GMAT Problem Solving Discussions

This is a solid q and it took me some time to arrive at the answer.

In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

(5/3)P + 5/3 (ROOT 3)

(5/3)P + 10/3 (ROOT 3)

(10/3)P + 5/3 (ROOT 3)

(10/3)P + 10/3 (ROOT 3)

(10/3)P + 20/3 (ROOT 3)

P - pie
Answer D


given that AB is parallel to CD and AB is diameter = 10 , problem is not tough ...

angle inscribed in a semicircle is 90...
and AB // CD, hence x = 30
So, triangle ACB is 30-60-90 triangle

CB=EB=root 3 /2 * 10 = 5root3

m(arc CAE) = 120 ( twice 2x)
Hence l (CAE) = 1/3*2pi*5 =10pi/3

Hence, p = CB+EB+EC
= 5root 3 + 5root3 + 10pi/3
=10pi/3 + 10root3 .. Ans D
ramviswa Says
How to find the perimeter of the shaded area? (See attachment)


AB//CD so angle x=30 degrees. from congruent triangle properties , BC=BE.

Let O is the centre then triangle OBC s isosceles , By sine rule-
Sin30/5=sin120/BC=>BC=(cos30*5/sin30)=5sqrt(3)

For arcCE, angle at centre will be twice the abgle on the arc so angle COE will be 120 degrees.
So CE=2*pi*5*120/360=10*pi/3
sp perimeter=10*pi/3+10 sqrt(3)

Thanks for all the posts. I agree with your explanations.

ramviswa Says
How to find the perimeter of the shaded area? (See attachment)

This is a solid q and it took me some time to arrive at the answer.

In the figure to the right, circle O has center O, diameter AB and a radius of 5. Line CD is parallel to the diameter. What is the perimeter of the shaded region?

(5/3)P + 5/3 (ROOT 3)

(5/3)P + 10/3 (ROOT 3)

(10/3)P + 5/3 (ROOT 3)

(10/3)P + 10/3 (ROOT 3)

(10/3)P + 20/3 (ROOT 3)

P - pie
Answer D


u did not quote while posting the question that the radius is 5.

a generalized solution to ramviswa's question

Suppore r is the radius of circle
first of all Length of an arc=2*pi*r*(theta/360 degrees)

theta=angle which an arc subtends at the center.(here it is 120 degrees)
The other two sides of the triangle can be computed easily and generalized as follows:-
Length of side=r*sqrt(3)

Perimeter=2(length of side)+Length of arc
=2*r*sqrt(3)+2*pi*r*(theta/360)

on substituting the values i got answer as
=(10*pi)/3 + 10*sqrt(3)

1.What is the highest prime factor of 4^17-2^28
2. What is the remainder when 3^(8n+3)+2 is divided by 5 , N is a natural number.

1.What is the highest prime factor of 4^17-2^28
2. What is the remainder when 3^(8n+3)+2 is divided by 5 , N is a natural number.


1. 4^17-2^28 = 2^34-2^28
= 2^28 (2^6 - 1)
= 2^28 * 63
= 2^28 * 3^2 * 7

Hence highest prime factor is 7 ...

2. To get the remainder , we need to find the units digit ..

units place for powers of 3 follows a cyclicity of 4 ...so we need to find the power of 3 in the form 3^(4k+p)

Hence,
3^(8n+3) = 3^(4*2*n+3) = 3^(4k+3)..so units digit is same as 3^3 i.e 7

hence units digit of 3^(8n+3)+2 is 9
so when divided by 5 remainder is 4 ...Ans

I am sorry if I have put you in trouble with missing details.

What is the value of x + y in the figure above?
(1) w = 95
(2) z = 125

Ignore my question above as it should not be posted here.

