# IIFT 2019 QA

Paper year paper questions for IIFT 2019 QA

**1. IIFT 2019 QA | Arithmetic - Percentage**

Joseph is in a dilemma. He has been offered a job which would pay him ₹ 80,000 per month for first three years and ₹ 1,20,000 per month for the next three years, and ₹ 1,50,000 per month for the remaining four years. He has also been offered an MBA at a prestigious place and he is considering whether to accept the job or go for the MBA. The first year tuition fee for the MBA program is ₹ 16,00,000 and the second year tuition fee for the MBA program is ₹ 20,00,000. After MBA, he'll get a salary of ₹ 2,00,000 per month for the first four years and then ₹ 2,50,000 per month for the remaining four years. What will be the approximate percentage gain for Joseph in opting for the MBA instead of the job in the 10 years horizon considering no discounting of money?

- A.
23%

- B.
25%

- C.
27%

- D.
29%

Answer: Option B

**Explanation** :

Total amount earned through job = [(80000 × 3) + (120000 × 3) + (150000 × 4)] × 12

= 1200000 × 12 = Rs. 144 lakhs

Net amount earned through MBA = salary earned in 8 years (after MBA) − MBA fees paid for 2 years

= [(200000 + 250000) × 4 × 12] − (1600000 + 2000000)

= 21600000 − 3600000

= Rs. 180 lakhs

∴ Percentage benefit = [(180 − 144)/144] × 100

= 25%

Hence, option (b).

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**2. IIFT 2019 QA | Arithmetic - Time & Work**

A group of women in a society decided to execute interior and exterior decoration of the society in a week’s time. Since 11 women dropped out every day from the second day, the entire decoration was completed on 12th day. How many women participated at the beginning? (Answer to the nearest integer)

- A.
137

- B.
141

- C.
145

- D.
148

Answer: Option C

**Explanation** :

Let, the number of women who started the work is N,

∴ Total work to be done = N × 7 woman days.

Day 1: Number of women worked = N

Day 2: Number of women worked = N - 11

Day 3: Number of women worked = N - 22 and so on till

Day 12: Number of women worked = N – 11 × 11 = 121

∴ Total work done = N + (N – 11) + (N – 22) + … + (N – 121) = 12N - 726 woman days.

⇒ 12N – 726 = 7N

⇒ N = 726/5 = 145.2 = 145 (nearest integer)

Hence, option (c).

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**3. IIFT 2019 QA | Arithmetic - Percentage**

KBC restaurant chain regularly conducts survey of its customers. The customers are asked to rate the food quality, service and price as Excellent, Good and Fair. Customers are also asked whether they would come back. It was found that 76% of customers say that they will come back. Amongst those who say that they will come back, 57% rate the restaurant as Excellent, 36% rate it as Good and remainder rate it as Fair. Of those who say that they will not return, the respective values are 14%, 32% and 54%. What percentage of customers rated the restaurant as good?

- A.
27.4%

- B.
35%

- C.
51%

- D.
30.7%

Answer: Option B

**Explanation** :

Let the total number of customers be 100x.

So, customers who said they will return and who said that they will not return is 76x and 24x respectively.

Of the 76x customers, the number of customers who rate it as good = 36% of 76x = 0.36 × 76x = 27.36x.

Of the 24x customers, the number of customers who rate it as good = 32% of 24x = 0.32 × 24x = 7.68x.

So, total customers who rated the restaurant as good = 27.36x + 7.68x = 35.04x.

Required percentage = (35.04x/100x) × 100 = 35.04 ≈ 35%.

Hence option (b).

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**4. IIFT 2019 QA | Modern Math - Permutation & Combination**

Four couples are to be seated in a circular table such that each couple sits together. In how many ways they can sit such that two males sit to the right of their female partners and the other two males sit to the left of their female partners?

