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rectangular wall

by jainrahul1985 » Tue Dec 06, 2011 9:44 pm
A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficncy of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C
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by GMATGuruNY » Tue Dec 06, 2011 10:45 pm
jainrahul1985 wrote:A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficncy of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C
We can plug in the answers, which represent A's rate : M's rate.

Answer choice C: 2:1.
A working with M:
Let A's rate = 2 units per hour and M's rate = 1 unit per hour.
Combined rate for A+M = 3 units per hour.
Since A completes 2 of these 3 units, the fraction of the work produced by A = 2/3.

A working with N and K:
Since N and K are twice as fast as M, N's rate = 2 units per hour and K's rate = 2 units per hour.
Combined rate for A+N+K = 6 units per hour.
Since A completes 2 of these 6 units, the fraction of the work produced by A = 2/6 = 1/3.

Let each side of the wall = 6 units, so that the total job = 12 units.

When A+M complete 3 units of one side, the work produced by A = (2/3)3 = 2 units.
When A+N+K complete the remaining 3 units of one side, the work produced by A = (1/3)3 = 1 unit.
When A completes the other side of the wall, the work produced by A = 6 units.

(Work produced by A)/(total work) = (2+1+6)/12 = 9/12 = 3/4 = 75%.
Success!

The correct answer is C.
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by Anurag@Gurome » Tue Dec 06, 2011 10:50 pm
jainrahul1985 wrote:A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficncy of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C
Out of 2 sides of a wall, A completed painting other side of wall all alone, which means A completed 50% of the work and A's share is 75% of the total amount given to all four workers. So, 25% of the work is done by the remaining workers.
Let us assume that work efficiency of M = M
Then work efficiency of N and K = 2M
Also let us assume the work efficiency of A = A
Then work done by A and M = (A/4)/(A + M)
Work done by A, N and K = (A/4)/(A + 4M)
So, (A/4)/(A + M) + (A/4)/(A + 4M) = 1/4
A[A + 4M + A + M] = (A + M)(A + 4M)
2A² + 5AM = A² + 4M² + 5AM
A² = 4M²
A = 2M
A:M = [spoiler]2:1[/spoiler]

The correct answer is C.
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by batmannavneet » Wed Dec 07, 2011 8:57 am
Anurag@Gurome wrote:
jainrahul1985 wrote:A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficncy of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C
Out of 2 sides of a wall, A completed painting other side of wall all alone, which means A completed 50% of the work and A's share is 75% of the total amount given to all four workers. So, 25% of the work is done by the remaining workers.
Let us assume that work efficiency of M = M
Then work efficiency of N and K = 2M
Also let us assume the work efficiency of A = A
Then work done by A and M = (A/4)/(A + M)
Work done by A, N and K = (A/4)/(A + 4M)
So, (A/4)/(A + M) + (A/4)/(A + 4M) = 1/4
A[A + 4M + A + M] = (A + M)(A + 4M)
2A² + 5AM = A² + 4M² + 5AM
A² = 4M²
A = 2M
A:M = [spoiler]2:1[/spoiler]

The correct answer is C.
Can you please tell me how did you come up with A/4 in the highlighted section ?
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by karthikpandian19 » Wed Dec 07, 2011 5:45 pm
I am also confused with this part?
Can you explain?
batmannavneet wrote:
Anurag@Gurome wrote:
jainrahul1985 wrote:A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficncy of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C
Out of 2 sides of a wall, A completed painting other side of wall all alone, which means A completed 50% of the work and A's share is 75% of the total amount given to all four workers. So, 25% of the work is done by the remaining workers.
Let us assume that work efficiency of M = M
Then work efficiency of N and K = 2M
Also let us assume the work efficiency of A = A
Then work done by A and M = (A/4)/(A + M)
Work done by A, N and K = (A/4)/(A + 4M)
So, (A/4)/(A + M) + (A/4)/(A + 4M) = 1/4
A[A + 4M + A + M] = (A + M)(A + 4M)
2A² + 5AM = A² + 4M² + 5AM
A² = 4M²
A = 2M
A:M = [spoiler]2:1[/spoiler]

The correct answer is C.
Can you please tell me how did you come up with A/4 in the highlighted section ?
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by karthikpandian19 » Wed Dec 07, 2011 5:57 pm
Is there any other method to find the answer other than the trial and error method ?
GMATGuruNY wrote:
jainrahul1985 wrote:A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficncy of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C
We can plug in the answers, which represent A's rate : M's rate.

