SSAT Upper Level Quantitative Reasoning - 2011 - Free SSAT Practice Test

This question isn't as hard as it seems: Remember, a cube has six faces. Since you're asked which shape can be folded into a cube with no overlapping flaps, the answer must contain exactly six faces. The only choice that does so is (B).

You know $20.05 is close to $20. Twenty-five percent of $20 would be $5.

Five will divide evenly into numbers that end in five or zero. You are asked to divide 63 by 5. The largest number less than 63 that 5 divides into evenly is 60. This means that 5 will divide into 63 with a remainder of 3.

This question is essentially an algebra question. Just isolate the \(x\) and solve.

The perimeter of a rectangle is equal to \(2(l + w)\), where \(l\) and \(w\) represent the length and width, respectively. The length of the rectangle is 9, so you need to find its width in order to solve. You're also told that the width of the rectangle is one-third of its length, so \(w = 3\). Plugging in the formula, the perimeter is equal to: \(2(9 + 3) = 2(12) = 24\).

Angles about a point add up to \(360^circ\), so you can write the following equation to solve for \(a\): \(45 + 75 + a + 45 + 75 + a = 360\). This simplifies to \(2a + 240 = 360\), leading to \(2a = 120\) and \(a = 60\).

The easiest way to solve is to compare each answer choice, looking for the largest digit in each place holder. The largest tenths digit, for example, is 7. Eliminate (A) since its tenths digit is 0. In the hundredths place, the largest digit is 9, so (B) can be eliminated since its hundredths digit is 8. (E) doesn't have a thousandths digit, so it is understood to be 0, which is less than the 2 that appears in the thousandths places in (C) and (D). (D) doesn't have a digit in the ten-thousandths place, so it is understood to be 0. It can be eliminated since it is less than the 3 in the ten-thousandths place. Thus, (C) is the largest.

Break this question down into parts, translating as you go. You're told that 6 added to 3 times a number \(N\) results in 48.

The statement indicates that adding 7 to \(N\) results in an odd number. Therefore, \(N\) must be an even number.

You are looking for the choice that best represents the area within a 7-foot radius around the center point.

First, calculate the total harvest: \(60 + 100 + 80 = 240\). Then, find the percentage of corn: \(frac{60}{240} times 100 = 25%\). Thus, the percent of the total harvest that is corn is \(25%\).

Multiples result when you multiply a number by an integer. For example, \(4 times 2 = 8\), so 8 is a multiple of 4.

First, convert all measurements to feet. A 3-foot, 2-inch board is \(3 + frac{2}{12} = frac{38}{12}\) feet. The comparison is then \(frac{38/12}{2} = frac{19}{12}\), which simplifies to approximately 1.58.

Use the distance formula: \(d = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\). Here, \(x_1 = -10, y_1 = -13, x_2 = -16, y_2 = -9\). The calculation becomes \(d = sqrt{(-16 + 10)^2 + (-9 + 13)^2} = sqrt{(-6)^2 + (4)^2} = sqrt{36 + 16} = sqrt{52} = 2sqrt{13}\).

A pentagon has five sides. To find the perimeter, multiply the length of each side by the number of sides: \(4 times 5 = 20\).

To break a number into its prime factors, start by finding factors of 60. \(60 = 2 times 30\), \(30 = 2 times 15\), and \(15 = 3 times 5\). Thus, the prime factorization is \((2)(2)(3)(5)\), which simplifies to \((2)(3)(5)(2)\).

First, calculate the amount Mike paid for the shares: \(120 times 10 = 1200\). Then find the amount Mike sold the shares for: \(40 times 10 = 400\). Finally, subtract: \(1200 - 400 = 800\).

Use the Pythagorean theorem: \(a^2 + b^2 = c^2\). Here, \(3^2 + b^2 = 5^2\). Thus, \(9 + b^2 = 25\), leading to \(b^2 = 16\) and \(b = 4\).

To find the area of a hexagon, divide it into 6 triangles. Calculate the area of one triangle and multiply by 6. The area of one triangle is \(frac{1}{2} times AB times OP = frac{1}{2} times 4 times 2sqrt{3} = 4sqrt{3}\), and thus the area of the hexagon is \(6 times 4sqrt{3} = 24sqrt{3}\).

To solve the absolute value equation, set up two equations: \(4a - 3 = 5\) and \(4a - 3 = -5\). Solving the first gives \(4a = 8 Rightarrow a = 2\), and solving the second gives \(4a = 2 Rightarrow a = -frac{1}{2}\). Therefore, the possible values for a are \(2\) or \(-frac{1}{2}\).

The sum of the first six grades is \(83 times 6 = 498\). To find the average grade of 83, divide the sum of the six grades by 6. The average with seven students is \(frac{498 + x}{7} = 85\). This gives \(498 + x = 85 times 7 = 595\); hence, \(x = 595 - 498 = 97\).

All factors of 6 are also factors of the number. The factors of 6 are: \(1, 2, 3,text{ and } 6\).

Since this is an isosceles triangle, the angles opposite the congruent sides are also congruent. The sum of the angles in a triangle equals 180°. Thus, \(65° + 65° + x° = 180°\). Solving gives \(x = 180° - 130° = 50°\).

Let \(p\) equal the price of the item. The equation is \(p times 0.20 = 8.00\). Solving for \(p\) gives \(p = frac{8.00}{0.20} = 40.00\).

On Monday, Jerry ate \( frac{1}{8} \) of the pie. After Monday, \( 1 - frac{1}{8} = frac{7}{8}\) of the pie was left. On Tuesday, she ate \( frac{1}{4}\) of the remaining pie: \( frac{1}{4} times frac{7}{8} = frac{7}{32}\). Thus, the total pie eaten is \( frac{1}{8} + frac{7}{32} = frac{4}{32} + frac{7}{32} = frac{11}{32}\).