IIFT 2010 QA questions and solutions

1. IIFT 2010 QA | Algebra - Simple Equations

Find the sum of the following series;

$\frac{2}{1!}$ + $\frac{3}{2!}$ + $\frac{6}{3!}$ + $\frac{11}{4!}$ + $\frac{18}{5!}$ + ....

A.
3e - 1
B.
3(e - 1)
C.
3(e + 1)
D.
3e + 1

Answer: Option B

Explanation :

Let S = $\frac{2}{1!}$ + $\frac{3}{2!}$ + $\frac{6}{3!}$ + $\frac{11}{4!}$ + $\frac{18}{5!}$ + .... ∴ S = $\sum_{n = 1}^{\infty}$ $\frac{n^{2} - 2 n + 3}{n!}$

∴ S = $\sum_{n = 1}^{\infty}$ $\frac{n}{(n - 1)!}$ - 2 $\sum_{n = 1}^{\infty}$ $\frac{1}{(n - 1)!}$ + $\sum_{n = 1}^{\infty}$ $\frac{3}{n!}$

$\sum_{n = 1}^{\infty}$ $\frac{1}{n!}$ = e - 1 ....(by definition of exponential series)

$\sum_{n = 1}^{\infty}$ $\frac{1}{(n - 1)!}$ = e

∴ $\sum_{n = 1}^{\infty}$ = 1 + $(1 + \frac{1}{1!})$ + $(\frac{1}{1!} + \frac{1}{2!})$ + $(\frac{1}{2!} + \frac{1}{3!})$ + $(\frac{1}{3!} + \frac{1}{4!})$ + ..

= 2 + 2 $(\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \frac{1}{5!} + . . .)$

= 2 + 2(e - 1)

= (e - 1) + $\sum_{n = 1}^{\infty}$ $\frac{1}{(n - 2)!}$ + $\sum_{n = 1}^{\infty}$ $\frac{1}{(n - 1)!}$

= 2 + 2(e – 1) −2(e) +3(e −1) = 3(e – 1)

Hence, option (b).

2. IIFT 2010 QA | Algebra - Number Theory

How many positive integers ‘n’ can we form using the digits 3, 4, 4, 5, 6, 6, 7 if we want ‘n’ to exceed 6,000,000?

A.
320
B.
360
C.
540
D.
720

Answer: Option C

Explanation :

As n has to exceed 6,000,000, the first digit of n can be 6 or 7.

Case (i):

The first digit of n is 6. Then the other six digits are to be chosen from 3, 4, 4, 5, 6, 7.

They can be arranged in $\frac{6!}{2!}$ = 360 ways

Case (ii):

The first digit of n is 7.

Then the other six digits are to be chosen from 3, 4, 4, 5, 6, 6.

They can be arranged in $\frac{6!}{2! \times 2!}$ = 180 ways

∴ Total number of arrangements = 360 + 180 = 540

Hence, option (c).

3. IIFT 2010 QA | Arithmetic - Percentage

A Techno company has 14 machines of equal efficiency in its factory. The annual manufacturing costs are Rs. 42,000 and establishment charges are Rs. 12,000. The annual output of the company is Rs. 70,000. The annual output and manufacturing costs are directly proportional to the no. of machines. The share holders get 12.5% profit, which is directly proportional to the annual output of the company. If 7.14% machines remain closed throughout the year, then the percentage decrease in the amount of profit of the share holders would be:

A.
12%
B.
12.5%
C.
13.0%
D.
None of these

4. IIFT 2010 QA | Arithmetic - Percentage

Sun Life Insurance Company issues standard, preferred, and ultra-preferred policies. Among the company’s policy holders of a certain age, 50% are standard with a probability of 0.01 of dying in the next year, 30% are preferred with a probability 0.008 of dying in the next year, and 20% are ultra-preferred with a probability of 0.007 of dying in the next year. If a policy holder of that age dies in the next year, what is the probability of the deceased being a preferred policy holder?

A.
0.1591
B.
0.2727
C.
0.375
D.
None of these

Answer: Option B

Explanation :

Required probability

= $\frac{0.3 \times 0.008}{0.5 \times 0.01 + 0.3 \times 0.008 + 0.2 \times 0.007}$

= $\frac{0.0024}{0.005 + 0.0024 + 0.0014}$

= $\frac{0.0024}{0.0088}$ ≈ 0.2727

Hence, option (b).

