The ratio of incomes of a and b is 3:4 and ratio of their expenditure is 4:5 . what can be their possible ratio of savings 9:10 or 3:4 or 4:5 or 13:20.
I don't find anything suitable in the options I think the answer is 2:3 but it has to be from the choices.
[Math] finding the ratio when two other ratios are given
ratio
Related Solutions
It depends on what you want to do. Below I formalize the problem and show some related estimation theory results. You have $n$ samples of data as ordered pairs $$(X_1, Y_1), (X_2, Y_2), ..., (X_n, Y_n)$$ where $$(X_i,Y_i) = (\mbox{area of sample $i$ (square foot)},\mbox{savings of sample $i$ (gallons)})$$ The $X_i$ value may be correlated with the $Y_i$ value. Assume $X_i>0$ for all $i \in \{1, 2, 3, ...\}$. You are considering these metrics:
Ratio of averages: $$ R_n = \frac{\sum_{i=1}^n Y_i}{\sum_{i=1}^n X_i} = \frac{\frac{1}{n}\sum_{i=1}^n Y_i}{\frac{1}{n}\sum_{i=1}^n X_i} $$ If $\{(X_i,Y_i)\}_{i=1}^{\infty}$ are i.i.d. vectors (with possibly correlated $X_i, Y_i$ values) then $R_n\rightarrow \frac{E[Y_1]}{E[X_1]}$ with prob 1.
Average of ratios: $$ M_n = \frac{1}{n}\sum_{i=1}^n \frac{Y_i}{X_i} $$ If $\{(X_i,Y_i)\}_{i=1}^{\infty}$ are i.i.d. vectors then $M_n\rightarrow E[\frac{Y_1}{X_1}]$ with prob 1.
The $R_n$ value represents an aggregate ratio (total area of all samples)/(total gallons over all samples). The $M_n$ value is an average of individual ratios.
Suppose a new (independent) person has an area $X$ and wants to predict his/her savings $Y$ using your data via some prediction $\hat{Y}$ that is a function of $X$ (where the function is formed from your data). Assume $(X,Y)$ has the same distribution as your data vectors $(X_i,Y_i)$, but independent of it. Also assume we have i.i.d. data sample vectors. Let's use simple linear predictors of the form $\hat{Y} = aX + b$.
Predictor $\hat{Y} = R_n X$. To understand performance, suppose $n$ is large so $R_n \approx \frac{E[Y]}{E[X]}$. Then $$\hat{Y} \approx \frac{E[Y]}{E[X]}X \implies E[\hat{Y}] \approx \frac{E[Y]}{E[X]}E[X] = E[Y]$$ and so this predictor is "approximately unbiased," meaning that $E[\hat{Y}]\approx E[Y]$. For the mean-square error we have \begin{align} &(\hat{Y}-Y)^2 \approx \left(\frac{E[Y]}{E[X]}X - Y\right)^2 \\ &\implies E[(\hat{Y}-Y)^2] \approx \frac{E[Y]^2E[X^2]}{E[X]^2} + E[Y^2] - \frac{2E[XY]E[Y]}{E[X]} \end{align} and you can approximate the statistics $E[Y^2], E[X^2], E[XY]$ via \begin{align} E[X^2] &\approx \frac{1}{n}\sum_{i=1}^n X_i^2\\ E[Y^2] &\approx \frac{1}{n}\sum_{i=1}^n Y_i^2\\ E[XY] &\approx \frac{1}{n}\sum_{i=1}^n X_iY_i \end{align}
Predictor $\hat{Y} = M_n X$. Suppose $n$ is large so $M_n \approx E[\frac{Y}{X}]$. Then $$ \hat{Y} \approx E[\frac{Y}{X}]X \implies E[\hat{Y}] \approx E[\frac{Y}{X}]E[X]$$ and so this is a biased estimator. You can calculate the mean square error in the same way and then compare.
