Hello saberteeth Originally Posted by

**saberteeth** **1)** The income of A,B and C are in the ratio of 7:9:12 and their spending are in the ratio of 8:9:15. If A saves 1/4th of his income, then the savings of A, B,C are in the ratio of?

**2)** A diamond falls down and breaks into three pieces whose weights are in the ratio 2:3:5. The value of diamond is proportionate to the square of it's weight. If the value of the original diamond is $20,000, what is the loss in value due to breakage?

How do i solve these?

1) Consider a total income of £$\displaystyle 28\;(=7+9+12)$. Of this, A's income is £$\displaystyle 7$ and he saves $\displaystyle \frac14$ of it. So he saves £$\displaystyle \frac74$. He spends $\displaystyle \frac34$ of it; so he spends £$\displaystyle \frac{21}{4}$.

B's income is £$\displaystyle 9$, and his spend is $\displaystyle \frac98$ of A's spend. So he spends £$\displaystyle \frac98\times\frac{21}{4}=$ £$\displaystyle \frac{189}{32}$. Therefore he saves £$\displaystyle \left(9-\frac{189}{32}\right) =$ £$\displaystyle \frac{99}{32}$.

I'll leave you to work out C's spend and saving in the same way.

When you've done that, write the savings as a ratio $\displaystyle \frac{7}{4}:\frac{99}{32}: ...$, and simplify the result. I reckon the final answer is $\displaystyle 56:99:279$.

2) $\displaystyle 2 + 3 + 5 = 10$. So the weight of the smallest piece is $\displaystyle \frac{2}{10}=0.2$ of the weight of the original single diamond. So its value is the square of this fraction multiplied by the original value, which is: $$\displaystyle 0.2^2 \times 20,000 =$ $$\displaystyle 800$

Work out the value of the other two pieces in the same way. Add these values together, and subtract from $$\displaystyle 20,000$ to find the total loss in value. I reckon the answer is $$\displaystyle 12,400$.

Grandad