- A.
144

- B.
288

- C.
1440

- D.
720

Answer: Option B

**Explanation** :

First of all, we will have to select 2 couples where the male sits to the right of the female in ^{4}C_{2} = 6 ways (and the other two couples select them automatically)

Now, we can fix anyone of the couple in 4 ways and in these couples, females may be seated to the left/right of the males

Total ways of fixing a couple = 8

The other three couples can be arranged in 3! ways = 6 ways

∴ Total number of ways = 6 × 8 × 8 = 288

Hence, option (b).

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**5. IIFT 2019 QA | Geometry - Coordinate Geometry**

If the co-ordinates of orthocentre and the centroid of a triangle ABC are (–5, 7) and (5, 5), then the circumcentre of the triangle ABC is:

- A.
(25, 1)

- B.
(10, 4)

- C.
(-5. 2)

- D.
(0, 6)

Answer: Option B

**Explanation** :

We know that Centroid divides the line joining the Orthocenter and Circumcenter in the ratio 2 : 1.

Let the coordinates of Circumcenter be(x, y)

∴ Orthocenter ((–5, 7)) ; Centroid (5, 5); Circumcenter (x, y).

Using the section formula;

5 = (2x − 5)/3 ⇒ x = 10.

5 = (2y + 7)/3 ⇒ y = 4.

So, the coordinates of the circum-center are (10, 4).

Hence option (b).

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**6. IIFT 2019 QA | Algebra - Quadratic Equations | Algebra - Inequalities & Modulus**

Consider the equation: |x – 5|^{2} + 5|x – 5| – 24 = 0. The sum of all the real roots of the above equation is:

- A.
2

- B.
3

- C.
8

- D.
10

Answer: Option D

**Explanation** :

Let |x − 5| = k.

Hence, the equation can be rewritten as: k^{2} + 5k − 24 = 0

Solving the equation, we get k = 3, −8.

But, |x − 5| cannot be negative, so k ≠ −8.

∴ |x − 5| = 3

So, x − 5 = ±3.

∴ x = 8 and x = 2

Required sum = 8 + 2 = 10.

Hence, option (d).

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**7. IIFT 2019 QA | Arithmetic - Average**

A cricket team has 11 players and each of them has played 20 matches till date. Virat, Rohit, Mahendra, Rahul and Shikhar have scored runs at an average of 60, 55, 50, 45 and 40 respectively. Rest of the players have scored at an average of 25 each. In the next 10 matches, Virat and Rohit each scored 900 runs whereas Mahendra scored twice that of Rahul. After 30 matches, if Virat’s new average score is twice that of Rahul, what is the approximate average score of Mahendra?

- A.
49

- B.
41

- C.
43

- D.
45

Answer: Option C

**Explanation** :

Let Rahul's average after 30 matches be A, so Virat's average after 30 matches is 2A.

Virat's score after 30 matches = (60 × 20) + 900 = 2A × 30

Solving this equation, we get; A = 35.

Rahul's score after 30 matches = (45 × 20) + x = A × 30 = 35 × 30 (where x is the runs scored by Rahul in the last 10 matches)

Solving this equation, we get; x = 150.

Mahendra's score after 30 matches = (50 × 20) + 2x = B × 30 (where B is the Mahendra's average after 30 matches)

Putting x = 150 in the above equation, we get B = 43.33 ≈ 43.

Hence option (c).

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**8. IIFT 2019 QA | Arithmetic - Simple & Compound Interest**

Ms. Debjani after her MBA graduation wants to have a start-up of her own. For this, she uses ₹ 8,00,000 of her own savings and borrows ₹ 12,00,000 from a public sector bank under MUDRA Scheme. As per the agreement with the bank, she is supposed to repay the principle of this loan equally over the period of the loan which is 25 years. Two years after taking the first loan, she borrowed an additional loan of ₹8,00,000 to finance expansion plan of her start-up. If Ms. Debjani clears all her loans in 25 years from the date of taking the first loan, how much total interest she has to pay on her initial borrowing ? Assume simple interest rate at 8 percent per annum.

