GMAT Problem Solving Discussions

aakash82 Says
Here are a couple one from the same set ... since all these questions are appearing right at the start, I assume tht these should be relatively simpler .. but i haven't seen such kind of questions previously and not sure about the approach to follow .... experts please guide me .. may be it will benefit few others

For ur first question.......reasoning is

h(100)=2*3*........*100
=2^50(1*2*......*50)
so prime no:s from 1 to 50 are factors of h(100)
therefore for p to be factor of h(100)+1 it should be greater than 50

ans is E.

Second question my reasoning is........

let no:of 7's in 350 be x & 77's be (n-x)

=> 7x+77(n-x)=350
=>x=(11n - 50)/10 or x=11n/10 - 5

since x can only be a whole number we can check options which is completely divisible by 10 and only option is C


Hope this will clear ur doubt........criticisms are welcome........

Hi.. can someone plz help me understand how to arrive at the solution to the problem in the attached word document?!

Thanks a ton!

Hi,
join OF and it makes OFCD a square.
Now OD=OF=OE=r(lets take)
=>OC=r*sqrt(2)

OC+OE=2.4(given)
=>r+r*sqrt(2)=2.4
=>2.4r(approx as sqrt(2)=1.41)=2.4
=> r=1
so closest circumference=2pi*r=2*pi*1=6.28=6.3(approx)

folks need ur help on these probs

1. The perimeter of a certain isosceles right triangle is 16+16 sqr root 2.
What is the lenght of the hypotenuse of the triangle.

a 8
b 16
c 4 sqr root 2
d 8 sqr root 2
e 16 sqr root 2

2. Which of the following inequalities has a soln set that, when graphed on the number line, is a single line segment of finite length?

a x^4>=1
b x^3=c x^2>=16
d 2e 2

folks need ur help on these probs

1. The perimeter of a certain isosceles right triangle is 16+16 sqr root 2.
What is the lenght of the hypotenuse of the triangle.

a 8
b 16
c 4 sqr root 2
d 8 sqr root 2
e 16 sqr root 2

2. Which of the following inequalities has a soln set that, when graphed on the number line, is a single line segment of finite length?

a x^4>=1
b x^3=c x^2>=16
d 2e 2
1. Ans B

sides of isosceles right triangle are in ratio 1:1:sqr rt 2
so x+x+x*(sqr rt 2)=16+16 sqr root 2

2.I think ans D

Correct me if i'm wrong.........
folks need ur help on these probs

1. The perimeter of a certain isosceles right triangle is 16+16 sqr root 2.
What is the lenght of the hypotenuse of the triangle.

a 8
b 16
c 4 sqr root 2
d 8 sqr root 2
e 16 sqr root 2

2. Which of the following inequalities has a soln set that, when graphed on the number line, is a single line segment of finite length?

a x^4>=1
b x^3=c x^2>=16
d 2e 2

1. ans is d.
2. ans is e.
folks need ur help on these probs

1. The perimeter of a certain isosceles right triangle is 16+16 sqr root 2.
What is the lenght of the hypotenuse of the triangle.

a 8
b 16
c 4 sqr root 2
d 8 sqr root 2
e 16 sqr root 2

2. Which of the following inequalities has a soln set that, when graphed on the number line, is a single line segment of finite length?

a x^4>=1
b x^3=c x^2>=16
d 2e 2

My take:

1) B
2) E
1. Ans B

sides of isosceles right triangle are in ratio 1:1:sqr rt 2
so x+x+x*(sqr rt 2)=16+16 sqr root 2

2.I think ans D

Correct me if i'm wrong.........

Sorry, 2nd Q my take is E.
1. ans is d.
2. ans is e.



The correct answers are

1. B (Use = 16 units)
2. E (-2/3

You are correct ..The OG answers are...
2 B
3.E
These q's are from GMAT Prep.

ashishjha100 Says
Sorry, 2nd Q my take is E.
Pls help to solve this problem........
ashishjha100 Says
Pls help to solve this problem........


Step 1: Calculate area of triangle ABE:
AB = 1 unit,
BE/ AB = tan(60 degrees) ==> BE = 1*tan(60 degrees) = sq.root(3)
==> Area(triangle ABE) = (1/2)*(1)*(sq.root(3)) = 1/

Step 2: Area(BCDE) = Area(triangle ACD) - Area(triangle ABE)
= * - 1/
= 9/4 sq. units.

