Problem Solving Territory: 700+ questions only..............

Here comes our next question.......

3. Fresh Grapes contain 90% water and 10% grape pulp whereas Dry Grapes
contain 20% water and 80% grape pulp. How many kgs of Dry Grapes can
be obtained from 20 kgs of Fresh Grapes?

Amount of grape pulp will remain same.
80Z/100 =10/100*20
z=2.5

Answer is 2.5 kgs

As simple as that....................

Here comes our next question

4. A group of men can complete a work in certain number of days. After they
worked for 12 days 1/4 of them left. To complete the remaining work the
remaining men have taken as many days as the initial number of men
would have taken to complete the entire work. In how many days is the
work completed?

I got DC = 5 and CE = 11/5. Also used similarity of triangles by a side and two adjacent angles.

EDIT: whooops, missed the second page Sorry, this was for question 2.

EDIT2: here goes number 4...

S = total job
v = total speed of all men
t = original time

S = v*t
S = 12*v + 0.75v*t

vt = 12v + 0.75vt
0.25t = 12
t = 48

ANSWER: the work will be done in (48 + 12) = 60 days.

That's the right answer......

Any other approaches.........

Here comes our next question.........

5. When a clock indicates completion of 24 hours it actually loses one hour. How much time does it loose in 5 days?

I get 5 hrs .. I know it is not that simple but that's what I can see.

Amit

A little less than that I would say. How about 4.8 hours? (It loses 1 hour every 25 hours, if I understand correctly).

Yed Honghu, that's the right answer

Folks, it's time again to get into some fight............

Here comes our next question.........

6.

very tough quest indeed
I find that the radius of the bigger circle is 5
I think this is not a kind of GMAT quest (easy math prob but take too much time and paper also)

I got 5 as well It's true that it's not a usual GMAT question

I worked it out using an XY plan.

The figure is symetrical. So, we can have the same kind of rectangle at the bottom left of the square (1 vertex on C and the other on the circle).

By defining the origin 0(0,0) = C(0,0), the equation of the circle is:
(1) (x-r)^2 + (y-r)^2 = r^2 with r the radius of the circle.

As the vertex of the rectangle on the circle has the coordinates (2,1), we can plug the values of x & y in (1).

(2-r)^2 + (1-r)^2 = r^2
<=> 4 - 4*r + r^2 + 1 - 2*r + r^2 = r^2
<=> r^2 - 6*r + 5 = 0

We can use the "reduiced" Delta = 3^2 - 5*1 = 4 = 2^2

Thus, we have 2 possibilies :
o r= (3+2)/1 = 5
o r= (3-2)/1 = 1

For the latter radius, the vertex lying on the circle is represented by a point on the "other side" of the circle. In the case of this picture, the radius is 5.

Hi Fig,

I m not clear how you can take the cordinates of vertex as (2 , 1) since 2 and 1 are the length of the rectangle and you are taking the center of the circle as the origin. Even by Visual observation I am not able to understand the login.. Can you please explain ..?


Regards,
Amit

sorry c as the Origin

I take the origine O(0,0) at the point C : the vertex of the scare at the bottom left

Since this origine definied,
o the vertex of a silimar little rectangle from C and on the circle has the coordinates (2,1) (lenght, width).
o the center of the cicle Z is at (r,r)

Hope this help

Ok

I tried like this:

r (12) = w
.75r (t - 12) = w ---> .75rt - 9r = 12r ---> rt = 28r ---> t = 28

t = 28 + 12 = 40

I see this is the wrong answer. Should I not have set the two equations equal to one another? Can someone suggest how to correct this