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What is the value of x + y in the figure above?
(1) w = 95
(2) z = 125

Is the answer to the Question 140?

sum of x,y,z,w is 360
x+y+z+w=360
x+y=360-125-95=360-220
=>140:clap:

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Train A leaves Delhi for Noida at 3 PM and travels at the constant speed of 100 Kph. An hour later, it passes Train B, which is making the trip from Noida to Delhi at a constant speed. If Train B left Noida at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in Delhi?
(1) Train B arrived in Delhi before Train A arrived in Noida .
(2) The distance between Delhi and Noida is greater than 140 miles.

-------------------

Train A leaves Delhi for Noida at 3 PM and travels at the constant speed of 100 Kph. An hour later, it passes Train B, which is making the trip from Noida to Delhi at a constant speed. If Train B left Noida at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in Delhi?
(1) Train B arrived in Delhi before Train A arrived in Noida .
(2) The distance between Delhi and Boston is greater than 140 miles.


wrong thread buddy !!

Speed distance & time is my weakness ....anyways my attempt

Let distance between delhi and noida be D kms and speed of second train be y kms per hr ...

Now, we know total travel time is 2 hrs ..
Hence, D/100 + D/y = 2

We need another eqn in D and y ...

St 1 : it only tells us that time taken by train 1 is more than 50 min than time taken by train 2 i.e D /100 - D/y > 5/6 ....not sufficient ...i.e indirectly distance should be between 800/6 and 200 for this to happen ...

St 2 : I dont know if its a typo error, but distance between delhi and boston has got nothin to with delhi and noida ...
Even if it had been greater than 140 miles between delhi and noida, still not suff ....

Even , combining 2 statements and assuming, D>140 miles between delhi and noida still not suff...

Hence, Ans E ..

Not really sure, if we need to use some relative speed concept ...

Pls correct if am wrong, wats d OA ?
-------------------

Train A leaves Delhi for Noida at 3 PM and travels at the constant speed of 100 Kph. An hour later, it passes Train B, which is making the trip from Noida to Delhi at a constant speed. If Train B left Noida at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in Delhi?
(1) Train B arrived in Delhi before Train A arrived in Noida .
(2) The distance between Delhi and Boston is greater than 140 miles.

My answer is A. whats the OA?
-------------------

Train A leaves Delhi for Noida at 3 PM and travels at the constant speed of 100 Kph. An hour later, it passes Train B, which is making the trip from Noida to Delhi at a constant speed. If Train B left Noida at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in Delhi?
(1) Train B arrived in Delhi before Train A arrived in Noida .
(2) The distance between Delhi and Boston is greater than 140 miles.


Okey lets see.

Lets say that the distance between the two statations is D and speed train of the departing train from noida be n

first eqation:
d/100 + d/n = 2hours ------ (1)

second eqation:
as the two trains met after 1 hour

D = 100 + n* 1/6 ------ (2)

Two eqations two variables.
We are able to get the value of both variables without actually using any of the given statements.

I am not smarter then the book or this question is wrong. Please comment on my approach so that I can find the error I comminted.

Guys try this.I feel this is one of the toughest question in DS , I have met.

-------------------

Train A leaves Delhi for Noida at 3 PM and travels at the constant speed of 100 Kph. An hour later, it passes Train B, which is making the trip from Noida to Delhi at a constant speed. If Train B left Noida at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in Delhi?
(1) Train B arrived in Delhi before Train A arrived in Noida .
(2) The distance between Delhi and Boston is greater than 140 miles.

My Answer is B. Whats the OA?
-------------------

Train A leaves Delhi for Noida at 3 PM and travels at the constant speed of 100 Kph. An hour later, it passes Train B, which is making the trip from Noida to Delhi at a constant speed. If Train B left Noida at 3:50 PM and if the combined travel time of the two trains is 2 hours, what time did Train B arrive in Delhi?
(1) Train B arrived in Delhi before Train A arrived in Noida .
(2) The distance between Delhi and Boston is greater than 140 miles.


Hey.. u sure you have the correct option posted in here.. you sure it says "Boston" and not "Noida" in the option 2..??..:|

As you would also understand, depending upon whether its Boston or Noida the answer might/would change..