GMAT Problem Solving Discussions

MBA
1720
Simmy Says
OA's are C, B, D. So now you can give reasoning for 2nd question. Thanks in advance.

2. let length of larger side of rectangle = x and smaller side = y

And side of square = a

From fig diagonal of square , a*sqrt(2)= x/2 = y

so 2(x+y)= 2(y+2y) = 18sqrt(2)

y=3*sqrt(2)

a=3 so perimeter= 4*3=12
1721
Hi,

Attached is a doc having 3 questions. Last question i could not make head and tail of it :(

Question 1 square on both sides and solve for x
You get x=1/3 and 3
so inconclusive.
With the second question we learn that x is not 3
so both put together we can ans this question.
1722

Hi,

Few more questions along with my logic which turned out to be wrong :D

So help 😃

1723
Hi,

Few more questions along with my logic which turned out to be wrong :D

So help :)


Question 1:

Join PR
you get an equilateral triangle pqr as all are diagonals of one face of a cube all of em are root 2 * a

Therefore angle q=p=r=60 deg
1724
If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

for all values of n which are even n(n + 1)(n + 2) will be divisible by 8(try by picking no:s) ---> 48 no:s

All values of n which are odd and just 1 above the multiples of 8(n=7,15,23...95) --->12

so req probability=(48+12)/96 = 5/8;)
1725
The annual rent collected by a corporation from a certain building was x percent more in 1998 than in 1997 and y percent less in 1999 than in 1998. Was the annual rent collected by the corporation from the building more in 1999 than in 1997?
(1) x > y


(2)(xy/100) x y

Rent in 1997 = a
1998 = (100+x)*a/100
1999 = (100-y)(100+x)*a/(100*100)

Is "(100-y)(100+x)*a/(100*100) > a"?

simplifiying 100(x-y) > xy
x-y> xy/100

St 2 : provides this option ......so B;)












1726

Some more questions guys 😃

1727

Answer to the 3rd question is E: r2 + (10 r)2

Area of the circle + ^2

Note* : Since a square has four sides.

Q3:
A thin piece of wire 40 meters long is cut into two pieces. One piece is used to form a circle with radius r, and the other is used to form a square. No wire is left over. Which of the following represents the total area, in square meters, of the circular and the square region in terms of r?
1) r2
2) r2 + 10
3) r2 + 2r2
4) r2 + (40 - 2r)2
5) r2 + (10 r)2


Still trying the rest!

1728
All of the stocks on the over the counter market are designated by either a 4-letter or a 5-letter code that is created by using the 26 letters of the alphabet. Which of the following gives a maximum number of different stocks that can be designated with their codes?
1) 2(265)
2) 26(264)
3) 27(264)
4) 26(265)
5) 27(265)

Max no: of 4 letter code = 26*26*26*26= 26^4 (Repetition allowed)
Max no: of 5 letter code = 26*26*26*26*26= 26^5 (Repetition allowed)

So total codes = 26^4+26^5 = 26^4(1+26) = 27(26^4)

Ans C.......is this the OA?
1729
Q2: My pick would be B.
Radius of circle = sqrt(3+1)= 2
so 2 = sqrt(s^2+t^2) (since (s,t) forms the radius) -------> eqn 1
Also from figure
two sides of triangle is the radius (2 units) and hypotenuse is 2*sqrt(2)
So PQ forms a line with length 2*sqrt(2)
2*sqrt(2) = sqrt{(s+sqrt(3))^2 + (t - 1)^2)} ------> eqn 2
Solving eqn 1 & 2 we get s = +/-1
But s is in first quadrant so s=1
Hope this is the OA....:idea:
1730
Simmy Says
Some more questions guys :)


1) D
consider triangle PRS,
let rps=x
prs=y
x+y+90=180
x+y =90 ------------1

consider triangle qps
qps=x+30
pqs=z

x+30+z+90=180
x+z=60-------------2

1 -2

y-z= 30
which is the req ans

Consider the second condition ,
with it we can infer that Qpr=30
which leads us to the same conclusion as 1

Hence D
1731

S is 1 indeed!!!

1732
Simmy Says
Some more questions guys :)


Thanks guys,

OA's are D, B, E, C
1733
mockingabird Says
Is the ans 4) sqrt3 ??? If Yes, I ll give ye de explanation for it


3) Ans is the 5th option

2*pi*r+40-2*pi*r

circle area= pi*r^2
side of the square = (40-2*pi*r)/4
Hence area is (10 -pi/2*r)^2
1734

At a certain hospital, 75% of the interns receive fewer than 6 hours of sleep and report feeling tired during their shifts. At the same time, 70% of the interns who receive 6 or more hours of sleep report no feelings of tiredness. If 80% of the interns receive fewer than 6 hours of sleep, what percent of the interns report no feelings of tiredness during their shifts?

A. 6
B. 14
C. 19
D. 20
E. 81
:huh:

1735

The y-intercept of a line l is 4. If the slope of l is -ve, which of the following could be the x-intercept of l
1) -1
2) 0
3) 6

1736

Two triangles are similar with angles x,y,z. Smaller triangle has a base 's' and larger one has 'S'. If the area of the larger triangle is twice the area of the smaller one, then in terms of s, S=
1) (2/2)s
2) (3/2)s
3) 2s
4) 3s
5) 2s

1739

I think the answer is (C) --> 19.

Assume there are 100 interns;

80 receive 20 receive > 6 hours sleep

75 receive
This means 80-75 = 5 ARE NOT TIRED.

Out of the 20 who receive > 6 hours sleep, 70% => 14 are NOT TIRED

In total 14 + 5 = 19 interns (or in this case 19% of the interns) are NOT TIRED.

At a certain hospital, 75% of the interns receive fewer than 6 hours of sleep and report feeling tired during their shifts. At the same time, 70% of the interns who receive 6 or more hours of sleep report no feelings of tiredness. If 80% of the interns receive fewer than 6 hours of sleep, what percent of the interns report no feelings of tiredness during their shifts?

A. 6
B. 14
C. 19
D. 20
E. 81
:huh:
1740
For similiar triangles with corresponding sides 'a' and 'b'

Ar of triangle with side 'a' / Ar of triangle with side 'b' = a^2 / b^2

Using this, answer shud be C

@ Ankit or Simmy
Can either of you please explain this solution, i am kinda lost. Maybe i m missing some funda.

Thanks in advance.
1741
@ Ankit or Simmy
Can either of you please explain this solution, i am kinda lost. Maybe i m missing some funda.

Thanks in advance.


Hi -- In two similar triangles, the ratio of their areas is the square of the ratio of their sides.

Let A1 = Area with small side 's'
A2 = Area with small side 'S'
Given A2/A1 = 2

A2/A1 = S^2 / s^2

=> S^2 / s^2 = 2

=> S/ s = 2
=> S = 2 s => so 'C' it is..

Hope this helps