# IIFT 2013 QA | Previous Year IIFT Paper

**1. IIFT 2013 QA | Modern Math - Probability**

Suppose there are 4 bags. Bag 1 contains 1 black and *a*^{2} – 6*a* + 9 red balls, bag 2 contains 3 black and *a*^{2} – 6*a* + 7 red balls, bag 3 contains 5 black and *a*^{2} – 6*a* + 5 red balls and bag 4 contains 7 black and *a*^{2} – 6*a* + 3 red balls. A ball is drawn at random from a randomly chosen bag. The maximum value of probability that the selected ball is black, is

- A.
$\frac{16}{{a}^{2}-6a+10}$

- B.
$\frac{20}{{a}^{2}-6a+10}$

- C.
$\frac{1}{16}$

- D.
None of the above

Answer: Option D

**Explanation** :

Each bag has *a*^{2} – 6*a* + 10 balls.

Bags 1, 2, 3 and 4 contain 1, 3, 5 and 7 black balls respectively.

Probability of selecting a black ball from a specific bag is

$\frac{n}{{a}^{2}-6a+9}$

where n is the number of black balls in that bag.

A bag is selected at random.

∴ Probability of selecting a particular bag = $\frac{1}{4}$

∴ Probability that the ball selected from that randomly chosen bag is black

= $\frac{1}{4}$$\left(\frac{1}{{a}^{2}-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{3}{{a}^{2}-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{5}{{a}^{2}-6a+10}\right)$ + $\frac{1}{4}$$\left(\frac{7}{{a}^{2}-6a+10}\right)$

=$\frac{1}{4}$$\left(\frac{16}{{a}^{2}-6a+10}\right)$ = $\left(\frac{4}{{a}^{2}-6a+10}\right)$

Hence, option (d).

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**2. IIFT 2013 QA | Algebra - Number Theory**

If the product of the integers a, b, c and d is 3094 and if 1 < a < b < c < d, what is the product of b and c?

- A.
26

- B.
91

- C.
133

- D.
221

Answer: Option B

**Explanation** :

3094 = 2 × 7 × 13 × 17 = a × b × c × d

As 1 < a < b < c < d

a = 2, b = 7, c = 13, d = 17

∴ b × c = 7 × 13 = 91

Hence, option (b).

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**3. IIFT 2013 QA | Modern Math - Permutation & Combination**

Mrs. Sonia buys Rs. 249.00 worth of candies for the children of a school. For each girl she gets a strawberry flavoured candy priced at Rs. 3.30 per candy; each boy receives a chocolate flavoured candy priced at Rs. 2.90 per candy. How many candies of each type did she buy?

- A.
21, 57

- B.
57, 21

- C.
37, 51

- D.
27, 51

Answer: Option B

**Explanation** :

Let the number of strawberry and chocolate flavoured candies be a and b respectively.

∴ 3.3a + 2.9b = 249

Since there is only one equation with two unknowns, substitute the given options into the equations.

By substitution, a = 57 and b = 21 satisfy the given equation.

Hence, option (b).

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**4. IIFT 2013 QA | Geometry - Triangles**

There is a triangular building (ABC) located in the heart of Jaipur, the Pink City. The length of the one wall in east (BC) direction is 397 feet. If the length of south wall (AB) is perfect cube, the length of southwest wall (AC) is a power of three, and the length of wall in southwest (AC) is thrice the length of side AB, determine the perimeter of this triangular building.

- A.
3209 feet

- B.
3213 feet

- C.
3773 feet

- D.
3313 feet

Answer: Option D

**Explanation** :

Let AB = *a*^{3} and AC = 3^{n}

Also

AC = 3 × AB

∴ 3* ^{n}* = 3 ×

*a*

^{3}

∴ 3^{(n – 1)} = *a*^{3}

Now, let the perimeter be equal to *p*

*p *= BC + AC + AB

= 397 + 3* ^{n}* + 3

^{(n – 1)}

(*p* – 397) = 3^{(n – 1)} × (3 + 1)

= 3^{(n – 1)} × 4

Thus the LHS of the above equation should be a multiple of 3 and 4. Substitute the value of perimeter given in the options and verify this. Among the options, only (3313 – 397) is divisible by 3 and 4..

Hence, option (d).

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**5. IIFT 2013 QA | Modern Math - Permutation & Combination**

Out of 8 consonants and 5 vowels, how many words can be made, each containing 4 consonants and 3 vowels?

- A.
700

- B.
504000

- C.
3528000

- D.
7056000

Answer: Option C

**Explanation** :

Out of 8 consonants, 4 consonants can be selected in ^{8}C_{4 } = 70 ways.

