Math Problem Statement

Janet saved R400R400 during the first month of her working life. In each subsequent month, she saved 11%11% more than what she had saved in the previous month.

What amount did she save during the nineteenth month?

Formula

(ignore subscripts in this step but not in written work)

Stocktake

Type in the list of known variables, example x=1,y=2...:

Substitute

(use subscripts if necessary in this step)

Solve

(use subscripts if necessary in this step)

(3)

(b) How much did she save all together in her first 17 working months?

Formula

(ignore subscripts in this step but not in written work)

Stocktake

Type in the list of known variables, example x=1,y=2...:

Substitute

(use subscripts if necessary in this step)

Solve

(use subscripts if necessary in this step)

(3)

(c) In which month of her working life was her total savings more than R20100R20100, for the first time?

Formula

(ignore subscripts in this step but not in written work)

Stocktake

Type in the list of known variables, example x=1,y=2...:

Substitute

(use subscripts if necessary in this step)

Solve

(use subscripts if necessary in this step)

That means after  months Janet's total saving will be more than R20100R20100 for the first time.

Solution

(a) What amount did she save during the nineteenth month?

Formula:

The monthly savings form a geometric sequence. The nn-th term of a geometric sequence is given by: an=arn1a_n = a \cdot r^{n-1} Where:

  • aa: the first term
  • rr: the common ratio
  • nn: the term number

Stocktake:

a=400,r=1.11,n=19a = 400, \quad r = 1.11, \quad n = 19


Substitute:

a19=400(1.11)191a_{19} = 400 \cdot (1.11)^{19-1}


Solve:

a19=400(1.11)18a_{19} = 400 \cdot (1.11)^{18}

Using a calculator: (1.11)186.143(1.11)^{18} \approx 6.143 a194006.143=2457.2a_{19} \approx 400 \cdot 6.143 = 2457.2


Final Answer:

Janet saved R2457.20 in the nineteenth month.


(b) How much did she save all together in her first 17 working months?

Formula:

The sum of the first nn terms of a geometric sequence is: S_n = a \cdot \frac{1-r^n}{1-r} \quad \text{(for r > 1)}


Stocktake:

a=400,r=1.11,n=17a = 400, \quad r = 1.11, \quad n = 17


Substitute:

S17=4001(1.11)1711.11S_{17} = 400 \cdot \frac{1 - (1.11)^{17}}{1 - 1.11}


Solve:

(1.11)175.534(1.11)^{17} \approx 5.534 S17=40015.5340.11S_{17} = 400 \cdot \frac{1 - 5.534}{-0.11} S17=4004.5340.11S_{17} = 400 \cdot \frac{-4.534}{-0.11} S17=40041.2216488S_{17} = 400 \cdot 41.22 \approx 16488


Final Answer:

Janet saved R16,488 in her first 17 months.


(c) In which month of her working life was her total savings more than R20,100 for the first time?

Formula:

We solve for nn using the sum formula for a geometric sequence: Sn=a1rn1rS_n = a \cdot \frac{1 - r^n}{1 - r} Rearranging to find nn, we solve: Sn(1r)a+1=rn\frac{S_n \cdot (1 - r)}{a} + 1 = r^n


Stocktake:

Sn=20100,a=400,r=1.11S_n = 20100, \quad a = 400, \quad r = 1.11


Substitute:

20100(11.11)400+1=(1.11)n\frac{20100 \cdot (1 - 1.11)}{400} + 1 = (1.11)^n 201000.11400+1=(1.11)n\frac{20100 \cdot -0.11}{400} + 1 = (1.11)^n 552.75+1=(1.11)n-552.75 + 1 = (1.11)^n 552.75=(1.11)n552.75 = (1.11)^n

Taking the logarithm: n=log(552.75)log(1.11)n = \frac{\log(552.75)}{\log(1.11)}


Solve:

Using a calculator: log(552.75)2.743,log(1.11)0.04576\log(552.75) \approx 2.743, \quad \log(1.11) \approx 0.04576 n=2.7430.0457659.92n = \frac{2.743}{0.04576} \approx 59.92


Final Answer:

Janet's total savings exceeded R20,100 during the 60th month.


Summary:

  • (a) Janet saved R2457.20 in the 19th month.
  • (b) Janet saved R16,488 in the first 17 months.
  • (c) Janet’s total savings exceeded R20,100 for the first time in the 60th month.

Would you like further clarification or detailed breakdowns? 😊

Here are five related questions for exploration:

  1. How would her savings pattern change if the growth rate was reduced to 8% instead of 11%?
  2. How much would she save in total after 24 months?
  3. What is the average monthly savings over the first 17 months?
  4. How long would it take for her to save a total of R50,000?
  5. What is the percentage increase in savings from the 19th month to the 20th month?

Tip: For geometric growth problems, logarithms are essential when solving for the term count!

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Math Problem Analysis

Mathematical Concepts

Sequences and Series
Geometric Progression
Logarithms

Formulas

nth term of a geometric sequence: an = a * r^(n-1)
Sum of the first n terms of a geometric sequence: Sn = a * (1 - r^n) / (1 - r) (for r > 1)
Logarithmic equation to solve for n: n = log(value) / log(base)

Theorems

Properties of geometric sequences and series

Suitable Grade Level

Grades 10-12