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AMC8 2012 #13 © MAA

Jamar bought some pencils costing more than a penny each at the school bookstore and paid

textdollar 1.43

. Sharona bought some of the same pencils and paid

textdollar 1.87

. How many more pencils did Sharona buy than Jamar?

AMC8 2015 #16 © MAA

In a middle-school mentoring program, a number of the sixth graders are paired with a ninth-grade student as a buddy. No ninth grader is assigned more than one sixth-grade buddy. If

frac{1}{3}

of all the ninth graders are paired with

frac{2}{5}

of all the sixth graders, what fraction of the total number of sixth and ninth graders have a buddy?

Solution

Let the number of sixth graders be

s

, and the number of ninth graders be

n

. Thus,

frac{n}{3}=frac{2s}{5}

, which simplifies to

n=frac{6s}{5}

. Since we are trying to find the value of

frac{frac{n}{3}+frac{2s}{5}}{n+s}

, we can just substitute

frac{6s}{5}

for

n

into the equation. We then get a value of

frac{frac{frac{6s}{5}}{3}+frac{2s}{5}}{frac{6s}{5}+s} = frac{frac{6s+6s}{15}}{frac{11s}{5}} = frac{frac{4s}{5}}{frac{11s}{5}} = boxed{textbf{(B)}~frac{4}{11}}

AMC8 2012 #23 © MAA

An equilateral triangle and a regular hexagon have equal perimeters. If the triangle’s area is 4, what is the area of the hexagon?

Solution

Let the perimeter of the equilateral triangle be

3s

. The side length of the equilateral triangle would then be

s

and the sidelength of the hexagon would be

frac{s}{2}

. A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio

1 : 4

, since the sidelength of the small equilateral triangle is half the side length of the large one. Thus, the area of one of the small equilateral triangles is

1

. The area of the hexagon is then

1 times 6 = boxed{textbf{(C)} 6}

AMC8 2012 #3 © MAA

On February 13 ”The Oshkosh Northwester” listed the length of daylight as 10 hours and 24 minutes, the sunrise was $6:57textsc{am}$ , and the sunset as $8:15textsc{pm}$ . The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?

Solution

The problem wants us to find the time of sunset and gives us the length of daylight and time of sunrise. So all we have to do is add the length of daylight to the time of sunrise to obtain the answer. Convert 10 hours and 24 minutes into

10:24

in order to add easier. Adding, we find that the time of sunset is

6:57textsc{am} + 10:24 implies 17:21 implies  boxed{textbf{(B)} 5:21textsc{pm}}

AMC8 2013 #23 © MAA

Angle $ABC$ of $triangle ABC$ is a right angle. The sides of $triangle ABC$ are the diameters of semicircles as shown. The area of the semicircle on $overline{AB}$ equals $8pi$ , and the arc of the semicircle on $overline{AC}$ has length $8.5pi$ . What is the radius of the semicircle on $overline{BC}$ ?
[figure]

Solution

If the semicircle on

overline{AB}

were a full circle, the area would be

16pi

pi r^2=16 pi Rightarrow r^2=16 Rightarrow r=+4

, therefore the diameter of the first circle is

8

. The arc of the largest semicircle is

8.5 pi

, so if it were a full circle, the circumference would be

17 pi

. So the

text{diameter}=17

. By the Pythagorean theorem, the other side has length

15

, so the radius is

boxed{textbf{(B)} 7.5}

~Edited by Theraccoon to correct typos. ==Brief Explanation== SavannahSolver got a diameter of

17

because the given arc length of the semicircle was

8.5pi

. The arc length of a semicircle can be calculated using the formula

pi r

, where

r

is the radius. let’s use the full circumference formula for a circle, which is

2pi r

. Since the semicircle is half of a circle, its arc length is

pi r

, which was given as

8.5pi

. Solving for

r

, we get

r=8.5

. Therefore, the diameter, which is

2r

, is

2cdot8.5=17

Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver the answer is

boxed{textbf{(B)} 7.5}

. - TheNerdWhoIsNerdy. Minor edits by -Coin1

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