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Re: The flowchart shows an algorithm that takes two positive integers as i [#permalink]
Abinash34 wrote:
Q1- m=(40+40)/2=40
Q2- m= 12+10=22


Can u explain how 12+10 part
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Re: The flowchart shows an algorithm that takes two positive integers as i [#permalink]
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Milli19 wrote:
Abinash34 wrote:
Q1- m=(40+40)/2=40
Q2- m= 12+10=22


Can u explain how 12+10 part



We are looking for intermediary value of n, that is the value of n that we found somewhere in middle of the process.

The value was m=x+10, but x itself changes depending on it being odd or even.

The value of x is 4, so even. x=3*4=12 and m=12+10=22.

Hope your query is addressed.
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Re: The flowchart shows an algorithm that takes two positive integers as i [#permalink]
1. ­If 5 and 8 are entered as values of x and y respectively, the output value m is [40]

x=5; y=8

x not even => x = 6 * 5 = 30

m = 30 + 10 = 40

y even => y = 5 * 8 = 40

m = \(\frac{(40+40)}{2}=40\)

2. If 4 and 3 are entered as values of x and y respectively, the intermediary value of m in the process is [22]

x=4; y=3

x even => x = 3 * 4 = 12

m = 12 + 10 = 22

Stop here since the question asks for the intermediary value of m­
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Re: The flowchart shows an algorithm that takes two positive integers as i [#permalink]
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