GMAT Problem Solving Discussions

MBA
2781
hi,
one eighteenth of a kilogram means 1000/18 grams...
1 pound means 1000/2.2 grams
so number of eighteenths of a kilogram are in one pound
= (1000/2.2)/(1000/18 )
= 18 / 2.2
= 8.1818 (8 approx)
hope i am correct...

An Empty Metal box weighs ten percent of its total weight when filled with paint. If the weight of a partly filled box is one half that of a completely filled box, what fraction of the box is filled ?

A) 3/5
B) 1/5
C) 1/2
D) 4/9
E) 2/5

Answer in Bold :biggrin:

let total weight (metal box + paint) is 100%
weight of metal box is 1000/10 = 10%
so weight of paint = 90%gms

in other case, the weight of the partly filled box = 1/2 * 100% = 50% gms
the paint in this case = 40% gms
fraction of the box that is filled is = 40/90 = 4/9


P.S : Thanks raghav for the format :biggrin:


Indeed !! Raghav, @nee! - totally on the mark.

The OAs are : 8 and 4/9 respectively !
2782
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?


I know the answer.. and also the means to calculate the same.. this jlt ques ;)


cHANCES OF HEAD EVERY TIME=1/8
CHANCES OF TAIL =1/8
SO EITHER HEAD OR TAIL=2/8=1/4
2783
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?

I know the answer.. and also the means to calculate the same.. this jlt ques ;)


Hi,

For either side 1/8.
Same side either one is 1/4
2784
When tossed, a certain coin has equal probability of landing on either side. If the coin is tossed 3 times, what is the probability that it will land on the same side each time?

I know the answer.. and also the means to calculate the same.. this jlt ques ;)

Okay... My approach is different.

Remember: we DONT care for 3 tosses...but only 2 tosses.who cares what the first toss is we just have to match the second and third to it.

Lets say first comes H/T.

Now ,probability of second to be same is 1/2 and third to be same is also 1/2....hence 1/2*1/2 = 1/4.
2785

Need help...
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

2786
Need help...
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160


my ans is in bold.
2787

Hey Vyomb... need little expln..

2788
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160

explanation

v-av speed
d-total distance
t-time
x-more distance

d=vt--eqn1

from condition 1
70+d=(v+5)(t+1)--eqn2

we get 65=v+5t
or 130=2v+10t

from condition 2
x+d=(t+2)(v+10)--eqn3

x=2v+10t+20
x=130+20
x=150
2789
Need help...
If a motorist had driven 1 hour longer on a certain day and at an average rate of 5 miles per hour faster, he would have covered 70 more miles than he actually did. How many more miles would he have covered than he actually did if he had driven 2 hours longer and at an average rate of 10 miles per hour faster on that day?
(A) 100 (B) 120 (C) 140
(D) 150 (E) 160



extra time=1 hour
extra distance=70 miles
so speed=70
this speed is 5 more than normal so normal speed=70-5=65

so for two hours and 10 miles faster,distance=2*(65+10)=2*75=150
2790
extra time=1 hour
extra distance=70 miles
so speed=70
this speed is 5 more than normal so normal speed=70-5=65

so for two hours and 10 miles faster,distance=2*(65+10)=2*75=150

Guys with Guts..I solved this in same fashion.
Answer should be 150.
2791
extra time=1 hour
extra distance=70 miles
so speed=70
this speed is 5 more than normal so normal speed=70-5=65

so for two hours and 10 miles faster,distance=2*(65+10)=2*75=150


Kool, this is really the way to kill these crappy problems...nice job man
2792
extra time=1 hour
extra distance=70 miles
so speed=70
this speed is 5 more than normal so normal speed=70-5=65

so for two hours and 10 miles faster,distance=2*(65+10)=2*75=150


Smart Approach, Liked it..
2793
Hi All,

I was going through some article on permutations and combination and the question below confused me a bit...

There are 8 people from class A and 7 from class B. Assume they are required to sit in a row. In how many ways can the two sets be seated if no two from the same class sit near one another? Please help me with some explanation.

Thanks
2794
Hi All,
I was going through some article on permutations and combination and the question below confused me a bit...
There are 8 people from class A and 7 from class B. Assume they are required to sit in a row. In how many ways can the two sets be seated if no two from the same class sit near one another? Please help me with some explanation.
Thanks


draw a 15 open space

_ _ _ _ _ _ _ _ _ _ _ _ _ _
A B A B A B A B A B A B A B A
Start fiiling these places

1) We can't start with the class who has 7 person else the last 2 seat will have repeated numbers from people of class 8, fill and SEE

2) So we'll have start with class of 8 people

answer shall be 8! * 7!

Cheers
2795

Q How many numbers beloe 200 have exactly 6 devisors?

Q A number has got only 12 factors and it consist of 3 prime factors. the sum of these prime factors is 20, find the minimum possible number??

2796

Q. Ho many integers are divisible by 3 between 10! and 10!+20 inclusive?????

A. 6
B. 7
C. 8
D. 9
E. 10

???????????????????????????????????????????????????




*confused !

2797
Hi All,

I was going through some article on permutations and combination and the question below confused me a bit...

There are 8 people from class A and 7 from class B. Assume they are required to sit in a row. In how many ways can the two sets be seated if no two from the same class sit near one another? Please help me with some explanation.

Thanks

Always start from the smaller number.

7 Class B students can be seated in 7 ! ways.

Now when these 7 are seated , there will be 8 spaces between them.

_B_B_B_B_B_B_B_

8 students from Class A can be seated now in 8! ways.

Answer : 7!*8!
2798


Q A number has got only 12 factors and it consist of 3 prime factors. the sum of these prime factors is 20, find the minimum possible number??


My ans is 308

3 prime factors add upto 20 means the factors can be 2,7,11 = 20
so rest 9 factors are composite.

this can be derived as (2^2)*7*11 = 308
and the factors will be 1,2,4,7,11,14,22,28,44,77,154,308
2799
Q. Ho many integers are divisible by 3 between 10! and 10!+20 inclusive?????

A. 6
B. 7
C. 8
D. 9
E. 10



Is the ans for the above q is option B)7

the nos will 10!
10!+1
10!+2.......10!+20

and the nos divisible will be
10!
10!+3
10!+6......10!+18

So total no of nos is 7
2800


Q A number has got only 12 factors and it consist of 3 prime factors. the sum of these prime factors is 20, find the minimum possible number??

My ans is 308

3 prime factors add upto 20 means the factors can be 2,7,11 = 20
so rest 9 factors are composite.

this can be derived as (2^2)*7*11 = 308
and the factors will be 1,2,4,7,11,14,22,28,44,77,154,308


for a given sum, product is min when they as apart as poss.

Prime factors should be 2,5,13
No. Should be 2^2 * 5 * 13 = 260 ..

Have done it in a hurry ..pls correct fr errors ..