OMPT drill — Algebra

Exponents and roots

Power rules, fractional exponents, and rationalising denominators. The single most-reused skill on the whole test. Get it automatic before touching anything else. Work through all six below before moving on; algebra slips compound into every other strand.

Tested inOMPT-A OMPT-B OMPT-C OMPT-D OMPT-E OMPT-F OMPT-G

Lesson

Every power rule comes from one idea: an exponent counts repeated multiplication. $a^{3} \cdot a^{2}$ is five $a$ s multiplied together, so $a^{3} \cdot a^{2} = a^{5}$ : you add exponents when multiplying like bases. From that one seed the rest grows: dividing subtracts exponents, a power of a power multiplies them, $a^{0} = 1$ and $a^{- n} = \frac{1}{a ^{n}}$ are what keep the pattern consistent, and $a^{1/ n}$ has to mean $n a$ because raising it to the $n$ must give $a$ back.

I put this topic first in every study plan I write, whatever the variant, because it is the most reused skill on the test. Logarithms are these rules read backwards. Differentiation of $x$ starts by rewriting it as $x^{1/2}$ . Exponential equations collapse the moment both sides share a base. If the rules here are automatic, four other chapters get easier at once; if they are shaky, every one of those chapters will bill you for it separately.

Problem 1Numeric answer

Simplify and evaluate: $\frac{2 ^{3} \cdot 2 ^{5}}{2 ^{6}}$ .

Show the worked solution

Answer: 4

Multiplying powers with the same base means adding exponents: $2^{3} \cdot 2^{5} = 2^{8}$ . Dividing subtracts them: $2^{8 - 6} = 2^{2} = 4$ . I never compute the big numbers first — exponent rules do the work.

Problem 2Multiple choice

Simplify $(3 a^{2} b)^{3}$ .

A $27 a^{6} b^{3}$
B $9 a^{6} b^{3}$
C $27 a^{5} b^{3}$
D $3 a^{6} b^{3}$

Show the worked solution

Answer: A — $27 a^{6} b^{3}$

The cube applies to every factor inside the brackets: $3^{3} = 27$ , $(a^{2})^{3} = a^{6}$ (a power of a power multiplies exponents), and $b^{3}$ . Together: $27 a^{6} b^{3}$ .

Problem 3Numeric answer

Solve for $m$ : $5^{2 m - 1} = 125$ .

Show the worked solution

Answer: 2

I write both sides with base 5: $125 = 5^{3}$ , so $5^{2 m - 1} = 5^{3}$ . Equal bases mean equal exponents: $2 m - 1 = 3$ , so $m = 2$ . Checking: $5^{4 - 1} = 5^{3} = 125$ .

Problem 4Multiple choice

Simplify $50 + 18$ .

A $82$
B $68$
C $152$
D $417$

Show the worked solution

Answer: A — $82$

I pull out the largest square factors: $50 = 25 \cdot 2 = 52$ and $18 = 9 \cdot 2 = 32$ . Now they are like terms: $52 + 32 = 82$ .

Problem 5Spot the error

A student computes the hypotenuse of a right triangle with legs 3 and 4. Step 1: $c = 3^{2} + 4^{2}$ . Step 2: $c = 9 + 16$ . Step 3: $c = 3 + 4 = 7$ . Which step contains the error?

Show the worked solution

Answer: Step 2

Step 1 is the correct Pythagorean setup. Step 2 splits the root over the sum, and that is illegal: $a + b \neq = a + b$ . The right computation adds first, then takes the root: $9 + 16 = 25 = 5$ . The famous 3-4-5 triangle confirms it.

Problem 6Numeric answer

Evaluate $1 6^{3/4}$ .

Show the worked solution

Answer: 8

A fractional exponent is a root and a power combined: $1 6^{3/4} = (1 6^{1/4})^{3}$ . The fourth root of 16 is 2, and $2^{3} = 8$ . Taking the root first keeps the numbers small — cubing 16 first would give 4096 and only then a fourth root.