Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?
A. 4
B. 6
C. 10
D. 20
E. 24
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Source: Beat The GMAT — Problem Solving |
M = {-6, -5, -4, -3, -2}
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?
A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5
T = {-2, -1, 0, 1, 2, 3}
If an integer is to be randomly selected from set M above and an integer is to be randomly selected from set T above, what is the probability that the product of the two integers will be negative?
A. 0
B. 1/3
C. 2/5
D. 1/2
E. 3/5
150, 200, 250, n
Which of the following could be the median of the 4 integers listed above?
I. 175
II. 215
III. 235
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
Which of the following could be the median of the 4 integers listed above?
I. 175
II. 215
III. 235
A. I only
B. II only
C. I and II only
D. II and III only
E. I, II, and III
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2^100 - 2^96Sanket05 wrote:What is the greatest prime factor of 2^100 - 2^96?
A. 2
B. 3
C. 5
D. 7
E. 11
2^96*( 2^4 - 1)
2^96 ( 16-1)
2^96 * 15
2^96 * 5 * 3
C. 5
Good ole weight average problems!Sanket05 wrote:Each of the 30 boxes in a certain shipment with either 10 pounds or 20 pounds, and the average (arithmetic mean) weight of the boxes in the shipment is 18 pounds. If the average weight of the boxes in the shipment is to be reduced to 14 pounds by removing some of the 20-pound boxes, how many 20-pound boxes must be removed?
A. 4
B. 6
C. 10
D. 20
E. 24
x = # of 10 pound boxes
y = # of 20 pound boxes
We know we have (30) total.
10x + 20y = 18(30)
10x + 20y = 540
Let r = number of boxes removed
10x + 20(y-r)/ 30-r = 14
10x + 20y - 20r =420 - 14r
Substitute, 540 for 10x + 20y
540 - 20r = 420 - 14r
120 = 6r
r = 20
D. 20 boxes?
For the median qn, answer is (C)I and II
Explanation:
n could be anywhere among the 3 other integers.
If n < 150, then median is (150+200)/2 = 175 (I possible)
Take option II -->215, 200 will be another integer competing for median if n will be either in 2nd place or 3rd place. Considering option II - 215*2=430. Since 200 is another integer for median, n will be 230 (3rd place) (II also possible)
Take option III --> n goes to 4th place which has no control over median calculation (III not possible)
Explanation:
n could be anywhere among the 3 other integers.
If n < 150, then median is (150+200)/2 = 175 (I possible)
Take option II -->215, 200 will be another integer competing for median if n will be either in 2nd place or 3rd place. Considering option II - 215*2=430. Since 200 is another integer for median, n will be 230 (3rd place) (II also possible)
Take option III --> n goes to 4th place which has no control over median calculation (III not possible)
I'll take the last one standing, the set M and T.
Set M has 5 integers in it, set T has 6. The question is asking for the probability that the product of any 2 integers is negative.
Total possibilities of products = 5 * 6 = 30
A negative product for 2 integers only can happen when a negative integer is multiplied by a positive one. There are 3 positive integers in set T, and 5 in set M, so total number of possibilities = 5 * 3 = 15.
Hence, probability that 2 numbers chosen will have negative product = 15/30 = 1/2. Choice D.
Set M has 5 integers in it, set T has 6. The question is asking for the probability that the product of any 2 integers is negative.
Total possibilities of products = 5 * 6 = 30
A negative product for 2 integers only can happen when a negative integer is multiplied by a positive one. There are 3 positive integers in set T, and 5 in set M, so total number of possibilities = 5 * 3 = 15.
Hence, probability that 2 numbers chosen will have negative product = 15/30 = 1/2. Choice D.
Speaking of boxes.
Jeff, you made a mistake!
10x + 20y = 540 - that's right.
Then, we also know that x+y=30
Solving these two we get:
x=6, y=24 - before reduction of the average weight.
We are asked how many 20-pound boxes should be taken?
use again the formula for an average:
6*10+(24-r)*20 = 14*30
r - a number of bosex that has to be taken away in order to reduce average weight from 18 to 14 pounds
Solving it, we get r=6
Jeff, you made a mistake!
10x + 20y = 540 - that's right.
Then, we also know that x+y=30
Solving these two we get:
x=6, y=24 - before reduction of the average weight.
We are asked how many 20-pound boxes should be taken?
use again the formula for an average:
6*10+(24-r)*20 = 14*30
r - a number of bosex that has to be taken away in order to reduce average weight from 18 to 14 pounds
Solving it, we get r=6
Hey mate!mp2437 wrote:I'll take the last one standing, the set M and T.
Set M has 5 integers in it, set T has 6. The question is asking for the probability that the product of any 2 integers is negative.
Total possibilities of products = 5 * 6 = 30
A negative product for 2 integers only can happen when a negative integer is multiplied by a positive one. There are 3 positive integers in set T, and 5 in set M, so total number of possibilities = 5 * 3 = 15.
Hence, probability that 2 numbers chosen will have negative product = 15/30 = 1/2. Choice D.
I think it is a bit easier =)
There are only negatives in set M. So there is no chance that the product of these integers would be negative if the integer from set T is negative or "0". So we need just to know the probability of picking positive integer from T. It is (3/6=0.5)
Jeff, I am sorry
You were actually right and I made a mistake myself
it should be
10*6 + (24-r)20=14(30-r)
then, r=20
You were actually right and I made a mistake myself
it should be
10*6 + (24-r)20=14(30-r)
then, r=20
I actually did it this way also, but for purposes of explaining it, I tried to show all possibilities.NikolayZ wrote: Hey mate!
I think it is a bit easier =)
There are only negatives in set M. So there is no chance that the product of these integers would be negative if the integer from set T is negative or "0". So we need just to know the probability of picking positive integer from T. It is (3/6=0.5)
Cheers!
