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1,501
|
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_9
| 1
|
A farmer divides his herd of $n$ cows among his four sons so that one son gets one-half the herd, a second son, one-fourth, a third son, one-fifth, and the fourth son, $7$ cows. Then $n$ is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240$
|
The first three sons get $\frac{1}{2}+\frac{1}{4}+\frac{1}{5}=\frac{19}{20}$ of the herd, so that the fourth son should get $\frac{1}{20}$ of it. But the fourth son gets $7$ cows, so the size of the herd is $n=\frac{7}{\frac{1}{20}} = 140$ . Then our answer is $\boxed{140}$ , and we are done.
| 140
|
1,502
|
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_15
| 1
|
In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$
|
WLOG, by scaling, that the hypotenuse has length 1. Let $\theta$ be an angle opposite from some leg. Then the two legs have length $\sin\theta$ and $\cos\theta$ respectively, so we have $2\sin\theta\cos\theta = 1^2$ . From trigonometry, we know that this equation is true when $\theta = 45^{\circ}$ , so our answer is $\boxed{45}$ and we are done.
| 45
|
1,503
|
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_15
| 2
|
In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$
|
Look at the options. Note that if $\textbf{(A)}$ is the correct answer, one of the acute angles of the triangle will measure to $15$ degrees. This implies that the other acute angle of the triangle would measure to be $75$ degrees, which would imply that $\textbf{(E)}$ is another correct answer. However, there is only one correct answer per question, so $\textbf{(A)}$ can't be a correct answer. Using a similar argument, neither $\textbf{(B)}$ $\textbf{(D)}$ , nor $\textbf{(E)}$ can be a correct answer. Since $\textbf{(C)}$ is the only answer choice left and there must be one correct answer, the answer must be $\boxed{45}$
| 45
|
1,504
|
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_19
| 1
|
With the use of three different weights, namely $1$ lb., $3$ lb., and $9$ lb., how many objects of different weights can be weighed, if the objects is to be weighed and the given weights may be placed in either pan of the scale? $\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 7$
|
The heaviest object that could be weighed with this set weighs $1 + 3 + 9 = 13$ lb., and we can weigh any positive integer weight at most that. This means that $13$ different objects could be weighed, so our answer is $\boxed{13}$ and we are done.
| 13
|
1,505
|
https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_48
| 1
|
Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$ , where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$ 's are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$ . [See example 25 for the meaning of $|x|$ .] The number of polynomials with $h=3$ is: $\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9$
|
We perform casework by the value of $n$ , the degree of our polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$
Case $n = 0$ : In this case we are forced to set $a_0 = 3$ . This contributes $1$ possibility.
Case $n = 1$ : In this case we must have $a_0 + |a_1| = 2$ , so our polynomial could be $1 + 1x, 0 + 2x, -1 + 1x$ . This contributes $3$ possibilities.
Case $n = 2$ : In this case we must have $a_0 + |a_1| + |a_2| = 1$ . However, because $a_0$ must be positive, it has to be $1$ , so our polynomial can only be $0 + 0x + 1x^2$ . This contributes $1$ possibility.
Case $n\geq 3$ : This case is impossible because $h = n+a_0+|a_1|+|a_2|+\cdots+|a_n|\geq n + a_0\geq 3 + 1 = 4 > 3$ , so it contributes $0$ possibilities.
Adding the results from all four cases, we find that there are $1 + 3 + 1 + 0 = 5$ possibilities in total, so our answer is $\boxed{5}$
| 5
|
1,506
|
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_6
| 1
|
The arithmetic mean between $\frac {x + a}{x}$ and $\frac {x - a}{x}$ , when $x \not = 0$ , is:
$\textbf{(A)}\ {2}\text{, if }{a \not = 0}\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ {1}\text{, only if }{a = 0}\qquad \textbf{(D)}\ \frac {a}{x}\qquad \textbf{(E)}\ x$
|
We have $\frac{1}{2}\cdot \left(\frac{x + a}{x} + \frac{x - a}{x}\right) = \frac{2}{2} = \boxed{1}$
| 1
|
1,507
|
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_12
| 1
|
If $P = \frac{s}{(1 + k)^n}$ then $n$ equals:
$\textbf{(A)}\ \frac{\log{\left(\frac{s}{P}\right)}}{\log{(1 + k)}}\qquad \textbf{(B)}\ \log{\left(\frac{s}{P(1 + k)}\right)}\qquad \textbf{(C)}\ \log{\left(\frac{s - P}{1 + k}\right)}\qquad \\ \textbf{(D)}\ \log{\left(\frac{s}{P}\right)} + \log{(1 + k)}\qquad \textbf{(E)}\ \frac{\log{(s)}}{\log{(P(1 + k))}}$
|
\[P=\frac{s}{(1+k)^n}\]
\[(1+k)^n=\frac{s}{P}\]
Take the $\log$ of each side.
\[n \log(1+k) = \log\left(\frac{s}{P}\right)\]
\[n = \frac{\log\left(\frac{s}{P}\right)}{\log(1+k)} \to \boxed{1}\]
| 1
|
1,508
|
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_13
| 1
|
The sum of two numbers is $10$ ; their product is $20$ . The sum of their reciprocals is:
$\textbf{(A)}\ \frac{1}{10}\qquad \textbf{(B)}\ \frac{1}{2}\qquad \textbf{(C)}\ 1\qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$
|
$x+y=10$
$xy=20$
$\frac1x+\frac1y=\frac{y}{xy}+\frac{x}{xy}=\frac{x+y}{xy}=\frac{10}{20}=\boxed{12}$
| 12
|
1,509
|
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_36
| 1
|
The sides of a triangle are $30$ $70$ , and $80$ units. If an altitude is dropped upon the side of length $80$ , the larger segment cut off on this side is:
$\textbf{(A)}\ 62\qquad \textbf{(B)}\ 63\qquad \textbf{(C)}\ 64\qquad \textbf{(D)}\ 65\qquad \textbf{(E)}\ 66$
|
Let the shorter segment be $x$ and the altitude be $y$ . The larger segment is then $80-x$ . By the Pythagorean Theorem \[30^2-y^2=x^2 \qquad(1)\] and \[(80-x)^2=70^2-y^2 \qquad(2)\] Adding $(1)$ and $(2)$ and simplifying gives $x=15$ . Therefore, the answer is $80-15=\boxed{65}$
| 65
|
1,510
|
https://artofproblemsolving.com/wiki/index.php/1958_AHSME_Problems/Problem_49
| 1
|
In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:
$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 55\qquad \textbf{(D)}\ 66\qquad \textbf{(E)}\ 132$
|
Expand the binomial $((a+b)+c)^n$ with the binomial theorem. We have:
\[\sum\limits_{k=0}^{10} \binom{10}{k} (a+b)^k c^{10-k}\]
So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:
\[\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{66}\]
| 66
|
1,511
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_1
| 1
|
The number of distinct lines representing the altitudes, medians, and interior angle bisectors of a triangle that is isosceles, but not equilateral, is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 7\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 3$
|
[asy] size(2cm); draw((-3,0)--(0,4)--(3,0)--cycle); draw((0,0)--(0,4), red); draw((-3,0)--(0.84, 2.88), green); draw((-3,0)--(1.5, 2), green); draw((-3,0)--(1.636, 1.818), green); draw((3,0)--(-0.84, 2.88), blue); draw((3,0)--(-1.5, 2), blue); draw((3,0)--(-1.636, 1.818), blue); [/asy]
As shown in the diagram above, all nine altitudes, medians, and interior angle bisectors are distinct, except for the three coinciding lines from the vertex opposite to the base. Thusly, there are $7$ distinct lines, so our answer is $\boxed{7}$ , and we are done.
| 7
|
1,512
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_7
| 1
|
The area of a circle inscribed in an equilateral triangle is $48\pi$ . The perimeter of this triangle is:
$\textbf{(A)}\ 72\sqrt{3} \qquad \textbf{(B)}\ 48\sqrt{3}\qquad \textbf{(C)}\ 36\qquad \textbf{(D)}\ 24\qquad \textbf{(E)}\ 72$
|
[asy] draw((-3,-sqrt(3))--(3,-sqrt(3))--(0,2sqrt(3))--cycle); draw(circle((0,0),sqrt(3))); dot((0,0)); draw((0,0)--(0,-sqrt(3))); [/asy] We can see that the radius of the circle is $4\sqrt{3}$ . We know that the radius is $\frac{1}{3}$ of each median line of the triangle; each median line is therefore $12\sqrt{3}$ . Since the median line completes a $30$ $60$ $90$ triangle, we can conclude that one of the sides of the triangle is $24$ . Triple the side length and we get our answer, $\boxed{72}$
| 72
|
1,513
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_8
| 1
|
The numbers $x,\,y,\,z$ are proportional to $2,\,3,\,5$ . The sum of $x, y$ , and $z$ is $100$ . The number y is given by the equation $y = ax - 10$ . Then a is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ \frac{5}{2}\qquad \textbf{(E)}\ 4$
|
In order to solve the problem, we first need to find each of the three variables. We can use the proportions the problem gives us to find the value of one part, and, by extension, the values of the variables (as $x$ would have $2$ parts, $y$ would have $3$ , and $z$ would have $5$ ). One part, after some algebra, equals $10$ , so $x$ $y$ , and $z$ are $20$ $30$ , and $50$ , respectively.
