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Sagot :
a=Anne's cards
D= Devi's cards
d+24=a
3d/5=a/2
6d/5=a
6d/5=d+24
6d/5-d=24
6d-5d=120
d=120
a=144
D= Devi's cards
d+24=a
3d/5=a/2
6d/5=a
6d/5=d+24
6d/5-d=24
6d-5d=120
d=120
a=144
LET ANNE HAS X NUMBER OF CARDS
LET DEVI HAS Y NUMBER OF CARDS
X-Y= 24 ------EQ 1
3/5Y=1/2X
6Y=5X (cross multiplying)-------EQ 2
FROM EQ 1 WE GET
X=24+Y
substituting the value of x in eq 2, we get
6y=5(24+y)
6y =120+5y
y= 120
NOW, PUTTING THE VALUE OF Y IN EQ 1
X-120=24
X=24+120
X=144
THEREFORE, ANNE HAS 144 CARDS
LET DEVI HAS Y NUMBER OF CARDS
X-Y= 24 ------EQ 1
3/5Y=1/2X
6Y=5X (cross multiplying)-------EQ 2
FROM EQ 1 WE GET
X=24+Y
substituting the value of x in eq 2, we get
6y=5(24+y)
6y =120+5y
y= 120
NOW, PUTTING THE VALUE OF Y IN EQ 1
X-120=24
X=24+120
X=144
THEREFORE, ANNE HAS 144 CARDS
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