by ISP 11 Replies latest jw friends
Greetings Apostopeople..........
My Daughter, 10 yrs, will be sitting her 11+ exam in January 2005 and I am going through some past papers. Some of the questions are pretty tasking....eek for me!!!
Here's one for you to try out.............
'Write down a number which leaves a remainder of 3 when divided by 7 and which also leaves a remainder of 4 when divided by 5.'
I got the answer through trial and error...anyone got any neat ways of getting the answer?
Best
ISP
24 ?
Go up in 7's until you get to 1 more than a multiple of 5
... add 3 and this number will leave 3 when divided by 7 and 4 when divided by 5
There isn't a formula, per se, that will solve this. But it may be shorter to write the problem like this:
r = 3 (mod 7) 3, 10, 17, 24, 31, 38
r = 4 (mod 5) 4, 9, 14, 19, 24, 29, 34
This first rewords your problem ... remainder = 3 when divided by 7 (mod 7), and remainder is 4 when devided by 5 (mod 5). Next to each line start with the remainder (3) and keep adding the mod value (7) so that you end up with 3, 10, 17, 24, etc. Do the same for the next line and see where they match. 24!
I got the answer to be 24.....I guess following the method u did. i thought there maybe a short cut involving equations or something. ISP
Heres another one from same paper.... In this question we are using the whole numbers from 1-9 Set A consists of the odd numbers (1,3,5,7,9) Set B consists of the suare numbers (1,4,9) The inners are the numbers which are in both sets, i.e (1.9) The incount is the number of members of the inners set, i.e.2 The inproduct is the product of the inners, i'e 1x9=9 The outers are the numbers which are in neither of the two sets, i.e. (2,6,8) The outcount is the number of members of the outcount set, i.e.3 The outproduct is the product of the outers, i.e. 2x6x8=96 For sets C and D incount =3 inproduct=42 outcount=3 outproduct=30 fill in one possible answer for sets C and D C= D= (Question 14c MGS paper1999) ISP
For that problem you basically need to find three unique numbers (from 1 - 9) that have a product of 42, and then three separate numbers (all different from the first three) that have a product of 30.
Then you can fill in the blanks to form two sets of numbers (C and D).
Here's one example:
1 x 6 x 7 = 42
2 x 3 x 5 = 30
So set C and D both contain 1, 6, and 7. Neither set contains 2, 3, or 5. The remaining numbers are only found in one set:
Set C = {1, 4, 6, 7}
Set D = {1, 6, 7, 8, 9}
In answer to the first question (haven't looked at the 2nd yet!)
Kinda reiterating what was already explained but in simple terms:
Count by 7's starting at the number 3:
3, 10, 17, 24, 31...
Then count by 5's starting at the number 4:
4, 9, 14, 19, 24, 29...
Whenever the numbers from each series match this will be your answer.
additional answers:
59, 94, 129, 164, 199...
Which is an unending series of numbers spaced by 35. So start with 24 and add 35 and you get 59. Add 35 and you get 94, etc.
Nice going drwtsn! I like yr style! I might ask u a few more, if necessary. Best ISP
Cheers ISP! Hope you and the fam are well...
Sincerely,
District Overbeer of the "Anti_Math" class
Thx Valis.........all are well here!
U still as cool as ever? I thought yr hair was great! We're of to Jamaica on the 14/12 for 2 weeks...so we are all quite excited. hope there are no hurricanes etc.
Best
ISP
