Indian Institute of Technology Kanpur (IIT-K) 2006 JT Quant - Question Paper

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Directions: 104 and 105 are based on the subsequent information:-
A 300 room motel can rent all its rooms at Rs. 150 per room per day. For every 1 rupee increase in tariff,
the occupancy falls by two rooms. Normal maintenance, independent of occupancy is Rs. 120 per room per
day. Occupancy entails additional costs of Rs. 16 per room per day on the units occupied.
104. The optimal tariff (in Rupees) to maximize annual profits is:
a. 150 b. 162 c. 158 d. 160
105. Now, suppose that rooms of the motel can be permanently shut down in blocks of 5, if desired,
resulting in saving of normal maintenance of Rs. 120 per room per day. The optimal tariff (in Rupees)
to maximize annual profits in this case would be:
a. 158 b. 160 c. 162 d. 156
For ques. 104 and 105:
Revenue = (300 – 2x) (150 + x)
Maintenance = 120 × 300 + 16 (300 – 2x)
Þ Profit = Revenue – Maintenance
= 4200 + 32x – 2x2
= 4328 – two (8 – x)2
Þ Maximum profit occurs when x = 8
104. c Optimal tariff = 150 + eight = Rs. 158
105. b Now,
Profit = 4200 + 32x – 2x2 + 600 [2x/5]
From the options, x = 10 provide us maximum profit Þ optimal tariff = 150 + 10 = Rs. 160
Page 12 / SET A JMET / true PAPER
106. An investor desires to invest a certain sum of money in 2 securities A and B. The risk and return
of A and B are:
A B
Risk (b) 3.00 6.00
Return in %(R) 9.00 12.00
Measures of both risk and return are additive, i.e. bP = XAbA + XBbB, RP = XARA + XBRB, where XA, XB are
the proportions of the money invested in the securities A & B in the portfolio P. The investor has a
maximum risk tolerance of 4.00. The return that he can earn (in percent) is:
a. nine b. 12 c. 10 d. 16
106. c Here three XA + six XB = 4
XA + XB = 1, one > x > 0 and one > y > 0
Solving both equations, we get
XA =
2
3 and XB =
1
3
So, P
2 1
R nine 12 six four 10

Earning:   Approval pending.