# IIFT 2014 QA | Previous Year IIFT Paper

**Read the following information and Tables and answer the questions.**

BHUBANESWAR, CHENNAI, KANYAKUMARI, KOCHI, MUMBAI and VIZAG are 6 major Indian cities. For some reason people use only a certain mode of transport between a pair of cities. The modes of transport are provided in Table 1, while in Table 2 the distances between different pairs of cities are given. Table 3 provides the speed of the mode of transport and the cost associated with each of them.

**1. IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

For which of the following options, travel time is the least?

- A.
MUMBAI – KANYAKUMARI

- B.
BHUBANESWAR - CHENNAI

- C.
CHENNAI - KOCHI

- D.
MUMBAI - CHENNAI

Answer: Option A

**Explanation** :

Time taken to travel = $\frac{\mathrm{Distance}}{\mathrm{Speed}\mathrm{of}\mathrm{available}\mathrm{mode}\mathrm{of}\mathrm{transport}}$

Considering the options, we have the travel times for (Mumbai – Kanyakumari, Bhubaneshwar – Chennai, Chennai – Kochi, Mumbai – Chennai) as,

$\frac{950}{40}$, $\frac{950}{30}$, $\frac{901}{30}$, $\frac{1000}{30}$

It can be observed that $\left(\frac{950}{40}\right)$ has the least value.

Hence, option (a).

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**2. IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

Mr. Ranjith lives in MUMBAI and wants to travel to KOCHI. However, the train services are on halt due to laying of track for bullet trains across the country. In this scenario, which of the following is the least cost route to reach KOCHI?

- A.
MUMBAI - BHUBANESWAR – KOCHI

- B.
MUMBAI – CHENNAI - KOCHI

- C.
MUMBAI – KANYAKUMARI – KOCHI

- D.
MUMBAI - VIZAG - KOCHI

Answer: Option B

**Explanation** :

Cost of travel = Distance × Cost per KM of available mode of transport

Considering the cost of travel in each option.

Option 1: (701 × 2) + (798 × 5)

≈ 1400 + 4000 = 5400

Option 2: (1000 × 1.5) + (901 × 1.5)

≈ 1900 × 1.5 = 2850

Option 3: (950 × 2) + (1100 × 2.5)

= 1900 + 2750 > 2850

Option 4: (500 × 5) + (600 × 5) > 2850

Hence, option (b).

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**3. IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

A school in Chennai is planning for an excursion tour for its students. They want to show them KANYAKUMARI, VIZAG, and BHUBANESWAR, not necessarily in the same order. What is the minimum travel cost (in RS.) the school should charge from each of the student for the entire tour?

- A.
Rs. 4300

- B.
Rs. 5000

- C.
Rs. 7500

- D.
Rs. 6800

Answer: Option A

**Explanation** :

Observe that on a per-km basis, ship is the cheapest and airplane is the costliest. Hence, between the given cities, try to select the order of travel such that most of the journey is done by ship, followed by bus, then train and so on.

The overall route to be followed is: Chennai – city 1 – city 2 - city 3 - Chennai.

Chennai is connected to Kanyakumari and Vizag by train and to Bhubaneshwar by ship.

Among the three given cities, Kanyakumari is also the farthest from Chennai.

Hence, Chennai to Kanyakumari is always costlier than Chennai to the other two cities.

Hence, those two cities should be the first and last to be visited (in any order, due to the symmetry of the journey) while Kanyakumari should be visited between the two, as shown below.

Chennai → Vizag/Bhubaneshwar → Kanyakumari → Bhubaneshwar/ Vizag → Chennai

Now, find individual costs:

Chennai – Vizag: (300 × 2.5) = 750

Chennai – Bhubaneshwar: (950 × 1.5) = 1425

Vizag – Kanyakumari: (250 ×1.5) = 375

Kanyakumari – Bhubaneshwar: (700 ×2.5) = 1750

The total cost is Rs. 4300/-.

Hence, option (a).

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**4. IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

Which of the following cities can be reached from BHUBANESWAR in least time?

- A.
CHENNAI

- B.
KANYAKUMARI

- C.
MUMBAI

- D.
VIZAG

Answer: Option C

**Explanation** :

Considering the options, we have the travel times as,

$\frac{950}{40}$, $\frac{700}{25}$, $\frac{701}{40}$, $\frac{1002}{25}$

It can be observed that $\left(\frac{701}{40}\right)$ has the least value.

Hence, option (c).

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**5. IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

What is the least cost way to reach to VIZAG from KOCHI?

