Problem Solving

1. A woman has seven cookies--four chocolate chip and three oatmeal. She gives one cookie to each of her six children: Nicole, Ronit, Kim, Deborah, Mark, and Terrance. If Deborah will only eat the kind of cookie that Kim eats, in how many different ways can the cookies be distributed?

(The leftover cookie will be given to the dog.)

a) 40
b) 50
c) 25
d) 15
e) 12


2. There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf ?

a) 20!
b) 5!
c) 20!/4!
d) 20!/(4!)^^5
e) 20!/5(4!)


3. At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time ?

a) 20 13/17 minutes past 7:00
b) 21 3/23 minutes past 7:00
c) 20 13/21 minutes past 7:00
d) 22 4/9 minutes past 7:00
e) 21 9/11 minutes past 7:00

4. Q) A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

a)y > ROOT2
b) y < ROOT3/4
c) ROOT3/4 < y < ROOT2/3
d) ROOT3/2 < y < ROOT2
e) ROOT2/3 < y < ROOT3/2

IMO:
1.D
Possibility that Deborah and Kim both eat chocolate chips
P = 4/7 *3/7 = 2/7
Possibility that D&K both eat oatmeal:
P = 3/7 *2/7 = 1/7
Total ways of delivering cookies to the children : 7!/(4!*3!) = 35
Ways of delivering that D&K can eat the same kind of cookie: (2/7+1/7)*35 = 15

2.E
total ways : 20P20/ /(4!*4!*4!*4!*4!)

3.E
Minute hand moves at a speed of 12 no. per hour
Hour hand moves at a speed of 1 no. per hour
Discrepancy in speed: 11 no. per hour
At 7:00 the distance (in clockwise direction) from the minute hand to the hour hand is : 7 no.
when minute and hour hands of an accurate working clock be precisely perpendicular to each other the distance is 3 no.
So the minute hand has to go faster than hour hand a distance of: 7 -3 = 4 no.
The time for the minute hand to do so: 4/11 (hour), or (4/11)*60 = 21 minutes 9/11 after 7:00

4.A
The area of the triangle : xy/2 = 1 ( why not xz/2 or yz/2, because x<y<z so x&y is the length of two sides that are perpendicular to each other)
so xy=2
while x<y, then y^2> x*y, it means y^2>2 or y>root(2)

OA pls.

Yep..Agree with all of the answers above from @limestone...

1. My approach is slightly different
If Deb & Kim both have chocolate chips, then the remaining 4 kids can have the following combinations of cookies
1 Chocolate Chip + 3 Oatmeal = 4 ways
2 Chocolate Chip + 2 Oatmeal = 6 ways

If Deb & Kim both have Oatmeal, then the remaining 4 kids can have the following combination of cookies
1 Oatmeal + 3 Chocolate Chips = 4 ways
4 Chocolate Chips = 1 way

Hence total = 15 ways

2. Approach and answer same as above

3. Slightly different approach.. I considered the degree value instead of numbers..

Speed of hour hand = 30 degrees/hr
Speed of min hand = 360 degrees/hr
Effective speed = 330 degree/hr (because they move in same dir.)

Initial separation (at 7 o'clock) = 210 degrees
Final separation desired = 90 degrees
Hence distance to be covered = 210-90 = 120 degrees.

Time to cover distance = Distance / velocity = 120 / 330 hrs = 4/11 hrs = 240/11 mins = 21+9/11 minutes

4. Approach and answer same as above..


Cheers!!

for the clocks problem we can use formula
30H- (11/2)M =angle between hands
h= hours m=min

perpendicular means 90 degrees
so
90= 30H-(11/2)M
H=7
90=210-(11/2)M

After solving this we get 11M=240
therefore M=240/11 = 21 * 9/11

limestone wrote:IMO:
1.D
Possibility that Deborah and Kim both eat chocolate chips
P = 4/7 *3/7 = 2/7
Possibility that D&K both eat oatmeal:
P = 3/7 *2/7 = 1/7
Total ways of delivering cookies to the children : 7!/(4!*3!) = 35
Ways of delivering that D&K can eat the same kind of cookie: (2/7+1/7)*35 = 15

2.E
total ways : 20P20/ /(4!*4!*4!*4!*4!)

