3.1 Arithmetic Operations, Fractions, Decimals, and Percentages

3.1 Arithmetic Operations, Fractions, Decimals, and Percentages

The Numerical Reasoning subtest of the General Scholastic Aptitude (GSA) assesses your quantitative fluency, mathematical logic, and practical problem-solving efficiency under strict time limits. BEA does not publish the time allowed per question, so build speed by moving beyond slow long division and column-by-column multiplication. You must develop mental agility, master benchmark percentage decompositions, simplify fractions by cross-cancellation, and leverage strategic estimation to eliminate implausible distractors instantly.

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Mental Arithmetic and Rapid Estimation Strategies

In a non-calculator testing environment, the fastest path to the correct option frequently relies on algebraic properties of numbers rather than brute-force mechanical arithmetic.

1. Distributive Splitting and Decomposition

When multiplying a multi-digit number by a composite integer, decompose the multiplier into friendly round components:

• Multiplying by 15: Multiply by 10, then add half of that product. For example, $36 \times 15 = (36 \times 10) + (36 \times 5) = 360 + 180 = 540$.
• Multiplying by 25: Treat 25 as $\frac{100}{4}$. Divide the number by 4 and append two zeros. For example, $48 \times 25 = \frac{48}{4} \times 100 = 12 \times 100 = 1,200$.
• Multiplying by 50: Divide by 2 and multiply by 100. For example, $64 \times 50 = 32 \times 100 = 3,200$.
• Multiplying by 9 or 11: Use $(10 - 1)$ or $(10 + 1)$. For example, $43 \times 9 = 43 \times (10 - 1) = 430 - 43 = 387$.

2. Difference of Two Squares Shortcut

When multiplying two numbers that are equidistant from an easy benchmark number, apply the algebraic identity $(x - y)(x + y) = x^2 - y^2$:

• Compute $49 \times 51$: The midpoint is 50, with a difference of 1. Thus, $(50 - 1)(50 + 1) = 50^2 - 1^2 = 2,500 - 1 = 2,499$.
• Compute $38 \times 42$: The midpoint is 40, with a difference of 2. Thus, $(40 - 2)(40 + 2) = 40^2 - 2^2 = 1,600 - 4 = 1,596$.

3. Front-End Estimation and Bounding

When test answer options are spaced far apart (for instance: PHP 420, PHP 580, PHP 750, PHP 1,100), computing exact values is a waste of precious examination minutes. Round numbers to their leading significant digits to establish lower and upper bounds:

• To estimate $49.2 \times 12.1$, round to $50 \times 12 = 600$. Any option under 500 or over 700 can be immediately eliminated.

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Operations with Fractions, Decimals, and Mixed Numbers

Fractions represent parts of a whole and division relationships. Numerical reasoning items often require moving fluidly between fractions, decimals, and percentages.

Fraction Arithmetic Foundations

1. Addition and Subtraction: Requires a least common denominator (LCD). Find the LCD by identifying the least common multiple of the denominators using prime factorization: $\frac{a}{b} \pm \frac{c}{d} = \frac{ad \pm bc}{bd}$ Example: $\frac{5}{6} - \frac{3}{8}$. The prime factors of 6 are $2 \times 3$, and for 8 are $2^3$. The LCD is $2^3 \times 3 = 24$. Converting: $\frac{20}{24} - \frac{9}{24} = \frac{11}{24}$.

2. Multiplication and Cross-Cancellation: Never multiply large numerators and denominators directly before simplifying. Always cancel common prime factors between any numerator and any denominator first: $\frac{14}{25} \times \frac{15}{21} = \frac{2 \times 7}{5 \times 5} \times \frac{3 \times 5}{3 \times 7} = \frac{2}{5} \times \frac{1}{1} = \frac{2}{5}$

3. Division via Reciprocals: Dividing by a fraction is mathematically identical to multiplying by its reciprocal (the "invert and multiply" rule): $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}$

4. Mixed Numbers: Always convert mixed numbers to improper fractions before multiplying or dividing: $3\frac{1}{4} = \frac{(3 \times 4) + 1}{4} = \frac{13}{4}$. For addition and subtraction, you may process the whole numbers and fractional parts separately, taking care when borrowing across the whole unit.

Essential Fraction-Decimal Equivalencies

Memorizing these foundational equivalents saves substantial scratch paper computation:

The Cross-Multiplication Comparison Test

To determine which of two positive fractions $\frac{a}{b}$ or $\frac{c}{d}$ is greater without finding a common denominator, cross-multiply numerators with opposite denominators: $\frac{a}{b} > \frac{c}{d} \iff a \times d > b \times c$ Example: Compare $\frac{7}{11}$ and $\frac{5}{8}$. Compute $7 \times 8 = 56$ and $11 \times 5 = 55$. Since $56 > 55$, it follows that $\frac{7}{11} > \frac{5}{8}$.

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Percentages, Multipliers, and Commercial Arithmetic

The word "percent" derives from the Latin per centum, meaning "out of one hundred." Every percentage problem can be resolved using the primary equation: $\text{Percentage } (P) = \text{Base } (B) \times \text{Rate } (R)$ where the Base represents the whole quantity, the Rate is the ratio expressed as a decimal or fraction, and the Percentage represents the specific portion.

