# permutation with repetition allowed

When the order doesmatter it is a Permutation. The formula is written: n r. where, n is number of things to choose from; r is number of things we choose of n; repetition is allowed; order matters; Permutation without Repetition In this post, we will see how to find all lexicographic permutations of a string where repetition of characters is allowed. 216. Another definition of permutation is the number of such arrangements that are possible. Permutations with Repetition Permutation when repetition is allowed. When a permutation can repeat, we just need to raise n to the power of however many objects from n we are choosing, so . For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. Male or Female ? "With repetition" means that repetition is allowed. First position can have N choices The second position can have ( N-1 ) choices. Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time. To improve this 'Permutation with repetition Calculator', please fill in questionnaire. There are basically two types of permutation: Repetition is Allowed: It could be “333”. A permutation is an arrangement of objects, without repetition, and order being important. Repeating of characters of the string is allowed. Permutations are items arranged in a given order meaning […] List permutations with repetition and how many to choose from. Like combinations, there are two types of permutations: permutations with repetition, and permutations without repetition. They have sometimes been referred to as permutations with repetition, although they are not permutations in general. Permutations. The number of permutations of ‘n’ things taken ‘r’ at a time is denoted by n P r It is defined as, n P r A permutation with repetition of n chosen elements is also known as an "n-tuple". Total […] No Repetition: for example the first three people in a running race. That was an $$r$$-permutation of $$n$$ items with repetition allowed. Permutations without Repetition In this case, we have to reduce the number of available choices each time. All the different arrangements of the letters A, B, C. All the different arrangements of the letters A, A, B A permutation is an ordering of a set of objects. Print all permutations with repetition of characters, Given a string of length n, print all permutation of the given string. In some cases, repetition of the same element is allowed in the permutation. Please update your bookmarks accordingly. (Repetition allowed, order matters) Ex: how many 3 litter words can be created, if Repetition is allowed? We should print them in lexicographic order. D. 320. For this case, n and k happens to be the same. Permutation formulas. Permutation without Repetition: for example the first three people in a running race. 1. There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. C. 120. When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions. With permutations, every little detail matters. These are the easiest to calculate. This is a permutation with repetition. It could be "333". Permutation with Repetition. You can’t be first and second. When the number of object is “n,” and we have “r” to be the selection of object, then; Choosing an object can be in n different ways (each time). For example, what order could 16 pool balls be in? Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. Permutations with repetition. Permutation with repetition occurs when a set has r different objects, and there are n choices every time. Repetition of characters is allowed. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Permutations without repetition A permutation is an arrangement, or listing, of objects in which the order is important. For the given input string, print all the possible permutations. Compare the permutations of the letters A,B,C with those of the same number of letters, 3, but with one repeated letter $$\rightarrow$$ A, A, B. After choosing, say, number "14" we can't choose it again. Permutations: There are basically two types of permutation: Repetition is Allowed: such as the lock above. Technically, there's no such thing as a permutation with repetition. n different things taking r at a time without repetition - definition The number of permutations of n different things, taking r at a time without repetition is denoted by n P r . We can actually answer this with just the product rule: $$10^5$$. It has following lexicographic permutations with repetition of characters - AAA, AAB, AAC, ABA, ABB, ABC, … 7.1.5 When repetition of objects is allowed The number of permutations of n things taken all at a time, when repetion of objects is allowed is nn. Permutations with repetition. OR 125. 1. Ordered arrangements of n elements of a set S, where repetition is allowed, are called n-tuples. There are two types of permutations: Repetition is Allowed: For the number lock example provided above, it could be “2-2-2”. For example, locks allow you to pick the same number for more than one position, e.g. Post, we have 26 choices of letter: $26^ { }! 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# Conditional Probability: probability of having a girl.

## Homework Statement

I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?

## The Attempt at a Solution

I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.

But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.

Which is right?

1/2
it doesn't matter

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My vote's 1/3. This is a different question to 'My eldest child of two is a girl, what is the probability that they are both girls', which is 1/2. In your case, the piece of information provided is not 'ordered' but, instead, excludes the BB possibility.

Using Bayes theorem, the probability of event A, given event B, (with events independent) is the probability of A intersect B divided by probability of B.

So, the probability of the kid being a girl, given that there is already a boy, is the probability of a girl intersected with a boy divided by the probability of a boy.

To find A intersect B for independent events, just multiply the probability of each. (1/2)(1/2). So, (1/4) divided by (1/2).

It's the probability that it's a girl, given that at least one is a girl - nothing involving a boy. Does that change things?

I think you're right. The outcome of one boy and one girl should be twice as likely as the other 2 outcomes, since this is a http://en.wikipedia.org/wiki/Binomial_distribution" [Broken] with n=2 and p=1/2.

More formally, let $p = 1/2$ be the probability a girl is born and on our universe U define the random variables X = number of girls, Y = number of boys. Define the events $C = \{X = 2 \}$ and $D = \{X \geq 1 \}$. Certainly $C \subseteq D$ so $C \cap D = C$. Since $p = 1/2$ and the two births are independent events, then $\mathbf{P}(C) = (1/2)^2 = 1/4$. The probability of the complement of D, $\mathbf{P}(U \setminus D)$ is the same as $\mathbf{P}\{X=0\}$, which is the probability of 2 boys. Similarly, since $1-p = 1/2$ and the two births are independent events, then $\mathbf{P}(U \setminus D) = (1/2)^2 = 1/4$. So $\mathbf{P}(D) = 1 - \mathbf{P}(U \setminus D) = 3/4$. Finally
$$\mathbf{P}(C|D) = \frac{\mathbf{P}(C \cap D)}{\mathbf{P}(D)} = \frac{\mathbf{P}(C)}{\mathbf{P}(D)} = \frac{1/4}{3/4} = 1/3 \, .$$

I think your proof using the reduced sample space is correct as well--and shorter, too!

