Need your help in the answers for the following questions:
Q1. 6 hrs from now, the population of a colony of bacteria will be 2.56 x 10 to the power 7. If the population of the bacteria colony doubles every 3 hrs, how many hours ago did the population surpass 100,000 ? A. 9 B. 18 C. 21 D. 27 E. 33
Q2. If r+s+p > 1, is p > 1 ? (1) p > r+s-1 (2) 1 -(r+S) > 0
i seem to be contradicting with the answers proposed and would like your help in confirming that.:bigear:
See if this makes sense to you guys.... Eqn. of a circle is (x-a)^2 + (y-b)^2 = R^2 where (a,b) is the center of the circle. (N.B. - ^ denotes to thepower) now (4,0) and (-4,0) are pts. on the circle. Therefore, (4-a)^2 + (0-b)^2 = R^2 ........................(1) (-4-a)^2 + (0-b)^2 = R^2 .......................(2)
(1) - (2) gives : a = 0. therefore . eqn. 1 or 2 reduces to 16 + b^2 = R^2. Since b^2 is always a positive quantity, R will have a bigger value , bigger the value of b.
See if this makes sense to you guys.... Eqn. of a circle is (x-a)^2 + (y-b)^2 = R^2 where (a,b) is the center of the circle. (N.B. - ^ denotes to thepower) now (4,0) and (-4,0) are pts. on the circle. Therefore, (4-a)^2 + (0-b)^2 = R^2 ........................(1) (-4-a)^2 + (0-b)^2 = R^2 .......................(2)
(1) - (2) gives : a = 0. therefore . eqn. 1 or 2 reduces to 16 + b^2 = R^2. Since b^2 is always a positive quantity, R will have a bigger value , bigger the value of b.
Need your help in the answers for the following questions:
Q1. 6 hrs from now, the population of a colony of bacteria will be 2.56 x 10 to the power 7. If the population of the bacteria colony doubles every 3 hrs, how many hours ago did the population surpass 100,000 ? A. 9 B. 18 C. 21 D. 27 E. 33
Q2. If r+s+p > 1, is p > 1 ? (1) p > r+s-1 (2) 1 -(r+S) > 0
i seem to be contradicting with the answers proposed and would like your help in confirming that.:bigear:
Q. A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the bushes at random, one at a time, and plant them i a row, what is the probability that the 2 rosebushes in the middle of the row will be red?
A. 1/12 B.1/6 C.1/5 D.1/3 E.1/2
The OA is B i.e 1/6 However, I derived 1/12 as the answer. 4 rosebushes will have 4! ways of arranging. There can be 2 possible arrangements for the red rosebushes to be in the middle of the row i.e. W1, R1,R2, W2 or W1, R2, R1, W2. Hence, number of possibilities is 2/4! = 1/12 Can anyone please point out my mistake?
Help needed with the following data suffienciency questions.
Q1. What is the median no. of employees assigned per project for projects at Company Z 1.25% of the projects at Company Z have 4 or more employees assigned to each project 2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project
Q2.In the xy plane, point(r,s) lies on the circle with its center at the origin. What is the value of r2+s2 ? 1.The circle has radius 2 2.The point (v2, -v2) lies on the circle
Help needed with the following data suffienciency questions.
Q1. What is the median no. of employees assigned per project for projects at Company Z 1.25% of the projects at Company Z have 4 or more employees assigned to each project 2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project
Q2.In the xy plane, point(r,s) lies on the circle with its center at the origin. What is the value of r2+s2 ? 1.The circle has radius 2 2.The point (v2, -v2) lies on the circle
Regards MSD
1.C
let that company has 20 projects.....
1.25% of the projects at Company Z have 4 or more employees assigned to each project ie 5 projects have 4 or more employees. Not sufficient so BCE
2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project ie 5 projects have 2 or fewer employees. Not sufficient so CE
Combining, the rest of the projects have 3 employees each......ie projects from 6-15 so median is the average no: of employees in 10& 11th project and it is 3 employees...... SO ans C
Q. A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the bushes at random, one at a time, and plant them i a row, what is the probability that the 2 rosebushes in the middle of the row will be red?
A. 1/12 B.1/6 C.1/5 D.1/3 E.1/2
The OA is B i.e 1/6 However, I derived 1/12 as the answer. 4 rosebushes will have 4! ways of arranging. There can be 2 possible arrangements for the red rosebushes to be in the middle of the row i.e. W1, R1,R2, W2 or W1, R2, R1, W2. Hence, number of possibilities is 2/4! = 1/12 Can anyone please point out my mistake?
Regards MSD
OA is correct......
Your mistake is ....u dont considered 2 more possible arrangements ie total 4 arrangements....(W1, R1,R2, W2), (W1, R2, R1, W2), (w2,R1,R2,W1) & (w2,R2,R1,W1)
I think the arrangement we are looking for here is W R R W
Probability of picking 1 W out of the 4 rose bushes = 2/4 Probability of picking 1 R out of the remaining 3 rose bushes = 2/3 Probability of picking 1 R out of the remaining 2 rose bushes = 1/2 Probability of picking 1 W out of 1 rose bush left = 1/1
Therefore total probability of this arrangement occurring =
W R R W = (2/4) * (2/3) * (1/2) * (1/1) = 1/6
Plz correct me if i'm wrong!
