Problem Solving

mannuunnam wrote:If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?

A. 4 : 1

B. 2 : 1

C. 10 : 1

D. 3 : 1

E. 1 : 2

My approach is

let D = dave; Di diana

Given that,

D = (D+Di) + 20 and

Di = (D+Di) + 5

Pick a number for (D+Di), let say 10 which is easy to calculate.

D = 10 + 20 => 30

Di = 10 + 5 => 15

D/Di = 30/15 => 2:1

I hope it can be done this way...

If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?


A. 4 : 1


B. 2 : 1


C. 10 : 1


D. 3 : 1


E. 1 : 2

dinesh19aug wrote:Is the answer 1:2??

There are two ways to solve this problem.

Approach 1: This question is a good candidate for a guess question.
If you see that Dave take 20 hours more
Diana takes 5 hours more to complete the same task.
This means that Diana rate is faster than Dave's rate
So Dave rate : Dian rate = x : y, which means that x < y
The only choice that reflects this is Answer choice E, (1:2) All other choice indicate that Dave is faster than Diana which is not true.

Approach 2:
Draw the matrix
==============================
| R | T | W
Dave | | |
Diana | | |
Together | | |

Now pick a number say 10 and assumes that this is the amount of time taken when both work together. Now fill in the values in matrix above


==============================
| R | T | W
Dave | 1/30| 10+20 = 30 | 1
Diana | 1/15| 10+5 = 15 | 1
Together | 1/10 | 10 | 1


So you can see that Dave rate : Diana Rate = 1:2 Answer

Note: This answer can also be solved by basic algebra by taking values as t1, t2 t etc and solving it, but that is too cumbersome and lengthy way. Don't waste your time on it.

- Dinesh
Dinesh

chris@veritasprep wrote:This is a great (and more difficult) work/rate question that is typical of what you might see on the test. When I teach work/rate problems I try to avoid any gimmicks and have students use the following approach, which will work for any work/rate problem:

1. Always convert times to rates. If times are given for a complete job then the rate is always equal to the inverse of the time: r = 1/t While many of these problems can be solved logically using times, it is often problematic.

2. Rates are additive. The combined rate is always equal to the sum of all the rates of the machines/people/etc. working in the problem. Rate Combined = r1 +r2 +r3.....

3. Use the Work = Rate x Time (W=RT) formula to solve for the necessary unknown in the problem at hand.

4. On more complicated problems where conditions are changing throughout the problem (machines/workers are being added or taken away) do not try to solve the problem with one equation. Treat it as multiple problems and account for changes with each new equation.

For this problem you only need the first two steps. While there are many different ways you could approach this, I prefer Asif's approach, which relies on an understanding that rates are additive. As there were some questions about the details of his approach I will elaborate.

As is the case with most work/rate problems, the information in the problem is given in times, which we must convert to rates.

Let t = the time together to complete the job. That means that time dave = t +20 and time diana = t + 5.

If we invert each of these to get the rates of dave and diana we can see that rate dave = 1/(t +20) and rate diana = 1/(t+5) and rate together = 1/t Using our understanding that rates are additive, we know that the rate dave + rate diana = rate combined. This gives us the equation that asif used in his post:

1/(t+20) + 1/(t+5) = 1/t

Now there is a little tedious algebra to isolate t and complete the problem, but it is not too bad.

As with any algebraic equation involving fractions and variables, multiply through the equation with the LCM of the three denominators to get rid of the fractions. Here that involves multiplying through by the product of the three denominators which is (t+20)(t+5)t

After this manipulation, you are left with (t)(t+5) + (t)(t+20) = (t+5)(t+20)

Simplify to get t^2 + 5t +t^2 +20t = t^2 +25t +100

Combine/Subtract like terms to get 2t^2 + 25t = t^2 + 25t + 100

Simplify again to see that t^2 = 100 and t = 10 (t cannot be -10 as it must be positive here).

If t = 10 then time dave = 30 and time diana = 15. Ratio of time dave to time diana = 30: 15 or 2:1 and the answer is B. Hope that helps!