- A.
Rs. 12,48,000

- B.
Rs. 12,84,000

- C.
Rs. 14,20,000

- D.
Rs. 12,96,000

Answer: Option A

**Explanation** :

Since the principal is to be paid equally over 25 years, principal amount paid per year = (12 lakhs)/25 = Rs. 48000.

For the first year, interest to be paid = (12 lakhs) × 0.08 ....(I)

For the second year, amount remaining = (12 lakhs) − 48000 = Rs. 11,52,000.

So interest for the second year = (Rs. 11,52,000) × 0.08 .... (II)

On similar lines, interest for the third year = (48000) × 0.08

Total interest = [(12 lakhs) × 0.08] + [(Rs. 11,52,000) × 0.08] + ...... [(48000) × 0.08]

This is an AP with first term as (48000) × 0.08 = 3840 with 25 terms and common difference of Rs. 3840. [{(12 lakhs) × 0.08} − {(Rs. 11,52,000) × 0.08} = Rs. 3840]

∴ Total interest = (25/2)[(2 × 3840) + {(25 − 1) × 3840}]

= Rs. 12,48,000.

Hence option (a).

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**9. IIFT 2019 QA | Algebra - Number Theory**

The number 37^{371} – 26^{371} is divisible by:

- A.
10

- B.
11

- C.
12

- D.
15

Answer: Option B

**Explanation** :

37^{371} − 26^{371} is of the form a^{n} − b^{n}

and, a^{n} − b^{n} is always divisible by a − b

Thus, the expression will be divisible by 37 − 26 = 11

Hence, option (b).

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**10. IIFT 2019 QA | Geometry - Triangles**

In a triangle, the two longest sides are 13 cm and 12 cm. The angles of the triangle are in arithmetic progression. The radius of the circle inscribed in this triangle is:

- A.
√3 cm

- B.
√3 - 1 cm

- C.
2 cm

- D.
1 cm

Answer: Option D

**Explanation** :

Let the three angles be a − d, a and a + d.

∴ (a − d) + a + (a + d) = 180

∴ a = 60.

Since (a + d) is the largest angle it will be opposite the largest side i.e., 13 and a will be opposite second largest side i.e., 12.

The triangle along with the angles are shown below.

Applying the cosine rule, we get;

Cos 60 = (x^{2} + 13^{2} − 12^{2})/(2 × x × 13)

∴ x = 10.65, 2.34.

Inradius = Area/Semiperimeter.

When x = 10.65, the semiperimeter = (10.65 + 13 + 12)/2 = 17.825.

∴ Inradius = [(1/2) × 10.65 × 13 × Sin60]/17.825 = 3.36.

When x = 2.34, the semiperimeter = (2.34 + 13 + 12)/2 = 13.67.

∴ Inradius = [(1/2) × 2.34 × 13 × Sin60]/13.67 = 0.963 ≈ 1.

Hence option (d).

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**11. IIFT 2019 QA | Geometry - Circles**

AB is the tangent on the circle at point A. The line BC meets the circle at points C and E. Line AD bisects the angle EAC. If angle EAC = 60° and angle BAC : angle ACB = 2: 5. Find angle ABC.

- A.
40°

- B.
60°

- C.
30°

- D.
None of these

Answer: Option A

**Explanation** :

Let ∠CAB and ∠ACB be 2x and 5x respectively.

From the alternate segment thorem (In any circle, the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment), we can deduce that ∠AEC = ∠CAB = 2x.

In ∆AEC, ∠AEC = 2x, ∠EAC = 30 × 2 = 60° and ∠ACE = 180 − 5x.

∴ 2x + 60 + (180 − 5x) = 180.

∴ x = 20.

∠ABC = 180 − (∠BAC + ∠ACB) = 180 − (2x + 5x) = 180 − 7x = 180 − (7 × 20) = 40°.

