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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Nov 24, 2009 5:19 pm

At a high school assembly, the principal plans to give out 3 awards for extraordinary achievement. The nominees for the awards are the 3 most highly rated teachers and the 5 students with the most improved GPAs. How many different groups of award recipients can be chosen in which at least one recipient is a teacher? (Two groups of award recipients are considered different if at least one group member is different.)

(A) 20
(B) 30
(C) 46
(D) 60
(E) 120

OA is C

Last edited by Abdulla on Tue Nov 24, 2009 6:42 pm, edited 1 time in total.

Abdulla

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Source: Beat The GMAT — Problem Solving | Tue Nov 24, 2009 5:19 pm

heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Nov 24, 2009 5:25 pm

are you sure the question is not missing information ? How many student and teachers are there ?

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Nov 24, 2009 6:42 pm

heshamelaziry wrote:are you sure the question is not missing information ? How many student and teachers are there ?

I am so sorry ... typo.. check it now

Abdulla

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Nov 24, 2009 6:54 pm

whenever you see at least in these types of questions do the following (if my answer is correct):

1- total possible 3 groups of 3 out of 8 -----> 8!/3!(8-3)! = 56

2- total number of groups of 3 without any teacher-------> 5!/3!(5-3)! = 10

S0, 56-10 = 46

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Nov 24, 2009 7:40 pm

heshamelaziry wrote:whenever you see at least in these types of questions do the following (if my answer is correct):

1- total possible 3 groups of 3 out of 8 -----> 8!/3!(8-3)! = 56

2- total number of groups of 3 without any teacher-------> 5!/3!(5-3)! = 10

S0, 56-10 = 46

Cool man .. you're doing well now.. tell me what else do you know in Comb., Perm. ,and Prob?

Abdulla

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Nov 24, 2009 8:57 pm

With these 4 problems that show 3 concepts, this is all I know about per, prob, comb

Q) John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter ?

1/8

1/6

2/9

5/18

1/3
ANSWER: Total Players to choose from : 9

No of players to choose : 5

Restriction : Team with John and Peter

- - - - -

The order in which the team members are selected doesn't matter.

1) No of ways of selecting John : 1
2) No of ways of selecting Peter : 1

3) No of ways of selecting the remaining 7i / 3i( 7-3i) = 35

4) 1 *2*3 = TOTAL DESIRED OUTCOME = 35
5) TOTAL POSSIBLE OUTCOMES = 9i / 5i( 9-5i ) = 126
SOLUTION = 4 / 5 = 5/18

Q) 9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
The guards can be chosen in 6C3 ways.
The forwards can be chosen in 3C2 ways.

6C3 * 3C2

6C3 = 20
3C2 = 3

20 * 3 = 60

Q) "If you flip a coin five times, what's the probability you get exactly three heads ?
The answer can be found in ten seconds- it's 5C3/2^5 = 5/16

Q) A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

Answer:

We can treat this exactly as we would a coin flip question; think of it as:

"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"

There's an easy to use formula for coin flip (and pseudo-coin flip) questions:

Probability of getting exactly k results out of n flips = nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula.)

In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:

4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Nov 24, 2009 9:15 pm

heshamelaziry wrote:With these 4 problems that show 3 concepts, this is all I know about per, prob, comb

Q) John and Peter are among the nine players a basketball coach can choose from to field a five-player team. If all five players are chosen at random, what is the probability of choosing a team that includes John and Peter ?

1/8

1/6

2/9

5/18

1/3
ANSWER: Total Players to choose from : 9

No of players to choose : 5

Restriction : Team with John and Peter

- - - - -

The order in which the team members are selected doesn't matter.

1) No of ways of selecting John : 1
2) No of ways of selecting Peter : 1

3) No of ways of selecting the remaining 7i / 3i( 7-3i) = 35

4) 1 *2*3 = TOTAL DESIRED OUTCOME = 35
5) TOTAL POSSIBLE OUTCOMES = 9i / 5i( 9-5i ) = 126
SOLUTION = 4 / 5 = 5/18

Q) 9 basketball players are trying out to be on a newly formed basketball team. Of these players, 5 will be chosen for the team. If 6 of the players are guards and 3 of the players are forwards, how many different teams of 3 guards and 2 forwards can be chosen?
The guards can be chosen in 6C3 ways.
The forwards can be chosen in 3C2 ways.

6C3 * 3C2

6C3 = 20
3C2 = 3

20 * 3 = 60

Q) "If you flip a coin five times, what's the probability you get exactly three heads ?
The answer can be found in ten seconds- it's 5C3/2^5 = 5/16

Q) A couple decides to have 4 children. If they succeed in having 4 children and each child is equally likely to be a boy or a girl, what is the probability that they will have exactly 2 girls and 2 boys?

Answer:

We can treat this exactly as we would a coin flip question; think of it as:

"If a fair coin is flipped 4 times, what's the probability of getting exactly 2 heads?"

There's an easy to use formula for coin flip (and pseudo-coin flip) questions:

Probability of getting exactly k results out of n flips = nCk/2^n

(nCk = n!/k!(n-k)!, the combinations formula.)

In this question, we have 4 flips and we want 2 heads, so n=4 and k=2:

4C2/2^4 = (4!/2!2!)/16 = (24/4)/16 = 6/16 = 3/8

Thanks hesham it was very helpful ..

