SAT · Math · Algebra
Desigualdades lineales y sistemas de desigualdades, en abstracto o en contextos (presupuestos, límites de peso…).
Tienes 50 \$ y cada entrada cuesta 7 \$: $7n\le50$ → $n\le7{,}14$ → como máximo 7 entradas.
Intenta responder antes de abrir la solución. Las preguntas están en el idioma del examen.
Which of the following values of $x$ satisfies the inequality $5x-3>17$?
Respuesta correcta: D. $5$
Add 3 to both sides: $5x>20$. Divide by 5: $x>4$. Of the choices, only 5 is greater than 4.
4 makes the two sides equal ($5(4)-3=17$), and a strict inequality (>) does not include equality. 2 and 3 give 7 and 12, which are less than 17.
$$-2(x-4)\le 3x-7$$
What is the least possible value of $x$ that satisfies the given inequality?
Respuesta numérica (escribes el número).
Respuesta: 3
Distribute: $-2x+8\le 3x-7$. Add $2x$ and $7$ to both sides: $15\le 5x$. Divide by $5$: $3\le x$, so $x\ge 3$. The least possible value is $3$.
Check: at $x=3$, $-2(-1)=2$ and $3(3)-7=2$, so both sides are equal.
An event planner needs at least $30$ centerpieces for a banquet. Small centerpieces cost \$3 each, and large centerpieces cost \$8 each. The planner can spend at most \$200 on centerpieces. What is the maximum number of large centerpieces the planner can buy?
Respuesta numérica (escribes el número).
Respuesta: 22
To buy as many large centerpieces as possible, buy only as many small ones as needed to reach $30$: if $\ell$ is the number of large centerpieces, buy $30-\ell$ small ones. The cost must satisfy $3(30-\ell)+8\ell\le 200$, so $90+5\ell\le 200$, $5\ell\le 110$, and $\ell\le 22$.
Check: $22$ large and $8$ small cost $176+24=200$ dollars. With $23$ large and $7$ small the cost is $184+21=205$ dollars, which is too much. Ignoring the $30$-centerpiece requirement gives $\frac{200}{8}=25$, which is not allowed.