# Time & Work - Swetabh Kumar

• Q1) 5 men or 8 women can reap a field in 12 days. Find the number of days (approximately) taken by 3 men and 4 women to reap the same field.

OA: 120/11
5 men in 12 days, so 1 man in 5 * 12=60 days
similarly one woman in 8 * 12 = 96 days
so in 1 day, work = (3/60+4/96) = 1/20+1/24 = 11/120
so number of days = 1/(11/120) = 120/11 = approx 11 days

Q2) Some students can complete an assignment in 12 days. How many days will be taken by two times the number of such students for 1/3rd of this assignment ?

Fundamentally, Let each student take P days. so 1 student in 1 day does 1/P work.
Say there are N such students.
so 12(N/P) = 1 so N/P = 1/12
twice students => 2N say X days needed.
X(2N/P) = 1/3
so x * 2 * 1/12=1/3 so X=2 days.

Q3) X is thrice as good a workman as Y and therefore is able to finish a job in 60 days less than Y. What is the time taken to do twice the work when they are working together ?

OA: 45 days.
Days reqd by X,Y respectively : X, (X+60)
and X+60 = 3X so X=30, so Y=X+60=90.
so n (1/30+1/90) = 2
n ( 4/90) = 2 so n= 45.

Q4) 15 workers working 4 hours a day for 25 days can build a platform of width 120 meters, length 10 meter and height 14 meters. How many days will 12 workers working 5 hours a day will take if they have to build a platform of width 600 meters, length 14 metres and height 12 metres?

OA: 150
Work-Time Equivalence.
M1 * D1 * H1/W1 = M2 * D2 * H2/W2
M,D, H, W are Men, Days, Hours and Work reqd respectively in the two scenarios.
so MDH/W remains constant for a FIXED task even if one or more variables of these get changed.
Memory trick: MDH Masala.
Here W=Work will refer to the Volume that has to be built.

Q5) (x – 2) men can do a job in x days and (x+ 7) men can do a job in 3/4 of the same work in (x – 10) days. Calculate the number of days in which (x + 4) men can finish the job.

OA: 15 days
Step 1: from 1st part, 1 man does it in x(x-2) = x^2-2x days
Step 2: from 2nd part, (x+7) can do whole work in 4/3 *(x-10) days
so 1 man does it in 4/3 (x-10) (x+7) days
Step 3: so x^2-2x = 4/3 (x-10)(x+7)
3x^2-6x = 4x^2-12x -280
x^2-6x-280 = 0
x=20
Step 4: so ques. asks for 20+4=24 men. when 1 man does in x^2-2x= 400-40=360.
so 360/24 = 15 days.

Q6) 4 men and 2 boys can do a job in 6 2/3 days. 3 women and 4 boys can finish the same job in 5 days. Also 2 men and 3 women can finish the same job in 4 days. In how many days can 1 man, 1 woman and 1 boy finish the work at their double efficiency?

OA: 5 days.
Step 1: 4/M + 2/B = 3/20
and 3/M + 4/B = 1/5
so 4a+2b = 3/20 and 3a+4b=1/5
solving: a=1/200, M=200
so B=200/7
Step 2 : 2/M + 3/W=1/4
7/100 +3/W= 1/4
W=300/18 = 50/3
Step 3: On doubling efficiency, M= 100, B=100/7, W=25/3
so in 1 day: (1/100+7/100+3/25) = 1/5 so 5 days.

Q7) 20 men can complete a work in 20 days. After how many days should the number of men be increased by 50% so that the work is completed in 75% of the actual Time?

OA: 5 days
one man in 20 * 20=400 days so 1/400 in 1 day
so X * 20 * 1/400 + 30 * (15-x)/400 = 1
20X+450-30x=400
so X=5

Q8) One Indian, one Chinese and one Japanese worked for a company for the same period. The Indian is twice as efficient as Chinese and the Chinese is thrice as efficient as a Japanese. If Rs. 10,00,000 were given to all the three together, then calculate the amount received by the Chinese and the Japanese together.

OA: 4 lakhs.
Concept: Amount is divided in ratio of total work done by person / total work done in aggregate
so Japanese: Chinese: Indian = 1:3:6 so reqd: (1+3)/10 * total

Q9) Amit takes 20 days to complete a certain work. Amit started the work and Suraj joined him 4 days before the work was completed. Find out the number of days for which Amit worked alone if Suraj’s efficiency is 25% more than that of Amit’s.

OA: 11 days
Amit : 20 days
Suraj: 4/5 * 20 = 16 days
so N/20 + 4 * (1/20+1/16) = 1
N/20 + 4 * 9/80 = 1
4N+36=80 so N=11 days

Looks like your connection to MBAtious was lost, please wait while we try to reconnect.