GMAT Problem Solving Discussions

jpmadhav · July 2, 2008, 6:21am

I guess 3rd questions is tricky...need someone to solve this out....

jha16june · July 2, 2008, 7:48am

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..
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)

I have adoubt in the above question:
Is it n = (p^2)*q or is it n = p^(2*q), I mean q is in the index or is it just a multiplicant with p^2?

getneonow · July 2, 2008, 9:48am

can someone help me to solve this problem?

If a committee of 3 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

A. 20

B. 40
C. 50
D. 80
E. 120

another soln..

first select 3 ppl from 10 members = 10C3

remove the possibility of forming a couple in those selection of 3
= 5C1 * 8 (selecting one couple and the third member as any remaining one)

so 10C3- 5*8 = 120-40 = 80

Neo

getneonow · July 2, 2008, 10:07am

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..

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)

so here's your soln..

p,q are prime integers and n is a multiple of 5=> atleast one of p or q must be 5 . mathematically,

p^2 * q = 5k (k is integer) => k = (p^2 * q)/5 = (p^2/5)*q = p^2*(q/5)

for k to be a integer you can see that either p^2 must be a multiple of 5 or q must be a multiple of 5 (since p,q are primes)

If p^2 is multiple of 5 => p is a multiple of 5 (since p is prime).

so you can eliminate options A, B, C clearly. coz we are not sure whether p or q is a multiple of 5.

Now p^2 * q^2 = (p^2 * q) * q

so if both cases where p or q can be a multiple of 5, the above will always be a multiple of 25. Got it?? You can just give some values to p,q and check..

Neo

bunty4billi · July 2, 2008, 7:14pm

Pls help in solving these probs...highlighted ones are OG's answers..

1. The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent.Which of the following is closest to the present change in the chemical A required to keep the reation rate unchanged?

a 100% decrease
b 50% decrease
c 40% decrease
d 40% increase
e 50% increase

2. The number 75 can be written as the sum of the squares of 3 different positive intergers.What is the sum of these 3 integers.

a 17
b 16
c 15
d 14
e 13

3. Three grades of milk are 1 percent ,2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of 2 percent grade,and z gallons of the 3 percent grade are mixed to give x+Y+z gallons of a 1.5 percent grade.What is x in terms of y and z?

a. y+3z
b (y+z)/4
c 2y+3z
d 3y+z
e 3y+4.5z

mockingabird · July 3, 2008, 1:51am

. There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900

ashishjha100 · July 3, 2008, 2:31am

. There were 36.000 hardback copies of a certain novel sold before the paperback version was issued. From the time the first paperback copy was sold until the last copy of the novel was sold. 9 times as many paperback copies as hardback copies were sold. If a total of 441.000 copies of the novel were sold in all, how many paperback copies were sold?
(A) 45.000
(B) 360.000
(C) 364.500
(D) 392.000
(E) 396.900

p=9h

36000+h+9h=441000
h=40500
so p=364500

reddy.harikisho · July 3, 2008, 2:42am

so here's your soln..

p,q are prime integers and n is a multiple of 5=> atleast one of p or q must be 5 . mathematically,

p^2 * q = 5k (k is integer) => k = (p^2 * q)/5 = (p^2/5)*q = p^2*(q/5)

for k to be a integer you can see that either p^2 must be a multiple of 5 or q must be a multiple of 5 (since p,q are primes)

If p^2 is multiple of 5 => p is a multiple of 5 (since p is prime).

so you can eliminate options A, B, C clearly. coz we are not sure whether p or q is a multiple of 5.

Now p^2 * q^2 = (p^2 * q) * q

so if both cases where p or q can be a multiple of 5, the above will always be a multiple of 25. Got it?? You can just give some values to p,q and check..