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diazona
Homework Helper
Yep, 1/3. For a less technical explanation, if it helps anyone:

Imagine surveying a large number of families with 2 children. In half of them, the older child will be a boy. Within that half, half again (or a quarter of the total) will have the younger child be a boy, and the rest (another quarter of the total) will have the younger child be a girl. The same goes for the remaining half of the families whose older child is a girl. So you have four possibilities: BB, BG, GB, GG (in order of age), each representing a quarter of the total population.

Now, the statement "given that at least one of them is a girl" excludes one of those possibilities, BB. There are three equally likely possibilities left (BG, GB, GG), of which exactly one has both children being girls. So the probability is 1/3.

Imagine surveying a large number of families with 2 children...

Yes, that was the the means I convinced myself of my analysis.

I think the comment about Bayes has actually taken the conclusion the wrong way. What we learn in the statement 'there is a girl' is a piece of information that teaches us to exclude 1/4 of the population set of 2-child families.

@AN; You have concluded Bayes says 'To have a boy is twice as likely as to have a girl'. Therefore probability of two girls is 1/3 and a boy and a girl is 2/3, because 2/3 is twice 1/3!!! (Because we've excluded the BB option from the P=1 outcome.)

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Oops, I was talking about girls and boys. That doesn't change the answer though. The Bayes theorem is sound. The answer is 1/2.

But what's the question? '1/2' is not the answer to the op, but is the answer to 'what is the probability of GG compared with not GG' (conditional on 'there is a girl').

Ray Vickson
Homework Helper
Dearly Missed

## Homework Statement

I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?

## The Attempt at a Solution

I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.

But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.

Which is right?

This is the same problem as: we toss a (fair) coin twice and get at least one head; what is the probability we get another head as well? The results LOOK a bit paradoxical when written out as follows: events are C = {two heads}, A = {first toss is heads}, B = {second toss is heads}. We have P{C|A} = P{C|B} = 1/2, but P{C|A or B} = 1/3. That is your problem in a nutshell. This looks paradoxical because the statement that we get at least one head is the same as saying that A or B occur, and each of these separately implies a 1/2 probability of C; however, when taken together they imply a 1/3 probability of C!

RGV

I like Serena
Homework Helper
Hi everyone!

Here's my view on the matter.

There is a difference between saying:
"given the first (or youngest) kid is a girl" (chance 2/4)
and "given that at least one kid is a girl" (chance 3/4).

In the first case the chance would be 1/2 as Arcana claims.
We can also write this as:
$$P(\text{2 girls | the youngest kid is a girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 2}$$

In the second case it is:
$$P(\text{2 girls | at least 1 girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 3}$$

So I believe the reasoning of pearapple is sound for this problem and the chance is 1/3.

vela
Staff Emeritus
Homework Helper
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Since the others have explained how to get the correct answer, I'll point out what mistake your friend is making:

You have four possible outcomes: BB, BG, GB, and GG. BG and GB are distinct outcomes. The order matters when you're enumerating outcomes.

When you combine them to form events, then the order may or may not matter depending on how the event is defined. The event "exactly 1 girl" would be {BG, GB}. Whether the girl comes first or last doesn't really matter.

Note that even though the order doesn't matter in determining what outcomes are included in an event, you still have to add up the probabilities for each outcome to get the probability of the event. That is, P({BG, GB}) = P({BG})+P({GB}) = 1/2. It's incorrect to say: order doesn't matter, so BG=GB and P({BG, GB})=P({BG})=1/4.

The event "at least one girl" is {BG, GB, GG}, which corresponds to a probability of 3/4. If you incorrectly reduce this to {BG, GG}, as your friend has done, you get a probability of 1/2. If you take the probability GG, which is 1/4, and divide it by 3/4, you get the correct answer of 1/3. If instead you divide it by the incorrect probability of 1/2, you get 1/2, the answer your friend got.

BruceW
Homework Helper
Its amazing that even such a simple example has an answer that is not immediately obvious.

The answer to this question is ambiguous because the question itself is ambiguous.

I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.

So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.

If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB

BruceW
Homework Helper
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?

Is different to:

I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?

I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.

I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.

HallsofIvy
Homework Helper
Pretty much what others are saying: there are two different questions that can be confused here:
1) A person has two children, one of which is a girl. What is the probability that the other child is a girl?

2) A person has two children, the older of which (or the younger or the one we happen to know- anything that distinguishes one child from another) is a girl. What is the probability that the other child is a girl?

With two children, and assuming that "boy" and "girl" for any given child are equally likely, There are four equally likely possibilities: BB, BG, GB, GG where, of course, "B" means "boy", "G" means "girl" and the order depends upon which was born first, or was the one we knew, etc.

With problem (1) above, knowing that one of the children is a girl removes "BB" from the list, leaving "BG", "GB", and "GG". In only one of those is the other child also a girl. Probability of two girls, 1/3.

With problem (2) above, knowing that the older child (or younger child or whichever is distinguished in some way) we remove both "BB" and "BG". We are left with "GB" and "GG" of which 1 has both girls. Probability of two girls, 1/2.

I like Serena
Homework Helper
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?

Is different to:

I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?

I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.

I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.

Uhh :uhh:. Aren't those the same?

The answer to this question is ambiguous because the question itself is ambiguous.

I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.

So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.

If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB

Uhh :uhh:. Aren't those the same?

Is there anything that you don't understand specifically about the above explanation?

I like Serena
Homework Helper
Is there anything that you don't understand specifically about the above explanation?

It seems to me that BruceW's explanation is not correct.

It seems to me that BruceW's explanation is not correct.

I feel similar about his explanation, not that it is incorrect, just maybe incomplete.

vela
Staff Emeritus
Homework Helper
I think his point might have been clearer if he had changed the question in the second example to "What's the probability the other one is a girl?" to make it clear he's picked one child and that she is a girl and now we're only wondering about the other one.

I don't think the question as originally stated (the one in the original post) is ambiguous. The mistake occurs because of a misconception. It boils down to the fact that the phrase at least one is a girl doesn't mean you can single one girl out and then worry about the rest.

