Plz help me to solve this 2 sums..!!

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1\. If a:b= 3:4 and c:d= 2:3 then find a(cube)* c(square) + b(cube)* d(square) divided by a*b(square)* d(square) +a(square)*b*c*d 2\. In a three digits number the tens digit is the average of the other two digits. The ratio of the number for...
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plz can u solve this 2 sum step by step if u dt mind

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hi chakshu06
for 1ques. divide the numerator and the denominator by b(cube)
also divide the numerator and denominator by d(square)
simplify and u'll get..
(a/b)^3 + 1
_________________
(a/b)^2 + (a/b)*(c/d)

substitute the values.... and u'll get 19/18.

for ques 2.
see let the three digit number be abc
it's given tens digit, i.e b=(a+c)/2

also,
10b+c 10a+b
_____ = _____
b+c a+b

put b=(a+c)/2

u'll get a=c
hence
numbers can be of the form ...
111
222
333
444
555
666
777
888
999

Correct me, if i am wrong :)

Only Time will tell.
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ans 1) 19/18

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1. If a:b= 3:4 and c:d= 2:3 then find a(cube)* c(square) + b(cube)* d(square) divided by a*b(square)* d(square) +a(square)*b*c*d


2. In a three digits number the tens digit is the average of the other two digits. The ratio of the number formed by its first two digits and their sum equal the ratio of the number formed by its last two digits and their sum. How many three digit numbers satisfy this condition.
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