GMAT Problem Solving Discussions

MBA
2096
Hi guys, it feels like ages when i started preparing for the GMAT and today I stand nowhere, you would get to know this by the level of problems I am posted below. However, I have thought that this time when I give the GMAT I really wanna focus on only my weak areas , so here they are some of them ;)

I really hope some angel would help me with them. I know the answer to many of them but I dont know the correct reasoning and explanation behind them so if someone can help me with it, it would be great !

Q1. If ba. x > -3
b. x c. x=3
d. x e. x > 3

lets take b=1 , so x=3/2
So C

Q2. A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^4, what was the number in the population 1 hour later ?
a. 2(10^4)
b. 6(10^4)
c. (2^6)(10^4)
d. (10^6)(10^4)
e. (10^4)^6

first 10 minute bacteria=2*10^4
second =2*2*10^4
third.........................=8*10^4
Fourth.......................=16*10^4
fifth...........................=32*10^4
sixth...........................=64*10^4

so C

Q3. A certain clock marks every hour by striking a no of times equal to the hour, and the time required for a stroke is exactly equal to the time interval between the strokes. At 6:00 the time lapse between the beginning of the first stroke and the end of the last stroke is 22 seconds. At 12:00, how many seconds elapse between the beginning of the first stroke and the end of the last stroke ?
a. 72
b. 50
c. 48
d. 46
e. 44

at 6 o clock, six strokes will be made-
1 stroke gap 2 stroke gap 3stroke..gap 4 stroke gap 5stroke gap sixth stroke.

so total six strokes +five gaps
total time for one stroke is say X
so total time=11x=22=>x=2

now for 12 O clock
12 strokes+11 gaps is equivalent to 23 strokes
so 23*2=46 seconds

so D

Q4. The addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3 and 4 exactly once in each integer. What is the sum of these 24 integers ?

1,234
1,243
1,324
....
....
+ 4,321
--------

a. 24,000
b. 26,664
c. 40,440
d. 60,000
e. 66,660

The chances of each digit appearing at the last didgit is equal so the sum of 1+2+3+4=10 will appear six times so last disgit sum=6*10=60 means zero and 6 carryover
similiar holds for 10,100,1000th digit
so sum will be
0
60
660
66660
so E


Q5. Four hours from now, the population of the a colony of bacteria will reach 1.28x10^6. If the population of the colony doubles every 4 hours, what was the population 12 hours ago ?
a. 6.4 x 10^2
b. 8.0 x 10^4
c. 1.6 x 10^5
d. 3.2 x 10^5
e. 8.0 x 10^6

B

Q6. At a certain pizzeria, 1/8 of the pizzas sold in one week were mushroom and 1/3 of the remaining pizzas sold were pepperoni. If n of the pizzas sold were pepperoni. How many were mushroom ?
a. 3/8n
b. 3/7n
c. 7/16n
d. 7/8n
e. 3n

Let total pizzas sold were 24
so 1/8 of 24=3 and balance of 24-3=21
and 1/3 of 21 =7=N
so if 7=n, then 3 is equal to 3n/7

so B


Q7. If it is true that x > -2 and x a. x > 2
b. x > -7
c. x d. -7 e. None of the above

B

Q.8. A rectangular box is 10 inches wide, 10 inches long and 5 inches high. What is the greatest possible (straight-line) distance, in inches, between any two points on the box ?
a. 15
b. 20
c. 25
d. 10 (2)^1/2
e. 10 (3)^1/2

diagonal= sqrt(100+100+25+=15

So A
Many Thanks for all your help and support.



Regards

Vikas
2097

From january 1 1991 to january 1 1993, the number of people enrolled in health maintainance organization increased by 15 percent. the enrollment on january 1 1993 was 45 million. how many million people to the nearest million, were enrolled in health maintainance organizations on january 1 1991?

A. 38
B. 39
c. 40
d. 41
E. 42

My main query here is: how does 15% become 1.15 i just dont get...

Someone please explain...

Thanks in advance.

2098
Hi Puys , i encountered the below mentioned questions while doing the earlier paper format tests and was really stuck in these-
Plz see the attachement.

Hi,Any taker for thiese probs?
2099
Hi Puys , i encountered the below mentioned questions while doing the earlier paper format tests and was really stuck in these-
Plz see the attachement.


Hey,

Don't think too deep while solving these problems, just keep it real.

I can explain you if still face problem in solving this, I can post the explanation.

Let me know.

Thx.
2100
Hey,

Don't think too deep while solving these problems, just keep it real.

I can explain you if still face problem in solving this, I can post the explanation.

Let me know.

Thx.

Hi Vikas ,
Though I solved the first one and was tsuck in second.However my approach is lengthy.So will request you to post explanation.

Regards
2101
From january 1 1991 to january 1 1993, the number of people enrolled in health maintainance organization increased by 15 percent. the enrollment on january 1 1993 was 45 million. how many million people to the nearest million, were enrolled in health maintainance organizations on january 1 1991?

A. 38
B. 39
c. 40
d. 41
E. 42

My main query here is: how does 15% become 1.15 i just dont get...

Someone please explain...

