150 SAT Subject Test Math Level Practice Questions - Free Printable

Let's solve each problem step by step.

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Problem 1:

A game has 2 spinners. Spinner #1 has a probability of landing red of 2/3. And, spinner #2 has a probability of landing red of 1/5. What is the probability spinner #1 lands red AND spinner #2 does NOT land red?

We are asked for the probability that:
- Spinner #1 lands red: $ P_1 = \frac{2}{3} $
- Spinner #2 does not land red: $ P_2^c = 1 - \frac{1}{5} = \frac{4}{5} $

Since the spinners are independent:
$$
P(\text{#1 red AND #2 not red}) = \frac{2}{3} \times \frac{4}{5} = \frac{8}{15}
$$

✔ Answer: b) 8/15

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Problem 2:

For some positive real number 'b', the first 3 terms in a geometric sequence are:
$ b - 1, \quad b + 4, \quad 3b + 2 $

In a geometric sequence, the ratio between consecutive terms is constant:
$$
\frac{b+4}{b-1} = \frac{3b+2}{b+4}
$$

Cross-multiply:
$$
(b+4)^2 = (b-1)(3b+2)
$$

Expand both sides:

Left:
$$
(b+4)^2 = b^2 + 8b + 16
$$

Right:
$$
(b-1)(3b+2) = 3b^2 + 2b - 3b - 2 = 3b^2 - b - 2
$$

Set equal:
$$
b^2 + 8b + 16 = 3b^2 - b - 2
$$

Bring all terms to one side:
$$
0 = 2b^2 - 9b - 18
$$

Solve quadratic:
$$
2b^2 - 9b - 18 = 0
$$

Use quadratic formula:
$$
b = \frac{9 \pm \sqrt{(-9)^2 - 4(2)(-18)}}{2(2)} = \frac{9 \pm \sqrt{81 + 144}}{4} = \frac{9 \pm \sqrt{225}}{4} = \frac{9 \pm 15}{4}
$$

So:
- $ b = \frac{24}{4} = 6 $
- $ b = \frac{-6}{4} = -1.5 $ → discard (since $ b > 0 $)

So $ b = 6 $

Now compute the first three terms:
- $ b - 1 = 5 $
- $ b + 4 = 10 $
- $ 3b + 2 = 20 $

Common ratio: $ r = \frac{10}{5} = 2 $

Fourth term: $ 20 \times 2 = 40 $

✔ Answer: e) 40

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Problem 3:

Which equation best models the following data?

| x | -6.7 | -1.3 | 3.2 | 8.8 |
|-------|------|------|------|------|
| y | 1.30 | 3.47 | 7.89 | 21.89 |

Try each option by plugging in values.

Let’s test option a): $ y = 1.2(4.4)^x $

Try $ x = -6.7 $:
$$
y = 1.2 \cdot (4.4)^{-6.7} \approx 1.2 / (4.4^{6.7})
$$
$ 4.4^6 \approx 7250 $, $ 4.4^{6.7} \approx 10^{\log_{10}(4.4)\cdot6.7} \approx 10^{0.643}\cdot6.7 \approx 10^{4.3} \approx 20000 $
So $ y \approx 1.2 / 20000 \approx 0.00006 $ — way too small.

Try option b): $ y = 4.4(1.2)^x $

At $ x = -6.7 $:
$$
y = 4.4 \cdot (1.2)^{-6.7} \approx 4.4 / (1.2^{6.7})
$$
$ 1.2^6 \approx 2.986 $, $ 1.2^{6.7} \approx 3.2 $, so $ y \approx 4.4 / 3.2 \approx 1.375 $ — close to 1.30

Now at $ x = -1.3 $:
$$
y = 4.4 \cdot (1.2)^{-1.3} \approx 4.4 / (1.2^{1.3}) \approx 4.4 / 1.25 \approx 3.52 $ — very close to 3.47

At $ x = 3.2 $:
$ y = 4.4 \cdot (1.2)^{3.2} \approx 4.4 \cdot 1.8 $ ≈ 7.92 — close to 7.89

At $ x = 8.8 $:
$ (1.2)^{8.8} \approx ? $

$ \log_{10}(1.2) \approx 0.07918 $, so $ 8.8 \cdot 0.07918 \approx 0.7 $, so $ 10^{0.7} \approx 5.01 $

So $ y \approx 4.4 \cdot 5.01 \approx 22.04 $ — close to 21.89

Perfect fit.

