Hi Rajat, 
It is given that $$V∝\ D^2$$ (When T is constant)
Hence, we can say that $$V=kT^xD^2$$ .... Eq(1) (Where kT^x is constant, and the power of T is unknown, which is x)
Similarly, It is given that $$V∝T\ $$ (When D is constant)
Hence, we can say that $$V=lD^yT$$ ...... Eq(2) (Where $$kD^y$$ is constant, and the power of D is unknown, which is y)
Similarly, we also know that Both T and D is related to V in a certain way.
Hence, $$V=mT^aD^b$$ .... Eq(3)
Multiplying Eq(1) with Eq(2), we get:
$$V^2\ =\ n\ D^{y+2}T^{x+1}$$ => $$V=nT^{\frac{\ x+1}{2}}D^{\ \frac{\ y+2}{2}}$$ ... Eq(4)
Comparing Eq(4) to Eq(3), we get:$$a=\ \frac{\ x+1}{2}$$, and $$b=\ \frac{\ y+2}{2}$$
Hence, Eq(3) can be rewritten as $$V=mT^{\ \frac{\ x+1}{2}}D^{\ \frac{\ y+2}{2}}$$
Comparing Eq(1) and Eq(3), we get:
x = (x+1)/2 => x = 1, and 2 = (y+2)/2 => y = 2
Hence, $$V\ =\ kTD^2$$


Or alternatively, 
Once we got Value = k(t)(d^2)
We are given that the d of the first coin is 4x and the d of the second coin is 3x, and we are also given that the value of the second coin is four times that of the first coin.
Dividing the first coin's value by the second coin's value, we would get:
 
$$\frac{4V_2}{V_2}=\frac{k\times\ 16x^2\times\ t_1}{k\times\ 9x^2\times\ t_2}$$
$$\frac{4V_2\times\ 9x^2\times\ k}{V_2\times\ k\times\ 16x^2}=\frac{\ t_1}{t_2}$$
$$\frac{4\times\ 9}{16}=\frac{\ t_1}{t_2}$$
$$\frac{t_1}{t_2}=\frac{9}{4}$$
Hope this would have helped. 
Please feel free to ask any further doubts you might have. 
Regards

Hi Dehati, we're not combining the two functions. You need to write the equation of 1 function and the equation of the other. 1 will be substituted in the other to help us solve it. It is not combined.

Value - v ; diameter - d ; Thickness - T
V is propotional to d sq when T is constant
V is propotional to T when d is constant
How can u say v is propotional to T and D with k as constant.
I cannot understand pls explain in detail

Hi,
It is given that $$V∝\ D^2$$ (When T is constant)
Hence, we can say that $$V=kT^xD^2$$  .... Eq(1) (Where kT^x is constant, and the power of T is unknown, which is x)
Similarly, It is given that $$V∝T\ $$ (When D is constant)
Hence, we can say that $$V=lD^yT$$ ...... Eq(2) (Where $$kD^y$$ is constant, and the power of D is unknown, which is y)
Similarly, we also know that Both T and D is related to V in a certain way.
Hence, $$V=mT^aD^b$$ .... Eq(3)
Multiplying Eq(1) with Eq(2), we get:
$$V^2\ =\ n\ D^{y+2}T^{x+1}$$  => $$V=nT^{\frac{\ x+1}{2}}D^{\ \frac{\ y+2}{2}}$$ ... Eq(4)
Comparing Eq(4) to Eq(3), we get:$$a=\ \frac{\ x+1}{2}$$, and $$b=\ \frac{\ y+2}{2}$$
Hence, Eq(3) can be rewritten as $$V=mT^{\ \frac{\ x+1}{2}}D^{\ \frac{\ y+2}{2}}$$
Comparing Eq(1) and Eq(3), we get:
x = (x+1)/2 => x = 1, and 2 = (y+2)/2 => y = 2
Hence, $$V\ =\ kTD^2$$
I hope this helps! Please let us know if the doubt still persists

Hi,

see last line of question "if the value of the first is four times that of the second"

Hi,
It is given in the question that "2 faulty weighing machines- A and B are used to measure grapes and raisins respectively."
Please check the solution once and let us know if the doubt still persists.