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GMAT CD ....HELP
Source: Beat The GMAT — Problem Solving |
-
codesnooker
- Legendary Member
- Posts: 543
- Thanked: 43 times
- GMAT Score:580
For Question I
----------------
Let say wire is divided into two parts: X and 40 - X.
According to question lets assume, X = Perimeter of square = 4a (a is the side of the square).
=> a = X/4 ----------> equation 1.
Therefore, the circumference of circle would be = 2Pr (I don't know how to make pie symbol in this, so marking it with letter P).
i.e. 2Pr = (40-X)
=> X = 40 - 2Pr -----> equation 2.
Now area of square = (a^2) = (X/4)^2. ----------> equation 3.
Now replace the value of X in equation 3 from equation 2.
there modified equation 3 will be
((40 - 2Pr)/4)^2 -----------> equation 4
Now taking 4 out of the equation 4, it can be easily reduced to
(10 - (Pr/2))^2 --------------> equation 5
Now area of circle would be P(r^2) ------------> 6
Now as we need to get the total area in terms of r, so add the equation 5 and 6 to get the desired answer.
For Question II
-----------------
We need to factorize the 450.
i.e. (3^2).(5^2).2.y = x^3
With all the equation I, II, and III none is reducing to the above equation to clear integer value, therefore none is satisfied the result. So, according to me the answer should be A rather B or something else.
If anyone know the correct procedure, then let me know.
----------------
Let say wire is divided into two parts: X and 40 - X.
According to question lets assume, X = Perimeter of square = 4a (a is the side of the square).
=> a = X/4 ----------> equation 1.
Therefore, the circumference of circle would be = 2Pr (I don't know how to make pie symbol in this, so marking it with letter P).
i.e. 2Pr = (40-X)
=> X = 40 - 2Pr -----> equation 2.
Now area of square = (a^2) = (X/4)^2. ----------> equation 3.
Now replace the value of X in equation 3 from equation 2.
there modified equation 3 will be
((40 - 2Pr)/4)^2 -----------> equation 4
Now taking 4 out of the equation 4, it can be easily reduced to
(10 - (Pr/2))^2 --------------> equation 5
Now area of circle would be P(r^2) ------------> 6
Now as we need to get the total area in terms of r, so add the equation 5 and 6 to get the desired answer.
For Question II
-----------------
We need to factorize the 450.
i.e. (3^2).(5^2).2.y = x^3
With all the equation I, II, and III none is reducing to the above equation to clear integer value, therefore none is satisfied the result. So, according to me the answer should be A rather B or something else.
If anyone know the correct procedure, then let me know.
The answer for the first question shud be F .
The second one the answer is B .
450.y = n^3
(3^2)(5^2)(2).y=n^3
Therefore to satisfy that y and n are + numbers y=(3)(5)(2^2) so that n is perfect positive number. Going by options B will only give an integer result
hope its clear
Senthil
The second one the answer is B .
450.y = n^3
(3^2)(5^2)(2).y=n^3
Therefore to satisfy that y and n are + numbers y=(3)(5)(2^2) so that n is perfect positive number. Going by options B will only give an integer result
hope its clear
Senthil
-
codesnooker
- Legendary Member
- Posts: 543
- Thanked: 43 times
- GMAT Score:580