Now check option c): $ y = -1.2(4.4)^x $ → negative → but y is always positive → discard

d) $ y = -4.4(1.2)^x $ → negative → discard

e) $ y = 1.2x^{4.4} $ → this is a power function, not exponential. Try $ x = -6.7 $: $ (-6.7)^{4.4} $ → complex or undefined for non-integer exponents with negative base → invalid

So only b) makes sense.

✔ Answer: b) $ y = 4.4(1.2)^x $

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Problem 4:

Square ABCD divided into 4 rectangles. Area of ABCD = 100. Find area of MOC P?

From diagram:
- ARMS: area = $ x^2 $
- RBOM: area = $ 4x $
- SMDP: area = $ 4x $
- MOC P: ??? Let's call it $ A $

But total area = 100

Also, from labeling:
- ARMS = $ x^2 $
- RBOM = $ 4x $
- SMDP = $ 4x $
- MOC P = ?

Wait — notice that:
- ARMS and RBOM share the same height (from A to B), and are adjacent horizontally.
- Also, ARMS has area $ x^2 $, RBOM has area $ 4x $

Let’s suppose:
- Let horizontal segment be split into two parts: let AR = a, RB = b
- Vertical: AS = c, SD = d

But better idea: since ARMS has area $ x^2 $, and RBOM has area $ 4x $, and they are adjacent vertically?

Wait — from figure:
- ARMS: top-left rectangle, area $ x^2 $
- RBOM: top-right, area $ 4x $
- SMDP: bottom-left, area $ 4x $
- MOC P: bottom-right — unknown

So:
Total area = $ x^2 + 4x + 4x + A = x^2 + 8x + A = 100 $

But we need another relation.

Notice: ARMS and SMDP are vertically stacked on left side.

ARMS: area $ x^2 $, SMDP: area $ 4x $

They share the same width (AS), say width = w

Then:
- Height of ARMS = h₁ → $ w \cdot h₁ = x^2 $
- Height of SMDP = h₂ → $ w \cdot h₂ = 4x $

But total height = h₁ + h₂

Similarly, right side:
- RBOM: area $ 4x $, width = w₂, height = h₁ → $ w₂ \cdot h₁ = 4x $
- MOC P: area = ?, width = w₂, height = h₂ → $ w₂ \cdot h₂ = ? $

But also, top row:
- ARMS: width = w, height = h₁ → $ w h₁ = x^2 $
- RBOM: width = w₂, height = h₁ → $ w₂ h₁ = 4x $

So divide:
$$
\frac{w₂ h₁}{w h₁} = \frac{4x}{x^2} \Rightarrow \frac{w₂}{w} = \frac{4}{x}
\Rightarrow w₂ = \frac{4}{x} w
$$

Now look at left column:
- ARMS: $ w h₁ = x^2 $
- SMDP: $ w h₂ = 4x $

So $ h₁ = \frac{x^2}{w}, h₂ = \frac{4x}{w} $

Now bottom right rectangle MOC P has:
- width = w₂ = $ \frac{4}{x} w $
- height = h₂ = $ \frac{4x}{w} $

So area:
$$
A = w₂ \cdot h₂ = \left( \frac{4}{x} w \right) \cdot \left( \frac{4x}{w} \right) = \frac{4}{x} \cdot \frac{4x}{1} = 16
$$

✔ So area of MOC P is 16

Answer: a) 16

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Problem 5:

$ \sin(\tan^{-1} 3) = ? $

Let $ \theta = \tan^{-1} 3 $. Then $ \tan \theta = 3 $

We want $ \sin \theta $

Draw a right triangle:
- Opposite = 3
- Adjacent = 1
- Hypotenuse = $ \sqrt{3^2 + 1^2} = \sqrt{10} $

So $ \sin \theta = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10} $

✔ Answer: b) $ \frac{3\sqrt{10}}{10} $

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✔ Final Answers:

1) b) 8/15
2) e) 40
3) b) $ y = 4.4(1.2)^x $
4) a) 16
5) b) $ \frac{3\sqrt{10}}{10} $