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Author Topic: IGCSE MATHS Doubts  (Read 130661 times)

Offline haris94

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Re: IGCSE MATHS Doubts
« Reply #375 on: May 11, 2010, 03:10:22 pm »
Quote from: astarmathsandphysics on May 11, 2010, 02:59:07 pm
(15.9128-15.2978)/15.2978*100
(new-original)/original*100%

i tried that but the ms says

(15.9128-15.2978)/15.9128*100    ???   ???
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Offline haris94

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Re: IGCSE MATHS Doubts
« Reply #376 on: May 11, 2010, 03:15:54 pm »
Quote from: Fastal on May 11, 2010, 03:00:18 pm
percentage decrease=old value-newvalue/oldvalue*100

so here

%=0.615/15.9128*100
thus % decrease=3.86%

u r saying percentage decrease=old value-newvalue/oldvalue*100

the old value is 15.2978
so shouldnt it be
(15.9128-15.2978)/15.2978*100

the ms answer is the same as yours
but the formula disagrees with the answer
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Offline astarmathsandphysics

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Re: IGCSE MATHS Doubts
« Reply #377 on: May 11, 2010, 03:16:50 pm »
Then ms is wrong
alwys divide by original
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Offline haris94

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Re: IGCSE MATHS Doubts
« Reply #378 on: May 11, 2010, 03:24:26 pm »
ok Thanks
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Offline Malak

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Re: IGCSE MATHS Doubts
« Reply #379 on: May 11, 2010, 03:26:03 pm »
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Offline Adzel

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Re: IGCSE MATHS Doubts
« Reply #380 on: May 11, 2010, 03:29:44 pm »
this one please..
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Offline PeaceMaker

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Re: IGCSE MATHS Doubts
« Reply #381 on: May 11, 2010, 03:32:45 pm »
Quote from: Adzel on May 11, 2010, 03:29:44 pm
this one please..

I would love to know part (a) these questions always come and i always get confuse by them so, anyone care to explain?
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Offline Vin

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Re: IGCSE MATHS Doubts
« Reply #382 on: May 11, 2010, 03:36:41 pm »
Q 20)

once u kno the area beneath 8-10 which is 19.4

A (trapezium) = 1/2 * ( sum of parallel sides ) * h

so 19.4 = 1/2 ( x + 12.4 ) * 2

19.4 = x + 12.4

x = 19.4 - 12.4

x= 7
  
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Offline Adzel

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Re: IGCSE MATHS Doubts
« Reply #383 on: May 11, 2010, 03:38:51 pm »
Quote from: PeaceMaker on May 11, 2010, 03:32:45 pm
I would love to know part (a) these questions always come and i always get confuse by them so, anyone care to explain?

Easy...just find the upper bound and the lower bound, as it said to the nearest km i.e. 1/2 = 0.5

So, 22 - 0.5 = 21.5 (lower bound)
     22 + 0.5 = 22.5 (upper bound)

    i.e. 21.5 < d < 22.5

....but im only concerned about part (b)
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Offline Malak

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Re: IGCSE MATHS Doubts
« Reply #384 on: May 11, 2010, 03:42:09 pm »
Quote from: Adzel on May 11, 2010, 03:38:51 pm
Easy...just find the upper bound and the lower bound, as it said to the nearest km i.e. 1/2 = 0.5

So, 22 - 0.5 = 21.5 (lower bound)
     22 + 0.5 = 22.5 (upper bound)

    i.e. 21.5 < d < 22.5

....but im only concerned about part (b)

Its least amount...so just multiply ur lower bound with 5 to get the distance for 5 days n thn multiply with the amount of money...
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Offline haris94

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Re: IGCSE MATHS Doubts
« Reply #385 on: May 11, 2010, 03:44:22 pm »
@adzel

lower bound=22-0.5 = 21.5km

1 day he walks 21.5km
5 days = 21.5*5 = 107.5km

1 km = $1.6
107.5km = 1.6*107.5 = $172
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Offline Adzel

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Re: IGCSE MATHS Doubts
« Reply #386 on: May 11, 2010, 03:44:49 pm »
Quote from: ~VIN1094~ on May 11, 2010, 03:36:41 pm
Q 20)

once u kno the area beneath 8-10 which is 19.4

A (trapezium) = 1/2 * ( sum of parallel sides ) * h

so 19.4 = 1/2 ( x + 12.4 ) * 2

19.4 = x + 12.4

x = 19.4 - 12.4

x= 7
  

how did you get 19.4???
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Offline diva_sparkles

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Re: IGCSE MATHS Doubts
« Reply #387 on: May 11, 2010, 03:46:10 pm »
Quote from: haris94 on May 11, 2010, 03:15:54 pm
u r saying percentage decrease=old value-newvalue/oldvalue*100

the old value is 15.2978
so shouldnt it be
(15.9128-15.2978)/15.2978*100

the ms answer is the same as yours
but the formula disagrees with the answer

it shld be (15.9128-15.2978)/15.2978*100 cos it it says in 2006 it wz 15.9128 n in 2007 it wz 15.2978 so ms is rite!!! :)

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Offline Adzel

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Re: IGCSE MATHS Doubts
« Reply #388 on: May 11, 2010, 03:46:25 pm »
Thanx haris94 and ang3l
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Offline haris94

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Re: IGCSE MATHS Doubts
« Reply #389 on: May 11, 2010, 03:46:36 pm »
Quote from: Adzel on May 11, 2010, 03:44:49 pm
how did you get 19.4???

total distance - distance in first 8s
so
100m-80.6
=19.4m
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