We can plug $x$ and $y$ into the equation given to us: $30 = 20a-10$ , and then solve to get $a = \boxed{2}$
| 2
|
1,514
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_9
| 1
|
The value of $x - y^{x - y}$ when $x = 2$ and $y = -2$ is:
$\textbf{(A)}\ -18 \qquad \textbf{(B)}\ -14\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ 18\qquad \textbf{(E)}\ 256$
|
Just plug in the numbers and follow the order of operations: \[2-(-2)^{2-(-2)}\] \[2-(-2)^4\] \[2-16\] \[\boxed{14}\]
| 14
|
1,515
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_19
| 1
|
The base of the decimal number system is ten, meaning, for example, that $123 = 1\cdot 10^2 + 2\cdot 10 + 3$ . In the binary system, which has base two, the first five positive integers are $1,\,10,\,11,\,100,\,101$ . The numeral $10011$ in the binary system would then be written in the decimal system as:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 40\qquad \textbf{(C)}\ 10011\qquad \textbf{(D)}\ 11\qquad \textbf{(E)}\ 7$
|
Numbers in binary work similar to their decimal counterparts, where the multiplier associated with each place is multiplied by two every single place to the left. For example, $1111_2$ $1111$ in base $2$ ) would equate to $1 * 2^3 + 1 * 2^2 + 1 * 2^1 + 1 * 2^0 = 8+4+2+1 = 15$
Using this same logic, $10011_2$ would be $1*2^4 + 1*2^1 + 1 * 2^0 = \boxed{19}$
| 19
|
1,516
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_23
| 1
|
The graph of $x^2 + y = 10$ and the graph of $x + y = 10$ meet in two points. The distance between these two points is:
$\textbf{(A)}\ \text{less than 1} \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \sqrt{2}\qquad \textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{more than 2}$
|
We can merge the two equations to create $x^2+y=x+y$ . Using either the quadratic equation or factoring, we get two solutions with $x$ -coordinates $0$ and $1$
Plugging this into either of the original equations, we get $(0,10)$ and $(1,9)$ . The distance between those two points is $\boxed{2}$
| 2
|
1,517
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_42
| 1
|
If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$
|
We first use the fact that $i^{-n}=\frac1{i^n}=\left(\frac1i\right)^n=(-i)^n$ . Note that $i^4=1$ and $(-i)^4=1$ , so $i^n$ and $(-i)^n$ have are periodic with periods at most 4. Therefore, it suffices to check for $n=0,1,2,3$
For $n=0$ , we have $i^0+(-i)^0=1+1=2$
For $n=1$ , we have $i^1+(-i)^1=i-i=0$
For $n=2$ , we have $i^2+(-i)^2=-1-1=-2$
For $n=3$ , we have $i^3+(-i)^3=-i+i=0$
Hence, the answer is $\boxed{3}$
| 3
|
1,518
|
https://artofproblemsolving.com/wiki/index.php/1957_AHSME_Problems/Problem_42
| 2
|
If $S = i^n + i^{-n}$ , where $i = \sqrt{-1}$ and $n$ is an integer, then the total number of possible distinct values for $S$ is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{more than 4}$
|
Notice that the powers of $i$ cycle in cycles of 4. So let's see if $S$ is periodic.
For $n=0$ : we have $2$
For $n=1$ : we have $0$
For $n=2$ : we have $-2$
For $n=3$ : we have $0$
For $n=4$ : we have $2$ again. Well, it can be seen that $S$ cycles in periods of 4. Select $\boxed{3}$
| 3
|
1,519
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_2
| 1
|
Mr. Jones sold two pipes at $\textdollar{ 1.20}$ each. Based on the cost, his profit on one was
$20$ % and his loss on the other was $20$ %.
On the sale of the pipes, he:
$\textbf{(A)}\ \text{broke even}\qquad \textbf{(B)}\ \text{lost }4\text{ cents} \qquad\textbf{(C)}\ \text{gained }4\text{ cents}\qquad \\ \textbf{(D)}\ \text{lost }10\text{ cents}\qquad \textbf{(E)}\ \text{gained }10\text{ cents}$
|
For the first pipe, we are told that his profit was 20%. In other words, he sold the pipe for 120% or $\frac{6}{5}$ of its original value. This tells us that the original price was $\frac{5}{6}\cdot1.20 = $1.00$
For the second pipe, we are told that his loss was 20%. In other words, he sold the pipe for 80% or $\frac{4}{5}$ of its original value. This tells us that the original price was $\frac{5}{4}\cdot1.20 = $1.50$
Thus, his total cost was $$2.50$ and his total revenue was $$2.40$
Therefore, he $\boxed{10}$
| 10
|
1,520
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_5
| 1
|
A nickel is placed on a table. The number of nickels which can be placed around it, each tangent to it and to two others is:
$\textbf{(A)}\ 4 \qquad\textbf{(B)}\ 5 \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\ 12$
|
Arranging the nickels in a hexagonal fashion, we see that only $\boxed{6}$ nickels can be placed around the central nickel.
| 6
|
1,521
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_6
| 1
|
In a group of cows and chickens, the number of legs was 14 more than twice the number of heads. The number of cows was:
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 10 \qquad\textbf{(D)}\ 12 \qquad\textbf{(E)}\ 14$
|
Let there be $x$ cows and $y$ chickens. Then, there are $4x+2y$ legs and $x+y$ heads. Writing the equation: \[4x+2y=14+2(x+y)\] \[4x+2y=14+2x+2y\] \[2x=14\] \[x=\boxed{7}\]
| 7
|
1,522
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_9
| 1
|
When you simplify $\left[ \sqrt [3]{\sqrt [6]{a^9}} \right]^4\left[ \sqrt [6]{\sqrt [3]{a^9}} \right]^4$ , the result is:
$\textbf{(A)}\ a^{16} \qquad\textbf{(B)}\ a^{12} \qquad\textbf{(C)}\ a^8 \qquad\textbf{(D)}\ a^4 \qquad\textbf{(E)}\ a^2$
|
This simplifies to \[(a^{\frac{9}{6}/3})^4 \cdot (a^{\frac{9}{3}/6})^4 = (a^{\frac{1}{2}})^4 \cdot (a^{\frac{1}{2}})^4 = (a^2)(a^2) = \boxed{4}.$
| 4
|
1,523
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_13
| 1
|
Given two positive integers $x$ and $y$ with $x < y$ . The percent that $x$ is less than $y$ is:
$\textbf{(A)}\ \frac{100(y-x)}{x}\qquad \textbf{(B)}\ \frac{100(x-y)}{x}\qquad \textbf{(C)}\ \frac{100(y-x)}{y}\qquad \\ \textbf{(D)}\ 100(y-x)\qquad \textbf{(E)}\ 100(x - y)$
|
Suppose that $x$ is $p$ percent less than $y$ . Then $x = \frac{100 - p}{100}y$ , so that $y - x = \frac{p}{100}y$ . Solving for $p$ , we get $p = \frac{100(y-x)}{y}$ , or $\boxed{100}$
| 100
|
1,524
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_16
| 1
|
The sum of three numbers is $98$ . The ratio of the first to the second is $\frac {2}{3}$ ,
and the ratio of the second to the third is $\frac {5}{8}$ . The second number is:
$\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 32 \qquad\textbf{(E)}\ 33$
|
Let the $3$ numbers be $a,$ $b,$ and $c$ . We see that \[a+b+c = 98\] and \[\frac{a}{b} = \frac{2}{3} \Rrightarrow 3a = 2b\] \[\frac{b}{c} = \frac{5}{8} \Rrightarrow 8b = 5c\] Writing $a$ and $c$ in terms of $b$ we have $a = \frac{2}{3} b$ and $c = \frac{8}{5} b$ .
Substituting in the sum, we have \[\frac{2}{3} b + b + \frac{8}{5} b = 98\] \[\frac{49}{15} b = 98\] \[b = 98 \cdot \frac{15}{49} \Rrightarrow b = 30\] $\boxed{30}$
| 30
|
1,525
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_25
| 1
|
The sum of all numbers of the form $2k + 1$ , where $k$ takes on integral values from $1$ to $n$ is:
$\textbf{(A)}\ n^2\qquad\textbf{(B)}\ n(n+1)\qquad\textbf{(C)}\ n(n+2)\qquad\textbf{(D)}\ (n+1)^2\qquad\textbf{(E)}\ (n+1)(n+2)$
|
The sum of the odd integers $2k-1$ from $1$ to $n$ is $n^2$ . However, in this problem, the sum is instead $2k+1$ , starting at $3$ rather than $1$ . To rewrite this, we note that $2k-1$ is $2$ less than $2k+1$ for every $k$ we add, so for $n$ $k$ 's, we subtract $2n$ , giving us $n^2+2n$ ,which factors as $n(n+2) \implies \boxed{2}$
| 2
|
1,526
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_27
| 1
|
If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$
|
Let the angle be $\theta$ and the sides around it be $a$ and $b$ .
The area of the triangle can be written as \[A =\frac{a \cdot b \cdot \sin(\theta)}{2}\] The doubled sides have length $2a$ and $2b$ , while the angle is still $\theta$ . Thus, the area is \[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\] \[\Rrightarrow \frac{4ab \sin \theta}{2} = 4A\] \[\boxed{4}\]
| 4
|
1,527
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_28
| 1
|
Mr. J left his entire estate to his wife, his daughter, his son, and the cook.
His daughter and son got half the estate, sharing in the ratio of $4$ to $3$ .
