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Tricky problem sets [#permalink]
04 Oct 2013, 21:47

2

This post received KUDOS

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This post was BOOKMARKED

Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals:

A) \(\frac{5}{2}\) B) \(\frac{3}{2}\) C) \(\frac{4}{3}\) D) \(\frac{5}{4}\) E) \(\frac{3}{4}\)

Re: Two non-zero real numbers, a and b, satisfy ab = a-b. which [#permalink]
05 Oct 2013, 00:18

Qoofi wrote:

fozzzy wrote:

Two non-zero real numbers, a and b, satisfy ab = a-b. which of the following is a possible value of \(\frac{a}{b} + \frac{b}{a} - ab\)?

a) \(-2\) b) \(\frac{-1}{2}\) c) \(\frac{1}{3}\) d) \(\frac{1}{2}\) e) \(2\)

Hi Fozzzy,

Given : ab = a-b

Solution :

a/b + b/a - ab

(a^2+ b^2)/ab - ab

[(a-b)^2 + 2ab]/ab - ab (We know a^2 + b^2 can be written as (a-b)^2 + 2ab)

Now, [(ab)^2 + 2ab]/ab - ab

ab + 2 - ab

Answer = 2

Can you explain that step are you adding 2ab? if you are then you have subtract that variable too isn't it? or this is something else... _________________

Re: Three men, Alpha,Beta, and Gamma, working together, do a job [#permalink]
05 Oct 2013, 02:46

fozzzy wrote:

Three men, Alpha,Beta, and Gamma, working together, do a job in 6 hours less time than Alpha alone, in 1 hour less time than Beta alone, and in one-half the time needed by Gamma when working alone. let h be the number of hours needed by Alpha and Beta, working together to do the job. Then h equals:

A) \(\frac{5}{2}\) B) \(\frac{3}{2}\) C) \(\frac{4}{3}\) D) \(\frac{5}{4}\) E) \(\frac{3}{4}\)

This one is bit of a toughie. Or I can say it required computation. Lets begin.

Let x be the time required for (A, B, G) together to complete the work. Their rate = \(1/x\) ( Work done per time )

now Time for A to complete work = x + 6, its rate = \(1 / (x+6)\)

Similarly Time for B = x + 1 and rate = \(1/(x+1)\)

& Time for G = 2x and rate = \(1/2x\) - Eqn (1)

Work done by Alpha A & Beta B per hour = \(1/(x+6) + 1/(x+1)\) Now, we can say work done by Gamma G by subtracting the work done per time of all 3 together with work done per time by Alpha A & Beta B

Which is, \((1/x) - ( 1/(x+6) + 1/(x+1) ) - Eqn (2)\)

Now equate Eqn (1) & (2)

\(1/2x = 1/x - ( 1/(x+6) + 1/(x+1) )\)

After cross multiplying, the above equation will come down to \((3x+2)(x+3)\) \(x= 2/3 or -3\) Leaving the negative solution, substitute x = 2/3 in \(1/(x+6) + 1/(x+1)\)

We get rate of work done by A & B together = 3/4.

Time req = Work / rate = 1/ (3/4) = 4/3

Tough one. I hope it is clear and really looking forward for a simpler solution.

Cheers Qoofi _________________

I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?

Re: Tricky problem sets [#permalink]
05 Oct 2013, 05:05

A teacher gave a test to a class in which 10% of the students are juniors and 90% are seniors. the average score on the test was 84. The juniors all received the same score, and the average score of the seniors was 83. what score did each of the juniors receive on the test?

Re: Tricky problem sets [#permalink]
05 Oct 2013, 07:38

1

This post received KUDOS

fozzzy wrote:

A teacher gave a test to a class in which 10% of the students are juniors and 90% are seniors. the average score on the test was 84. The juniors all received the same score, and the average score of the seniors was 83. what score did each of the juniors receive on the test?

A) 85 B) 88 C) 93 D) 94 E) 98

Lets say there are 100 students and there are 10 Juniors & 90 Seniors.

X is the score of Juniors.

(10 X + 83*90) / 100 = 84

Solving you get X = 93

Hope it is clear.

Cheers Qoofi _________________

I'm telling this because you don't get it. You think you get it which is not the same as actually getting it. Get it?

Re: Tricky problem sets [#permalink]
05 Oct 2013, 08:31

New problem added! Kudos for correct solutions

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