Word Problems - At A Glance

Answer

If Mrs. Baker reduces the amount of sugar by  each week, that means she keeps  of the previous amount of sugar. So on the second week she uses  cups of sugar, and so on. Let's make a table.

(a) From the table, we can see that on the fifth week Mrs. Baker uses

cups of sugar.

(b) The amount of sugar Mrs. Baker uses each week is described by a geometric series with a = 2 and . We want to know how much sugar she uses over half a year, or 26 weeks. This means we need to take the partial sum of the geometric series with n = 26. We get

She uses almost 6 cups of sugar.

(c) To find the amount of sugar Mrs. Baker uses if she bakes cookies every week for all eternity, we need to find the infinite sum of the geometric series:

Mrs. Baker uses 6 cups of sugar if she bakes cookies every week for all eternity, and from (b) we know that almost all that sugar is used in the first half year of baking.

Answer

The question says for each exam Mr. Vold writes "half again as many questions as were on the previous exam."

This means the number of questions for an exam is

or

(a) The first exam had 10 questions. The second exam has

questions.

(b) The third exam has

which rounds up to 23 questions.

(c) There are (at least) two ways to interpret this question, depending on when we deal with rounding.

In part (b) we found that there were 23 questions on the third exam. In this case there are

which rounds up to 35 questions on the fourth exam.

Then there are

which rounds up to 53 questions on the fifth exam.

We could leave the rounding until the end, instead. If the first exam has 10 questions then the second exam has

the third exam has

and the fifth exam has

This rounds up to 51 questions on the fifth exam.

(d) For this question we don't want to deal with partial rounding. We're looking at a geometric series with a = 10 and . To find how many questions Mr. Vold writes over 20 exams we need to add up the first 20 terms of this geometric series:

Rounding up, we see Mr. Vold will write 66486 questions over the semester.

Answer

For n ≥ 5, let

a_n be the number of books Mo reads when he's n years old. We're told

a₅ = 4

to start off, and that to get from one number to the next we need to multiply by 2 and then add 3.

This means for n > 5,

a_{n + 1} = 2a_n + 3.

(a) Take a₅, multiply by 2, then add 3:

a₆ = 2a₅ + 3

= 2(4) + 3

= 11.

Mo reads 11 books when he's 6 years old.

(b) When Mo is seven he reads

a₇ = 2(11) + 3 = 25 books.

When he's eight he reads

a₈ = 2(25) + 3 = 53 books.

(c) Since we're asked to give a formula for the number of books Mo reads when he's 5 + i years old, we'll write a₆ as a_{5 + 1},

a₇ as a_{5 + 2}, and so on.

Following the hint, look at how we got the answers to (a) and (b) without simplifying so much.

The first one was simple:

a_{5 + 1} = 2(4) + 3

Instead of simplifying, put this answer straight into the equation for a₇ = a_{5 + 2}:

a_{5 + 2} = 2a_{5 + 1} + 3

= 2(2(4) + 3) + 3

= 2²(4) + 2 × 3 + 3

Now we can put the non-simplified expression

a_{5 + 2} = 2²(4) + 2 × 3 + 3

into the equation for a_{5 + 3}:

a_{5 + 3} = 2a_{5 + 2} + 3

= 2(2²(4) + 2 × 3 + 3) + 3

= 2³(4) + 2² × 3 + 2 × 3 + 3

Writing all our expressions next to each other, we have

a_{5 + 0} = 4

a_{5 + 1} = 2(4) + 3

a_{5 + 2} = 2²(4) + 2 × 3 + 3

a_{5 + 3} = 2³(4) + 2² × 3 + 2 × 3 + 3

We can generalize to get an expression for a_{5 + i}:

a_{5 + i} = 2ⁱ(4) + 2^{i – 1} × 3 + 2^{i – 2} × 3 + ... 2 × 3 + 3

This expression has 2 parts. The first part is 4 multiplied by a power of 2. The second part is itself a series. Writing the series in summation notation, we can rewrite the expression as

Don't be confused by the use of j in the summation. We need some letter for the index of summation, and n and i have already been used in this problem.

(d) We want to find

a₁₅ = a_{5 + 10}.

Using the formula from (c),

The second piece of the expression is a geometric series with first term

a = 2⁰ × 3 = 3

and ratio 2. There are 10 terms from j = 0 to j = 9, so take n = 10. This partial sum is

Adding up the pieces of the expression,

a₁₅ = 4096 + 3069 = 7165.

