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Mathematics: Post your doubts here!

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lower bound for area =23.25*19.65
lower bound for perimeter=2(23.25+19.65)

about the question 12
edc=180 - 109 (angles between parallel lines add up to 180)
=71
FAB= ABC
sum of internal angles = 720
angle EDC = 109( parallel lines)
let fab = abc= x

2x+95+109+71+109=720
2x=720-384
x=168
Can u kindly tell me from where 720 came
 
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A = 23.3 B =19.7
Correct to 1 decimal Lower bound A1 = 23.3 -0.5 = 22.8 B1 = 19.7-0.5= 19.2

Lower bound for the permeter = 2 * A1 + 2 * B1 = 2 * 22.8 + 2 * 19.2 = 2 * 42 = 84 cm

Lower bound for the area = A1 * B1 = 22.8 * 19.2 = 437.76 = 437.8 cm
thank you so much :)
 
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Can u kindly tell me from where 720 came
well its the formula for the sum of internal angles where n is the number of sides
since the given diagram has six sides
(6-2)*180=720
the general formula is (n-2)*180
you may need it for some questions in p4 :)
 
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A = 23.3 B =19.7
Correct to 1 decimal Lower bound A1 = 23.3 -0.5 = 22.8 B1 = 19.7-0.5= 19.2

Lower bound for the permeter = 2 * A1 + 2 * B1 = 2 * 22.8 + 2 * 19.2 = 2 * 42 = 84 cm

Lower bound for the area = A1 * B1 = 22.8 * 19.2 = 437.76 = 437.8 cm
thank you so much :)
the accuracy is to 1 decimal place so = 0.1/2 = 0.05
Lower bound A1 = 23.3-0.05=23.25
Lower bound for b1=19.7-0.05=19.65
 
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plz kindly help me i need thses answers plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
May June 2008 question 10
May June 2009 question 3/6and 8b/11
May June 2006 question 22 AND 23
 
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plz kindly help me i need thses answers plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
May June 2008 question 10
May June 2009 question 3/6and 8b/11
May June 2006 question 22 AND 23
sorry i am busy right now i ll try to reply later if no one answers your post
 
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plz kindly help me i need thses answers plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
May June 2008 question 10
May June 2009 question 3/6and 8b/11
May June 2006 question 22 AND 23
 
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mm few simple questions is 0.5 x base x height is only for right angled triangles ? or for any triangle? and the inverse function o x^3 or 3^x ?? how do we get that :/
 
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plz kindly help me i need thses answers plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz
May June 2008 question 10
May June 2009 question 3/6and 8b/11
May June 2006 question 22 AND 23
 
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At 05 06 Mr Ho bought 850 fish at a fish market for $2.62 each.
95 minutes later he sold them all to a supermarket for $2.86 each.
(a) What was the time when he sold the fish?
(b) Calculate his total profit

In 2005 there were 9 million bicycles in Beijing, correct to the nearest million.
The average distance travelled by each bicycle in one day was 6.5 km correct to one decimal place.
Work out the upper bound for the total distance travelled by all the bicycles in one day

In January Sunanda changed £25 000 into dollars when the exchange rate was $1.96 = £1.
In June she changed the dollars back into pounds when the exchange rate was $1.75 = £1.
Calculate the profit she made, giving your answer in pounds (£).


A statue two metres high has a volume of five cubic metres.
A similar model of the statue has a height of four centimetres
(a) Calculate the volume of the model statue in cubic centimetres
(b) Write your answer to part (a) in cubic metres.


A car manufacturer sells a similar, scale model of one of its real cars.
The fuel tank of the real car has a volume of 64 litres and the fuel tank of the model has a
volume of 0.125 litres.
Show that the length of the real car is 8 times the length of the model car.
The area of the front window of the model is 0.0175 m2.
Find the area of the front window of the real car.
 
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92
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1
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8
At 05 06 Mr Ho bought 850 fish at a fish market for $2.62 each.
95 minutes later he sold them all to a supermarket for $2.86 each.
(a) What was the time when he sold the fish?
(b) Calculate his total profit

In 2005 there were 9 million bicycles in Beijing, correct to the nearest million.
The average distance travelled by each bicycle in one day was 6.5 km correct to one decimal place.
Work out the upper bound for the total distance travelled by all the bicycles in one day

In January Sunanda changed £25 000 into dollars when the exchange rate was $1.96 = £1.
In June she changed the dollars back into pounds when the exchange rate was $1.75 = £1.
Calculate the profit she made, giving your answer in pounds (£).


