Math GRE - #33

How many integers from 1 to 1000 are divisible by 30 by not by 16?

1. 29
2. 31
3. 32
4. 33
5. 38

Solution :

Choice 1 is the answer.

There are 33 multiples of 30 under 1000 (1000/30 = 33.33..). So we only need to focus our attention on choices 1 to 4.

Note that the factorization of 30 is: $2\cdot 3\cdot 5.$
Now lets consider multiples of 30 in the form: $30n$. For $30n$ to be divisible by 16, $n$ must be divisible by 8 (since we will have a factor of 2 from 30 and another factor of 8 from $n$ to make a factor of 16).
Since $n$ can only go from 1 to 33 (as there are only 33 multiples of 30 under 1000), we only need to find numbers under 33 that are divisible by 8.

There are four such numbers: 8, 16, 24, 32. These numbers correspond to the multiples: 240, 480, 720, 960.

All the multiples remaining are not divisible by 16. Thus there are 33 - 4 = 29 multiples of 30 under 1000 that are divisible by 30 but not 16.

Physics GRE - #33

A radioactive nucleus decays, with the activity shown in the plot below. The half-life of the nucleus is:

1. 2 min
2. 7 min
3. 11 min
4. 18 min
5. 23 min

Solution :

Choice 2 is the answer.

The initial activity is $6\cdot 10^3$ (yes, 6 since the line marked $10^3$ is really $1\cdot 10^3$ and you start counting up from there). At $3\cdot 10^3,$ the time is around 5 to 10 minutes. Thus the only possible answer is choice 2.

Math GRE - #32

Let $\mathbb{R}$ be the set of real numbers and let $f$ and $g$ be functions from $\mathbb{R}$ into $\mathbb{R}$. Which of the following is the negation of the statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

1. For each $s$ in $\mathbb{R}$, there does not exist an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$.
2. For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
3. There exists an $s$ in $\mathbb{R}$ such that for each $r$ in $\mathbb{R}$ such that $f(r)>0$ and $g(s)\leq0$.
4. There exists an $s$ in $\mathbb{R}$ and there exists an $r$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.
5. For each $r$ in $\mathbb{R}$, there exists an $s$ in $\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.

Solution :

Choice 3 is the answer.

If you've got an innate talent for logic, you could probably tell that choice 3 is the answer after a bit of thinking. 

Another way is to use quantifiers to translate the statement into formal logic.

The statement:

"For each $s$ in $\mathbb{R}$, there exists an $r$ in $\mathbb{R}$ such that if $f(r)>0$, then $g(s)>0$."

can be translated as: \[\forall s\in\mathbb{R},\;\exists r\in\mathbb{R},\; f(r)>0\rightarrow g(s)>0.\]

We can negate this statement (imagine the negation as an arrow, flipping each quantifier as it flies past it) to be: \[\exists s\in\mathbb{R},\;\forall r\in\mathbb{R},\; f(r)>0\wedge g(s)\leq0.\]

This is choice 3 precisely.

Physics GRE - #32

As shown below, a coaxial cable having radii $a$, $b$, and $c$ carries equal and opposite currents of magnitude $i$ on the inner and outer conductors. What is the magnitude of the magnetic induction at a point $P$ outside the cable at a distance $r$ from the axis?

1. $0$
2. $\dfrac{\mu_0 ir}{2\pi a^2}$
3. $\dfrac{\mu_0 i}{2\pi r}$
4. $\dfrac{\mu_0 i}{2\pi r}\dfrac{c^2-r^2}{c^2-b^2}$
5. $\dfrac{\mu_0 i}{2\pi r}\dfrac{r^2-b^2}{c^2-b^2}$

Solution :

Choice 1 is the answer.

The currents cancel out the magnetic fields created by each other, resulting in a net field of zero. More formally, this is an application of Ampere's Law.

Math GRE - #31

Let $\text{F}$ be a constant force that is given by the vector $\left(\begin{array}{c}-1\\0\\1\end{array}\right)$. What is the work done by $\text{F}$ on a particle that moves along the path given by $\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)$ between time $t=0$ and $t=1$?

1. $-\dfrac{1}{4}$


2. $-\dfrac{1}{4\sqrt{2}}$


3. $0$


4. $\sqrt{2}$


5. $3\sqrt{2}$

Solution :

Choice 3 is the answer.

Recall that work can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot \left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.

Physics GRE - #31

Which of the following is nearly the mass of the Earth? The radius of the Earth is about $6.4\cdot10^6\,\text{m}$.

1. $6\cdot10^{24}\,\text{kg}$
2. $6\cdot10^{27}\,\text{kg}$
3. $6\cdot10^{30}\,\text{kg}$
4. $6\cdot10^{33}\,\text{kg}$
5. $6\cdot10^{36}\,\text{kg}$

Solution :

Choice 1 is the answer.

If you don't remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]

Math GRE - #30

What is \[\int_{-3}^3{|x+1|\,dx}?\]

1. 0
2. 5
3. 10
4. 15
5. 20

Solution :

Choice 3 is the answer.

This is a very basic question. The usual method of dealing with absolute values in integrals is to split it into a sum of two integrals and remove the absolute values. Since $|x+1|$ is negative on $[-3, 1)$ and positive on $[-1, 3]\,\,\,$ , we can split the integral into: \[\int_{-3}^3{|x+1|\,dx}=-\int_{-3}^{-1}{(x+1)\,dx}+\int_{-1}^3{(x+1)\,dx}=10.\] Another way of doing this problem is to imagine the graph of $|x+1|$ and realize that the integral is the sum of two 45-45-90 triangles (one with base 2 and height 2,  the other with base 4 and height 4). Simply add the area of triangles up and we're done.

Bonus question: Given that $c$ is a constant, what is \[\int_{-\infty}^{\infty}{e^{-|x+c|}\,dx}?\]