Question

A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

Answer 1

Let's denote the prices of the cups as: 
- Small size = \(s\) 
- Medium size = \(m\) 
- Large size = \(l\)

Given: 
1) \(s \times m \times l = 800\)
2) \(s : m = 2 : 5\) or \(m = \frac{5}{2} s\)

Substituting the value of \(m\) in the first equation: 

\(s \times \frac{5}{2} s \times l = 800\)

\(⇒\) \(5s^2l = 1600\)

\(⇒\) \(s^2l = 320\) -----(i) 

Now, when the prices of the smallest and the medium ones are increased by 6: 
New price of small size = \(s + 6\)
New price of medium size = \(m + 6\)

Given: 
\((s + 6) \times (m + 6) \times l = 3200\)

Substituting the value of \(m\): 
\((s + 6) \times (\frac{5}{2} s + 6) \times l = 3200\)

Expanding and rearranging: 
\(5s^2l + 15s^2 + 5s^2 + 15sl + 12l = 3200\) 

From equation (i): 
\(s^2l = 320\)
\(⇒ \)\(5s^2l = 1600\)

Thus: \(30sl + 12l = 1600\)
\(⇒ \)\(l(30s + 12) = 1600\)
\(⇒ \) \(l(30s + 12) = 2 \times 800\)

From this, since \(s^2l = 320\) is constant from the first scenario, the only way \(l(30s + 12)\) could become twice of 800 is if \(l\) is halved.  

So, the new \(l = \frac{l}{2}\)

Old \(l = 2 \times \frac{l}{2} = l\)

Now, using \(s \times m \times l = 800\): 

\(s \times \frac{5}{2} s \times l = 800\)
 \(⇒\) \(5s^2l = 1600\)

Given \(s^2l = 320\): 

\(⇒\)\(s^2 = \frac{320}{l}\)

\(⇒\) \(5s^2 = \frac{1600}{l}\)

From which \(l = 5\)

Substituting in the ratio of \(s : m\) : 
\(s = 10\) and \(m = 25\)
 Sum of the original prices =\(s + m + l = 10 + 25 + 5 = 40\)

So, the sum of the original prices of the three different sizes is INR 40.

Answer 2

Let the little cup's price be two times, the medium's five times, and the large's y. 
According to condition 1, we now have \(2x × 5x × y = 800\), which gives us \(x ^2 y = 80\). 
Using condition 2, we now get\( (2x+6)× (5x+6) y = 3200. \)

Subtracting (2) from (1) yields \(\frac{{(2x + 6) \times (5x + 6)}}{{x^2}} = 40\)
We have \( 10 x^ 2 + 42 x + 36 = 40 x ^2, \)and \(30 x ^2 − 42 x − 36 = 0.\)
 \(5x ^2 −7x−6=0 \)
\(x=2 \) is what we obtain. 
Thus, \(2x=4 \) and \(5x= 10.\)
After changing in (1), we obtain \(y = 20\). Thus, sum =\( 4+10+20 = 34\) at this point.