IIFT 2011 QA | Previous Year IIFT Paper
Mandeep and Jagdeep had gone to visit Ranpur, which is a seaside town and also known for the presence of the historical ruins of an ancient kingdom. They stayed in a hotel which is exactly 250 meters away from the railway station. At the hotel, Mandeep and Jagdeep learnt from a tourist information booklet that the distance between the sea-beach and the gate of the historical ruins is exactly 1 km. Next morning they visited the sea-beach to witness sunrise and afterwards decided to have a race from the beach to the gate of the ruins. Jagdeep defeated Mandeep in the race by 60 meters or 12 seconds. The following morning they had another round of race from the railway station to the hotel. How long did Jagdeep take to cover the distance on the second day?
- A.
53 seconds
- B.
47 seconds
- C.
51 seconds
- D.
45 seconds
Answer: Option B
Explanation :
Jagdeep defeated Mandeep in a 1000 m race by 60 m or 12 s.
This means that Mandeep would travel 60 m in 12 s.
∴ Mandeep’s speed = 5 m/s
∴ Mandeep covered 1000 m in 1000/5 = 200 s
∴ Jagdeep covered 1000 m in 200 – 12 = 188 s
∴ Jagdeep covered 250 m in 47 s.
Hence, option (b).
Workspace:
Sujoy, Mritunjoy and Paranjoy are three friends, who have worked in software firms Z Solutions, G Software’s and R Mindpower respectively for decade. The friends decided to float a new software firm named XY Infotech in January 2010. However, due to certain compulsions, Mritunjoy and Paranjoy were not able to immediately join the start-up in the appointed time. It was decided between friends that Sujoy will be running the venture as the full time director during 2010, and Mritunjoy and Paranjoy will be able to join the business only in January 2011. In order to compensate Sujoy for his efforts, it was decied that he will receive 10 percent of the profits and in the first year will invest lesser amount as compared to his friends. The remaining profit will be distributed among the friends in line with their contribution. Sujoy invested Rs. 35,000/- for 12 months, Mritunjoy invested Rs. 1,30,000/- for 6 months and Paranjoy invested Rs. 75,000/- for 8 months. If the total profit earned during 2010 was Rs. 4,50,000/-, then Paranjoy earned a profit of:
- A.
Rs. 1, 75, 500
- B.
Rs. 1, 35, 500
- C.
Rs. 1, 39, 500
- D.
None of the above
Answer: Option D
Explanation :
Sujoy receives 10% of the profit for his efforts.
Remaining 90% of the profit is distributed among the friends in the line with their contribution.
Now, Sujoy invested Rs. 35,000 for 12 months, Mritunjoy invested Rs. 1,30,000 for 6 months, and Paranjoy invested Rs. 75,000 for 8 months.
∴ Sujoy’s, Mritunjoy’s and Paranjoy’s investments are in the ratio
35000 × 12 : 130000 × 6 : 75000 × 8 ≡ 7 : 13 : 10
∴ Paranjoy's share = × 10 = Rs. 135000
Hence, option (d).
Workspace:
In Bilaspur village, 12 men and 18 boys completed construction of a primary health center in 60 days, by working for 7.5 hours a day. Subsequently the residents of the neighbouring Harigarh village also decided to construct a primary health center in their locality, which would be twice the size of the facility build in Bilaspur. If a man is able to perform the work equal to the same done by 2 boys, then how many boys will be required to help 21 men to complete the work in Harigarh in 50 days, working 9 hours a day?
- A.
45 boys
- B.
48 boys
- C.
40 boys
- D.
42 boys
Answer: Option D
Explanation :
A man is able to perform the work equal to the same done by 2 boys.
∴ 18 boys ≡ 9 men
Work done by 12 men and 18 boys, working 7.5 hours a day, in 60 days = (12 + 9) × 7.5 × 60
= 21 × 7.5 × 60 man hours
As the facility in Harigarh is twice that in Bilaspur, man hours required to build health facility in Harigarh
= 2 × 21 × 7.5 × 60
Let a boys be required to complete the work in 50 days, working 9 hours a day.
∴ we have,
(21 + a/2) × 9 × 50 = 2 × 21 × 7.5 × 60
∴ a = 42
Hence, option (d).
Workspace:
÷ = ?
- A.
- B.
- C.
- D.
None of the above
Answer: Option A
Explanation :
= ×
=
Hence, option (a).
Workspace:
If = , find the value of
- A.
- B.
- C.
- D.
None of the above
Answer: Option C
Explanation :
=
∴ =
∴ =
Hence, option (c).
Workspace:
While preparing for a management entrance examination Romit attempted to solve three paper, namely Mathematics, Verbal English and Logical Analysis, each of which have the full marks of 100. It is observed that one-third of the marks obtained by Romit in Logical Analysis is greater than half of his marks obtained in Verbal English By 5. He has obtained a total of 210 marks in the examination and 70 marks in Mathematics. What is the difference between the marks obtained by him in Mathematics and Verbal English?
- A.
40
- B.
10
- C.
20
- D.
30
Answer: Option C
Explanation :
Let Romit score l and v marks in Logical Analysis and Verbal English respectively.
∴ l + v + 70 = 210
∴ l + v = 140 …(i)
Now, = + 5 ...(ii)
Solving i and ii, we get,
l = 90, and v = 50
Hence, required difference = 70 – 50 = 20
Hence, option (c).
Workspace:
Aniket and Animesh are two colleagues working in PQ Communications, and each of them earned an investible surplus of Rs. 1, 50, 000/- during a certain period. While Animesh is a risk-averse person, Aniket prefers to go for higher return opportunities. Animesh uses his entire savings in Public Provident Fund (PPF) and National Saving Certificates (NSC). It is observed that one-third of the savings made by Animesh in PPF is equal to one-half of his savings in NSC. On the other hand, Aniket distributes his investible funds in share market, NSC and PPF. It is observed that his investments in share market exceeds his savings in NSC and PPF by Rs. 20,000/- and Rs. 40,000/- respectively. The difference between the amount invested in NSC by Animesh and Aniket is:
- A.
Rs. 25,000/-
- B.
Rs. 15,000/-
- C.
Rs. 20,000/-
- D.
Rs. 10,000/-
Answer: Option D
Explanation :
Let Animesh invest Rs. a in PPF and Rs. b in NSC.
∴ =
Also, a + b = 150000
∴ 5b/2 = 150000
∴ b = 60000
Let Aniket invest Rs. x in PPF.
∴ He invests Rs. (x + 40000) in shares and Rs. (x + 20000) in NSC.
Also, x + x + 40000 + x + 20000 = 150000
∴ x = 30000
∴ Aniket’s investment in NSC = Rs. 50000
Animesh’s investment in NSC = Rs. 60000
∴ Difference = Rs. 10000
Hence, option (d).
Workspace:
In March 2011, EF Public Library purchased a total of 15 new books published in 2010 with a total expenditure of Rs. 4500. Of these books, 13 books were purchased from MN Distributors, while the remaining two were purchased from UV Publishers. It is observed that one-sixth of the average price of all the 15 books purchased is equal to one-fifth of the average price of the 13 books obtained from MN Distributors. Of the two books obtained from UV Publishers, if one-third of the price of one volume is equal to one-half of the price of the other, then the price of the two books are
- A.
Rs. 900/- and Rs. 600/-
- B.
Rs. 600/- and Rs. 400/-
- C.
Rs. 750/- and Rs. 500/-
- D.
None of the above
Answer: Option C
Explanation :
Average price of 15 books = 4500/15 = Rs. 300
Let the average price of 13 books purchased from MN distributors be a.
∴ 300/6 = a/5
∴ a = 250
∴ Price of books purchased from MN distributirs = 250 × 13 = 3250
∴ Price of books purchased from UV Publishers = 4500 – 3250 = Rs. 1250
Let the two books purchased from UV publishers cost Rs. x and Rs. y.
∴ x/3 = y/2
As x + y = 1250
∴ x = 750 and y = 500
Hence, option (c).
