## SAT MAth Practice questions – all topics

**Advanced Math**Weightage: 35% Questions: 13-15- Equivalent expressions
- Nonlinear equations in one variable and systems of equations in two variables
- Nonlinear functions

## SAT MAth and English – full syllabus practice tests

**[Calc]**** ****Question*** *** Hard**

\(f(x))= 3^{(-2)(x+1)}\)

Which of the following equivalent forms of the given function *f* displays, as the base or the coefficient, the *y*-coordinate of the *y*-intercept of the graph of *y* = *f*(*x*) in the *xy*-plane?

A) \(f(x) ={( \frac{1}{3})}^{(2x+2)}\)

B) \(f(x) =\frac{1}{9}{( \frac{1}{3})}^{(2x)}\)

C) \(f(x) = 81^{(\frac{-1}{2}-\frac{1}{2})}\)

D) \(f(x)= 3^{(-2x-2)}\)

**▶️Answer/Explanation**

**B) \(f(x) =\frac{1}{9}{( \frac{1}{3})}^{(2x)}\)**

To find the \(y\)-intercept of the function \(f(x)\), we set \(x\) to \(0\) and solve for \(y\).

Given:

\[ f(x) = 3^{2(x+1)} \]

Let’s find \(f(0)\):

\[ f(0) = 3^{2(0+1)} = 3^2 = 9 \]

So, when \(x = 0\), \(y = 9\).

**[Calc]**** ****Question** ** Hard**

The interstate route from Los Angeles, California, to Jacksonville, Florida, cost about \(\$ 5\) billion to build and has a total distance of 2,500 miles. Each mile of the interstate route cost about \(\$ 2\) million to build. If the linear relationship between the distance \(x\), in thousands of miles, and the cost \(y\), in billions of dollars, is represented in the xy-plane, what is the \(y\)-intercept of the graph?

A) \((0,0)\)

B) \((0,200,000)\)

C) \((200,000,0)\)

D) \((200,000,200,00)\)

**▶️Answer/Explanation**

Ans:A

We need to express the relationship between distance \(x\) (in thousands of miles) and cost \(y\) (in billions of dollars) in the \(xy\)-plane and find the \(y\)-intercept of this graph.

The total distance in thousands of miles is:

\[

\frac{2500}{1000} = 2.5 \text{ thousands of miles}

\]

Given that each mile costs \$2 million, we convert this to billions:

\[

2 \text{ million dollars} = 0.002 \text{ billion dollars}

\]

The cost per thousand miles is:

\[

0.002 \text{ billion dollars} \times 1000 = 2 \text{ billion dollars per thousand miles}

\]

Thus, the linear relationship is:

\[

y = 2x

\]

At \(x = 0\) (the starting point), the cost \(y\) is also 0. Therefore, the \(y\)-intercept is:

\[

(0, 0)

\]

**[Calc]**** ****Question** **Hard**

The populations, in thousands, of Alaska and Hawaii from 1960 to 2016 can be modeled by the functions \(A\) and \(H\), where \(x\) is the number of years since January 1, 1960, and \(0 \leq \mathrm{x} \leq 55\)

Alaska: \(A(x)=221+9.78 x\)

Hawaii: \(H(x)=645+14.5 x\)

Based on the model, what is the predicted population of Alaska, on January 1, 1960?

A) 9.78

B) 221

C) 9,780

D) 221,000

**▶️Answer/Explanation**

Ans:B

The population of Alaska at \(x = 0\) (January 1, 1960) is given by the function:

\[

A(x) = 221 + 9.78x

\]

Substitute \(x = 0\):

\[

A(0) = 221 + 9.78(0) = 221

\]

The predicted population of Alaska on January 1, 1960, is:

\[

\boxed{221}

\]

**[Calc]**** ****Question** **Hard**

The populations, in thousands, of Alaska and Hawaii from 1960 to 2016 can be modeled by the functions \(A\) and \(H\), where \(x\) is the number of years since January 1, 1960, and \(0 \leq \mathrm{x} \leq 55\)

Alaska: \(A(x)=221+9.78 x\)

Hawaii: \(H(x)=645+14.5 x\)

Based on the model, in which year does the predicted population of Hawaii first exceed 900,000?

A) 1966

B) \(\quad 1967\)

C) 1976

D) \(\quad 1977\)

**▶️Answer/Explanation**

Ans:D

The function for Hawaii’s population is given by:

\[ H(x) = 645 + 14.5x \]

We need to solve for \(x\) when \(H(x) > 900\):

\[ 645 + 14.5x > 900 \]

Subtract 645 from both sides:

\[ 14.5x > 255 \]

Divide both sides by 14.5:

\[ x > \frac{255}{14.5} \]

\[ x > 17.5862 \]

Since \(x\) must be an integer (as it represents the number of years), we round up to the next whole number:

\[ x = 18 \]

Now, \(x = 18\) corresponds to 18 years after January 1, 1960:

\[ 1960 + 18 = 1978 \]

However, the correct options given are 1966, 1967, 1976, and 1977. We can check when the population first exceeds 900,000 by checking the values at \(x = 17\) and \(x = 18\).

For \(x = 17\):

\[ H(17) = 645 + 14.5 \times 17 \]

\[ H(17) = 645 + 246.5 \]

\[ H(17) = 891.5 \]

For \(x = 18\):

\[ H(18) = 645 + 14.5 \times 18 \]

\[ H(18) = 645 + 261 \]

\[ H(18) = 906 \]

So, the population of Hawaii first exceeds 900,000 in the year 1978. The closest year in the given options is 1977, which means there might be a typographical error in the choices. Based on the correct calculation, the predicted population first exceeds 900,000 in 1978, which is closest to the year 1977 given in the options.