# Crack IBPS PO : Time and Work : Day 49

D.1-5) Study the following passage and answer the questions accordingly.

Five members of a family live in Mumbai namely A, B, C, D and E. A and B together can do a piece of work in 80 days. B and C together can do a piece of work in 60 days. C and D together can do a piece of work in 40 days. D and E together can do a piece of work in 20 days. A alone can do a piece of work in 120 days.

(A+B)’s one day work=1/80;

(B+C)’s one day work=1/60;

(C+D)’s one day work=1/40;

(D+E)’s one day work=1/20

A’s one day work=1/120

Now, According to question,

B’s one day work=1/80-1/120=1/240

C’s one day work=1/60-1/240=1/80

D’s one day work=1/40-1/80=1/80

E’s one day s work=1/20-1/80=3/80

Q.1) If A, B and C together can do a piece of work in ‘x’ days then how much work could be done in the same days when E do the same work?

a) 2/5

b) 1/3

c) 2/3

d) 4/3

e) None of these

According to question,

(A+B+C)’s one day work=1/120+1/240+1/80=(2+1+3)/240=6/240=1/40

Required Answer, E alone works to finish,

E=(1/40)/(3/80)=1/40×(80/3)=2/3  of the work

Ans=c)2/3

Q.2) B, C and D can complete a piece of work in ‘x’ days. If all of them work together and after three days B left and the remaining work be completed by C and D with help of E. In how many days can C, D and E do the remaining work?

a) 11 3/5 days

b) 12 3/5 days

c) 13 3/5 days

d) 14 3/5 days

e) None of these

(B+C+D)’s one days work=1/240+1/80+1/80=7/240

According to quesiton,

(B+C+D)’s three days work=7/240×3=7/80

Then, remaining work, =1-7/80=73/80

(73/80)/(1/80+1/80+3/80)=73/80×80/5=73/5=14 3/5  days

Ans=d) 14 3/5  days

Q.3) A, C and D can do a piece of work in x, y and z days, respectively. They work alternately in a way that first day A , second day C and third day D, fourth day A and so on. How many days will be needed to complete the work in this way?

a) 90 days

b) 80 days

c) 70 days

d) 60 days

e) None of these

A’s one day work=1/120;

C’s one day work=1/80;

D’s one day work=1/80

According to question,

work done in first 3 days,

=1/120+1/80+1/80=(2+3+3)/240=8/240=1/30

Time taken to complete 1/30  part of work=30 days

Required Answer, (Time taken to complete the whole work)

=3×30=90 days

Ans=a) 90 days

Q.4) A, B and C can do a piece of work in ‘x’ days, ‘y’ days and ‘z’ days respectively. As they were ill, they could do 90% , 75% and 80% of their efficiency, respectively. How many days will they take to do the work together?

a) 43 16/33 days

b) 33 1/33 days

c) 48 16/33 days

d) 12 12/33 days

e) None of these

According to question,

A’s one day work=90% of 1/120=90/100×1/120=3/400

B’s one day work=75% of 1/240=75/100×1/240=1/320

C’s one day work=80% of 1/80=80/100×1/80=1/100

(A+B+C)’s one day’s work=3/400+1/320+1/100=(12+5+16)/1600=33/1600

Hence, time taken by them to complete the work=1600/33=48 16/33

Ans=c) 48 16/33  days.

Q.5) C can do 1/4  of a work in 80 days, D can do 40% of the same work in 80 days and E can do 1/3 of a work in 800/3 days. Who will complete the work first?

a) C

b) D

c) E

d) Both C and D

e) None of these

Time taken to complete the work by C

=80×4=320 days

Time taken to complete the work by

D=80×100/40=200 days

Time taken to complete the work by E

=800/3×3=800 days

D.6-10) Study the following information carefully and answer the question given below.

Given table shows percentage of number of students from 5 different schools attended 5 different tournaments.

Q.6) Number of students from C who attended chess, hockey and cricket is equal to the number of students who attended football, cricket and tennis from school E. Total number of students in school E is 160. Find the number of students who attended tennis from school C.

a) 50

b) 31

c) 48

d) 45

e) 43

Number of students attended chess, hockey and cricket from school C =14%+41%+12%=67%

Number of students attended football, cricket and tennis from school E= 20%+25%+20%=65%

Total students in school E= 160

67% of School C = 65% of school E

67% of school C=65/100×160=104

Total students in school C =104/67×100=155.22≈155

Students attended tennis from School C=20% of 155

=20/100 ×155=31

Q.7) Number of students who attended hockey tournament from school A is the sum of the number of students who attended football tournament from school D and the number of students who attended cricket tournament from the same school. Then find the number of students from school D is what per cent of number of students from school A?

a) 40 20/37%

b) 41%

c) 40 5/9%

d) 20 20/37%

e) 44%

Number of students attended hockey tournament from school A=15% of school A

Number of students attended football and

Cricket tournament from school D  = 37% of school D

15% of school A= 37% of school D

Required percentage =(school D total )/(school A total)×100

=(15/37)A /A×100=15/37×100=40 20/37%

Q.8) Number of students from school B who attended hockey tournament is approximately what percentage less than the number of students from school C who attended tennis tournament. If the number of students who attended football from school B is 52 and the number of students who attended chess from school C is 15 more than the number of students who attended chess from school B.

a) 24%

b) 23%

c) 19%

d) 15%

e) 17%

Number of students attended football from B =52

Number of students attended hockey from B=52/25×21=44

Number of students attended chess tournament from school C=15+52/25×11=38

Number of students attended tennis from school C=38/14×20=54

Required percentage =(54-44)/54×100=19% (approx.)

Q.9) If the difference between the number of students who attended cricket from school E and number of students who attended football from same school is 25, and the total number of students from school B is 40% more than the total number of students from school E. In school E percentage of number of students who attended tennis is equal to percentage of number of students who attended cricket. Find the number of students who attended football from school B

a) 150

b) 125

c) 180

d) 140

e) 175

Difference between the number of students who attended football and number of students who attended cricket from school E = (25%-20%) = 5% of school E

Total students from school E=25/5×100=500

Total students from school B = 500×1.4=700

Number of students attended football from school B=25/100×700=175

Q.10) Find the number of students who attended cricket from school C, if the number of students who attended tennis from school C is 80?

a) 46

b) 38

c) 15

d) 48

e) 20