- A.
Take a flight from KOCHI to VIZAG

- B.
Take a ship from KOCHI to CHENNAI and then take a train to VIZAG

- C.
Take a train from KOCHI to KANYKUMARI and then take a ship to VIZAG

- D.
Take a train from KOCHI to MUMBAI and then take a flight to VIZAG

Answer: Option B

**Explanation** :

Considering the cost in each option.

Option 1: (600 × 5) = 3000

Option 2: (901 × 1.5) + (300 × 2.5) ≈ 2100

Option 3: (1100 × 2.5) + (250 × 1.5) > 2100

Option 4: (300 × 2.5) + (500 × 5) > 2100

Hence, option (b).

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**Answer the following question based on the information given below.**

Read the following information on ‘Sectoral Trends in Mergers & Acquisitions in India (2001-02 to 2006-07)’ given in Tables below and answer the questions.

**6. IIFT 2014 QA | DI - Tables & Graphs**

What is the approximate proportion of ‘mergers’ to ‘acquisitions’ for the entire period (2001 - 02 to 2006 - 07)?

- A.
26%

- B.
36%

- C.
30%

- D.
20%

Answer: Option B

**Explanation** :

The given question requires the proportion of ‘mergers’ to ‘acquisitions’ for the entire period. Hence, the acquisitions table must be constructed and the sum of the mergers and acquisitions must be calculated. It can be shown as follows,

It must be noted that the overall sum can be calculated by calculating the individual sums either year wise or sector wise for this question. There is no need to calculate both.

$\frac{\mathrm{Total}\mathrm{Mergers}}{\mathrm{Total}\mathrm{Acquisitions}}$ = $\frac{1499}{4147}$ × 100 = 36%

Hence, option (b).

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**7. IIFT 2014 QA | DI - Tables & Graphs**

For how many sectors is the proportion of ‘mergers’ to mergers & acquisitions’ greater than 20% for the entire period (2001 - 02 to 2006 - 07)?

- A.
2

- B.
3

- C.
4

- D.
5

Answer: Option D

**Explanation** :

Considering the sector-wise sums of Mergers and Acquisitions and Mergers, the following relation must be considered.

$\frac{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{Mergers}\mathrm{for}\mathrm{the}\mathrm{sector}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{Merger}\mathrm{and}\mathrm{Acquisitions}\mathrm{for}\mathrm{the}\mathrm{sector}}$ ≥ 0.2

It can be expressed as, 5(Total number of Mergers for the sector) ≥ (Total number of Mergers and Acquisitions for the sector)

The above relation is satisfied for 5 sectors i.e. Food and Beverages, Textile, Drugs and Pharma, Financial Services and Other Services.

Hence, option (d).

**Note:**

It can be seen that the acquisitions table was not required for this question. Hence, this question could have been attempted even without constructing the acquisitions table.

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**8. IIFT 2014 QA | DI - Tables & Graphs**

For how many sectors merger activity (measured by number of mergers) is more in the first 3 years as compared to the last 3 years?

- A.
7

- B.
3

- C.
6

- D.
5

Answer: Option D

**Explanation** :

This is completely an observation based question. The values for each sector from the first three columns and the last three columns of the ‘Mergers’ table only need to be considered. If the sum of the values of the first three columns is higher than that of the last columns, that sector showed greater merger activity in the first three years. If each individual value for a sector in a year is greater for the first three years than for the last three years, then the sums also need not be calculated.

It can be seen that the values increased for 5 sectors including Food and Beverages, IT and Telecom, Diversified, Financial Services and Other Services.

Hence, option (d).

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**9. IIFT 2014 QA | DI - Tables & Graphs**

If the turbulence over a period is defined by the sum of each of the differences (in absolute terms) in number of mergers & acquisitions on a year-on-year basis, then which sector is considered most turbulent for the entire period (2001 - 02 to 2006 - 07)?

- A.
Financial Services

- B.
IT & Telecom

- C.
Food & Beverages

- D.
Other services

Answer: Option D

**Explanation** :

Turbulence from 2001-02 to 2006-07

= |Number of mergers in 2001-02 – Number of acquisitions in 2001-02| + |Number of mergers in 2002-03 – Number of acquisitions in 2002-03| + … + |Number of mergers in 2006-07 – Number of acquisitions in 2006-07|

It can be seen that the overall sum of mergers and acquisitions can be calculated to find the required value of turbulence.

Turbulence from 2001-02 to 2006-07

= |Total Number of mergers in 2001-07 – Total Number of acquisitions in 2001-07|

Considering the values of Mergers for Other services, it can be seen that the number of acquisitions is 1197. The corresponding sum for mergers is 512. The difference between the two is 685.

