Percent Increase and Decrease - ISEE Upper Level Quantitative Reasoning

  1. What is the percent change from 3030 to 4545?
    50%50\%. The change of 1515 relative to the original 3030 results in a 50%50\% increase.
  2. What is the final value after multiplying 150150 by a 0.90.9 factor (a 10%10\% decrease)?
    135135. The factor 0.90.9 reduces the original value by 10%10\%, yielding the new amount.
  3. Identify the net percent change for a decrease of 20%20\% followed by an increase of 25%25\%.
    0%0\%. The combined multipliers 0.80.8 and 1.251.25 yield 1.001.00, resulting in no net change.
  4. What is the original value if a 20%20\% increase gives a new value of 9696?
    8080. Dividing 9696 by 1.21.2 reverses the 20%20\% increase to find the original value.
  5. What is the new value after increasing 120120 by 15%15\%?
    138138. Multiplying 120120 by the factor 1.151.15 accounts for the 15%15\% increase.
  6. State the formula for the percent change from original OO to new NN.
    N−OO×100%\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.
  7. What is the new value after decreasing 200200 by 12%12\%?
    176176. Multiplying 200200 by the factor 0.880.88 accounts for the 12%12\% decrease.
  8. Identify the multiplier for a p%p\% decrease (in terms of pp).
    1−p1001-\frac{p}{100}. Represents the factor by which the original value is multiplied to achieve the decreased amount.
  9. What is the percent decrease from 6060 to 4545?
    25%25\%. The difference of 1515 divided by the original 6060 yields 0.250.25, or 25%25\%.
  10. What is the percent change from 5050 to 4040?
    −20%-20\%. The change of −10-10 relative to the original 5050 results in a −20%-20\% change.
  11. What is the percent of the original remaining after a p%p\% decrease?
    (100−p)%(100-p)\%. The remaining portion after subtracting p%p\% from 100%100\% of the original.
  12. Identify the multiplier for a p%p\% increase (in terms of pp).
    1+p1001+\frac{p}{100}. Represents the factor by which the original value is multiplied to achieve the increased amount.
  13. What is the overall multiplier for successive changes of +a%+a\% then +b%+b\%?
    (1+a100)(1+b100)\left(1+\frac{a}{100}\right)\left(1+\frac{b}{100}\right). Combines the individual growth factors for each successive percentage increase.
  14. What is the percent decrease from 8080 to 6060?
    25%25\%. The difference of 2020 divided by the original 8080 yields 0.250.25, or 25%25\%.
  15. What is the percent increase from 4545 to 6060?
    1003%\frac{100}{3}\%. The difference of 1515 divided by the original 4545 yields 13\frac{1}{3}, or 1003%\frac{100}{3}\%.
  16. What is the original value if a 25%25\% decrease gives a new value of 7575?
    100100. Dividing 7575 by 0.750.75 reverses the 25%25\% decrease to find the original value.
  17. What is the percent increase from 4040 to 5050?
    25%25\%. The difference of 1010 divided by the original 4040 yields 0.250.25, or 25%25\%.
  18. State the formula for the new value after a p%p\% increase of original OO.
    N=O(1+p100)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.
  19. What is the percent of the original after a p%p\% increase?
    (100+p)%(100+p)\%. The increased portion by adding p%p\% to 100%100\% of the original.
  20. State the formula for the new value after a p%p\% decrease of original OO.
    N=O(1−p100)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.
  21. What percent decrease is needed to go from 100100 to 8080?
    20%20\%. The difference of 2020 divided by the original 100100 yields 0.20.2, or 20%20\%.
  22. What percent increase is needed to go from 8080 to 100100?
    25%25\%. The difference of 2020 divided by the original 8080 yields 0.250.25, or 25%25\%.
  23. What is the new value after decreasing 9090 by 13\frac{1}{3}?
    6060. Decreasing by 13\frac{1}{3} means multiplying by 23\frac{2}{3}, yielding 6060.
  24. Identify the net percent change for an increase of 10%10\% followed by a decrease of 10%10\%.
    −1%-1\%. The combined multipliers 1.11.1 and 0.90.9 yield 0.990.99, equivalent to a −1%-1\% net change.

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