5. IIFT 2010 QA | Arithmetic - Profit & Loss

A metro train from Mehrauli to Gurgoan has capacity to board 900 people. The fare charged (in RS.) is defined by the function

f = ${(54 - \frac{x}{32})}^{2}$

where ‘x’ is the number of the people per trip. How many people per trip will make the marginal revenue equal to zero?

A.
1728
B.
576
C.
484
D.
364

Answer: Option B

Explanation :

Fare per person f = ${(54 - \frac{x}{32})}^{2}$

∴ Total fare f₁ = x ${(54 - \frac{x}{32})}^{2}$

∴ Marginal Revenue is the change in total fare due to change in number of people by one unit. It is given by

$\frac{d f_{1}}{d x}$ = x $[2 (54 - \frac{x}{32}) (- \frac{1}{32})]$ + ${(54 - \frac{x}{32})}^{2}$

= $\frac{- x}{16}$ $(54 - \frac{x}{32})$ + ${(54 - \frac{x}{32})}^{2}$

= $(54 - \frac{x}{32})$ $(\frac{- x}{16} + 54 - \frac{x}{32})$

= $(54 - \frac{x}{32})$ $(54 - \frac{3 x}{2})$

Now $\frac{d f_{1}}{d x}$ = 0

∴ $(54 - \frac{x}{32})$ $(54 - \frac{3 x}{2})$ = 0

∴ x = 1728 or x = 576

But x ≤ 900

∴ x = 576

Hence, option (b).

6. IIFT 2010 QA | Geometry - Trigonometry

If each α, β, γ is a positive acute angle such that

sin(α + β + γ) = $\frac{1}{\sqrt{2}}$ , cosec(β + γ - α) = $\frac{2}{\sqrt{3}}$ and tan(γ + α - β) = $\frac{1}{\sqrt{3}}$

What are the values of α, β, γ?

A.
$(37 \frac{1}{2}, 52 \frac{1}{2}, 45)$
B.
(37, 53, 45)
C.
$(45, 37 \frac{1}{2}, 52 \frac{1}{2})$
D.
$(34 \frac{1}{2}, 55 \frac{1}{2}, 45)$

Answer: Option A

Explanation :

We evaluate options:

Option (1) Let α = 37.5, β = 52.5, γ = 45.

∴ α + β – γ = 45

∴ sin (α + β – γ) = $\frac{1}{\sqrt{2}}$

β + γ – α = 60

cosec(β + γ – α) = $\frac{2}{\sqrt{3}}$

γ + α – β = 30

tan(γ + α – β) = $\frac{1}{\sqrt{3}}$

Hence, option (a).

7. IIFT 2010 QA | Arithmetic - Profit & Loss

Shyam, Gopal and Madhur are three partners in a business. Their capitals are respectively Rs 4000, Rs 8000 and Rs 6000. Shyam gets 20% of total profit for managing the business. The remaining profit is divided among the three in the ratio of their capitals. At the end of the year, the profit of Shyam is Rs 2200 less than the sum of the profit of Gopal and Madhur. How much profit, Madhur will get?

A.
Rs.1600
B.
Rs.2400
C.
Rs.3000
D.
Rs.5000

Answer: Option B

Explanation :

Capitals of Shyam, Gopal and Madhur are in the ratio 2 : 4 : 3.

Let the total profit be x .

∴ By conditions,

0.8x × $\frac{7}{9}$ - $[0.8 x \times \frac{2}{9} + 0.2 x]$ = 2200

∴ x = 9000

∴ Madhur’s share in the profit

= $\frac{3}{9}$ × 9000 × 0.8 = Rs. 2400

Hence, option (b).

8. IIFT 2010 QA | Modern Math - Permutation & Combination

In how many ways can four letters of the word ‘SERIES’ be arranged?

A.
24
B.
42
C.
84
D.
102

9. IIFT 2010 QA | Geometry - Triangles

The area of a triangle is 6, two of its vertices are (1, 1) and (4, –1), the third vertex lies on y = x + 5. Find the third vertex.

