Rather than listing out all of the prime numbers up to 43, stay focused on the unique number, 2, the only even prime number. All of the choices are odd, and two odd numbers would yield an even sum, so you’ll only be able to eliminate answers by adding 2 to an odd number.Each of the incorrect answers, therefore, is the sum of 2 and the previous prime number:Choice (A) is 2 + 5; choice (B) is 2 + 17; choice (D) is 2 + 29; and choice (E) is 2 + 41.The answer is choice (C).

Try Plugging In on this one. If x = 4, then choice (A) is ¹⁄₄ or 0.25, choice (B) is ¹⁄₅ or 0.2, choice (C) is ⁴⁄₅ or 0.8, choice (D) is \(\frac{4}{\frac{1}{5}}=4\div \frac{1}{5}=4\times 5=20\), and choice (E) is \(\frac{4}{\frac{4}{5}}=4\div \frac{5}{4}=4\times \frac{5}{4}=5\). Choice (D) is the greatest.

It is easier to work with the factors of Tasha’s favorite number, rather than with the number itself. Write out the number as 3 × 3 × 17 × 17 and make a list of the divisors—or factors—in pairs.The pairs are: 1 and 3 × 3 × 17 × 17, 3 and 3 × 17 × 17, 17 and 3 × 3 × 17, 3 × 3 and 17 × 17, and 3 × 17 and 3 × 17.The final pair contains only one distinct factor, giving you a total of 9 factors.

The prime numbers less than 10 are 2, 3, 5, and 7—don’t forget, 1 is not prime.Their sum is 17, and 3 × 17 = 51.The only prime numbers between 20 and 30 are 23 and 29, and their sum is 52. Quantity B is greater.

Try rounding your values before you calculate.The expression can be estimated as

\(\sqrt{\frac{(100)\left ( \frac{1}{2} \right )^{2}}{\frac{1}{4}}}= \sqrt{\frac{100\left ( \frac{1}{4} \right )}{\frac{1}{4}}}=\sqrt{\frac{25}{\left ( \frac{1}{4} \right )}}=\sqrt{25\times\frac{4}{1}}=\sqrt{100}=10\)

Time to Plug In! If you make m = 2 and n = 3, then Quantity A becomes \(\frac{2}{2\times 3}=\frac{1}{3}\), and Quantity B becomes . Quantity B is bigger; eliminate choices (A) and (C). However, if you make and n = 2, then the situation is reversed: Quantity A will be , and Quantity B will be .Eliminate choice (B); the answer must be choice (D).

Plug in values for the variables, such as u = 2, v = 4, x = 6, y = 8, and z = 10. With these values, the expression equals ¹⁄₆.Try halving each of the values to find which one would change the value of the expression to ¹⁄₃. Halving v to 2 works; the answer is choice (B).

Solve this problem by brute force, but be systematic about it. Set Y has a finite number of elements, so list them out and start finding the products when those elements are multiplied by positive multiples of 5. Set Y = {3, 5, 7, 11, 13, 17, 19}, so multiplying by 5— the first positive multiple of 5—yields 15, 25, 35, 55, 65, 85, and 95; that’s 7 elements for set Z thus far. Multiplying by 10—the next positive integer multiple of 5—yields 3 more products, 30, 50, and 70. Multiplying by 15 yields two new products, 45 and 75; multiplying by 20 yields only one new product, 60.That’s a total of 13 elements for set Z so far.You already have 75 as a member of set Z, so multiplying by 25 yields no new products; multiplying by 30 yields the final new product, 90. Set Z thus consists of 14 elements: set Z = {15, 25, 30, 35, 45, 50, 55, 60, 65, 70, 75, 85, 90, 95}. If you got choice (E), you may have mistakenly included 2 as an element of set Y.

The two positive integers must have a product of 28, so find the factor pairs of 28: 1 and 28, 2 and 14, and 4 and 7. Only choice (B) gives a possible difference: 7 − 4 = 3.

When y = −1, Quantity A is −2 and Quantity B is −20.Eliminate choices (B) and (C). Plug in another value for y. When y = −100, Quantity A is −200 and Quantity B is −2,000. Quantity A is always greater.