His wife got twice as much as the son. If the cook received a bequest of $\textdollar{500}$ , then the entire estate was:
$\textbf{(A)}\ \textdollar{3500}\qquad \textbf{(B)}\ \textdollar{5500}\qquad \textbf{(C)}\ \textdollar{6500}\qquad \textbf{(D)}\ \textdollar{7000}\qquad \textbf{(E)}\ \textdollar{7500}$
|
The wife, daughter, son, and cook received estates in the ratio $6:4:3:1.$ The estate is worth $6+4+3+1 = 14$ units of $$500.,$ which is $\boxed{7000}$
| 0
|
1,528
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_31
| 1
|
In our number system the base is ten. If the base were changed to four you would count as follows: $1,2,3,10,11,12,13,20,21,22,23,30,\ldots$ The twentieth number would be:
$\textbf{(A)}\ 20 \qquad\textbf{(B)}\ 38 \qquad\textbf{(C)}\ 44 \qquad\textbf{(D)}\ 104 \qquad\textbf{(E)}\ 110$
|
The $20^{\text{th}}$ number will be the value of $20_{10}$ in base $4$ . Thus, we see \[20_{10} = (1) \cdot 4^2 + (1) \cdot 4^1 + 0 \cdot 4^0\] \[= 110_{4}\]
$\boxed{110}$
| 110
|
1,529
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_34
| 1
|
If $n$ is any whole number, $n^2(n^2 - 1)$ is always divisible by
$\textbf{(A)}\ 12\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ \text{any multiple of }12\qquad \textbf{(D)}\ 12-n\qquad \textbf{(E)}\ 12\text{ and }24$
|
Suppose $n$ is even. So, we have $n^2(n+1)(n-1).$ Out of these three numbers, at least one of them is going to be a multiple of 3. $n^2$ is also a multiple of 4. Therefore, this expression is always divisible by $\boxed{12}.$
| 12
|
1,530
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_43
| 1
|
The number of scalene triangles having all sides of integral lengths, and perimeter less than $13$ is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2 \qquad\textbf{(C)}\ 3 \qquad\textbf{(D)}\ 4 \qquad\textbf{(E)}\ 18$
|
We can write all possible triangles adding up to 12 or less \[(2, 4, 5)=11\] \[(3, 4, 5)=12\] \[(2, 3, 4)=9\]
This leaves $\boxed{3}$ scalene triangles.
| 3
|
1,531
|
https://artofproblemsolving.com/wiki/index.php/1956_AHSME_Problems/Problem_49
| 1
|
Triangle $PAB$ is formed by three tangents to circle $O$ and $\angle APB = 40^{\circ}$ ; then $\angle AOB$ equals:
$\textbf{(A)}\ 45^{\circ}\qquad \textbf{(B)}\ 50^{\circ}\qquad \textbf{(C)}\ 55^{\circ}\qquad \textbf{(D)}\ 60^{\circ}\qquad \textbf{(E)}\ 70^{\circ}$
|
First, from triangle $ABO$ $\angle AOB = 180^\circ - \angle BAO - \angle ABO$ . Note that $AO$ bisects $\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\angle BAO = \angle BAT/2$ . Similarly, $\angle ABO = \angle ABR/2$
Also, $\angle BAT = 180^\circ - \angle BAP$ , and $\angle ABR = 180^\circ - \angle ABP$ . Hence,
$\angle AOB = 180^\circ - \angle BAO - \angle ABO = 180^\circ - \frac{\angle BAT}{2} - \frac{\angle ABR}{2} = 180^\circ - \frac{180^\circ - \angle BAP}{2} - \frac{180^\circ - \angle ABP}{2}= \frac{\angle BAP + \angle ABP}{2}.$
Finally, from triangle $ABP$ $\angle BAP + \angle ABP = 180^\circ - \angle APB = 180^\circ - 40^\circ = 140^\circ$ , so \[\angle AOB = \frac{\angle BAP + \angle ABP}{2} = \frac{140^\circ}{2} = \boxed{70}.\]
| 70
|
1,532
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_9
| 1
|
A circle is inscribed in a triangle with sides $8, 15$ , and $17$ . The radius of the circle is:
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 7$
|
We know that $A = sr$ , where $A$ is the triangle's area, $s$ its semiperimeter, and $r$ its inradius. Since this particular triangle is a right triangle (which we can verify by the Pythagorean theorem), the area is half of $8*15 = 120$ , and the semiperimeter is half of $8 + 15 + 17 = 40$ . Therefore, the inradius is $\frac{120}{40} = 3$ , so our answer is $\boxed{3}$ and we are done.
| 3
|
1,533
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_12
| 1
|
The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$
|
First, square both sides. This gives us
\[\sqrt{5x-1}^2+2\cdot\sqrt{5x-1}\cdot\sqrt{x-1}+\sqrt{x-1}^2=4 \Longrightarrow 5x-1+2\cdot\sqrt{(5x-1)\cdot(x-1)}+x-1=4 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}+6x-2=4\] Then, adding $-6x$ to both sides gives us
\[2\cdot\sqrt{5x^2-6x+1}+6x-2-6x=4-6x \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}-2 =-6x+4\] After that, adding $2$ to both sides will give us
\[2\cdot\sqrt{5x^2-6x+1}-2+2=-6x+4+2 \Longrightarrow 2\cdot\sqrt{5x^2-6x+1}=-6x+6\] Next, we divide both sides by 2 which gives us
\[\frac{2\cdot\sqrt{5x^2-6x+1}-2+2}{2}=\frac{-6x+4+2}{2} \Longrightarrow \sqrt{5x^2-6x+1}=-3x+3\] Finally, solving the equation, we get
\[5x^2-6x+1=(-3x+3)^2 \Longrightarrow 5x^2-6x+1=9x^2-18x+9\] \[\Longrightarrow 5x^2-6x+1-(9x^2-18x+9)=9x^2-18x+9-(9x^2-18x+9)\] \[\Longrightarrow -4x^2+12x-8=0 \Longrightarrow -4(x-1)(x-2)=0\] \[\Longrightarrow x-1=0 \text{or}\ x-2=0 \Longrightarrow x=1 \text{or}\ x=2\] Plugging 1 and 2 into the original equation, $\sqrt{5x-1}+\sqrt{x-1}=2$ , we see that when $x=1$
\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot1-1}+\sqrt{1-1}=2 \Longrightarrow \sqrt4+\sqrt0=2 \Longrightarrow 2+0=2 \Longrightarrow 2=2\] the equation is true. On the other hand, we note that when $x=2$
\[\sqrt{5x-1}+\sqrt{x-1}=2 \Longrightarrow \sqrt{5\cdot2-1}+\sqrt{2-1}=2 \Longrightarrow \sqrt9+\sqrt1=2 \Longrightarrow 3+1=2 \Longrightarrow 4=0\] the equation is false.
Therefore the answer is $\boxed{1}$
| 1
|
1,534
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_12
| 2
|
The solution of $\sqrt{5x-1}+\sqrt{x-1}=2$ is:
$\textbf{(A)}\ x=2,x=1\qquad\textbf{(B)}\ x=\frac{2}{3}\qquad\textbf{(C)}\ x=2\qquad\textbf{(D)}\ x=1\qquad\textbf{(E)}\ x=0$
|
Let us test the answer choices, for it is in this case simpler and quicker. $x=0$ and $x=2/3$ obviously doesn't work, since square roots of negative numbers are not real. $x=2$ doesn't work either, because the first term is already bigger than 2, and the second term cannot be negative. By the process of elimination, the answer is $\boxed{1}$
| 1
|
1,535
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_23
| 1
|
In checking the petty cash a clerk counts $q$ quarters, $d$ dimes, $n$ nickels, and $c$ cents. Later he discovers that $x$ of the nickels were counted as quarters and $x$ of the dimes were counted as cents. To correct the total obtained the clerk must:
$\textbf{(A)}\ \text{make no correction}\qquad\textbf{(B)}\ \text{subtract 11 cents}\qquad\textbf{(C)}\ \text{subtract 11}x\text{ cents}\\ \textbf{(D)}\ \text{add 11}x\text{ cents}\qquad\textbf{(E)}\ \text{add }x\text{ cents}$
|
If the clerk mistook $x$ nickels as quarters, then every mistake inflates the total by $20$ cents. In order to correct this, we have to subtract $20$ cents $x$ times, for a total of $20x$ cents. We can do the same for the $x$ dimes that were turned into pennies (or cents). This exchange would increase the total value of the coins by $9x$
As a total, you get $-20x + 9x = -11x$ . In order to correct the amount, the clerk should $\boxed{11}$
| 11
|
1,536
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_30
| 1
|
Each of the equations $3x^2-2=25, (2x-1)^2=(x-1)^2, \sqrt{x^2-7}=\sqrt{x-1}$ has:
$\textbf{(A)}\ \text{two integral roots}\qquad\textbf{(B)}\ \text{no root greater than 3}\qquad\textbf{(C)}\ \text{no root zero}\\ \textbf{(D)}\ \text{only one root}\qquad\textbf{(E)}\ \text{one negative root and one positive root}$
Solution
|
Since the question asks us about the unifying characteristic of all three equations' roots, we have to first determine them.
$3x^2-2 = 25$ can be rewritten as $3x^2 - 27 = 0$ , which gives the following roots $+3$ and $-3$
$(2x-1)^2 = (x-1)^2$ can be expanded to $4x^2-4x+1=x^2-2x+1$ , which in turn leads to $3x^2-2x=0$ . The roots here are $0$ and $\frac{2}{3}$
$\sqrt{x^2-7}=\sqrt{x-1}$ , when squared, also turns into a quadratic equation: $x^2 - x - 6 = 0$ . Binomial factoring gives us the roots $-2$ and $3$
We can clearly see that, between all of the equations, there is $\boxed{3}$
| 3
|
1,537
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_38
| 1
|
Four positive integers are given. Select any three of these integers, find their arithmetic average,
and add this result to the fourth integer. Thus the numbers $29, 23, 21$ , and $17$ are obtained. One of the original integers is:
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 23 \qquad \textbf{(D)}\ 29 \qquad \textbf{(E)}\ 17$
|
Define numbers $a, b, c,$ and $d$ to be the four numbers. In order to satisfy the following conditions, the system of equation should be constructed. (It doesn't matter which variable is which.) \[\frac{a+b+c}{3}+d=29\] \[\frac{a+b+d}{3}+c=23\] \[\frac{a+c+d}{3}+b=21\] \[\frac{b+c+d}{3}+a=17\] Adding all of the equations together, we get: $2(a+b+c+d)=90$ . This means that $a+b+c+d=45$
We can determine that $\frac{a+b+c}{3}+d+16=a+b+c+d$ . This, with some algebra, means that $\frac{1}{3}(a+b+c)=8$ $d$ must be $\boxed{21}$
| 21
|
1,538
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_46
| 1
|
The graphs of $2x+3y-6=0, 4x-3y-6=0, x=2$ , and $y=\frac{2}{3}$ intersect in:
$\textbf{(A)}\ \text{6 points}\qquad\textbf{(B)}\ \text{1 point}\qquad\textbf{(C)}\ \text{2 points}\qquad\textbf{(D)}\ \text{no points}\\ \textbf{(E)}\ \text{an unlimited number of points}$
|
We first convert each of the lines into slope-intercept form ( $y = mx + b$ ):
$2x+3y-6=0 ==> 3y = -2x + 6 ==> y = -\frac{2}{3}x + 2$
$4x - 3y - 6 = 0 ==> 4x - 6 = 3y ==> y = \frac{4}{3}x - 2$
$x = 2$ stays as is.
