Algebra

Polynomials

Medium

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:

Entered answer:

✅ Correct Answer: 34

Since the smallest and medium sizes are in the ratio $2:5$ , we'll use:

Price of smallest cup $= 2x$

Price of medium cup $= 5x$

Price of largest cup $= p$

When two quantities are in a ratio, expressing them as multiples of a common variable makes the math much cleaner.

The product of all three prices equals 800:

$2x \times 5x \times p = 800$

$10x^2 \times p = 800$

$p = \tfrac{800}{10x^2} = \tfrac{80}{x^2}$

We've expressed the largest cup's price in terms of $x$ . This will help us solve the problem with just one unknown.

After increasing the smallest and medium prices by INR 6:

New smallest price $= 2x + 6$

New medium price $= 5x + 6$

Largest price remains $= p$

The new product equals $3200$ :

$(2x + 6)(5x + 6) \times p = 3200$

$\Rightarrow (10x^2 + 42x + 36) \times p = 3200$

Substituting $p = \tfrac{80}{x^2}$ from our first condition:

$(10x^2 + 42x + 36) \times \tfrac{80}{x^2} = 3200$

Dividing both sides by 80:

$\tfrac{10x^2 + 42x + 36}{x^2} = \tfrac{3200}{80} = 40$

$10x^2 + 42x + 36 = 40x^2$

$40x^2 - 10x^2 - 42x - 36 = 0$

$30x^2 - 42x - 36 = 0$

Dividing by 6:

$5x^2 - 7x - 6 = 0$

To factor $5x^2 - 7x - 6 = 0$ , we need two numbers that multiply to give $(5)(-6) = -30$ and add to give $-7$ .

These numbers are $-10$ and $+3$ (since $-10 \times 3 = -30$ and $-10 + 3 = -7$ ).

$5x^2 - 10x + 3x - 6 = 0$

$5x(x - 2) + 3(x - 2) = 0$

$(5x + 3)(x - 2) = 0$

This gives us: $x = 2$ or $x = -\tfrac{3}{5}$

Since prices must be positive, $x = 2$ .

With $x = 2$ :

Smallest cup price $= 2x = 2(2) = 4$ INR

Medium cup price $= 5x = 5(2) = 10$ INR

Largest cup price $= p = \tfrac{80}{x^2} = \tfrac{80}{4} = 20$ INR

Sum of original prices $= 4 + 10 + 20 = 34$

CAT 2022 Slot 1

Let $a, b, c$ be non-zero real numbers such that $b^2 < 4ac$ , and $f(x) = ax^2 + bx + c$ . If the set S consists of all integers m such that $f(m) < 0$ , then the set S must necessarily be

CAT 2020 Slot 2

Let $f(x)=x^{2}+a x+b$ and $g(x)=f(x+1)-f(x-1)$ . If $f(x) \geq 0$ for all real $x$ , and $g(20)=72$ , then the smallest possible value of $b$ is

CAT 2017 Slot 1

If $f_{1}(x)=x^{2}+11 x+n$ and $f_{2}(x)=x$ , then the largest positive integer $n$ for which the equation $f_{1}(x)=f_{2}(x)$ has two distinct real roots, is

CAT 2022 Slot 3

Suppose $k$ is any integer such that the equation $2 x^{2}+k x+5=0$ has no real roots and the equation $x^{2}+(k-5) x+1=0$ has two distinct real roots for $x$ . Then, the number of possible values of $k$ is

CAT 2022 Slot 3

If $(3+2\sqrt{2})$ is a root of the equation $ax^2 + bx + c = 0$ , and $(4+2\sqrt{3})$ is a root of the equation $ay^2 + my + n = 0$ , where $a, b, c, m$ and $n$ are integers,
then the value of $(\frac{b}{m} + \frac{c-2b}{n})$ is

2025 Slot 2

The equations $3x^2 - 5x + p = 0$ and $2x^2 - 2x + q = 0$ have one common root. The sum of the other roots of these two equations is

Solution

Related questions:

Keyboard Shortcuts

CAT Past Year Questions (Topic-Wise)