IGCSE/GCSE/O & A Level/IB/University Student Forum

Qualification => IGCSE/ GCSE => Queries => Topic started by: spookyjaransh on November 03, 2009, 03:20:15 pm

Title: maths problem
Post by: spookyjaransh on November 03, 2009, 03:20:15 pm
a gardner has 357 tulips to plant. if she plants x rows with y bulbs in each row, the equation fr the bulbs left over would be 357-xy (right?). if 10<x<20 and y>20, what r x and y ??
Title: Re: maths problem
Post by: Ghost Of Highbury on November 03, 2009, 03:30:36 pm
which past paper q is this?
Title: Re: maths problem
Post by: holychalice on November 03, 2009, 03:30:54 pm
y = 21 nd x = 17 ...
im nt sure ... is it frm a past ppr ?
Title: Re: maths problem
Post by: nid404 on November 03, 2009, 03:33:44 pm
The question seems incomplete....

The only way i thought we can solve this was by taking factors of 357

357

the only way to write 357 for x between 10 and 20 and y greater than 20 would be 17X21(since they have to be whole no.s)

hence x=17 and y=21

but i still thing there's summin missin.....

Title: Re: maths problem
Post by: @d!_†oX!© on November 03, 2009, 03:34:49 pm
this is from june 1990
http://www.scribd.com/doc/3182857/19982003-Paper-4-Maths
page 26
Title: Re: maths problem
Post by: spookyjaransh on November 03, 2009, 03:44:11 pm
not a linear prog. question, but still thanx.....
Title: Re: maths problem
Post by: nid404 on November 03, 2009, 03:51:28 pm
It's takin time to load...I'll check and reply
Title: Re: maths problem
Post by: nid404 on November 03, 2009, 03:56:17 pm
you missed a part of the question....as i said summin is missin :D

It said find x and y such that no bulbs are left over

then 17 and 21 is correct...caue it exactly divides 357....no remainders
Title: Re: maths problem
Post by: spookyjaransh on November 03, 2009, 03:57:27 pm
oh, ya sorry.......... :P :-[
Title: Re: maths problem
Post by: nid404 on November 03, 2009, 03:58:58 pm
Quote from: spookyjaransh on November 03, 2009, 03:57:27 pm
oh, ya sorry.......... :P :-[

it's k....so u got this right?
Title: Re: maths problem
Post by: spookyjaransh on November 03, 2009, 04:00:40 pm
ya, thanx.........i was trying to get the total of 257 :-[ :-[ :-[ :-[ :-[ (not intentionally, just a little worked out now...), so thats wat was sooo confusing........ :P :P :P :P :P :P
Title: Re: maths problem
Post by: holychalice on November 03, 2009, 04:02:28 pm
heyy .. so hw did u do ?? hit and trial ?? or by sum formula ...
can sum1 confirm this answer ... cuz i think it cud be done graphically
is this a correct method ??
Title: Re: maths problem
Post by: nid404 on November 03, 2009, 04:08:48 pm
no graph required......it has got to be using factors.

no graph for a 2mark question
Title: Re: maths problem
Post by: wassup on November 03, 2009, 04:09:30 pm
can ne1 explain me d b part???
Title: Re: maths problem
Post by: nid404 on November 03, 2009, 04:12:48 pm
square root of 357 is 18.89....but there can't be fraction of bulbs....In the square number of rows and bulbs will be same...hence u take root...hope u got that... So the answer would be 18...the nearest whole number lesser than the number...it can't be 19 cuz 192>357
Title: Re: maths problem
Post by: holychalice on November 03, 2009, 04:13:49 pm
square will be no of rows=no of bulbs to make a square
answer 18

18x18=324
19x19=361
Title: Re: maths problem
Post by: spookyjaransh on November 03, 2009, 04:15:50 pm
that one was easy, but the c part was fr which i had to use hit and trial....
Title: Re: maths problem
Post by: wassup on November 03, 2009, 05:01:15 pm
Quote from: nid404 on November 03, 2009, 04:12:48 pm
square root of 357 is 18.89....but there can't be fraction of bulbs....In the square number of rows and bulbs will be same...hence u take root...hope u got that... So the answer would be 18...the nearest whole number lesser than the number...it can't be 19 cuz 192>357

thanks