Found problems: 4
2023 AMC 12/AHSME, 2
The weight of $\frac 13$ of a large pizza together with $3 \frac 12$ cups of orange slices is the same as the weight of $\frac 34$ of a large pizza together with $\frac 12$ cup of orange slices. A cup of orange slices weighs $\frac 14$ of a pound. What is the weight, in pounds, of a large pizza?
$\textbf{(A)}~1\frac45\qquad\textbf{(B)}~2\qquad\textbf{(C)}~2\frac25\qquad\textbf{(D)}~3\qquad\textbf{(E)}~3\frac35$
2008 ITest, 2
One day while Tony plays in the back yard of the Kubik's home, he wonders about the width of the back yard, which is in the shape of a rectangle. A row of trees spans the width of the back of the yard by the fence, and Tony realizes that all the trees have almost exactly the same diameter, and the trees look equally spaced. Tony fetches a tape measure from the garage and measures a distance of almost exactly $12$ feet between a consecutive pair of trees. Tony realizes the need to include the width of the trees in his measurements. Unsure as to how to do this, he measures the distance between the centers of the trees, which comes out to be around $15$ feet. He then measures $2$ feet to either side of the first and last trees in the row before the ends of the yard. Tony uses these measurements to estimate the width of the yard. If there are six trees in the row of trees, what is Tony's estimate in feet?
[asy]
size(400);
defaultpen(linewidth(0.8));
draw((0,-3)--(0,3));
int d=8;
for(int i=0;i<=5;i=i+1)
{
draw(circle(7/2+d*i,3/2));
}
draw((5*d+7,-3)--(5*d+7,3));
draw((0,0)--(2,0),Arrows(size=7));
draw((5,0)--(2+d,0),Arrows(size=7));
draw((7/2+d,0)--(7/2+2*d,0),Arrows(size=7));
label("$2$",(1,0),S);
label("$12$",((7+d)/2,0),S);
label("$15$",((7+3*d)/2,0),S);
[/asy]
2023 AMC 10, 2
The weight of $\frac 13$ of a large pizza together with $3 \frac 12$ cups of orange slices is the same as the weight of $\frac 34$ of a large pizza together with $\frac 12$ cup of orange slices. A cup of orange slices weighs $\frac 14$ of a pound. What is the weight, in pounds, of a large pizza?
$\textbf{(A)}~1\frac45\qquad\textbf{(B)}~2\qquad\textbf{(C)}~2\frac25\qquad\textbf{(D)}~3\qquad\textbf{(E)}~3\frac35$
1959 AMC 12/AHSME, 9
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 $