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A jar contains 30 marbles, of which 20 are red and 10 are blue. If 9 of the marbles are removed, how many of the marbles left in the jar are red?
(1) Of the marbles removed, the ratio of the number of red ones to the number of blue ones is 2 : 1.
(2) Of the first 6 marbles removed, 4 are red.
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
Vamsi1981 SaysAnswer is E
E 😲 :O
From statement 1, if 9 marbles are removed, and the ratio of the number of red ones to the number of blue ones is 2 : 1, then 6 red marbles and 3 blue marbles are removed.
So, 20-6 =14 red marbles left in the jar , Hence A/D are valid.
From Statement 2, number of red/blue marbles removed after all are removed is not given, so B is insuffiient.
Hence, A is correct
answer is E. It happens in quant, so i started working for quant regularly now... here is another good one:
A cook went to a market to buy some eggs and paid $12. But since the eggs were quite small, he talked the seller into adding two more eggs, free of charge. As the two eggs were added, the price per dozen went down by a dollar. How many eggs did the cook bring home from the market?
A. 8
B. 12
C. 15
D. 16
E. 18
Let the cook buy X eggs. Now the price per dozen = (12/x) * 12.
After the bargain , p-1 = (12/x+2) * 12.
Solving,we get x = 16.So the cook bought home 18 eggs.
1) 1, 4, a, 5, b, 4
If a and b are both integers > 0 and the average (arithmetic mean) of the values above is 3, what is a possible value for the median?
1.5
2
2.5
3
4
2) Together, Mary and Joe have x dollars. If Mary has $80 less than Joe, how much money, in terms of x will Joe have if he receives x dollars?
Choices
A x + 40
B 2x + 40
C 3x/2+ 40
D x/2 + 80
E 3x+8
1) 1, 4, a, 5, b, 4
If a and b are both integers > 0 and the average (arithmetic mean) of the values above is 3, what is a possible value for the median?
1.5
2
2.5
3
4
2) Together, Mary and Joe have x dollars. If Mary has $80 less than Joe, how much money, in terms of x will Joe have if he receives x dollars?
Choices
A x + 40
B 2x + 40
C 3x/2+ 40
D x/2 + 80
E 3x+8
First answer is 3, and second answer is C
1) 1, 4, a, 5, b, 4
If a and b are both integers > 0 and the average (arithmetic mean) of the values above is 3, what is a possible value for the median?
1.5
2
2.5
3
4
2) Together, Mary and Joe have x dollars. If Mary has $80 less than Joe, how much money, in terms of x will Joe have if he receives x dollars?
Choices
A x + 40
B 2x + 40
C 3x/2+ 40
D x/2 + 80
E 3x+8
1) got it. average = 3. so a+b = 4
lets say a=1 so b=3, series is 1,1,3,4,4,5 median = 3.5
a=2so b=2, series is 1,2,2,4,4,5 median = 3
3.5 is not in choices so median has to be 3. answer is D.
2) let Joe has J.
Mary has M = J - 80
and M+J = x
So J = (x+80)/2
now J get x dollars more so J has : (x+80)/2 + x = (3x+80)/2 = option C.
chintamani.kurs SaysFirst answer is 3, and second answer is C
Correct ... plz share the reasons behind your answer ..
Please solve this problem. I calculated it via counting method but I would like to know if there is/are any other method/methods to solve this.
A photographer will arrange 6 people of 6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing in front of someone in the second row. The heights of the people within each row must increase from left to right, and each person in the second row must be taller that hte person standing in front of him or her. How many such arrangements of the people are possible?
(A) 5
(B) 6
(C) 9
(D) 24
(E) 36
Thanks,
Anurag...
Please solve this problem. I calculated it via counting method but I would like to know if there is/are any other method/methods to solve this.
A photographer will arrange 6 people of 6 different heights for photograph by placing them in two rows of three so that each person in the first row is standing in front of someone in the second row. The heights of the people within each row must increase from left to right, and each person in the second row must be taller that hte person standing in front of him or her. How many such arrangements of the people are possible?
(A) 5
(B) 6
(C) 9
(D) 24
(E) 36
Thanks,
Anurag...
Is the answer (A).
couldn;t think of any logic except taking cases.
Is the answer (A).
couldn;t think of any logic except taking cases.
Yes, It is A.
Thanks,
Anurag...
Yes, It is A.
Thanks,
Anurag...
Hi
Could you elaborate on this answer
Hi
Could you elaborate on this answer
I can try. Suppose 1,2,3,4,5,6 are the different people with heights increasing in the same order. then according to the question arrangements could be:
456 256 356
123 134 124 and so on ...So there can be 5 possible arrangements.
Thanks,
Anurag...
I can try. Suppose 1,2,3,4,5,6 are the different people with heights increasing in the same order. then according to the question arrangements could be:
456 256 356
123 134 124 and so on ...So there can be 5 possible arrangements.
Thanks,
Anurag...
Agree that counting in this case is the smartest way. I tried combinations, turned out to be lengthier and complicated. BTW I was unable to solve using P&C; .
Gail.Wynand SaysAgree that counting in this case is the smartest way. I tried combinations, turned out to be lengthier and complicated. BTW I was unable to solve using P&C; .
Yes, even I could not figure out any other method other than basic counting.
Thanks,
Anurag...
Try this one guys
Danny,Doris and Dolly flipped a coin 5 times and each time the coin landed on " heads " ,Dolly bet that the coin will land up as "tails " on the 6th time.What is the probablity that she is right
a) 1
b) 1/2
c) 3/4
d) 1/4
e) 1/3
Try this one guys
Danny,Doris and Dolly flipped a coin 5 times and each time the coin landed on " heads " ,Dolly bet that the coin will land up as "tails " on the 6th time.What is the probablity that she is right
a) 1
b) 1/2
c) 3/4
d) 1/4
e) 1/3
Is it 1/2? Although I hate to solve P&C; questions, would still take my guess at B.
Thanks,
Anurag...
Try this one guys
Danny,Doris and Dolly flipped a coin 5 times and each time the coin landed on " heads " ,Dolly bet that the coin will land up as "tails " on the 6th time.What is the probablity that she is right
a) 1
b) 1/2
c) 3/4
d) 1/4
e) 1/3
The probability of getting a heads or tail of a coin will never change i.e. always 0.5 or 50%. Mutually exclusive event so the 5 outcomes don't affect the 6th. what' s the OA?
Gail.Wynand SaysThe probability of getting a heads or tail of a coin will never change i.e. always 0.5 or 50%. Mutually exclusive event so the 5 outcomes don't affect the 6th. what' s the OA?
Aha!!! I think I am doing good in probability now. Time to change my status again......
Thanks,
Anurag...
Aha!!! I think I am doing good in probability now. Time to change my status again......
Thanks,
Anurag...
yes its 1/2 . Essentially the probability will remain 1/2