Growth and decay GCSE Questions and Worked Answers

Each hour leaves 112% of the previous mass, so multiply by 1.12.

$$250\times1.12^3$$ $$=351.232$$

The mass is 351.2 g to 1 decimal place. Tip: the power is 3 because three hours have passed.

The multiplier is $1-0.15=0.85$. After year 1:

$$800\times0.85=680$$

After year 2:

$$680\times0.85=578$$

The value is £578. This is not a 30% loss of the original value: the second loss uses £680.

Let $B_n$ be the balance in pounds after n complete years, including their deposits. The starting value is $B_0=500$. One year multiplies by 1.04, then adds 60:

$$B_{n+1}=1.04B_n+60$$

Year 1:

$$B_1=1.04(500)+60$$ $$=520+60$$ $$=580$$

Year 2 uses this new balance:

$$B_2=1.04(580)+60$$ $$=603.20+60$$ $$=663.20$$

The answer is £663.20. The first £60 deposit earns interest in year 2; the second does not earn interest before this endpoint.

(a) Interest first:

$$900\times1.02=918$$ $$918-100=818$$

The balance is £818. (b) Repayment first:

$$900-100=800$$ $$800\times1.02=816$$

The balance is £816. In (b), the repaid £100 does not incur that month's 2% interest, saving £2. Tip: preserve the order with brackets: $1.02(B_n-100)$ for repayment first.

Let $V_n$ be litres after n complete hours, so $V_0=50$. Losing 20% leaves 80%, then add 4:

$$V_{n+1}=0.8V_n+4$$

Hour 1:

$$V_1=0.8(50)+4=44$$

Hour 2:

$$V_2=0.8(44)+4=39.2$$

Since 44 is not below 40 but 39.2 is, the first crossing is after 2 complete hours.

Month 1 interest:

$$250\times1.02=255$$

After the £100 payment:

$$255-100=155$$

Month 2 interest:

$$155\times1.02=158.10$$

After payment:

$$158.10-100=58.10$$

Month 3 interest:

$$58.10\times1.02=59.262$$

This is less than £100. Pay the remaining £59.262 in the model, reported as £59.26 to the nearest penny. The loan clears with the third payment. Tip: do not subtract another full £100 and call a negative balance a debt.