Workspace:
2 years ago, one-fifth of Amita’s age was equal to one-fourth of the age of Sumita, and the average of their age was 27 years. If the age of Paramita is also considered, the average age of three of them declines to 24. What will be the average age of Sumita and Paramita 3 years from now?
- A.
25 years
- B.
26 years
- C.
27 years
- D.
Cannot be determined
Answer: Option B
Explanation :
By conditions,
=
∴ 4a = 5s – 2 …(i)
Also, = 27
∴ a + s = 58 …(ii)
Solving (i) and (ii),
a = 32 and s = 26
Also, = 24
∴ p = 20
∴ Average age of Sumita and Paramita 3 years from now = (s + p + 6)/2 = 26 years.
Hence, option (b).
Workspace:
An old lady engaged a domestic help on the condition that she would pay him Rs. 90 and a gift after service of one year. He served only 9 months and received the gift and Rs. 65. Find the value of the gift.
- A.
Rs. 10
- B.
Rs. 12
- C.
Rs. 15
- D.
None of the above
Answer: Option A
Explanation :
Let the price of the gift be x.
Hence, total value to be paid to domestic help for one year = 90 + x
∴ Value of domestic help for 9 months = 3/4 × (90 + x)
∴ 65 + x = 3/4 × (90 + x)
∴ x = 10
Hence, option (a).
Workspace:
There are four prime numbers written in ascending order of magnitude. The product of the first three is 7429 and last three is 12673. Find the first number.
- A.
19
- B.
17
- C.
13
- D.
None of the above
Answer: Option B
Explanation :
7429 = 17 × 19 × 23
17, 19 and 23 are consecutive primes.
Also, 12673 = 19 × 23 × 29
Hence, first number = 17
Hence, option (b).
Workspace:
A rectangular piece of paper is 22 cm. long and 10 cm. wide. A cylinder is formed by rolling the paper along its length. Find the volume of the cylinder.
- A.
175 cm3
- B.
180 cm3
- C.
185 cm3
- D.
None of the above
Answer: Option D
Explanation :
Cylinder is formed by rolling the paper along its length.
Hence, circumference of cylinder = 22
∴ 2πr = 22
∴ r = 7/2
∴ Volume of cylinder = πr2h = × × 10
= 385 cm3
Hence, option (d).
Workspace:
Find the value of x from the following equation:
Log103 + log10 (4x+1) = log10 (x+1) + 1
- A.
2/7
- B.
7/2
- C.
9/2
- D.
None of the above
Answer: Option B
Explanation :
log10 3 + log10 (4x + 1) = log10 (x + 1) + 1
∴ log10 3 + log10 (4x + 1) – log10 (x + 1) = 1
∴ log10 [3(4x + 1)/(x + 1)] = 1
∴ (12x + 3)/(x + 1) = 10
∴ x = 7/2
Hence, option (b).
Workspace:
Consider the volumes of the following objects and arrange them in decreasing order:
i. A parallelepiped of length 5 cm, breadth 3 cm and height 4 cm
ii. A cube of each side 4 cm.
iii. A cylinder of radius 3 cm and length 3 cm
iv. A sphere of radius 3 cm
- A.
iv, iii, ii, i
- B.
iv, ii, iii, i
- C.
iv, iii, i, ii
- D.
None of the above
Answer: Option A
Explanation :
i. Volume of parallelepiped = 5 × 4 × 3 = 60 cm3
ii. Volume of cube = 4 × 4 × 4 = 64 cm3
iii. Volume of cylinder = πr2h = π × 3 × 3 × 3 > 81 cm3
iv. Volume of sphere = 4/3 × πr3 = 36π cm3 > 81 cm3
∴ Required order is iv, iii, ii, i.
Hence, option (a).
Workspace:
If x satisfies the inequality |x – 1| + |x - 2| + |x - 3| ≥ 6, then:
- A.
0 ≤ x ≤ 4
- B.
x ≤ 0 or x ≥ 4
- C.
x ≤ -2 or x ≥ 3
- D.
None of the above
Answer: Option B
Explanation :
|x – 1| + |x – 2| + |x – 3| ≥ 6
Now, consider the following cases,
Case I:
(x < 1)
In this case, x – 1, x – 2 and x – 3 all are negative.