Thanks in advance.


let those enrolled in 1991 be x

x+15x/100=45

1.15x=45

x=900/23=39.1 million.

nearest is option (B)
2102
Hi Puys , i encountered the below mentioned questions while doing the earlier paper format tests and was really stuck in these-
Plz see the attachement.


1)let length of the inner rectangle be x inches.

breadth is 15x/18=5x/6 inches

area=5x^2/6

but,

2*5x^2/6=18*15

x=9(sqrt 2)

option (A)

2)10 ways.just try all the ways of doing it.its easy.

a formula,not sure,have seen it somewhere:

(h+v-2)C(h-1)

h=number of horizontal lines.
v=number of vertical lines.

here,

(4+3-2)C(4-1)=5C3=10
2103
Hi guys, it feels like ages when i started preparing for the GMAT and today I stand nowhere, you would get to know this by the level of problems I am posted below. However, I have thought that this time when I give the GMAT I really wanna focus on only my weak areas , so here they are some of them ;)

I really hope some angel would help me with them. I know the answer to many of them but I dont know the correct reasoning and explanation behind them so if someone can help me with it, it would be great !

Q1. If ba. x > -3
b. x c. x=3
d. x e. x > 3

Q2. A certain population of bacteria doubles every 10 minutes. If the number of bacteria in the population initially was 10^4, what was the number in the population 1 hour later ?
a. 2(10^4)
b. 6(10^4)
c. (2^6)(10^4)
d. (10^6)(10^4)
e. (10^4)^6

Q4. The addition problem below shows four of the 24 different integers that can be formed by using each of the digits 1,2,3 and 4 exactly once in each integer. What is the sum of these 24 integers ?

1,234
1,243
1,324
....
....
+ 4,321
--------

a. 24,000
b. 26,664
c. 40,440
d. 60,000
e. 66,660

Q6. At a certain pizzeria, 1/8 of the pizzas sold in one week were mushroom and 1/3 of the remaining pizzas sold were pepperoni. If n of the pizzas sold were pepperoni. How many were mushroom ?
a. 3/8n
b. 3/7n
c. 7/16n
d. 7/8n
e. 3n

.


Ans for the above set is

1. x2. (2^6)(10^4)
4.Did find the option among the ans (33,330)
6.3n/7

Please correct me if I'm wrong.

-Deepak.
2104

H(n) is defined as the product of all even numbers from 2 to n, inclusive.If P is the smallest prime factor of H(100)+1 , then P is
a.between 2-10
b.between 10-20
c..............20-30
d..............30-40
e >40

Hi Puys , this is one of the question from GMATprep quants.

2105
H(n) is defined as the product of all even numbers from 2 to n, inclusive.If P is the smallest prime factor of H(100)+1 , then P is
a.between 2-10
b.between 10-20
c..............20-30
d..............30-40
e >40

Hi Puys , this is one of the question from GMATprep quants.


I think # has to be greater than 49.

H(100) = 2*4*6*8*10*.....*100
=> H(100) = 2^50(1*2*3*.....*50)
=> H(100) = 2^50(50!)

if you add 1 to this then all of the #s less than 50 are prime factors of H(100). if you add 1 to H(100)+1 isn't divisible by any of these numbers.

e.g. for a number N to be divisible by P it has to be in following form. n is divisible by P

N = n+ap (where a is an integer)

that's only thing came to mind. If anybody has better way to do it, pls post your explanation.
2106
Ans for the above set is

1. x2. (2^6)(10^4)
4.Did find the option among the ans (33,330)
6.3n/7

Please correct me if I'm wrong.

-Deepak.


1. D

b 2x=3b => x=3b/2 now if b is less than 2 there's no way x can be greater than 3. Hence x
2. C

0th sec (intial count) -> 10^4
after 10 secs -> 2 X 10^4
after 20 secs -> 2X2X10^4
after 30 secs -> 2X2X2X10^4
after 40 secs -> 2X2X2X2X10^4
after 50 secs -> 2X2X2X2X10^4
after 60 secs -> 2X2X2X2X10^4 => 2^6 X 10^4

3. e

now you don't do calculate exact sum. there are 24 such numbers. so there are going to be 6 numbers which have 1 at 1000s digit, 6 numbers with 2 at 1000s digit and so on.

so 1000*6+2000*6+3000*6+4000*6 = 60000

we hacen't added values of digits at 100s places.

100*6+200*6+300*6+400*6 = 6000

same for 10s

10*6+20*6+30*6+40*6 = 600

and for unit digits

1*6+2*6+3*6+4*6 = 60

add them up and you get your number i.e. 66,660

Obviously you don't need to calculate beyond thousands, since you know answer is going to be 60000+ only D is possible answer.

Q6. C

otal pizzas sold= x
mushroom Pizzas => x/8
remaining => 7x/8
pepperoni => 7x/8*3 i.e. 7x/24 = n so x= 24n/7

hence mushroom Pizzas = 3n/7
2107
I think # has to be greater than 49.

