find the percentage of type X in the new blend, given that

[Chikis]

A blend was made from types of tea, X and Y, so that the mixure contains 20% of type X. If the quantity of X is doubled, find the percentage of type X in the new blend, given that;

(i) the same quantity of Y was used;

(ii) the total quantity of the mixure was unaltered.

 

How can I tackle this?

Edited October 28, 2014 by Chikis

[imatfaal]

Chikis - you have asked a lot of questions and prompted some good threads in which you and others will have learnt; but I think you need to start to make a few guesses of your own about how to start these problems. The way you deal with the questions once given a start leaves me in no doubt that you can do this well if you put your mind to it.

 

No one is going to mock or criticise you on this forum for trying and failing - so put a few ideas out there and the members will help out. Even though I said you should start I cannot resist a hint; write down what you know from the question in algebraic form, and then attempt to create a new formula for what you need to answer

[Chikis]

I think I have come up with something.

From the first sentence, we know that the blend = x + y

 

When the mixure contains 20% of x, the percentage of y in the blend was 80%.

 

When the quantity of x is doubled, the percentage of x in the new blend becomes 40%.

.

(i) if the same quantity of y was used the percentage x = 40%

 

(ii) the total quantity of the mixure = 40 + 80 = 120%, [/math]\therefore[math] x remains 40%

 

Am also thinking that in the first sentence

x/5 + 4y/5 = [math]\frac{20x}{100}+\frac{80y}{100}

The ratio of x to y is 20 : 80

= 1 : 4

 

 

When x is doubled, we have 2x/5 + 4y/5 = [\math]\frac{40x}{100}+\frac{80y}{100}[math]

the ratio of x to y becomes 1 : 2

[Chikis]

I think I have come up with something.

From the first sentence, we know that the blend = x + y

When the mixure contains 20% of x, the percentage of y in the blend was 80%.

When the quantity of x is doubled, the percentage of x in the new blend becomes 40%.

.

(i) if the same quantity of y was used the percentage x = 40%

(ii) the total quantity of the mixure = 40 + 80 = 120%, [math]\therefore[/math] x remains 40%

Am also thinking that in the first sentence

x/5 + 4y/5 = [math]\frac{20x}{100}+\frac{80y}{100}[/math]

The ratio of x to y is 20 : 80

= 1 : 4

When x is doubled, we have 2x/5 + 4y/5 = [math]\frac{40x}{100}+\frac{80y}{100}[/math]

the ratio of x to y becomes 1 : 2

Edited November 8, 2014 by Chikis

[Chikis]

Chikis - you have asked a lot of questions and prompted some good threads in which you and others will have learnt; but I think you need to start to make a few guesses of your own about how to start these problems. The way you deal with the questions once given a start leaves me in no doubt that you can do this well if you put your mind to it.

 

No one is going to mock or criticise you on this forum for trying and failing - so put a few ideas out there and the members will help out. Even though I said you should start I cannot resist a hint; write down what you know from the question in algebraic form, and then attempt to create a new formula for what you need to answer

I think I have tried my best. Could you please come to my aid now?

[LaurieAG]

A blend was made from types of tea, X and Y, so that the mixure contains 20% of type X. If the quantity of X is doubled, find the percentage of type X in the new blend, given that;

(i) the same quantity of Y was used;

(ii) the total quantity of the mixure was unaltered.

 

How can I tackle this?

 

(i) 20 x + 80 y + 20 x gives x = 40, y = 80, total = 120, x = 1/3 of total and y = 2/3 of total

 

(ii) 20 x + (80 - 20) y + 20 x gives x = 40, y = 60, total = 100, x = 2/5 of total and y = 3/5 of total