A statue two metres high has a volume of five cubic metres.
A similar model of the statue has a height of four centimetres
(a) Calculate the volume of the model statue in cubic centimetres
(b) Write your answer to part (a) in cubic metres.

lz kindly help me i need thses answers plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz







A car manufacturer sells a similar, scale model of one of its real cars.
The fuel tank of the real car has a volume of 64 litres and the fuel tank of the model has a
volume of 0.125 litres.
Show that the length of the real car is 8 times the length of the model car.
The area of the front window of the model is 0.0175 m2.
Find the area of the front window of the real car.
 
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f in verse is log(base10)X
take x as 1 answer will be zero
because anything with the power zero answer is 1


BUT HOW WE WILL WRITE IT IN EXAM BECAUSE ACCORDING TO MY TEACHER YOU CANNOT USE OR WRITE LOG IN EXAM B/C TS NOT INCLUDED IN IGCSE
 
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At 05 06 Mr Ho bought 850 fish at a fish market for $2.62 each.
95 minutes later he sold them all to a supermarket for $2.86 each.
(a) What was the time when he sold the fish?
(b) Calculate his total profit

In 2005 there were 9 million bicycles in Beijing, correct to the nearest million.
The average distance travelled by each bicycle in one day was 6.5 km correct to one decimal place.
Work out the upper bound for the total distance travelled by all the bicycles in one day

In January Sunanda changed £25 000 into dollars when the exchange rate was $1.96 = £1.
In June she changed the dollars back into pounds when the exchange rate was $1.75 = £1.
Calculate the profit she made, giving your answer in pounds (£).


A statue two metres high has a volume of five cubic metres.
A similar model of the statue has a height of four centimetres
(a) Calculate the volume of the model statue in cubic centimetres
(b) Write your answer to part (a) in cubic metres.


A car manufacturer sells a similar, scale model of one of its real cars.
The fuel tank of the real car has a volume of 64 litres and the fuel tank of the model has a
volume of 0.125 litres.
Show that the length of the real car is 8 times the length of the model car.
The area of the front window of the model is 0.0175 m2.
Find the area of the front window of the real car.
Q1:
a) 06 41
b) Total profit = difference/cost price * 100
= 0.24/2.62 * 100
= 9.16 %

Q2:
Absolute error of the number of bicycles = 1000000/2 = 500000
Upper bound of number of bicycles = 9000000 + 500000 = 9500000
Absolute error of distance traveled by 1 bicycle in one day = 0.1/2 = 0.05
Upper bound of of distance traveled by 1 bicycle in one day = 6.5 + 0.05 = 6.55

Upper bound of the total distance traveled by all the bicycles in one day = 9500000 * 6.55 = 62225000 km

Q3:
a) £1 = $1.96
£25000 = $49000
Later:
$1 = £(1/1.75)
$49000 = £[(1/1.75)* 49000]
$49000 = £28000

Profit = difference/initial cash * 100 = (28000-25000)/25000 * 100 = 12 %

Q4:

a) V of statue/V of model = (H of Statue/H of model)^3
(5*100*100*100)/Vm = (2*100/4)^3
Simplify:
Vm = (5000000)/(125000)
Vm = 40 cm^3
b)
40 cm^3 = ? m^3
1 cm^3 = (1/1000000) m^3
40 cm^3 = [(1/1000000)*40] m^3
40 cm^3 = 4*10^-5 m^3

Q5:
a)
Vs/Vl = (Ls/Ll)^3
(Vs/Vl)^(1/3) = Ls/Ll
Convert litres to cubic cm
(125/64000)^(1/3) = Ls/Ll
(5/40) = Ls/Ll

Ls:Ll = 5:40
Ls:Ll = 1 : 8
b)
(Vl/Vs)^(1/3) = (Al/As)^(1/2)
(6.4/0.0125)^(1/3) = (Al/0.0175)^(1/2) {notice that i changes the cm^2 to m^2}
Simplify:
Al = 1.12 m^2 :)
 
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