H(100) = 2*4*6*8*10*.....*100
=> H(100) = 2^50(1*2*3*.....*50)
=> H(100) = 2^50(50!)

if you add 1 to this then all of the #s less than 50 are prime factors of H(100). if you add 1 to H(100)+1 isn't divisible by any of these numbers.

e.g. for a number N to be divisible by P it has to be in following form. n is divisible by P

N = n+ap (where a is an integer)

that's only thing came to mind. If anybody has better way to do it, pls post your explanation.



After arriving at 2^50(50!) + 1,jus check for some of the factors for small numbers.

take 2^2(2!)+1 = 9,the prime factor will be 3.
Take 2^3(3!)+1 = 49,the prime factor will be 7
Take 2^4(4!)+1 = 385,the prime factor will be 5.

So the ans for the above prblm can be 1.between 2 and 10.

Any thoughts on the above approach?

-Deepak.
2108
After arriving at 2^50(50!) + 1,jus check for some of the factors for small numbers.

take 2^2(2!)+1 = 9,the prime factor will be 3.
Take 2^3(3!)+1 = 49,the prime factor will be 7
Take 2^4(4!)+1 = 385,the prime factor will be 5.

So the ans for the above prblm can be 1.between 2 and 10.

Any thoughts on the above approach?

-Deepak.


2^2(2!)+1 = 9 prime factor is greater than 2
2^3(3!)+1 = 49 prime factor is greater than 3
2^4(4!)+1 = 385 prime factor is greater than 5

so it follows that prime factor of 2^50(50!) + 1 has to be greater than 50.

Hope this helps.
2109
After arriving at 2^50(50!) + 1,jus check for some of the factors for small numbers.

take 2^2(2!)+1 = 9,the prime factor will be 3.
Take 2^3(3!)+1 = 49,the prime factor will be 7
Take 2^4(4!)+1 = 385,the prime factor will be 5.

So the ans for the above prblm can be 1.between 2 and 10.

Any thoughts on the above approach?

-Deepak.


but if u check for 5!

2^5(5!)+1 = 3841 then smallest prime factor for 3841 is 23

so ans is option E
2110

the ans is obviously E as we dnt knw abt cd

A----------B---------C----------D


Is
CD > BC ?
(1) AD = 20
(2) AB = CD


A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH Statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.


2111
Q. Six mobsters have arrived at the theater for the premiere of the film Goodbuddies. One of the mobsters, Frankie, is an informer, and he's afraid that another member of his crew, Joey, is on to him. Frankie, wanting to keep Joey in his sights, insists upon standing behind Joey in line at the concession stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Frankies requirement is satisfied?
(A) 6
(B) 24
(C) 120
(D)360
(E)720

Hi Pguys,

Could you help with the above problem?

Thanks,
Rohit.

P = -2(S 4)2 + 32
friend this bold part is correct?

because -2 before bracket and 2 after?

Two questions :

The number of passengers on a certain bus at any given time is
given by the equation P = -2(S 4)2 + 32, where P is the number of
passengers and S is the number of stops the bus has made since
beginning its route. If the bus begins its route with no passengers,
how many passengers will be on the bus two stops after the stop
where it has its greatest number of passengers?


1. 32
2. 30
3. 24
4. 14
5. 0
In a room filled with 7 people, 4 people have exactly 1 friend in
the room and 3 people have exactly 2 friends in the room. If two
individuals are selected from the room at random, what is the
probability that those two individuals are NOT friends?

5/21
3/7
4/7
5/7
16/21


answer to first question (as it is given here), should be A:32

for second it should be C: 4/21

let me know if these are correct, I can post explanation.

Could help with the delayed response. I was just going through the old posts and saw that nobody replied to this, so thought of posting.

Thx.
2112

I am getting the wrong answer for this question...please see if u can solve ..please post the method u followed

A certain law firm of 4 senior partners and 6 junior partners.How many different groups of 3 partners can be formed in which at least one member of the group is senior partner ? (Two groups are considered different if at least one group member is different.

1) 48
2) 100
3) 120
4) 288
5) 600

2113
I am getting the wrong answer for this question...please see if u can solve ..please post the method u followed

A certain law firm of 4 senior partners and 6 junior partners.How many different groups of 3 partners can be formed in which at least one member of the group is senior partner ? (Two groups are considered different if at least one group member is different.

1) 48
2) 100
3) 120
4) 288
5) 600


Ans :100

Total groups of 3 partners with atleast 1 senior member = total possible groups of 3 partners - total groups of 3 members comprising of all junior partners

i.e 10C3 - 6C3
= 120 -20
= 100 ....Ans

Alternatively,

Total possible grps = selection of
(1 senior partner & 2 junior partners) OR ( 2 senior partners & 1 junior partner ) OR ( all 3 senior partners )

i.e (4C1 * 6C2) + (4C2 * 6C1) + (4C3)
= 60 + 36 + 4
=100 ...Ans
2114

100 is 100% Correct !!

Thanks!!

2115

Please solve this one....Please give the solution also...



Two couples and one single person are seated at random in a row of five chairs. What is the probability that neither of the couples sits together in adjacent chairs?



1) 1/5
2) 1/4
3) 3/8
4) 2/5
5) 1/2