TOUGH & TRICKY SET Of PROBLEMS

This problem can be solved much quicker using the right triangles.
Note that line y = 3/4*x - 3, x-axis and y-axis form a 3:4:5 type triangle.
now, there are 2 quick ways to solve it from here:
1) using similar triangles - perpendicular to line y = 3/4*x - 3 creates two triangles similar to the original one , hence the length of the perpendicular can be calculated
or 2) using formula for the area of the triangle (a=1/2hieght x base) - we know area is 6 (4x3/2=6) and we know base is 5, so plug this into the formula and we get 12/5 or 2,4 for the height, which is also the length of the perpendicular we were looking for.
Hope this helps

Expanding is not a good idea. Below is a solution for this problem:

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?
A. 1/4
B. 3/8
C. 1/2
D. 5/8
E. 3/4

\(n(n+1)(n+2)\) is divisible by 8 in two cases:

1. When \(n\) is even:
\(n=2k\) --> \(n(n+1)(n+2)=2k(2k+1)(2k+2)=4k(2k+1)(k+1)\) --> either \(k\) or \(k+1\) is even so 8 is a multiple of \(n(n+1)(n+2)\).

# of even numbers between 1 and 96, inclusive is \(\frac{96-2}{2}+1=48\) (check this: totally-basic-94862.html?hilit=last%20first%20range%20multiple)

AND

2. When \(n+1\) is divisible by 8. --> \(n+1=8p\) (\(p\geq{1}\)), \(n=8p-1\) --> \(8p-1\leq{96}\) --> \(p\leq{12.1}\) --> 12 such numbers.

Also note that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.

Total=48+12=60

Probability: \(\frac{60}{96}=\frac{5}{8}=62.5%\)

Answer: D.

Again that's not correct. Let me ask you a question: what is the probability that a person in born on Monday? Is it 1/2? No, it's 1/7.

No, in that case we would be asked "what is the the probability of the first ball being red and the second ball being white?"

If an integer n is to be chosen at random from the integers 1 to 96, inclusive, what is the probability that n(n + 1)(n + 2) will be divisible by 8?

A. 25%
B. 50%
C. 62.5%
D. 72.5%
E. 75%

\(n(n + 1)(n + 2)\) is divisible by 8 in two cases:

A. \(n=even\), in this case \(n+2=even\) too and as \(n\) and \(n+2\) are consecutive even integers one of them is also divisible by 4, so their product is divisible by 2*4=8;
B. \(n+1\) is itself divisible by 8;

(Notice that these two sets have no overlaps, as when \(n\) and \(n+2\) are even then \(n+1\) is odd and when \(n+1\) is divisible by 8 (so even) then \(n\) and \(n+2\) are odd.)

Now, in EACH following groups of 8 numbers: {1-8}, {9-16}, {17-24}, ..., {89-96} there are EXACTLY 5 numbers satisfying the above two condition for n, for example in {1, 2, 3, 4, 5, 6, 7, 8} n can be: 2, 4, 6, 8 (n=even), or 7 (n+1 is divisible by 8). So, the overall probability is 5/8=0.625.

Answer: C.

Similar question: divisible-by-12-probability-121561.html

Hope it helps.

1 leap year in 4 --> the probability of being born in leap year = 1/4 --> the probability of not being born in leap year = 3/4.

The probability that among 3 people at least 1 of them is born in a leap year is:

P(at least 1) = P(exactly 1) + P(exactly 2) + P(all 3) = 3*1/4*3/4*3/4 + 3*1/4*1/4*3/4 + 1/4*1/4*1/4 = 37/64.

Or P(at least 1) = 1 - P(none) = 1 - (3/4)^3 = 37/64.

