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Explanation:

1. Let number of SE’s (special skilled employees) in T5 in the first month be ‘x’. Then number of SE’s in T4, T3, T2 and T1 will be x + 1, x + 2, x + 3 and x + 4 respectively. As per given data, number of SE’s is 10.

∴ x + x + 1 + x + 2 + x + 3 + x + 4 = 10

∴ x = 0

So number of SE’s in T1, T2, T3, T4 and T5 will be 4, 3, 2, 1 and 0 respectively. Now T1, T2, T3, T4 and T5 get the challenging projects in months.

1, 2, 3, 4 and 5 respectively.

Also as total number of employees in the team having the challenging projects in a particular month is 1 more than the other teams the standard projects in the first month, T1 will have a total of 5 employees in the 1st month and hence 1 RE (Regular Skilled Employee). Since the other teams have a total of 4 employees each in the 1st month, the number of RE’s in T2, T3, T4 and T5 will be 1, 2, 3 and 4 respectively in the first month. So in the 1st month break up of employees in each of the teams will be as follows

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In the second month, the number of team members following condition (a) will be as follows

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Following condition (b) number of employees in month 2 will be

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For month 3 following condition (a)

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Following condition (b)

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For month 4, following condition (a)

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his is the final number of SE’s and RE’s in month4 as after following condition (a) as T1 has no SE and T4 has no RE, no further interchange can happen
For month 5, following condition (a)

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Following condition (b)

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Using this information let us answer the questions.

As 200 points are divided equally between 5 team members working on a challenging project and 100 points are divided equally between 4 team members working on a standard Project, each team 

member working on a challenging project gets 2005 or 40 points and each team member working on a standard project gets 1004 or 25 points. Now each member works on exactly 5 projects.

Let us assume that ‘x’ out of these 5 projects are standard projects and the remaining ‘5 – x’ are challenging projects, so total points in terms of x will be 40(50 – x) + 25x

⇒ 200 – 40 + 25x

⇒ 200 – 15x

Putting values of x as 0, 1, 2, 3, 4 and 5, we get total points as 200, 185, 170, 155, 140 and 125. Looking at the options we can see options (1), (3) and (4) are possible.

Option 2 i.e., 150 is not possible.

Hence, option (b).

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