Target Test Prep GRE quant challenge

Solution:

We're looking for the number of musicians who play exactly three instruments.

$\Rightarrow$ Total # of Unique Elements = # in (Group A) + # in (Group B) + # in (Group C) – # in (Groups of Exactly Two) – 2[#in (Group of Exactly Three)] + # in (Neither)

$\Rightarrow$ Let T = # in (Group of Exactly Three)

$\Rightarrow$ 120 = 50 + 70 + 60 – 30 – 2(T) + 0

$\Rightarrow$ 120 = 150 – 2T

$\Rightarrow$ 2T = 30

$\Rightarrow$ T= 15

Thus, 15 people play exactly all three instruments. Notice that since each musician must play one or more of the three instruments, the number of people in the Neither region is zero.

Solution:

The sum of the exterior angles of any polygon is 360°. However, this statement is only true when we take only one exterior angle per vertex and add up only the measures of each of those angles. Here there are actually two exterior angles per vertex, a situation that will yield two sets of exterior angles. We can consider angles a, c, e, g, i, k as one set and angles b, d, f, h, j, and l as the other. Since each of these two sets of exterior angles adds up to 360°, the two sets together give us a total of 2 × 360° =  720°.

Solution:

We should recognize that we have subtraction of bases with exponents. This means before we can combine the equation's terms, we need to factor out common factors. However, to help see what we can factor out, we can rewrite the equation.

$\begin{array}{l}\Rightarrow 5^{\mathrm{x}}-5^{\mathrm{x}-1}=500\\ \Rightarrow 5^{\mathrm{x}}-{\left(5\right)}^{\mathrm{x}}{\left(5\right)}^{-1}=500\end{array}$

Now we can easily see to factor out the common term of 5^x. We now have:

$\begin{array}{l}\Rightarrow 5^{\mathrm{x}}-{\left(5\right)}^{\mathrm{x}}{\left(5\right)}^{-1}=500\\ \Rightarrow 5^{\mathrm{x}}\left(1-5^{-1}\right)=500\\ \Rightarrow 5^{\mathrm{x}}\left(1-\frac{1}{5}\right)=500\\ \Rightarrow 5^{\mathrm{x}}\times \frac{4}{5}=500\end{array}$

Our next step is to break down all the values into prime factors. This will make canceling out much easier.

$\begin{array}{l}\Rightarrow 5^{\mathrm{x}}\times \frac{4}{5}=500\to 5^{\mathrm{x}}\times \frac{2^2}{5}=5^3\times 2^2\\ \Rightarrow 5^{\mathrm{x}}=\frac{5^3\times 2^2\times 5}{2^2}\\ \Rightarrow 5^{\mathrm{x}}=5^4\to \mathrm{x}=4\end{array}$

Finally, we have to solve for (x – 1)², so (4 – 1)² = 3² = 9

Solution:

The first step is to prime factorize the number  ${50}^8\times 8^3\times {11}^2$. This becomes:

$\begin{array}{l}\Rightarrow {50}^8\times 8^3\times {11}^2\\ \Rightarrow {\left(5\times 5\times 2\right)}^8\times \left(2^9\right)\times {11}^2\\ \Rightarrow 5^{16}\times 2^8\times 2^9\times {11}^2\\ \Rightarrow 5^{16}\times 2^{17}\times {11}^2\end{array}$

The sixteen (5 × 2) pairs contribute a total of sixteen trailing zeros to the number. One 2 and two 11's remain. The product of (2 × 11 × 11) = 242, which is 3 digits. Thus, the number has (16 + 3) = 19 total digits.

Solution:

This is an if/then rate question. We're given a hypothetical “if” scenario, and we need to use this to determine the actual scenario. We are told that the distance traveled was 60 miles. Most importantly, we are told that the hypothetical speed was 2 mph faster than the actual speed. Since we do not have any values for the actual speed, we can express the actual speed as r and the faster speed as (r+2).

Now we have enough information to determine the faster and slower times:

$\begin{array}{l}\Rightarrow {\text{time}}_{\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{u}\mathrm{a}\mathrm{l}\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{ miles}}{\mathrm{r}\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{ miles}}{1}\times \frac{\text{1 }\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\mathrm{r}\text{ miles}}=\frac{60}{\mathrm{r}}\text{ hours}\\ \Rightarrow {\text{time}}_{\text{faster }\mathrm{s}\mathrm{p}\mathrm{e}\mathrm{e}\mathrm{d}}=\frac{\text{distance}}{\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}}=\frac{60\text{ miles}}{\left(\mathrm{r}+2\right)\frac{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{e}\mathrm{s}}{\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}}=\frac{60\text{ miles}}{1}\times \frac{\text{1 }\mathrm{h}\mathrm{o}\mathrm{u}\mathrm{r}}{\left(\mathrm{r}+2\right)\text{ miles}}=\frac{60}{\left(\mathrm{r}+2\right)}\text{ hours}\end{array}$

We can now set up an equation using the travel time. We know that if Thomas had ridden at the faster speed, he would have arrived 1 hour earlier. Hence:

$\begin{array}{l}\text{⇒ Faster Time}+\text{1 Hour}=\text{ Slower Time}\\ \text{⇒}\frac{60}{\mathrm{r}+2}+1=\frac{60}{\mathrm{r}}\\ \text{⇒}\mathrm{r}(\mathrm{r}+2)(\frac{60}{\mathrm{r}+2}+1)=\frac{60}{\mathrm{r}}\\ \text{⇒}60\mathrm{r}+\mathrm{r}(\mathrm{r}+2)=60(\mathrm{r}+2)\\ \text{⇒}60\mathrm{r}+{\mathrm{r}}^2+2\mathrm{r}=60\mathrm{r}+120\\ \text{⇒}{\mathrm{r}}^2+2\mathrm{r}-120=0\\ \text{⇒}(\mathrm{r}+12)(\mathrm{r}-10)=0\\ \text{⇒}\mathrm{r}=-12,\text{ r}=10\end{array}$

Since we cannot have a negative rate, Thomas's actual rate was 10 mph.

(Note: If you thought that factoring this quadratic was difficult, keep in mind that once you have r² + 2r - 120 = 0, you can look at the answer choices to help you determine r. Notice that 10 and 12 are 2 units apart and also multiply to 120. Thus, we can quickly see that r² + 2r - 120 = 0= 0 will factor down to (r + 12)(r - 10). This is a useful method to use anytime you have a tricky quadratic in a multiple choice question.)

Solution:

We can assume that there are 100 guests at the dinner party since no specific number of guests is defined. This assumption results in having 40 people wear both jackets and ties. The second sentence takes some care. Since we don't know the number of guests who wore jackets, we can let the variable J represent that number. This means that the guests who did wear a jacket but not a tie can be represented as 0.50J.

At this point, we've produced an equation in the matrix. We know that

$\begin{array}{l}\Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->40+0.50\mathrm{J}=\mathrm{J}\\ \Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->40=0.50\mathrm{J}\\ \Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->400=5\mathrm{J}\\ \Rightarrow \; <!--Rightarrow,\; Implies,\; rArr,\; DoubleRightArrow-->\mathrm{J}=80\end{array}$

Thus, 80 guests wore jackets, which means that 80 percent of the guests wore jackets.