It is exactly like
(x+y+80)/3= [(y+z+80)/3 + 6 ]
Solving RHS
(x+y+80)/3= (y+z+80+18)/3
how did you get 18 when the 3 is still present ?


i simply did it like multiply both side by 3 to eliminate 3
then subtract 80 from both sides
then subtract y from both sides you will get x=z+6

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My dear, it is [(y+z+80)/3]+ 6. It is not (y+z+80+6)/3
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Why?
Because average of three numbers y, z and 80 would be simply y+z+80/3
.hence "6 more than average of y,z,80 would be.
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[(y+z+80)/3]+ 6.
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For example 4 more than average of 2,4,6 would be
(2+4+6/3 + 4)
= 12/3 +4
= 4+4=8

It would no be (2+4+6+4)/3= 16/3= 15.33 No
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And simplifying [(y+z+80)/3]+ 6 , take 3 as LCM, 6 would be multifplied with 3 and then 3 would be common divider and hence as a common divider of a full side of an equation can be cancelled with other side's 3. But with out making it under the 6, we can not do it.
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how did you get 18 when the 3 is still present ?

I got 18 becuase 3 was not under 6 actually. It was under other 3 (y,z,80)
i simply did it like multiply both side by 3 to eliminate 3
You can not do this because 3 was uder half equation of a side. You have to make 3 a common divisor.

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sood hota kyaa hay? zyaan yani asal say zada. when you give money to someone on sood (interest), he returns with zyaan and you get it without any of your own struggle. soo zyaan is mutradif. but faeda is its opposite. you get faeda (profit) when you exert yourself in some business etc. Without exerting no profit. Actually in my opinion if you see it through Islamic principle, it will be clear to you.

Does any one have any idea when FIA Test dated 7 sep results will be announced.

ooh meray bhayi, even i said that it is not y+z+80+6/3

lets solve it ur way
taking lcm ;
(3x+3y+240)/3 = (3y+3z+240+18)/3
multiplying both sides by 3 will cancel 3s' on both sides ?
we will have ;
3x+3y+240 = 3y+3z+240+18
by simplifying further eliminate 3y and 240
we will have 3x = 3z + 18
taking 3z to the left ;
we will have 3x - 3z = 18
taking 3 as common on both sides ;
3(x-z)=3(6)
divide both sides by 3
x-z = 6
hence proved ?

The average of five consecutive even numbers is 38. The largest number among them is?

What was the answer for the following analogy

As stubborn as mule::

Dear brother, Again you are doing the same mistake. RHS means (3y+3z+240+18)/3 is right. But on the LHS side , 3x+3y+240)/3 is not right. Why are you taking LCM of LHS. It is already LCM 3.
Start your calculation from (x+y+80)/3 instead of (3x+3y+240)/3..
Why you have multiplied x, y and 80 with 3 when they already have common divisor 3.
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But on the other side, as 3 was not under 18, you multiplied it right way. Because it was (y+z+80)/3 + 18.... And 3 was not common divisor. While LHS already had 3 as common divisor so calculations will be start from
(x+y+80)/3= (3y+3z+240+18)/3

There is no point of multiplying with 3 . It is a very simple question.

Answer is 38

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Solving method
Lets first even number= x
Then second consecutive even number= x+2
3rd consecutive even number= x+4
4th consecutive even number= x+6
5th (highest) consecutive even number= x+8.
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Now as per statement
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[x+(x+2)+(x+4)+(x+6)+(x+8)]/5= 38 // (dividing 5 because total number are 5)

x+(x+2)+(x+4)+(x+6)+(x+8)= 38*5
x+(x+2)+(x+4)+(x+6)+(x+8)= 190

5x+ 20=190
5x=170
x= 34
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So first number is 34
Hence highest number 34+8= 42
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Verification
Five consecutive numbers are; 34, 36, 38, 40, 42

Add them and take average 34+36+38+40+42= 190
Average=190/5
= 38

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The examiner asks "what is the highest number".The highest number among them is 42 not 38.

*u can't suppose x as even number, as x could be of any value i.e, 1,2,3,4 etc.
*u can't suppose x+2 as even number because if x=1, then number=3 and 3 is not an even number.

You are right. Highest number is 42....Due to typing mistake, I wrote the average one... thank you very much dear, for correction.

Yes, x can be any number 1,2,3 or even -0.00007. But putting that into equation as per the given condition will give you the only one and real answer. As it has given here, and has been verified and proved. No other number, 1, 3, or 9.99999 fulfils the criteria that was told in question.

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