1989 AJHSME Problems

Problem 1

$(1+11+21+31+41)+(9+19+29+39+49)=$

$\text{(A)}\ 150 \qquad \text{(B)}\ 199 \qquad \text{(C)}\ 200 \qquad \text{(D)}\ 249 \qquad \text{(E)}\ 250$

Solution

Problem 2

$\frac{2}{10}+\frac{4}{100}+\frac{6}{1000} =$

$\text{(A)}\ .012 \qquad \text{(B)}\ .0246 \qquad \text{(C)}\ .12 \qquad \text{(D)}\ .246 \qquad \text{(E)}\ 246$

Solution

Problem 3

Which of the following numbers is the largest?

$\text{(A)}\ .99 \qquad \text{(B)}\ .9099 \qquad \text{(C)}\ .9 \qquad \text{(D)}\ .909 \qquad \text{(E)}\ .9009$

Solution

Problem 4

Estimate to determine which of the following numbers is closest to $\frac{401}{.205}$ .

$\text{(A)}\ .2 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$

Solution

Problem 5

$-15+9\times (6\div 3) =$

$\text{(A)}\ -48 \qquad \text{(B)}\ -12 \qquad \text{(C)}\ -3 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 12$

Solution

Problem 6

Solution

Problem 7

If the value of $20$ quarters and $10$ dimes equals the value of $10$ quarters and $n$ dimes, then $n=$

$\text{(A)}\ 10 \qquad \text{(B)}\ 20 \qquad \text{(C)}\ 30 \qquad \text{(D)}\ 35 \qquad \text{(E)}\ 45$

Solution

Problem 8

$(2\times 3\times 4)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right) =$

$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 9 \qquad \text{(D)}\ 24 \qquad \text{(E)}\ 26$

Solution

Problem 9

There are $2$ boys for every $3$ girls in Ms. Johnson's math class. If there are $30$ students in her class, what percent of them are boys?

$\text{(A)}\ 12\% \qquad \text{(B)}\ 20\% \qquad \text{(C)}\ 40\% \qquad \text{(D)}\ 60\% \qquad \text{(E)}\ 66\frac{2}{3}\%$

Solution

Problem 10

What is the number of degrees in the smaller angle between the hour hand and the minute hand on a clock that reads seven o'clock?

$\text{(A)}\ 50^\circ \qquad \text{(B)}\ 120^\circ \qquad \text{(C)}\ 135^\circ \qquad \text{(D)}\ 150^\circ \qquad \text{(E)}\ 165^\circ$

Solution

Problem 11

Solution

Problem 12

$\frac{1-\frac{1}{3}}{1-\frac{1}{2}} =$

$\text{(A)}\ \frac{1}{3} \qquad \text{(B)}\ \frac{2}{3} \qquad \text{(C)}\ \frac{3}{4} \qquad \text{(D)}\ \frac{3}{2} \qquad \text{(E)}\ \frac{4}{3}$

Solution

Problem 13

$\frac{9}{7\times 53} =$

$\text{(A)}\ \frac{.9}{.7\times 53} \qquad \text{(B)}\ \frac{.9}{.7\times .53} \qquad \text{(C)}\ \frac{.9}{.7\times 5.3} \qquad \text{(D)}\ \frac{.9}{7\times .53} \qquad \text{(E)}\ \frac{.09}{.07\times .53}$

Solution

Problem 14

Solution

Problem 15

Solution

Problem 16

In how many ways can $47$ be written as the sum of two primes?

$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ \text{more than 3}$

Solution

Problem 17

The number $\text{N}$ is between $9$ and $17$ . The average of $6$ , $10$ , and $\text{N}$ could be

$\text{(A)}\ 8 \qquad \text{(B)}\ 10 \qquad \text{(C)}\ 12 \qquad \text{(D)}\ 14 \qquad \text{(E)}\ 16$

Solution

Problem 18

Many calculators have a reciprocal key $\boxed{\frac{1}{x}}$ that replaces the current number displayed with its reciprocal. For example, if the display is $\boxed{00004}$ and the $\boxed{\frac{1}{x}}$ key is depressed, then the display becomes $\boxed{000.25}$ . If $\boxed{00032}$ is currently displayed, what is the fewest number of times you must depress the $\boxed{\frac{1}{x}}$ key so the display again reads $\boxed{00032}$ ?

$\text{(A)}\ 1 \qquad \text{(B)}\ 2 \qquad \text{(C)}\ 3 \qquad \text{(D)}\ 4 \qquad \text{(E)}\ 5$

Solution

Problem 19

Solution

Problem 20

Solution

Problem 21

Jack had a bag of $128$ apples. He sold $25\%$ of them to Jill. Next he sold $25\%$ of those remaining to June. Of those apples still in his bag, he gave the shiniest one to his teacher. How many apples did Jack have then?