SPEED and VELOCITY can complete a piece of work respectively in FEW days and LITTLE days while working separately. The amount of work done by SPEED in 21 days exceeds the amount of work done by VELOCITY in 8 days by one-sixth of the amount of work done by SPEED in FEW days.
SPEED and VELOCITY began the work together but SPEED left after LITTLE/2 days forcing VELOCITY to complete the remaining work all alone in FEW/9 days.
Determine FEW and LITTLE.
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Submitted by K Sengupta
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Rating: 3.6667 (3 votes)
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Solution:
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(FEW, LITTLE) = (54,36) constitutes the required solution to the problem under reference.
EXPLANATION:
Let us denote FEW and LITTLE respectively by x and y.
By the problem:
In one day SPEED and VELOCITY working separately can respectively complete (1/x)th portion and (1/y)th portion of the work and working together they can complete (1/x + 1/y)th portion of the work(in one day).
Accordingly, by conditions of the problem, we obtain:
(i) 21/x - 8/y = 1/6
(ii)(y/2)*(1/x + 1/y) + (x/9)*(1/y) = 1
From (ii), we obtain:
y/(2x) + x/(9y) = 1 - 1/2 =1/2
or, 9*(y^2) - 9*xy + 2*(x^2)= 0
or, (3y - 2x)(3y - x) = 0
or, y/x = 2/3, 1/3
If possible, let us assume y/x = 1/3.
From (i), we obtain:
21/x - 8/y = 1/6
or, 21/x > 8/y, so that,y/x > 8/21 > 1/3 , which is a contradiction.
Hence, y/x = 2/3 ( which is greater than 8/21); giving :
y = 2x/3.
Now, 21/x - 8/y = 1/6
giving, 9/x = 1/6 or, x =54 , so that, y = 2*(54)/3 = 36.
Consequently, (FEW, LITTLE) = (54,36) constitutes the required solution to the problem under reference.
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Also refer to Bractals' solution to this problem which is given in this location.
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