Out of 5 vowels, 3 vowels can be selected in ^{5}C_{3}

= 10 ways.

These 7 selected letters can be arranged among themselves in 7! ways.

Thus total number of ways = 70 ´ 10 ´ 7! = 3528000

Hence, option (c).

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**6. IIFT 2013 QA | Algebra - Simple Equations**

If *x*^{2} + 3*x* – 10 is a factor of 3*x*^{4} + 2*x*^{3} – *ax*^{2} + *bx* – *a* + *b* – 4, then the closest approximate values of *a* and *b* are

- A.
25, 43

- B.
52, 43

- C.
52, 67

- D.
None of the above

Answer: Option C

**Explanation** :

*x*^{2} + 3*x* – 10 has 2 factors 2 and −5

*x*^{2} + 3*x* – 10 is a factor of 3*x*^{4} + 2*x*^{3 }– *ax*^{2} + *bx *– *a* + *b* – 4

Therefore 2 and −5 are also the factors of this expression.

Substitute 2 and −5, to get 2 equations

For *x* = 2,

48 + 16 – 4*a* + 2*b* – *a* + *b* – 4 = 0

∴ 5*a* – 3*b* = 60…… (I)

For *x* = −5,

1875 – 250 – 25*a* – 5*b* – *a* + *b* – 4 = 0

26*a* + 4*b* = 1621……… (II)

Solving (I) and (II)

*a* = 52 and *b* = 67

Hence, option (c).

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**7. IIFT 2013 QA | Algebra - Number Theory**

If the product of *n* positive integers is *n ^{n}*, then their sum is

- A.
a negative integer

- B.
equal to n

- C.
equal to n + $\frac{1}{n}$

- D.
never less than

*n*^{2}

Answer: Option D

**Explanation** :

As all the numbers are positive integers their sum cannot be negative. Thus option 1 is not possible.

Also, as all the numbers are positive integers their sum cannot be a fraction. Thus option 3 is not possible.

Hence, option (d).

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**8. IIFT 2013 QA | Arithmetic - Time, Speed & Distance**

A tennis ball is initially dropped from a height of 180 m. After striking the ground, it rebounds (3/5)^{th} of the height from which it has fallen. The total distance that the ball travels before it comes to rest is

- A.
540 m

- B.
600 m

- C.
720 m

- D.
900 m

Answer: Option C

**Explanation** :

Initial distance travelled = 180 m

Distance travelled after 1st rebound

Upward = $\frac{3}{5}$ × 180 = 108 m

Downward = 108 m

Total = 108 + 108 = 216 m

Distance travelled after 2nd rebound (upward and downward)

= $\frac{3}{5}$ × 108 × 2 = $\frac{3}{5}$ × 216 m

This gives an infinite G.P. with

a = 216 and r = $\frac{3}{5}$.

For an infinite G.P. with r <>

s = $\frac{a}{1-r}$ = $\frac{216}{1-{\displaystyle \frac{3}{5}}}$ = 540 m

Since initial distance was 180 m,

Total distance = (180 + 540) m = 720 m

Hence, option (c).

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**9. IIFT 2013 QA | Modern Math - Permutation & Combination**

In a sports meet for senior citizens organized by the Rotary Club in Kolkata, 9 married couples participated in Table Tennis mixed double event. The number of ways in which the mixed double team can be made, so that no husband and wife play in the same set, is

- A.
1512

- B.
1240

- C.
960

- D.
640

Answer: Option A

**Explanation** :

Two men can be selected in ^{9}C_{2} ways.

After selecting two men, two women can be selected in ^{7}C_{2} ways from (9 – 2 = 7) women, so that no husband and wife play in the same set.

Also, these selected 4 people can be grouped in 2 ways.

∴ The total number of mixed double teams

= ^{9}C_{2 }×^{7}C_{2 }× 2 = 1512

Hence, option (a).

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**10. IIFT 2013 QA | Arithmetic - Percentage**

A survey was conducted to test relative aptitudes in quantitative and logical reasoning of MBA applicants. It is perceived (prior to the survey) that 80 percent of MBA applicants are extremely good in logical reasoning, while only 20 percent are extremely good in quantitative aptitude. Further, it is believed that those with strong quantitative knowledge are also sound in data interpretation, with conditional probability as high as 0.87. However, some MBA applicants who are extremely good in logical reasoning can be also good in data interpretation, with conditional probability 0.15. An applicant surveyed is found to be strong in data interpretation. The probability that the applicant is also strong in quantitative aptitude is

- A.
0.4

- B.
0.6

- C.
0.8

- D.
0.9

Answer: Option B

**Explanation** :

Suppose 100 MBA applicants were surveyed.

