10-11 Bond maths paper help please

[QUEENMA]

Hi all

Need help with 2 questions from this book. Please could someone explain ?

Question1) Paper 21 - questions 11-16

Roast chicken must be cooked for 50 mins per kg and then for extra 20 minutes.
Complete this table of cooking time in hours and minutes

weight of chicken(kg) cooking time
1
1.5
2
2.5
3
3.5

DS found proportion of the the cooking time and the extra time for half an hour and then applied it to the full kgs. Answers don't match at all and I can't think of where he is going wrong. e.g.he came up with 1 hr 45 mins for 1.5 kgs whereas the answer given is 1 hr 35 mins.
Ans :
2 kg - 2 hrs
2.5kg - 2h 25 mins
3kg - 2h 50 mins
3.5kg - 3h 15 mins

What are we missing ?

Question 2) paper 21 - sum 43
Scott has 1/3 as many computer games as Anish , and Anish has 1/2 as many games as Nick. together they have 140 games. How many do they each have ?

Please can someone explain how to do these types without algebra ? Or is algebra the best method ?

Many thanks in advance !

[Guest55]

Question1) Paper 21 - questions 11-16

Roast chicken must be cooked for 50 mins per kg and then for extra 20 minutes.
Complete this table of cooking time in hours and minutes

weight of chicken(kg) cooking time
1
1.5
2
2.5
3
3.5

I'll do one of them:

1.5 kg: cooking time = 50 + 25 + 20 = 95 mins which is 1 hr 35 mins

I can't work out from what you've said where his error is coming from ...

[Guest55]

Question 2) paper 21 - sum 43
Scott has 1/3 as many computer games as Anish , and Anish has 1/2 as many games as Nick. together they have 140 games. How many do they each have ?

Who has the most games? Nick does

You could use algebra like this - let the number of games Nick has be n
Anish has n/2
Scott has a third of Anish ie n/6

Then .... does that help?

[QUEENMA]

Many Thanks Guest55 !

For question 1, the 20 mins extra is standard across whatever the amount of chicken ? Isn't that a bit odd ?

For question 2 - thanks again - so best to use algebra for these types. Is there any other way ?

[Guest55]

Yes - it's the 'normal' cooking times for cooking chicken or any meat.

You could try a trial and improvement method?

140 games Nick has the most, so guess 70

Nick 70, Anish 35, doesn't divide by 3 so no good so we know we need a number divisible by 6

Nick 72, Anish 36, Scott 12 but 72 + 36 + 12 = 120 not 140

Nick 78, Anish 39, Scott 13 but 78 + 39+ 13 = 130 so we can see the answer?

Nick 84, Anish 42, Scott 14

[QUEENMA]

Thanks once again !

[ToadMum]

Many Thanks Guest55 !

For question 1, the 20 mins extra is standard across whatever the amount of chicken ? Isn't that a bit odd ?

For question 2 - thanks again - so best to use algebra for these types. Is there any other way ?

I do hope that either you are vegetarian, or someone else in your household is responsible for cooking roast dinners .

From a food safety point of view, the actual calculation is weight in pounds x cooking time per pound, plus another hour's worth of minutes (per pound) for that meat type. So for pork, it would be:

Pounds weight x 35, plus an extra 35 minutes (or kilo weight x 77, plus an extra 35).

See here
https://www.bbcgoodfood.com/howto/tools/roast-timer

[russet]

For the games question I would have started by supposing Scott had only one game, Anish would have 3 (3 times as many) and Nick 6 (twice as many as Anish). That accounts for 10 games: 1+3+6.
But there are 140 games to account for; 14 times as many. So each child must have 14 times as many as the proportions stay the same

[QUEENMA]

Dear ToadMum

Although this is getting pedantic now, but check your own response - "plus another hour's worth of minutes (per pound)" - my interpretation to that would be that the extra coking time is also related to 'per pound' meaning it is proportional - that is exactly the question I had .

Queenma

[QUEENMA]

Dear russet.

Thank you so much ! Need to tell DS to start thinking about solutions in different ways.
Many thanks.

Queenma