At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
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
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.