Q. A gardener is going to plant 2 red rosebushes and 2 white rosebushes. If the gardener is to select each of the bushes at random, one at a time, and plant them i a row, what is the probability that the 2 rosebushes in the middle of the row will be red?
A. 1/12 B.1/6 C.1/5 D.1/3 E.1/2
The OA is B i.e 1/6 However, I derived 1/12 as the answer. 4 rosebushes will have 4! ways of arranging. There can be 2 possible arrangements for the red rosebushes to be in the middle of the row i.e. W1, R1,R2, W2 or W1, R2, R1, W2. Hence, number of possibilities is 2/4! = 1/12 Can anyone please point out my mistake?
1) There are 4^3 ways of posting 3 letters in 4 letter boxes = 64 ways. But 64 includes the number of ways by which all the letters can be put in 1 letter box. There are 4 ways in which all letters can be put into each of the letter boxes. Therefore 64-4 = 60 is your answer.
2) Probability of selecting a red marble first = 4/16 Probability of selecting a blue marble next = 3/15
Therefore total probability of selecting a red marble first followed by a blue marble = (4/16)*(3/15) = 1/20
3) The number of unique signals that can be made using 4 flags - refers to the number of possible unique arrangements of the 4 flags. This can be given by 4! = 24. Therefore there are 24 ways of arranging 4 flags.
4) --> Given: Y takes "d" days to produce "w" --> Given: X takes "d+2" days to produce "w" --> Given: X+Y take 3 days to produce 5w/4 --> Therefore, X+Y take 12/5 days to produce "w" (ensure that the quantity produced is the same by X & Y working together and alone)
1/(X+Y) = (1/X) + (1/Y) (Solve to get d = 4 days)
X days "d+2" days to produce "w" --> 6 days to produce "w" Therefore to produce "2w" it will take 6*2 = 12 days.
5) Plz refer to page 104 in this forum where the question has been discussed in detail.
I also tend to agree with B as the answer for Question 1, but for q2, why not E. E.g. if r+s =0.7 which satisfies the condition r+s 1 or p could be 1.5 which again satisfied p+r+s > 1. So we cannot say for sure that condition 2 is sufficient. What do you think ?
I also tend to agree with B as the answer for Question 1, but for q2, why not E. E.g. if r+s =0.7 which satisfies the condition r+s 1 or p could be 1.5 which again satisfied p+r+s > 1. So we cannot say for sure that condition 2 is sufficient. What do you think ?
Quote: Originally Posted by msd_2008 View Post Help needed with the following data suffienciency questions.
Q1. What is the median no. of employees assigned per project for projects at Company Z 1.25% of the projects at Company Z have 4 or more employees assigned to each project 2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project
Q2.In the xy plane, point(r,s) lies on the circle with its center at the origin. What is the value of r2+s2 ? 1.The circle has radius 2 2.The point (v2, -v2) lies on the circle
Regards MSD
1.Ans is C
lets consider this situation with 20 projects so
1.25% of the projects at Company Z have 4 or more employees assigned to each project ie 5 projects have 4 or more employees . Not sufficient so BCE
2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project ie 5 projects have 2 or fewer employees. Not sufficient so CE
Combining the rest(middle) 15 projects of company has 3 employees each.... So median value is the employee in 10th project it is obviously 3
So ans is C
ashishjha100, I guess the answer should be E as if take your case as total number is even the median will be mean of 10th and 11th project which will be (3+(4+n))/2. as we dont know n so we can not find median number of employees
Quote: Originally Posted by msd_2008 View Post Help needed with the following data suffienciency questions.
Q1. What is the median no. of employees assigned per project for projects at Company Z 1.25% of the projects at Company Z have 4 or more employees assigned to each project 2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project
Q2.In the xy plane, point(r,s) lies on the circle with its center at the origin. What is the value of r2+s2 ? 1.The circle has radius 2 2.The point (v2, -v2) lies on the circle
Regards MSD
1.Ans is C
lets consider this situation with 20 projects so
1.25% of the projects at Company Z have 4 or more employees assigned to each project ie 5 projects have 4 or more employees . Not sufficient so BCE
2.25 % of the projects at Company Z have 2 or fewer employees assigned to each project ie 5 projects have 2 or fewer employees. Not sufficient so CE
Combining the rest(middle) 15 projects of company has 3 employees each.... So median value is the employee in 10th project it is obviously 3
So ans is C
ashishjha100, I guess the answer should be E as if take your case as total number is even the median will be mean of 10th and 11th project which will be (3+(4+n))/2. as we dont know n so we can not find median number of employees
Please correct me if i am not
Dear Inder, i used an example for easier understanding.......consider case as such....from the combination of 2 statements we can know the no: of employees in remaining 50% projects is 3 (b/c 25% has 2 or fewer and 25 % has 4 or more) By considering any ways the median comes in the group of 50% projects and the only no: of employees in this 50% group is 3 so median is 3.....just consider with odd or even no: of projects.....
A certain quantity is measured on two different scales, the R-scale and the S-scale, that are related linearly. Measurements on the R-scale of 6 and 24 correspond to measurements on the S-scale of 30 and 60, respectively. What measurement on the R-scale corresponds to a measurement of 100 on the S-scale? A. 20 B. 36 C. 48 D. 60 E. 84