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I like Serena
Homework Helper
BruceW might have said: "I have two kids. I know the youngest of them, which is a girl. What is the probability that they are both girls?"

Then it would be right.
Now it's not.

BruceW
Homework Helper
The age doesn't matter. I agree with HallsofIvy's post. As long as there is some way to distinguish between the two children, then there are 2 very similar questions. (But the questions are different, and the answer is different).
And the question asked was the first one, but pearapple's friend mistook it to mean the second one. (I am guessing that is where his mistake came from).

I like Serena
Homework Helper
Right. In retrospect I concur.


# Assume that $(a_n)^{\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \ge 0$, what is the formula for $a_n$?

Assume that $(a_n)^{\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \ge 0$, what is the formula for $a_n$?

we know that $F(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots$
and our sequence is $1, 2, 5, 12, 27, 58, \dots$
to find the generating function for the sequence I started with
$F(x)=a_0+(\underline{2a_0}+\widetilde0)x+(\underline{2a_1}+\widetilde1)x^2+(\underline{2a_2}+\widetilde2)x^3+\dots$
after distributing the variables into the parenthesis, i separated the terms with $\underline{}$ and $\widetilde{}$
$\underline{} = 2a_0x+2a_1x^2+2a_2x^3+2a_3x^4+\dots$
$\underline{} = 2x(a_0+a_1x^1+a_2x^2+a_3x^3+\dots)$
$\underline{} = 2x \times F(x)$

$\widetilde{} = x^2 + 2x^3 + 3x^4 + 4x^5$
now I don't know what to do with this part, if I had used $n$ earlier instead of putting in the numbers I could just factor out $nx^2$ and get $\widetilde{} = nx^2(\frac{1}{1-x})$ but I don't think I can have $n$ when solving for the generating function of the sequence?

Any ideas of how to get past just this part?

• Pull an $x^2$ out and note that $$\frac{d}{dx}\left(\frac{1}{1-x}\right)=1+2x+3x^2+4x^3+...$$ Oct 22, 2016 at 1:56
• @Eleven-Eleven If I use your idea then I would get $F(x)=1+2xF(x)+\frac{1}{(1-x)^2}$, this equals $F(x)=\frac{(1-x)^2+1}{(1-2x)(1-x)^2}$ then I could use partial fractions to separate this into three terms with a $1-\underline{}x$ in the denominators, find a summation and find $a_n$ Oct 22, 2016 at 2:21
• Yes except you have $x^2$ in the numerator, not 1. Oct 22, 2016 at 2:27
• For all things generating functions, look up Wilf's Generatingfuntionology... it is free to download online and you get all the tricks to solve such expression... Oct 22, 2016 at 2:32

From the recursion formula, you can guess a linear combination of $a_n$ and $n$ will be a geometric sequence, that is, you can assume $a_{n+1}-\lambda(n+1)+M=2(a_n-\lambda n+M)$ for some $\lambda$ and $M$. Put it into the the recursion formula, we can take $\lambda=-1$ and $M=1$. So we get $$a_{n+1}+(n+1)+1=2(a_n+n+1).$$

Let $b_n=a_n+n+1$, then $b_{n+1}=2b_n$, which implies that $b_n=2^nb_0$. Since $b_0=a_0+0+1=2$, then $b_n=2^{n+1}$, which implies that $$a_n=b_n-n-1=2^{n+1}-n-1.$$

• Your method works. However, the op is trying to use generating functions to solve... Oct 22, 2016 at 2:02

My preferred approach using generating functions starts by assuming that $a_n=0$ for $n<0$ and rewriting the recurrence so that it hold for all $n\ge 0$:

$$a_n=2a_{n-1}+n-1+2[n=0]\;.\tag{1}$$

Here $[n=0]$ is an Iverson bracket, and the last term of $(1)$ ensures that $a_0=1$. Now multiply $(1)$ by $x^n$ and sum over $n\ge 0$; if we let $g(x)=\sum_{n\ge 0}a_nx^n$ be the generating function, we have

\begin{align*} g(x)&=\sum_{n\ge 0}a_nx^n\\ &=2\sum_{n\ge 0}a_{n-1}x^n+\sum_{n\ge 0}nx^n-\sum_{n\ge 0}x^n+2\\ &=2x\sum_{n\ge 0}a_nx^n+x\sum_{n\ge 0}nx^{n-1}-\frac1{1-x}+2\\ &=2xg(x)+\frac{x}{(1-x)^2}-\frac1{1-x}+2\;. \end{align*}

Solving for $g(x)$, decomposing the resulting rational function into partial fractions, and rewriting the resulting terms as formal power series, we get

\begin{align*} g(x)&=\frac{x}{(1-x)^2(1-2x)}-\frac1{(1-x)(1-2x)}+\frac2{1-2x}\\ &=\frac{x-(1-x)+2(1-x)^2}{(1-x)^2(1-2x)}\\ &=\frac{1-2x+2x^2}{(1-x)^2(1-2x)}\\ &=\frac2{1-2x}-\frac1{(1-x)^2}\\ &=2\sum_{n\ge 0}(2x)^n-\sum_{n\ge 0}(n+1)x^n\\ &=\sum_{n\ge 0}\left(2^{n+1}-n-1\right)x^n\;, \end{align*}

which yields the closed form

$$a_n=2^{n+1}-n-1\;.$$

You must note that $a_0=1$ which is unaccounted for...
Using the fact that

$$\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2}=1+2x+3x^2+...$$

you now should get

$$F(x)=1+2xF(x)+x^2\frac{1}{(1-x)^2}$$ $$F(x)(1-2x)=1+\frac{x^2}{(1-x)^2}$$ $$F(x)=\frac{1-2x+2x^2}{(1-2x)(1-x)^2}$$

Now use partial fraction decomposition to obtain three power series and look at the coefficients.