A.
$(\frac{2}{5}, \frac{27}{5})$
B.
$(- \frac{3}{5}, \frac{22}{5})$
C.
$(\frac{3}{5}, \frac{28}{3})$
D.
None of these

Answer: Option A

Explanation :

If the coordinates of the vertices of a triangle are (x₁, y₁), (x₂, y₂) and (x₃, y₃) then its area is given by

A = $\frac{1}{2}$ $|\begin{matrix} x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \\ 1 & 1 & 1 \end{matrix}|$

∴ If the third vertex of the given triangle is (x, y)

6 = $\frac{1}{2}$ $|\begin{matrix} 1 & 4 & x \\ 1 & - 1 & y \\ 1 & 1 & 1 \end{matrix}|$

∴ 12 = 1(−1 − y) − 4(1 − y) + x(1 + 1)

∴ 12 = −1 – y − 4 + 4y + 2x

∴ 2x + 3y = 17 …(i)

But (x, y) also lies on y = x + 5 …(ii)

Solving (i) and (ii)

x = $\frac{2}{5}$ , y = $\frac{27}{5}$

Hence, option (a).

10. IIFT 2010 QA | Arithmetic - Ratio, Proportion & Variation

A small confectioner bought a certain number of pastries flavoured pineapple, mango and black-forest from the bakery, giving for each pastry as many rupees as there were pastry of that kind; altogether he bought 23 pastries and spent Rs.211; find the number of each kind of pastry that he bought, if mango pastry are cheaper than pineapple pastry and dearer than black-forest pastry.

A.
(10, 9, 4)
B.
(11, 9, 3)
C.
(10, 8, 5)
D.
(11, 8, 4)

11. IIFT 2010 QA | Algebra - Simple Equations

Find the root of the quadratic equation bx²– 2ax + a = 0

A.
$\frac{\sqrt{b}}{\sqrt{b \pm \sqrt{a - b}}}$
B.
$\frac{\sqrt{a}}{\sqrt{b \pm \sqrt{a - b}}}$
C.
$\frac{\sqrt{a}}{(\sqrt{a} \pm \sqrt{a - b})}$
D.
$\frac{\sqrt{a}}{\sqrt{a \pm \sqrt{a + b}}}$

Answer: Option C

Explanation :

bx² – 2ax + a = 0

Using the quadratic formula,

x = $\frac{2 a \pm \sqrt{4 a^{2} - 4 a b}}{2 b}$

= $\frac{2 a \pm 2 \sqrt{a^{2} - a b}}{2 b}$

= $\frac{a \pm \sqrt{a (a - b)}}{b}$

= $\frac{(a \pm \sqrt{a} \sqrt{a - b}) (a \mp \sqrt{a} \sqrt{a - b})}{b (a \mp \sqrt{a} \sqrt{a - b})}$

= $\frac{(a^{2} - a (a - b))}{b (a \pm \sqrt{a} \sqrt{a - b})}$

= $\frac{(a^{2} - a^{2} + a b)}{b (a \pm \sqrt{a} \sqrt{a - b})}$

= $\frac{a}{\sqrt{a} (\sqrt{a} \pm \sqrt{a - b})}$

= $\frac{\sqrt{a}}{(\sqrt{a} \pm \sqrt{a - b})}$

Hence, option (c).

12. IIFT 2010 QA | Arithmetic - Time & Work

Three Professors Dr. Gupta, Dr. Sharma and Dr. Singh are evaluating answer scripts of a subject. Dr. Gupta is 40% more efficient than Dr. Sharma, who is 20% more efficient than Dr. Singh. Dr. Gupta takes 10 days less than Dr. Sharma to complete the evaluation work. Dr. Gupta starts the evaluation work and works for 10 days and then Dr. Sharma takes over. Dr. Sharma evaluates for next 15 days and then stops. In how many days, Dr. Singh can complete the remaining evaluation work.

A.
7.2 days
B.
9.5 days
C.
11.5 days
D.
None of these

Answer: Option A

Explanation :

Let Dr. Gupta take x days to complete the evaluation work.

∴ Dr. Sharma takes x + 10 days

As Dr. Gupta is 40% more efficient than Dr. Sharma, we have

$\frac{1}{x}$ = $\frac{1.4}{x + 10}$

∴ x = 25

∴ x + 10 = 35

Also, Dr. Sharma is 20% more efficient than Dr. Singh.

