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Let the CP of 1 horse be Rs. x
Cp of 1 Cow be Rs. y
According to the First Condition
4x + 9y = 1390
i.e., 4x + 9y - 1340 = 0 (i)
According to the Second Condition
4*10x/100 + 9*20y/100 = 188
40x + 180y = 18800
2x + 9y - 940 = 0 (ii)
On solving i and ii, we get
x/(-8460 + 12060) = y/(-2680 + 3760) = 1/(36-18)
x/3600 = y/1080 = 1/18
=> x = 200, y = 60
i.e., the Cost of 1 horse = Rs.200
& cost of 1 cow = Rs.60
Cp of 1 Cow be Rs. y
According to the First Condition
4x + 9y = 1390
i.e., 4x + 9y - 1340 = 0 (i)
According to the Second Condition
4*10x/100 + 9*20y/100 = 188
40x + 180y = 18800
2x + 9y - 940 = 0 (ii)
On solving i and ii, we get
x/(-8460 + 12060) = y/(-2680 + 3760) = 1/(36-18)
x/3600 = y/1080 = 1/18
=> x = 200, y = 60
i.e., the Cost of 1 horse = Rs.200
& cost of 1 cow = Rs.60
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