Some "at least" probability questions to practice:
leila-is-playing-a-carnival-game-in-which-she-is-given-140018.html
a-fair-coin-is-tossed-4-times-what-is-the-probability-of-131592.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-132074.html
a-string-of-10-light-bulbs-is-wired-in-such-a-way-that-if-131205.html
a-shipment-of-8-tv-sets-contains-2-black-and-white-sets-and-53338.html
on-a-shelf-there-are-6-hardback-books-and-2-paperback-book-135122.html
in-a-group-with-800-people-136839.html
the-probability-of-a-man-hitting-a-bulls-eye-in-one-fire-is-136935.html
for-each-player-s-turn-in-a-certain-board-game-a-card-is-141790.html
the-probability-that-a-convenience-store-has-cans-of-iced-128689.html
triplets-adam-bruce-and-charlie-enter-a-triathlon-if-132688.html
a-manufacturer-is-using-glass-as-the-surface-144642.html
the-probability-is-1-2-that-a-certain-coin-will-turn-up-head-144730.html (OG13)
a-fair-coin-is-to-be-tossed-twice-and-an-integer-is-to-be-148779.html
in-a-game-one-player-throws-two-fair-six-sided-die-at-the-151956.html
in-a-drawer-of-shirts-8-are-blue-6-are-green-and-4-are-24418.html

How is n = (k-1)/2??

If the last term is K, and the 1st term is 1, the no. of terms should be k-1+1
Hence, here the no. of terms should be =>(k-1+1)/2 => k/2 since there should be half as many even numbers...
Am I missing something?

Excellent! +1.

For those who don't know the distance formula (which is in fact very rarely tested) this is the easiest and most elegant solution.

I've some problem to understand this ; as the two events are mutually exclusive , independent.
So why not the answer is 3/9*2/9 = 2/27

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.

How do you make this Bunuel , I dont get it

Yes it's kind of confusing, partly because I didn't use the formula formatting. Edited the original solution. So here it is:


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: \(3.2+5xy=5.8-0.41xy\), solve for \(xy\) --> \(xy=0.48\);

The cost of a bushel of corn in \(y\) days (the # of days when these two bushels will have the same price) will be \(3.2+5xy=3.2+5*0.48=5.6\).

Answer: E.

Hope it's clear. Tell me if it needs more clarification.

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \(\frac{3}{9}*\frac{2}{9}\), so \(P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}\).

One more thing worth of mentioning: \(\frac{3}{9}*\frac{2}{9}\) 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.

Hope it's clear.

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):



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

Answer: D.[/quote]

The probability of drawing a red and a white ball has to be( 2/9)*(3/9) and then the solution follows. How come it is (2*3/9) / (2/9), i dont get it again .

Will continue again with rest of the questions.

P.S. I have never liked questions who have just been tough but no brainerbut the one in the list here are some of the best questions to train on many tricks and fundamentals I believe. More I see the stuff on quant , more I admire the work of the people engaged with the site. Bravo.. and cheers from cold Paris but grâce à GMAT for keeping me hot [/quote]

We want out of two balls one to be red (R) and another white (W).

There are two ways this could happen: we can draw FIRST ball red and SECOND white OR FIRST ball white and SECOND red.

Probability of each case is: \(\frac{3}{9}*\frac{2}{9}\), so \(P=\frac{3}{9}*\frac{2}{9}+\frac{2}{9}*\frac{3}{9}=2*\frac{3}{9}*\frac{2}{9}\).

One more thing worth of mentioning: \(\frac{3}{9}*\frac{2}{9}\) 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.

Hope it's clear
Perfect. That is how I solved it.

Next attack : Inequality : Will be going through your links to flex "inequality muscles" . Can't thank enough for the efforts you guys have put.

9. PROBABILITY OF INTEGER BEING DIVISIBLE BY 8:
If n is an integer from 1 to 96 (inclusive), what is the probability for n*(n+1)*(n+2) being divisible by 8?

A. 25%
B 50%
C 62.5%
D. 72.5%
E. 75%

N=n*(n+1)*(n+2)

N is divisible by 8 in two cases:
When n is even:
No of even numbers (between 1 and 96)=48
AND
When n+1 is divisible by 8. -->n=8p-1 --> 8p-1<=96 --> p=12.3 --> 12 such nembers

Total=48+12=60

Probability=60/96=0.62


my doubt is total even numbers is 48 + 12 , but the 12 numbers divisible by 8 already included in 48 . so why de we need to add 12

thanks in advance