problem stringlengths 42 632 | level stringclasses 5
values | solution stringlengths 89 1.55k | type stringclasses 7
values |
|---|---|---|---|
Compute $\tan 20^\circ + 4 \sin 20^\circ.$ | Level 2 | We can write
\begin{align*}
\tan 20^\circ + 4 \sin 20^\circ &= \frac{\sin 20^\circ}{\cos 20^\circ} + 4 \sin 20^\circ \\
&= \frac{\sin 20^\circ + 4 \sin 20^\circ \cos 20^\circ}{\cos 20^\circ}.
\end{align*}By double angle formula,
\[\frac{\sin 20^\circ + 4 \sin 20^\circ \cos 20^\circ}{\cos 20^\circ} = \frac{\sin 20^\circ... | Precalculus |
Subtract $111.11$ from $333.33.$ Express the result as a decimal to the nearest hundredth. | Level 3 | We can organize the subtraction concisely using columns as follows: \[
\begin{array}{@{}c@{}c@{}c@{}c@{}c@{}c}
& 3 & 3 & 3. & 3 & 3 \\
- & 1 & 1 & 1. & 1 & 1
\\ \cline{1-6}
& 2 & 2 & 2. & 2 & 2 \\
\end{array}
\] The answer is $\boxed{222.22}$. | Prealgebra |
Each of $a_1,$ $a_2,$ $\dots,$ $a_{100}$ is equal to $1$ or $-1.$ Find the minimum positive value of
\[\sum_{1 \le i < j \le 100} a_i a_j.\] | Level 5 | Let $S$ denote the given sum. Then
\begin{align*}
2S &= (a_1 + a_2 + \dots + a_{100})^2 - (a_1^2 + a_2^2 + \dots + a_{100}^2) \\
&= (a_1 + a_2 + \dots + a_{100})^2 - 100.
\end{align*}To find the minimum positive value of $2S,$ we want $(a_1 + a_2 + \dots + a_{100})^2$ to be as close to 100 as possible (while being gre... | Intermediate Algebra |
Compute $\displaystyle \frac{2+4-8+16+32-64}{4+8-16+32+64-128}$. | Level 2 | Factoring the numerator and denominator, we have:
$\displaystyle \frac{2+4-8+16+32-64}{4+8-16+32+64-128}=\frac{2(1+2-4+8+16-32)}{4(1+2-4+8+16-32)}=\frac{2}{4}=\boxed{\frac{1}{2}}$. | Algebra |
How many non-empty subsets of $\{ 1 , 2, 3, 4, 5, 6, 7, 8 \}$ consist entirely of odd numbers? | Level 4 | We consider the subset $\{ 1, 3, 5, 7 \}$ which consists only of the odd integers in the original set. Any subset consisting entirely of odd numbers must be a subset of this particular subset. And, there are $2^4 - 1 = \boxed{15}$ non-empty subsets of this 4-element set, which we can easily see by making the choice of ... | Counting & Probability |
Let $x$ and $y$ be positive real numbers such that $x + y = 10.$ Find the minimum value of $\frac{1}{x} + \frac{1}{y}.$ | Level 2 | By AM-HM,
\[\frac{x + y}{2} \ge \frac{2}{\frac{1}{x} + \frac{1}{y}}.\]Hence,
\[\frac{1}{x} + \frac{1}{y} \ge \frac{4}{x + y} = \frac{4}{10} = \frac{2}{5}.\]Equality occurs when $x = y = 5,$ so the minimum value is $\boxed{\frac{2}{5}}.$ | Intermediate Algebra |
Melinda has three empty boxes and $12$ textbooks, three of which are mathematics textbooks. One box will hold any three of her textbooks, one will hold any four of her textbooks, and one will hold any five of her textbooks. If Melinda packs her textbooks into these boxes in random order, the probability that all three ... | Level 5 | The total ways the textbooks can be arranged in the 3 boxes is $12\textbf{C}3\cdot 9\textbf{C}4$, which is equivalent to $\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6}{144}=12\cdot11\cdot10\cdot7\cdot3$. If all of the math textbooks are put into the box that can hold $3$ textbooks, there are $9!/(4!\cdot 5!)=9\... | Counting & Probability |
In the equation $|x-7| -3 = -2$, what is the product of all possible values of $x$? | Level 3 | We rearrange the given equation to $|x-7| = 1$. Thus either $x-7 = 1$, meaning $x = 8$, or $x-7 = -1$, meaning $x=6$. Our answer is therefore $6\cdot 8 = \boxed{48}$. | Algebra |
