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The sum of four consecutive even numbers is 107 more than the sum of three consecutive odd numbers. If the sum of smallest odd number and the smallest even number is 55. What is the smallest even number ?
- A36
- B40
- C32
- D38
Answer not set
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CCommunity
Solution
Let the four consecutive even numbers are = $x , (x+2) , (x+4) , (x+6)$
and three consecutive odd numbers = $y , (y+2) , (y+4)$
Acc to ques,
=> $x + y = 55$ ------------Eqn(1)
and $[(x) + (x+2) + (x+4) + (x+6)] - [(y) + (y+2) + (y+4)] = 107$
=> $(4x + 12) - (3y + 6) = 107$
=> $4x - 3y = 107 - 6 = 101$ -----------Eqn(2)
Multiplying eqn(1) by 3 and add it to eqn(2), we get :
=> $3x + 4x = 55 \times 3 + 101$
=> $7x = 165 + 101 = 266$
=> $x = \frac{266}{7} = 38$
Let the four consecutive even numbers are = $x , (x+2) , (x+4) , (x+6)$
and three consecutive odd numbers = $y , (y+2) , (y+4)$
Acc to ques,
=> $x + y = 55$ ------------Eqn(1)
and $[(x) + (x+2) + (x+4) + (x+6)] - [(y) + (y+2) + (y+4)] = 107$
=> $(4x + 12) - (3y + 6) = 107$
=> $4x - 3y = 107 - 6 = 101$ -----------Eqn(2)
Multiplying eqn(1) by 3 and add it to eqn(2), we get :
=> $3x + 4x = 55 \times 3 + 101$
=> $7x = 165 + 101 = 266$
=> $x = \frac{266}{7} = 38$
0
CCommunity
38