Difference between revisions of "2019 AMC 8 Problems"

Revision as of 01:21, 20 November 2019

Problem 1

Ike and Mike go into a sandwich shop with a total of $$30.00$ to spend. Sandwiches cost $$4.50$ each and soft drinks cost $$1.00$ each. Ike and Mike plan to buy as many sandwiches as they can and use the remaining money to buy soft drinks. Counting both soft drinks and sandwiches, how many items will they buy?

$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad\textbf{(E) }10$

Solution

Problem 2

$[asy] draw((0,0)--(3,0)); draw((0,0)--(0,2)); draw((0,2)--(3,2)); draw((3,2)--(3,0)); dot((0,0)); dot((0,2)); dot((3,0)); dot((3,2)); draw((2,0)--(2,2)); draw((0,1)--(2,1)); label("A",(0,0),S); label("B",(3,0),S); label("C",(3,2),N); label("D",(0,2),N); [/asy]$ Three identical rectangles are put together to form rectangle , as shown in the figure below. Given that the length of the shorter side of each of the smaller rectangles is $5$ feet, what is the area in square feet of rectangle $ABCD$ ? (A) 45 (B) 75 (C) 100 (D) 125 (E) 150

$\textbf{(A) }45\qquad\textbf{(B) }75\qquad\textbf{(C) }100\qquad\textbf{(D) }125\qquad\textbf{(E) }150$

Solution

Problem 3

3. Which of the following is the correct order of the fractions $\frac{15}{11}$ , $\frac{19}{15}$ , and $\frac{17}{13}$ , from least to greatest?

$\textbf{(A) }\frac{15}{11} < \frac{17}{13} < \frac{19}{15}\qquad\textbf{(B) }\frac{15}{11} < \frac{19}{15} < \frac{17}{13}\qquad\textbf{(C) }\frac{17}{13} < \frac{19}{15} < \frac{15}{11}\qquad\textbf{(D) }\frac{19}{15} < \frac{15}{11} < \frac{17}{13}\qquad\textbf{(E) }\frac{19}{15} < \frac{17}{13} < \frac{15}{11}$

Solution

Problem 4

Quadrilateral $ABCD$ is a rhombus with perimeter $52$ meters. The length of diagonal $\overline{AC}$ is $24$ meters. What is the area in square meters of rhombus $ABCD$ ?

$[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); [/asy]$

Solution

$[asy] draw((-13,0)--(0,5)); draw((0,5)--(13,0)); draw((13,0)--(0,-5)); draw((0,-5)--(-13,0)); draw((0,0)--(13,0)); draw((0,0)--(0,5)); draw((0,0)--(-13,0)); draw((0,0)--(0,-5)); dot((-13,0)); dot((0,5)); dot((13,0)); dot((0,-5)); label("A",(-13,0),W); label("B",(0,5),N); label("C",(13,0),E); label("D",(0,-5),S); label("E",(0,0),SW); [/asy]$

Because it is a rhombus all sides are equal. Implies all sides are 13. In a rhombus diagonals are perpendicular and bisect each other. Which means $\overline{AE}$ = $12$ = $\overline{EC}$ .

Consider one of the right triangles.

$[asy] draw((-13,0)--(0,5)); draw((0,0)--(-13,0)); draw((0,0)--(0,5)); dot((-13,0)); dot((0,5)); label("A",(-13,0),W); label("B",(0,5),N); label("E",(0,0),SE); [/asy]$

$\overline{AB}$ = $13$ . $\overline{AE}$ = $12$ . Which means $\overline{BE}$ = $5$ .

Thus the values of the two diagonals are $\overline{AC}$ = $24$ and $\overline{BD}$ = $10$ . Which means area = $\frac{d_1*d_2}{2}$ = $\frac{24*10}{2}$ = $120$

@@ Line 55: / Line 55: @@
 label("D",(0,-5),S);
 </asy>
+[[2019 AMC 8 Problems/Problem 4|Solution]]
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Difference between revisions of "2019 AMC 8 Problems"

Revision as of 01:21, 20 November 2019

Contents

Problem 1

Problem 2

Problem 3

Problem 4

Solution

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25