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Explanations for Problem solving questions of OG

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by arun.uict » Sat Apr 11, 2009 11:16 pm
Maths Official guide is great but it lacks the explanation as in Verbal OG. Need of the hour is explanations for PS & DS quesiton. Althouhg many explanation guides exist to solve & explain OG questions, self realization or discussion in forums presents better learning expereience.

Okay, I just searched this site. Although, many isolated & scattered threads discussing about individual OG problem solving questions, a one common thread which would have the explanation for almost all OG questions is not there. Over the period of time, such a thread itself can be compiled & a neat database can be made.

Mods can point out if such a thread already exist & can delete this thread also if necessary.

If not, I propose following details to be mentioned by the seekers of the explanation!!

1) OG version. (10th, 11th, 12th etc)
2) Question no
3) Reproduce atleast a part of question (this saves the time to refer back OG)
3) Brief explanation of difficulty they face with that particular question
4) Brief explanation of their methodology (this is for those looking for faster method)

If this is successful, then we can go for similair thread for DS also.
Last edited by arun.uict on Sat Apr 11, 2009 11:29 pm, edited 1 time in total.
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Source: — Problem Solving |
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by arun.uict » Sat Apr 11, 2009 11:27 pm
I will just start with one of my problems. I use OG 11th edition.

Q. 193) In a certain calculus dass, the ratio of the mathematics majors to the number of students who are not mathematics majors is 2 to 5. If 2
more mathematics majors were to enter the class, the ratio would become 1 to 2. How many students are in class?

For this question, I consider initial ratio of maths to non-maths as 0.4 and final ratio becomes 0.5, that is, 25% increase in ratio. This 25% increase is equivalent to 2 maths students. Hence originally 100% = 8 maths students were there & 20 non maths students. Hence total stregnth orginially was 28.

Now look for the another similair question in 170.

Q. 170) In ia certain mmpany, the ratio of the number of managers to the number of produetion-line workers is 5 to 72. If 8 additional production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

Now, If I apply the same methodology I used of 193, I would be struck here great time since ratio involved are not as simple as before. So, do somebody have any better method to solve this problem? I solve this 170 thruogh options & trial n error.

Thanks for your help....
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by sureshbala » Sun Apr 12, 2009 3:55 am
arun.uict wrote:I will just start with one of my problems. I use OG 11th edition.

Q. 193) In a certain calculus dass, the ratio of the mathematics majors to the number of students who are not mathematics majors is 2 to 5. If 2 more mathematics majors were to enter the class, the ratio would become 1 to 2. How many students are in class?

For this question, I consider initial ratio of maths to non-maths as 0.4 and final ratio becomes 0.5, that is, 25% increase in ratio. This 25% increase is equivalent to 2 maths students. Hence originally 100% = 8 maths students were there & 20 non maths students. Hence total stregnth orginially was 28.

Now look for the another similair question in 170.

Q. 170) In ia certain mmpany, the ratio of the number of managers to the number of produetion-line workers is 5 to 72. If 8 additional production-line workers were to be hired, the ratio of the number of managers to the number of production-line workers would be 5 to 74. How many managers does the company have?

Now, If I apply the same methodology I used of 193, I would be struck here great time since ratio involved are not as simple as before. So, do somebody have any better method to solve this problem? I solve this 170 thruogh options & trial n error.

Thanks for your help....
These problems from ratios can be answered quickly by making use of the following concept.....

Problem 193:

Initial ratio on Maths: Non-maths = 2:5

2 maths students joined the class.

Now the ratio of Maths:Non-maths = 1:2

Since only math students joined, the number of non-math students remain same. Try to reflect this even in ratios.

So take initial ratio as 4:10

Later ratio as 5:10.

So the increase is 1 part which is given as 2 students.

Now the original strength is represented by 14 parts (as per our new ratio).

So if 1 part is 2 students, 14 parts must be 28 students.


Probelm 170:

As discussed above,

Ratio of managers:workers = 5:72

8 workers joined...

Ratio of manager : workers = 5:74

There is no need to change this ratio as the number of managers is constant and the same info is reflected by these ratios.

So the increase in workers is 2 parts and this is given as 8 workers.

We have the number of managers as 5 parts and hence it is equal to 20.
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by arun.uict » Sun Apr 12, 2009 7:07 am
Thanks, sureshbala. That is bit of similair & I missed it.

Another OG question....

173) The probability is 1/2 that a certain coin will turn up heads on any given toss. If the coin is to be tossed three times what is the probabality that on atleast one tosses, the coil will turn up tails?

The way I procedded to solve this problem: I listed out different ways in which result of tossing coins thrice can be shown.

{H, H, H}, {H, H, T}, {H, T, T}, {T, T, T}

Here, in atleast three ways, I get minimum one tail. So the probablity is required event is 3/4. Simple, isnt? But OG lists the answer as 7/8. Can somebody confirm? What mistake I'm making?
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by kapsii » Sun Apr 12, 2009 12:37 pm
arun,

The mistake you made was in counting, total possible outcomes are 8:
{H,H,H},{H,H,T},{H,T,H},{T,H,H}, {H,T,T}, {T,H,T}, {T,T,H}, {T,T,T}

For probability, there are two simple rules to remember:
Probability of an event happening = (number of favorable events) / (total number of events)

and Probability of an event happening = 1 - Probability of that event NOT happening.

Total possible results you can have in 3 coin tosses.
There are three tosses and each toss can have two outcomes, either Head or Tails. Hence total possible outcomes = 2*2*2 = 8.
To find the probability of at least one tail occurring, it would be easy to go the other way round.
How many outcomes are possible where there is not a single tails in the result? Ans - There is only one such outcome. {H,H,H}.
Thus probability that NO tails is obtained in the three tosses is 1/8.
Thus probability that there is at least one tails in the three tosses = 1 - (1/8) = 7/8.

HTH.
Cheers,
Dubes
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