Free SSAT Upper Level Math Practice Test 1 - 2011 - Math Achievement

Translate from English into math. Let Bob's current age be \(x\), and let Jerry's current age be \(x + 5\). To find their ages 4 years ago, subtract 4 years from each current age. 4 years ago, Bob was \(x - 4\) and Jerry was \(x + 5 - 4 = x + 1\). The sum of Bob's and Jerry's ages 4 years ago was: \((x - 4) + (x + 1) = 2x - 3\).

The contestant answered a total of 5 questions correctly. Using our percent formula, \( ext{Percent} imes ext{Whole} = ext{Part} \), we have: \( 20% imes ext{total number of questions} = 5 \). Multiply both sides of the equation by \( frac{100}{20} \), and the total number of questions equals 25. Thus, statement I is incorrect.

For statement II, there were \( 25 - 15 = 10 \) questions remaining, and 1 of these 10 questions was answered correctly. So, he answered \( 10% \) of the remaining questions correctly, making statement II true.

Finally, statement III is true because 1 of the remaining 10 was answered correctly, so 9 of these 10 were not answered correctly. Therefore, the correct answer is (D).

Let \( C = A imes B \). Pick two consecutive numbers for \( A \) and \( B \), such as 2 and 3. Their product is \( 6 \), which is positive. However, if we select 1 and 0, the product is 0, which is neither positive nor negative.

Because the integers are consecutive, one of the integers must be even. Hence, the product of any two consecutive integers must be even. Therefore, the correct answer is (D).

The sweaters were sold for \( 100% - 40% \) of their original price. Using the percent formula, \( ext{Part} = ext{Percent} imes ext{Whole} \), we have \( 54 = 60% imes ext{old price} \). Converting \( 60]% \) to \( frac{60}{100} \) and solving gives an original price of \( $90 \) for the cotton sweater. Similarly, for the wool sweater, the original price was \( $120 \). Therefore, the difference in price before the discount is \( $120 - $90 = $30 \).

Each car carries \( 18 - frac{5}{9} = frac{157}{9} \) tons of freight. Multiplying this by 40 cars gives the total freight: \( frac{157}{9} imes 40 = 697 frac{7}{9} \) tons.

In one hour, the minute hand completes a full revolution of \( 360^circ \). In two hours, it revolves through two full circles, equating to \( 720^circ \).

A fraction is not more than \(frac{1}{3}\) if three times the numerator is not more than the denominator. Of the fractions listed, only \(frac{15}{46}\) has a numerator that is not more than \(frac{1}{3}\) of the denominator.

The figure is a square, so all four sides are equal in length. The perimeter is the sum of the lengths of the four sides. You could just multiply \(frac{2x+1}{3}\) by 4 to get the right answer.

The face of the cube is a square, 1 inch by 1 inch. Use the Pythagorean Theorem to find the length of the diagonal of the square: \(sqrt{2}\).

The average speed for the entire trip is the total distance (360 miles) divided by the total time (5 hours), which yields 72 mph.

This is a proportion problem. Set up the proportion as follows: \(frac{1 frac{1}{4}}{2 frac{3}{8}} = frac{4}{x}\). Solving the proportion yields the enlarged shorter dimension as \(7.6 \) inches.

To find the ratio of the circle to the area of the square, first find the area of each. The diameter of the circle equals the width of the square, which is 6 inches. The area of the circle is \(pi r^2\) and the area of the square is \(s^2\). The ratio is approximately \(frac{7}{9}\).

The difference is 0.80 inches, but the outside diameter consists of two thicknesses of metal (one on each side). Therefore, the thickness of the metal is \(0.80 div 2 = 0.40\) inches.

If 40% of the games were predicted accurately, 60% of the games were predicted inaccurately. Let x = games played, then 0.60x = 30, giving x = 50 games played. Therefore, 50 - 30 = 20 games were won. Thus, the sportswriter was accurate for 20 games.

Forty percent of the boys are from New York State, and 10% of them (0.10 of them) are from New York City. Therefore, 4% (0.40 x 0.10) of the boys in the camp are from New York City.

This can be set up as a proportion where x is the unknown number: \(55:x = 110:0.55\). Note that 5 is one half of 110, therefore the denominators must also have the same relationship. One half of 0.55 is \(0.275\).

The square root of 30 is less than the square root of 36 (which is 6) and greater than the square root of 25 (which is 5). Therefore, \(n\) is between 5 and 6.

To find the difference, you should move from the larger number to the smaller number.

To find how many sixths are in \(frac{3}{5}\), divide \(frac{3}{5}\) by \(frac{1}{6}\). This is equivalent to multiplying by the reciprocal, resulting in \(frac{3}{5} times 6 = frac{18}{5}\) or \(3frac{3}{5}\).

Four games averaging 34,800 people per game total to 139,200 attendance. The total for the first three games was 105,500. Thus, the fourth game attracted \(139,200 - 105,500 = 33,700\) people.