GMAT Problem Solving Discussions

MBA
703
"On Monday a certain animal shelter housed 55 cats and dogs. By Friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday"?
A) 11
B) 12
C) 13
D) 14
E) 20

C) 13

Is this correct ...
704

Vicky could you explain it in detail? tks.

705
"On Monday a certain animal shelter housed 55 cats and dogs. By Friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday"?
A) 11
B) 12
C) 13
D) 14
E) 20

d = dogs adopted, c = cats adopted
4d + 5c = 55
consider all the cases here
(c,d) = (11,0) (7,5) (3,10)

max value of c+d = 13.

i think this should suffice. lemme know.

slam.
706
"On Monday a certain animal shelter housed 55 cats and dogs. By Friday, exactly 1/5 of the cats and 1/4 of the dogs had been adopted; no new cats or dogs were brought to the shelter during this period. What is the greatest possible number of pets that could have been adopted from the animal shelter between Monday and Friday"?
A) 11
B) 12
C) 13
D) 14
E) 20


the simplest explanation is 1/4 is greater than 1/5...so max. possible no. of dogs should be taken into consideration i.e. the no. of dogs should be as large as possible to giv the largest value of adopted pets and the largest integer values for dogs and cats that will suffice is 40 and 15 so that the no. of adopted dogs and cats also turn out to be in integers and not fraction.. hence, answer is C...13
707
"In a particular state, 70 percent of the counties received some rain on Monday, and 65 percent of the counties received some rain on Tuesday. NO rain fell either day in 25 percent of the counties in the state. What percent of the counties received some rain on Monday and Tuesday?
A) 12,5%
B) 40%
C) 50%
D) 60%
E) 67,5%

Tks in advance for the detailed explanation.

Is this ok ? 70+65-x+25=100
x=60, so D)
708
"In a particular state, 70 percent of the counties received some rain on Monday, and 65 percent of the counties received some rain on Tuesday. NO rain fell either day in 25 percent of the counties in the state. What percent of the counties received some rain on Monday and Tuesday?
A) 12,5%
B) 40%
C) 50%
D) 60%
E) 67,5%

Tks in advance for the detailed explanation.

Is this ok ? 70+65-x+25=100
x=60, so D)


let countries receiving rain only on Monday be M and for countries receiving only on Tuesday be T. x be the countries receving rain on both days.
M + x=70
T + x=65
adding both the eqn we get
M+T+2x=135
agn M+T+x+25=100
M+T+x=75
equating both the eqn
we get x=60
709
"In a particular state, 70 percent of the counties received some rain on Monday, and 65 percent of the counties received some rain on Tuesday. NO rain fell either day in 25 percent of the counties in the state. What percent of the counties received some rain on Monday and Tuesday?
A) 12,5%
B) 40%
C) 50%
D) 60%
E) 67,5%

Tks in advance for the detailed explanation.

Is this ok ? 70+65-x+25=100
x=60, so D)

25% had no rain...so 75% had over 2 days...

70% on Monday
65% on Tuesday -- which has to cover 5% from 75% (75-70 on monday)

Rained on both days : 65 - 5 = 60

Ans D
710

Vic could you explain why if "25% had no rain...so 75% had over 2 days...?
Tks in advance.

711
Vic could you explain why if "25% had no rain...so 75% had over 2 days...?
Tks in advance.

NO rain fell either day in 25 percent of the counties in the state.

which means 75 percent of the countries had rain during these 2 days
712

ok, but I cant make the link with those 75% with the 70% of Monday and with the 60% of Tuesday. How do you conclude "70% on Monday 65% on Tuesday -- which has to cover 5% from 75% (75-70 on monday)?

713
paki Says
ok, but I cant make the link with those 75% with the 70% of Monday and with the 60% of Tuesday. How do you conclude "70% on Monday 65% on Tuesday -- which has to cover 5% from 75% (75-70 on monday)?

Maybe I wrote it too short ...

If 25% countries had no rainfall => 75% countries had rainfall over 2 days

70% had rainfall on monday and 65% on tuesday (given)

70% countries had rainfall on monday out of which 65% can again receive rainfall on tuesday. But if we do such a mapping we are left with 5% countries which received rainfall but we didnt consider them on any of the days.

