SSAT Upper Quantitative I practice test

Because exponents come first, $a^2=16$; then $2a^2=32$, and $32+(-3)=29$.

The trays hold $20\times 5=100$ sandwiches. After 12 are removed, $100-12=88$ remain.

B is correct because markers make up $100\mathrm{\%}-40\mathrm{\%}-35\mathrm{\%}=25\mathrm{\%}$ of the items, and $25\mathrm{\%}$ of 80 is 20; A subtracts the known percentage numbers directly from the item count; C treats the missing percentage as a count; D uses the pencils’ percentage; E counts brushes and pencils instead of markers.

The garden’s perimeter is $2(14+9)=46$ feet. At 3 per foot, the edging costs $46(3)=138$ dollars.

The first four scores total 48. Adding 22 gives 70; 70 divided by 5 is 14.

The ratio $a:b=3:4$ already has $b=4$ parts. Rewrite $b:c=2:5$ as $b:c=4:10$. Then $a:b:c=3:4:10$, so $a:c=3:10$.

E is correct because the equal red and blue counts are unspecified, so several green counts are possible; A assumes there are no green tiles; B assumes all three colors have equal counts; C assumes half the tiles are green; D assumes there are no red or blue tiles.

The two acute angles total $90^{\circ }$. The ratio $2:3$ has 5 equal parts, so each part is $18^{\circ }$. The smaller angle is $2\times 18^{\circ }=36^{\circ }$.

The probability of heads is ${\displaystyle \frac{1}{2}}$, and the probability of an even number is ${\displaystyle \frac{3}{6}}={\displaystyle \frac{1}{2}}$. Multiplying gives ${\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{4}}$.

Using $V={\displaystyle \frac{1}{3}}Bh$, we get $40={\displaystyle \frac{1}{3}}(24)h=8h$. Therefore, the height is 5 meters.

From $a-b=4$, $a=b+4$. Substituting gives $2(b+4)+b=17$, so $3b=9$ and $b=3$.

Factor 75 as $25\cdot 3$. Then $\sqrt{75}=\sqrt{25}\sqrt{3}=5\sqrt{3}$.

Corresponding angles are congruent, so $5x-30=3x+10$. This gives $2x=40$, and therefore $x=20$.

The numbers are $-0.75$, $-0.72$, $-0.80$, $-0.70$, and $-0.69$. Among negative numbers, the number closest to 0 is greatest, so $-0.69$ is greatest.

The difference between $70\mathrm{\%}$ and $40\mathrm{\%}$ is $30\mathrm{\%}$. Thus, $0.30n=21$, so $n=70$.

After a 20% markup the price is $25\cdot 1.2=30$. A 20% discount then gives $30\cdot 0.8=24$.

The quotient is greater than $7,800÷42$, which exceeds 180, and less than $8,200÷38$, which is below 220.

Equate squared distances: $4^2+(q-2{)}^2=2^2+(q-8{)}^2$. Expanding and simplifying gives $12q=48$, so $q=4$.

The equation gives ${\displaystyle \frac{p^2+q^2}{pq}}=5$, so $p^2+q^2=5pq$. Then $(p-q{)}^2=p^2-2pq+q^2=5pq-2pq=3pq$, so ${\displaystyle \frac{(p-q{)}^2}{pq}}=3$.

The average weekly increases for the consecutive intervals are 18, 16, 13, and 21 centimeters. The greatest rate is 21 centimeters per week from Week 7 to Week 8.

Let the original ticket counts be $5k$ and $8k$. The refund condition gives ${\displaystyle \frac{5k-10}{8k-2}}={\displaystyle \frac{1}{2}}$, so $10k-20=8k-2$ and $k=9$. The original total was $13k=117$.

There are $0.40(150)=60$ used pages. If these are $60\mathrm{\%}$ of the pages left, then $60\mathrm{/}0.60=100$ pages remain, so $150-100=50$ pages were removed.

A is correct because the product is negative for $-3<x<1$, while the distance condition requires $x\le -1$ or $x\ge 3$, leaving $-3<x\le -1$. B interprets at least 2 units as at most 2 units; C ignores the distance condition; D selects a positive product before applying the distance condition; E takes the union of the two solution sets instead of their intersection.

Because $M$ and $N$ are the two midpoints, $MN={\displaystyle \frac{JK+LP}{2}}$. Thus $19={\displaystyle \frac{(2x-3)+(x+8)}{2}}$, which gives $x=11$.

C is correct because the first condition gives $-4\le x\le 10$, while the second gives $x<1$ or $x>9$, leaving $-4\le x<1$ or $9<x\le 10$. A misses the second solution region beyond 9; B selects the region between the equality boundaries instead of the regions outside them; D ignores the factor of two and selects numbers closer to $-3$ than to 3; E ignores the restriction that $x$ must be at most 7 units from 3.