1996 AJHSME Problems/Problem 5

Problem 5

The letters $P$, $Q$, $R$, $S$, and $T$ represent numbers located on the number line as shown.

$[asy] unitsize(36); draw((-4,0)--(4,0)); draw((-3.9,0.1)--(-4,0)--(-3.9,-0.1)); draw((3.9,0.1)--(4,0)--(3.9,-0.1)); for (int i = -3; i <= 3; ++i) { draw((i,-0.1)--(i,0)); } label("-3",(-3,-0.1),S); label("-2",(-2,-0.1),S); label("-1",(-1,-0.1),S); label("0",(0,-0.1),S); label("1",(1,-0.1),S); label("2",(2,-0.1),S); label("3",(3,-0.1),S); draw((-3.7,0.1)--(-3.6,0)--(-3.5,0.1)); draw((-3.6,0)--(-3.6,0.25)); label("P",(-3.6,0.25),N); draw((-1.3,0.1)--(-1.2,0)--(-1.1,0.1)); draw((-1.2,0)--(-1.2,0.25)); label("Q",(-1.2,0.25),N); draw((0.1,0.1)--(0.2,0)--(0.3,0.1)); draw((0.2,0)--(0.2,0.25)); label("R",(0.2,0.25),N); draw((0.8,0.1)--(0.9,0)--(1,0.1)); draw((0.9,0)--(0.9,0.25)); label("S",(0.9,0.25),N); draw((1.4,0.1)--(1.5,0)--(1.6,0.1)); draw((1.5,0)--(1.5,0.25)); label("T",(1.5,0.25),N); [/asy]$

Which of the following expressions represents a negative number?

$\text{(A)}\ P-Q \qquad \text{(B)}\ P\cdot Q \qquad \text{(C)}\ \dfrac{S}{Q}\cdot P \qquad \text{(D)}\ \dfrac{R}{P\cdot Q} \qquad \text{(E)}\ \dfrac{S+T}{R}$

Solution 1

First, note that $P < Q < 0 < R < S < T$. Thus, $P$ and $Q$ are negative, while $R$, $S$, and $T$ are positive.

Option $A$ is the difference of two numbers. If the first number is lower than the second number, the answer will be negative. In this case, $P < Q$, so $P - Q < 0$, and $P-Q$ is thus indeed a negative number. So option $\boxed{A}$ is correct.

Option $B$ is the product of two negatives, which is always positive.

Option $C$ starts with the quotient of a positive and a negative. This is negative. But this negative quotient is then multiplied by a negative number, which gives a positive number.

Option $D$ has the product of two negatives in the denominator. Thus, the denominator is positive. Additionally, the numerator is (barely) positive. A positive over a positive will be positive.

Option $E$ has the sum of two positives in the numerator. That will be positive. The denominator is also positive. Again, the quotient of two positives is positive.

Solution 2

Estimate the numbers, and calculate.

$P\approx -3.5$

$Q\approx -1.1$

$R\approx 0.1$

$S\approx 0.9$

$T\approx 1.5$

For option $A$, the value of $P - Q = -3.5 - (-1.1) = -3.5 + 1.1 = -2.4$. Thus $\boxed{A}$ is the right answer.

For option $B$, $PQ = -3.5 \cdot -1.1 = +3.85$

For option $C$, $\frac{S}{Q}\cdot P = \frac{0.9}{-1.1}\cdot -3.5 \approx -0.818 \cdot -3.5 \approx +2.836$

For option $D$, $\frac{R}{PQ} = \frac{0.1}{3.85} \approx +0.026$

For option $E$, $\frac{S+T}{R} = \frac{0.9 + 1.5}{0.1} = \frac{2.4}{0.1} = +24$