Baker's Dozen

If \(x<0\) then \(|x|=-x\) and \(\frac{|x|}{x}=\frac{-x}{x}=-1\), (\(-\frac{x}{x}=-1\) no matter whether \(x\) is negative or positive).

You are right that |x|=-x since x is negative but tell me, what is -x, negative or positive? Negative of negative gives you positive, right? So -x must be positive. Now, if x is negative,
-x/x must be positive/negative giving you -1.

You are confusing yourself too much with negatives and positives. Just think of it this way:

|x|/x = -x/x (By definition, since |x| = -x when x < 0)
-x/x = -1 ( x and x get canceled here leaving you with -1)

Did this way- prob of choosing 6 will be 1/10 on each of he three times (and these choices can be in \(5C3\)=10 ways) and the rest two digits can be any thing from set 0-9 (except 6) so in total we have 9 choices (i..e a prob of 9/10 for the rest of the 2 positions). So total prob will be\(10*(1/10*1/10*1/10*9/10*9/10)= 810/10^5\)(B)

Hi Bunuel,

I didn't quite understand this ..

X and Y are odd then how can |x| = -x it has to be +ive.

x and y are negative numbers, not odd numbers.

If \(x\geq{0}\) then \(|x|=x\);
If \(x<{0}\) then \(|x|=-x\). So if \(x\) is negative then \(|x|=-x=-negative=positive\). For example, if \(x=-2\) then \(|x|=|-2|=2=-x\).

Check Absolute Values chapter of Math Book for more: math-absolute-value-modulus-86462.html

Hope it helps.

Although there is hardly anything left after your's explanations but still I'll post my approach, which might help few people out here (esp. on the # trailing 0's as you people have kept in the Math book)--

first I simplified the whole x value as it looks so clumsy/heavy: given
\(x = (8!)^6 [(8!)^4 - 1]/ (8!)^3[(8!)^2 - 1] \\
= (8!)^3[(8!)^2 + 1][(8!)^2 - 1]/[(8!)^2 - 1]\\
so, (x/ (8!)^3) - 39 = [(8!)^2 + 1] -39\\
= (8!)^2 -38\)
Now as per the Q we need to know unit & tenth digit of the above number- as we know factorials of numbers > 4 have trailing zero's so here we can determine how many trailing 0's in 8!. Since \(8 < 5^2\) so that means we have only 1 trailing 0 in here (alternately you can have manual calc in this case as well because it's small number). So \((8!)^2\) has 2 trailing 0's. Now it does not matter what hundredth position is becasue we need to subtract only 38- this would result in 6 in tenth pos and 2 in unit pos. so product is = 12...

P.S.- just for refrence # trailing 0's in factorial N is determined by- \(N/5^1+N/5^2+....+N/5^k (where 5^k < N)\)for each division we take integer part (quotient) and then add them up. this should give # trailing 0's

I tried feagure out this way first but thought to calculate little differently-

Since there are a toatl of 12 Marbels and after picking the required # marbels we are left with 4 marbels in the Jar- so our task is determine out of these 4 in how many ways we can pick R, B marbels so that there exists at least 1 marbel of each type- (R-1,B-3), (R-2,B-2),(R-3,B-1)

So the required # ways = \(7C1*5C3 + 7C2*5C2 + 7C3*5C1\)
= \((7 * 10) + (21 * 10)+ (35 * 5)\)
= \(70 + 210 + 175\)
= 455 (D)

Hi Bunuel,
Please help me in understanding the below text. I think I am lacking this concept.I always try to solve by taking LCM in order to get rid of the fractions. However, I found your approach short n simple. If you could shed some light on this, would be really helpful

Fritz owns \(\frac{2}{2+3}=\frac{2}{5}\)th of all shares

Thanks
H

It's quite simple: A has $2 and B has 3$ --> A has 2/3rd of B's amount and also 2/(2+3)=2/5th of total amount of $5.

Hope it's clear.