Compute $\arcsin (-1).$ Express your answer in radians. | Level 1 | Since $\sin \left( -\frac{\pi}{2} \right) = -1,$ $\arcsin (-1) = \boxed{-\frac{\pi}{2}}.$ | Precalculus |
If $x@y=xy-2x$, what is the value of $(5@3)-(3@5)$? | Level 2 | $5@3=5\cdot3-2\cdot5=5$ and $3@5=3\cdot5-2\cdot3=9$, so $(5@3)-(3@5)=5-9=\boxed{-4}$. Another way to solve this problem is to realize that the expression $(5@3)-(3@5)$ is of the form $(x@y)-(y@x)=xy-2x-yx+2y=-2x+2y$, so the expression is just equal to $-2\cdot5+2\cdot3=\boxed{-4}$. | Algebra |
Two parabolas are the graphs of the equations $y=3x^2+4x-5$ and $y=x^2+11$. Give all points where they intersect. List the points in order of increasing $x$-coordinate, separated by semicolons. | Level 5 | Setting the right-hand sides of the given equations equal gives $3x^2+4x-5=x^2+11$. Combining like terms gives $2x^2+4x=16$. Dividing by $2$ gives $x^2+2x=8$, and rearranging gives $x^2 +2x - 8=0$. Factoring gives $(x+4)(x-2)=0$, so our solutions are $x=-4$ and $x=2$. Substituting these into either of the original eq... | Algebra |
If the odds for pulling a prize out of the box are $3:4$, what is the probability of not pulling the prize out of the box? Express your answer as a common fraction. | Level 3 | If the odds for pulling a prize out of the box are $3:4$, that means that 3 out of 7 times will result in a prize, while 4 out of 7 times will not. So the probability of not pulling the prize out of the box is $\boxed{\frac{4}{7}}$. | Counting & Probability |
If $(x,y)$ is a solution to the system
\begin{align*}
xy &= 6, \\
x^2 y + xy^2 + x + y &= 63,
\end{align*}find $x^2 + y^2.$ | Level 2 | The second equation factors as $(xy + 1)(x + y) = 63,$ so $7(x + y) = 63,$ or $x + y = 9.$ Then
\[x^2 + y^2 = (x + y)^2 - 2xy = 9^2 - 2 \cdot 6 = \boxed{69}.\] | Intermediate Algebra |
A bridge is built by suspending a plank of wood between two triangular wedges with equal heights, as in the following diagram: [asy]
import olympiad;
import math;
// Draw triangles
pair A = (0, 1);
pair B = (-cos(1.3962), 0);
pair C = (cos(1.3962), 0);
pair D = (2, 1);
pair E = (2-cos(1.3089), 0);
pair F = (2+cos(1.30... | Level 2 | There are several ways to proceed, and here is one. Since $\triangle ABC$ and $\triangle DEF$ are both isosceles, it should be easy to find that $\angle B = \angle C = 80^\circ$ and $\angle E = \angle F = 75^\circ.$ Now, connect $C$ and $E$:
[asy]
import olympiad;
import math;
// Draw triangles
pair A = (0, 1);
pair ... | Geometry |
Base prime representation of a natural number is defined using the exponents of its prime factorization as follows. Each place in a base prime represents a prime number, and it is occupied by the corresponding exponent of that prime, starting on the right side with the smallest prime number and proceeding to the left w... | Level 4 | The prime factorization of $225$ is $225 = 15^2 = 3^2 \times 5^2$. Since $2$ does not divide into $225$, we treat $2$ as having a $0$ exponent; the next two primes are $3$ and $5$. Thus, the answer is $\boxed{220}.$ | Number Theory |
The product of the base seven numbers $24_7$ and $30_7$ is expressed in base seven. What is the base seven sum of the digits of this product? | Level 4 | We can ignore the $0$ digit for now, and find the product of $24_7 \times 3_7$. First, we need to multiply the units digit: $4_7 \times 3_7 = 12_{10} = 15_7$. Hence, we write down a $5$ and carry-over the $1$. Evaluating the next digit, we need to multiply $2_7 \times 3_7 + 1_7 = 7_{10} = 10_{7}$. Thus, the next digit ... | Number Theory |
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