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Question

Hi,

Can someone help to solve this question?

Thank.you !

Answer

“Model is provided’ in Method 1 as follows :

Method 1 :
At first —
20c coins : (1 unit) + (1 unit) + (1 unit) = 3 units
50c coins : (1 unit) + (1 unit) + (1 unit) + (1 unit) = 4 units

When 40 fifty-cent coins were taken out and replaced by the same value of twenty-cent coins —
20c coins : (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) = 16 units – 160
50c coins : (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) = 4 units – 40

(40 x 0.50)/0.20 = 100 (twenty-cent coins obtained from 40 fifty-cent coins)
16 units – 160 = 3 units + 100
1 unit = (100 + 160)/(16 – 3) = 20
3 x 20 x 0.20 + 4 x 20 x 0.50 = 52

1 Reply 0 Likes ✔Accepted Answer

Oh…appreciate so much for ur help!!

I got it. Will explain to her.

Thanks again for simplifying out for us. 👍👍👍

0 Replies 0 Likes

Method 1 :
At first —
20c coins : (1 unit) + (1 unit) + (1 unit) = 3 units
50c coins : (1 unit) + (1 unit) + (1 unit) + (1 unit) = 4 units

When 40 fifty-cent coins were taken out and replaced by the same value of twenty-cent coins —
20c coins : (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) = 16 units – 160
50c coins : (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) = 4 units – 40

(40 x 0.50)/0.20 = 100 (twenty-cent coins obtained from 40 fifty-cent coins)
16 units – 160 = 3 units + 100
1 unit = (100 + 160)/(16 – 3) = 20
3 x 20 x 0.20 + 4 x 20 x 0.50 = 52

Method 2 :
At first —
twenty-cent coins : fifty-cent coins
3u : 4u

When 40 fifty-cent coins were taken out and replaced by the same value of twenty-cent coins —
twenty-cent coins : fifty-cent coins
4p : 1p

(40 x 0.50)/0.20 = 100 (twenty-cent coins obtained from 40 fifty-cent coins)
3u + 100 = 4p
4u – 40 = 1p
3u + 100 ——-> 4 x (4u – 40) = 16u – 160
1u ——-> (100 + 160)/(16 – 3) = 20
3 x 20 x 0.20 + 4 x 20 x 0.50 = 52

Ans : $52.

1 Reply 0 Likes

Hi, can explain to me why this part

(4U-40) must multiply by 4 and I do not understand it from this part onward  ?

Do you mind to simplify it or have a model to work for this type of question ?

Thank you ✌

2 Replies 0 Likes

It is a substitution.

The working given:

3u + 100 = 4p (means 4 times of 1p)              ===== equation [a]
1p = 4u – 40 (1p is equal  to 4u – 40)              ===== equation [b]

3u + 100 = 4 x (4u – 40)      This is a re-write of equation [a]. However, instead of writing 1p, he substituted  (4u-40) for 1p. He wrote (4u-40) in place of 1p. So, 4p becomes four times of (4u – 40).

What happens next is a opening of brackets.

E.g. 4 x ($3 – $2) = 4 x $1 = $4 ===== solve the part in brackets first, then multiply.

gives the same results as (4 x $3) – (4 x $2) = $12 – $8 = $4 ===== take 4 outside and multiply by first term, put in the order of operation (-), then take 4 outside and multiply by the second term.

However, this approach is more for the secondary level.

.

.

.

1u ——-> (100 + 160)/(16 – 3) = 20
3 x 20 x 0.20 + 4 x 20 x 0.50 = 52

0 Replies 0 Likes

‘(4U-40) must multiply by 4’ as the question stated “When 40 fifty-cent coins were taken out and replaced by the same value of twenty-cent coins, the ratio of the number of twenty-cent coins to the number of fifty-cent coins became 4 : 1.” ‘From this part onward,’ it is simply solving for u or p whichever is easier.

I will look into simplifying it or having a model to work for this question.

Don’t mention.

1 Reply 0 Likes

Thank you so much!

Will wait for your simplifying methods as I still cannot understand. Need to explain to my DD. She will be able to visualize it if model is provided. 

Sorry to trouble you.

1 Reply 0 Likes

“Model is provided’ in Method 1 as follows :

Method 1 :
At first —
20c coins : (1 unit) + (1 unit) + (1 unit) = 3 units
50c coins : (1 unit) + (1 unit) + (1 unit) + (1 unit) = 4 units

When 40 fifty-cent coins were taken out and replaced by the same value of twenty-cent coins —
20c coins : (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) = 16 units – 160
50c coins : (1 unit – 10) + (1 unit – 10) + (1 unit – 10) + (1 unit – 10) = 4 units – 40

(40 x 0.50)/0.20 = 100 (twenty-cent coins obtained from 40 fifty-cent coins)
16 units – 160 = 3 units + 100
1 unit = (100 + 160)/(16 – 3) = 20
3 x 20 x 0.20 + 4 x 20 x 0.50 = 52

1 Reply 0 Likes ✔Accepted Answer

Oh…appreciate so much for ur help!!

I got it. Will explain to her.

Thanks again for simplifying out for us. 👍👍👍

0 Replies 0 Likes
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