CAT 2025 Lesson : Interpretative Tables - Case 7

Details for Questions 24 to 27 are provided below

A health-drink company’s R&D department is trying to make various diet formulations, which can be used for certain specific purposes. It is considering a choice of 5 alternative ingredients (O, P, Q, R, and S), which can be used in different proportions in the formulations. The table below gives the composition of these ingredients. The cost per unit of each of these ingredients is O: 150, P: 50, Q: 200, R: 500, S: 100.


[CAT 2007]

24) The company is planning to launch a balanced diet required for growth needs of adolescent children. This diet must contain at least 30% each of carbohydrate and protein, no more than 25% fat and at least 5% minerals. Which one of the following combinations of equally mixed ingredients is feasible?

(1) O and P
(2) R and S
(3) P and S
(4) Q and R
(5) O and S

25) For a recuperating patient, the doctor recommended a diet containing 10% minerals and at least 30% protein. In how many different ways can we prepare this diet by mixing at least two ingredients?

(1) One
(2) Two
(3) Three
(4) Four
(5) None

26) Which among the following is the formulation having the lowest cost per unit for a diet having 10% fat and at least 30% protein? (The diet has to be formed by mixing two ingredients.)

(1) P and Q
(2) P and S
(3) P and R
(4) Q and S
(5) R and S

27) In what proportion should P, Q and S be mixed to make a diet having at least 60% carbohydrate at the lowest cost per unit?

(1) 2 : 1 : 3
(2) 4 : 1 : 2
(3) 2 : 1 : 4
(4) 3 : 1 : 2
(5) 4 : 1 : 1

Solution

24)

Since the ingredients are mixed equally, we can take the simple average of the ingredient. Alternatively, we can check whether the ingredients adds up to twice the given condition e.g., the diet should have at least 30% carbohydrate which means $\dfrac{a + b}{2} \geq 30 \%$ or $a + b \geq 60 \%$



Based on the above table, only option (5) satisfies all the conditions.

Answer: (5) O and S

25)

The diet should contain 10% minerals. From the table only two ingredients contain 10% minerals, ingredient O and Q. Other ingredients have less than 10% minerals. Also, ingredients O and Q have 30% protein, so if we mix ingredient O and Q in any ratio the diet will definitely have 30% protein. Hence this is the only way possible.

Answer: (1) One

26)

Since the diet should have 10% fat, we can eliminate option (2) (P and S) and option (3) (P and R). This is because the fat percentage in these ingredients is less than or equal to 10%, so if we mix these ingredient in any ratio the fat percent will be less than 10%.

We can eliminate option (1) (P and Q), as the diet should contain at least 30% protein. The protein percentage in ingredients P and Q is 20% and 30% respectively, so if we mix these ingredients in any ratio the protein percentage will be between 20% and 30%.

Considering option (4) (Q and S)

⇒ To get 10% fat, let's take x and y as the ratio in which the ingredients Q and S are to be mixed respectively.

Fat % in Q = 50% , S = 0%

⇒ $\dfrac{x(50 \%) + y(0 \%)}{x + y} = 10 \%$
⇒ $50x = 10x + 10y$
⇒ $40x = 10y$, $\dfrac{x}{y} = \dfrac{1}{4}$

Protein % in Q = 50%, S = 50%

Protein % in (Q and S) = $\dfrac{1(30 \%) + 4(50 \%)}{1 + 4} = \dfrac{230 \%}{5} = 46 \%$, protein % is atleast 30%.

Cost per unit of (Q and S) = $\dfrac{1(200) + 4(100)}{1 + 4} = \dfrac{600 \%}{5} = 120$

Considering option (5) (R and S)

⇒ To get 10% fat, let's take x and y as the ratio in which the ingredients R and S are to be mixed respectively.

Fat % in R = 40%, S = 0%

⇒ $\dfrac{x(40 \%) + y(0 \%)}{x + y} = 10 \%$
⇒ $40x = 10x + 10y$
⇒ $30x = 10y$, $\dfrac{x}{y} = \dfrac{1}{3}$

Protein % in R = 50%, S = 50%
Protein % in (R and S) = $\dfrac{1(50 \%) + 3(50 \%)}{1 + 3} = \dfrac{200 \%}{4} = 50 \%$, protein % is atleast 30%.

Cost per unit of (R and S) = $\dfrac{1(200) + 4(100)}{1 + 4} = \dfrac{600}{5} = 120$

As the cost per unit of Q and S is less than R and S, we can select option (4).

Answer: (4) Q and S

27)

Let us form a table for the options.



From the above table, only options (2) and (5) satisfy the carbohydrate condition. Now we have to check for the lowest cost per unit for these two options.

Cost per unit of option (2) = $\dfrac{4(50 + 1(200) + 2(100)}{4 + 1 + 2} = \dfrac{600}{7} = 85.7$

Cost per unit of option (5) = $\dfrac{4(50) + 1(200) + 1(100)}{4 + 1 + 1} = \dfrac{500}{6} = 83.33$

Answer: (5) 4:1:1

Answer:
24) (5) O and S
25) (1) One
26) (4) Q and S
27) (5) 4 : 1 : 1