Check the answer by direct computation. We left off before at a₈ = 53, so

a₉ = 2(53) + 3 = 109

a₁₀ = 2(109) + 3 = 221

a₁₁ = 2(221) + 3 = 445

a₁₂ = 2(445) + 3 = 893

a₁₃ = 2(893) + 3 = 1789

a₁₄ = 2(1789) + 3 = 3581

a₁₅ = 2(3581) + 3 = 7165

Thankfully, this answer agrees with what we got using the formula. We conclude that Mo reads an absolutely ludicrous number of books the year he's 15: 7165 books.

Answer

Here's the timeline for the money going into Komi's account:

(a) By the end of February Komi has made deposits of $100 (in January) and $10 (in February) so his account contains $110.

(b) By the end of March Komi's account contains $10 more than it did at the end of February, for a total of $120.

(c) At the end of one month (end of January), Komi's account contains $100. At the end of two months (end of February), the account contains

100 + 10.

At the end of three months (end of March), the account contains

100 + 10 × 2.

Generalizing, at the end of n months Komi's account contains

100 + 10(n – 1) dollars.

And in case you were wondering, no, this problem doesn't have much to do with arithmetic or geometric series. You could look at the amount of money in Komi's account as $100 plus a partial sum of an arithmetic series with d = 0, but why bother?

Answer

(a) Here's the timeline:

At the beginning of March, Kendra has $131.25. In the middle of March she withdraws 75% of that amount, leaving 25% in the bank.

Rounded to the nearest cent, her bank account holds

0.25(131.25) = $32.81

at the end of March.

(b) This question is similar to (b) of the example, but it asks about what's happening at a different point in the month.

At the end of the month, Kendra has 25% of the money she had at the beginning of the month. We'll call the amount she has left at the end of the month L_n. There are two ways to do this problem: we could use the answer to (b) of the example, or we could start from scratch. It's a good idea to understand both ways.

Using the example:

At the beginning of month n, Kendra has M_n dollars in her account.

At the end of month n, she has 25% of that amount left.

So

Starting from scratch:

Let L_n be the amount of money in Kendra's account at the end of n months. To get from L_{n – 1} to L_n she deposits $100, then keeps 25% of the resulting amount.

This means

L_n = 0.25(100 + L_{n – 1} ).

This tells us how to build each L_n from the previous one, so look at some values of L_n.

The value L_n is the sum of a geometric series with r = 0.25,

a = 0.25(100) = 25,

and n terms (since the exponents go from 1 to n). So

Thankfully, this is equivalent to the formula we found when we used the example instead of starting from scratch.

Answer

This exercise is very similar to the previous one. In January, Lizette deposits $50 and earns 0.25% interest. After the interest calculation, she has

1.0025(50) = 50.125

which we truncate to $50.12.

Remember that to calculate interest we multiply by (1 + ( interest rate)) since we have to account for the initial amount of money as well as the interest earned.

(a) At the beginning of February, Lizette adds $50 to the $50.12 she had from January:

Since interest isn't calculated until the end of the month, in the middle of February she still has

50 + 50.12 = 100.12.

(b) Extend the timeline one more month:

Lizette earns interest on her $100.12, then deposits another $50. In the middle of March she has

50 + 1.0025(100.12) = 150.37.

(c) Let M_n be the amount of money in Lizette's account in the middle of the nth month. Then M_{n – 1} was the amount of money in her account in the middle of the (n – 1)st month.

To find M_n we calculate interest on M_{n – 1}, then add another $50. So

M_n = 50 + 1.0025M_{n – 1}.

Here are the values of M_n for some small n:

Since we're so good at recognizing these patterns by now, we see that M_n is the sum of a finite geometric series with a = 50,

r = 1.0025, and n terms:

(d) There are two ways to do this problem. We could use the formula from (c) to find M₁₂, or we could calculate M_n for n = 1, n = 2, all the way up to n = 12. Since n = 12 is pretty small and we want to make sure we get the right answer, we'll do it both ways.

Using the formula from (c):

All we have to do is plug in n = 12 to the formula we found in (c). We get

which, since we're truncating our answers, works out to $608.31 in the bank.

Using direct calculation:

We found that, in March, Lizette had

M₃ = 150.37.

Now we apply the formula

M_n = 50 + 1.0025M_{n – 1}

nine times:

M₄ = 200.74

M₅ = 251.24

M₆ = 301.86

M₇ = 352.61

M₈ = 403.49

M₉ = 454.49

M₁₀ = 505.62

M₁₁ = 556.88

M₁₂ = 608.27

We're 4 cents off from the number we got using the formula, because here we truncated each value of M_n along the way. The answers are close enough for us to be confident that the formula is correct.