Neo

I think we can solve this very simply. if n is multiple of 5 then n^2 (n square) would be multiple of 25.therefore, n^2= (p^2q)^2, which is answer D.

jpmadhav · July 3, 2008, 4:39am

Hi Bunty,

For the first Question
---------------------------
Let CA be concentration of chemical A and CB be concentration of chemical B .....initially
Hence rate of a certain chemical reaction => x proportional to (CA^2)/CB

Consider CA and CB as 100 and 100 before start....

so x proportional to (100*100)/100

After the concentration change in B .suppose CBF....present change in the chemical A required to keep the reation rate unchanged be CAF....

now.... x proportional to (CAF^2)/(CBF)
Since the concentration change in B increased is 100 percent...
CBF = CB+100....

so x will be proportional to (CAF^2)/(CB+100)

i.e ..(CAF^2)/200

if we equal both cases.....(CAF^2)/200 = (100*100)/100
...................................(CAF^2) = 200
...................................CAF = square root of 200 = 141... = CA+41 percent.

So change in the chemical A required to keep the reation rate unchanged is approx 40% increase.......

Hope you got the clear explanation....

vicky.verma · July 3, 2008, 5:10am

Pls help in solving these probs...highlighted ones are OG's answers..

1. The rate of a certain chemical reaction is directly proportional to the square of the concentration of chemical A present and inversely proportional to the concentration of chemical B present. If the concentration of chemical B is increased by 100 percent.Which of the following is closest to the present change in the chemical A required to keep the reation rate unchanged?

a 100% decrease
b 50% decrease
c 40% decrease
d 40% increase
e 50% increase

2. The number 75 can be written as the sum of the squares of 3 different positive intergers.What is the sum of these 3 integers.

a 17
b 16
c 15
d 14
e 13

3. Three grades of milk are 1 percent ,2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of 2 percent grade,and z gallons of the 3 percent grade are mixed to give x+Y+z gallons of a 1.5 percent grade.What is x in terms of y and z?

a. y+3z
b (y+z)/4
c 2y+3z
d 3y+z
e 3y+4.5z

1. R=kA^2/B

If B increases by 100% we get R=kA^2/2*B

To keep this same are before we need A=Root(2)*A => A 41% increase for A

2. Hit and trial : The three numbers are 1,7,5

1 + 49 + 25 = 75 => Ans=1+7+5=13

3. 0.01x + 0.02y + 0.03z = 0.015(x+y+z)

Solving the above eqn. we get x=y+3z

ashishjha100 · July 3, 2008, 5:47am

Pls help with reasoning...........

bunty4billi · July 3, 2008, 6:35am

Hi Madhav,

Its crystal clear ! Thank u very much...i was making mistake at this step
" if we equal both cases.....(CAF^2)/200 = (100*100)/100"

Cheers
Bunty..

Hi Bunty,

For the first Question
---------------------------
Let CA be concentration of chemical A and CB be concentration of chemical B .....initially
Hence rate of a certain chemical reaction => x proportional to (CA^2)/CB

Consider CA and CB as 100 and 100 before start....

so x proportional to (100*100)/100

After the concentration change in B .suppose CBF....present change in the chemical A required to keep the reation rate unchanged be CAF....

now.... x proportional to (CAF^2)/(CBF)
Since the concentration change in B increased is 100 percent...
CBF = CB+100....

so x will be proportional to (CAF^2)/(CB+100)

i.e ..(CAF^2)/200

if we equal both cases.....(CAF^2)/200 = (100*100)/100
...................................(CAF^2) = 200
...................................CAF = square root of 200 = 141... = CA+41 percent.

So change in the chemical A required to keep the reation rate unchanged is approx 40% increase.......

Hope you got the clear explanation....

jpmadhav · July 3, 2008, 6:46am

1. Cost he purchased = (0.5x)/3...Where x is the number of pounds....
Cost he sold = x/4
Profit => x/6 - x/4 = 10
.....................x=120

3. Consider there are 100 toys in total
So we have 25 Red toys and 75 Blue Toys
.... 50 of which are Size A and 50 Size B

If 10 out of 100 are Red and Size A ...
remaining 40 (50-10) of Size A are Blue....
from this we have (75-40) 35 blue toys remaining which are of Size B....
So the answer is 35...

4. From the first sentence
(S+5)(T+1) = ST+70
from (2) we have
(S+10)(T+2) = ST+P ...where P is the more miles would he have covered

From 1 & 2 we get P = 150..

6. PR^2 + RS^2 = 16
PR^2 + QR^2 = 9
QR+QS = 5 ....So QR = 5-QS

Solving the above equations we get QR = 9/5...
By using QR...we get PR = 12/5

jpmadhav · July 3, 2008, 7:19am

For 2nd Question
--------------------------
AB=BC=CD=DA= r ..This becomes a rhombus...
Take centre of BD as O....

