## Algebra for 12 - 16 year olds

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## Information

These questions are intended to help illustrate the benefit of using algebra. Parts ‘a’ to ‘d’ may be solved intuitively, or algebraically; but part ‘e’ may only be [easily] solved using algebra.

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Question 1 of 5## 1. Question

10 points

(A)– A man borrows donkeys to use for transporting goods.To re-pay the loan the man must give the lender deben of copper every month per donkey.

The man uses each donkey for days per month

for transporting goods and earns deben of copper per donkey per day for this work.

How many deben of copper does the man make per month?

Correct

## Intuitive solution:

The man has to spend 30 x 200 = 6000 deben of copper per month in loan repayments.

The man receives 16 x 20 = 320 deben of copper per month for one donkey. For thirty donkeys he therefore receives 320 x 30 = 9600 deben of copperThe man therefore makes 9600 – 6000 = 3600 deben of copper per month

## Algebraic solution:

Algebraic solution:

Define:

N = number of donkeys = 30

X = monthly cost per donkey = 200

D = number of days a donkey works per month = 16

Y = amount man charges per donkey per day = 20Cost to the man = N x X

Earnings = N x D x YThe man keeps the earnings – the costs, therefore:

The man makes (N x D x Y) – (N x X) = (30 x 16 x 20) – (30 x 200) = 3600 deben of copper per monthIncorrect

Question 2 of 5## 2. Question

25 points

(B)The donkeys take some looking after, though.The man has to spend deben of copper per donkey per day for feed. When the donkeys are working they need twice as much feed as they do when they’re resting.

How much does the man have to spend per month to keep the donkeys?

(There were days in the

ancient Egyptian month and the donkeys are working as described in part in Question 1)

Correct

## Intuitive solution:

A resting donkey needs 2 deben of copper worth of food per day. Therefore a working donkey needs 2 x 2 = 4 deben of copper worth of food per day.

Each donkey works for 16 days, which is 16 x 4 = 64 deben of copper worth of food.

The rest of the time the donkey is resting so it needs (30 – 16) x 2 = 14 x 2 = 28 deben of copper worth of food.Each donkey therefore needs 64 + 28 = 92 deben of copper worth of food per month.

For 30 donkeys this is 30 x 92 =

2760 deben of copper per month## Algebraic solution:

Define:

N = number of donkeys = 30

C = cost (at rest) per day = 2

D = number of days a donkey works per month = 16

M = number of days in a month = 30Cost at work per donkey = D x 2 x C

Cost at rest per donkey = (M – D) x CCost per month per donkey = (D x 2 x C) + ((M – D) x C)

Cost for all the donkeys = N x ((D x 2 x C) + ((M – D) x C)) = 30 x ((16 x 2 x 2) + ((30 – 16) x 2)) =

2760 deben of copper per monthIncorrect

Question 3 of 5## 3. Question

30 points

(C)Occasionally the man has to get the donkey doctor to visit if the donkeys get sick.Over a five-month period the donkey doctor has to visit times

the first month, in the second, times in the

third, times the fourth and in the fifth month.

What is the average number of visits the donkey doctor makes per month?

Correct

Total number of visits over 5 month period = 3 + 1 + 4 + 6 + 1 = 15

Average number of visits = total visits ÷ number of months = 15 ÷ 5 =

3 visits per monthIncorrect

Question 4 of 5## 4. Question

40 points

(D)The donkey doctor charges deben of copper per visit.Taking into account the amount the man must spend on the loan (Question A) and the feed (Question B), and the amount that the man makes from hiring out the donkeys (Question A) –

how much does the man get to keep each month?

Correct

## Intuitive solution:

Donkey doctors costs per month = number of visits per month x cost per visit = 3 [Question C] x 150 = 450 deben of copper per month.

Cost of feed per month = answer to part ‘B’ = 2760

Man’s earnings not including cost of feed and doctor’s bills = answer to part ‘A’ = 3600

The man gets to keep = earnings – costs = 3600 – 2760 – 450 =

390 deben of copper per month

## Algebraic solution:

V = number of doctor’s visits per month = answer to ‘C’ = 3

Z = doctor’s charge = 150

E = earnings = answer to ‘A’ = 3600

F = cost of feed = answer to ‘B’ = 2760The doctor’s charges = V x Z

The total costs are therefore F + (V x Z)

The man therefore earns E – (F + (V x Z)) = 3600 – (2760 + (3 x 150)) =

390 deben of copper per monthIncorrect

Question 5 of 5## 5. Question

60 points(E) If the man needs to earn at least deben of copper per month to support his family then what is the minimum number of whole days he needs each donkey to work to make enough money?

(assume the donkey doctor makes the same number of visits per month)?

Correct

## Algebraic solution: All definitions remain as before.

From part ‘D’ we know that the man’s ‘salary’, which we will call S = E – (F + (V x Z))

From previous definitions we can expand the compound parts of the equation (in other words, the previous answers) to break the equation down to the figures given in the questions. Therefore:

E = (N x D x Y) – (N x X)

F = N x ((D x 2 x C) + ((M – D) x C))

V = 3

So, S = (N x D x Y) – (N x X) – (N x ((D x 2 x C) + ((M – D) x C)) + (3 x Z))Now, we want S to be greater than or equal to 1000 and we want to know the value of D that satisfies this.

So let’s set S = 1000 and the equation becomes:

1000 = (N x D x Y) – (N x X) – (N x ((D x 2 x C) + ((M – D) x C)) + (3 x Z))We want to solve this for D, so, to make the equation easier to read let’s remove the multiplication signs, so it becomes:

1000 = NDY – NX – (N(2DC + (M – D)C) + 3Z)Now expand the brackets out:

1000 = NDY – NX – 2NDC – NMC +NDC – 3ZNow re-arrange so all terms including D are on the right:

1000 + NX + NMC + 3Z = NDY – 2NDC + NDCSimplify the right hand side by adding common terms:

1000 + NX + NMC + 3Z = NDY – NDCSwitch the equation round (left to right) so it’s easier to resolve:

NDY – NDC = 1000 + NX + NMC + 3ZNow separate out the D term:

D(NY – NC) = 1000 + NX + NMC + 3ZAnd divide both sides by (NY – NC) to resolve for D:

D = (1000 + NX + NMC + 3Z)/ (NY – NC)Substitute the values:

D = (1000 + 30 x 200 + 30 x 30 x 2 + 3 x 150)/(30 x 20 – 2 x 30)And solve the equation:

D = 9250 / 540 = 17.12 (2 decimal places)Rounding this up to the nearest whole number shows that the man must hire

out his donkeys for at least 18 days per month to be able to support his family.Incorrect

#### Ancient Egyptian Algebra for 12 – 16 year olds

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