On calculating the total number of acquisitions for the remaining options, it can be seen that the rest of the values are < 685, and will decrease further when the number of mergers is subtracted from each of them.

Hence, Other Services had the maximum turbulence.

Hence, option (d).

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**10. IIFT 2014 QA | DI - Tables & Graphs**

In which year maximum sectors have exhibited higher number of acquisitions compared to previous year?

- A.
2003 - 04

- B.
2004 - 05

- C.
2005 - 06

- D.
2006 - 07

Answer: Option C

**Explanation** :

Observe the acquisitions table. It can be seen that maximum number of sectors exhibited higher number of acquisitions compared to the previous year in 2005 – 2006.

Hence, option (c).

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**Answer the following question based on the information given below.**

Charts given below describe the energy scenario of a country. Assume that the country does not export any form of energy and whatever is produced and imported is consumed in the same year. Go through the Charts and answer the questions.

**11. IIFT 2014 QA | DI - Tables & Graphs**

What was the approximate total import of energy in 2010?

- A.
400 MTOE

- B.
300 MTOE

- C.
360 MTOE

- D.
430 MTOE

Answer: Option C

**Explanation** :

Note that “nuclear energy” and “others” has not been imported.

From Chart 1, the consumption of coal, crude oil, natural gas and hydro electricity are 540, 290, 80 and 50 MTOE respectively.

From Chart 2, for the year 2010, the corresponding growth rates in consumption are 15%, 10%, 5% and 10% respectively.

And from Chart 3, for the year 2010, the corresponding percentages of imports are 25%, 50%, 50% and 5% respectively.

∴ The total import of energy in the year 2010 = 540 × 1.15 × 0.25 + 290 × 1.1 × 0.5 + 80 × 1.05 × 0.5 + 50 × 1.1 × 0.05

= 359.5 MTOE

Hence, option (c).

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**12. IIFT 2014 QA | DI - Tables & Graphs**

The import of natural gas in 2012, when compared to 2010, is approximately:

- A.
Reduced by 10 MTOE

- B.
Reduced by 13 MTOE

- C.
Increased by 10 MTOE

- D.
Increased by 5 MTOE

Answer: Option B

**Explanation** :

As seen in the previous question, the import of natural gas in the year 2010

= 80 × 1.05 × 0.5 = 42 MTOE.

From Chart 2, the consumption of natural gas in the year 2012

= 80 × 1.05 × 1.1 × 1.05 = 97.02 MTOE.

Using Chart 3, the import of natural gas in the year 2012

= 97.02 × 0.3 = 29.11 MTOE

Thus, the import of natural gas has reduced by approximately 13 MTOE.

Hence, option (b).

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**13. IIFT 2014 QA | DI - Tables & Graphs**

What is the approximate domestic production of crude oil in 2011?

- A.
220 MTOE

- B.
190 MTOE

- C.
160 MTOE

- D.
280 MTOE

Answer: Option B

**Explanation** :

The total consumption of crude oil in 2011 = 290 × 1.1 × 1.1 = 350.09 MTOE

From Chart 3, we can see that crude oil imported was 45% of the total consumption in the year 2011.

∴ Domestic production = 55% of the total consumption

⇒ 350.09 × 0.55 = 192.55 MTOE

Hence, option (b).

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**14. IIFT 2014 QA | DI - Tables & Graphs**

What is the approximate proportion of coal in the domestic consumption of energy in 2012?

- A.
52

- B.
54

- C.
58

- D.
56

Answer: Option D

**Explanation** :

The total consumption of energy in the year 2012

= 540 × 1.15 × 1.1 × 1.15 + 290 × 1.1 × 1.1 × 1.15 + 80 × 1.05 × 1.05 × 1.1 + 20 × 1.2 × 1.15 × 1.1 + 50 × 1.1 × 1.05 × 1.1 + 20 × 1.15 × 1.15 × 1.1

= 1409.1 MTOE

The consumption of coal in the year 2012

= 540 × 1.15 × 1.1 × 1.15 = 785.56 MTOE

The proportion of coal in the total domestic consumption

= 785.56/1409.1 × 100 = 55.75%

Hence, option (d).

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**15. IIFT 2014 QA | DI - Tables & Graphs**

What is the sum of the approximate domestic production of nuclear energy and hydroelectricity in 2011?

- A.
75 MTOE

- B.
80 MTOE

- C.
90 MTOE

- D.
100 MTOE

Answer: Option B

**Explanation** :

The domestic production of nuclear energy in 2011 = 20 × 1.2 × 1.15

= 27.6 MTOE

The domestic production of hydro electricity in 2011 = 50 × 1.1 × 1.05 × 0.95 = 54.86 MTOE

∴ Sum of the approximate domestic production of nuclear energy and hydro electricity in 2011

= 27.6 + 54.86 = 82.46 MTOE

Hence, option (b).