Best linear estimator. Consider a predictor of the form $\hat{Y} = aX$. Using standard estimation theory, we want to optimize the coefficient $a$ to get the smallest mean-square error $E[(\hat{Y}-Y)^2]$. We have $$ E[(\hat{Y}-Y)^2] = E[(aX-Y)^2] = a^2E[X^2] -2aE[XY] + E[Y^2]$$ Taking $\partial/\partial a = 0$ gives $$ 2aE[X^2] - 2E[XY] =0 \implies a^* = \frac{E[XY]}{E[X^2]}$$ so the best linear predictor is: $$ \hat{Y} = \left(\frac{E[XY]}{E[X^2]}\right)X$$ and you can approximate the statistics $E[XY]$ and $E[X^2]$ from the data samples $(X_i,Y_i)$ as above. This introduces a third metric $Z_n$ that can be compared to $R_n$ and $M_n$: $$ \boxed{Z_n = \frac{\frac{1}{n}\sum_{i=1}^n X_iY_i}{\frac{1}{n}\sum_{i=1}^nX_i^2} \approx \frac{E[XY]}{E[X^2]}}$$
Best affine estimator. Consider predictors of the form $\hat{Y} = aX + b$. Optimizing the coefficients $a$ and $b$ gives: \begin{align} E[(aX+b-Y)^2] = a^2E[X^2] + b^2 + E[Y^2] + 2abE[X] - 2aE[XY] - 2bE[Y] \end{align} Taking $\partial/\partial a = \partial/\partial b = 0$ gives \begin{align} 2aE[X^2] + 2bE[X] -2E[XY]&=0 \\ 2b + 2aE[X] - 2E[Y] &= 0 \end{align} from which I get (assuming $Var(X)>0$ so $E[X^2]>E[X]^2$): $$ \boxed{a^* = \frac{E[XY]-E[X]E[Y]}{E[X^2]-E[X]^2} = \frac{Cov(X,Y)}{Var(X)}}$$ $$ \boxed{b^* = E[Y] - a^*E[X]}$$ So this estimator $\hat{Y} = a^*X + b^*$ is unbiased and has the smallest mean square error over all predictors of the form $\hat{Y} = aX+b$. The result $Cov(X,Y)/Var(X)$ is quite famous. You can compute approximate values $\tilde{a}^*$, $\tilde{b}^*$ by approximating the statistics $E[X^2], E[XY]$ from the data samples $(X_i,Y_i)$ as described above.
If I compare both of them physically, they are not equivalent
If you compare which physically?
because 80 Ml larger than 8 Ml
There's no reason to compare the $80$ Ml to the $8$ Ml. $80$ Ml is $\frac 83$ times bigger than $30$ Ml. So they are in ratio of $8:3$. And $8$ Ml is $\frac 83$ times bigger than $3$ Ml. So they are in ratio of $8:3$. And the Pacific Ocean at $704,000,000$ cubic kilometers is $\frac 83$ times bigger than the Indian Ocean at $264,000,000$ cubic kilometers. So there are in ration of $8:3$.
A ratio compares the sizes of two different things in proportion to each other. The absolute size doesn't matter.
If you are trying to compare $80$ ml to $8$ ml they are in $10:1$ proportional. And that is the same proportion that $30$ ml is to $3$ ml.
If you compare the oil to vinegar there is always $\frac 83$ more oil than vinegar no matter what size your recipe is.
And if you are comparing the two different recipes: the bigger recipe is $10$ times bigger than the smaller recipe. So the bigg recipe will have $10$ times as much oil, or $10$ times as much vinegar or $10$ times as many eggs, etc.
and even tho It's amazing that they are equivalent.
Actually it's very dull and mundane and would be very weird if they weren't.
Best Answer
We can scale everything in terms of A's income, so call that $1$. B's income is then $\frac 43$. Let A's expenses be $x$, which makes B's expenses $\frac 54x$. A's savings are $1-x$, while B's are $\frac 43-\frac 54x$, giving a ratio of $\frac {1-x}{\frac 43-\frac 54x}=\frac {12-12x}{16-15x}$. We are now asked which of the given ratios can match this fraction. If $x=0$, so there are no expenses, the ratio is $3:4$. I think you are supposed to think this is not possible. As $x$ increases toward $1$, the ratio decreases toward zero. The only choice that is less than $3:4$ is $13:20$ so that must be the intended answer.
However, if I solve $\frac {12-12x}{16-15x}=\frac 9{10}$ I get a perfectly good answer of $x=\frac 85$. In that cases both peoples expenses exceed their income. A is saving $-\frac 35$ and B is saving $-\frac 23$, which gives the proper ratio. We cannot get a ratio of $\frac 45$, but can get as close as we want if $x$ gets very large.