sum of three numbers is 60 such that twice the first, 3 times the second and 4 times the third together make 180. first number is 3 more than the second one. find the numbers
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Let the numbers be b+ 3, b and a
According to first condition
a + b + b + 3 = 60
=> a + 2b = 57
=> 4a + 8b = 228 -----(1)
According to second condition,
2 ( b + 3) + 3b + 4a = 180
=> 2b + 6 + 3b + 4a = 180
=> 4a + 5b = 174 -----(2)
On subtracting equation 2 from 1, we get
3b = 54
=> b = 18
Now,
On putting the value of b in equation 1, we get
=> a + 36 = 57
=> a = 21
First number = b + 3 = 18 + 3 = 21
Second number = 18
Third number = 21
According to first condition
a + b + b + 3 = 60
=> a + 2b = 57
=> 4a + 8b = 228 -----(1)
According to second condition,
2 ( b + 3) + 3b + 4a = 180
=> 2b + 6 + 3b + 4a = 180
=> 4a + 5b = 174 -----(2)
On subtracting equation 2 from 1, we get
3b = 54
=> b = 18
Now,
On putting the value of b in equation 1, we get
=> a + 36 = 57
=> a = 21
First number = b + 3 = 18 + 3 = 21
Second number = 18
Third number = 21
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