∴ If Dr. Singh takes y days to complete the evaluation work,

$\frac{1}{35}$ = $\frac{1.2}{y}$

∴ y = 42

Now, let Dr. Singh complete the evaluation work in n days after Dr. Gupta has worked for 10 days and Dr. Sharma has worked for 15 days.

∴ $\frac{10}{25}$ + $\frac{15}{35}$ + $\frac{n}{42}$ = 1

∴ n = 7.2

Hence, option (a).

13. IIFT 2010 QA | Algebra - Simple Equations

If [x] is the greater integer less than or equal to ‘x’, then find the value of the following series

[ $\sqrt{1}$ ] + [ $\sqrt{2}$ ] + [ $\sqrt{3}$ ] + [ $\sqrt{4}$ ] + ... + [ $\sqrt{361}$ ]

A.
4389
B.
4839
C.
3498
D.
3489

Answer: Option A

Explanation :

[ $\sqrt{1}$ ] = 1

[ $\sqrt{2}$ ] = 1

[ $\sqrt{3}$ ] = 1

[ $\sqrt{4}$ ] = 1

[ $\sqrt{5}$ ] = 1

[ $\sqrt{8}$ ] = 1

[ $\sqrt{9}$ ] = 1

and so on.

Thus, [ $\sqrt{n}$ ] = k

where k² is the greatest perfect square less than or equal to n.

Also, the difference between two consecutive perfect squares = (k + 1)² – k² = 2k + 1

∴ The required sum is

$[\sum_{k = 1}^{18} k (2 k + 1)]$ + $\sqrt{361}$

= $[\sum_{k = 1}^{18} 2 k^{2} + k]$ + 19

= 2 $\frac{(18) (18 + 1) (2 \times 18 + 1)}{6}$ + $\frac{18 (18 + 1)}{2}$ + 9

= 4408

Hence, option (a).

Note: Correct answer option was not present in actual paper.

14. IIFT 2010 QA | Algebra - Logarithms

What is the value of $\sqrt{\frac{a}{b}}$ ,

If log₄ log₄ 4^{a - b} = 2 log₄ ( $\sqrt{a}$ - $\sqrt{b}$ ) + 1

A.
-5/3
B.
2
C.
5/3
D.
1

Answer: Option C

Explanation :

log₄ log₄ 4^{a - b} = 2log₄ $(\sqrt{a} - \sqrt{b})$ + 1

∴ log₄ (a - b) log₄ 4 = 2log₄ $(\sqrt{a} - \sqrt{b})$ + 1

∴ log₄ (a - b) = 2log₄ $(\sqrt{a} - \sqrt{b})$ + 1

∴ log₄ (a - b) = 2log₄ ${(\sqrt{a} - \sqrt{b})}^{2}$ + 1

∴ log₄ $[\frac{a - b}{{(\sqrt{a} - \sqrt{b})}^{2}}]$ = 1

∴ $\frac{a - b}{a + b - 2 \sqrt{a b}}$ = 4

∴ a - b = 4a + 4b - 8 $\sqrt{a b}$

∴ 8 $\sqrt{a b}$ = 3a + 5b

∴ 8 = 3 $\sqrt{\frac{a}{b}}$ + 5 $\sqrt{\frac{b}{a}}$

Let $\sqrt{\frac{a}{b}}$ = x

∴ 8 = 3x + $\frac{5}{x}$

∴ 8x = 3x² + 5

∴ 3x² - 8x + 5 = 0

∴ 3x² – 3x – 5x + 5 = 0

∴ 3x(x – 1) – 5(x – 1) = 0

∴(3x – 5)(x – 1) = 0

∴ x = $\frac{5}{3}$ or x = 1

But x ≠ 1 as $\sqrt{a}$ ≠ $\sqrt{b}$

∴ x ≠ 5/3

∴ $\sqrt{\frac{a}{b}}$ = $\frac{5}{3}$

Hence, option (c).