$y = \frac{2}{3}$ stays as is
We can graph the four lines here: [1]
When we do that, the answer turns out to be $\boxed{1}$
| 1
|
1,539
|
https://artofproblemsolving.com/wiki/index.php/1955_AHSME_Problems/Problem_50
| 1
|
In order to pass $B$ going $40$ mph on a two-lane highway, $A$ , going $50$ mph, must gain $30$ feet.
Meantime, $C, 210$ feet from $A$ , is headed toward him at $50$ mph. If $B$ and $C$ maintain their speeds,
then, in order to pass safely, $A$ must increase his speed by:
$\textbf{(A)}\ \text{30 mph}\qquad\textbf{(B)}\ \text{10 mph}\qquad\textbf{(C)}\ \text{5 mph}\qquad\textbf{(D)}\ \text{15 mph}\qquad\textbf{(E)}\ \text{3 mph}$
|
Let $V_A, V_B, V_C$ be $A, B, C$ 's velocity, respectively. We want to pass $B$ before we collide with $C$ . Since $A$ and $B$ are going in the same direction and $V_A>V_B$ $A$ will pass $B$ in $\frac{30\mathrm{ft}}{V_A-V_B}$ time. Since $A$ and $C$ are going in opposite directions, their relative velocity is $V_A+V_C$ , so the amount of time before $A$ will collide with $C$ is given by $\frac{210\mathrm{ft}}{V_A+V_C}$ . We want to pass $B$ before we collide with $C$ , so $V_A$ must satisfy the inequality $\frac{30\mathrm{ft}}{V_A-V_B}<\frac{210\mathrm{ft}}{V_A+V_C}$ . We can eliminate all the units, simplifying the inequality to $\frac{1}{V_A-V_B} < \frac{7}{V_A+V_C}$ . Solving this and substituting our known values of $V_B$ and $V_C$ yields $330<6V_A$ , so $A$ must increase his speed by $\boxed{5}$ miles per hour.
| 5
|
1,540
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_9
| 1
|
A point $P$ is outside a circle and is $13$ inches from the center. A secant from $P$ cuts the circle at $Q$ and $R$ so that the external segment of the secant $PQ$ is $9$ inches and $QR$ is $7$ inches. The radius of the circle is:
$\textbf{(A)}\ 3" \qquad \textbf{(B)}\ 4" \qquad \textbf{(C)}\ 5" \qquad \textbf{(D)}\ 6"\qquad\textbf{(E)}\ 7"$
|
Using the Secant-Secant Power Theorem, you can get $9(16)=(13-r)(13+r)$ , where $r$ is the radius of the given circle. Solving the equation, you get a quadratic: $r^2-25$ . A radius cannot be negative so the answer is $\boxed{5}$
| 5
|
1,541
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_22
| 1
|
The expression $\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)}$ cannot be evaluated for $x=-1$ or $x=2$ ,
since division by zero is not allowed. For other values of $x$
$\textbf{(A)}\ \text{The expression takes on many different values.}\\ \textbf{(B)}\ \text{The expression has only the value 2.}\\ \textbf{(C)}\ \text{The expression has only the value 1.}\\ \textbf{(D)}\ \text{The expression always has a value between }-1\text{ and }+2.\\ \textbf{(E)}\ \text{The expression has a value greater than 2 or less than }-1.$
|
$\frac{2x^2-x}{(x+1)(x-2)}-\frac{4+x}{(x+1)(x-2)} = \frac{2x^2-2x-4}{(x+1)(x-2)}$
This can be factored as $\frac{(2)(x^2-x-2)}{(x+1)(x-2)} \implies \frac{(2)(x+1)(x-2)}{(x+1)(x-2)}$ , which cancels out to $2 \implies \boxed{2}$
| 2
|
1,542
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_30
| 1
|
$A$ and $B$ together can do a job in $2$ days; $B$ and $C$ can do it in four days; and $A$ and $C$ in $2\frac{2}{5}$ days.
The number of days required for A to do the job alone is:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 2.8$
|
Let $A$ do $r_A$ of the job per day, $B$ do $r_B$ of the job per day, and $C$ do $r_C$ of the job per day. These three quantities have unit $\frac{\text{job}}{\text{day}}$ . Therefore our three conditions give us the three equations: \begin{align*} (2\text{ days})(r_A+r_B)&=1\text{ job},\nonumber\\ (4\text{ days})(r_B+r_C)&=1\text{ job},\nonumber\\ (2.4\text{ days})(r_C+r_A)&=1\text{ job}.\nonumber \end{align*} We divide the three equations by the required constant so that the coefficients of the variables become 1: \begin{align*} r_A+r_B&=\frac{1}{2}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_B+r_C&=\frac{1}{4}\cdot\frac{\text{job}}{\text{day}},\nonumber\\ r_C+r_A&=\frac{5}{12}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align*} If we add these three new equations together and divide the result by two, we obtain an equation with left-hand side $r_A+r_B+r_C$ , so if we subtract $r_B+r_C$ (the value of which we know) from both equations, we obtain the value of $r_A$ , which is what we wish to determine anyways. So we add these three equations and divide by two: \[r_A+r_B+r_C=\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{5}{12}\right)\cdot\frac{\text{job}}{\text{day}}=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}.\] Hence: \begin{align*} r_A &= (r_A+r_B+r_C)-(r_B+r_C)\nonumber\\ &=\frac{7}{12}\cdot\frac{\text{job}}{\text{day}}-\frac{1}{4}\cdot\frac{\text{job}}{\text{day}}\nonumber\\ &=\frac{1}{3}\cdot\frac{\text{job}}{\text{day}}.\nonumber \end{align*} This shows that $A$ does one third of the job per day. Therefore, if $A$ were to do the entire job himself, he would require $\boxed{3}$ days.
| 3
|
1,543
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_31
| 1
|
In $\triangle ABC$ $AB=AC$ $\angle A=40^\circ$ . Point $O$ is within the triangle with $\angle OBC \cong \angle OCA$ .
The number of degrees in $\angle BOC$ is:
$\textbf{(A)}\ 110^{\circ} \qquad \textbf{(B)}\ 35^{\circ} \qquad \textbf{(C)}\ 140^{\circ} \qquad \textbf{(D)}\ 55^{\circ} \qquad \textbf{(E)}\ 70^{\circ}$
|
Since $\triangle ABC$ is an isosceles triangle, $\angle ABC = \angle ACB = 70^{\circ}$ . Let $\angle OBC = \angle OCA = x$ . Since $\angle ACB = 70$ $\angle OCB = 70 - x$ . The angle of $\triangle OBC$ add up to $180$ , so $\angle BOC = 180 - (x + 70 - x) = \boxed{110}$
| 110
|
1,544
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_33
| 1
|
A bank charges $\textdollar{6}$ for a loan of $\textdollar{120}$ . The borrower receives $\textdollar{114}$ and repays the loan in $12$ easy installments of $\textdollar{10}$ a month. The interest rate is approximately:
$\textbf{(A)}\ 5 \% \qquad \textbf{(B)}\ 6 \% \qquad \textbf{(C)}\ 7 \% \qquad \textbf{(D)}\ 9\% \qquad \textbf{(E)}\ 15 \%$
|
The borrower pays $\textdollar{120}$ in a single year for a loan of $\textdollar{114}$ . This means that the bank charges an interest of $\textdollar{6}$ for a loan of $\textdollar{114}$ over a single year, so that the annual interest rate is $100*\frac{\textdollar{6}}{\textdollar{114}} = \frac{100}{19} \approx 5$ percentage points. Therefore, our answer is $\boxed{5}$ and we are done.