Reading the discussion on this question made me realize how I end up retracting on the suggested strategies again and again. Since we are talking about Quant, you would expect to have a definite set of rules but no sir, you don't.

I repeat this quite often: When you have a question using variables and the answer is in terms of those variables, your life is easier than if you get a question using numbers. The reason - you can assign your own 'cool' values to the variables for which the relations hold. Then just go on and check the option which gives you the right answer.
I retract it with the following - But if the number of variables is high, say 3 or 4 or more, it might be too cumbersome to plug in values for each variable and keep in mind what stands for what. This could lead to a lot of confusion and errors.
I retract it again - But if you can give some very convenient values to the variables, go ahead and plug it in.

This question has variables and the answer is in terms of the variables so plugging in values is a good option. But the number of variables is 3 which could make it cumbersome. But, you can give the variables such values that you get your answer quickly.

I need to find the rate of A. There are no constraints on the values x, y, and z can take except z > x (drain C empties slower than pipe A fille)

Let's say, x = 4, y = 8, z = 8
What did I do here? I made the rate of B same as the rate of C. This means, whenever both of them are working together, drain C cancels out the work of pipe B. So the entire pool will be filled by pipe A and the amount of water pumped in by A will be the entire pool. Hence, if y = z, pipe A fills the entire pool i.e. the amount of water in terms of fraction of the pool pumped by A is 1.
In the options, put y = z and see which option gives you 1.
Only options (B) and (E) do.

Now let's say, x = 8, y = 4, z = 8.00001 ( z should be greater than x but let's assume it is infinitesimally greater than x such that we can approximate it to 8 only)
Rate of work of C is half the rate of work of B. Rate of work of C is same as rate of work of A. All the work done by pipe A is removed by drain C. So if pipe B fills the pool, drain C empties half of it in that time (this is the water pumped by pipe A)
Put x = z in the options B and E. You get y/z = 4/8 = 1/2 in option (B).

Answer (B)

Even if you end up feeling that this method is complicated, try and wrap your head around it. It might give you some ideas of logical solutions in some other questions.

Can any one please explain: why it is (9*9*1*1*1* 5C3) in Question no. 1 (password problem)?
Why not : 9*9*1*1*1* 5!/2! ??

That's a tough 700+ problem.

We are interested 666XX numbers. Now, two this can be arranged in 10 ways;
666XX;
66X6X;
6X66X;
X666X;

66XX6;
6X6X6;
X66X6;

6XX66;
XX666;

X6X66.

So, basically with \(C^3_5=10\) we are choosing which 3 places out of 5 will be occupied by X's. Or we could do the other way around: \(C^2_510\) choosing which 2 places out of 5 will be occupied by 6's.

Next, each X can take any value from 0 to 9, inclusive but 6, so total of 9 values. Notice that the first digit can be zero here since the 5-digit password unlike 5-digit number can start with 0.

Total \(9*9*C^3_5=810\).

Hope it's clear.

Can you explain how you got from (3^5-3^2)^2 to 3^4*(3^3-1)^2? When extracting 3^2 from (3^5-3^2), do we also square the 3^2? I know it may be simple algebra but its a bit confusing, I got 3^2(3^3-1)^2. I know thats wrong but can you explain why? Thanks.

Consider this: \((ab-ac)^2=(a(b-c))^2=a^2*(b-c)^2\), so \((3^5-3^2)^2=(3^2*(3^3-1))^2=(3^2)^2*(3^3-1)^2=3^4*(3^3-1)^2\).

Hope it helps.

Bunuel,
I did a mistake while calculating 10^5 - I calculated 9*10^4. I have a doubt : when the question says "5 digit" code, doesn't it mean that numbers start from 10000? I have seen some GMATClub questions were three digit codes start from 100. Can you please help?
Thanks

Since we are talking about a 5-digit password and not a 5-digit number, then it can start with zero. For example a password can be 00123 or 00001.

Hope it's clear.