SSAT Upper Quantitative I practice test

If $x$ is the original price, then $x-7=18$. Adding 7 to both sides gives $x=25$.

D is correct because the distance is $2-(-5)=7$ units. A counts only the distance from zero to $Q$; B subtracts the absolute values of the coordinates; C counts only the integer points strictly between the endpoints; E counts both endpoints instead of the intervals between them.

First find $\dfrac23$ of 36, which is 24. Then subtract 5 to get 19.

The angles labeled $64°$ and $x°$ are vertical angles, so they have equal measures and $x=64$.

The coding total is $10+13=23$. The total number of students is $12+8+10+15+7+13=65$, so the requested fraction is $\dfrac{23}{65}$.

E is correct because the scoop’s capacity in tablespoons is not given; A assumes each scoop holds one teaspoon, or one-third of a tablespoon; B assumes each scoop holds one tablespoon; C assumes each scoop holds two tablespoons; D assumes each scoop holds four tablespoons.

Substitute 4 for $n$: $f(4)=2(4\text{ squared})-3=2(16)-3=29$.

The scale factor is $15/6=2.5$. The corresponding larger side is $8\times2.5=20$.

The relevant group is the $7+11=18$ portrait-session participants. Of these, 11 attended in the afternoon, so the probability is $\dfrac{11}{18}$.

Since $6^3=216$, each edge of the cube is 6 centimeters long. A cube has 12 edges, so the total edge length is $12\cdot6=72$ centimeters.

Let the smaller number be $s$. Then $s + (s + 2) = 14$, so $2s = 12$ and $s = 6$. The larger number is 8.

The even power gives $(-5)^2=25$, while the odd power gives $(-5)^3=-125$. Their sum is $25-125=-100$.

The square has side length 8. The inscribed circle has diameter 8, so its radius is 4 and its area is $\pi(4)^2=16\pi$.

There are $7\times18=126$ tiles, and $126-30=96$ remain. Each table receives $96\div6=16$ tiles.

$15\%$ of 80 is 12, so the sale price is $80-12=68$ dollars.

Subtracting 3 gives $\dfrac{x}{4} = 8$, so $x = 8 \times 4 = 32$. Dividing instead of multiplying (8 ÷ 4) gives the trap 2; adding before multiplying ($(11+3)\times 4$) gives 56.

Use $365 \approx 360$. Then $360 \times 24 \times 3{,}600$ is about 31,000,000, or $3 \times 10^7$.

Opposite sides of a parallelogram have the same displacement. Since $\overrightarrow{BC}=(3,-5)$, add $(3,-5)$ to $A(-4,1)$ to get $D=(-1,-4)$.

Their combined rate is $\dfrac{1}{6}+\dfrac{1}{12}=\dfrac{1}{4}$ crate per hour, so one crate takes 4 hours.

The bar for the first option is labeled 12, so 12 students chose it.

Let the original amounts of salt and water be $2k$ and $9k$. Then $\dfrac{2k}{9k-18}=\dfrac{1}{3}$, so $6k=9k-18$ and $k=6$. The original amount was $2k+9k=11k=66$ liters.

Let $x$ be the number of new members. Then $x=\dfrac14(36+x)$. This gives $4x=36+x$, so $3x=36$ and $x=12$.

The number of factors is $(a+1)(b+1)=12$. With positive $a$ and $b$, possible exponent pairs include $(1,5)$, $(2,3)$, $(3,2)$, and $(5,1)$. The corresponding values are 486, 108, 72, and 96, so the least is 72.

Since $\pi r^2=18\pi$, $r^2=18$. The square’s diagonal is $2r$, so $d^2=4r^2=72$. The area of a square is $\dfrac{d^2}{2}$, so the area is 36.

Since $31-3y$ must be even, $y$ must be odd. The positive pairs are $(14,1),(11,3),(8,5),(5,7),(2,9)$, with products 14,33,40,35,18. The maximum occurs at $(8,5)$, so $x+y=13$.