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**16. IIFT 2014 QA | Geometry - Circles**

Let P_{1 }be the circle of radius r. A square Q_{1} is inscribed in P_{1} such that all the vertices of the square Q_{1} lie on the circumference of P_{1}. Another circle P_{2} is inscribed in Q_{1}. Another Square Q_{2} is inscribed in the circle P_{2}. Circle P_{3} is inscribed in the square Q_{2} and so on. If S_{N} is the area between Q_{N} and P_{N+1}, where N represents the set of natural numbers, then the ratio of sum of all such S_{N} to that of the area of the square Q_{1} is:

- A.
$\frac{4-\pi}{2}$

- B.
$\frac{2\pi -4}{\pi}$

- C.
$\frac{\pi -2}{2}$

- D.
None of the above

Answer: Option A

**Explanation** :

Radius of P_{1} = r ⇒A(P_{1}) = πr^{2}

Diameter = d = 2r

Side of Q_{1} = $\frac{2r}{\sqrt{2}}$ = r$\sqrt{2}$ ⇒ A(Q_{1}) = 2r^{2}

Radius of P_{2} = $\frac{r\sqrt{2}}{2}$ = $\frac{r}{\sqrt{2}}$ ⇒ A(P_{1}) = $\frac{\pi {r}^{2}}{2}$

Side of Q_{2} = $\left(\frac{r}{\sqrt{2}}\right)$ × $\sqrt{2}$ = r ⇒ A(Q_{2}) = r^{2 }... and so on.

i.e., Areas of circles are in G.P. with common ratio = $\frac{1}{2}$

Also, areas of squares are in G.P. with common ratio = $\frac{1}{2}$

S_{N} = Q_{1} – P_{2} + Q_{2} – P_{3} + Q_{3} – P_{4} + …

= (Q_{1} + Q_{2} + Q_{3} + … ) – (P_{2} + P_{3} + P_{4} + … )

= 2r^{2} $\left(1+\frac{1}{2}+\frac{1}{4}+...\right)$ - $\frac{\pi {r}^{2}}{2\left(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}+...\right)}$

= $\left(1+\frac{1}{2}+\frac{1}{4}+...\right)$$\left(2{r}^{2}-\frac{\pi {r}^{2}}{2}\right)$

= r^{2} × $\left[\frac{1}{1-{\displaystyle \frac{1}{2}}}\right]$ × $\frac{4-\pi}{2}$

= (4 - π)r^{2}

S_{N}: A(Q_{1}) = (4 – π)r^{2 }: 2r^{2} = (4 – π)/2

Hence, option (a).

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**17. IIFT 2014 QA | Modern Math - Permutation & Combination**

In a school, students were called for the Flag Hoisting ceremony on August 15. After the ceremony, small boxes of sweets were distributed among the students. In each class, the student with roll no. 1 got one box of sweets, student with roll number 2 got 2 boxes of sweets, student with roll no. 3 got 3 boxes of sweets and so on. In class III, a total of 1200 boxes of sweets were distributed. By mistake one of the students of class III got double the sweets he was entitled to get. Identify the roll number of the student who got twice as many boxes of sweets as compared to his entitlement.

- A.
22

- B.
24

- C.
28

- D.
30

Answer: Option B

**Explanation** :

(1 + 2 + 3 + … + n) + x = 1200

Where, x is the roll number of the student who got twice as many as compared to his entitlement.

n (n + 1)/2 = 1200 – x ⇒ n (n + 1)

= 2400 – 2x

Substituting values of x from options, only for x = 24, the RHS can be expressed as a product of two natural numbers.

Hence, option (b).

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**18. IIFT 2014 QA | Geometry - Triangles**

A Boat is being rowed away in still water, from a 210 metres high cliff at the speed of 3 km/hr. What is the approximate time taken for the angle of depression of the cliff at the boat to change from 60 deg. To 45 deg.?

- A.
5 min

- B.
4 min

- C.
1 min

- D.
2 min

Answer: Option D

**Explanation** :

When the boat was at C, the angle of depression = 60° ⇒ m∠ BAC = 30°

When the boat was at D, the angle of depression = 45° ⇒ m∠DAB = 45° ⇒ BD

= AB = 210 m

In ∆ABC, by theorem of 30° – 60° – 90°,

BC = 70√3 ≈ 121.1

DC ≈ 210 – 121.2 = 88.8 m

Speed of the boat = 3 km/ hr = 50 m/min

Time taken = 88.8/ 50 = 1.776 minutes

Hence, option (d).

