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ira [324]
3 years ago
11

three numbers are in the ratio of 4:5:6. if the sum of the largest and smallest exceeds the third number by 55 find the number​

Mathematics
2 answers:
lilavasa [31]3 years ago
7 0
  • Let numbers be 4x,5x,6x

ATQ

\\ \sf\longmapsto 4x+6x=5x+55

\\ \sf\longmapsto 10x=5x+55

\\ \sf\longmapsto 10x-5x=55

\\ \sf\longmapsto 5x=55

\\ \sf\longmapsto x=11

  • 5x=55
Ivenika [448]3 years ago
4 0

Answer:

Let us assume the ratio as 4x,5x and 6x

<em><u>A.T.Q.</u></em>

6x + 4x = 5x + 55

=> 10x = 5x + 55

=> 10x − 5x = 55

=> 5x = 55

=> x = \frac{55}{5}

x = 11

<em><u>Therefore</u></em><em><u>-</u></em><em><u>:</u></em>

First no. =4×11=44

Second no. =5×11=55

Third no. =6×11=66

Hence the three numbers are 44,55 and 66.

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