Hence option (a).

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**12. IIFT 2019 QA | Geometry - Mensuration**

A square of length 1 m is inside a square of length 2 m and four quarter circles are joined as shown in the figure. The value of y – x is given by,

- A.
(8 - π)/10

- B.
(4 - π)/5

- C.
(2π - 1)/8

- D.
(π - 3)/4

Answer: Option D

**Explanation** :

Area of the shaded region = (Area of the bigger square) − (4 × Area of the quarter of radius = 1) − 4x.

= 22 − [4 × π × 1^{2}/4)] − 4x ...(1)

Also, area of the same shaded region = (Area of the smaller square) − 4y.

= 12 − 4y ...(2)

Equating (1) and (2), we get,

4 -π – 4x = 1 – 4y

⇒ (y − x) = (π − 3)/4

Hence option (d).

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**13. IIFT 2019 QA | Geometry - Mensuration**

Consider a cuboidal underground tunnel of length 500 m whose cross-section is given in the figure. If 1 m^{3} of concrete costs 1000 rupees, find the amount of money needed to build the tunnel.

- A.
8(4 - π) × 10

^{6}rupees - B.
64(4 - π) × 10

^{6}rupees - C.
16(4 - π) × 106 rupees

- D.
32(4 - π) × 106 rupees

Answer: Option C

**Explanation** :

Total volume of Tunnel = 8 × 16 × 500 = 64000 m^{3}

Radius of the given semicircle = 8 m

Volume of the semi-circle = (π × 8 × 8)/2 × 500 = 32π × 500 = 16000π m^{3}

∴ Volume of the required cross-section = 64000 - 16000π = 16000(4 − π) m^{3}

Total amount of money needed = 16000(4 − π) × 10^{3} = 16(4 − π) × 10^{6} rupees

Hence, option (c).

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**14. IIFT 2019 QA | Arithmetic - Simple & Compound Interest**

Monika buys a Samsung’s 360 litre refrigerator from M/s Coldrush Agencies for ₹42,000. She makes a down payment of ₹12,000 and the remaining amount in 4 equal half yearly instalments. If M/s Coldrush Agencies charge an interest of 10% per annum, approximately what amount Monika has to pay every six months?

- A.
Rs. 8230

- B.
Rs. 8600

- C.
Rs. 8460

- D.
Rs. 8620

Answer: Option C

**Explanation** :

Cost of refrigerator = Rs. 42000.

Down payment = Rs. 12000.

Loan amount for Monika = 42000 − 12000 = Rs. 30,000.

Interest charged every six months = 10/2 = 5%

∴ 30000 × 1.05^{4} = x × 1.05^{3} + x × 1.05^{2} + x × 1.05 + x.

∴ 36465 = 4.31x

∴ x ≈ 8460.

Hence option (c).

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**15. IIFT 2019 QA | Ratio, Proportion & Variation**

An E-rickshaw owner makes 24 trips a month with 4 passengers per trip. If his interest cost for purchase of E-rickshaw is ₹ 120/month, he earns 15 percent profit a month (Profit is the difference between revenue and cost). What will be the approximate percentage profit for the same month if the owner undertakes 20 trips a month with 5 passengers and his interest cost is reduced by 10 percent for the month?

Assume: (a) Total cost to be proportional to the interest cost;

(b) Revenue per passenger is the same in both cases.

- A.
33.33

- B.
66.67

- C.
72

- D.
100

Answer: Option A

**Explanation** :

Let the revenue per passenger be R.

24 trips with 4 passengers

Total revenue = 24 × 4 × R = 96R.

Total cost = 120k (∵ Total cost is proportional to the interest cost)

Profit % = 15%

∴ 120k × 1.15 = 96R

⇒ R/k = 138/96

20 trips with 5 passengers

Total revenue = 20 × 5 × R = 100R.

