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 By: Rohana (offline)  Monday, June 17 2013 @ 12:59 AM CDT (Read 699 times)
Rohana

Hi there, can someone help me with this problem.

Q1. A group of friends met for a tennis game. Each of them played with everyone else. Ann played
With four times as many girls as boys. Bob played with five times as many girls as boys.
(a). How many people were there altogether ?
(b). We're there more boys or girls? How many more?

Junior

Registered: 01/24/12
Posts: 18

 By: echeewh (offline)  Monday, June 17 2013 @ 12:04 PM CDT
echeewh

Hi Rohana,

Following pls find my worked solution:

<A played>

G : B
4 : 1

So, G = 4p + 1; B = 1p

<B played>

G : B
5 : 1

So, G = 5u; B = 1u + 1

Since the number of girls (G) and boys (B ) are unchanged, we have ...

4p + 1 = 5u
1p = 1u + 1

Rearranging these equations,

5u - 4p = 1 -- (1)
1p - 1u = 1 -- (2)

Eliminate part (p) qty,

(2)x4:
4p - 4u = 4 -- (3)

(1)+(3):
5u - 4u = 1 + 4
1u --> 5

(a)
Total (B + G) = 6u + 1
6u + 1 --> (6 × 5) + 1 = 31

(b)
G = 5u --> 5 × 5 = 25
B = 1u + 1 --> 5 + 1 = 6
G - B = 25 - 6 = 19

There were 19 more Girls (G)

===========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward.

Active Member

Registered: 04/21/11
Posts: 627

 By: echeewh (offline)  Monday, June 24 2013 @ 06:59 PM CDT
echeewh

Hey there,

Following is an <Alternative method> to solving the problem sum given.

<A played - 1G played>

G : B ** Total
4 : 1 *** 5 ** -- (1)

<B played - 1B played>

G : B ** Total
5 : 1 *** 6 ** -- (2)

Given that A, B played with everyone else, the total number of girls (G) and boys (B ) each played is the same. Hence , the <Unchanged Total> concept ...

(1)x6:
G : B
24 : 6 -- (3)

(2)x5:
G : B
25 : 5 -- (4)

(a)
In (1) ...
Total (G + B ) = 24 + 6 + 1 = 31

(b)
In (1) ...
G = 24 + 1 = 25
G - B = 25 - 6 = 19

Hence there were 19 more Girls (G).

=========

Trust this helps.

Thank you and kind regards,
Edward

Active Member

Registered: 04/21/11
Posts: 627

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