3.E
Minute hand moves at a speed of 12 no. per hour
Hour hand moves at a speed of 1 no. per hour
Discrepancy in speed: 11 no. per hour
At 7:00 the distance (in clockwise direction) from the minute hand to the hour hand is : 7 no.
when minute and hour hands of an accurate working clock be precisely perpendicular to each other the distance is 3 no.
So the minute hand has to go faster than hour hand a distance of: 7 -3 = 4 no.
The time for the minute hand to do so: 4/11 (hour), or (4/11)*60 = 21 minutes 9/11 after 7:00

4.A
The area of the triangle : xy/2 = 1 ( why not xz/2 or yz/2, because x<y<z so x&y is the length of two sides that are perpendicular to each other)
so xy=2
while x<y, then y^2> x*y, it means y^2>2 or y>root(2)

OA pls.

Not catching on as fast. Limestone, can you further explain how you came to your answers for one and two, more specifically the factorial portions?
For question 1, how come total ways is 7!/3!*4!?
For question 2, how did 20!/(5*4!) come about?

Thank you

Sure,

1. From the given question, we need to pick six cookies from a pool of seven cookies to give to six children, so total ways of pick is: 7P6 = 7x6x5x4x3x2 = 7!
However, these seven cookies are not different from each other. There're only two kinds: chocolate chip (4 pieces) and oatmeal (3 pieces). That's why I used 4! & 3! to eliminate duplicated cases. Then the total ways must be : 7!/ (4!*3!) = 35

2. There are 4 copies of 5 different books, so the total no. of books is 20. Total ways of putting those books on a shelf is :
20P20 = 20!, however, there're only 5 different books, each of them is duplicated 4 folds. So I used 4! to elimate duplicated cases for each of 5 different books.Thus, I must use 4! five times. So the total ways is: 20!/ (5*4!)

To make it clearer, take this example:
A,B,C,D : 4!= 24 ways of arranging
A,B,B,B : ABBB, BABB, BBAB, BBBA : 4 ways or 4!/3!
A,B,B,C : ABBC, ACBB, BBAC, BBCA, CABB, CBBA, BACB, BCAB, CBAB, ABCB, BCBA, BABC - a long list to be named lol - 12 ways or 4!/2!

That's the rule.

Can you please explain for question 1 why it isn't a 7C6 instead of 7P6. the order of distributing the cookies does not matter so why did you use a permutation ?

Thanks

The order of distributing cookies does matter here. For example, 6 cookies (3 chocolates & 3 oatmeals) for 6 children, Kim & Debohra received chocolates, the other four received 3 oatmeals and 1 chocolate. If we don't consider order of cookies, the case that Kim received 1 oatmeal, Debohra received 1 chocolate, the other four received 2 oatmeals, 2 chocolates will be considered the same as the case mentioned above (in which Kim& Debohra received chocolates while the remains received 1 chocolate & 3 oatmeals). The first case is suitable to the requiring condition;the second one, however, is not. So, they must be different cases, and, order of distributing does really matter.

For my understanding, how come a problem like this does not need to account for duplicates:

If a jury of 12 people is to be selected randomly from a pool of 15 potential jurors, and the jury pool consists of 2/3 men and 1/3 women, what is the probability that the jury will comprise at least 2/3 men?

To see how many ways to choose 12 people from 15 is simply 15c12 (15!/(3!*12!)). How do we account that there are 10 men and 5 females to choose from? Even though they are different people, they are represented by M,M,M,F,F,F....

Thank you

Dear Limestone,

Can you please explain me the method to solve the clock question. I am not understanding the logic behind the use of numbers. It would be grateful if you can elaborate the same.

saurabhmahajan wrote:for the clocks problem we can use formula
30H- (11/2)M =angle between hands
h= hours m=min

perpendicular means 90 degrees
so
90= 30H-(11/2)M
H=7
90=210-(11/2)M

After solving this we get 11M=240
therefore M=240/11 = 21 * 9/11

From where did we got this formula ? How did we landed up with (11/2)M ???

2.E
total ways : 20P20/ /(4!*4!*4!*4!*4!)


FYI, 4!*4!*4!*4!*4! does NOT equal 5(4!)

5(4!) = 4!+4!+4!+4!+4!

IMO, correct answer for #2 is D not E

I don't think clock sums are under the scope of the GMAT. I haven't seen an official problem testing the concept.

Anyways, At exactly 7 they hour and minute hands are 35 mins apart. They will be right angles when there is a space of 15 mins between them.

The minute hand will gain a 5 min lead to the hour hand in (5 * 60)/55 = 5 5/11 minutes.

They will make right angles at 2 position when they are 15 mins apart, or when the minute hand has gained (35-15) = 20 or when minute hand has gained (35+15) = 50 mins.

