TOUGH & TRICKY SET Of PROBLEMS

SOLUTION:
2. THE PRICE OF BUSHEL:
The price of a bushel of corn is currently $3.20, and the price of a peck of wheat is $5.80. The price of corn is increasing at a constant rate of \(5x\) cents per day while the price of wheat is decreasing at a constant rate of \(\sqrt{2}*x-x\) cents per day. What is the approximate price when a bushel of corn costs the same amount as a peck of wheat?

(A) $4.50
(B) $5.10
(C) $5.30
(D) $5.50
(E) $5.60

Note that we are not asked in how many days prices will cost the same.

Let \(y\) be the # of days when these two bushels will have the same price.

First let's simplify the formula given for the rate of decrease of the price of wheat: \(\sqrt{2}*x-x=1.41x-x=0.41x\), this means that the price of wheat decreases by \(0.41x\) cents per day, in \(y\) days it'll decrease by \(0.41xy\) cents;

As price of corn increases \(5x\) cents per day, in \(y\) days it'll will increase by \(5xy\) cents;

Set the equation: \(320+5xy=580-0.41xy\), solve for \(xy\) --> \(xy=48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(320+5xy=320+5*48=560\) or $5.6.

Answer: E.

Bunuel,

While solving for xy i get xy = 56

5xy - 0.41xy = 580-320
4.59xy = 260
xy = 56.6 cents

Let me know if i missed something!

The following answers may not be correct. If you find some discrepancy in the solution, please PM me or post it under the same topic. I took a cursory look into the OA's but am yet to match all my answers and compare the logic.


Let A's speed: A m/s
Let B's speed: B m/s

1/10th of a minute = 6sec
1/30th of a minute = 2sec

Total distance of the race track = 480m

First heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs

B's distance = 480-48=432m
B's speed = B m/s
B's time = t+6 secs

A's time = B's time - 6 (if A took 1000 seconds; B took 1006 seconds)
\(\frac{480}{A}=\frac{432}{B}-6\) ------ 1st (\(rate*time=distance \quad or \quad time,t=D/r\))

Second heat;
A's distance = 480m
A's speed = A m/s
A's time = t secs

B's distance = 480-144=336m
B's speed = B m/s
B's time = t-2 secs

A's time = B's time + 2
\(\frac{480}{A}=\frac{336}{B}+2\) ------ 2nd

Using 1st and 2nd,
\(\frac{432}{B}-6=\frac{336}{B}+2\)

\(\frac{432}{B}-\frac{336}{B}=8\)

\(\frac{96}{B}=8\)

\(B=12\)

B's speed : 12m/s

Ans: "A"

I dont have any Question on calculations made above.
But i do have a clarification to ask.

While calculating the time of B , it appears that we have not accounted the time taken by B to cover the first 48m(first heat) or 144m (second heat).
( As if the start time taken into consideration is the point from when both A and B are in action ).
Why is that we are going by this. I mean why are we not considering the head start time provided to B.

Please correct me if i missed anything.

5. RACE:
A and B ran, at their respective constant rates, a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10th of a

minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30th of a minute. What is B’s speed in m/s?
(A) 12
(B) 14
(C) 16
(D) 18
(E) 20

Isnt the correct answer A. it says 45 is the lowest temperature. Doest that statement imply that all other numbers have to be greater than 45. Hence at least 45,00001?

6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):



We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.

Hi Bunuel,
I understand the below solution. However, I was wondering if it could be solved using combination approach. Please tell me where I am going wrong.

Probability = favorable/ total outcomes.
Total outcomes = 9C2 (i.e total ways if drawing 2 balls)
Favorable = 3C2*4C1
Hence Probability = 3C2*4C1/9C2 = 1/6. which is completely wrong.
Please reply.
Thanks
H

Hi Bunuel - Regarding this question, I believe the answer should be 2 i.e. (B).

It is because the probability of a person's birth year to be a leap yr = 1/2, this is identical to a rainy/non-rainy day situation, where a year is either a leap or normal year.


Case I - No of people = 1
Probability of atleast 1 person's birth yr falling on a leap year = 1/2 = 50%
.....INCORRECT

Case II - No of people = 2
Probability of none of guys' birth year falling on a leap year = 1/2*1/2 = 1/4
Probability of atleast one of them was born on a leap year = 1 - 1/4 = 3/4 = 75%
.....CORRECT

Hence (B).

What do you think? Thanks!

That's not correct out of 4 years 1 is leap, so the probability of a randomly selected person NOT to be born in a leap year is 3/4.

But the question doesn't say that it's a set of 4 consecutive years. What if the years chosen are 2011, 2010, 2009, 2007?
IMO, a person's birth year to fall on a leap year should be a binary value - Y or N, hence a 50% probability.

SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.

The question says that red and white balls are selected in two Successive draws. Doesn't this imply that white is selected AFTER red? Thus no need for x2?

Thanks,
Diana

SOLUTION:
6. PROBABILITY OF DRAWING:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?
(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case (and was solved incorrectly by some of you):

\(P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}\)

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

Answer: D.

Hi Bunuel

I guess the question should be re-worded to restrict the number of draws to 2.

I think i made mistake since 6 can be also a probable number but is not a multiple of 4..............Bunuel can u pls help me solve this one using multiple principle

Winning scenario consists of TWO cases RW and WR. Probability of each case is: 3/9*2/9, so 3/9*2/9+3/9*2/9=2*3/9*2/9.

Only 3/9*2/9 would be correct answer if we were asked to determine the probability of FIRST ball being red and SECOND white, OR FIRST white and SECOND red. These are the cases in which order of drawing matters.