O-Level Additional Math

[jieheng]

Michaelia0816:

4. Rashid had sweets of 3 different flavours. There were 99 strawberry flavoured, 42 orange flavoured and 141 lemon flavoured sweets. He packed an equal number of each flavour into small packet but 1 strawberry and 1 lemon flavoured sweets were left out.
(a) (i) What was the maximum number of packets?
(ii) How many sweets were there in each small packets?

(b) If all sweets were included,
(i) what was the maximum number of packets
(Ii) how many sweets were there in each packet

4)

Strawberry 98 = 2*7*7
Orange 42 = 2*7*3
Lemon 140 = 2*7*10

HCF = 2*7 =14

a)
i) the max no of packets = 14

ii) No of sweets in each small packets = 7+3+10 = 20

For part b , you can try yourself

[Michaelia0816]

4. Rashid had sweets of 3 different flavours. There were 99 strawberry flavoured, 42 orange flavoured and 141 lemon flavoured sweets. He packed an equal number of each flavour into small packet but 1 strawberry and 1 lemon flavoured sweets were left out.


(a) (i) What was the maximum number of packets?
(ii) How many sweets were there in each small packets?

(b) If all sweets were included,
(i) what was the maximum number of packets
(Ii) how many sweets were there in each packet

Help me!

[jieheng]

Michaelia0816:

4. Rashid had sweets of 3 different flavours. There were 99 strawberry flavoured, 42 orange flavoured and 141 lemon flavoured sweets. He packed an equal number of each flavour into small packet but 1 strawberry and 1 lemon flavoured sweets were left out.


(a) (i) What was the maximum number of packets?
(ii) How many sweets were there in each small packets?

(b) If all sweets were included,
(i) what was the maximum number of packets
(Ii) how many sweets were there in each packet

Help me!

4)

Strawberry 98 = 2*7*7
Orange 42 = 2*7*3
Lemon 140 = 2*7*10

HCF = 2*7 =14

a)
i) the max no of packets = 14

ii) No of sweets in each small packets = 7+3+10 = 20


Strawberry 99 = 3^2*11
Orange 42 = 2*3*7
Lemon 141 = 3*47

HCF =3

b)

i) the max no of packets = 3

ii) No of sweets in each small packets = 3*11+2*7+47 = 94

[jieheng]

quekcc:

jieheng:

[quote=\"Michaelia0816\"]4. Rashid had sweets of 3 different flavours. There were 99 strawberry flavoured, 42 orange flavoured and 141 lemon flavoured sweets. He packed an equal number of each flavour into small packet but 1 strawberry and 1 lemon flavoured sweets were left out.


(a) (i) What was the maximum number of packets?
(ii) How many sweets were there in each small packets?

(b) If all sweets were included,
(i) what was the maximum number of packets
(Ii) how many sweets were there in each packet

Help me!

4)

Strawberry 98 = 2*7*7
Orange 42 = 2*7*3
Lemon 140 = 2*7*10

HCF = 2*7 =14

a)
i) the max no of packets = 7+3+10 = 20

ii) No of sweets in each small packets = 14

Strawberry 99 = 3^2*11
Orange 42 = 2*3*7
Lemon 141 = 3*47

HCF =3

b)

i) the max no of packets = 3*11+2*7+47 = 94

ii) No of sweets in each small packets = 3

Hi, your answer for (i) and (ii) should exchange.

The HCF will give you the max number of packets for both cases.

a) i) max no of packets = 2 x 7 = 14
ii) no sweets in each small packet = 7 + 3 + 10 = 20
b) i) max no of packets 3
ii) no sweets in each small packet = 3 x 11 + 2 x 7 + 47 = 94[/quote]Hi ,

You are right . Thanks for pointing out the error.

[LOL123]

pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

[Augmum]

LOL123:

pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

Post on behalf on my gal....

(a) subst x = -1 into the equation
y = 1 - 2(-1)
y = 3
Substx = 3 into the equation
y = 1 - 2(3)
y = -5

Thus, the range of f is : { y: -5 < y ≤ 3}

(b) qn means that: find the range of values of x such that f(x) = 0
1 -2x = 0
2x = 1
x = 0.5

1 - 2x = 5
2 x = -4
x = - 2

{x : -2 < x ≤ 0.5 }

Note: The domain of the function refers to the x value while the range of the function refers to the y value.

[LOL123]

Augmum:

LOL123:

pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

Post on behalf on my gal....

(a) subst x = -1 into the equation
y = 1 - 2(-1)
y = 3
Substx = 3 into the equation
y = 1 - 2(3)
y = -5

Thus, the range of f is : { y: -5 < y ≤ 3}

(b) qn means that: find the range of values of x such that f(x) = 0
1 -2x = 0
2x = 1
x = 0.5

1 - 2x = 5
2 x = -4
x = - 2

{x : -2 < x ≤ 0.5 }

Note: The domain of the function refers to the x value while the range of the function refers to the y value.

:thankyou:

i have another question

The diagram shows the graph of the curve y=4/x-1 for x>1

With the aid of the graph http://i47.tinypic.com/2w6w4sk.jpg\"> , find

(a) the range of f(x)=4/x-1 for the domain 2<x<5,
(b) the domain of f defined by f(x)=4/x-1 corresponding to the range 1<f(x)<3.

[Augmum]

LOL123:

Augmum:

[quote=\"LOL123\"]pls help me


The function f is defined by f(x)=1-2x.
(a) Find the range of f corresponding to the domain -1≤x<3
(b) Find the domain of f corresponding to the range 0≤f(x)<5

thanks

Post on behalf on my gal....

(a) subst x = -1 into the equation
y = 1 - 2(-1)
y = 3
Substx = 3 into the equation
y = 1 - 2(3)
y = -5

Thus, the range of f is : { y: -5 < y ≤ 3}

(b) qn means that: find the range of values of x such that f(x) = 0
1 -2x = 0
2x = 1
x = 0.5

1 - 2x = 5
2 x = -4
x = - 2

{x : -2 < x ≤ 0.5 }

Note: The domain of the function refers to the x value while the range of the function refers to the y value.

:thankyou:

i have another question

The diagram shows the graph of the curve y=4/x-1 for x>1

With the aid of the graph http://i47.tinypic.com/2w6w4sk.jpg\"> , find

(a) the range of f(x)=4/x-1 for the domain 2<x<5,
(b) the domain of f defined by f(x)=4/x-1 corresponding to the range 1<f(x)<3.[/quote](a) sub in x= 2
y = 4/2-1
y = 4

Sub in x= 5
y= 4/5-1
y= 1

Range of f(x) : { 1<f(x)<4 }

(b) since f(x) = 4/x-1
And y= f(x),
So, y= 4/x-1
1 < f(x)
1 < 4/x-1
Multiply x-1 on both sides...
x-1 < 4
x < 5

f (x)
So, 4/x-1
Multiply x-1 on both sides...
4 < 3(x-1)
4 < 3x-3
4+3 < 3x
7/3 < x

As x<5 and 7/3<x, the domain is {x: 7/3<x<5}

Hope there is no typo error....

Btw, LOL123, which sec level are u in now ? Is this topic being taught currently ?

[Michaelia0816]

Ty jieheng

[LOL123]

Thank you so much! I have 1 last question


The arrow diagram shows part of the mapping h(x)=12/ax+b, x≠-b/a
Calculate the values of a and b. Hence find

(i) the element that has an image of -2 under this mapping,
(ii) the values of x for which h(x)=x

I am in sec 2 this year. and yup, i am currently doing functions http://i48.tinypic.com/25kq4hs.jpg\">