Total cost reduces by 10% = 120k × 0.9 = 108k

∴ New profit % = [(Revenue − Cost)/Cost] × 100 = [(Revenue/Cost) − 1] × 100

∴ P = [(100R/108k) − 1] × 100 = [13800/(108×96) - 1] × 100 = 33.1%

Hence option (a).

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**16. IIFT 2019 QA | Venn Diagram**

400 students were admitted to the 2018-19 MBA batch. 200 of them did not choose “Business Statistics”. 100 of them did not choose “International Management’. There were 80 students who did not choose any of the two subjects. Find the number of students who chose both Business Statistics and International Management.

- A.
180

- B.
220

- C.
280

- D.
300

Answer: Option A

**Explanation** :

Number of students who chose Business statistics = 400 − 200 = 200

Number of students who chose International Management = 400 − 100 = 300

Number of students who chose at least one of the two subjects = 400 − 80 = 320

∴ Number of students who chose both the subjects = 200 + 300 − 320 = 500 − 320 = 180

Hence, option (a).

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**17. IIFT 2019 QA | Arithmetic - Time, Speed & Distance**

A motorboat takes the passengers from Rishikesh to Haridwar and back. Both the cities, Rishikesh and Haridwar are located on the banks of River Ganga. During Kumbh Mela, to earn more money, the owner of the motorboat decided to have more trips from Rishikesh to Haridwar and back, so he increased the speed of the motorboat in still water, by 50%. By increasing the speed, he was able to cut down the travel time from Rishikesh to Haridwar and back, by 60%. What is the ratio of the speed of motorboat in still water to that of the speed of river Ganga?

- A.
√(11/6)

- B.
√11/6

- C.
√(3/2)

- D.
√3/2

Answer: Option A

**Explanation** :

Let the speed of boat in still water be ‘b’ and speed of river be ‘r’ and the one-way distance between the two cities be ‘d’.

Time taken initially to go and come back = $\frac{d}{b+r}$ + $\frac{d}{b-r}$

When the speed of boat increases by 50%, time taken to go and come back = $\frac{d}{1.5b+r}$ + $\frac{d}{1.5b-r}$

New time taken is 40% of the initial time taken

⇒ $\frac{d}{1.5b+r}+\frac{d}{1.5b-r}$ = $\frac{2}{5}\left[\frac{d}{b+r}+\frac{d}{b-r}\right]$

⇒ $\frac{3b}{2.25{b}^{2}-{r}^{2}}$ = $\frac{2}{5}\left[\frac{2b}{{b}^{2}-{r}^{2}}\right]$

⇒ 15b^{2} – 15r^{2 }= 9b^{2} – 4r^{2}

⇒ 6b^{2} = 11r^{2}

⇒ $\frac{b}{r}$ = $\sqrt{\frac{11}{6}}$

Hence option (a).

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**18. IIFT 2019 QA | Arithmetic - Time, Speed & Distance**

You travel by Delhi Metro everyday from Botanical Garden, Noida to Hauz Khas, Delhi. At Hauz Khas metro station, you use an escalator to get out from the station. The escalator takes 80 seconds. One day, escalator was not working and you walk up the escalator in 50 seconds. How many minutes does it approximately take you to walk down the working escalator?

- A.
1.5 minutes

- B.
2.2 minutes

- C.
2.8 minutes

- D.
2.6 minutes

Answer: Option B

**Explanation** :

Let, the number of steps on the escalator be LCM(80, 50) = 400

∴ Speed of escalator = 400/80 = 5 steps/ second, and

Speed of the person = 400/50 = 8 steps/ second

When the person tries to go up a moving down escalator, the effective speed of the person = (8 − 5) steps /second = 3 steps/second

∴ Time taken = 400/3 = 133.33 steps/ second = 2.2 minutes (approx.)

Hence, option (b).

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**19. IIFT 2019 QA | Geometry - Trigonometry**

A man standing on the line joining the two poles finds that the top of the poles make an angle of elevation of 60° and 45° respectively. After walking for sometime towards the other pole, the angles change to 30° and 60° respectively. The ratio of the height of the poles is:

- A.
[(√3) − 1]/2

- B.
[(√3) + 1]/2

- C.
[(√3) − 1]/4

- D.
[(√3) + 1]/4

Answer: Option A

**Explanation** :

Consider the image below where A and B are the initial and final position of the man. PQ and LM are the two poles of heights a and b respectively.

From ∆PQA, QA = a/Tan60 = a/√3. From ∆ALM, AM = b/Tan45 = b.

From ∆PQB, QB = a/Tan30 = a√3. From ∆BLM, BM = b/Tan60 = b/√3.

∴ [a/√3] + b = a√3 + [b/√3]

Solving we get,

a/b = [(√3) − 1]/2

Hence option (a).

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**20. IIFT 2019 QA | Geometry - Triangles**

In ∆MNL, line NP bisects the angle MNL. If NP : NL= 2 : 3 and angle MNL = 120°. Then NP : NL: MN is:

- A.
2 : 3 : 4

- B.
2 : 3 : 6

- C.
2 : 3 : 5

- D.
2 : 3 : 9

Answer: Option B

**Explanation** :

Applying cosine rule in ∆NPL, we get;

Cos60 = 1/2 = [NP^{2} + NL^{2} − PL^{2}]/(2 × NP × NL)

∴ 1/2 = [(2x)^{2} + (3x)^{2 }− PL^{2}]/(2 × 2x × 3x)

∴ PL = x√7.

Using angle bisector theorem, MN/NL = MP/PL

So, MN/3x = MP/x√7

∴ MN = MP × (3/√7) ...(I)

Applying cosine rule in ∆MNP, we get;

Cos60 = 1/2 = [MN^{2} + NP^{2} − MP^{2}]/(2 × MN × NP)

Let MN = a, so using (I) MP = (√7/3) a or MP^{2} = (7/9)a^{2}.

∴ 1/2 = [a^{2} + 4x^{2}− (7/9)a^{2} ]/(2 × a × 2x)

∴ a^{2} − 9ax + 18x^{2} = 0

∴ a = 6x or 3x.

If a = 3x, then MN = NL = 3x, so ∆MNL becomes isosceles and hence ∠NPL = 90°, which means that NL^{2} = NP^{2} + PL^{2}.

NL^{2} = 9x^{2} and NP^{2} + PL^{2} = 4x^{2} + (x^{2}/7) = (29/7)x^{2 }≠ 9x^{2}. Hence a ≠ 3x.

∴ a = 6x.

NP : NL: MN = 2x : 3x : 6x = 2 : 3 : 6.

Hence option (b).

Note: We can use the direct formula for the length of the angle bisector L. Consider the triangle drawn below

L = (2abCosQ)/(a + b)

So, 2x = (2 × MN × 3x × Cos60)/(MN + 3x)

Solving, we get; MN = 6x.

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**21. IIFT 2019 QA | Algebra - Surds & Indices**

If x = 8 - $\sqrt{32}$ and y = 2 + √2, then ${\left(x+\frac{1}{y}\right)}^{2}$ is given by:

- A.
16x

^{2}/25 - B.
64x

^{2}/81 - C.
25y

^{2}/16 - D.
81x

^{2}/64

Answer: Option D

**Explanation** :

x = 8 − √32 = 8 − 4√2 = 4(2 − √2)

$\frac{1}{y}$ = $\frac{1}{2+\sqrt{2}}$

On rationalising, we get

$\frac{1}{y}$ = $\frac{1}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$ = $\frac{2-\sqrt{2}}{2}$ = $\frac{x}{8}$

∴ x + 1/y = x + x/8 = 9x/8

⇒ ${\left(x+\frac{1}{y}\right)}^{2}$ = ${\left(x+\frac{x}{8}\right)}^{2}$ = ${\left(\frac{9x}{8}\right)}^{2}$ = $\frac{81{x}^{2}}{64}$

Hence option (d).