The Benchmark Percentage Method

Rather than multiplying by cumbersome decimals, break any percentage into quick mental building blocks:

• $10\%$ = Shift the decimal point one place to the left.
• $1\%$ = Shift the decimal point two places to the left.
• $5\%$ = Take half of $10\%$.
• $20\%$ = Double $10\%$.
• $2.5\%$ = Take half of $5\%$.

Application: Find $35\%$ of PHP 480:

1. $10\%$ of $480 = 48$.
2. $30\% = 48 \times 3 = 144$.
3. $5\% = \frac{48}{2} = 24$.
4. $35\% = 144 + 24 = \text{PHP }168$.

Percentage Increase, Decrease, and Multipliers

To compute percentage change, always divide the absolute difference by the original starting value: $\text{Percentage Change} = \frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%$

Using decimal multipliers dramatically accelerates multi-step calculations:

• A $15\%$ increase corresponds to a multiplier of $1 + 0.15 = 1.15$.
• A $20\%$ decrease corresponds to a multiplier of $1 - 0.20 = 0.80$.

Successive (Compound) Discounts

Retail word problems often involve successive trade or promotional discounts. When a retailer offers "$20\%$ off plus an additional $10\%$ off for loyalty cardholders," the net discount is never $30\%$. The second discount applies strictly to the already discounted price.

Let the original price be $P_0$. After discount $d_1$, the price is $P_1 = P_0(1 - d_1)$. After discount $d_2$, the final price is: $P_{\text{final}} = P_0(1 - d_1)(1 - d_2)$ The single equivalent discount rate ($d_{\text{eq}}$) is: $d_{\text{eq}} = 1 - (1 - d_1)(1 - d_2) = d_1 + d_2 - d_1 d_2$ with the discounts written as decimals. Written as percentages $D_1$ and $D_2$, the same rule is $D_{\text{eq}} = D_1 + D_2 - \frac{D_1 D_2}{100}$. For successive discounts of $20\%$ and $10\%$: $D_{\text{eq}} = 20\% + 10\% - \frac{20 \times 10}{100}\% = 30\% - 2\% = 28\%$ The customer pays $100\% - 28\% = 72\%$ of the original retail price.

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Step-by-Step Worked Word Problems

Worked Example 1: Successive Retail Discounts

Problem: A complete Senior High School technical drawing kit is listed at an original retail price of PHP 2,400. During an academic opening sale, the bookstore offers an initial $25\%$ storewide discount. A student presents an active scholar voucher granting an additional $10\%$ discount off the reduced price. What is the final amount the student pays?

Step-by-Step Solution:

1. Identify the base and discount rates: Initial base $B = \text{PHP }2,400$; primary discount $d_1 = 25\% = 0.25$; voucher discount $d_2 = 10\% = 0.10$.
2. Method A — Sequential Reductions:
   • Calculate the first discount: $25\%$ of $2,400 = \frac{1}{4} \times 2,400 = \text{PHP }600$.
   • Intermediate price after first discount: $2,400 - 600 = \text{PHP }1,800$.
   • Calculate the second discount: $10\%$ of $1,800 = \text{PHP }180$.
   • Final price paid: $1,800 - 180 = \text{PHP }1,620$.
3. Method B — Compounded Multiplier:
   • Net payment multiplier: $(1 - 0.25)(1 - 0.10) = 0.75 \times 0.90 = 0.675$.
   • Final price: $2,400 \times 0.675 = 2,400 \times \frac{27}{40} = 60 \times 27 = \text{PHP }1,620$.

Worked Example 2: Fractional Remainder Budget Problem

Problem: Maria receives a monthly educational allowance. She spends $\frac{2}{5}$ of her total allowance on city transport commuting to school. Of the remaining funds, she spends $\frac{1}{3}$ on canteen lunches. If she has PHP 720 remaining for school supplies and emergency savings, what was her total monthly allowance?

Step-by-Step Solution:

1. Determine the transport allocation: Transport $= \frac{2}{5}$ of total allowance $T$.
2. Determine the remaining fraction after transport: Remainder $= 1 - \frac{2}{5} = \frac{3}{5} T$.
3. Determine the canteen meal allocation: Lunch $= \frac{1}{3} \times \left(\frac{3}{5} T\right) = \frac{1}{5} T$.
4. Calculate total fractional expenditure: Total spent $= \frac{2}{5} T + \frac{1}{5} T = \frac{3}{5} T$.
5. Set up the equation for the remaining balance: $\text{Remaining Fraction} = 1 - \frac{3}{5} = \frac{2}{5}$ $\frac{2}{5} T = \text{PHP }720$
6. Solve for $T$: $T = 720 \times \frac{5}{2} = 360 \times 5 = \text{PHP }1,800$ Verification: $\frac{2}{5}$ of $1,800 = 720$ (transport), leaving $1,080$. $\frac{1}{3}$ of $1,080 = 360$ (lunch). Remaining: $1,080 - 360 = 720$. The solution is verified.