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**19. IIFT 2014 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

X and Y are the two alloys which were made by mixing Zinc and Copper in the ratio 6 : 9 and 7 : 11 respectively. If 40 grams of alloy X and 60 grams of alloy Y are melted and mixed to form another alloy Z, what is the ratio of Zinc and Copper in the new alloy Z?

- A.
6 : 9

- B.
59 : 91

- C.
5 : 9

- D.
59 : 90

Answer: Option B

**Explanation** :

Zinc in alloy Z = 40 × $\frac{6}{15}$ + 60 × $\frac{7}{18}$

= $\frac{118}{3}$

Copper in alloy Z = 40 × $\frac{9}{15}$ + 60 × $\frac{11}{18}$

= $\frac{182}{3}$

Ratio of Zinc and Copper in Z = 118 : 182 = 59 : 91

Hence, option (b).

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**20. IIFT 2014 QA | Geometry - Circles**

ABCDEF is a regular hexagon and PQR is an equilateral triangle of side a. The area of the shaded portion is X and CD : PQ : : 2 : 1. Find the area of the circle circumscribing the hexagon in terms of X.

- A.
$\frac{16\pi}{23\sqrt{3}}$x

- B.
$\frac{42\pi}{5\sqrt{3}}$x

- C.
$\frac{2\pi}{3\sqrt{3}}$x

- D.
2$\sqrt{3}$πx

Answer: Option A

**Explanation** :

Let the length of each side of the hexagon be y.

Area of the hexagon = $\frac{3\sqrt{3}}{2}$y^{2}

Area of the triangle = $\frac{\sqrt{3}}{16}$y^{2}

Area of the shaded region = $\frac{23\sqrt{3}}{16}$ y^{2} = X

y^{2} = $\frac{16}{23\sqrt{3}}$X

Area of the circle = πy^{2}

Thus, area of the circle in terms of X

= $\frac{16\pi}{23\sqrt{3}}$X

Hence, option (a).

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**21. IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

Ravindra and Rekha got married 10 years ago, their ages were in the ratio of 5 : 4. Today Ravindra’s age is one sixth more than Rekha’s age. After marriage, they had 6 children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of 3 : 2 : 1. What is the largest possible value of the present total age of the family?

- A.
79

- B.
93

- C.
101

- D.
107

Answer: Option D

**Explanation** :

Let the present age of Ravindra and Rekha be R_{v} and R_{e} respectively.

$\frac{{R}_{v}-10}{{R}_{e}-10}$ = $\frac{5}{4}$

R_{v} = (1 + 1/6)R_{e}

Solving these equations, we get R_{v} = 35 and R_{e} = 30

Maximum age of a child = 9 years (10 year after marriage)

To maximize the age of family,

Age of triplets = 9 years, twins = 6 years and sixth child = 3 years

Age of the family = 35 + 30 + (9 × 3) + (6 × 2) + 3 = 107

Hence, option (d).

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**22. IIFT 2014 QA | Modern Math - Permutation & Combination**

In the MBA Programme of a B – School, there are two sections A and B. 1/4th of the students in Section A and 4/9th of the students in section B are girls. If two students are chosen at random, one each from section A and Section B as class representative, the probability that exactly one of the students chosen is a girl, is:

- A.
23/72

- B.
11/36

- C.
5/12

- D.
17/36

Answer: Option D

**Explanation** :

Selecting a girl from section A and section B is 1/4 and 4/9 respectively.

Selecting a boy from section A and section B is 3/4 and 5/9 respectively.

Case 1: A girl from section A and a boy from section B.

P_{1} = (1/4) × (5/9) = 5/36

Case 2: A boy from section A and a girl from section B.

P_{2} = (3/4) × (4/9) = 12/36

Required probability = P_{1} + P_{2} =17/36

Hence, option (d).

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**23. IIFT 2014 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

A milk vendor sells 10 litres of milk from a can containing 40 litres of pure milk to the 1^{st} customer. He then adds 10 litres of water to the milk can. He again sells 10 litres of mixture to the 2^{nd} customer and then adds 10 litres of water to the can. Again he sells 10 litres of mixture to the 3^{rd} customer and then adds 10 litres of water to the can and so on. What amount of pure milk will the 5^{th} customer receive?

- A.
$\frac{510}{128}$ litres

- B.
$\frac{505}{128}$ litres

- C.
$\frac{410}{128}$ litres

- D.
$\frac{405}{128}$ litres

Answer: Option D

**Explanation** :

Total volume, V = 40 litres

Volume replaced, v = 10 litres

Number of replacements = 4

Pure milk in the mixture after 4 replacements = [(40 – 10)/ 40]^{4}

Pure milk in 10 litres of the mixture

= 10 × (3/4)^{4} = 405/128

Hence, option (d).