| 5
|
1,545
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_43
| 1
|
The hypotenuse of a right triangle is $10$ inches and the radius of the inscribed circle is $1$ inch. The perimeter of the triangle in inches is:
$\textbf{(A)}\ 15 \qquad \textbf{(B)}\ 22 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 26 \qquad \textbf{(E)}\ 30$
|
To begin, let's notice that the inscribed circle of the right triangle is its incircle , and that the radius of the incircle is the right triangle's inradius . In this case, the hypotenuse is 10, and the inradius is 1. The formula for the inradius of a right triangle is $r = (a+b-c)/2$ , where $r$ is the length of the inradius of the triangle, $c$ is the length hyptotenuse the the right triangle, and $a$ and $b$ are the lengths of the legs of the right triangle. From this, we can plug in values to notice that:
\begin{align*} r & = \frac{a+b-c}{2}\\\\ 1 & = \frac{a+b-10}{2}\\\\ 2 & = a+b-10\\\\ 12 &= a+b\\\\ \end{align*}
From this, we arrive at $a+b+c = 12+10 = 22$ . The answer is clearly $\boxed{22}$
| 22
|
1,546
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_44
| 1
|
A man born in the first half of the nineteenth century was $x$ years old in the year $x^2$ . He was born in:
$\textbf{(A)}\ 1849 \qquad \textbf{(B)}\ 1825 \qquad \textbf{(C)}\ 1812 \qquad \textbf{(D)}\ 1836 \qquad \textbf{(E)}\ 1806$
|
If a man born in the 19th century was $x$ years of in the year $x^2$ , it implies that the year the man was born was $x^2-x$ . So, if the man was born in the first half of the 19th century, it means that $x^2-x < 1850$ . Noticing that $40^2 - 40 = 1560$ and $50^2-50 = 2450$ , we see that $40 < x < 50$ . We can guess values until we hit a solution. $43^2-43 = 1806$ , so we see that the man had to have been $43$ years old in the year $1849=43^2$ , so the answer is $\boxed{1806}$
| 806
|
1,547
|
https://artofproblemsolving.com/wiki/index.php/1954_AHSME_Problems/Problem_48
| 1
|
A train, an hour after starting, meets with an accident which detains it a half hour, after which it proceeds at $\frac{3}{4}$ of its former rate and arrives $3\tfrac{1}{2}$ hours late. Had the accident happened $90$ miles farther along the line, it would have arrived only $3$ hours late. The length of the trip in miles was:
$\textbf{(A)}\ 400 \qquad \textbf{(B)}\ 465 \qquad \textbf{(C)}\ 600 \qquad \textbf{(D)}\ 640 \qquad \textbf{(E)}\ 550$
|
Let the speed of the train be $x$ miles per hour, and let $D$ miles be the total distance of the trip, where $x$ and $D$ are unit-less quantities. Then for the trip that actually occurred, the train travelled 1 hour before the crash, and then travelled $D-x$ miles after the crash. In other words, the train travelled for $\frac{(D-x)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x}{3x/4}\text{hr}$ after the crash. So, not including the half hour detention after the crash, the entire trip took \[\left(1+\frac{D-x}{3x/4}\right)\text{hours}.\] Now let us consider the alternative trip, where the accident happened $90$ miles farther along the line. Originally, the accident happened $x\frac{\text{mi}}{\text{hr}}\cdot(1\text{ hr})=x\text{ mi}$ down the line, but in this situation it happens $x+90$ miles from the line. Therefore the accident happens $\frac{(x+90)\text{ mi}}{x\frac{\text{mi}}{\text{hr}}}=\frac{x+90}{x}\text{hr}$ into the trip. The length of the remaining part of the trip is now $D-x-90$ , so the remaining trip takes $\frac{(D-x-90)\text{mi}}{\frac{3x}{4}\frac{\text{mi}}{\text{hr}}}=\frac{D-x-90}{3x/4}\text{hr}$ . So, not including the half hour detention after the crash, the entire trip took \[\left(\frac{x+90}{x}+\frac{D-x-90}{3x/4}\right)\text{hours}.\] We are given that the trip that actually happened resulted in being $3.5$ hours late, while the alternative trip would have resulted in being only $3$ hours late. Therefore the first trip took exactly $\frac{1}{2}$ hour more: \[1+\frac{D-x}{3x/4}=\frac{x+90}{x}+\frac{D-x-90}{3x/4}+\frac{1}{2}.\] After simplification and cancellation, we get the equation \[0=\frac{90}{x}-\frac{90}{3x/4}+\frac{1}{2}.\] Therefore $(90/x)+(1/2)=(120/x)$ , so $1/2=30/x$ . We solve for $x$ and have the original speed of the train: $x=60$
We must now solve for the length of the actual trip. If the train had gone $60\frac{\text{mi}}{\text{hr}}$ for the entire trip, then it would have taken $\frac{D}{60}\text{hr}$ , and the train would have been on time. But the trip actually took $1+\frac{D-60}{45}$ hours, not including the half-hour detention, and the train was $3.5$ hours late. Therefore \[\frac{D}{60}+3.5=1+\frac{D-60}{45}+0.5\Rightarrow 3D+630=180+4D-240+90\Rightarrow D=600.\] The length of the trip in miles was $\boxed{600}$
| 600
|
1,548
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_1
| 1
|
A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$
|
The boy buys $3$ oranges for $10$ cents or $1$ orange for $\frac{10}{3}$ cents. He sells them at $\frac{20}{5}=4$ cents each.
That means for every orange he sells, he makes a profit of $4-\frac{10}{3}=\frac{2}{3}$ cents.
To make a profit of $100$ cents, he needs to sell $\frac{100}{\frac{2}{3}}=\boxed{150}$
| 150
|
1,549
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_1
| 2
|
A boy buys oranges at $3$ for $10$ cents. He will sell them at $5$ for $20$ cents. In order to make a profit of $$1.00$ , he must sell:
$\textbf{(A)}\ 67 \text{ oranges} \qquad \textbf{(B)}\ 150 \text{ oranges} \qquad \textbf{(C)}\ 200\text{ oranges}\\ \textbf{(D)}\ \text{an infinite number of oranges}\qquad \textbf{(E)}\ \text{none of these}$
|
The boy buys $3$ oranges for $10$ cents. He sells them at $5$ for $20$ cents. So, he buys $15$ for $50$ cents and sells them at $15$ for $60$ cents, so he makes $10$ cents of profit on every $15$ oranges. To make $100$ cents of profit, he needs to sell $15 \cdot \frac{100}{10} = \boxed{150}$ oranges.
| 150
|
1,550
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_9
| 1
|
The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
|
Say we add $N$ ounces of water to the shaving lotion. Since half of a $9$ ounce bottle of shaving lotion is alcohol, we know that we have $\frac{9}{2}$ ounces of alcohol. We want $\frac{9}{2}=0.3(9+N)$ (because we want the amount of alcohol, $\frac{9}{2}$ , to be $30\%$ , or $0.3$ , of the total amount of shaving lotion, $9+N$ ). Solving, we find that \[9=0.6(9+N)\implies9=5.4+0.6N\implies3.6=0.6N\implies6=N.\] So, the total amount of water we need to add is $\boxed{6}$
| 6
|
1,551
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_9
| 2
|
The number of ounces of water needed to reduce $9$ ounces of shaving lotion containing $50$ % alcohol to a lotion containing $30$ % alcohol is:
$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 6 \qquad \textbf{(E)}\ 7$
|
The concentration of alcohol after adding $n$ ounces of water is $\frac{4.5}{9+n}$
To get a solution of 30% alcohol, we solve $\frac{4.5}{9+n}=\frac{3}{10}$
$45=27+3n$
$18=3n$
$6=n \implies \boxed{6}$
| 6
|
1,552
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_10
| 1
|
The number of revolutions of a wheel, with fixed center and with an outside diameter of $6$ feet, required to cause a point on the rim to go one mile is:
$\textbf{(A)}\ 880 \qquad\textbf{(B)}\ \frac{440}{\pi} \qquad\textbf{(C)}\ \frac{880}{\pi} \qquad\textbf{(D)}\ 440\pi\qquad\textbf{(E)}\ \text{none of these}$
|
We know that the radius of the wheel is $3$ feet, so the total circumference of the wheel is $6\pi$ feet. We also know that one mile is equivalent to $5280$ feet. It takes $\frac{5280}{6\pi}$ revolutions for any one point on the wheel to travel a mile. Simplifying, we find that the answer is $\boxed{880}$
| 880
|
1,553
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_29
| 1
|
The number of significant digits in the measurement of the side of a square whose computed area is $1.1025$ square inches to
the nearest ten-thousandth of a square inch is:
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 1$
|
There are 5 significant digits, $1$ $1$ $0$ $2$ , and $5$ . The answer is $\boxed{5}$
| 5
|
1,554
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_31
| 1
|
The rails on a railroad are $30$ feet long. As the train passes over the point where the rails are joined, there is an audible click.
The speed of the train in miles per hour is approximately the number of clicks heard in:
$\textbf{(A)}\ 20\text{ seconds} \qquad \textbf{(B)}\ 2\text{ minutes} \qquad \textbf{(C)}\ 1\frac{1}{2}\text{ minutes}\qquad \textbf{(D)}\ 5\text{ minutes}\\ \textbf{(E)}\ \text{none of these}$
|
We assume that the clicks are heard at the head of the train. Then if the train's speed in miles per hour is $x$ , we can convert it to clicks per minute: \[\frac{x\text{ mile}}{\text{hr}}\cdot\left(\frac{1\text{ hr}}{60\text{ min}}\right)\cdot\left(\frac{5280\text{ ft}}{1\text{ mile}}\right)\cdot\left(\frac{1\text{ click}}{30\text{ ft}}\right)=\frac{5280x}{1800}\cdot\frac{\text{click}}{\text{min}}.\] Therefore every minute, on average, $5280x/1800$ clicks are heard. The number of clicks heard in $y$ minutes is $y\cdot 5280x/1800$ , so the number of clicks heard in $1800/5280$ minutes is equal to $x$ . In other words, the speed of the train in miles per hour is equal to the number of clicks heard in $1800/5280$ minutes, which is approximately one-third of a minute, or $\boxed{20}$
| 20
|
1,555
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_34
| 1
|
If one side of a triangle is $12$ inches and the opposite angle is $30^{\circ}$ , then the diameter of the circumscribed circle is:
$\textbf{(A)}\ 18\text{ inches} \qquad \textbf{(B)}\ 30\text{ inches} \qquad \textbf{(C)}\ 24\text{ inches} \qquad \textbf{(D)}\ 20\text{ inches}\\ \textbf{(E)}\ \text{none of these}$
|
By the Extended Law of Sines, the diameter, or twice the circumradius $R$ , is given by \[2R=\frac{12\text{ inches}}{\sin30^\circ}=\boxed{24}.\]
| 24
|
1,556
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_36
| 1
|
Determine $m$ so that $4x^2-6x+m$ is divisible by $x-3$ . The obtained value, $m$ , is an exact divisor of:
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 48 \qquad \textbf{(E)}\ 64$
|
Since the given expression is a quadratic, the factored form would be $(x-3)(4x+y)$ , where $y$ is a value such that $-12x+yx=-6x$ and $-3(y)=m$ . The only number that fits the first equation is $y=6$ , so $m=-18$ . The only choice that is a multiple of 18 is $\boxed{36}$
| 36
|
1,557
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_38
| 1
|
If $f(a)=a-2$ and $F(a,b)=b^2+a$ , then $F(3,f(4))$ is:
$\textbf{(A)}\ a^2-4a+7 \qquad \textbf{(B)}\ 28 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 11$
|
We find $f(4)=(4)-2=2$ , so $F(3,f(4))=F(3,2)=(2)^2+3=\boxed{7}$
| 7
|
1,558
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_39
| 1
|
The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$
|
\[a^x=b\] \[b^y=a\] \[{a^x}^y=a\] \[xy=1\] \[\log_a b\log_b a=1\] As a result, the answer should be $\boxed{1}$
| 1
|
1,559
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_39
| 2
|
The product, $\log_a b \cdot \log_b a$ is equal to:
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ a \qquad \textbf{(C)}\ b \qquad \textbf{(D)}\ ab \qquad \textbf{(E)}\ \text{none of these}$
|
Apply the change of base formula to $\log_a b$ and $\log_b a$ . For simplicity, let us convert to base-10 log.