• just making sure, it's probably a typo but shouldn't the numerator be $1-2x+2x^2$? Oct 22, 2016 at 2:34
• Yes.... on an iPhone so mistakes are bound... Oct 22, 2016 at 2:36


# Speed of two trains travelling side by side

I'm a high school student, and I have come across a problem that I cannot solve. I feel there must be something obvious that I'm not seeing.

Problem: The distance between two train stations is $96$ km. One train covers this distance in $40$ minutes less time than another one. The second train is $12$ km/h faster than the first one. Find both trains' speeds.

What I have done: Set $v_1+12 = v_2$ (the speed of train $2$ is $12$km/h more than speed of train $2$), and $96/(v_1) = (96/v_2)-40$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train $2$) Now, from here I get to: $v_1 = v_2-12$.

\begin{align} &\frac{96}{v_2-12} = \frac{96}{v_2}-40 \\ &\qquad\implies \frac{96}{v_2-12} = \frac{96-40v_2}{v_2} \\ &\qquad\implies v_2\cdot 96 = (v_2-12)\cdot (96-40v_2) \\ &\qquad\implies v_2\cdot 96 = v_2\cdot 96-40v_2^2-1152-380v_2 \\ &\qquad\implies 0 = -40v_2^2-380v_2-1152 \end{align}

Solving this quadratic equation yields no real roots.

Could you please suggest the right way to go?

• You are not going to get a sensible solution unless the faster train takes less time – Henry Jan 10 '18 at 14:34
• You wrote this: $96/(v_1)=(96/v_2)−40$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train 2). The equation and the parenthetical remark contradict each other. If your equation is correct, $96/(v_1)$ is smaller ($40$ less) than $96/(v_2)$, so your equation says train 1 is faster. – Steve Kass Jan 10 '18 at 14:39
• +1 for first time poster a.) showing work, b.) not flat out asking for the answer – jameselmore Jan 10 '18 at 14:40
• If you find one or more of the answers helpful, it is customary to "Accept" one of them. Either accept the answer that you think is most helpful, or (if there are several that are all just as good), you might consider accepting the answer from the user with the least reputation (as accepting an answer gives a small reputation reward). – Xander Henderson Jan 10 '18 at 15:53
• The moral of the story is: write down your units! Now that you have found your mistake, here is a follow up problem. The two trains are at opposite ends of the 96km route, heading towards each other on the same track at the speeds you have deduced. How far apart are they when they collide? – Eric Lippert Jan 10 '18 at 19:53

Let $v_1$ km/hr denote the speed of the faster train, and $v_2$ km/hr denote the speed of the slower train (I seem to have reversed your notation—sorry, $v_1$ feels like faster variable to me than $v_2$). First off, we can relate the amount of time it takes for each train to travel the 96 km to the speed of each train. So, let $$t_1 \text{ hrs} = \frac{96 \text{ km}}{v_1 \ \frac{\text{km}}{\text{hr}}} = \frac{96}{v_1}\text{ hrs} \qquad\text{and}\qquad t_2 \text{ hrs} = \frac{96 \text{ km}}{v_2 \ \frac{\text{km}}{\text{hr}}} = \frac{96}{v_2}\text{ hrs}\tag{1}$$ denote these two times. We know that the faster train arrives 40 minutes (that is, $\frac{2}{3}$ of an hour—watch the units! (this is an easy mistake to make—I messed it up, too!)) earlier than the slower train, which implies that $$t_1 \text{ hrs} = t_2 \text{ hrs} - \frac{2}{3} \text{ hrs} = \left( t_2 - \frac{2}{3} \right)\text{ hrs},$$ and we know that the faster train is 12 kph faster than the slower train, hence $$v_1 \ \frac{\text{km}}{\text{hr}} = v_2 \ \frac{\text{km}}{\text{hr}} + 12 \ \frac{\text{km}}{\text{hr}} = \left(v_2 + 12\right) \ \frac{\text{km}}{\text{hr}}.$$ It should be noted that the only major mistake that I see in your work is in the above step—in your model the faster train takes more time to cover the distance, which is a problem. Substituting these into the equations at (1) (and eliding units—the units of time are hours, the unit of distance are kilometers, and the units of speed are km/hr), we get the system $$\begin{cases} t_2 - \dfrac{2}{3} = \dfrac{96}{v_2 + 12} \\ t_2 = \dfrac{96}{v_2}. \end{cases}$$ Can you solve it from here?

• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25
• Xander, I wonder if you would consider actually including the units in your equations? I consider writing units to be such an important best practice (and, in particular, one that would have made one of the errors obvious) that I'd like to see it represented among the answers. But I don't think I could improve on your explanation, which is why I didn't just write my own answer. – David Z Jan 11 '18 at 0:38
• @DavidZ Indeed, and I probably would have done that in the first place if the "obvious" sign error hadn't made me complacent. I've edited the answer to include units in the setup. – Xander Henderson Jan 11 '18 at 6:29
• Thanks! (Your edit looked a little strange at first and I took a while to realize it's because I would omit units after the variables, e.g. let $v_1$ represent a speed, not just a number, and then write things like $v_1 = v_2 + 12\ \mathrm{km/hr.}$, but I suppose that's just a matter of preference.) – David Z Jan 11 '18 at 7:08

The question states that "One train covers this distance in 40 mins less than the other". Although it does not tell you which train, it is quite obvious that the faster train (i.e. train 2) takes 40 mins less.

So instead, it should be $96/(v2)=(96/v1)−40$.

• Also, you've mixed up hours and minutes. It should be 2/3 not 40. – Michael Behrend Jan 10 '18 at 14:46
• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25

I'll post this as an answer, since i'm not yet allowed to write comments. So, as mentioned in the comments (and the answer given by glowstonetrees) you might want to use $96/(v2)=(96/v1)−40$ instead.

Also keep in mind that you are subtracting minutes from hours. Your final formula should be $96/(v2)=(96/v1)−(40/60)$.

• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25

A method that doesn't require the quadratic formula:

Once you've corrected the errors mentioned in the other answers, you should have

96/(v+12) = 96/v – 2/3

You can rewrite that as

8/(v/12 +1) = 8/(v/12) -2/3

12/(v/12+1)=12/(v/12) – 1

This is in some ways a more complicated form, but if we assume that each term is an integer, then v/12 and v/12+1 must be factors of 12. And what factors of twelve are there that differ by 1? Just 3,4. If we plug those in and check, 12/4 = 12/3 -1 => 3 = 4-1. So v/12 = 3 => v = 36 and the other speed is 4*12 = 48.

Also, when you get an equation like

v2⋅96=(v2−12)⋅(96−40v2)

You should divide out by the common factor of 8 and get

v2⋅12=(v2−12)⋅(12−5v2)

before multiplying it out.


# The approximate uncertainty in $r$

The volume of a cylinder is given by the expression

$V=\pi r^2 h$

The uncertainties for $V$ and $h$ are as shown below

\begin{align} V&\pm 7\%\\ h&\pm 3\% \end{align}

What is the uncertainty in $r$?

Now, the obvious answer would be $2\%$, from $$\frac{dV}{V}=\frac{dh}{h}+2\frac{dr}{r}$$

However, rearranging to $r^2=\frac{V}{\pi h}$ gives $$2\frac{dr}{r}=\frac{dV}{V}+\frac{dh}{h}$$ which gives a different answer of $5\%$. Thus, by simply rearranging the formula, we get different values for uncertainty in $r$.

How do you explain this?

(The mark scheme lists $5\%$ as the correct answer)

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@Helder, thats fine! –  0sh Sep 22 '11 at 13:09

You're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated from the $x_{i}$.

In the example you give, you have two independent measurements of $V$ and $h$ and are expected to calculate the uncertainty in $r$. Well, to use the above formula, you need to write $r$ as a dependent variable of $V$ and $h$. Therefore, it's only correct to solve for $r$ first, and then calculate the uncertainty.

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Makes sense, thanks! –  0sh Apr 20 '11 at 12:36
Ted Bunn's answer is a very important addition to this answer. –  Qmechanic Apr 20 '11 at 17:56
@Qmechanic: agreed. I typed this up to get the jist of an answer before I had to leave for work. –  Jerry Schirmer Apr 20 '11 at 18:26

Jerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right.

But the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the "errors" mean standard deviations as usual, then the correct procedure is to add the errors in quadrature (i.e., to add the squares and take the square root): $${\delta r\over r}={1\over 2}\sqrt{\left(\delta V\over V\right)^2+\left(\delta h\over h\right)^2}.$$ See, e.g., any of the first few Google hits for "propagation of errors."

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the measurements of V and h are independent? –  Helder Velez Apr 26 '11 at 2:35
That's what one is apparently meant to assume in this problem. Without more detail about how the measurements were made, there's no way to tell if this is reasonable, but if $h$ is measured with a ruler and $V$ is measured by dunking the thing in water and watching the water level rise, for instance, this would be correct. If $V$ is measured by measuring $r$ and $h$ and calculating $\pi r^2h$, then of course the measurements wouldn't be independent, but the question would make no sense if that were the case, so I think we're meant to assume it's not. –  Ted Bunn Apr 26 '11 at 2:52

This should be a comment to Jerry Schirmer's right answer.

The solutions you give solve different problems.

In the first calculation you measure V with its uncertainty $\Delta V$, you know it depends with r and h, you know the uncertainty you have in the measurement of h, $\Delta h$ and thus you infer the uncertainty you had associated with measuring r, $\Delta r$.

In the second calculation you want to measure r from V and h. Knowing the uncertainties in V and h you predict the uncertainty that you will have in r.

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Your first formula is correct, but it looks like you substituted in the incorrectly signed value for (dh/h). Your second formula is missing a sign change for the (dh/h) term.

In the first equation, substituting, to get the MAXIMUM (dr/r) --->

+7% = (-3%) + (2 x (+5%)) -OR- -7% = (+3%) + (2 x (-5%))

So, your text is correct: +/-5% is the (approximate) uncertainty for r.

If you solve this problem algebraically, you will find that the (corrected) differential equations you constructed gives a very close but not exact answer. The correct uncertainty for r is something like + SQRT(1.07/0.97) and - SQRT(0.93/1.03) = +5.0282% / -4.9783%. This is because the differential eqns are valid only for infinitesimal deltas; but we are using deltas on the order of +/-5% (hardly infinitesimal).

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I will use two ways to derive the uncertainity on r
1 - very simple method: the 'interval arithmetic' present in the Euler math package (free)
2 - derive the formula of dr(V,h, dV,dh)
Both answers : 5% (for any value V or h that I choose for V and h)

// 'using interval arithmetic'
>dV=0.07 dh=0.03         // deltas
>V=1000 h=40             //arbitrary choosen values
>r0=sqrt(1/pi*V/h)       //central value for r
2.82094791774
>r=sqrt(1/pi*((V±V*dV)/(h±h*dh))) //interval values for r
~2.6,3~                          //    from 2.6 to 3
>(r-r0)
~-0.14,0.14~
>(r-r0)/r0
**~-0.05,0.05~                     // 5% for dr/r0 for any V, h**

//calc dr(V,h, dV,dh) (seeResistance_measurement Example and adapting the derivatives)

dr=1/2/sqrt(pi)*sqrt(dV^2/V/h+dh^2*V/h^3)
0.00106245302667
1-(r-dr)/r0
~-0.05,0.05~ // the result is again 5%

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Summation of series 12,40,90,168,280,432,…? [closed]

I have been stuck in finding the general term and the sum of $n$ terms of the series

$$12,\,40,\,90,\,168,\,280,\,432, ...$$

I am not able to see any relationship between the successive terms of the series.

Is there a pattern which I am not able to see?

