# Question

40%J + 60%R = 470
80%J + 120%R = 940

60%J + 40%R = 600
180%J + 120%R = 1800

180%J – 100%J = 1800 – 940
100%J = 860
40%J = 344
(470 – 32) / 2 = 219
470 – 344 = 126
219 – 126 = 93 as the answer.

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part 1 of the solution: tough needs to solve simultaneous eq…

JM- Red:Blue = 2u:3u

Rudy- Red:Blue =3p:2p

thus 2u + 3p = 470 —(1)

3u+2p = 600 —-(2)

(1) +(2): 5u +5p = 1070

u +p = 214

2u +2p =428—(3)

(2) – (1): 1u =42

therefore 1p = 172

JM has 344 red marbles.

Rudy has 126 red marbles.

part 2 of the solution – easy

total =600 red marbles

2 units = 600 -32 =438

1 unit = 219

rudy has 219 red marbles in the end

red marbles given to her 219 – 126=93

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You may want to try using symbols or “visual algebra” to represent the unknowns.  Some examples here:

cheers

This is Matrix question.

480 × 20% = 96 children,  384 adult

380 × 70% = 266 children, 114 adult

Total = 362 children, 498 adult

 Adult Children Total Group X 498 – 1u 362 – 1u 860 – 2u Group Y 1u 1u 2u 498 362 860

(498 – 1u) / 60 = (362 – 1u) / 40

1992 – 4u = 2172 – 6u

6u – 4u = 2172 – 1992

2u = 180

860 – 180 = 680

680 – 480 = 200

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This is a Matrix question.

 Red Blue Total Jie Min 40u  = 344 60u 100u Rudy 470 – 40u 600  – 60u 470 600

(470 – 40u ) / 6 = (600 – 60u) / 4

1880 – 160u = 3600 – 360u

360u – 160u = 3600 – 1880

200u = 1720

1u = 1720 ÷ 200 = 8.6

40u = 40 × 8.6 = 344 (JM red)

470 – 344 = 126 (Rudy red)

344 – 1u = 126 + 1u + 32

344 – 158 = 1u + 1u

2u = 186

1u = 186 ÷ 2 = 93

Do you have a simpler method. This method is too long and hard to understand for my boy.

Yes. This method is for advance level question.

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Let x = jiemin red marbles

y = rudy red marbles

a = jiemin blue marbles

b = rudy blue marbles

x+y = 470 red marbles in total, equation(1)
a+b = 600 blue marbles in total, equation(2)

as 40% of jiemin’s marbles are red, 60 % are blue,

0.4x = 0.6a

as 60% of rudy’s marbles are red, 40% are blue,

0.6y = 0.4b

Thus two equations can be derived,

1. a = 2/3x

2. b= 3/2y

Sub both equations a and b into equation(2),

a+b = 600

2/3x + 3/2y = 600, equation(3)

Let x be the subject in equation(1),

x+y = 470

x = 470 -y

Sub equation x in to equation(3),

2/3(470 – y) + 3/2y = 600

5y = 1720

y=344

Therefore X = 470 – 344 = 126

Let number of red marbles given to Rudy by jiemin by C,

(y-c) – (x+c) = 32, equation(4)

Sub both x and y into equation(4)

(344-c) – (126+c) = 32
218-2c=32
2c=186
c=93

Therefore 93 red marbles were given to Rudy by Jiemin