By change of base, the expression becomes $\frac{\log b}{\log a} * \frac{\log a}{\log b} = \boxed{1}$
| 1
|
1,560
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_42
| 1
|
The centers of two circles are $41$ inches apart. The smaller circle has a radius of $4$ inches and the larger one has a radius of $5$ inches.
The length of the common internal tangent is:
$\textbf{(A)}\ 41\text{ inches} \qquad \textbf{(B)}\ 39\text{ inches} \qquad \textbf{(C)}\ 39.8\text{ inches} \qquad \textbf{(D)}\ 40.1\text{ inches}\\ \textbf{(E)}\ 40\text{ inches}$
|
[asy] size(400); draw((0,0)--(41,0)); draw((0,0)--(45/41,200/41)--(1645/41,-160/41)); draw((0,0)--(1600/41,-360/41)--(41,0)); draw(circle((0,0),5)); draw(circle((41,0),4)); label("$A$",(0,0),W); label("$B$",(41,0),E); label("$C$",(45/41,200/41),N); label("$D$",(1645/41,-160/41),SE); label("$E$",(1600/41,-360/41),E); [/asy]
Let $A$ be the center of the circle with radius $5$ , and $B$ be the center of the circle with radius $4$ . Let $\overline{CD}$ be the common internal tangent of circle $A$ and circle $B$ . Extend $\overline{BD}$ past $D$ to point $E$ such that $\overline{BE}\perp\overline{AE}$ . Since $\overline{AC}\perp\overline{CD}$ and $\overline{BD}\perp\overline{CD}$ $ACDE$ is a rectangle. Therefore, $AC=DE$ and $CD=AE$
Since the centers of the two circles are $41$ inches apart, $AB=41$ . Also, $BE=4+5=9$ . Using the Pythagorean Theorem on right triangle $ABE$ $CD=AE=\sqrt{41^2-9^2}=\sqrt{1600}=40$ . The length of the common internal tangent is $\boxed{40}$
| 40
|
1,561
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_45
| 1
|
The lengths of two line segments are $a$ units and $b$ units respectively. Then the correct relation between them is:
$\textbf{(A)}\ \frac{a+b}{2} > \sqrt{ab} \qquad \textbf{(B)}\ \frac{a+b}{2} < \sqrt{ab} \qquad \textbf{(C)}\ \frac{a+b}{2}=\sqrt{ab}\\ \textbf{(D)}\ \frac{a+b}{2}\leq\sqrt{ab}\qquad \textbf{(E)}\ \frac{a+b}{2}\geq\sqrt{ab}$
|
Since both lengths are positive, the AM-GM Inequality is satisfied. The correct relationship between $a$ and $b$ is $\boxed{2}$
| 2
|
1,562
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_48
| 1
|
If the larger base of an isosceles trapezoid equals a diagonal and the smaller base equals the altitude,
then the ratio of the smaller base to the larger base is:
$\textbf{(A)}\ \frac{1}{2} \qquad \textbf{(B)}\ \frac{2}{3} \qquad \textbf{(C)}\ \frac{3}{4} \qquad \textbf{(D)}\ \frac{3}{5}\qquad \textbf{(E)}\ \frac{2}{5}$
|
[asy] draw((0,0)--(1,3)--(4,3)--(5,0)--cycle); draw((0,0)--(4,3)); draw((4,3)--(4,0)); draw((3.8,0)--(3.8,0.2)--(4,0.2)); label("$A$",(0,0),W); label("$B$",(1,3),NW); label("$C$",(4,3),NE); label("$D$",(5,0),E); label("$E$",(4,0),S); label("1",(2,1.5),NW); [/asy]
Let $a$ be the length of the smaller base of isosceles trapezoid $ABCD$ , and $1$ be the length of the larger base of the trapezoid. The ratio of the smaller base to the larger base is $\frac a1=a$ . Let point $E$ be the foot of the altitude from $C$ to $\overline{AD}$
Since the larger base of the isosceles trapezoid equals a diagonal, $AC=AD=1$ . Since the smaller base equals the altitude, $BC=CE=a$ . Since the trapezoid is isosceles, $DE=\frac{1-a}{2}$ , so $AE = 1-\frac{1-a}{2} = \frac{a+1}{2}$ . Using the Pythagorean Theorem on right triangle $ACE$ , we have \[a^2+\left(\frac{a+1}{2}\right)^2=1\] Multiplying both sides by $4$ gives \[4a^2+(a+1)^2=4\] Expanding the squared binomial and rearranging gives \[5a^2+2a-3=0\] This can be factored into $(5a-3)(a+1)=0$ . Since a must be positive, $5a+3=0$ , so $a=\frac 35$ . The ratio of the smaller base to the larger base is $\boxed{35}$
| 35
|
1,563
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_50
| 1
|
One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$
|
Let the triangle have side lengths $14, 6+x,$ and $8+x$ . The area of this triangle can be computed two ways. We have $A = rs$ , and $A = \sqrt{s(s-a)(s-b)(s-c)}$ , where $s = 14+x$ is the semiperimeter. Therefore, $4(14+x)=\sqrt{(14+x)(x)(8)(6)}$ . Solving gives $x = 7$ as the only valid solution. This triangle has sides $13,14$ and $15$ , so the shortest side is $\boxed{13}$
| 13
|
1,564
|
https://artofproblemsolving.com/wiki/index.php/1953_AHSME_Problems/Problem_50
| 2
|
One of the sides of a triangle is divided into segments of $6$ and $8$ units by the point of tangency of the inscribed circle. If the radius of the circle is $4$ , then the length of the shortest side is
$\textbf{(A) \ } 12 \mathrm{\ units} \qquad \textbf{(B) \ } 13 \mathrm{\ units} \qquad \textbf{(C) \ } 14 \mathrm{\ units} \qquad \textbf{(D) \ } 15 \mathrm{\ units} \qquad \textbf{(E) \ } 16 \mathrm{\ units}$
|
Label the tangent points on $BC, CA, AB$ as $D, E, F$ respectively. Let $AF=AE=6$ $BF=BD=8$ , and $CE=CD=x.$ The problem is a matter of solving for $x$ . To this, we use the fact that if $A,B,C$ are the angles of a triangle, then $\tan{\frac{A}{2}}\tan{\frac{B}{2}}+\tan{\frac{B}{2}}\tan{\frac{C}{2}}+\tan{\frac{C}{2}}\tan{\frac{A}{2}} = 1.$ We know that $\tan{\frac{A}{2}} = \frac{2}{3}$ $\tan{\frac{B}{2}} = \frac{1}{2}$ , and $\tan{\frac{C}{2}} = \frac{4}{x},$ so we have the equation $\frac{1}{2}\cdot \frac{2}{3} + \frac{1}{2}\cdot \frac{4}{x} + \frac{4}{x}\cdot \frac{2}{3} = 1.$ Solving this equation yields $x=7$ , so the shortest side has length $\boxed{13}$
| 13
|
1,565
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_2
| 1
|
Two high school classes took the same test. One class of $20$ students made an average grade of $80\%$ ; the other class of $30$ students made an average grade of $70\%$ . The average grade for all students in both classes is:
$\textbf{(A)}\ 75\%\qquad \textbf{(B)}\ 74\%\qquad \textbf{(C)}\ 72\%\qquad \textbf{(D)}\ 77\%\qquad \textbf{(E)\ }\text{none of these}$
|
The desired average can be found by dividing the total number of points earned by the total number of students. There are $20\cdot 80+30\cdot 70=3700$ points earned and $20+30=50$ students. Thus, our answer is $\frac{3700}{50}$ , or $\boxed{74}$
| 74
|
1,566
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_6
| 1
|
The difference of the roots of $x^2-7x-9=0$ is:
$\textbf{(A) \ }+7 \qquad \textbf{(B) \ }+\frac{7}{2} \qquad \textbf{(C) \ }+9 \qquad \textbf{(D) \ }2\sqrt{85} \qquad \textbf{(E) \ }\sqrt{85}$
|
Denote the $2$ roots of this quadratic as $r_1$ and $r_2$ . Note that $(r_1-r_2)^2=(r_1+r_2)^2-4r_1r_2$ . By Vieta's Formula's $r_1+r_2=7$ , and $r_1r_2=-9$ . Thus, $r_1-r_2=\sqrt{49+4\cdot 9}=\boxed{85}$
| 85
|
1,567
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_8
| 1
|
Two equal circles in the same plane cannot have the following number of common tangents.
$\textbf{(A) \ }1 \qquad \textbf{(B) \ }2 \qquad \textbf{(C) \ }3 \qquad \textbf{(D) \ }4 \qquad \textbf{(E) \ }\text{none of these}$
|
Two congruent coplanar circles will either be tangent to one another (resulting in $3$ common tangents), intersect one another (resulting in $2$ common tangents), or be separate from one another (resulting in $4$ common tangents).