So at the max out of 70% we can only take 60% countries...

hope this helps 😃
714

Hi All,

A problem that I encoutered in one of the practice tests.

Shipment Defective chips Total no. of chips

S1 2 5,000
S2 5 12,000
S3 6 18,000
S4 4 16,000

A computer manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ration for shipments S1, S2, S3 and S4 combined, as shown in the table above. What is the expected number of defective chips in a shipment of 60,000 chips.

1. 14
2. 20
3. 22
4. 24
5. 25

Please suggest the approach as well.

Thanks in advance.

715
Hi All,

A problem that I encoutered in one of the practice tests.

Shipment Defective chips Total no. of chips

S1 2 5,000
S2 5 12,000
S3 6 18,000
S4 4 16,000

A computer manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ration for shipments S1, S2, S3 and S4 combined, as shown in the table above. What is the expected number of defective chips in a shipment of 60,000 chips.

1. 14
2. 20
3. 22
4. 24
5. 25

Please suggest the approach as well.

Thanks in advance.

Ans.=> 2) 20
716
vicky.verma Says
Ans.=> 2) 20


Approach and method please.
717
Hi All,

A problem that I encoutered in one of the practice tests.

Shipment Defective chips Total no. of chips

S1 2 5,000
S2 5 12,000
S3 6 18,000
S4 4 16,000

A computer manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ration for shipments S1, S2, S3 and S4 combined, as shown in the table above. What is the expected number of defective chips in a shipment of 60,000 chips.

1. 14
2. 20
3. 22
4. 24
5. 25

Please suggest the approach as well.

Thanks in advance.


in S1,S2,S3, and S4
defective chips=2+5+6+4=17
total chips=5000+12000+8000+16000=51000.
let d be the no of defective chips in the shipment of 60,000 chips.
hence by condition
17/51000=d/60000
=>d=20
i hope you got that
718
radhika5481 Says
Approach and method please.

I guess you got confused by this sentence :

"A computer manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ration for shipments S1, S2, S3 and S4 combined, as shown in the table above."

"combined" --> that's the catch...

Approach same as that of carobee's
719
I guess you got confused by this sentence :

"A computer manufacturer expects the ratio of the number of defective chips to the total number of chips in all future shipments to equal the corresponding ration for shipments S1, S2, S3 and S4 combined, as shown in the table above."

"combined" --> that's the catch...

Approach same as that of carobee's



Precisely... I missed the word "combined". Thanks a lot.
720

Hi All,

Another problem which I got stuck with :

0

* *
* * *
* * * *
* * * * *
1 2 | 3 4


The figure shown represents a board with 4 rows of pegs and at teh bottom of the board are 4 cells numbered 1 to 4. Whenever a ball (shown with 0) passes thru the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has probablity 1/2 of passing thru the opening immediately to the left of that peg and probability of 1/2 of passing thru the opening immediately to the right. What is the probability that when the ball passes thru the first two pegs at the top, it will end up in cell 2.

1. 1/16

2. 1/8

3. 1/4

4. 3/8

5. 1/2

Approach and solution please....

Thanks is advance.

721
Hi All,

Another problem which I got stuck with :

0

* *
* * *
* * * *
* * * * *
1 2 | 3 4


The figure shown represents a board with 4 rows of pegs and at teh bottom of the board are 4 cells numbered 1 to 4. Whenever a ball (shown with 0) passes thru the opening between two adjacent pegs in the same row, it will hit the peg directly beneath the opening. The ball then has probablity 1/2 of passing thru the opening immediately to the left of that peg and probability of 1/2 of passing thru the opening immediately to the right. What is the probability that when the ball passes thru the first two pegs at the top, it will end up in cell 2.

1. 1/16

2. 1/8

3. 1/4

4. 3/8

5. 1/2

Approach and solution please....

Thanks is advance.



Please note tha the figure has a triangular shape... not able to give the right indentation.
722
radhika5481 Says
Please note tha the figure has a triangular shape... not able to give the right indentation.

1/8

do tell if the ans is correct.

i ll post the soln there after

regards
rock