Abdulla

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Nov 24, 2009 9:34 pm

Could you help with this:

Q. What is the units digit of a^36?

a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit

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2010gmat: Master | Next Rank: 500 Posts; Posts: 371; Joined: Fri Mar 06, 2009 2:48 am; Thanked: 27 times; GMAT Score:740

by 2010gmat » Tue Nov 24, 2009 9:34 pm

great ques hesham...

if the question says at least 3 heads in 5 tosses...

then??

will it be

5c0 + 5c1 + 5c2 + 5c3)/2^5 ???i.e 30/32

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Nov 24, 2009 9:42 pm

2010gmat wrote:great ques hesham...

if the question says at least 3 heads in 5 tosses...

then??

will it be

5c0 + 5c1 + 5c2 + 5c3)/2^5 ???i.e 30/32

I don't know 2010gmat. I am poor in this subject. I just memorize question patterns, unfortunately. I never took a class that covers all this stuff.

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Nov 24, 2009 9:56 pm

heshamelaziry wrote:Could you help with this:

Q. What is the units digit of a^36?

a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit

Sure, I think the answer is D

All what you need to do is to find the pattern. from the first statement we know that a^2 = 9 , so try to continue to get the pattern ..
If the unit digit of a^2 = 9 then the unit digit of a^3 = 1 ( 9 * 9 = 81) and so on.. you will notice that the unit digit will be either 9 or 1 ... so we can get the answer.. Sufficient.

Statement 2 ... same steps .. Sufficient.

Abdulla

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Tue Nov 24, 2009 10:07 pm

Abdulla wrote:
heshamelaziry wrote:Could you help with this:

Q. What is the units digit of a^36?

a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
Sure, I think the answer is D

All what you need to do is to find the pattern. from the first statement we know that a^2 = 9 , so try to continue to get the pattern ..
If the unit digit of a^2 = 9 then the unit digit of a^3 = 1 ( 9 * 9 = 81) and so on.. you will notice that the unit digit will be either 9 or 1 ... so we can get the answer.. Sufficient.

Statement 2 ... same steps .. Sufficient.

Apologize for the bother. I know how to do this if i have a number, but what is the concept behind a^3 = 81 ? a^2 = 9 means that a =3 so why not a^3 = 27 ?

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Abdulla: Master | Next Rank: 500 Posts; Posts: 328; Joined: Thu Aug 07, 2008 5:25 pm; Location: Philadelphia; Thanked: 4 times; GMAT Score:550

by Abdulla » Tue Nov 24, 2009 11:05 pm

heshamelaziry wrote:
Abdulla wrote:
heshamelaziry wrote:Could you help with this:

Q. What is the units digit of a^36?

a) a^2 has 9 as the units digit
b) a^3 has 3 has the units digit
Sure, I think the answer is D

All what you need to do is to find the pattern. from the first statement we know that a^2 = 9 , so try to continue to get the pattern ..
If the unit digit of a^2 = 9 then the unit digit of a^3 = 1 ( 9 * 9 = 81) and so on.. you will notice that the unit digit will be either 9 or 1 ... so we can get the answer.. Sufficient.

Statement 2 ... same steps .. Sufficient.

Apologize for the bother. I know how to do this if i have a number, but what is the concept behind a^3 = 81 ? a^2 = 9 means that a =3 so why not a^3 = 27 ?

Sorry I missed it up, but still the concept is the same and the correct answer is the same.

Just try to get the pattern .. the key is that the unit digit will keep repeated .. for example..

3^1=3 3^2= 9 3^3=27 3^4 = 81 3^5= 243 ... so If we continue we definitely will find the unit digit for a^36... sorry for my bad explanations..

Abdulla

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Pooja Bhula: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Tue Nov 24, 2009 12:48 am

by Pooja Bhula » Wed Nov 25, 2009 1:28 am

Hi Hesham,
Im gonna help you with the sum to find a^36. Now see we are only concerned with the units digit here.. so lets see which square(s) gives us the units digit 9, n which cube(s) give us 3 in the units digit. So here go the squares - 2 = 4, 3 = 9, 4 = 16, 5 = 25, 6 = 36, 7 = 49.thus only 3 and 7 end in units digit 9, when they are squared. Therefore, lets see the cubes of 3 and 7 to decide which no. it is. 3^3 = 27, so this does not work coz we need 3 in the units place. Now take 7, 7^3 = 49 x 7 = 343. So bingo.. 7 is our no.

Now the next step is to find a pattern.... for 7^n... 7^1 = 7, 7^2 = 49, 7^3 = 343, 7 ^4 = here the units digit ends in 1 as 7 x 3 = 21, next multiply the units digit by 7..so 7^5 = units digit is 7. Thus every 4th one ends in 1, and 36 being a multiple of 4 also ends in 1, or has 1 in the units place.

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heshamelaziry: Legendary Member; Posts: 869; Joined: Wed Aug 26, 2009 3:49 pmFollowed by: 3 members; Location: California; Thanked: 13 times

by heshamelaziry » Wed Nov 25, 2009 1:32 am

Pooja Bhula wrote:Hi Hesham,
Im gonna help you with the sum to find a^36. Now see we are only concerned with the units digit here.. so lets see which square(s) gives us the units digit 9, n which cube(s) give us 3 in the units digit. So here go the squares - 2 = 4, 3 = 9, 4 = 16, 5 = 25, 6 = 36, 7 = 49.thus only 3 and 7 end in units digit 9, when they are squared. Therefore, lets see the cubes of 3 and 7 to decide which no. it is. 3^3 = 27, so this does not work coz we need 3 in the units place. Now take 7, 7^3 = 49 x 7 = 343. So bingo.. 7 is our no.

Now the next step is to find a pattern.... for 7^n... 7^1 = 7, 7^2 = 49, 7^3 = 343, 7 ^4 = here the units digit ends in 1 as 7 x 3 = 21, next multiply the units digit by 7..so 7^5 = units digit is 7. Thus every 4th one ends in 1, and 36 being a multiple of 4 also ends in 1, or has 1 in the units place.

Is the answer b or d in your opinion ?

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