Percentage: Level 5
1.
The kinetic energy \(K\) of an object is given by the formula \(K = \frac{1}{2}mv^2\). If the mass \(m\) is increased by 20% and the velocity \(v\) is decreased by 10%, what is the percent change in the kinetic energy?
Solution (B):
Let original \(K = 1\).
New \(m’ = 1.20m\).
New \(v’ = 0.90v\).
New \(K’ = \frac{1}{2}(1.20m)(0.90v)^2\).
We look at the coefficients: \(1.20 \times (0.90)^2\).
\(1.20 \times 0.81 = 0.972\).
The new energy is 97.2% of the old energy.
Change = \(97.2\% – 100\% = -2.8\%\). (Decrease of 2.8%).
2.
Fresh watermelon contains 90% water by weight. Dried watermelon (chips) contains 10% water by weight. How many kilograms of fresh watermelon are required to produce 1 kilogram of watermelon chips?
Solution (D):
The “solid/pulp” weight is constant.
In Chips (1 kg): Water is 10%, so Solid is 90%.
Solid weight = \(0.90 \times 1 \text{ kg} = 0.9 \text{ kg}\).
In Fresh Watermelon (\(W\) kg): Water is 90%, so Solid is 10%.
We need the solid content to match:
\(0.10 \times W = 0.9 \text{ kg}\).
\(W = \frac{0.9}{0.1} = 9 \text{ kg}\).
3.
A specific bacteria population decreases by 40% every hour due to an antibiotic. However, at the end of every hour (after the decrease), 100 new bacteria are spontaneously generated. If the population stabilizes (stops changing) at a value \(P\), what is \(P\)?
Solution (C):
“Stabilizes” means the population at the start of the hour equals the population at the end.
Let population be \(P\).
1. Decrease by 40%: Remaining is \(0.60P\).
2. Add 100: \(0.60P + 100\).
3. Set equal to original \(P\): \(P = 0.60P + 100\).
4. \(0.40P = 100\).
5. \(P = \frac{100}{0.40} = 250\).
Score: 0 / 0