∴ |x – 1| + |x – 2| + |x – 3| = 1 – x + 2 – x + 3 – x = 6 – 3x ≥ 6
∴ x ≤ 0, which is possible.
Case II:
(1 ≤ x ≤ 2)
In this case, only x – 2, and x – 3 are negative.
Hence, |x – 1| + |x – 2| + |x – 3| = x – 1 + 2 – x + 3 – x = 4 – x ≥ 6
∴ x ≤ –2
This is not possible.
Hence there is no solution in this range.
Case III:
(2 ≤ x ≤ 3)
In this case, only x – 3 is negative.
Hence, |x – 1| + |x – 2| + |x – 3| = x – 1 + x – 2 + 3 – x = x ≥ 6
This also is not possible.
Case IV:
(x ≥ 3)
In this case, all are positive.
Hence, |x – 1| + |x – 2| + |x – 3| = 3x – 6 ≥ 6
∴ x ≥ 4
This is possible.
Hence, the required range is x ≤ 0 and x ≥ 4
Hence, option (b).
Workspace:
A five digit number divisible by 3 is to be formed using the numerals 0, 1, 2, 3, 4 and 5 without repetition. The total number of ways in which this can be done is:
- A.
220
- B.
600
- C.
240
- D.
None of the above
Answer: Option D
Explanation :
The five digit number will either have 0 or it will not have 0.
Case i:
0 is included.
The five digits must be 0, 1, 2, 4, 5 (as the number has to be divisible by 3)
The count of five-digit numbers = 4 × 4 × 3 × 2 × 1 = 96
Case ii:
0 is excluded.
In this case, the five digits should be 1, 2, 3, 4 and 5.
Number of five digit numbers = 5 × 4 × 3 × 2 × 1 = 120
∴ Required count = 120 + 96 = 216
Hence, option (d).
Workspace:
If 2, a, b, c, d, e, f and 65 form an arithmetic progression, find out the value of ‘e’.
- A.
48
- B.
47
- C.
41
- D.
None of the above
Answer: Option B
Explanation :
65 is the eighth term of the AP.
∴ 2 + (8 – 1)r = 65, where r is the common difference.
∴ r = 9
e is the sixth term.
∴ e = 2 + 5(9) = 47
Hence, option (b).
Workspace:
A contract is to be completed in 56 days and 104 men are set to work, each working 8 hours a day. After 30 days, 2/5th of the work is finished. How many additional men may be employed so that work may be completed on time, each man now working 9 hours per day?
- A.
56 men
- B.
65 men
- C.
46 men
- D.
None of the above
Answer: Option A
Explanation :
Work done by 104 men, working 8 hours a day, in 30 days = 104 × 8 × 30 man hours.
This is 2/5th of the total work.
= 104 × 8 × 30 × × = 104 × 4 × 90 man hours.
∴ Remaining work, which is 3/5th of the total work
Let a more men be employed to complete this work.
∴ (104 + a) × 9 × (56 – 30) = 104 × 4 × 90
∴ a = 56
Hence, option (a).
Workspace:
A bag contains 8 red and 6 blue balls. If 5 balls are drawn at random, what is the probability that 3 of them are red and 2 are blue?
- A.
80/143
- B.
50/143
- C.
75/143
- D.
None of the above
Answer: Option D
Explanation :
5 balls can be drawn in 14C5 ways.
Out of these 5 balls, 3 will be red in 8C3 × 6C2 ways
∴ Required probability = =
Hence, option (d).
Workspace:
In a circle, the height of an arc is 21 cm and the diameter is 84 cm. Find the chord of ‘half of the arc’
- A.
45 cm
- B.
40 cm
- C.
42 cm
- D.
None of the above
Answer: Option C
Explanation :
The given scenario can be depicted in the diagram below.
As, OC ⊥ AB and OC = AO/2,
ΔAOC is a 30-60-90 triangle.
∴ Angle subtended by the arc AB = 2 × 60 = 120°
∴ Angle subtended by half the arc is 60°
∴ Triangle formed by the chord of ‘half the arc’ and two radii is equilateral.
