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[MathIzzzFun]

2DMommy:

Need help with these two percentage questions :


1) The ratio of Alan's total marks to that of Benji's total marks in their Mid-Year Examination was 4 : 5 . Alan's marks increased by 50% in the End-of-Year Examination. By what percentage must Benij's marks be increased so that their total marks would be equal ?

2) Timothy had a box of strawberry sweets and pineapple sweets. When he added in 30 strawberry sweets, 20% of the sweets were pineapple sweets. When he added in another 100 pineapple sweets, 60% of the sweets were pineapple sweets. How many strawberry sweets were there in the container at first ?\t\t\t\t\t\t\t\t\t

Thanks !!

Hi

1) Alan:Benji = 4u:5u
Alan's marks increased by 50% = 4u x 150% = 6u
So, in order for Benji to get the same marks, percentage increase
= (6u-5u)/5u x 100% = 20%

2) After 30 strawberry sweets added,
Strawberry sweets: Pineapple sweets= 80% : 20% = 4u : 1u

When another 100 pineapple sweets added,
Strawberry sweets : Pineapple sweets = 40% : 60% =4u : 6u

so 6u-1u = 5u = 100, 1u = 20

Number of strawberry sweets after 30 added = 4u = 80
so, number of strawberry sweets at first = 80 - 30 = 50

cheers.

[2DMommy]

MathIzzzFun:

2DMommy:

Need help with these two percentage questions :


1) The ratio of Alan's total marks to that of Benji's total marks in their Mid-Year Examination was 4 : 5 . Alan's marks increased by 50% in the End-of-Year Examination. By what percentage must Benij's marks be increased so that their total marks would be equal ?

2) Timothy had a box of strawberry sweets and pineapple sweets. When he added in 30 strawberry sweets, 20% of the sweets were pineapple sweets. When he added in another 100 pineapple sweets, 60% of the sweets were pineapple sweets. How many strawberry sweets were there in the container at first ?\t\t\t\t\t\t\t\t\t

Thanks !!

Hi

1) Alan:Benji = 4u:5u
Alan's marks increased by 50% = 4u x 150% = 6u
So, in order for Benji to get the same marks, percentage increase
= (6u-5u)/5u x 100% = 20%

2) After 30 strawberry sweets added,
Strawberry sweets: Pineapple sweets= 80% : 20% = 4u : 1u

When another 100 pineapple sweets added,
Strawberry sweets : Pineapple sweets = 40% : 60% =4u : 6u

so 6u-1u = 5u = 100, 1u = 20

Number of strawberry sweets after 30 added = 4u = 80
so, number of strawberry sweets at first = 80 - 30 = 50

cheers.

Thank you so much !

[Neat]

Hi! Morning,


Please help to solve the following question:

There were 2/9 as many red beads as blue beads in a bag. When some blue beads were taken out and an equal number of red beads were added into the bag, the number of red beads became 1/4 of the total number of beads in the bag. If there were 176 beads in the bag at first, how many blue beads were taken out from the bag?

Thanks you in advance.

[MathIzzzFun]

Neat:

Hi! Morning,


Please help to solve the following question:

There were 2/9 as many red beads as blue beads in a bag. When some blue beads were taken out and an equal number of red beads were added into the bag, the number of red beads became 1/4 of the total number of beads in the bag. If there were 176 beads in the bag at first, how many blue beads were taken out from the bag?

Thanks you in advance.

Hi

The total number of beads remained unchanged.

Initially, number of blue beads = 9/(2+9) x 176 = 144
In the end, number of blue beads = 3/4 x 176 = 132
Number of blue beads taken out = 144 - 132 = 12

cheers.

[Andrew Lee]

Adam Ma:

May I know what level of class this problem comes from?

It from Red Swastika,P6,2010,Prelim

[Andrew Lee]

MadScientist:

Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

[Andrew Lee]

Andrew Lee:

MadScientist:

Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

Can help me solve Q20 :?
http://i53.tinypic.com/2ld831i.jpg\">

[whywhy]

Tin A contains some 20-cents coins and Tin B contains some 50-cents coins.

There are 9 more coins in Tin A than in Tin B.
The amount of money in Tin A is $2.70 less than that in Tin B.
How many 50-cents coins are there in Tin B ?

Please help, Thank u

[Andrew Lee]

Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm[/quote]

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:[/quote]

Can help me solve this question, Q3 ? tks :?
http://i55.tinypic.com/2jg0x2c.jpg\">

[Andrew Lee]

Andrew Lee:

Andrew Lee:

[quote=\"MadScientist\"]Here goes...


A) [ 5 / (5+9) ] = 35.71% of the circle is shaded.

B) 168 * 35.71% = 60 sqcm
Since AE = EB, Triangle is an isoceles triangle (ie. two equal sides)
And shaded part is 40% of triangle, therefore, 60 sqcm is 40% of triangle.
Area of triangle DEC = (100/40) x 60 = 150 sqcm

But the answer for Q15a is 20% is shaded & 15b is 75sqcm :slapshead:

[/quote]