AC^2 = AD^2 + CD^2 = r^2 + r^2 = 2* (r^2)

From this...... AC=root(2) * r...AO becomes (root(2)*r)/2 = r/root(2)...

Similarly..BO = r/root(2)....

Tan(x) = AO/BO = 1..x = Tan Inverse of(1) = 45degrees

For Question 5
-----------------------------
Rectangle Area = 30*40 = 1200

Semi-circle area excluding shaded area = (Outer Circle Area - Inner Circle Area)
==> 20*20*pi - 10*10*pi = 300pi

Total Area = 1200+300pi....

getneonow · July 3, 2008, 7:37am

reddy.harikisho Says
I think we can solve this very simply. if n is multiple of 5 then n^2 (n square) would be multiple of 25.therefore, n^2= (p^2q)^2, which is answer D.

dude..option D given is p^2 * q^2 but not (p^2q)^2 ..

Neo

msd_2008 · July 3, 2008, 7:44am

Hello,

Can someone help me answer the following questions on data sufficiency?
Please also provide explanation on how you arrived at the answer.

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

Regards
MSD

msd_2008 · July 3, 2008, 7:47am

Hello,

Please help me with the following questions (with step by step explanation).

1. John and Mary were each paid x dollars in advance to do a certain job together. John worked on the job for 10 hrs and Mary worked for 2 hours less than him. If Mary gave John y dollars of her payment so that they receive the same hourly wage, what was the dollar amount, in terms of y, that John was paid in advance?

a. 4y b.5y c.6y d.8y e.9y

2. A certain junior class has 1000students and a senior class has 800 students. Among these students, there are 60 sibling pairs, each consisting of 1 junior, 1 senior. If 1 student is to be selected at random from each class, what is the probability that the 2 students selected will be a sibling pair?

a. 3/40,000 b.1/3,600 c.9/2,000 d.1/60 e.1/15

Thanks
MSD

bunty4billi · July 3, 2008, 8:26am

Ashish ..for 3rd q..with table u can solve quickly...

Red Blue | %
_________________
Size A 10 | 40 50
--------------------
Size B | ? 50
--------------------
% 25 |75 100

so, Blue & Size B = 75-40 = 35

|ashishjha100 Says
Pls help with reasoning...........

inder.14 · July 3, 2008, 9:54am

Hello,

Can someone help me answer the following questions on data sufficiency?
Please also provide explanation on how you arrived at the answer.

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

Regards
MSD

for the first question

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

n is the number of defective bulbs and nlets take first statement
1) P=total defective bulbs/Total number of bulbs
p=nC2/10C2=1/15(given in first statement)
n!*8!/(n-2)!*10! =1/15
n*(n-1)/10*9=1/15
n(n-1)=6
and on solving we will get n=3 and n=-2
hence it can be answered by first statement

Now lets take second statement
2) nC1 *10-nC1/10C2=7/15
on solving this we will get
n(10-n)=21
=>n=7,3 and as n n=3
it can be answered by second statemnet as well

so answere will be D i.e it can be answered 1 as well as second statement indipendently

Some please correct m eif i am wrong

inder.14 · July 3, 2008, 10:00am

Hello,

Can someone help me answer the following questions on data sufficiency?
Please also provide explanation on how you arrived at the answer.

Q.1 A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1)The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Q2. If 2 different representatives are to be selected at random from a group of 10 employees and if p is the probability that both representatives selected will be women, is p > 1/2?
(1)More than 1/2 of the 10 employees are women.
(2)The probability that both representatives selected will be men is less than 1/10

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

Regards
MSD

For third question

Q3. What is the probability that a student randomly selected from a class of 60 students will be a male who has brown hair?
(1)One-half of the students have brown hair.
(2)One-third of the students are males.

by using statement one and two independently we can not answer this question

lets try with bot the statemnets
Student with brown hair = 30 (from statement 1)
male student = 20(from statement 2)

probability of male student with brown hair = p1(probability of stdent is male)*p2(probability of stdent is having brown hair)
=20/60*30/60=1/3*1/2=1/6

hence it can be answered by using both the statement i.e C

correct me if i am wrong