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**exampundit**. Here is a set of Data Interpretation based on Time & Work for upcoming IBPS PO Mains. This DI also features Missing Data.

**Direction :-**A, B, C,D and E are persons employed to complete a job X. line graph shows the data regarding the time taken by these persons to complete the job X. Table graph shows the actual time for which every one of him worked on the job X.

**Note 1 :**- All the person worked on the job X for whole no days.

**Note 2:-**Two jobs Y and Z are similar to job X and require same effort as required by job X.

(1) if A, C and E worked on job Z for 2 days each and the remaining job was done by B and D. if the ratio of no of days for which B and D worked is in ration 20:21. then find no of days for which B worked. ?

a) 50 days

b) 4.5 days

c) 5.5 days

d) 4 days

e) None of these

(2) E worked on job Z for 5 days and the remaining job was completed by A,B and D who worked on alternate days starting with A and followed by B and D in that order. find the no. of days B worked for ?

a) 2 days

b) 4 days

c) 9 days

d) 3 days

e) None of these

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(3) A and C worked on job Y working alternatively for 10 days, then B and D worked together for X days. if 1/36 th of the job was still remained, then find the value of X ?

a) 2 days

b) 1.25 days

c) 1.33 days

d) 1 day

e) 1.28 days

(4) If C worked on job Y with 5/4 times his given efficiency and was assisted by B every 3rd day, then find the time taken by C to complete the job Y ?

a) 13 days

b) 12 + 1/6 days

c) 13 + 1/2 days

d) 12 days

e) None of these

(5) if the ratio of the no of days for which B and D worked on job X is 4:3, then find the difference between no of days for which B and D worked ?

a) 2 days

b) 3 days

c) 1 days

d) 4 days

e) None of these

Solutions:

**( Before solving these types of questions consider the Data carefully. It will help you a lot and save your time as well as boost up your confidence level.)**

Take LCM of no of days which takes A,B,C,D and E individually to complete the job. Which is 180 units(total work)

A = 10 days = 18unit@day

B = 12 days = 15unit@day

C = 15 days = 12unit@day

D = 18 days = 10unit@day

E = 20 days = 9 unit@day

**1). D.**

2A +2C + 2E = 2×39 = 78/180

Remaining work = 17/30

B + D = 20x/12 + 21x/18 = 17/30

x = 1/5

B = 20x = 20/5 = 4 days

**2). D.**

5E = 9×5 = 45

Remaining = 135

A +B+ D = 18+15+10 = 43 unit in 3 days

So, 9 days = 129 units

Remaining 6 unit will being done by A itself so, no need to find it..

No of days for which B worked = 9/3 = 3 days(3 participants work alternatively)

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**3). D.**

(In this types of questions one should remember note 2 as mentioned above.)

5A + 5C = 90+60 = 150units completed.

And at last remaining work is 1/36 of 180 units

= 5 units

So, work done by B+D = 25 units

B+D can do it in only 1 day. (See their efficiency which 15 and 10 respectively)

**4). E**. 9 days

C = 12×5/4 = 15

3days = 3C + B = 15×3 + 15 = 60

9days = 180

So, c takes 9 days to complete the job when he is assisted by B on each 3rd day.

**5).C.**

2A + 3C + 2E = 36+36+18 = 90 units

Remaining 1/2

B+D = 4x/12 + 3x/18 = 1/2

x = 1 day

4x -3x = x = 1 day

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**Team ExamPundit**

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