80 of them are good in logical reasoning and 20 are good in quantitative aptitude.

0.87 × 20 ≈ 17 are good in quantitative aptitude and in data interpretation as well.

0.15 × 80 = 12 are good in logical reasoning and in data interpretation as well.

Thus, there are 17 + 12 = 29 MBA applicants those are strong in data interpretation.

Of the 29 MBA applicants 17 are strong in quantitative aptitude.

The required probability = $\frac{17}{29}$ = 0.58

Hence, option (b).

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**11. IIFT 2013 QA | Geometry - Coordinate Geometry**

Your friend’s cap is in the shape of a right circular cone of base radius 14 cm and height 26.5 cm. The approximate area of the sheet required to make 7 such caps is

- A.
6750 sq cm

- B.
7280 sq cm

- C.
8860 sq cm

- D.
9240 sq cm

Answer: Option D

**Explanation** :

Let the base radius, height, slant height be r, h and l respectively.

r = 14 cm

h = 26.5 cm

∴ l = $\sqrt{{r}^{2}+{h}^{2}}$

∴ l = 29.97 cm ≈ 30 cm

To make a conical cap from a sheet of paper, the curved surface area of cap must be calculated.

Curved surface area = π × r × l

To calculate the area of the sheet required to make 7 caps,

total area = 7 × π × r × l = 7 × $\frac{22}{7}$ × 14 × 3

= 9240 sq.cm.

Hence, option (d).

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**12. IIFT 2013 QA | Geometry - Coordinate Geometry**

In an engineering college there is a rectangular garden of dimensions 34 m by 21 m. Two mutually perpendicular walking corridors of 4 m width have been made in the central part and flowers have been grown in the rest of the garden. The area under the flowers is

- A.
320 sq.m

- B.
400 sq.m

- C.
510 sq.m

- D.
630 sq.m

Answer: Option C

**Explanation** :

The following diagram displays the garden along with the paths,

The shaded area represents the flower beds,

For each rectangular flower bed

breadth = $\frac{21-4}{2}$ = 8.5 m

length = $\frac{34-4}{2}$ = 15 m

There are 4 such flower beds

Therefore, total area = 4 × 15 × 8.5 = 510 m^{2}

Hence, option (c).

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**13. IIFT 2013 QA | Arithmetic - Percentage**

If decreasing 70 by X percent yields the same result as increasing 60 by X percent, then X percent of 50 is

- A.
3.84

- B.
4.82

- C.
7.10

- D.
The data is insufficient to answer the question

Answer: Option A

**Explanation** :

70 - $\frac{x}{100}$ × 70 = 60 + $\frac{x}{100}$ × 60

∴ 10 = $\frac{70x}{100}$ + $\frac{60x}{100}$

∴ x = $\frac{100}{13}$

Calculating x % of 50,

$\frac{100}{13}$ × $\frac{1}{100}$ × 50 = 3.84

Hence, option (a).

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**14. IIFT 2013 QA | Geometry - Basics**

A rod is cut into 3 equal parts. The resulting portions are then cut into 12, 18 and 32 equal parts, respectively. If each of the resulting portions have integer length, the minimum length of the rod is

- A.
6912 units

- B.
864 units

- C.
288 units

- D.
240 units

Answer: Option B

**Explanation** :

As the three portions of the rod correspond to 12, 18 and 32 equal parts, their length will be equal to LCM of 12, 18 and 32.

LCM of 12, 18 and 32 = 288

Since there are three equal parts of the rod, total length = 288 × 3 = 864

Hence, option (b).

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**15. IIFT 2013 QA | Algebra - Logarithms**

If ${\mathrm{log}}_{10}$ x - ${\mathrm{log}}_{10}$$\sqrt[3]{x}$ = 6 ${\mathrm{log}}_{x}$ 10 then the value of x is

- A.
10

- B.
30

- C.
100

- D.
1000

Answer: Option D

**Explanation** :

${\mathrm{log}}_{10}$ x - ${\mathrm{log}}_{10}$ $\sqrt[3]{x}$ = $\frac{6}{{\mathrm{log}}_{10}x}$

∴ ${\mathrm{log}}_{10}$ x - $\frac{1}{3}$ ${\mathrm{log}}_{10}$ x = $\frac{6}{{\mathrm{log}}_{10}x}$

${\mathrm{log}}_{10}$ x = a

∴ a^{2} - $\frac{1}{3}$ a^{2} = 6

∴ a^{2} = 9

∴ a = ±3

As log cannot be negative a = 3

∴ ${\mathrm{log}}_{10}$ x = 3

∴ x = 1000

Hence, option (d).