Hence (a) is the correct answer.
Step 1: Calculate area of triangle ABE:
AB = 1 unit,
BE/ AB = tan(60 degrees) ==> BE = 1*tan(60 degrees) = sq.root(3)
==> Area(triangle ABE) = (1/2)*(1)*(sq.root(3)) = 1/

Step 2: Area(BCDE) = Area(triangle ACD) - Area(triangle ABE)
= * - 1/
= 9/4 sq. units.

Hence (a) is the correct answer.

But OA is B

Your reasonig is correct but some calculation mistake (Area of smaller triangle is sqrt(3)/2) .....thanks dude for ur help
But OA is B

Your reasonig is correct but some calculation mistake (Area of smaller triangle is sqrt(3)/2) .....thanks dude for ur help



The answer to this is 7 squareroot(3)/4
pls help me in solving this prob

Q16:
The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?

A. 1,104
B. 972
C. 672
D. 294
E. 192
pls help me in solving this prob
Q16:
The sum of three integers is 40. The largest integer is 3 times the middle integer, and the smallest integer is 23 less than the largest integer. What is the product of the three integers?
A. 1,104
B. 972
C. 672
D. 294
E. 192


let three numbers are S,M,L
S+M+L=40(given)
L=3M (given)and

S=L-23(given)>S=3M-23
3M-23+M+3M=40
=>7M=63
M=9
=>L=27 and S=4
hence S*M*L=972

hope m correct
Hey Guys,

attached are a couple of questions from gmatprep.. somehow im not able to get the solution for both of them .. Please guide me


for the 2nd ques according to me the answer should be 'Using both the statemenst' i.e C

whats the correct answer??

Hi...
Could someone help me in solving these problems. I have posted answers along with these. But haven't got a clue on how to arrive at these answers..

1) For every positive integer n, the function h(n) is defined to be the product of all event integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is
(a) between 2 and 10
(b) between 10 and 20
(c) between 20 and 30
(d) between 30 and 40
(e) greater than 40
Ans: e

2) On a certain sight-seeing tour, the ratio of the number of women to the number of children was 5 to 2. What was the number of men on the sight-seeing tour?
(1) On the sight-seeing tour, the ratio of the number of children to the number of men was 5 to 11
(2) The number of women on the sight-seeing tour was less than 30.
Ans (c)

3) If n is a multiple of 5 and n=p^2q, where p and q are prime numbers, which of the following must be a multiple of 25?
(a) p^2
(b) q^2
(c) pq
(d) p^2q^2
(e) p^3q
Ans (d)

4) Of the students who eat in a certain cafeteria, each student either likes or dislikes lima beans and each student either likes or dislikes brussel sprouts. Of these students, 2/3 dislike lima beans; and of those who dislike lima beans, 3/5 also dislike brussel sprouts. How many of teh students like brussels sprouts but dislike lima beans?
(1) 120 students eat in the cafeteria
(2) 40 of the students like lima beans
Ans (D)

5) If each of the following fractions were written as a repeating decimal, which would have the longest sequence of different digits?
(a) 2/11
(b) 1/3
(c) 41/99
(d) 2/3
(e) 23/37
Ans : e

Will go from bottom to top...
5th Question
Simply using division
2/11=0.222 (repeating decimal is 1 digit)
1/3=0.333 (repeating decimal is 1 digit)
41/99 = 0.414141 (repeating decimal is 2 digits)
2/3=0.6666 (repeating decimal is 1 digit)
23/37 =0.621621(repeating decimal is 3 digits)

so the number with longest sequence is option e

4th question
----------------
2/3 dislike lima beans ....
From (1)..of total 120 ..its 80 students (2/3 * 120)
out of 80 ...3/5 dislike brussel sprouts...so its 48 students ..
so students like brussels sprouts but dislike lima beans 80-48 = 32

From (2) ...40 students like lima beans
If 2/3 dislike lima beans of all students...1/3 like lima beans
so (1/3 * total students) = 40 ...so total students =120
and the from above we can achieve students like brussels sprouts but dislike lima beans....

So the answer is D..

2nd Question
---------------
women:children = 5:2

From(1)
children:men = 5:11

From(2)
women
Let children be 10....then women will be 25 ..this might hold true since women and men will be (11/5) * 10 = 22

If only (1) is given..we may get multiples of the above results...
If only (2) is provided, with out men ratio...it is difficult to calculate men

So the answer will be D