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jhick285
17.10.2021 •
Mathematics
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
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Ответ:
ATQ
![\\ \sf\longmapsto x=11](/tpl/images/2510/1636/d86d0.png)
5x=55Ответ:
Let us assume the ratio as 4x,5x and 6x
A.T.Q.
6x + 4x = 5x + 55
=> 10x = 5x + 55
=> 10x − 5x = 55
=> 5x = 55
=> x =![\frac{55}{5}](/tpl/images/2510/1636/1407b.png)
x = 11
Therefore-:
First no. =4×11=44
Second no. =5×11=55
Third no. =6×11=66
Hence the three numbers are 44,55 and 66.
Ответ:
The theory of personality
Explanation: