Tabular Data
Data Interpretation

 Back to Questions

Common Information

A health­drink 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.

Table below can be scrolled horizontally

Ingredient

Composition

Carbohydrate%

Protein%

Fat%

Minerals%

O

50

30

10

10

P

80

20

0

0

Q

10

30

50

10

R

5

50

40

5

S

45

50

0

5

Q.

Common Information Question: 4/4

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

 A.

2 : 1 : 3

 B.

4 : 1 : 2

 C.

2 : 1 : 4

 D.

3 : 1: 2

 E.

4 : 1 : 1

 Hide Ans

Solution:
Option(E) is correct

P, Q and S contain 80%, 10% and 45% carbohydrates.

To achieve 60% carbohydrates, proportion of P should be the maximum. Hence, options (A) and (C) are eliminated.

Option (B): Carbohydrate content:

$= ({320 + 10 + 90}/700)$

$= 420/700$

$= 60\%$

Cost per unit:

$ = ({200 + 200 + 200}/700)$

$= 6/7$

$= 0.857$

Option (D): Carbohydrate content:

$= ({240 + 10 + 90}/60)< 60%$

Option (E): Carbohydrate content:

$= (320 + 10 + 45)/600$

$= 62.5\%$

Cost per unit:

$= ({200 + 200 + 100}/600)$

$= 5/6$

$= 0. 833$

P, Q and S in the proportion 4 : 1 : 1 has the lowest costper unit.

Hence, option E.


(0) Comment(s)