15. IIFT 2010 QA | Arithmetic - Time & Work

Three pipes A, B and C are connected to a tank. These pipes can fill the tank separately in 5 hours, 10 hours and 15 hours respectively. When all the three pipes were opened simultaneously, it was observed that pipes A and B were supplying water at 3/4th of their normal rates for the first hour after which they supplied water at the normal rate. Pipe C supplied water at 2/3rd of its normal rate for first 2 hours, after which it supplied at its normal rate. In how much time, tank would be filled.

A.
1.05 Hours
B.
2.05 Hours
C.
3.05 Hours
D.
None of these

Answer: Option C

Explanation :

Let the pipes work for n hours.

By the given conditions,

$\frac{1}{5 \times \frac{4}{3}}$ + $\frac{1}{10 \times \frac{4}{3}}$ + (n - 1) $[\frac{1}{5} + \frac{1}{10}]$ + $\frac{2}{15 \times \frac{3}{2}}$ + $\frac{n - 2}{15}$ = 1

∴ $\frac{3}{20}$ + $\frac{3}{40}$ + (n - 1) $\frac{3}{10}$ + $\frac{4}{45}$ + $\frac{n - 2}{15}$ = 1

∴ $\frac{9}{40}$ + $\frac{3 n - 3}{10}$ + $\frac{4}{45}$ + $\frac{n - 2}{15}$ = 1

∴ $\frac{9 n - 9}{30}$ + $\frac{2 n - 4}{30}$ = 1 - $\frac{9}{40}$ - $\frac{4}{45}$

∴ $\frac{11 n - 13}{30}$ = $\frac{360 - 81 - 32}{360}$

∴ 11n - 13 = $\frac{247 \times 30}{360}$

∴ n ≈ 3.05

Hence, option (c).

16. IIFT 2010 QA | Geometry - Trigonometry

The minimum value of 3^sinx + 3^cosx is

A.
2
B.
2 $(3^{- \frac{1}{\sqrt{2}}})$
C.
$3^{1 - \frac{1}{\sqrt{2}}}$
D.
None of these

Answer: Option B

Explanation :

f(x) = 3^sinx + 3^cos x

Using the inequality, AM ≥ GM

f(x) will have a minimum

where 3 ^sin x = 3 ^cos x

∴ sin x = cos x

∴ sin x = cos x = - $\frac{1}{\sqrt{2}}$

f(x)_min = $3^{- \frac{1}{\sqrt{2}}}$ + $3^{- \frac{1}{\sqrt{2}}}$

= 2 $(3^{- \frac{1}{\sqrt{2}}})$

Hence, option (b).

17. IIFT 2010 QA | Arithmetic - Average

In a B-School there are three levels of faculty positions i.e. Professor, Associate Professor and Assistant Professor. It is found that the sum of the ages of all faculty present is 2160, their average age is 36; the average age of the Professor and Associate Professor is 39; of the Associate Professor and Assistant Professor is 32 $\frac{8}{11}$ ; of the Professor and Assistant Professor is 36 $\frac{2}{3}$ , Had each Professor been 1 year older, each Associate Professor 6 years older, and each Assistant Professor 7 years older, then their average age would increase by 5 years. What will be the number of faculty at each level and their average ages?

A.
(16, 24, 20 : 45, 35, 30 years)
B.
(18, 24, 20 : 42, 38, 30 years)
C.
(16, 20, 24 : 50, 30, 30 years)
D.
None of these

Answer: Option A

Explanation :

Let the number of professors, associate professors and assistant professors be a, b and c respectively and their average ages be p, q and r respectively.

∴ a + b + c = $\frac{2160}{36}$ = 60 ...(i)

$\frac{a p + b q}{a + b}$ = 39

$\frac{b q + c r}{b + c}$ = 32 $\frac{8}{11}$

and $\frac{c r + a p}{a + c}$ = 36 $\frac{2}{3}$

Also,

$\frac{a (p + 1) + b (q + 6) + c (r + 7)}{a + b + c}$ = 41

As $\frac{a p + b q + c r}{a + b + c}$ = 36,

∴ $\frac{a + 6 b + 7 c}{a + b + c}$ = 5

a + 6b + 7c = 5a + 5b + 5c

b + 2c = 4a

∴ b = 4a – c

Substituting in (i),

a + 4a – 2c + c = 60

∴ 5a – c = 60 …(ii)

We find that option 1 satisfies equations (i) and (ii).