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**22. IIFT 2019 QA | Arithmetic - Time & Work**

Rohit purchased a cistern which had a leakage. The cistern can be filled by two inlet pipes, which can individually fill the cistern in 12 min and 15 min respectively. Despite leakage, the two inlet pipes together can fill the cistern in 20 min. How long will it take to empty the completely full the cistern due to leakage?

- A.
10 mins

- B.
12 mins

- C.
15 mins

- D.
16 mins

Answer: Option A

**Explanation** :

Let, the total capacity of cistern be LCM (12,15,20) = 60 units

∴ Rate of the two inlet pipes are 60/12 = 5 units per minute and 60/15 = 4 units per minute.

Also, rate of filling of cistern along with the leakage = 60/20 = 3 units per minute

∴ Rate of leakage (L) = 3 – (5 + 4) = -6 units per minute. (-ve indicates that leakage is emptying the tank)

∴ Time taken by leakage to empty the cistern = 60/6 = 10 mins

Hence, option (a).

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**23. IIFT 2019 QA | Algebra - Simple Equations**

Nawab has two sons Saif and Amir who have export businesses. Nawab’s satisfaction/ utility level is given by adding twice of the satisfaction level of Saif with the satisfaction level of Amir. If Saif makes a profit of ₹100, his satisfaction level goes up by 10% and if he suffers a loss of ₹100, his satisfaction level goes down by 10%. If Amir makes a profit of ₹100, his satisfaction level goes up by 5% and if he suffers a loss of ₹100, his satisfaction level goes down by 15%. Currently, Nawab’s satisfaction level is 24 and the satisfaction level of Saif is same as the satisfaction level of Amir. If Saif makes a profit of ₹100 and Amir suffers a loss of ₹100, what is the approximate percentage change in Nawab’s satisfaction level?

- A.
1.25%

- B.
1.33%

- C.
1.5%

- D.
1.66%

Answer: Option D

**Explanation** :

Currently satisfaction level of Saif and Amir is same (let’s say) ‘s’

∴ The present satisfaction level of Nawab = 24 = 2s + s

⇒ s = 8

∴ Present satisfaction level of both Saif and Amir = 8.

∵ Saif makes a profit of 100, his satisfaction level goes up by 10% ⇒ Saif’s new satisfaction level= 8 × 1.1 = 8.8

∵ Amir suffers a loss of 100, his satisfaction level goes down by 15% ⇒ Amir’s new satisfaction level= 8 × 0.85 = 6.8

⇒ Nawab's new satisfaction level = 2 × 8.8 + 6.8 = 24.4

∴ Percentage change = (24.4 − 24)/24 × 100 = (0.4 × 100)/24 = 1.66 %.

Hence, option (d).

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**24. IIFT 2019 QA | Miscellaneous**

At what time between 2.00 pm and 3.00 pm, the two arms of a watch are completely opposite to each other?

- A.
2:40 pm

- B.
2:44 pm

- C.
2:45 pm

- D.
2:47 pm

Answer: Option B

**Explanation** :

Since, the two arms are completely opposite to each other, if the hour hand is between 2 and 3, the minute hand has to be between 8 and 9.

Thus, the time (in minutes) has to be between 40 and 45.

Hence, from among the options, the time should be 2:44 pm

Hence, option (b).

Workspace:

**25. IIFT 2019 QA | Modern Math - Probability**

According to birth registration data available with the South Delhi Municipal Corporation, 7 babies were born in a particular week in a private hospital. What is the probability that three babies were born on the same day of the week?

- A.
1800/7

^{5} - B.
1600/7

^{5} - C.
2100/7

^{5} - D.
2400/7

^{5}

Answer: Option A

**Explanation** :

The question is wrong and was discarded by IIFT.

Workspace:

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