Answer choice C: 2:1.
A working with M:
Let A's rate = 2 units per hour and M's rate = 1 unit per hour.
Combined rate for A+M = 3 units per hour.
Since A completes 2 of these 3 units, the fraction of the work produced by A = 2/3.

A working with N and K:
Since N and K are twice as fast as M, N's rate = 2 units per hour and K's rate = 2 units per hour.
Combined rate for A+N+K = 6 units per hour.
Since A completes 2 of these 6 units, the fraction of the work produced by A = 2/6 = 1/3.

Let each side of the wall = 6 units, so that the total job = 12 units.

When A+M complete 3 units of one side, the work produced by A = (2/3)3 = 2 units.
When A+N+K complete the remaining 3 units of one side, the work produced by A = (1/3)3 = 1 unit.
When A completes the other side of the wall, the work produced by A = 6 units.

(Work produced by A)/(total work) = (2+1+6)/12 = 9/12 = 3/4 = 75%.
Success!

The correct answer is C.
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by user123321 » Wed Dec 07, 2011 8:31 pm
There are many ways to do this one. But we have another way of doing it using time rather than efficiency.

let A takes y hrs
M takes 2x hrs
N takes x hrs ( x hrs because N's efficiency is twice of M's so he takes half the time)
K takes x hrs

if we assume total work as 1
1/4 th work was done by M & A
=> in 1 hr - 1/2x + 1/y = (y+2x)/2xy work will be done
=> 1 work will take 2xy/(y+2x) hrs
=> 1/4th work will take 1/4 * 2xy/(y+2x) hrs

1/4 th work was done by N,K & A
=> in 1 hr - 1/x + 1/x + 1/y = 2/x + 1/y = (2y+x)/xy
=> 1/4 th work will take 1/4 * xy/(2+x) hrs

But we know A took y/4 hrs on first side since he did 25% of the work on first side
=>(1/4 * 2xy/(y+2x)) + (1/4 * xy/(2+x)) = y/4
=>xy/4 * (2/(x+2y) + 1/(2x+y)) = y/4
=>y(5x+4y) = (x+2y)(2x+y)
=> x=y

we need to find ratio of efficiency of A & M = time taken by M/time taken by A = 2x/y = 2x/x = 2/1

hence C

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by GMATGuruNY » Thu Dec 08, 2011 12:23 am
karthikpandian19 wrote: A took a contract to paint a rectangular wall on both sides he did 1/2 of the work with M & remaining 1/2 with help of N & K each of whose efficency is twice that of M . A completed painting other side of wall working alone . If A's share is 75% of the total amount given to all four workers what is the ratio of efficiency of A:M?

1)3:1
2)4:1
3)2:1
4)1:4
5)1:2

OA C

Is there any other method to find the answer other than the trial and error method ?
Let each side of the wall = 2 units, for a total job of 4 units.
To produce 75% of the job, A must produce 3 units: 2 units when he paints one side of the wall by himself, 1 unit when he paints with the other workers.

Let M's rate = 1 unit per hour.
Since N and K work twice as fast, N's rate = 2 units per hour and K's rate = 2 units per hour, implying a combined rate of 4 units per hour when they work together.
Let A's rate = A.

A and M together:
A produces A units per hour and M produces 1 unit per hour.
Of the total amount of work produced, the fraction produced by A = A/(A+1).
A, N and K together:
A produces A units per hour and N and K produce 4 units per hour.
Of the total amount of work produced, the fraction produced by A = A/(A+4).

Since A must produce 1 unit when he paints with the other workers, the sum of the fractions above is 1:
A/(A+1) + A/(A+4) = 1.
(A² + 4A + A² + A)/(A² + 5A + 4) = 1
2A² + 5A = A² + 5A + 4
A² = 4
A = 2

Since A=2 and M=1, the ratio of A's rate to M's rate = 2:1.

The correct answer is C.

The approach above combines plugging in with algebra.
Plugging in is very helpful here: it provides an easy way to satisfy the conditions given -- that A completes 75% of the work and that N and K work twice as fast as M -- so that the algebra yields an equation in terms of a single variable.
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As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

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