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Bismuth83
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Say,

Initial number of workers = n

In the final 2 days, 60 units remain to be made

So, in the first 5 days, 90 - 60 = 30 units were made.

So,

(n)*(r)*(5) = 30
=>(1) -> (n)*(r) = 6

"a" workers were added. The work was completed on time.

So,

(n+a) *(r)*(2) = 60

Using (1),

(a)*(r) = 24

Also, we are given that a<r.

From the choices, the only possible answer choices are (a,r) = (4,6).


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How I approached the problem:

60 units are yet to be made in 2 days. So 30 units were made in 5 days.

30/5 = 6 units per day. Therefore 3 employees working at 2 units per day (given).

Problem statement: We need 60 units in 2 days. So 30 units per day.

We already have employees to make 6 units per day. So we're 24 units per day short.

The only combinations from the table that fit our description are (4,6) and (6,4)

But, as we know that a<r, we can eliminate (6,4).

So we're left with (4,6).
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1. The question asks us to find a working combination of a and r.

2. Let n be the number of original workers.

3. It is known that in the first 5 days n workers made 90 - 60 = 30 units, with each of them making 2 units per day. In other words, \(30 = (2 * n) * 5 \rightarrow n = 3\).

4. The final 2 days had \(n + a = a + 3\) workers that made 60 units, with a of them making r units per day. So, \(60 = (a * r + 3 * 2) * 2 \rightarrow ar = 24\).

5. Know we can look at the answer options with this knowledge and that a < r. \(4 * 4 = 16 < 24\) and 24 isn't divisible by 5, which means r = 6 and a = 4.

6. Our answer will be: a = 4 and r = 6.
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as per given condition in first 5 days only 30 units are produced i.e. 6 units/day
since rate of each worker=2 units/day. so, no. of workers initially=3
now a additional workers are added to work at the rate of r units/day for the next two days to meet the lapse of 60 units.
hence 2(3*2+a*r)=60 or 12+2a*r=60
2a*r=48 or a*r=24
since a<r so only values that satisfy are a=4 & r=6
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