Could you do these 11+ maths questions?

[willow]

I've had a go but would be interested to know what answers you lot come up with (not sure I'm right, tbh!)

1. Three fifths of my cattle are Jerseys; the rest are Herefords. 3/4 of my Jerseys are female, as are 3/8 of my Herefords.

a) What fraction of all my cattle are females.

b) What fraction of my male cows are Hereford?

2. In my flock of sheep, there are three breeds: Costwolds, Swaledales and Leicester Longwools. The ration of Cotswolds to Swaledales is 3:5 and Swaledales to Leicester Longwools is 7:4.

a) What is ratio of Cotswolds to Leicester Longwools?

b) There are more than 100 sheep in my flock. What is the smallest number of sheep there could be?

ALGEBRA

Write algebraic expressions for each of the following questions.

a) Billy had r pound coins and found another. How many pound coins did he then have?

b) Rob had q pound coins and gave h of them away. How many pound coins did he have then?

c) Sam had y pounds and his mother gave him g five-pound notes. How much did he have then, in pounds?

d) Tom had t pound coins and gave k of them away. Will had w pound coins and gave j of them away. How many more pound coins did Tom have than Will?

FINALLY CLOCK FACES

What would you say angle between hour and minute hand is when time is 10:20?

And if it is between 10 and 11 o'clock and hands of clock are 180 degrees apart, what time is it?

marking my place while I work them out! The algebra is easier for me than the sheep stuff.

1a- can't do fractions, are we allowed to convert to decimel to work it out?

2b- ditto

3a- 21:20

3b- 176

4a- r + 1

4b- q - h

4c- y + g(£5)

4d- (t - k) - (w - j)

clock- 180 deg (just under actually but i assume they want the simple answer)

10.20 (as above)

Feel free to laugh, my maths skills are shockingly bad

Yes, I'm sure I could but they look too trivial to be fun so I sha'n't bother

1a 3/5
1b 3/5
2a 21:20
2b 152

a r+1
b q-h
c y + 5g
d t - k - w + j

180 degrees (actually just a bit less)
10:20ish

Do they want the exact answers for the clock face ones?

clock is 170 degrees i think... hour hand moves 1/12 of 360 ndegrees each hour and at 20 past has moved 1/3 of way through hour = 10 degrees, so 180 - 10...
and yes, for 11+ they will expect the accurate answer.

1a) 3/5 female
1b) 5/8ths

Now given up!!!

3/5 of your cattle are female
5/8 of your males are hereford

Cot : ll = 21:20
420

r+1
q-h
y+5g
t+j-k-w

170 degrees

I get one B to be 1/4

1021 and 49 seconds

Time on clock: please tell me I am allowed to use a calculator!
I make it 10 hours 21 minutes and 49 seconds (rounded to nearest second).

ooh snap, talker.

1(a) 3/5
1(b) 5/8

2(a) 21:20
2(b) 152

Algebra ones all the same as AMumInScotland

Clock:
First angle is 170 degrees exactly.
I can't get an exact time in minutes for the second part, I make it almost 10:22 (10:21:49 to the nearest second).

amothersplace, the male Herefords make up 1/4 of the total sheep.
That was fun.

Didn't look, just did this in notepad.

40 cows (5x8) - 4 already goes into 8. 24 jerseys - 18 female, 6 male 16 herefords - 6 female, 12 male

A: proportion female: 18/24 = 3/5
fraction of male (18) hereford: 2/3

Cw:Sd:LL
21:35:20

ratio of Cw:LL = 21:20
smallest num of sheep = 152 (2x all orig numbers)

a. r+1
b. q-h
c. y+5g
d. impossible to tell? actual sum is (t-k)-(w-j)

180 degrees
20 past 10

Ah interesting answers for question 4, and 3d. I didn't think of that. I think I was deceived by the other simpler questions to think those two must have been simpler than they appeared!

I agree with the 152 ... my 420 was nonsense

Why does nobody else have 2/3 for the amount of herefords out of all males? How are you all working it out?

Also 1/4 of the cows aren't male herefords, 12/40 (6/20) are - by my calculations anyway. Did I read the question right?

Oh I see - 5/8 of the herefords are male. I still have 2/3 of the total males being hereford though.

Gah! I typed out a long answer to you, Bertie, but I'm having problems with my internet connection and now it's all disappeared. Anyway you will find if you have a herd of 40 you should only have 10 male Herefords. Hope that helps!

oh dear well here goes.

1A) 1/2
1B) 1/4

2A) 21:20
2B) 152

3A) r+1
3B) q-h
3C) y+(g5)
3d) (t-k) - (w-j)

4A) 180°
4B) 10:20

So how did I do? Off to check what others got now.

can't work out how everyone has got 170° for the clock face?

1/12 360° = 30° ??? Therefore each 5 mins = 30° ?????

Are you allowing for the hand to have moved past the 10??? So for each 5 minutes the hand moves x° for each 30° the other hand moves?

If thats the case I make it 172°????

But a herd of 40:

3/5 are Jerseys so 24 Jerseys

Which leaves 16 Herefords.

3/4 of jerseys are female - 3/4 of 24 is 18. Leaving 6 male jerseys.

3/8 of herefords are female, which is 6, leaving... oh, 10 Sorry! You are right there :)