2013 SSAT Quantitative Math Practice Questions

Call the number N. Write an inequality using the information given. Remember, "of" means multiply. 1/4XN<16. We need to isolate N, our unknown value. Multiplying both sides by the reciprocal of 1/4 which is 4 produces a result of N<16 x 4, and thus N < 64 . (A) is correct.

The perimeter of a square is given by \(P = 4s\), and the perimeter of a rectangle is given by \(P = 2(l + w)\). For the square:

\( P = 4 times 5 = 20 \). For the rectangle:

\( P = 2(8 + 6) = 28 \). The difference is \(28 - 20 = 8\).

Simplify the expression step by step. Group terms with the same denominators:

\( (frac{1}{5} - frac{1}{5}) + (frac{2}{7} - frac{1}{7} - frac{1}{7}) + (frac{3}{4} + frac{1}{4}) \) simplifies to:

\( 0 + 0 + 1 = 1 \).

To find the number of correct answers, multiply 72% by 50:

\( 72% = frac{72}{100} = 0.72 times 50 = 36 \). Mark answered 36 questions correctly.

Let the number be \(x\). The problem states that \(x - 5 = frac{1}{3}x\). Solving this:

\( x - frac{1}{3}x = 5 rightarrow frac{2}{3}x = 5 rightarrow x = frac{5 times 3}{2} = frac{15}{2} = 7.5 \).

To find the least common multiple (LCM) of 4 and 5, note that the LCM must be divisible by both numbers. The LCM of 4 and 5 is 20, so multiples of 20 will satisfy the condition. Among the given choices, 90 is divisible by both 4 and 5.

Set up a proportion:

\( frac{7}{4} = frac{35}{x} \). Solving for \(x\), we get:

\( 7x = 35 times 4 = 140 rightarrow x = 20 \).

There are 35 girls and 24 boys in the class, a total of 59 students. One-fifth of the girls, or \( frac{35}{5} = 7 \), and two-thirds of the boys, or \( frac{2}{3} times 24 = 16 \), participated. Thus, \(7 + 16 = 23\) students participated, and the fraction of the total class is \( frac{23}{59} \).

We are given that \(y @ z = y times z - 2 = 0\). Plugging in \(y = frac{1}{8}\), we have:

\( frac{1}{8} times z - 2 = 0 rightarrow frac{1}{8}z = 2 rightarrow z = 16 \).

Use the definition of \(y @ z = y times z - 2\) and substitute \(z = 3\) into the equation:

\( y times 3 - 2 = 10 \). Solving this gives:

\( y times 3 = 12 rightarrow y = 4 \).

To solve this, apply the given operation:

\( 3 @ 9 = 3 times 9 - 2 = 27 - 2 = 25 \). Therefore, \(3 @ 9 = 25\).

The average formula is:

\( text{Average} = frac{text{Sum of the terms}}{text{Number of terms}} \).

Adding the number of calls made on each day, we have:

\( 500 + 1,000 + 500 + 1,500 = 3,500 \). There are four days, so the average is:

\( frac{3,500}{4} = 875 \).

The remainder when 17 is divided by 4 is 1, since \( 17 - (4 times 4) = 1 \). Now, we need to find which number, when divided into 82, gives a remainder of 1. Since \( 9 times 9 = 81 \), dividing 82 by 9 also leaves a remainder of 1. Therefore, the answer is 9.

The pie chart shows that chocolate makes up 1/4 of the total cones sold. With 300 total cones sold, the number of chocolate cones is:

\( 1/4 times 300 = 75 \). Therefore, 75 chocolate cones were sold.

The temperature was initially 10 degrees. It then dropped 16 degrees. So, we subtract 16 from 10:

\( 10 - 16 = -6 \). Therefore, the temperature at midnight was -6 degrees, or 6 degrees below zero.

To solve \( frac{7}{8} - frac{6}{8} \), subtract the fractions to get:

\( frac{7 - 6}{8} = frac{1}{8} \). In decimal form, this is equal to 0.125.

The problem tells us that 10.75 out of 100 families live in rural areas, which can be interpreted as 10.75%. To find how many families that is in a population of 20 million, we calculate:

\( 10.75% times 20,000,000 = 2,150,000 \). Therefore, 2.15 million families live in rural areas.

Using the information given, isolate \(a\):

\( a = 3b - 3 + 7 = 3b + 4 \).

Thus, \(a = 3b + 4\). Next, add 5 to both sides of this equation:

\( a + 5 = 3b + 4 + 5 = 3b + 9 \).

The question asks which choice can be written as \( (J + 5) times 3 \). Since 3 is a factor of the expression, the answer must be a multiple of 3. The sum of the digits of 81 is 9, which is a multiple of 3. Therefore, (E) is correct.

The question asks for all the points where the book touches the table. Answer choice (A) is incorrect because it only includes the boundary of the points. Answer choice (E) shows all the points that touch the table, including those inside the rectangle. Therefore, (E) is correct.

Let \(BC = x\), then \(AB = 2x\), and since \(AB = CD\), we also have \(CD = 2x\). The total length of \(AD\) is:

\( AD = AB + BC + CD = 2x + x + 2x = 5x = 110 \).

Solving for \(x\), we get \(x = 22\). Therefore, the distance between points B and D is:

\( BD = BC + CD = x + 2x = 3x = 66 \).

The equation is \( (x - y) + 2 = 6 \). First, subtract 2 from both sides:

\( x - y = 4 \).

Adding \( y \) to both sides, we get \( x = y + 4 \). Since \(\) y < 3 [/latex], [latex] x < 7 [/latex]. Therefore, the only choice that is not less than 7 is 8, so [latex] x [/latex] cannot be 8.

To find how many times the cost of the house is compared to the land, we divide the cost of the house by the cost of the land:

\( frac{3,000,000}{60,000} = frac{300}{6} = 50 \).

Therefore, the house costs 50 times more than the land.

The sum of the three interior angles of any triangle is 180 degrees. In this case, angle A is 45 degrees, and angle B is 90 degrees (as it is a right triangle). So, the remaining angle C must also be 45 degrees. This means triangle ABC is a 45-45-90 triangle. In such triangles, the two legs are equal. Hence, side \(a\) is equal to the given side, which is 6.

The minimum number of points Team B could have scored is one more than Team A, or 40. Using the average formula, we have:

\( text{Average} = frac{text{Sum of terms}}{text{Number of terms}} \).

We are given that the sum is 40 points and the number of players is 8. Thus,

\( text{Average} = frac{40}{8} = 5 \).

The average score of the players on Team B must have been at least 5 points per player.