# Table 71

Directions (1 to 5): Answer the following questions based on the information given below.

An athletics coach was trying to make an energy drink which is best suited for the athletes. For this purpose, he took five of the best known energy drinks– A, B, C, D and E in the market with the idea of mixing them to get the result he desired. The table below gives the composition of these drinks. The cost of each of these energy drinks per litre is A – 150, B – 50, C – 200 D – 500 and E – 100.

 Energy Drink Composition Carbohydrate (%) Protein (%) Fat (%) Minerals (%) A 50 30 10 10 B 80 20 0 0 C 10 30 50 10 D 5 50 40 5 E 45 50 0 5

Q1.How many drinks contains at least 10% minerals and at least 30% protein.
1. 1
2. 2
3. 3
4. 4
5. 5

Q2. Which among the following is the combination having the lowest cost per unit for a drink having 10% fat and at least 30% proteins? The drink has to be formed by mixing two ingredients together.
1. B and C
2. B and E
3. B and D
4. E and C
5. D and E

Q3. In what proportion should B, C and E be mixed to make a drink 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

Q4. A drink containing 30% each of carbohydrate and protein, no more than 25% fat and at least 5% minerals is to be made. Which of the following two drinks must be mixed in equal quantities to obtain the required drink?
1. A and B
2. D and E
3. B and E
4. C and D
5. A and E

Q5. If a drink is prepared by mixing 1/5th of drink A, 10% of drink B, 20% of drink C, 1/10th of drink D and remaining drink E. What will be the cost of the drink prepared per unit?
1. 165
2. 185
3. 195
4. 145
5. 174

1.2

Drink A & C Contains more than or equals to 10% Minerals and 30% Protein.

2.2

By Mixing the following Drinks the composition of Fat and Protein will be,
B + C => Protein = (20+30)/2 = 25% & Fat = (0+50)/2 = 25%
=> Protein is less than 30%
B + E => Protein = (20+50)/2 = 35% & Fat = (0+0)/2 = 0%
=> Fat is less than 10%
B + D => Protein = (20+50)/2 = 35% & Fat = (0+40)/2 = 20%
Cost per unit = 50 + 500 = 550/2 = 275
E + C => Protein = (50+30)/2 = 40% & Fat = (0+50)/2 = 25%
Cost per unit = 100 + 200 = 300/2 = 150
D + E => Protein = (50+50)/2 = 50% & Fat = (40+0)/2 = 20%
Cost per unit = 500 + 100 = 600/2 = 300
Therefore, required mixture is C+E

3.5

Solution:From the given options:
If ratio is 2 : 1 : 3,
Cost = 2*50 + 1*500 + 3*100 = 900
Carbohydrate% = (2*80+1*10+3*45)/(2+1+3)=50.84%
If ratio is 4 : 1 : 2,
Cost = 4*50 + 1*500 + 2*100 = 900
Carbohydrate% = (4*80+1*10+2*45)/(4+1+2)=46.66%
If ratio is 2 : 1 : 4,
Cost = 2*50 + 1*500 + 4*100 = 1000
Carbohydrate% = (2*80+1*10+4*45)/(2+1+4)=50%
If ratio is 3 : 1 : 2,
Cost = 3*50 + 1*500 + 2*100 = 950
Carbohydrate% = (3*80+1*10+2*45)/(3+1+2)=56.6%
If ratio is 4 : 1 : 1,
Cost = 4*50 + 1*500 + 1*100 = 800
Carbohydrate% = (4*80+1*10+1*45)/(4+1+1)=62.5%
Therefore, require ratio is 4 : 1 : 1

4.5

 Products Comparative service cost in USA and Asian countries (in US Dollars) USA India Thailand Singapore Malaysia LCD TV 0 300 300 200 400 Home gym 0 500 400 500 500 Refrigerator 0 500 500 400 600 Air Conditioner 0 700 500 500 800 Washing machine 0 500 600 500 400 Music system 0 900 600 400 400 Digital camera 0 500 600 500 600

Required combination is A and E

5.1 