A number is divided into two parts in such a way that 30% of first part is 25 more than the 20% of second part. 50% of second part is 33.5 more than the 60% of first part. What is the number?

Let the number be $$100(x+y)$$ such that both the parts are $$100x$$ and $$100y$$

According to ques,

=> $$(\frac{30}{100}\times100x)-(\frac{20}{100}\times100y)=25$$

=> $$30x-20y=25$$

=> $$6x-4y=5$$ -----------(i)

Similarly, $$50y-60x=33.5$$

=> $$10y-12x=6.7$$ -------------(ii)

Multiplying first equation by '2' and adding both equations,

=> $$10y-8y=6.7+10$$

=> $$2y=16.7$$

=> $$y=\frac{16.7}{2}=8.35$$

Similarly, $$6x=5+4(8.35)=38.4$$

=> $$x=\frac{38.4}{6}=6.4$$

$$\therefore$$ Number = $$100(6.4+8.35)$$

= $$100\times14.75=1475$$

=> Ans - (A)