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**24. IIFT 2014 QA | Arithmetic - Time, Speed & Distance**

A fery carries passengers to Rock of Vivekananda and back from Kanyakumari. The distance of Rock of Vivekananda from Kanyakumari is 100 km. One day, the ferry started for Rock of Vivekananda with passengers on board, at a speed of 20 km per hour. After 90 minutes, the crew realized that there is a hole in the ferry and 15 gallons of sea water had already entered the ferry. Sea water is entering the ferry at the rate of 10 gallons per hour. It requires 60 gallons of water to sink the ferry. At what speed should the driver now drive the ferry so that it can reach the Rock of Vivekananda and return back to Kanyakumari just in time before the ferry sinks?

(Current of the sea water from Rock of Vivekananda to Kanyakumari is 2 km per hour.)

- A.
40 km/hr towards the Rock & 39 km/hr while returning to Kanyakumari

- B.
41 km/hr towards the Rock & 38 km/hr while returning to Kanyakumari

- C.
42 km/hr towards the Rock & 36 km/hr while returning to Kanyakumari

- D.
35 km/hr towards the Rock & 39 km/hr while returning to Kanyakumari

Answer: Option C

**Explanation** :

Since the current is from the Rock towards the Kanyakumari, the boat travels the first part against the stream and the second part with the stream.

Thus, in 90 minutes the boat has travelled 27 kms.

Since 15 gallons of water is already filled, there are 45 more gallons required to sink the ship which will take 4.5 hours at 10 gallons/hr.

Thus, the boat needs to cover 73 kms of first part and 100 kms of the second part within 4.5 hrs.

Let the speed of the boat be S_{B}

Since the first part is upstream, the time required will be = 73/(S_{B} − 2)

The time required for the second part will be = 100/(S_{B} + 2)

Thus,

Time taken by the ship ≤ 4.5 i.e.

$\frac{73}{{S}_{B}-2}$ + $\frac{100}{{S}_{B}+2}$ ≤ 4.5

We now put the values of SB from the options and calculate the time taken by the ship.

Thus, the calculated times from options (1), (2), (3), (4) would be 4.36, 4.37, 4.455, 4.65 hrs respectively.

Please note that in the question, the ferry needs to complete the journey just in time and hence the answer would be the option closest to 4.5 hrs i.e. 4.455 hrs.

Hence, option (c).

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**25. IIFT 2014 QA | Algebra - Simple Equations**

The sum of 1 - $\frac{1}{6}$ + $\left(\frac{1}{6}\times \frac{1}{4}\right)$ - $\left(\frac{1}{6}\times \frac{1}{4}\times \frac{5}{18}\right)$ + ... is

- A.
$\frac{2}{3}$

- B.
$\frac{2}{\sqrt{3}}$

- C.
$\sqrt{\frac{2}{3}}$

- D.
$\frac{\sqrt{3}}{2}$

Answer: Option D

**Explanation** :

1 - $\frac{1}{6}$ + $\frac{1}{6}$ × $\frac{1}{4}$ - $\frac{1}{6}$ × $\frac{1}{4}$ × $\frac{5}{18}$ .....

1 - $\frac{1}{6}$ = $\frac{5}{6}$

Observe that the sum of further terms will be positive.

Hence, the sum will be greater than 5/6 i.e. 0.833.

Also, if we add the 2^{nd} and 3^{rd} term, 3^{rd} and 4^{th} term, and so on, we can see that each of these terms are negative.

∴ The sum is less than 1.

The only option that satisfies this is $\frac{\sqrt{3}}{2}$.

Hence, option (d).

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**26. IIFT 2014 QA | Geometry - Trigonometry**

The value of log_{7} log_{7} $\sqrt{7\sqrt{7\sqrt{7}}}$ is equal to:

- A.
7

- B.
log

_{7}2 - C.
1 – 3 log27

- D.
1 – 3 log

_{7}2

Answer: Option D

**Explanation** :

log_{7} log_{7} $\sqrt{7\sqrt{7\sqrt{7}}}$ log_{7} log_{7} ${7}^{\frac{7}{8}}$

= log_{7} $\frac{7}{8}$ = 1 - 3log_{7} 2

Hence, option (d).

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**27. IIFT 2014 QA | Geometry - Coordinate Geometry**

A ladder just reaches a window that is 8 metres high above the ground on one side of the street. Keeping one end of the ladder at the same place, the ladder is moved to the other side of the street so as to reach a 12 metre high window. If the ladder is 13 metres long, what is the width of the street?

