MBA Questions of the day (27 Sep’13)


0

 Choose the appropriate alternative :

 

51. A company has a job to prepare certain number cans and there are three machines A, B & C for this job. A can complete the job in 3 days, B can complete the job in 4 days and C can complete the job in 6 days. How many days the company will take to complete the job if all the machines are used simultaneously?

(a) 4 days

(b) 4\3 days

(c) 3 days

(d) 12 days

 

52. n3 is odd. Which of the following statement(s) is(are) true.

I.   n is odd

II.  n2 is odd

III. n2 is even

(a) I only

(b) II only

(c) I and II only

(d) I and IIIonly

 

 

Q53 & 54: Production pattern for number of units (in cubic feet) per day.

 

Day

1

2

3

4

5

6

7

No. of units

150

180

120

250

160

120

150

 

For a truck that can carry 2000 cubic feet, hiring cost per day is Rs.1000. Storing cost per cubic feet is Rs.5 per day.

 

53. If all the units should be sent to the market, on which days should the trucks be hired to minimize the cost?

(a) 2nd , 4th , 6th , 7th

(b) 7th

(c) 2nd , 4th , 5th , 7th

(d) None of these.

 

54. If storage cost reduced to Rs.0.8 per cubic feet per day, then

(a) 4th

(b) 7th

(c) 4th and 7th

(d) None of these

 

55. One bacteria splits into eight bacteria of the next generation. But due to environment only 50% of one generation can produce the next generation. If the seventh generation number is 4096 million, what is the number in first generation?

(a) 1 million

(b) 2 million

(c) 4 million

(d) 8 million

 

 

Answers:

 

51 (b)
52 (c)
53 (c)
54 (b)
55 (a)

51. In one day A would do 1/3rd of the job, B would do 1/4th of the job and C would do 1/6th of the job. Hence if all three of them work simultaneously, in one day they would do (1/3 + ¼ +1/6) = 3/4th of the job. Hence to complete the entire job together they would take 4/3 days.

52. If n3 is odd then n should also be odd. Hence n should also be odd. And n2 will again be odd and not even. So only I and II are true.

53. We can see that there are two types of cost : (i) Transportation cost (viz. hiring truck) and (ii) storing cost. It can be seen that the daily production is far less than the capacity of the truck. So a truck can be hired to carry multiple days production at one go. So as long as the storing cost is less than the cost of hiring the truck (i.e. Rs.1000), it makes sense to store the production. When the storing cost exceeds Rs.1000, it is best that the entire lot be sent to the market. The cost pattern is as given in the following table:

Units Produced

Units to be stored

Cost of storing

Cost of sending to the market

Should you hire truck?

Cost Incurred

1

150

150

Rs.750

Rs.1000

No

Rs.750

2

180

(150+180) = 330

Rs.1650

Rs.1000

Yes

Rs.1000

3

120

120

Rs.600

Rs.1000

No

Rs.600

4

250

(120+250) = 370

Rs.1850

Rs.1000

Yes

Rs.1000

5

160

160

Rs.800

Rs.1000

No

Rs.800

6

120

(160+120) = 280

Rs.1400

Rs.1000

Yes

Rs.1000

7

150

150

Rs.750

Rs.1000

Yes*

Rs.1000

TOTAL COST

Rs.5900

* In spite of the fact that storing is cheaper than hiring truck on the last day, we have to with the latter option because everything that is manufacture has to sent to the market.

So according to this table if the truck is hired on 2nd , 4th , 6th and 7th day, total cost = Rs.5900. But is this the most cost effective scheme? It can be seen that we are hiring truck on two consecutive days (6th & 7th ). Hence since everything that is manufacture has to sent to the market, we have yet another option of hiring the truck on the 5th day and sending the 6th & 7th days production together on the last day. In that case the cost on 5th day would be Rs.1000 (i.e. Rs.200 more than the present cost), the cost on the 6th day would be (120×5) = Rs.600 (i.e. Rs.400 less than the present cost) and the cost on the 7th day would be Rs.1000 (The same as the present cost). Hence we can that the total cost would actually come down by (+200 – 400= – 200) Rs.200. Hence this becomes the most cost effective scheme. So we should hire trucks on 2nd , 4th , 5th and 7th day.

 

 

54. If the storage cost is reduced to Rs.0.8 per cubic feet then the cost pattern is as given in the following table :

Units Produced

Units to be stored

Cost of storing

Cost of sending to the market

Should you hire truck?

Cost Incurred

1

150

150

Rs.120

Rs.1000

No

Rs.120

2

180

(150+180) = 330

Rs.264

Rs.1000

No

Rs.264

3

120

(330+120) = 450

Rs.360

Rs.1000

No

Rs.360

4

250

(450+250) = 700

Rs.560

Rs.1000

No

Rs.560

5

160

(700+160) = 860

Rs.688

Rs.1000

No

Rs.688

6

120

(860+120) = 980

Rs.784

Rs.1000

No

Rs.784

7

150

(980+150) = 1130

Rs.904

Rs.1000

Yes*

Rs.1000

TOTAL COST

Rs.3776

* In spite of the fact that storing is cheaper than hiring truck on the last day, we have to with the latter option because everything that is manufacture has to sent to the market.

Hence the most cost effective scheme would be to send the entire production on the 7th day.

55. Let us try and find a pattern. Let there be x bacteria in the first generation. Hence n1 = x. But only x/2 among them will be able to produce the next generation. And they would give rise to 8(x/2) = 4x baxteria. Hence n2 = 4x. Of these only 2x will give rise to next generation. So number of bacteria in the third generation = 8(2x) = 16x. So n3 = 16x. Similarly we would find that n4 = 64x. So if we observe : n1 = x, n2 = 4x, n3 = 16x, n4 = 64x. So this will form a GP with a = x and r = 4. The seventh term of this GP = 4096. So we can write, 4096 = x(4)6 = x(2)12 = 4096x. Hence x = 1 i.e. 1 million.


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