Further, the values of p, q and r in option 1 also satisfy the other equations.

Hence, option (a).

18. IIFT 2010 QA | Algebra - Logarithms

log₅ 2 is

A.
An integer
B.
A rational number
C.
A prime number
D.
An irrational number

Answer: Option D

Explanation :

Let log₅ 2 be rational.

Then log₅ 2 = $\frac{m}{n}$

∴ $5^{\frac{m}{n}}$ = 2

∴ ${(5^{m / n})}^{n}$ = $2^{n}$

∴ 5^m = 2ⁿ

However, number 2 raised to any positive integer power must be even, but 5 raised to any positive integer power must be odd.

Hence, we have a contradiction.

∴ log₅ 2 is irrational.

Hence, option (d).

19. IIFT 2010 QA | Geometry - Basics

In a square of side 2 meters, isosceles triangles of equal area are cut from the corners to form a regular octagon. Find the perimeter and area of the regular octagon.

A.
$\frac{16}{2 + \sqrt{2}}$ ; $\frac{4 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$
B.
$\frac{8}{2 + \sqrt{2}}$ ; $\frac{2 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$
C.
$\frac{16}{1 + \sqrt{2}}$ ; $\frac{3 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$
D.
None of these

Answer: Option D

Explanation :

Let the side of isosceles triangle be x.

∴ The side of octagon x $\sqrt{2}$ .

∴ x + x + x $\sqrt{2}$ = 2

∴ 2x + x $\sqrt{2}$ = 2

∴ x = $\frac{2}{2 + \sqrt{2}}$

∴ Side of octagon = x $\sqrt{2}$ = $\frac{2 \sqrt{2}}{2 + \sqrt{2}}$ = $\frac{2}{1 + \sqrt{2}}$

∴ Perimeter of octagon = 8 × $\frac{2}{1 + \sqrt{2}}$ = $\frac{16}{1 + \sqrt{2}}$ units

Area of octagon = Area of square – 4 × Area of isosceles triangle

= 2² - 4 × $\frac{1}{2}$ x²

= 4 - 4 × $\frac{1}{2}$ × ${(\frac{\sqrt{2}}{1 + \sqrt{2}})}^{2}$

= 4 - $\frac{4}{3 + 2 \sqrt{2}}$ = $\frac{8 (1 + \sqrt{2})}{3 + 2 \sqrt{2}}$ sq. units

Hence, option (d).

20. IIFT 2010 QA | Algebra - Number Theory

The smallest perfect square that is divisible by 7!

A.
44100
B.
176400
C.
705600
D.
19600

21. IIFT 2010 QA | Arithmetic - Percentage

A survey shows that 61%, 46% and 29% of the people watched “3 idiots”, “Rajneeti” and “Avatar” respectively. 25% people watched exactly two of the three movies and 3% watched none. What percentage of people watched all the three movies?

A.
39%
B.
11%
C.
14%
D.
7%

22. IIFT 2010 QA | Geometry - Triangles

In a triangle ABC the length of side BC is 295. If the length of side AB is a perfect square, then the length of side AC is a power of 2, and the length of side AC is twice the length of side AB. Determine the perimeter of the triangle.

A.
343
B.
487
C.
1063
D.
None of these

23. IIFT 2010 QA | Algebra - Number Theory

In a Green view apartment, the houses of a row are numbered consecutively from 1 to 49. Assuming that there is a value of ‘x’ such that the sum of the numbers of the houses preceding the house numbered ‘x’ is equal to the sum of the numbers of the houses following it. Then what will be the value of ‘x’?

A.
21
B.
30
C.
35
D.
42

Answer: Option C

Explanation :

Sum of numbers before x = sum of numbers after x

∴ 1 + 2 + 3 + … + (x – 1) = (x + 1) + (x + 2) + … + 49

Adding (1 + 2 + 3 + 4 + … + (x – 1) + x) on both sides

∴ 2(1 + 2 + 3 + … +(x – 1)) + x = (1 + 2 + 3 + … + 49)

∴ $\frac{2 \times (x - 1) x}{2}$ + x = $\frac{49 \times 50}{2}$

∴ x² = 49 × 25

∴ x = 7× 5 = 35

Hence, option (c).