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**16. IIFT 2013 QA | Arithmetic - Time & Work**

A mother along with her two sons is entrusted with the task of cooking Biryani for a family get-together. It takes 30 minutes for all three of them cooking together to complete 50 percent of the task. The cooking can also be completed if the two sons start cooking together and the elder son leaves after 1 hour and the younger son cooks for further 3 hours. If the mother needs 1 hour less than the elder son to complete the cooking, how much cooking does the mother complete in an hour?

- A.
33.33%

- B.
50%

- C.
66.67%

- D.
None of the above

Answer: Option B

**Explanation** :

Let the number of days required to complete the work by mother, elder son and younger son be m, e and y hours respectively.

Therefore, work done by mother; elder son and younger son in one day is 1/m, 1/e and 1/n respectively.

Also m = e – 1 (∵ Mother take one hour less than the elder son)

Based on the conditions given,

$\frac{1}{m}$ + $\frac{1}{e}$ + $\frac{1}{y}$ = 1 ...(1)

$\frac{1}{e}$ + $\frac{1}{y}$ + $\frac{3}{y}$ = 1 ...(2)

$\frac{1}{y}$ = $\frac{e-1}{4e}$

Also m = e – 1

Substituting values of (1/m) and (1/y) in (1)

$\frac{1}{e-1}$ + $\frac{1}{e}$ + $\frac{e-1}{4e}$ = 1

Solving,

e = 3

Therefore, m = 2 and work done in 1 hour = 50%

Hence, option (b).

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**17. IIFT 2013 QA | Arithmetic - Time, Speed & Distance**

It was a rainy morning in Delhi when Rohit drove his mother to a dentist in his Maruti Alto. They started at 8.30 AM from home and Rohit maintained the speed of the vehicle at 30 Km/hr. However, while returning from the doctor’s chamber, rain intensified and the vehicle could not move due to severe water logging. With no other alternative, Rohit kept the vehicle outside the doctor’s chamber and returned home along with his mother in a rickshaw at a speed of 12 Km/hr. They reached home at 1.30 PM. If they stayed at the doctor’s chamber for the dental check-up for 48 minutes, the distance of the doctor’s chamber from Rohit’s house is

- A.
15 km

- B.
30 km

- C.
36 km

- D.
45 km

Answer: Option C

**Explanation** :

Let the distance between the doctor’s chamber and Rohit’s house be x km.

Total time spent = 5 hours

Time spent at the doctor’s chamber

= $\frac{48}{60}$ = $\frac{4}{5}$

From the given conditions,

$\frac{x}{30}$ + $\frac{x}{12}$ + $\frac{4}{5}$ = 5

∴ x = 36

Hence, option (c).

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**18. IIFT 2013 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two alloys of aluminium have different percentages of aluminium in them. The first one weighs 8 kg and the second one weighs 16 kg. One piece each of equal weight was cut off from both the alloys and first piece was alloyed with the second alloy and the second piece alloyed with the first one. As a result, the percentage of aluminium became the same in the resulting two new alloys. What was the weight of each cut-off piece?

- A.
3.33 kg

- B.
4.67 kg

- C.
5.33 kg

- D.
None of the above

Answer: Option C

**Explanation** :

Let weight of the cut-off piece = X kg

Let percentage of aluminium in 8 kg and 16 kg alloy be a and b respectively.

∴ $\frac{(8-X)a+Xb}{8\times 100}$ = $\frac{(16-X)b+Xa}{16\times 100}$

∴ 16a – 2Xa + 2Xb = 16b – Xb + Xa

∴ 16 – 2X = X and 2X = 16 – X

∴ X = $\frac{16}{3}$ = 5.33 kg

Hence, option (c).

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**19. IIFT 2013 QA | Arithmetic - Percentage**

Three years ago, your close friend had won a lottery of Rs. 1 crore. He purchased a flat for Rs. 40 lakhs, a car for Rs. 20 lakhs and shares worth Rs. 10 lakhs. He put the remaining money in a bank deposit that pays compound interest @ 12 percent per annum. If today, he sells off the flat, the car and the shares at certain percentage of their original value and withdraws his entire money from the bank, the total gain in his assets is 5%. The closest approximate percentage of the original value at which he sold off the three items is

- A.
60 percent

- B.
75 percent

- C.
90 percent

- D.
105 percent

Answer: Option C

**Explanation** :

Principal put in bank deposit

= 1,00,00,000 – 40,00,000 – 20,00,000 – 10,00,000

= Rs. 30,00,000

Amount after three years

= 3000000 × (1.12)^{3} = Rs. 42,14,784

Total gain after 3 years is 5%.