Having only $\boxed{1}$ common tangent is impossible, unless the circles are non-congruent and internally tangent.
| 1
|
1,568
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_14
| 1
|
A house and store were sold for $\textdollar 12,000$ each. The house was sold at a loss of $20\%$ of the cost, and the store at a gain of $20\%$ of the cost. The entire transaction resulted in:
$\textbf{(A) \ }\text{no loss or gain} \qquad \textbf{(B) \ }\text{loss of }\textdollar 1000 \qquad \textbf{(C) \ }\text{gain of }\textdollar 1000 \qquad \textbf{(D) \ }\text{gain of }\textdollar 2000 \qquad \textbf{(E) \ }\text{none of these}$
|
Denote the original price of the house and the store as $h$ and $s$ , respectively. It is given that $\frac{4h}{5}=\textdollar 12,000$ , and that $\frac{6s}{5}=\textdollar 12,000$ . Thus, $h=\textdollar 15,000$ $s=\textdollar10,000$ , and $h+s=\textdollar25,000$ . This value is $\textdollar1000$ higher than the current price of the property, $2\cdot \textdollar12,000$ . Hence, the transaction resulted in a $\boxed{1000}$
| 0
|
1,569
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_17
| 1
|
A merchant bought some goods at a discount of $20\%$ of the list price. He wants to mark them at such a price that he can give a discount of $20\%$ of the marked price and still make a profit of $20\%$ of the selling price. The per cent of the list price at which he should mark them is:
$\textbf{(A) \ }20 \qquad \textbf{(B) \ }100 \qquad \textbf{(C) \ }125 \qquad \textbf{(D) \ }80 \qquad \textbf{(E) \ }120$
|
Let $C$ represent the cost of the goods, and let $L$ $S$ , and $M$ represent the list, selling, and marked prices of the goods, respectively. Hence, we have three equations, which we need to manipulate in order to relate $M$ and $L$
$C=\frac{4}{5}L$
$S=C+\frac{1}{5}S$
$S=\frac{4}{5}M$
We find that $M=\frac{5}{4}L$ . Hence, the percent of the list price which should be marked is $\boxed{125}$
| 125
|
1,570
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_18
| 1
|
$\log p+\log q=\log(p+q)$ only if:
$\textbf{(A) \ }p=q=\text{zero} \qquad \textbf{(B) \ }p=\frac{q^2}{1-q} \qquad \textbf{(C) \ }p=q=1 \qquad$
$\textbf{(D) \ }p=\frac{q}{q-1} \qquad \textbf{(E) \ }p=\frac{q}{q+1}$
|
$\log p+\log q=\log(p+q)\implies \log pq=\log(p+q)\implies pq=p+q\implies \boxed{1}$
| 1
|
1,571
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_25
| 1
|
A powderman set a fuse for a blast to take place in $30$ seconds. He ran away at a rate of $8$ yards per second. Sound travels at the rate of $1080$ feet per second. When the powderman heard the blast, he had run approximately: $\textbf{(A)}\ \text{200 yd.}\qquad\textbf{(B)}\ \text{352 yd.}\qquad\textbf{(C)}\ \text{300 yd.}\qquad\textbf{(D)}\ \text{245 yd.}\qquad\textbf{(E)}\ \text{512 yd.}$
|
Let $p(t)=24t$ be the number of feet the powderman is from the blast at $t$ seconds after the fuse is lit, and let $q(t)=1080t-32400$ be the number of feet the sound has traveled. We want to solve for $p(t)=q(t)$ \[24t=1080t-32400\] \[1056t=32400\] \[t=\frac{32400}{1056}\] \[t=\frac{675}{22}=30.6\overline{81}\] The number of yards the powderman is from the blast at time $t$ is $\frac{24t}3=8t$ , so the answer is $8(30.6\overline{81})$ , which is about $245$ yards. $\boxed{245}$
| 245
|
1,572
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_27
| 1
|
The ratio of the perimeter of an equilateral triangle having an altitude equal to the radius of a circle, to the perimeter of an equilateral triangle inscribed in the circle is:
$\textbf{(A)}\ 1:2\qquad\textbf{(B)}\ 1:3\qquad\textbf{(C)}\ 1:\sqrt3\qquad\textbf{(D)}\ \sqrt3:2 \qquad\textbf{(E)}\ 2:3$
|
If the radius of the circle is $r$ , then the perimeter of the first triangle is $3\left(\frac{2r}{\sqrt3}\right)=2r\sqrt3$ , and the perimeter of the second is $3r\sqrt3$ . So the ratio is $\boxed{23}$
| 23
|
1,573
|
https://artofproblemsolving.com/wiki/index.php/1952_AHSME_Problems/Problem_44
| 1
|
If an integer of two digits is $k$ times the sum of its digits, the number formed by interchanging the digits is the sum of the digits multiplied by
$\textbf{(A) \ } 9-k \qquad \textbf{(B) \ } 10-k \qquad \textbf{(C) \ } 11-k \qquad \textbf{(D) \ } k-1 \qquad \textbf{(E) \ } k+1$
|
Let $n = 10a+b$ . The problem states that $10a+b=k(a+b)$ . We want to find $x$ , where $10b+a=x(a+b)$ . Adding these two equations gives $11(a+b) = (k+x)(a+b)$ . Because $a+b \neq 0$ , we have $11 = k + x$ , or $x = \boxed{11}$
| 11
|
1,574
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_1
| 1
|
The percent that $M$ is greater than $N$ is:
$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$
|
$M-N$ is the amount by which $M$ is greater than $N$ . We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\boxed{100}$
| 100
|
1,575
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_4
| 1
|
A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$
|
The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$
$((65+50)2)2+130=590 \Rightarrow \boxed{590}$
| 590
|
1,576
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_5
| 1
|
Mr. $A$ owns a home worth $ $10,000$ . He sells it to Mr. $B$ at a $10\%$ profit based on the worth of the house. Mr. $B$ sells the house back to Mr. $A$ at a $10\%$ loss. Then:
$\mathrm{(A) \ A\ comes\ out\ even } \qquad$ $\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}$ $\qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad$ $\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }$ $\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }$
|
Mr. $A$ sells his home for $(1 + 10$ $)$ $\cdot$ $10,000$ dollars $=$ $1.1$ $\cdot$ $10,000$ dollars $=$ $11,000$ dollars to Mr. $B$ . Then, Mr. $B$ sells it at a price of $(1-10$ $)$ $\cdot$ $11,000$ dollars $=$ $0.9$ $\cdot$ $11,000$ dollars $=$ $9,900$ dollars, thus $11,000 - 9,900$ $=$ $\boxed{1100}$
| 100
|
1,577
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_9
| 1
|
An equilateral triangle is drawn with a side of length $a$ . A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
$\textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$
|
The perimeter of the first triangle is $3a$ . The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing $3a+\frac{3a}{2}+\frac{3a}{4}+\cdots$
The starting term is $3a$ , and the common ratio is $1/2$ . Therefore, the sum is $\frac{3a}{1-\frac{1}{2}}=\boxed{6}$
| 6
|
1,578
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_15
| 1
|
The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$ , is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$
|
Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore $6$ must divide the given expression.
Plugging in $n=2$ yields $6$ . So the largest possibility is $6$
Clearly the answer is $\boxed{6}$
| 6
|
1,579
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_15
| 2
|
The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$ , is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$
|
In general, $r!$ $n(n+1)(n+2)...(n+r-1)$ were $r$ and $n$ are integers. So here $3!$ $n^3$ $n$ always for any integer $n$ .Hence,the correct answer is $6$ $\boxed{6}$
| 6
|
1,580
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30
| 1
|
If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$
|
The two poles formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\frac1{\frac1{20}+\frac1{80}}$ , or $\frac1{\frac1{16}}=\boxed{16}$
| 16
|
1,581
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_30
| 2
|
If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$
|
The two lines can be represented as $y=\frac{-x}{5}+20$ and $y=\frac{4x}{5}$ .
Solving the system,
$\frac{-x}{5}+20=\frac{4x}{5}$
$20=x.$
So the lines meet at an $x$ -coordinate of 20.
Solving for the height they meet,
\[y=\frac{4\cdot 20}{5}\] \[y=\boxed{16}.\]
| 16
|
1,582
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_32
| 1
|
If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$ , then $AC+BC$ must be
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$
|
Because $AB$ is the diameter of the semi-circle, it follows that $\angle C = 90$ . Now we can try to eliminate all the solutions except for one by giving counterexamples.
$\textbf{(A):}$ Set point $C$ anywhere on the perimeter of the semicircle except on $AB$ . By triangle inequality, $AC+BC>AB$ , so $\textbf{(A)}$ is wrong. $\textbf{(B):}$ Set point $C$ on the perimeter of the semicircle infinitesimally close to $AB$ , and so $AC+BC$ almost equals $AB$ , therefore $\textbf{(B)}$ is wrong. $\textbf{(C):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$ , it follows that $\textbf{(C)}$ is wrong. $\textbf{(E):}$ Because we proved that $AC+BC$ can be very close to $AB$ in case $\textbf{(B)}$ , it follows that $\textbf{(E)}$ is wrong.
Therefore, the only possible case is $\boxed{2}$
| 2
|
1,583
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_37
| 1
|
A number which when divided by $10$ leaves a remainder of $9$ , when divided by $9$ leaves a remainder of $8$ , by $8$ leaves a remainder of $7$ , etc., down to where, when divided by $2$ , it leaves a remainder of $1$ , is:
$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$
|
If we add $1$ to the number, it becomes divisible by $10, 9, 8, \cdots, 2, 1$ . The LCM of $1$ through $10$ is $2520$ , therefore the number we want to find is $2520-1=\boxed{2519}$
| 519
|
1,584
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_40
| 1
|
$\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$ $\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$
|
First, note that we can pull the exponents out of every factor, since they are all squared. This results in $\left(\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\right)^{4}\cdot\left(\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\right)^{4}$ Now, multiplying the numerators together gives $\left(\frac{x^3+1}{x^3+1}\right)^{4}\cdot\left(\frac{x^3-1}{x^3-1}\right)^{4}$ ,
which simplifies to $\boxed{1}$
| 1
|
1,585
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_44
| 1
|
If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$ , where $a, b, c$ are other than zero, then $x$ equals:
$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$ $\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$
|
Note that $\frac{1}{a}=\frac{1}{x}+\frac{1}{y}$ $\frac{1}{b}=\frac{1}{x}+\frac{1}{z}$ , and $\frac{1}{c}=\frac{1}{y}+\frac{1}{z}$ . Therefore
\[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{2}\]
Therefore
\[\frac{1}{x}=\frac{1}{2a}+\frac{1}{2b}+\frac{1}{2c}-\frac{1}{c}=\frac{1}{2a}+\frac{1}{2b}-\frac{1}{2c}\]
A little algebraic manipulation yields that
\[x=\boxed{2}\]
| 2
|
1,586
|
https://artofproblemsolving.com/wiki/index.php/1951_AHSME_Problems/Problem_45
| 1
|
If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$ , then the only logarithm that cannot be found without the use of tables is:
$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$
|
While $\log 17 = \log(8 + 9)$ , we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore $\boxed{17}$
| 17
|
1,587
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_2
| 1
|
Let $R=gS-4$ . When $S=8$ $R=16$ . When $S=10$ $R$ is equal to:
$\textbf{(A)}\ 11\qquad\textbf{(B)}\ 14\qquad\textbf{(C)}\ 20\qquad\textbf{(D)}\ 21\qquad\textbf{(E)}\ \text{None of these}$
|
Our first procedure is to find the value of $g$ . With the given variables' values, we can see that $8g-4=16$ so $g=\frac{20}{8}=\frac{5}{2}$
With that, we can replace $g$ with $\frac{5}{2}$ . When $S=10$ , we can see that $10\times\frac{5}{2}-4=\frac{50}{2}-4=25-4=\boxed{21}$
| 21
|
1,588
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_5
| 1
|
If five geometric means are inserted between $8$ and $5832$ , the fifth term in the geometric series:
$\textbf{(A)}\ 648\qquad\textbf{(B)}\ 832\qquad\textbf{(C)}\ 1168\qquad\textbf{(D)}\ 1944\qquad\textbf{(E)}\ \text{None of these}$
|
We can let the common ratio of the geometric sequence be $r$ $5832$ is given to be the seventh term in the geometric sequence as there are five terms between it and $a_1$ if we consider $a_1=8$ .