• There has to be. If there is no pattern, you can't come up with the other terms. – Laray Jan 17 '17 at 16:07
• There is a pattern, which clearly points at the next term being $$42$$ as every other term after it. Thus, for every $n\geqslant6$, the sum of the $n$ first terms is $$42n+770$$ – Did Jan 17 '17 at 16:12
• If you compute repeated differences, then you'll see that there is a simple cubic polynomials that explains these values: WA. – lhf Jan 17 '17 at 16:13

Clearly there is a pattern.

Firstly find the orders of differences.

They are

28, 50, 78, 112, 152,...

22,     28,     34,     40, ....

6,      6,      6, ....

0,      0,.....

Hence, the $n^{th}$ term can be written as

$12+28(n-1)+\frac{22n(n-1)(n-2)}{2!}+\frac{6n(n-1)(n-2)(n-3)}{3!}$

=$n^3+5n^2+6n$

The sum of n terms of the above series is

$\sum {n^3}+5 \sum {n^2}+6\sum {n}$

=$12n$+$\frac{28n(n-1)}{2!}+\frac{22n(n-1)(n-2)}{3!}+\frac{6n(n-1)(n-2)(n-3)}{4!}$

=$\frac{n(3n^2+26n+69n+46}{12})$

=$\frac{1}{12}n(n+1)(3n^2+23n+46)$

There is indeed. Observe that if you go on taking successive differences (at each step) then, you get

$$12,40,90,168,280,432,...$$ $$28,50,78,112,152,...$$ $$22,28,34,40,...$$ $$6,6,6,6,...$$

Perhaps, a constant sequence. Does this strike something?

Yes! You can assert that, the general term of the given sequence is of the form $x_n = an^3+bn^2+cn+d$. Solve for $(a,b,c,d)$ using the fact that $x_1=12, x_2 = 40, x_3 = 90$ and $x_4 = 168$.

The Online Encyclopedia of Integer Sequences thinks that the terms are given by $$a_n=n^3+5n^2+6n$$

• So does WA. – lhf Jan 17 '17 at 16:14
• Divide the entries by $2$ and you get oeis.org/A005564 – Barry Cipra Jan 17 '17 at 16:24

Per my solution here it can be shown that the $n$-th term ($n=0,1,2,3,\cdots$), $b_n$ is given by

$$b_n=\sum_{r=0}^{\min(3,n)}\binom nr a_r$$ where $a_r=12,28,22,6$ for $r=0,1,2,3$, i.e.

\begin{align} b_0&=\binom 00 12&&=12\\ b_1&=\binom 10 12+\binom 11 28&&=40\\ b_2&=\binom 20 12+\binom 21 28+\binom 22 22 &&=90\\ b_3&=\binom 30 12+\binom 31 28+\binom 32 22 +\binom 33 6&&=168\\ b_4&=\binom 40 12+\binom 41 28+\binom 42 22 +\binom 43 6&&=280\\ b_5&=\binom 50 12+\binom 51 28+\binom 52 22 +\binom 53 6&&=432\\ \vdots\\ \color{red}{b_n}&\color{red}{=\binom n0 12+\binom n1 28 + \binom n2 22 +\binom n3 6}\\ &\color{red}{=n^3+8n^2+19n+12}\\ &\color{red}{=(n+1)(n+3)(n+4)} \end{align}

Note that $a_r$ is the first term of the $r$-th difference series, with $r=0$ referring to the original series.

The sum of the first $n$ terms is given by

\begin{align} S_n &=\sum_{r=0}^n \binom r0 12 +\underbrace{\sum_{r=1}^n\binom r1 28 + \underbrace{\sum_{r=2}^n\binom r2 22 +\underbrace{\sum_{r=3}^n\binom r3 6}_{n\ge 3}}_{n\ge 2}}_{n\ge 1}\\\\ &=\color{red}{\binom {n+1}1 12 +\underbrace{\binom {n+1}2 28 +\underbrace{\binom {n+1}3 22 +\underbrace{\binom {n+1}4 6}_{n\ge 3}}_{n\ge 2}}_{n\ge 1}}\\ &\color{red}{=(n+1)(3n^3+13n^2-2n+12)\qquad [\text{for }n\ge 3]} \end{align}

NB: References above to the $n$-th term count from $n=0$.


# What is the probability that both cards are not aces?

Suppose two cards are drawn from a standard 52 card deck without replacement. Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces?

My Solution

A = Event that first card is an ace

B = Event that second card is an ace given that first is an ace

C = Event that both cards are aces

D = Event that both cards are not aces $$P(A) = \frac{4}{52}$$ $$P(B) = \frac{3}{51}$$ $$P(C) = \frac{4}{52}*\frac{3}{51} = \frac{1}{221}$$ $$P(D) = 1 - P(C) = 1 - \frac{1}{221} = \frac{220}{221}$$

Actual Solution

A = Event that first card is not an ace

B = Event that second card is not an ace given that first is not an ace

C = Event that both cards are not aces $$P(A) = \frac{48}{52}$$ $$P(B) = \frac{47}{51}$$ $$P(C) = \frac{48}{52}*\frac{47}{51} = \frac{188}{221}$$

Why my solution that P(D) = 1 - P(C) is wrong?

