Difficulty - Help

first 3 problems:

1. The present ratio of Students to Teachers at a certain school is 30 to 1. If the student enrollment were to increase by 50 students and the no. of teachers were to increase by 5, the ratio of students to teachers would then be 25 to 1. What is the present no. of teachers?

S/ T = 30/1
Hence S = 30 T

S + 50 / T + 5 = 25 / 1

S + 50 = 25T + 125

But S= 30T

So, 30T + 50 = 25T + 125

5T = 75

T = 15

Choice E - 15

2. In a certain company, the ratio of no. of managers to production line workers is 5 to 72. If 8 additional production line workers were to be hired, the ratio of the no. of managers to production line workers would be 5 to 74. How many managers does the company have?

M/ PW = 5/72

72M = 5PW

M/PW + 8 = 5/74

74M = 5PW + 40

But 5PW = 72M

SO, 74M = 72M + 40

2M = 40

M= 20

Choice D - 20

3. In a certain Calculus class, the ratio of the no. of Mathematics majors to the no. of students who are not
Mathematics majors is 2 to 5. If 2 more Mathematics majors were to enter the class, the ratio would be 1 to 2. How many students are in the class?

MM / S = 2/5

5MM = 2S

S = 5MM/2

MM + 2 / S = 1/2

2MM+ 4 = S

2MM + 4 = 5MM/2

4MM + 8 = 5MM

MM= 8

S= 5 * 8 /2 = 20

Total = 20 + 8 = 28.

Choice D - 28

Here hope these explanations help....


30x + 50 / x + 5 = 25/1

30x + 50 = 25x + 125

5x = 75

Therefore x = 75/5 = 15

Total number of teachers = 15 Hence (E)

2)

5x / 72x + 8 = 5 / 74

5x * 74 = 5 (72x + 8)

370x = 360x + 40

10 x = 40

Therefore x = 4

Hence the total number of managers with the company = 5*4 = 20

chk out the attachment for the rest of the answers ... hope they help u .. take care

tnks to all of you

Hi
I have some doubts. Please help

217.)in the first class the probability of choosing one of the 60 sibling pair is 60 /1000 … now the sibling of the kid selected can be selected in 1/ 800 ways .. so total probability is 60/1000*1/800 … 3 /40.000.. A


In the second class I was thinking that the kid can chosen only in 60/800 ways, since we need a sibling. You have stated that it would be 1/800 ways. What is the mistake I am making here?

248.) given that the side PR of the triangle is parallel to X axis so side PQ will have to be parallel to Y axis …. So P & R will have the same Y coordinates and P & Q will have the same X coordinates …… so for P ( & Q ) the X coordinates can be chosen in 10 ways and for R the no of choices for X coordinate wuld be 9 … similarly for P ( & R) the no. of ways of choosing the y coordinates is 11 and for Q it is 10 .. so total no of triangles that can be made is 10*9*11*10 = 9900 ways …C

I am not able to follow the solution. Would be grateful if you could throw some more light.

Thanks in advance

Prasanna

Prasanna wrote:Hi
I have some doubts. Please help

217.)in the first class the probability of choosing one of the 60 sibling pair is 60 /1000 … now the sibling of the kid selected can be selected in 1/ 800 ways .. so total probability is 60/1000*1/800 … 3 /40.000.. A


In the second class I was thinking that the kid can chosen only in 60/800 ways, since we need a sibling. You have stated that it would be 1/800 ways. What is the mistake I am making here?

248.) given that the side PR of the triangle is parallel to X axis so side PQ will have to be parallel to Y axis …. So P & R will have the same Y coordinates and P & Q will have the same X coordinates …… so for P ( & Q ) the X coordinates can be chosen in 10 ways and for R the no of choices for X coordinate wuld be 9 … similarly for P ( & R) the no. of ways of choosing the y coordinates is 11 and for Q it is 10 .. so total no of triangles that can be made is 10*9*11*10 = 9900 ways …C

I am not able to follow the solution. Would be grateful if you could throw some more light.

Thanks in advance

Prasanna

for the first q .. the condition is that 1 child is selected from each of the class .... and the probability needed is of the 2 kids picked being siblings ....

now starting with any class one element of the pair of siblings can be chosen in 60 ways .... so suppose we start with the senior class ... the probabilltiy of choosing one kid out of the 60 sibling pairs is 60/800


now comes the important part ... suppose u chose a kid named A whose sibling B is in the junior class ... now that u have already selected A the only way of completing the sibling pair is to choose B form the junior class ... so when u are choosing the 2nd kid u have effectively only one choice ... so that is the reason that the second time the probability is 1/ 1000 ... therefore the total prob= 60/800*1/1000 = 3 / 40,000...


2.) for the second question... the q says that for a right triangle PQR with angle p = 90 .... PR is parallel to the x axis ... now since y axis is perpendicular to the X axis .... pq ( the other leg of the right triangle) will be parallel to the y axis .....

now for any line parallel to the y axis ... the x coordinate will be same for all points on the line ... similarly for any line parallel to the x axis the y coordinate will be same for all th points on the line ...

now it is given that th X and Y coordinates of p,q,r needs to be integers and the range for them is -4< = x<= 5 and 6 <= y<= 16 ... so x can assume the integer values ... -4,-3,-2,-1,0,1,2,3,4,5 that is 10 values in all ... and y can assume the values 6,7,8,9,10,11,12,13,14,15,16 that is 11 values in all ....

now let us start choosing the values for the possible x coordinates for p,q,r.. starting with p ... so p will have 10 choices... now remember that PQ is parallel to y axis so Q will have the same x coordinate as P .... that leaves us with R which can obviusly have 9 choices for the x coordinate ( bcoz out of the 10 choices one value is already alloted to P and Q )... so total no of ways of choosing the x coordiante is ..10*9

similarly choosing the y coordinates .. starting with P ... y coordinate can be chosen for P in 11 ways and since PR is parallel to the X axis .... R will have the same y coordiante as P ... that leaves Q for which the y coordiante can be chosen in 10 ways ( bcoz out of the 11 choices one value has already been alloted to P and R )... so total no. of ways of choosing the y coordinate is 11*10

therefore the no. of ways of choosing x and y coordinates is 10*9*11*10 = 9900..

pheewww.. i know that was rather long ... but hope u understtand it this time.. take care

Hi, Gabriel---

Your answer was great! I understand PS # 248 of OG 11 now.

Prasanna, I'm not sure if you felt the same confusion I did after reading the explanation for this answer in the OG. What really baffled me was when it said that "Q has to be vertical to R". Wouldn't that make the right angle at R, instead of at P?

Anyway, I suppose it doesn't matter if Q is vertical to R or if it's vertical to P. We still have three points where two points have the same x-coordinate and two points have the same y-coordinate, with the same restrictions.

-Jen

Thanks a ton Gabriel for taking the pain to detail the answer. I understood it now.

Jen - I was trying to solve the problem in the post and I did not know that it was in OG. Now that I looked at it, the words 'Q has to be vertical with respect to R' is indeed confusing. But as you say we have three points anyway.