GMAT Problem Solving Discussions

MBA
1700
Consider a parallelogram ABCD , let AB be the base and CD be the side parallel to AB .

Drop a perpendicular from D to AB and name it X......and the height of this perprndicular is 3(=18/6)

So triangle AXD is a right triangle and we know AD=6(side of parallelogram) and DX=3(ht of parallelo)
sin A = DX/AD=3/6=1/2.......... so A = sin -1(1/2)=30 degree

Hope this helps...........



Thanks....I took the side as 9 instead of 6 to calculate sine !
Thats the kinda mistakes I make & it pisses me off....
1701
A string of 10 lightbulbs is wired in such a way that if any individual lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during time period T is 0.06, what is the probability that the string of lightbulbs will fail during time period T?
A.0.06
B.(0.06)^10
C.1-(0.06)^10
D.(0.94)^10
E.1-(0.94)^10
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216
1702
A string of 10 lightbulbs is wired in such a way that if any individual lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during time period T is 0.06, what is the probability that the string of lightbulbs will fail during time period T?
A.0.06
B.(0.06)^10
C.1-(0.06)^10
D.(0.94)^10
E.1-(0.94)^10
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216

My picks are E & D.
1703
A string of 10 lightbulbs is wired in such a way that if any individual lightbulb fails, the entire string fails. If for each individual lightbulb the probability of failing during time period T is 0.06, what is the probability that the string of lightbulbs will fail during time period T?
A.0.06
B.(0.06)^10
C.1-(0.06)^10
D.(0.94)^10
E.1-(0.94)^10
Of the three-digit positive integers that have no digits equal to zero, how many have two digits that are equal to each other and the remaining digit different from the other two?
A. 24
B. 36
C. 72
D. 144
E. 216


My take would be E & E

1. P(at least one bulb to fail) = 1 - P(all bulbs fail)
= 1 - (1-0.06) (1-0.06)...10 times
= 1-(0.94)^10

2. the numbers are : 1,2,3,4,5,6,7,8,9
formations: aba aab baa
no.of ways to pick 'b' is 9 ways, and 'a' can be picked in 8 ways
= 9* 8 = 72 ways.. and we have three ways so 72 * 3 = 216

What are the OA's?
1704
My take would be E & E

1. P(at least one bulb to fail) = 1 - P(all bulbs fail)
= 1 - (1-0.06) (1-0.06)...10 times
= 1-(0.94)^10

2. the numbers are : 1,2,3,4,5,6,7,8,9
formations: aba aab baa
no.of ways to pick 'b' is 9 ways, and 'a' can be picked in 8 ways
= 9* 8 = 72 ways.. and we have three ways so 72 * 3 = 216

What are the OA's?


Thanks a lot. OA's are E & E.
1705

Hey guys,

One more question on probablity, I am not good at it 😞 so please give me the explanation with ur answer.

A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
A) 0
B) 1/9
C) 2/9
D) 1/3
E) 1

1706
Hey guys,

One more question on probablity, I am not good at it 😞 so please give me the explanation with ur answer.

A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
A) 0
B) 1/9
C) 2/9
D) 1/3
E) 1

My pick is C

when the passenger comes first time, prob to get into any of the car=3C1/3C1=1
when the passenger comes second time, prob to get into any of other 2 car= 2C1/3C1=2/3
when the passenger comes third time, prob to get third car=1/3C1=1/3

Total prob=1*2/3*1/3=2/9
1707

If (10^50) - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A) 424
B) 433
C) 440
D) 449
E) 467

1708
If (10^50) - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A) 424
B) 433
C) 440
D) 449
E) 467

My pick would be C......
1709
If (10^50) - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A) 424
B) 433
C) 440
D) 449
E) 467


My pick is c)
1710
My take would be E & E

1. P(at least one bulb to fail) = 1 - P(all bulbs fail)
= 1 - (1-0.06) (1-0.06)...10 times
= 1-(0.94)^10

2. the numbers are : 1,2,3,4,5,6,7,8,9
formations: aba aab baa
no.of ways to pick 'b' is 9 ways, and 'a' can be picked in 8 ways
= 9* 8 = 72 ways.. and we have three ways so 72 * 3 = 216

What are the OA's?

why is the probability of all bulbs fail 1-0.06 and not just 0.06?
1711
Hey guys,

One more question on probablity, I am not good at it 😞 so please give me the explanation with ur answer.

A certain roller coaster has 3 cars, and a passenger is equally likely to ride in any 1 of the 3 cars each time that passenger rides the roller coaster. If a certain passenger is to ride the roller coaster 3 times, what is the probability that the passenger will ride in each of the 3 cars?
A) 0
B) 1/9
C) 2/9
D) 1/3
E) 1


My ans is 1/9
The first time he goes he can pick any one of the 3 cars probability is 1
2nd time he has to pick the same car so prob is 1/3 and the same with the 3rd time ...
so 1*1/3*1/3 = 1/9

what is the OA?
1712
My ans is 1/9
The first time he goes he can pick any one of the 3 cars probability is 1
2nd time he has to pick the same car so prob is 1/3 and the same with the 3rd time ...
so 1*1/3*1/3 = 1/9

what is the OA?


OA is C - 2/9
1713

Ashish and Italian, you are right. OA is C. Can u pls explain?

1714
If (10^50) - 74 is written as an integer in base 10 notation, what is the sum of the digits in that integer?
A) 424
B) 433
C) 440
D) 449
E) 467

10^50 contains 1 with 50 zeros to right....... 10^50-74 will consit of 48 nines followed by 26.......so total 48*9+2+6=440.
1715
Simmy Says
Ashish and Italian, you are right. OA is C. Can u pls explain?


Sorry read the question wrong. It is indeed 2/9

label the cars ABC
Number of ways in which a, b , c cal appear with repeatition
1st time 3
2nd time 3
3rd time 3

-----------------------------
Number of time abc,bca, etc occur is factorial 3 = 6
ans 6/27

2/9

Read the question as drives the same car each time last time around **Blush***
1716

Hi,

Attached is a doc having 3 questions. Last question i could not make head and tail of it 😞

1717
The sum of all the integers k such that -26
A. 0
B. -2
C. -25
D. -49
E. -51
I got it as D.

k={-25,-24,..........,23}

no: of terms=49
sum=(-25+23)*49/2= -49( sum of 49 consecutive integers from -25 to 23)
1718
Hi,

Attached is a doc having 3 questions. Last question i could not make head and tail of it :(

2nd question i got ans as B (not sure).if its the OA i'll give reasoning......
1719
ashishjha100 Says
2nd question i got ans as B (not sure).if its the OA i'll give reasoning......


OA's are C, B, D. So now you can give reasoning for 2nd question. Thanks in advance.