• Somewhat ambiguous question. Suppose the two drawn cards are called A and B. Do you mean A is not an ace and B is also not an ace? It could be interpreted as A and B are both not aces (as a pair) but one of them could be an ace. As with all good math word problems, it is important to understand the question clearly first before attempting to solve it. – David Jul 20 '16 at 11:18
• @David I am now confused with my question after seeing your comment. But the actual answer states its 188/221 so I assume A is not an ace and B is also not an ace. – jblixr Jul 20 '16 at 14:50
• Yes that seems to be the correct interpretation that neither of the $2$ drawn cards are aces so you drew $0$ aces. An "easy" way to compute this probability is imagine we remove all $4$ aces from the deck of $52$, leaving us with $48$ cards only. We have $48$ ways to draw the first card and $47$ ways to draw the 2nd card. Now imagine we put the deck back to $52$ cards and draw $2$ cards. We have $52$ ways to draw the 1st card and $51$ ways to draw the 2nd card. Therefore, the probability of drawing no aces for the first $2$ cards is $(48*47)/(52*51)$. – David Jul 21 '16 at 3:20
• In the above comment, we can ignore the "missing" divide by $2$ for both the numerator and the denominator since they will cancel out. Conceptually, this is about the simplest way I can explain the answer. That is, remove the aces to guarantee no aces drawn the draw $2$ non ace cards which is exactly what we want. Then put all the aces back and draw $2$ cards . Then compute the ratio and that is the probability (about $85$%). – David Jul 21 '16 at 3:45

The probability that both cards are not aces is the complement of the event that at least one of the cards selected is an ace. You overlooked the possibility that exactly one of the two cards selected is an ace.

The probability that both cards are aces is $$\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$$
The probability that the first card is an ace and the second card is not an ace is $$\frac{4}{52} \cdot \frac{48}{51} = \frac{16}{221}$$ The probability that the first cards is not an ace and the second card is an ace is $$\frac{48}{52} \cdot \frac{4}{51} = \frac{16}{221}$$ Therefore, the probability that at least one of the two cards is an ace is $$\frac{1}{221} + \frac{16}{221} + \frac{16}{221} = \frac{33}{221}$$ Hence, the probability that both cards are not aces is $$1 - \frac{33}{221} = \frac{188}{221}$$ which can be found more simply by using the method you included in your post.

This may be a language issue. You calculated the probability that not both cards are aces, whereas the problem asks for the probability that both cards are not aces.

Since you overloaded your event variables, I'll define new ones: Let $E$ be the event that the first card is an ace and $F$ the event that the second card is an ace; then the event that not both cards are aces is $\overline{E\cap F}=\overline E\cup\overline F$, and the event that both cards are not aces is $\overline E\cap\overline F=\overline{E\cup F}$.

If you want neither of the first $2$ drawn cards to be aces then it is simply $48 \choose 2$ / $52 \choose 2$ which is $188/221$. Here we are simply choosing $2$ cards from the $48$ non aces and are dividing by the total number of possible $2$ card pairs from the full deck of $52$.

If $C$ is the event that both cards are aces, then $1 - P(C)$ is the probability that at least one of the two cards is not an ace.

What you need to be calculating (using your A,B) is:

$$(1-P(A))(1-P(B))$$

which is the probability that the first card is not an ace, and the second is an ace given that the first is not.


# Show that $\sum\limits_{n\ge1}\frac1{n^2}=\sum\limits_{n\ge1}\frac3{n^2\binom{2n}n}$ without actually evaluating both series

$$\sum\limits_{n\ge1}\frac1{n^2}=\sum\limits_{n\ge1}\frac3{n^2\binom{2n}n}\tag1$$

Note that $$(1)$$ holds since the LHS is given by $$\zeta(2)$$ whereas the RHS by $$6\arcsin^2 1$$ which both equal $$\dfrac{\pi^2}6$$ as it is well-known. However I am interested in proving $$(1)$$ without actually evaluating both series.

I am aware of an elegant approach contributed by Markus Scheuer as an answer to Different methods to compute Basel problem. Although this answers my question partially I am looking for different attemps. For example within Jack D'Aurizio's notes there is a way proposed exploiting creative telescoping $$($$see page $$5$$f.$$)$$ which I am sadly speaking not able to understand completely yet. Hence I have come across a proof of a similiar equality concerning $$\zeta(3)$$ on AoPS given by pprime I am confident that there are in fact other possible methods.

I would like to see attempts of proving $$(1)$$ beside the mentioned which do not rely on actually showing that they both equal $$\dfrac{\pi^2}6$$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similiar by pprime.

EDIT I

I have found another interesting approach, again by Jack D'Aurizio, which can be found here utilizing harmonic sums and creative telescoping in combination.

EDIT II

As pointed out by Zacky on page $$31$$ of Jack's notes another methode can be found which makes it three possibilities provided by Jack alone. Quite impressive!

• Here is a similar type of series (just 2 hours ago, also answered by yourself), using $\arcsin$. But yes, here is the series evaluated directly. Seems pretty good to me. – Dietrich Burde Dec 24 '18 at 12:53
• @DietrichBurde I am not sure how this is of use. Could you elaborate on the utility of the linked post in order to answer my question? Furthermore I am aware of this post since I posted an answer there $2$ hours ago ^^ – mrtaurho Dec 24 '18 at 12:59
• I thought, it would be the best to show that both sides are equal to $\zeta(2)$, but you want a different solution (and I just do not know why this should be better or more interesting, but this is of course due to my missing understanding). – Dietrich Burde Dec 24 '18 at 13:08
• @DietrichBurde Ah okay. I have not considered it from this point of view. Regarding to the value of alternative proofs: I am just interested in different approaches. For sure it is quite convincing to show both series equal the same value but out of experience I have observed that is most likely to be a way more difficult to show the equality of the series all by themselves $($see for example here: math.stackexchange.com/q/2942630 where an elegant trick was needed to show that the integrals are equivalent$)$. – mrtaurho Dec 24 '18 at 13:10

