Hi..the simple formula for finding # of terms in an exp (a1+a2+a3....an)^m like this is:
n+(m-1))!
---------- = "(n+m-1) choose (m-1)"
(m!)(n-1)!
Hope this helps!!
Vikram
http://itsnotaboutthebook.blogspot.com/
It should be E
nairpraveenk SaysI guess 'A' is the answer.
I think it is D. The use of 'like' is a hint.Wotsay?
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.
Answer is (E) because there can be many combinations with the two statements.
Yes, answer is E.
But can you please post question in their respective threads? We have a seperate thread for DS questions:
http://www.pagalguy.com/discussions/gmat-data-sufficiency-discussions-25020702
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.
Answer is (E) because there can be many combinations with the two statements.
Hi Guys,
try this one out.
A batch of cookies was divided among 3 tins. 2/3 of all cookies were placed in either the blue tin or green tin and the rest were placed in the red tin.If 1/4 of all the cookies were placed in the blue tin what fraction of the cookies that were placed in the other tins were placed in the green tin.
a) 15/2
b) 9/4
c) 5/9
d) 7/5
e) 9/7
Hi Guys,
try this one out.
A batch of cookies was divided among 3 tins. 2/3 of all cookies were placed in either the blue tin or green tin and the rest were placed in the red tin.If 1/4 of all the cookies were placed in the blue tin what fraction of the cookies that were placed in the other tins were placed in the green tin.
a) 15/2
b) 9/4
c) 5/9
d) 7/5
e) 9/7
Should be 5/9.
yep thats correct
I guess it is 5/7... how did u get 5/9?
vasus_magic SaysI guess it is 5/7... how did u get 5/9?
@ vasus_magic,
The answer is 5/9. Here is the solution:
Total Cookies -> C
Cookies in Blue Tin + Cookies in green Tin -> (2/3)C
Cookies in Red Tin -> C - (2/3) C = C/3
Cookies in Blue Tin = C/4
Therefore, Cookies in Green Tin = (2/3)C - C/4 = (5/12)C
Cookies in Green Tin as a fraction of the Cookies in Red + Green Tins (other than Blue Tin) = (5/12)C / (3/4)C = 5/9
@ vasus_magic,
The answer is 5/9. Here is the solution:
Total Cookies -> C
Cookies in Blue Tin + Cookies in green Tin -> (2/3)C
Cookies in Red Tin -> C - (2/3) C = C/3
Cookies in Blue Tin = C/4
Therefore, Cookies in Green Tin = (2/3)C - C/4 = (5/12)C
Cookies in Green Tin as a fraction of the Cookies in Red + Green Tins (other than Blue Tin) = (5/12)C / (3/4)C = 5/9
wooo... I misinterpreted the question.. I attributed 'other tins' to 'green tin'
so I arrived at (5C/12)/(C/3 + C/4)=5/7
Thank you very much...
guy with guts SaysYour approach is correct.However the mistake was -12+-4=-16 instead of -8
This is pretty simple actually... the original equation can be arrived at (x+2)^2/(x-2)^2 = 3
=> x+2/x-2 = 3
=> x=4
Hi Guys
1 more problem
Fence x is twice as long as fence y , and fence y is 2 feet shorter than fence z .If 3 feet were added to each fence which of the following must be true
I x is twice as long as y
II y is 2 feet shorter than z
III x is longer than z
a I only
b II only
c III only
d I and II
e II and III
not sure if the answer if c or e or is the problem missing something
Hi Guys
1 more problem
Fence x is twice as long as fence y , and fence y is 2 feet shorter than fence z .If 3 feet were added to each fence which of the following must be true
I x is twice as long as y
II y is 2 feet shorter than z
III x is longer than z
a I only
b II only
c III only
d I and II
e II and III
not sure if the answer if c or e or is the problem missing something
The answer is b.
Suppose y= L
then x = 2L and z = L+2
Now 3 feet added to all
y = L+3, x = 2L +3 and z = L+5
Now from I ->
y = L+3 x= 2L+3 NOT TRUE
II ->
Z-y=2
L+5 - (L+3) = L+5-L-3=2 TRUE
III ->
check x>z
x= 2L+3
z= L+5
For L=5
x= 13 and z= 10 , x > z
For L=1
x= 5 and z= 6 , x
So, its not always true.
Hence the answer is b. II
Also wanted to about some good books on probability and speed/distance . I had done it in 12th but have pretty muc forgotten abt it
Heres 1 more .Try it out
During a season in a certain baseball league every team plays every other team in the league 10 times. If there are 10 teams in the league, how many games are played in the league in one season?
a 45
b 90
c 450
d 900
e 1000
Heres 1 more .Try it out
During a season in a certain baseball league every team plays every other team in the league 10 times. If there are 10 teams in the league, how many games are played in the league in one season?
a 45
b 90
c 450
d 900
e 1000
Total Teams = 10
Total opponents each team have = 9
Each team play opponent = 10 times.
Each team play total matches = 9x10 = 90
total matches = number of teams x total matches each team play
= 10x90 = 900
So answer should be D, whats the OA?
answer is B because
Each of 10 teams is playing the 9 other teams 10 times each; but each baseball game involves 2 teams. So 
the total number of games needed.
sorry C = 450
cgt2k1 Sayssorry C = 450
Oh yes, I missed that one..my bad !!
its 900/2 =450