maths-THX..^^

a) let the selling prices of an apple and an orange be X and Y respectively,

8X + 5Y = 31 --- (1)
5X + 8Y = 34 --- (2)

(1) X 5 - (2) X 8 :
40X + 25Y - ( 40X + 64Y ) = 155 - 272

25Y - 64Y = -117

Y = 3
Therefore, the selling price of an orange is $3 and that of a apple is [ 31-5(3) ] / 8 = $2

b) Let the no. of apples and oranges used be P and Q respectively,

P + Q = 5 --- (1)
P(2) + Q(3) = 12 --- (2)
from (1), we know that P = 5 - Q

sub P = 5 - Q into (2),
(5 - Q)(2) + 3Q = 12
10 - 2Q + 3Q = 12
Q = 2
Therefore, the no. of oranges used is 2 and that of apples is (5 - 2)=3.

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a ) Ex . Appe = a ; Orange = b
8 a + 5 b = 31
5 a + 8 b = 34
( 5 a + 8 b ) - ( 8 a + 5 b ) = 3
3 b - 3 a = 3
b - a = 1
b > a 1
8 a + 5 b = 1 3 b - 8 = 31
13 b = 31 + 8
13 b = 39
b = 3
a < b 1 = 2

Apple = $ 2 , Orange =$ 3

b ) 3 Apples , 2 Oranges

Let x be the unit price of an orange and y be the unit price of an apple

(a)
5x+8y=31
x=(31-8y)/5 ---(1)

5y+8x=31+3=34
x=(34-5y)/8 ---(2)

(1) = (2), hence,

(31-8y)/5=(34-5y)/8
8(31-8y)=5(34-5y)
248-64y=170-25y
39y=78
y=2 ---(3)

Substitute (3) into (1):

x=(31-8y)/5
=(31-16)/5
=3

Selling price of an apple is $2 and selling price for an orange is $3

(b) Let a apples and b oranges are required to make a fruit salad

a+b= 5
a=5-b --- (4)

2a+3b=12 ----(5)

Substitute (4) into (5):

2(5-b) + 3b = 12
10-2b + 3b = 12
b=2---- (6)

Put (6) into (4):
a=5-2 = 3

Therefore, 3 apples and 2 oranges are required to make fruit salad

(a)Let the selling price of apple and orange be a and r respectively,
8a+5r=31-------(1)
5a+8r=34-------(2)
(2)-(1)
3r-3a=3
r-a=1
r=1+a--------(3)
Sub (3)in (1)
8a+5(1+a)=31
a=2
r=3
∴The selling price of apple and orange are $2 and $3 respectively
(b)It is easily obserable that not all of the 5 fruits are of the same kind as 2x5=10=/=12 and 3x5=15=/=12
By trial and error,
2x1+3x4=14=/=12
2x2+3x3=13=/=12
2x3+3x2=12
2x4+3x1=11=/=12
∴It is found that 3 apples and 2 oranges are used to make the fruit salad