24. IIFT 2010 QA | Arithmetic - Simple & Compound Interest

To start a new enterprise, Mr. Yogesh has borrowed a total of Rs. 60,000 from two money lenders with the interest being compounded annually, to be repaid at the end of two years. Mr. Yogesh repaid Rs.38, 800 more to the first money lender compared to the second money lender at the end of two years. The first money lender charged an interest rate, which was 10% more than what was charged by the second money lender. If Mr. Yogesh had instead borrowed Rs. 30,000 from each at their respective initial rates for two years, he would have paid Rs.7,500 more to the first money lender compared to the second. Then money borrowed by Mr. Yogesh from first money lender is?

A.
20,000
B.
35,000
C.
40,000
D.
42,000

Answer: Option C

Explanation :

Let the rate of second lender be y%, then rate of first lender is (y + 10)%.

When they lend equal amounts

30000 ${(1 + \frac{y + 10}{100})}^{2}$ - 30000 ${(1 + \frac{y}{100})}^{2}$ = 7500

∴ ${(\frac{110 + y}{100})}^{2}$ - ${(\frac{100 + y}{100})}^{2}$ = $\frac{1}{4}$

∴ 110² - 100² + 220y - 200y = 2500

y = 20%

Now we evaluate options and find that option 3 satisfies conditions.

40000 ${(1 + \frac{30}{100})}^{2}$ - 20000 ${(1 + \frac{20}{100})}^{2}$ = 38800

Hence, option (c).

25. IIFT 2010 QA | Algebra - Number System

Find the coefficient of x¹² in the expansion of (1 – x⁶)⁴(1 – x)^{– 4}

A.
113
B.
119
C.
125
D.
132

Answer: Option C

Explanation :

(1 – x⁶)⁴ (1 – x)^–4

= (1 – 4x⁶ + 6x¹² – 4x¹⁸ + x²⁴)(1 – x)^–4

To find the coefficient of x¹² in the given expression, we need to find the coefficients of x¹², x⁶ and x⁰ terms in (1 – x)^–4

We use the Binomial Theorem for negative coefficients.

Coefficient of x¹²
= $\frac{(- 4) (- 4 - 1) (- 4 - 1 - 2) . . (- 4 - 11)}{12!}$

= $\frac{4 \times 5 \times 6 \times . . . \times 15}{12!}$ = $\frac{13 \times 14 \times 15}{1 \times 2 \times 3}$

= 35 × 13

Coefficient of x⁶
= $\frac{(- 4) (- 4 - 1) . . . (- 4 - 5)}{6!}$

= $\frac{7 \times 8 \times 9}{3!}$ = 84

The coefficient of x⁰ is 1.

∴ The coefficient of x¹² is 35 × 13 + (– 4) × 84 + 6

= 125

Hence, option (c).

26. IIFT 2010 QA | Arithmetic - Time, Speed & Distance

Mukesh, Suresh and Dinesh travel from Delhi to Mathura to attend Janmasthmi Utsav. They have a bike which can carry only two riders at a time as per traffic rules. Bike can be driven only by Mukesh. Mathura is 300Km from Delhi. All of them can walk at 15Km/Hrs. All of them start their journey from Delhi simultaneously and are required to reach Mathura at the same time. If the speed of bike is 60Km/Hrs then what is the shortest possible time in which all three can reach Mathura at the same time.

A.
8 $\frac{2}{7}$ Hrs
B.
9 $\frac{2}{7}$ Hrs
C.
10 Hrs
D.
None of these

Answer: Option B

Explanation :

Let Mukesh take Suresh on his bike till B and leave him there to walk till C(Mathura). In the meanwhile, Dinesh keeps walking to reach D, Mukesh comes back picks Dinesh and then both ride to Mathura.

When Mukesh comes back, let us say he meets Dinesh at E.

Let AB = x, then BC = 300 – x

Since Dinesh walks at 15 kmph and bike’s speed is 60 kmph, we have AD = x/4.

