jolenekoh:

Appreciate your help and explanation. Or any drawings to show? Thank you


Mr. Ong needs 2 boys from the volleyball team to be the group captain and assistant captain. 2 boys, Raju and Alan, and 2 girls May and Jun, volunteer for the positions. How may ways are there for Mr. Ong to pair them if he wants a boy and a girl in the positions?

Rita sells chocolate, strawberry and vanilla soft yogurt. There also are three toppings available: candies, chocolate chips and mixed fruits. How many different combinations can she offers if they can choose one flavour soft yogurt with one of the toppings?

ANSWER for 2nd problem: 9 combinations.

We can draw a diagram to determine the various combinations as below:

http://i61.tinypic.com/il9a2o.jpg\">

http://i61.tinypic.com/il9a2o.jpg

Count the number of lines/branches for each flavour of yougurt and
then add up all to get answer of 9

Try the same for Problem 1 to get answer as 4 ways.

Detailed explanation for Tap A, B C problem sum:


[IMG]http://i57.tinypic.com/2qi6zjn.jpg[/IMG]

http://i57.tinypic.com/2qi6zjn.jpg

Hope that was easy to understand.

Q : Justin had 420 more trading cards than Angus. Angus gave away 3/5 of his trading cards and Justin gave away 5/6 of his trading cards. In the end, Angus and Justin had the same number of trading cards left. How many trading cards did Justin have at first ?

Answer:

[IMG]http://i61.tinypic.com/29416py.jpg[/IMG]

After giving away the cards,
Justin's 1/6 (cards left) is equal to Angus' 2/5 (cards left)
Therefore the end model looks as above.
From the model,
7U=420
1U=420/7=60
No. of cards Justin had at first = 12 x 60 = 720

http://i62.tinypic.com/scykcm.jpg\"> -please help me with this question

ANSWER:

AE = 1/3 x 12m = 4m
AE:AF:DG = 1:3:2

Since AE = 4m, we use this to determine the unknown multiplier for the ratio to get,
1 unit = 4m/1 = 4m

Therefore,
AF = 4m x 3 = 12m
DG = 4m x 2 = 8m
Therefore BF = 18m – 12m = 6m
CG = 18m – 8m = 10m

Compute the shaded triangles as follows:
Area of triangle AEW = 1/2 x 9m x 4m = 18 m2
Area of triangle CGW = 1/2 x 10m x 8m = 40m2
Area of triangle BFW = 1/2 x 6m x 4m = 12m2

Total shaded area = 18m2 + 40m2 + 12m2 = 70m2

Total area of rectangle ABCD = 18m x 12m = 216m2

Fraction of rectangle shaded = 70/216 = 35/108

Hi,


Can someone please help with the following question :-

Lynn and Jane had an equal amount of money. Each day, Lynn spent $12 of her money and Jane spent $3 more than Lynn.

(a) How many days have passed when Lynn had $148 left and Jane had $106 left ?

(b) How much money did Lynn have at first ?

Thank you for your kind assistance.

Please see the detailed answer below:

NOTE:
1. Both started with same amount of money.
2. The number of days they had spen the money for is also same for both.
3. Total money = Money spent + Money left

a) Let N be the unknown number of days they spent the money.

Total money Lynn had at first = N days X $12 + $148
Total money Jane had at first = N days X $15 + $106

Since they both had the same amount at first we equate the two statements and determine the number of days.

N days X $12 + $148 = N days X $15 + $106

Bring like terms together to get,

$148 - $106 = N x $15 - N x $12

$42 = N x $3

Solve for N to get,

N = $42/$3 = 14

14 days have passed when Lynn had $148 left and Jane had $106 left

b) Total money Lynn had at first= 14 days X $12 + $148 = $316

Please see the detailed answer below:


ΔCAB and ΔCDB are between two parallel lines

∴ Height of ΔCAB = Height of ΔCDB

And

ΔCAB and ΔCDB share a common base CB.
∴ Area of ΔCAB = Area of ΔCDB
∴ Area of ΔCDB= 480cm2

1. Area of shaded part ΔCAE= Area of ΔCAB – Area of ΔCEB
= 480cm2 – 189cm2
= 291cm2

2. Area of shaded part ΔBED= Area of ΔCDB – Area of ΔCEB
= 480cm2 – 189cm2
= 291cm2

∴ Area of total shaded part = Area of shaded part ΔCAE + Area of shaded part ΔBED
= 291cm2 + 291cm2
= 582cm2

Hope this helps.

Alternate method of solution for Question ‘c’ above:


Total no. of squares = 2 + 5+ 8 +11 + 14 + 17…….+2003 + 2006 + 2009 + 2012

We can write each term in the series as below:

Total no. of squares = 2 + (2+1x3) + (2+2x3) + (2+3x3) + (2+4x3) + (2+5x3)…. + (2+667 x 3) + (2+668 x 3) + (2+669x3) + (2+670x3)

There are 671 2’s in the above series. We separate the 2’s from the 3’s as below:

Total no. of squares = (2 x 671) + (1x3 + 2x3 + 3x…+667x3 + 668x3 + 669x3 + 670x3)

= (2 x 671) +3(1+2+3+…+667+668+669+670)
= (2 x 671) +3 x (670 x (670+1)/2)
=1342 + 674355
=675697

a) Circle ( simply repeat the sequence circle, square star to see that 10th column will be have circle)


b) Star (because if you write the number series for the column number of each shape, we see that the column numbers of the star shape are multiples of 3 while that for circle and square are not. Therefore since 2013 is a multiple of 3, it must have stars)

c) See the detailed solution below:

Write the column numbers for the squares. We get a series of numbers with constant difference of 3. The 1st number will be 2. Since the 2013rd column will be a column of stars (answer for question b explained above), the previous column of 2012 will be the last column of squares.

Therefore,
Number series for the squares: 2, 5, 8, 11, 14, 17…….2003, 2006, 2009, 2012

We need to find the sum of the series to determine the total number of squares in 2013 columns.

No. of terms = ((2012-2)/3) +1 = 671 wherein,
2 is first term,
2012 is last term and
3 is the common difference between two adjacent numbers in the series.

If we add the 1st term to 671st term, the 2nd term to 670th term, the 3rd term to 669th term etc., we get a constant sum of 2014. We can thus determine how many 2014 are there in the series and add them up to find the sum of the series.

Since we are pairing the numbers to make a sum of 2014, the total number of 2014 in the series will be 335 (671 divided by 2). But this leaves behind 1 extra term that cannot be paired (since 671 is odd number) whose value we need to know.

This extra term is the 336th term in the series since the first 335 terms are paired up with the last 335 terms.

The formula for the nth term is {a+(n-1)d} wherein:

a = 1st term (in our problem, a=2)
d = constant difference (in our problem, d=3)
n = no. of terms (in our problem n=336)

Therefore 336th term = 2+(336-1)x3 = 1007

Therefore we calculate the sum of the series as,

Sum of series = Total no. of squares in 2013 columns = 335 x 2014 + 1007 = 675697

Hope it was clear.