Since we have $$\frac{1}{\binom{2n}{n}} =\frac{n!n!}{(2n)!}$$ by the binomial identity, we obtain $$\frac{1}{n^2\binom{2n}{n}} =\frac{(n-1)!(n-1)!}{(2n)!} =\frac{\color{purple}{\Gamma(n)\Gamma(n)}}{2n\color{purple}{\Gamma(n+n)}}=\frac{1}{2n}\color{purple}{B(n,n)}$$ Therefore we get \begin{align*} \sum_{n=1}^\infty \frac{1}{n^2\binom{2n}{n}} & =\frac12\sum_{n=1}^\infty \frac{1}{n}\color{purple}{B(n,n)} =\frac12\sum_{n=1}^\infty \frac{1}{n}\color{purple}{\int_0^1 (x(1-x))^{n-1}dx} \\ & = -\frac12\int_0^1 \frac{1}{x(1-x)} \color{blue}{\left(-\sum_{n=1}^\infty \frac{(x(1-x))^{n}}{n}\right)}dx = - \frac12 \int_0^1 \frac{\color{blue}{\ln(1-x(1-x))}}{x(1-x)}dx \\ & = -\frac12 \bigg(\int_0^1 \frac{\ln(1-x(1-x))}{x}dx+\underbrace{\int_0^1 \frac{\ln(1-x(1-x))}{1-x}dx}_{1-x\ \rightarrow \ x}\bigg) \\ & = - \frac12\left(\int_0^1 \frac{\ln(1-x(1-x))}{x}dx +\int_0^1 \frac{\ln(1-(1-x)x)}{x}dx\right) \\ & = - \int_0^1 \frac{\ln(1-x+x^2)}{x}dx = - \int_0^1 \frac{\ln\left(\frac{1+x^3}{1+x}\right)}{x}dx \\ & = \int_0^1 \frac{\ln(1+x)}{x}dx-\underbrace{\int_0^1 \frac{\ln(1+x^3)}{x}dx}_{x^3 \rightarrow x} \\ & = \int_0^1 \frac{\ln(1+x)}{x}dx -\frac13 \int_0^1 \frac{\ln(1+x)}{x}dx =\frac23\int_0^1 \frac{\ln(1+x)}{x}dx \end{align*} $$\quad \quad \quad \quad \quad \quad \displaystyle{ =\frac13 \int_0^1 \frac{\ln x}{x-1}dx}$$$$\displaystyle{=-\frac13\sum_{n=0}^\infty \int_0^1 x^n \ln xdx= \frac13 \sum_{n=0}^\infty \frac{1}{(n+1)^2}=\frac13 \sum_{n=1}^\infty \frac{1}{n^2}}$$
As an alternative just take $$x=1$$ in the following relation shown by Felix Marin: $$\sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}} =-\int_{0}^{1} \frac{\ln(1-(1-t)tx)}{t} dt.$$


# Solve algebraically: $\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}}$

$$\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}}$$

The professor says we can't use l'hopital's rule and must solve algebraically.

• Multiply top and bottom by $5+\sqrt{x^2+16}$. Then factor the resulting polynomial downstairs and cancel what you can. – David Mitra Feb 28 '14 at 15:17

## 2 Answers

Try multiplying the numerator and denominator by the conjugate of the denominator:

Multiply by $$\dfrac{5+ \sqrt{x^2 +16}}{5+\sqrt{x^2 + 16}}$$

$$\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}} \cdot \dfrac{5+ \sqrt{x^2 +16}}{5+\sqrt{x^2 + 16}} = \lim_{x\to 3}\dfrac{(3-x)(5 + \sqrt {x^2 + 16)}}{25 - (x^2 + 16)}$$

Then note that you have a difference of squares in the denominator: $$25 - (x^2 + 16) = 9 - x^2 = (3-x)(3+x)$$

Now, you can cancel the factor $3-x$, as it appears in both the numerator and denominator.

• Thank you so much for you quick reply! I understand and found the limit of 5/3! – dev Feb 28 '14 at 15:28
• Exactly! Nice work, dev. – Namaste Feb 28 '14 at 15:28

Hint $\ \ f\bar f = (a\!-\!x)(a\!+\!x)\ \Rightarrow\ \dfrac{a-x}f \,=\, \dfrac{\bar f}{a+x}\ \$ where $\ \ \bar f,\,f\, =\, 5\pm \sqrt{x^2+16}$

Remark $\$ This is a special case of rationalizing the denominator, which is often handy.

• Does that link imply those three accounts all belong to you? – TMM Feb 28 '14 at 15:49


# Category: Mathematics

## MCQs Mathematics 2

The GRE mathematical reasoning section tests your knowledge of arithmetic, basic algebra, applied mathematics, elementary geometry, and common graphs and charts. Let Starts with the GRE MCQs Mathematics quiz.

1. If the value of a piece of the property decreases by 10% while the tax rate on the property increases by 10%, what is the effect on the taxes?

2. In a cafeteria, the people are either faculty members or students. The number of faculty members is 15% of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50% of the original total. The number of students who left is what fractional part of the original number of students?

3. If $x=\frac{y}{7}$ and $7x=12$, then $y=?$

4. ABC works two part-time jobs. One week ABC worked 8 hours at one job, earning $\$150$, and$4.5$hours at the other job, earning$\$90$. What were his average hourly earnings for the week?

5. $2000 amount is deposited into a savings account that earns interest at the rate of 10% per year, compounded semi-annually. How much money will there be in the account at the end of one year? 6. A certain photo state machine produces 13 copies every 10 seconds. If the machine operates without interruption, how many copies will it produce in an hour? 7. If$3x-5=x+11$, then$x=$? 8. If$x=k+\frac{1}{2}=\frac{k+1}{2}$, then$x=$? 9. If$x+y=8$and$2x-y=10$then$x=$? 10. In a certain population, 40% of all people have biological characteristics$X$; the others do not. if 8000 people have characteristic$X$, how many people do not have$X$? 11. For how many 3-digit whole numbers is the sum of the digits equal to 3? 12. A car dealer who gives a customer a 20% discount on the list price of a car still realizes a net profit of 25% of the cost. If the dealer’s cost is$4800 what is the usual list price of the car?

13. If $7-x=0$, then $10-x=$?

GRE Mathematics-1

## GRE Mathematics 1

The GRE mathematical reasoning section tests your knowledge of arithmetic, basic algebra, applied mathematics, elementary geometry, and common graphs and charts.

Please go to GRE Mathematics 1 to view the test

The problem solving questions are typically word problem questions. These sort of questions are usually part of different competitive and ability tests.

Take another test: Sequence and Series