- A.
14.6 metres

- B.
15.8 metres

- C.
15.2 metres

- D.
15.5 metres

Answer: Option C

**Explanation** :

The length of the ladder is 13 metres.

Thus, for the cases given in the question, the height is given as 8 metres and 12 metres and the hypotenuse will be 13 metres for both cases.

Thus, by Pythagoras theorem, the width of the street = 5 + √105 = 15.2 metres

Hence, option (c).

Workspace:

**28. IIFT 2014 QA | Algebra - Number System**

The total number of eight – digit landline telephone numbers that can be formed having at least one of their digits repeated is:

- A.
98185600

- B.
97428800

- C.
100000000

- D.
None of the above

Answer: Option A

**Explanation** :

The total number of 8-digit landline telephone numbers that can be formed having at least one of their digits repeated = The total number of 8-digit landline numbers – The number of 8-digit landline numbers in which no digit is repeated.

The total number of 8-digit landline numbers = 10^{8}

The number of 8-digit landline numbers in which no digit is repeated = 10!/2

∴ Number of required landline numbers = 98185600

Hence, option (a).

Workspace:

**29. IIFT 2014 QA | Algebra - Number Theory**

The business consulting division of TCS has overseas operations in 3 locations: Singapore, New York and London. The Company has 22 analysts covering Singapore, 28 covering New York and 24 covering London. 6 analysts cover Singapore and New York but not London, 4 analysts cover Singapore and London but not New York, and 8 analysts cover New York and London but not Singapore. If TCS has a total of 42 business analysts covering at least one of the three locations: Singapore, New York and London, then the number of analysts covering New york alone is:

- A.
14

- B.
28

- C.
5

- D.
7

Answer: Option D

**Explanation** :

Using the data we can have following Venn diagram.

From the given data, we can form the following equations:

a + b + c + d = 42 – 18 = 24

⇒ a + b + c + d = 24 … (i)

a + b + c + 2(18) + 3d = 22 + 28 + 24 = 74 ⇒ a + b + c + 3d = 42 … (ii)

Solving (i) and (ii), d = 7

∴ 6 + 7 + 8 + b = 28 ⇒ b = 7

Hence, option (d).

Workspace:

**30. IIFT 2014 QA | Arithmetic - Simple & Compound Interest**

Eight years after completion of your MBA degree, you start a business of your own. You invest INR 30,00,000 in the business that is expected to give you a return of 6%, compounded annually. If the expected number of years by which your investment shall double is 72/r, where r is the percent interest rate, the approximate expected total value of investment (in INR) from your business 48 years later is:

- A.
2,40,00,000

- B.
3,60,00,000

- C.
4,80,00,000

- D.
None of the above

Answer: Option C

**Explanation** :

Investment doubles in 72/6 = 12 years

So, in 48 (=12 × 4) years, the investment becomes 2^{4} times.

∴ The approximate total value of the investment (in INR)

= 16 × 3000000 = 4,80,00,000

Hence, option (c).

Workspace:

**31. IIFT 2014 QA | Geometry - Circles**

A right circular cylinder has a radius of 6 and a height of 24. A rectangular solid with a square base and a height of 20, is placed in the cylinder such that each of the corners of the solid is tangent to the cylinder wall. If water is then poured into the cylinder such that it reaches the rim, the volume of water is:

- A.
288(π – 5)

- B.
288(2π – 3)

- C.
288(3π – 5)

- D.
None of the above

Answer: Option C

**Explanation** :

Volume of the cylinder = π × 6^{2} × 24 = 864π cu. units

Diameter of the cylinder = 2 × 6 = 12 units

Side of the square base = $\frac{12}{\sqrt{2}}$ = 6$\sqrt{2}$ units

Volume of the rectangular solid

= (6$\sqrt{2}$)^{2} × 20 = 1440 cu. units

∴ Volume of the cylinder = 844π – 1440 = 288(3π – 5) cu. units

Hence, option (c).

Workspace:

**32. IIFT 2014 QA | Fact, Inference & Judgement**

The student fest in an Engineering College is to be held in one month’s time and no sponsorship has yet been arranged by the students. Finally the General Secretary (GS) of the student body took the initiative and decided to go alone for sponsorship collection. In fact, he is the only student doing the fund raising job on the first day. However, seeing his enthusiasm, other students also joined him as follows: on the second day, 2 more students join him; on the third day, 3 more students join the group of the previous day; and so on. In this manner, the sponsorship collection is completed in exactly 20 days. If an MBA student is twice as efficient as an Engineering student, the number of days which 11 MBA students would take to do the same activity , is:

- A.
70

- B.
80

- C.
90

- D.
100

Answer: Option A

**Explanation** :

Let engineering student collects fund of Rs. X on one day.