By the formula for each term in a geometric sequence, we find that $a_n=a_1r^{n-1}$ or $(5382)=(8)r^6$ We divide by eight to find: $r^6=729$ $r=\pm 3$
Because $a_2$ will not be between $8$ and $5832$ if $r=-3$ we can discard it as an extraneous solution. We find $r=3$ and $a_5=a_1r^4=(8)(3)^4=\boxed{648}$
| 648
|
1,589
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_8
| 1
|
If the radius of a circle is increased $100\%$ , the area is increased:
$\textbf{(A)}\ 100\%\qquad\textbf{(B)}\ 200\%\qquad\textbf{(C)}\ 300\%\qquad\textbf{(D)}\ 400\%\qquad\textbf{(E)}\ \text{By none of these}$
|
Increasing by $100\%$ is the same as doubling the radius. If we let $r$ be the radius of the old circle, then the radius of the new circle is $2r.$
Since the area of the circle is given by the formula $\pi r^2,$ the area of the new circle is $\pi (2r)^2 = 4\pi r^2.$ The area is quadrupled, or increased by $\boxed{300}$
| 300
|
1,590
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_9
| 1
|
The area of the largest triangle that can be inscribed in a semi-circle whose radius is $r$ is:
$\textbf{(A)}\ r^{2}\qquad\textbf{(B)}\ r^{3}\qquad\textbf{(C)}\ 2r^{2}\qquad\textbf{(D)}\ 2r^{3}\qquad\textbf{(E)}\ \frac{1}{2}r^{2}$
|
The area of a triangle is $\frac12 bh.$ To maximize the base, let it be equal to the diameter of the semi circle, which is equal to $2r.$ To maximize the height, or altitude, choose the point directly in the middle of the arc connecting the endpoints of the diameter. It is equal to $r.$ Therefore the area is $\frac12 \cdot 2r \cdot r = \boxed{2}$
| 2
|
1,591
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_16
| 1
|
The number of terms in the expansion of $[(a+3b)^{2}(a-3b)^{2}]^{2}$ when simplified is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
|
Use properties of exponents to move the squares outside the brackets use difference of squares.
\[[(a+3b)(a-3b)]^4 = (a^2-9b^2)^4\]
Using the binomial theorem, we can see that the number of terms is $\boxed{5}$
| 5
|
1,592
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_18
| 1
|
Of the following
(1) $a(x-y)=ax-ay$
(2) $a^{x-y}=a^x-a^y$
(3) $\log (x-y)=\log x-\log y$
(4) $\frac{\log x}{\log y}=\log{x}-\log{y}$
(5) $a(xy)=ax \cdot ay$
$\textbf{(A)}\text{Only 1 and 4 are true}\qquad\\\textbf{(B)}\ \text{Only 1 and 5 are true}\qquad\\\textbf{(C)}\ \text{Only 1 and 3 are true}\qquad\\\textbf{(D)}\ \text{Only 1 and 2 are true}\qquad\\\textbf{(E)}\ \text{Only 1 is true}$
|
The distributive property doesn't apply to logarithms or in the ways illustrated, and only applies to addition and subtraction. Also, $a^{x-y} = \frac{a^x}{a^y}$ , so $\boxed{1}$
| 1
|
1,593
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_20
| 1
|
When $x^{13}+1$ is divided by $x-1$ , the remainder is:
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ -1\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ \text{None of these answers}$
|
Using synthetic division, we get that the remainder is $\boxed{2}$
| 2
|
1,594
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_21
| 1
|
The volume of a rectangular solid each of whose side, front, and bottom faces are $12\text{ in}^{2}$ $8\text{ in}^{2}$ , and $6\text{ in}^{2}$ respectively is:
$\textbf{(A)}\ 576\text{ in}^{3}\qquad\textbf{(B)}\ 24\text{ in}^{3}\qquad\textbf{(C)}\ 9\text{ in}^{3}\qquad\textbf{(D)}\ 104\text{ in}^{3}\qquad\textbf{(E)}\ \text{None of these}$
|
If the sidelengths of the cubes are expressed as $a, b,$ and $c,$ then we can write three equations:
\[ab=12, bc=8, ac=6.\]
The volume is $abc.$ Notice symmetry in the equations. We can find $abc$ my multiplying all the equations and taking the positive square root.
\begin{align*} (ab)(bc)(ac) &= (12)(8)(6)\\ a^2b^2c^2 &= 576\\ abc &= \boxed{24}
| 24
|
1,595
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_22
| 1
|
Successive discounts of $10\%$ and $20\%$ are equivalent to a single discount of:
$\textbf{(A)}\ 30\%\qquad\textbf{(B)}\ 15\%\qquad\textbf{(C)}\ 72\%\qquad\textbf{(D)}\ 28\%\qquad\textbf{(E)}\ \text{None of these}$
|
Without loss of generality, assume something costs $100$ dollars. Then with each successive discount, it would cost $90$ dollars, then $72$ dollars. This amounts to a total of $28$ dollars off, so the single discount would be $\boxed{28}$
| 28
|
1,596
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24
| 1
|
The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$
|
$x + \sqrt{x-2} = 4$ Original Equation
$\sqrt{x-2} = 4 - x$ Subtract x from both sides
$x-2 = 16 - 8x + x^2$ Square both sides
$x^2 - 9x + 18 = 0$ Get all terms on one side
$(x-6)(x-3) = 0$ Factor
$x = \{6, 3\}$
If you put down A as your answer, it's wrong. You need to check for extraneous roots.
$6 + \sqrt{6 - 2} = 6 + \sqrt{4} = 6 + 2 = 8 \ne 4$
$3 + \sqrt{3-2} = 3 + \sqrt{1} = 3 + 1 = 4 \checkmark$
There is $\boxed{1}$
| 1
|
1,597
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_24
| 2
|
The equation $x + \sqrt{x-2} = 4$ has:
$\textbf{(A)}\ 2\text{ real roots }\qquad\textbf{(B)}\ 1\text{ real and}\ 1\text{ imaginary root}\qquad\textbf{(C)}\ 2\text{ imaginary roots}\qquad\textbf{(D)}\ \text{ no roots}\qquad\textbf{(E)}\ 1\text{ real root}$
|
We can create symmetry in the equation: \[x+\sqrt{x-2} = 4\] \[x-2+\sqrt{x-2} = 2.\] Let $y = \sqrt{x-2}$ , then we have \[y^2+y-2 = 0\] \[(y+2)(y-1) = 0\] The two roots are $\sqrt{x-2} = -2, 1$
Notice, that the first root is extraneous as the range for the square root function is always the non-negative numbers (remember, negative numbers in square roots give imaginary numbers - imaginary numbers in square roots don't give negative numbers); thus, the only real root for $x$ occurs for the second root; squaring both sides and solving for $x$ gives $x=3 \Rightarrow \boxed{1}$
| 1
|
1,598
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25
| 1
|
The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$
|
$\log_{5}\frac{(125)(625)}{25}$ can be simplified to $\log_{5}\ (125)(25)$ since $25^2 = 625$ $125 = 5^3$ and $5^2 = 25$ so $\log_{5}\ 5^5$ would be the simplest form. In $\log_{5}\ 5^5$ $5^x = 5^5$ . Therefore, $x = 5$ and the answer is $\boxed{5}$
| 5
|
1,599
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_25
| 2
|
The value of $\log_{5}\frac{(125)(625)}{25}$ is equal to:
$\textbf{(A)}\ 725\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 3125\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ \text{None of these}$
|
$\log_{5}\frac{(125)(625)}{25}$ can be also represented as $\log_{5}\frac{(5^3)(5^4)}{5^2}= \log_{5}\frac{(5^7)}{5^2}= \log_{5} 5^5$ which can be solved to get $\boxed{5}$
| 5
|
1,600
|
https://artofproblemsolving.com/wiki/index.php/1950_AHSME_Problems/Problem_26
| 1
|
If $\log_{10}{m}= b-\log_{10}{n}$ , then $m=$
$\textbf{(A)}\ \frac{b}{n}\qquad\textbf{(B)}\ bn\qquad\textbf{(C)}\ 10^{b}n\qquad\textbf{(D)}\ b-10^{n}\qquad\textbf{(E)}\ \frac{10^{b}}{n}$
|
We have $b=\log_{10}{10^b}$ . Substituting, we find $\log_{10}{m}= \log_{10}{10^b}-\log_{10}{n}$ . Using $\log{a}-\log{b}=\log{\dfrac{a}{b}}$ , the left side becomes $\log_{10}{\dfrac{10^b}{n}}$ . Because $\log_{10}{m}=\log_{10}{\dfrac{10^b}{n}}$ $m=\boxed{10}$
| 10
|
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