GRE Diagnostic Test – Math

AC and BC are congruent, making C the midpoint of AB. AC is 7, so BC is also 7, making AB equal to 14. However, it is not given that B is the midpoint of AD, just that it lies between A and D. Therefore, AB and BD are not necessarily congruent. There is no sufficient information to determine the length of CD.

In the form y = mx + b, the slope is given by m and the y-intercept is given by b. For the equation y = x – 5, the y-intercept is -5. Therefore, the line crosses the y-axis at (0, -5).

To have an average of 93 after six tests, Eric’s total score would need to be 93 × 6 = 558. He already has a total score of 96 + 93 + 86 + 100 + 94 = 469. So, he needs 558 – 469 points, which is 89.

13^2 = 169 and 12^2 = 144, so 169 – 144 = 25.

The graph of 4 < x < 7 shows an open interval, indicating that x is greater than 4 and less than 7. The graph would have open circles at both 4 and 7 to show that these points are not included in the solution set.

If both dimensions of a rectangle are doubled, the new area is 4 times the original area, representing a 300% increase. For example, a 1″ by 2″ rectangle has an area of 2 square inches. Doubling both dimensions makes it 2″ by 4″, giving an area of 8 square inches. The increase from 2 to 8 is 300%.

None of the statements provided is always true. If necessary, try each of the answers to see that all are false in some cases. For example, (C) is untrue for the number 1 because the square of 1 equals 1.

The cost per pencil is \(c/p\) cents. Therefore, the cost for n pencils is \(n \times (c/p) = (cn)/p\) cents.

A floor measuring 12′ x 12′ has an area of 144 square feet. Using 1′ x 1′ tiles, 144 tiles are needed. For 9″ x 9″ tiles (which are 0.75′ x 0.75′), each tile covers 0.5625 square feet. Thus, to cover 144 square feet, we need 256 tiles. The difference is 256 – 144 = 112 more tiles.

The average price per pound of cake is \(\frac{25 + 30 + 35}{3} = 30\) cents. Therefore, the total number of pounds sold is \(\frac{18}{0.30} = 60\) pounds.

Solve the equation for \(x\): \(3x – 2 = 13\) gives \(3x = 15\), so \(x = 5\). Then, \(6x + 20 = 6(5) + 20 = 50\).

Use the formula \(D = R \times T\) to find the actual travel time: \(T = \frac{290}{50} = 5.8\) hours, or 5 hours 48 minutes. The scheduled time was 5 hours 43 minutes, so the train arrived about 5 minutes late.

The area of a circle is proportional to the square of its diameter. To increase the area by a factor of 9, the diameter must be increased by a factor of \(\sqrt{9} = 3\). Therefore, the new diameter is \(1 \times 3 = 3\) feet.

The volume of a cube is proportional to the cube of its side length. If the side length of the second cube is 4 times that of the first, its volume is \(4^3 = 64\) times as large.

As \(x\) increases, \(y\) decreases because they are inversely proportional. For example, \(\frac{1}{2} > \frac{1}{3} > \frac{1}{4}\).

Substitute \(t = 4\) into the formula \(S = 16t^2\). \(S = 16 \times (4^2) = 16 \times 16 = 256\) feet.

Each board is \(3 \text{feet} 9 \text{inches} = 3.75 \text{feet}\). For four boards, he needs \(4 \times 3.75 = 15\) feet of wood.

The circumference of a circle is given by \(C = 2 \pi r\). Therefore, to find the radius, divide the circumference by \(2 \pi\). In this case, \(r = \frac{30}{2 \pi}\).

A cubic foot contains \(12 \times 12 \times 12 = 1,728\) cubic inches. If each cubic inch weighs 1 pound, the total weight is \(1 \times 1,728 = 1,728\) pounds.

To compare fractions, we use cross-multiplication. First compare \( \frac{2}{3} \) and \( \frac{7}{5} \): the cross-products give \( 2 \times 5 = 10 \) and \( 7 \times 3 = 21 \), so \( \frac{7}{5} > \frac{2}{3} \). Repeat this process for the other fractions to arrange them in order.

First, solve for \( x \) by subtracting 5 from both sides: \( 9x = 27 \), so \( x = 3 \). Then substitute into the second expression: \( 18x + 5 = 18(3) + 5 = 54 + 5 = 59 \).

We find the least common multiple (LCM) by prime factorizing each number. The prime factorizations are: \( 28 = 2^2 \times 7 \), \( 24 = 2^3 \times 3 \), and \( 32 = 2^5 \). The LCM is the product of the highest powers of all prime factors: \( 2^5 \times 3 \times 7 = 672 \).

The area of a circle is \( A = \pi r^2 \) and the circumference is \( C = 2 \pi r \). Given that the area and circumference are equal, we have \( \pi r^2 = 2 \pi r \). Solving for \( r \), we find \( r = 2 \), so the diameter is \( 2r = 4 \) inches.

An area 1 foot long by 1 1/4 feet wide is 12 inches \times 15 inches, or 180 square inches. Each block covers an area of 6 square inches (since 1 \times 2 = 6). Therefore, the number of blocks needed is 180 \div 6 = 30 blocks. The height of each block is irrelevant to the solution of the problem.