IIFT 2014 Question Paper

The distance of boat from bottom of cliff when angle of depression is 60 deg is

tan $$60^{\circ}$$ = $$\frac{210}{d}$$

d = 121.3 m

The distance of boat from bottom of cliff when angle of depression is 45 deg is

tan $$45^{\circ}$$ = $$\frac{210}{d}$$

d = 210 m

The distance covered by boat is 210 - 121.3 = 88.7 m

Speed of boat given is 3 km/hr

In m/s it is 3*5/18 = 5/6 m/s = 0.83 m/s

Time taken to cover 88.7 m is 88.7/0.83 = 106.44 sec $$\cong$$ 2 min

Alloy X has zinc and copper in ratio 6:9

If 40 grams is taken then weights of zinc and copper are: 

Zinc = $$\frac{6}{15}$$ * 40 = 16 grams

Copper = $$\frac{9}{15}$$ * 40 = 24 grams

Alloy Y has zinc and copper in ratio 7:11

If 60 grams is taken then weights of zinc and copper are: 

Zinc = $$\frac{7}{18}$$ * 60 = $$\frac{70}{3}$$ grams

Copper = 60 - $$\frac{70}{3}$$ = $$\frac{110}{3}$$ grams

When mixed together, the weights of Zinc and Copper are:

Zinc = 16 + $$\frac{70}{3}$$ = $$\frac{118}{3}$$

Copper = 24 + $$\frac{110}{3}$$ = $$\frac{182}{3}$$

Ratio = 118/3:182/3 = 59:91

Let 'a' be the length of side of the hexagon. Therefore, the area of hexagon = $$6*\dfrac{\sqrt{3}}{4}\times a^2$$.

The area of triangle PQR = $$\dfrac{\sqrt{3}}{4}\times (a/2)^2$$

Therefore, the are of the shaded region = $$6*\dfrac{\sqrt{3}}{4}\times a^2$$ - $$\dfrac{\sqrt{3}}{4}\times (a/2)^2$$ = $$\dfrac{23\sqrt{3}a^2}{16} = X$$

$$\Rightarrow$$ $$a^2 = \dfrac{16X}{23\sqrt{3}}$$

The area of the circle = $$\pi*a^2$$

$$\Rightarrow$$ $$\dfrac{16 \pi}{23 \sqrt{3}}$$ X. Hence, option A is the correct answer. 

10 years ago, Let age of Ravindra be 5x and Rekha be 4x

At present, Ravindra is 7/6 times of Rekha's age.

5x + 10 = $$\frac{7}{6}$$ (4x + 10)

Solving, x =5

Ravindra was 25 years (10 years ago) and Rekha was 20 years (10 years ago)

Now, ages of their children is 3:2:1

Maximum possible ages of children is 9,6,3 years.

Total age of family is: 35 + 30 + 9*3 + 6*2 + 3 = 107 years.

Probability of choosing a boy from Section A = 3/4

Probability of choosing a boy from Section B = 5/9

Probability of choosing a girl from Section A = 1/4

Probability of choosing a girl from Section B = 4/9

There are two cases. 

Case 1: Boy from section A and girl from section B

Probabilty = $$\frac{3}{4}$$ x $$\frac{4}{9}$$

Case 1: Boy from section A and girl from section B

Probabilty = $$\frac{1}{4}$$ x $$\frac{5}{9}$$

Adding gives final probability = $$\frac{12}{36}$$ + $$\frac{5}{36}$$ = $$\frac{17}{36}$$ 

The amount of pure milk received by 5th customer = amount of pure milk after 4 replacements - amount of milk after 5 replacements

Formula for amount of milk after n replacements is : 40 $$(1 - \frac{10}{40})^{n}$$

Amount of milk received by 5th customer = 40 $$(1 - \frac{10}{40})^{4}$$ - 40 $$(1 - \frac{10}{40})^{5}$$

=> 40 * $$(\frac{3}{4})^{4} (1 - \frac{3}{4})$$

=> $$\frac{405}{128}$$ lts.

The ferry must have traveled 1.5*(20-2) = 27 kms already. The ferry needs to cover 73 kms to reach Rock of Vivekananda and 100 kms back to Kanyakumari. 15 gallons of water has already entered the ship. 45 more gallons will enter in order for the ship to sink. So it takes 4.5 hours. The ferry has to travel 70 km and return 100 km before it sinks. Speed of stream from Rock memorial to Kanniyakumari is 2 km/hr

Check the options.

Option 1: $$\frac{73}{38}+\frac{100}{41} = 4.36$$
Option 2: $$\frac{73}{39}+\frac{100}{40} = 4.37$$
Option 3 : $$\frac{73}{40}+\frac{100}{38} = 4.46$$
Option 4: $$\frac{73}{33}+\frac{100}{41} = 4.65$$
Thus, the closest option is option C.
Hence, option C is the correct answer.

We can see that the magnitude in each succeeding term is less than that of preceding term. 

Hence, we can say that for S = $$1 - \frac{1}{6} + (\frac{1}{6} \times \frac{1}{4})-(\frac{1}{6} \times \frac{1}{4} \times \frac{5}{18})+$$ ...

The value will lie between (5/6, 1). We can check with option choices. 

Option A: $$\frac{2}{3}$$ < $$\frac{5}{6}$$. Hence, this can't be the answer. 

Option B: $$\frac{2}{\sqrt{3}}$$ = 1.155 > $$1$$. Hence, this can't be the answer. 

Option C: $$\sqrt{\frac{2}{3}}$$ = 0.8164 < $$\frac{5}{6}$$. Hence, this can't be the answer. 

Option D: $$\frac{\sqrt{3}}{2}$$ = 0.866. This lies between (5/6, 1).Hence, this is the correct answer. 

$$\text{log}_{7} \text{log}_{7} \sqrt{7(\sqrt{7\sqrt{7}})}$$ 
= $$\text{log}_{7} [\frac{1}{2} (\text{log}_{7} 7+\text{log}_{7} \sqrt{7(\sqrt{7})})]$$
= $$\text{log}_{7} [\frac{1}{2} (1 +\frac{1}{2}\text{log}_{7}{7(\sqrt{7})})]$$
= $$\text{log}_{7} [\frac{1}{2} (1 +\frac{1}{2}(1+1/2))]$$
= $$\text{log}_{7}\frac{7}{8}$$
= $$\text{log}_{7}7 - \text{log}_{7}8$$
= $$1-3 \text{log}_{7}2$$
Hence, option D is the correct answer.

We can represent the given information in the diagram as follows:

The width of the street = BD = BC + CD

$$\Rightarrow$$ $$\sqrt{13^2 - 8^2}$$ + $$\sqrt{13^2 - 12^2}$$

$$\Rightarrow$$ $$10.24$$m + $$5$$m = 15.2 m. Hence, option C is the correct answer.