∴ BD = $\frac{3 x}{4}$

∴ BE = $(\frac{60}{75})$ $(\frac{3 x}{4})$ = $\frac{3 x}{5}$

Hence, $\frac{300 - x}{15}$ = $\frac{\frac{3 x}{5} + \frac{3 x}{5} + 300 - x}{60}$

∴ 4(300 – x) = 300 + x/5

∴ 900 = 4x + $\frac{x}{5}$

$\frac{4500}{21}$ = x

x = $\frac{1500}{7}$ km

Hence, minimum time

= $\frac{x}{60}$ + $\frac{300 - x}{15}$

= $\frac{1500}{60 \times 7}$ + 20 - $\frac{1500}{15 \times 7}$

= $\frac{25}{7}$ + 20 - $\frac{100}{7}$

= 20 - $\frac{75}{7}$

= $\frac{65}{7}$ = 9 $\frac{2}{7}$

Hence, option (b).

27. IIFT 2010 QA | Geometry - Circles

In a rocket shape firecracker, explosive powder is to be filled up inside the metallic enclosure. The metallic enclosure is made up of a cylindrical base and conical top with the base of radius 8 centimeter. The ratio of height of cylinder and cone is 5:3. A cylindrical hole is drilled through the metal solid with height one third the height of the metal solid. What should be the radius of the hole, so that volume of the hole (in which gun powder is to be filled up) is half of the volume of metal solid after drilling?

A.
4 $\sqrt{3}$ cm
B.
4.0 cm
C.
3.0 cm
D.
None of these

Answer: Option A

Explanation :

Let the height of the cylinder be 5h than that of the conical part be 3h.
∴ Height of hole = $\frac{1}{3}$ (3h + 5h)
= $\frac{8 h}{3}$

Radius of cone = radius of cylinder = 8

Let radius of hole = r

Now, 1/2(Total volume – Volume of hole) = Volume of hole

∴ Total volume = 3Volume of hole

∴ π × 8² × 5h + $\frac{1}{3}$ × π × 8² × 3h = 3πr² $(\frac{8 h}{3})$

∴ 8² × 6 = 8r²

∴ r² = 8 × 6

∴ r = 4 $\sqrt{3}$ cm

Hence, option (a).

28. IIFT 2010 QA | Arithmetic - Percentage

A small and medium enterprise imports two components A and B from Taiwan and China respectively and assembles them with other components to form a toy. Component A contributes to 10% of production cost. Component B contributes to 20% of the production cost. Usually the company sells this toy at 20% above the production cost. Due to increase in the raw material and labour cost in both the countries, component A became 20% costlier and component B became 40% costlier. Owing to these reasons the company increased its selling price by 15%. Considering that cost of other components does not change, what will be the profit percentage, if the toy is sold at the new price?

A.
15.5%
B.
25.5%
C.
35.5%
D.
40%

Answer: Option B

Explanation :

Let initial production cost be 100.

Then cost of A = 10 and cost of B = 20

Selling price = 120

∴ Cost of rest =100 – Cost of A – Cost of B

= 100 – 10 – 20 = 70

New cost of A = 1 + $\frac{20}{100}$ × 10 = 12

New cost of B = 1 + $\frac{40}{100}$ × 20 = 28

∴ New cost = 28 + 12 + 70= 110

New selling price = $(1 + \frac{15}{100})$ × original selling price

= $(1 + \frac{15}{100})$ × 120

= 138

New profit = $\frac{N e w S P - N e w C P}{N e w C P}$ × 100

= $\frac{138 - 110}{110}$ × 100

≈ 25.5%

Hence, option (b).

29. IIFT 2010 QA | Geometry - Circles

What is the value of c² in the given figure, where the radius of the circle is ‘a’ unit.

A.
c² = a² + b² – 2ab cos θ
B.
c² = a² + b² – 2ab sin θ
C.
c² = a² – b² + 2ab cos θ
D.
None of these

Answer: Option A

Explanation :

By the cosine rule, in a ΔABC with ∠ACB = θ and AB = c, BC = a and AC = b,

cos θ = $\frac{a^{2} + b^{2} - c^{2}}{2 a b}$

∴ 2 ab cos θ = a² + b² – c²

∴ c² = a² + b² - 2ab cos θ

Hence, option (a).

30. IIFT 2010 QA | Algebra - Number Theory

How many subsets of {1, 2, 3 … 11} contain at least one even integer?

A.
1900
B.
1964
C.
1984
D.
2048

IIFT 2010 QA | Previous Year IIFT Paper

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