Number of students collecting funds on:

First day = 1

Second Day = 1 + 2

…

Twentieth day = 1 + 2 + 3 + 4 … + 20

Total collection = [1 + (1 + 2) + (1 + 2 + 3) + … + (1 + 2 + 3 + … + 20)] X

= $\sum _{i=1}^{20}\frac{n(n+1)}{2}$ = $\frac{1}{2}$$\left\{\sum _{i=1}^{20}{n}^{2}+\sum _{i=1}^{20}n\right\}$ ...(i)

∑${n}^{2}$ = $\frac{n(n+1)(2n+1)}{6}$ and

∑n = $\frac{n(n+1)}{2}$

Thus, from equation (i),

Total collection

= $\frac{1}{2}$$\left\{\frac{20(20+1)(2\times 20+1)}{6}+\frac{20(20+1)}{2}\right\}$

= 1540X

An MBA student collects 2X on a single day.

∴ Number of days required for 11 MBA students to complete the activity

= 1540X / 11 × 2X = 70

Hence, option (a).

Workspace:

**33. IIFT 2014 QA | Arithmetic - Percentage**

A pharmaceutical company manufactures 6000 strips of prescribed diabetic drugs for Rs. 8,00,000 every month. In July 2014, the company supplied 600 strips of free medicines to the doctors at various hospitals. Of the remaining medicines, it was able to sell 4/5^{th} of the strips at 25 pecent discount and the balance at the printed price of Rs. 250. Assuming vendor’s discount at the rate of a uniform 30 percent of the total revenue, the approximate percentage profit / loss of the pharmaceutical company in July 2014 is:

- A.
5.5 percent (profit)

- B.
4 percent (loss)

- C.
5.5 percent (loss)

- D.
None of the above

Answer: Option C

**Explanation** :

600 strips were given free to doctors.

Of 5400 strips, (4/5) × 5400 = 4320 strips were sold at 25% discount.

Revenue generated from these strips

= 250 × 0.75 × 4320 = Rs. 8,10,000

Revenue generated from (5400 – 4320 =) 1080 strips = 250 × 1080 = Rs. 2,70,000

Total revenue = Rs. 10,80,000

Vendor’s discount = 30% of the total revenue.

∴ Total earning = 70% of 1080000 = Rs. 7,56,000

Loss = Rs. 44,000

% loss = 5.5

Hence, option (c).

Workspace:

**34. IIFT 2014 QA | Arithmetic - Time, Speed & Distance**

A bouncing tennis ball is dropped from a height of 32 metre. The ball rebounds each time to a height equal to half the height of the previous bounce. The approximate distance travelled by the ball when it hits the ground for the eleventh time, is:

- A.
64 metre

- B.
96 metre

- C.
128 metre

- D.
150 metre

Answer: Option B

**Explanation** :

Distance covered when the ball hits the ground the first time = 32 m

The ball covers 16 + 16 = 32 m more when it hits the ground 2^{nd} time.

The ball covers 8 + 8 = 16 m more when it hits the ground 3^{rd} time.

And so on i.e., at the nth hit, the ball covers 32/2^{(n – 2)} more distance.

Thus, at the 11^{th} hit, the ball covers 32/29 m more.

Thus, the total distance the ball covers when it hits the ground 11^{th} time

= 32 + 32(1 + ½ + ¼ + … 1/2^{9})

= 32 + 32(1-1/2^{10})/(1-1/2) ≈ 32 + 64

= 96 m

Hence, option (b).

Workspace:

**35. IIFT 2014 QA | Modern Math - Permutation & Combination**

In an Engineering College in Pune, 8 males and 7 females have appeared for Student Cultural Committee selection process. 3 males and 4 females are to be selected. The total number of ways in which the committee can be formed, given that Mr. Raj is not to be included in the committee if Ms. Rani is selected, is:

- A.
1960

- B.
2840

- C.
1540

- D.
None of the above

Answer: Option C

**Explanation** :

Selecting 3 males from 8 and 4 females from 7 can be done in ^{8}C_{3} × ^{7}C_{4} = 56 × 35 = 1960

Raj and Rani together cannot be in the committee.

Selection of both Raj and Rani can happen in ^{7}C_{2} × ^{6}C_{3} = 21 × 20 = 420

Required number of ways = 1960 – 420 = 1540

Hence, option (c).

Workspace:

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