ISEE Upper Level Quantitative Reasoning Flashcards: Percent Increase And Decrease

What this deck covers

This deck focuses on Percent Increase And Decrease, giving you a quick way to review the definitions, rules, and examples that matter most for ISEE Upper Level Quantitative Reasoning.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: What is the percent change from $30$ to $45$?

Answer: $50\%$. The change of $15$ relative to the original $30$ results in a $50\%$ increase.

Flashcard 2: What is the final value after multiplying $150$ by a $0.9$ factor (a $10\%$ decrease)?

Answer: $135$. The factor $0.9$ reduces the original value by $10\%$, yielding the new amount.

Flashcard 3: Identify the net percent change for a decrease of $20\%$ followed by an increase of $25\%$.

Answer: $0\%$. The combined multipliers $0.8$ and $1.25$ yield $1.00$, resulting in no net change.

Flashcard 4: What is the original value if a $20\%$ increase gives a new value of $96$?

Answer: $80$. Dividing $96$ by $1.2$ reverses the $20\%$ increase to find the original value.

Flashcard 5: What is the new value after increasing $120$ by $15\%$?

Answer: $138$. Multiplying $120$ by the factor $1.15$ accounts for the $15\%$ increase.

Flashcard 6: State the formula for the percent change from original $O$ to new $N$.

Answer: $\frac{N-O}{O}\times 100\%$. Calculates the relative change by dividing the difference by the original value and multiplying by 100 to express as a percentage.

Flashcard 7: What is the new value after decreasing $200$ by $12\%$?

Answer: $176$. Multiplying $200$ by the factor $0.88$ accounts for the $12\%$ decrease.

Flashcard 8: Identify the multiplier for a $p\%$ decrease (in terms of $p$).

Answer: $1-\frac{p}{100}$. Represents the factor by which the original value is multiplied to achieve the decreased amount.

Flashcard 9: What is the percent decrease from $60$ to $45$?

Answer: $25\%$. The difference of $15$ divided by the original $60$ yields $0.25$, or $25\%$.

Flashcard 10: What is the percent change from $50$ to $40$?

Answer: $-20\%$. The change of $-10$ relative to the original $50$ results in a $-20\%$ change.

Flashcard 11: What is the percent of the original remaining after a $p\%$ decrease?

Answer: $(100-p)\%$. The remaining portion after subtracting $p\%$ from $100\%$ of the original.

Flashcard 12: Identify the multiplier for a $p\%$ increase (in terms of $p$).

Answer: $1+\frac{p}{100}$. Represents the factor by which the original value is multiplied to achieve the increased amount.

Flashcard 13: What is the overall multiplier for successive changes of $+a\%$ then $+b\%$?

Answer: $\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right)$. Combines the individual growth factors for each successive percentage increase.

Flashcard 14: What is the percent decrease from $80$ to $60$?

Answer: $25\%$. The difference of $20$ divided by the original $80$ yields $0.25$, or $25\%$.

Flashcard 15: What is the percent increase from $45$ to $60$?

Answer: $\frac{100}{3}\%$. The difference of $15$ divided by the original $45$ yields $\frac{1}{3}$, or $\frac{100}{3}\%$.

Flashcard 16: What is the original value if a $25\%$ decrease gives a new value of $75$?

Answer: $100$. Dividing $75$ by $0.75$ reverses the $25\%$ decrease to find the original value.

Flashcard 17: What is the percent increase from $40$ to $50$?

Answer: $25\%$. The difference of $10$ divided by the original $40$ yields $0.25$, or $25\%$.

Flashcard 18: State the formula for the new value after a $p\%$ increase of original $O$.

Answer: $N=O\left(1+\frac{p}{100}\right)$. Applies the growth factor by multiplying the original value by 1 plus the percentage increase divided by 100.

Flashcard 19: What is the percent of the original after a $p\%$ increase?

Answer: $(100+p)\%$. The increased portion by adding $p\%$ to $100\%$ of the original.

Flashcard 20: State the formula for the new value after a $p\%$ decrease of original $O$.

Answer: $N=O\left(1-\frac{p}{100}\right)$. Applies the reduction factor by multiplying the original value by 1 minus the percentage decrease divided by 100.

Flashcard 21: What percent decrease is needed to go from $100$ to $80$?

Answer: $20\%$. The difference of $20$ divided by the original $100$ yields $0.2$, or $20\%$.

Flashcard 22: What percent increase is needed to go from $80$ to $100$?

Answer: $25\%$. The difference of $20$ divided by the original $80$ yields $0.25$, or $25\%$.

Flashcard 23: What is the new value after decreasing $90$ by $\frac{1}{3}$?

Answer: $60$. Decreasing by $\frac{1}{3}$ means multiplying by $\frac{2}{3}$, yielding $60$.

Flashcard 24: Identify the net percent change for an increase of $10\%$ followed by a decrease of $10\%$.

Answer: $-1\%$. The combined multipliers $1.1$ and $0.9$ yield $0.99$, equivalent to a $-1\%$ net change.