Hence, length of chord AM = 42 cm
Hence, option (c).
Workspace:
Mr. and Mrs. Gupta have three children – Pratik, Writtik and Kajol, all of whom were born in different cities. Pratik is 2 years elder to Writtik. Mr. Gupta was 30 years of age when Kajol was born in Hyderabad, while Mrs. Gupta was 28 years of age when Writtik was born in Bangalore. If Kajol was 5 years of age when Pratik was born in Mumbai, then what were the ages of Mr. and Mrs. Gupta respectively at the time of Pratik’s birth?
- A.
35 years, 26 years
- B.
30 years, 21 years
- C.
37 years, 28 years
- D.
None of the above
Answer: Option A
Explanation :
Mr Gupta was 30 years old when Kajol was born.
Kajol is 5 years older than Pratik.
∴ Mr. Gupta was 35 years old when Pratik was born.
Now, Mrs. Gupta was 28 years old when Writtik was born.
As Pratik is 2 years older than Writtk, Mrs, Gupta was 28 – 2 = 26 years old when Pratik was born.
Hence, option (a).
Workspace:
Mr. Sinha received a certain amount of money by winning a lottery contest. He purchased a new vehicle with 40 percent of the money received. He then gave 20 percent of the remaining amount to each of his two sons for investing in their business. Thereafter, Mr. Sinha spent half of the remaining amount for renovation of his house. One-fourth of the remaining amount was then used for purchasing a LCD TV and the remaining amount – Rs. 1,35,000/- was deposited in a bank. What was the amount of his cash prize?
- A.
Rs. 10,00,000/-
- B.
Rs. 9,00,000/-
- C.
Rs. 8,00,000/-
- D.
None of the above
Answer: Option A
Explanation :
Without loss of generality, we can assume that Mr. Sinha received Rs. 100 from the lottery.
After purchasing a vehicle he was left with Rs. 60.
After giving 20% of the remaining to each of the sons, he was left with 0.6 × 60 = Rs. 36
After spending half of the remaining amount on renovation, he is left with Rs. 18.
After spending on the LCD, he is left with Rs. (3 × 18/4) = 54/4
Now, this corresponds to Rs. 135000
∴ The original amount = 135000 × 100/(54/4) = Rs. 10,00,000
Hence, option (a).
Workspace:
The ratio of number of male and female journalists in a newspaper office is 5:4. The newspaper has two sections, political and sports. If 30 percent of the male journalists and 40 percent of the female journalists are covering political news, what percentage of the journalists (approx.) in the newspaper is currently involved in sports reporting?
- A.
65 percent
- B.
60 percent
- C.
70 percent
- D.
None of the above
Answer: Option A
Explanation :
Let there be 90 journalists in all.
∴ There are 50 males and 40 female journalists.
Now, 30% of males and 40% of females cover politics.
∴ 15 males and 16 females cover politics.
∴ (90 – 15 – 16) = 59 cover sports news.
∴ Required percentage = 59 × ≈ 65%
Hence, option (a).
Workspace:
The ratio of ‘metal 1’ and ‘metal 2’ in Alloy ‘A’ is 3:4. In Alloy ‘B’ same metals are mixed in the ratio 5:8. If 26 kg of Alloy ‘B’ and 14 kg of Alloy ‘A’ are mixed then find out the ratio of ‘metal 1’ and ‘metal 2’ in the new Alloy.
- A.
3 : 2
- B.
2 : 5
- C.
2 : 3
- D.
None of the above
Answer: Option C
Explanation :
Ratio of metal 1 and metal 2 in alloy A = 3 : 4
In 14 kg, there will be 3/7 × 14 = 6 kg of metal 1 and 4/7 × 14 = 8 kg of metal 2.
Similarly, alloy B contains 10 kg and 16 kg of metal 1 and metal 2 respectively.
∴ Metal 1 content in final mixture = 6 + 10 = 16 kg
Metal 2 content in final mixture = 8 + 16 = 24 kg
∴ Required ratio = 16 : 24 ≡ 2 : 3
Hence, option (c).
Workspace:
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