Since they ask for the first time, we need to calculate when minute hand gains 20 mins. (or) 4* 5 5/11 = 21 9/11 post 7.

or when they are 20 mins apart or when they are 50 mins apart.

Viren1808 wrote:Dear Limestone,

Can you please explain me the method to solve the clock question. I am not understanding the logic behind the use of numbers. It would be grateful if you can elaborate the same.

I agree, somehow all of them missed it

quiimari wrote:2.E
total ways : 20P20/ /(4!*4!*4!*4!*4!)


FYI, 4!*4!*4!*4!*4! does NOT equal 5(4!)

5(4!) = 4!+4!+4!+4!+4!

IMO, correct answer for #2 is D not E

I received a PM asking me to comment....

1. A woman has seven cookies--four chocolate chip and three oatmeal. She gives one cookie to each of her six children: Nicole, Ronit, Kim, Deborah, Mark, and Terrance. If Deborah will only eat the kind of cookie that Kim eats, in how many different ways can the cookies be distributed?

(The leftover cookie will be given to the dog.)

a) 40
b) 50
c) 25
d) 15
e) 12

Arrange the kids and dog in the following sequence: NRKDMTX where X represents the dog. Now, the problem is equivalent to finding the number of distinct rearrangements of CCCCOOO. This sequence would be equivalent to Nicole, Ronit, Kim, and Deborah getting chocolate chip cookies and mark, terrance, and the dog getting oatmeal cookies. Of course, we also have to account for the restriction that Deborah and Kim have to get the same kind of cookie, so we have two possible cases:

Case 1:

_ _ C C _ _ _

Now, we have to arrange CCOOO. There are 5!/3!2!=10 distinct ways to sequence these letters using the anagram method. Or we could reason that there are 5C3=10 ways to choose who gets the oatmeal cookies, and once we do that, the remaining people would have to get chocolate chip cookies.

Case 2:

_ _ O O _ _ _

Now, we have to arrange CCCCO. There are 5 spots for the one oatmeal cookie. Once we choose who gets that, the others automatically get chocolate chip cookies, so 5 total ways to arrange these remaining cookies.

Case 1 + Case 2=10+5=15

2. There are 4 copies of 5 different books. In how many ways can they be arranged on a shelf ?

a) 20!
b) 5!
c) 20!/4!
d) 20!/(4!)^^5
e) 20!/5(4!)

Let the different books be A,B,C,D, and E. With four copies of each, the problem is equivalent to determining the number of unique sequences of AAAABBBBCCCCDDDDEEEE. There are 20!/(4!4!4!4!4!) ways to do this, with the 4!4!4!4!4! in the denominator dividing out the repetitions that are counted in 20!. This is the same as 20!/(4!)^5.

3. At exactly what time past 7:00 will the minute and hour hands of an accurate working clock be precisely perpendicular to each other for the first time ?

a) 20 13/17 minutes past 7:00
b) 21 3/23 minutes past 7:00
c) 20 13/21 minutes past 7:00
d) 22 4/9 minutes past 7:00
e) 21 9/11 minutes past 7:00

A clock is circular, so it has a total of 360 degrees. We can arbitrarily label the 12 as zero degrees. There are there are 360/12=30 degrees between each number, so at 7:00, with the minute hand at the 12 and the hour hand at the 7, the minute hand is at 0 degrees and the hour hand is at 210 degrees. Each hour, the hour hand moves 30 degrees (from one number to the next), and the minute hand moves 360 degrees (one complete revolution). So t hours after 7:00, the hour hand is at h=210+30t degrees and the minute hand is at m=0+360t or m=360t degrees. The hands will be perpendicular when h-m=90, so (210+30t)-360t=90. Solving for t: 330t=120--->t=12/33. So 12/33 of an hour past 7:00, the hands will be perpendicular. The answer choices are all in minutes, so we multiply 12/33 by 60 to convert to minutes. 60*(12/33)=20*12/11=21 9/11 minutes past 7:00.

4. Q) A certain right triangle has sides of length x, y, and z, where x < y < z. If the area of this triangular region is 1, which of the following indicates all of the possible values of y ?

a)y > ROOT2
b) y < ROOT3/4
c) ROOT3/4 < y < ROOT2/3
d) ROOT3/2 < y < ROOT2
e) ROOT2/3 < y < ROOT3/2

If a right triangle has sides x<y<z, then z must be the hypotenuse. If the area is 1, then xy/2=1 so xy=2, or x=2/y. If y>x, then y>2/y. We know y is positive, so we can multiply both sides of the inequality by y to get y^2>2. Therefore, y>√2.