Total value = 10000000 × 1.05 = Rs. 1,05,00,000

10500000 − 4214784 = 6285216

Let *x* be the percentage at which he sold off the three items.

∴ $\frac{x\times 7000000}{100}$ = 6285216

Solving this, x ≈ 90%

Hence, option (c).

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**20. IIFT 2013 QA | Algebra - Logarithms**

If log_{13} log_{21} {$\sqrt{x+21}$ + $\sqrt{x}$} = 0 then the value of x is

- A.
21

- B.
13

- C.
81

- D.
None of the above

Answer: Option D

**Explanation** :

log_{13} log_{21} {$\sqrt{x+21}$ + $\sqrt{x}$} = 0

∴ log_{21} {$\sqrt{x+21}$ + $\sqrt{x}$} = 13° = 1

∴ {$\sqrt{x+21}$ + $\sqrt{x}$} = 21^{1} = 21

∴ $\sqrt{x+21}$= 21 - $\sqrt{x}$

Squaring both sides,

x + 21 = 441 + x - 42$\sqrt{x}$

∴ 42$\sqrt{x}$ = 420

∴ x = 100

Hence, option (d).

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**21. IIFT 2013 QA | Algebra - Simple Equations**

If *x* is real, the smallest value of the expression 3*x*^{2} – 4*x* + 7 is:

- A.
2/3

- B.
3/4

- C.
7/9

- D.
None of the above

Answer: Option D

**Explanation** :

The minimum value of expression *ax*^{2} + *bx* + *c*

= $\frac{(4ac-{b}^{2})}{4a}$ = $\frac{84-16}{12}$ = $\frac{17}{3}$

Hence, option (d).

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**22. IIFT 2013 QA | Arithmetic - Average**

The average of 7 consecutive numbers is P. If the next three numbers are also added, the average shall

- A.
remain unchanged

- B.
increase by 1

- C.
increase by 1.5

- D.
increase by 2

Answer: Option C

**Explanation** :

Since the question mention 7 consecutive natural numbers, the first 7 natural numbers can also be considered.

The first 7 natural numbers are (1, 2, 3, 4, 5, 6, 7) and their average is 4.

When the next 3 numbers i.e. 8, 9 and 10 are added, the new average is 5.5.

∴ Average increases by (5.5 – 4) = 1.5

Hence, option (c).

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**23. IIFT 2013 QA | Arithmetic - Time, Speed & Distance**

The duration of the journey from your home to the College in the local train varies directly as the distance and inversely as the velocity. The velocity varies directly as the square root of the diesel used per km., and inversely as the number of carriages in the train. If, in a journey of 70 km. in 45 minutes with 15 carriages, 10 litres of diesel is required, then the diesel that will be consumed in a journey of 50 km. in half an hour with 18 carriages is

- A.
2.9 litres

- B.
11.8 litres

- C.
15.7 litres

- D.
None of the above

Answer: Option B

**Explanation** :

T = $\frac{K\times D\times C}{\sqrt{A}}$

Where T is the time taken, D is the distance, C is the number of carriages, A is the diesel used per km and K is the proportionality constant.

Based on the data given:

45 = $\frac{K\times 70\times 15}{\sqrt{{\displaystyle \frac{1}{7}}}}$

K = $\frac{3}{70\sqrt{7}}$

T' = $\frac{3\times D\text{'}\times C\text{'}}{70\sqrt{7}\times \sqrt{A\text{'}}}$

30 = $\frac{3\times 50\times 18}{70\sqrt{7}\times \sqrt{A\text{'}}}$

Solving, A’ = 11.8 liters

Hence, option (b).

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**24. IIFT 2013 QA | Arithmetic - Time & Work**

Capacity of tap Y is 60% more than that of X. If both the taps are opened simultaneously, they take 40 hours to fill the rank. The time taken by Y alone to fill the tank is

- A.
60 hours

- B.
65 hours

- C.
70 hours

- D.
75 hours

Answer: Option B

**Explanation** :

Tap X does a units of work in 1 hour.

∴ Tap Y does 1.6a units of work in 1 hour.

∴ In 1 hour Tap X and Y together do 2.6a units of work.

∴ Work done by Tap X and Y in 40 hours = 2.6a × 40

∴ Time taken by Tap Y alone to do this work

= $\frac{2.6a\times 40}{1.6a}$ = 65 hours

Hence, option (b).

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