 people in each trip. One day an early morning B1 carried 50 people in few trips alone. When it stopped carrying the passengers B2 and B3 started carrying the people together. It took a total of 10 trips to carry 300 people by B1, B2 and B3. It is known that each day on an average 300 people cross the river using only one of the 3 boats B1, B2 and B3. How many trips it would take to B1, to carry 150 passengers alone?

A) 15 B) 30
C) 25 D) 10

Answer:   A) 15



Explanation:

Combined efficiency of all the three boats = 60 passenger/trip 

 

Now, consider option(a)

 

15 trips and 150 passengers means efficiency  of B1 =  10 passenger/trip 

 

which means in carrying 50 passengers B1 must has taken 5 trips. So the rest trips equal to 5 (10-5 = 5) in which B2 and B3 together carried remaining 250 (300 - 50 = 250) Passengers.

 

Therefore the efficiency of B2 and B3 = 250/5 = 50 passenger/trip  

 

Since, the combined efficiency of B1, B2 and B3 is 60. Which is same as given in the first statement hence option(a) is correct.

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
 
Answer & Explanation Answer: A) 10

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. 

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3 353
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
 
Answer & Explanation Answer: C) 3

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.

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0 528
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
 
Answer & Explanation Answer: C) 8 : 15

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.

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5 716
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
 
Answer & Explanation Answer: C) 2

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

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5 619
Q:

70000 a year is how much an hour?

A) 80 B) 8
C) 0.8 D) 0.08
 
Answer & Explanation Answer: B) 8

Explanation:

Given for year = 70000

=> 365 days = 70000

=> 365 x 24 hours = 70000

=>   1 hour = ?

70000/365x24 = 7.990 = 8

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2 681
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
 
Answer & Explanation Answer: B) 1/6

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.

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6 1842
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
 
Answer & Explanation Answer: B) 225 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.

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9 2352
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?

Answer

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.

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9 1045