# IIFT 2018 QA | Previous Year IIFT Paper

**1. IIFT 2018 QA | Arithmetic - Percentage**

A Business Group has 3 Companies X, Y, Z and a Trust P which is engaged in charitable activities. Each group company has to donate 5% of its own funds to the Trust, excluding the loan which the company has taken from other companies of the group. X has given a loan to Y which is equivalent to 10% of the funds of Y. After receiving the loan, Y has funds which are 2 times the funds of Z. If Z gave Rs. 10,000 as donation to the Trust P, how much is the approximate contribution of Y to the Trust P?

- A.
Rs. 17,000

- B.
Rs. 18,000

- C.
Rs. 19,000

- D.
Rs. 20,000

Answer: Option B

**Explanation** :

Let the initial amount (in Rs.) with companies X, Y and Z be x, y and z respectively.

X has given an amount equivalent to 10% of the amount originally with Y. Hence, X has given Rs. (0.1y) to Y.

∴ Amount now with Y = y + 0.1y = Rs. (1.1y)

The amount with Y is twice the amount with Z. Hence, the amount with Z is Rs. (0.55y).

Z gave Rs. 10,000 i.e. 5% of his funds to the Trust.

∴ 0.05 × 0.55y = 10000

Amount given by Y = 0.05y

= 0.05 × (10000)/(0.55 × 0.05)

= 10000/0.55

= 18181 i.e. approx. Rs. 18,000

Hence, option (b).

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**2. IIFT 2018 QA | Geometry - Mensuration**

A bucket contains 200 cc of liquid. A solid ball is dropped in the bucket resulting in the rise of liquid level to 1.3 times of its original level. If the radius of the base of the bucket is 3 cm and the radius of the surface of the liquid level is 1 cm more than the radius of the base of the bucket before the ball is dropped. Find the volume of the solid metal ball.

- A.
68cc

- B.
80cc

- C.
92cc

- D.
Can’t be determined

Answer: Option B

**Explanation** :

The radius of the bucket is 3 cm and the radius of the original water level is (3 + 1) = 4 cm

Original volume = 200 cc

Let the original height of water = h cm

Using the formula for volume of frustum of cone:

$200=\frac{h}{3}\pi [{4}^{2}+{3}^{2}+(4\left)\right(3\left)\right]$

∴ h = 600/37π

Now, the height of the liquid increases by 0.3 times itself i.e. it becomes 1.3h

In a frustum of cone, ratio of radii is the same as ratio of heights (applying the concepts of similarity).

∴ New radius = 4.3 cm

∴ Volume of ball = new volume – original volume

$=\frac{1.3h}{3}\pi [4.{3}^{2}+{3}^{2}+(4.3\left)\right(3\left)\right]-200\phantom{\rule{0ex}{0ex}}$

$=\frac{1.3h}{3}\times \pi \times \frac{600}{37\pi}\times 40.4-200$

= 283.9 - 200 = 83.9 cc

The closest value in the options is 80 cc.

Hence, option (b).

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**3. IIFT 2018 QA | Modern Math - Permutation & Combination**

P . . . . . Q

R . . . . . S

T . . . . . U

V . . . . . W

Using 5 dots in each of the lines PQ, RS, TU and VW as the vertices, how many triangles can be drawn such that the base is on any one of the above lines?

- A.
120

- B.
150

- C.
200

- D.
600

Answer: Option D

**Explanation** :

The base has to be on one of the four lines.

Hence, the base can first be chosen in 4 ways.

Now, on this base, two points need to chosen from five to form the actual base. This can be done in ^{5}C_{2} ways i.e. 10 ways

For the third point, one of three lines has to be chosen and then one of five points from each line has to be chosen.

∴ Number of possible triangles = 4 × 10 × 3 × 5 = 600

Hence, option (d).

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**4. IIFT 2018 QA | Geometry - Triangles**

In the triangle PQR, S is the midpoint of QR. X is any point on PR. T is the point on QR such that PT‖SX. If the area of triangle PQR is 5.8 sq. cm, then the area of triangle RTX is

- A.
2.9 sq. cm

- B.
3.2 sq. cm

- C.
5.8 sq. cm

- D.
2.45 sq. cm

Answer: Option A

**Explanation** :

Since S is the midpoint of QR, A(∆PSR)

= A(∆PQR)/2

= 5.8/2 = 2.9 sq.cm

Now A(∆PSR) = A(∆PSX) + A(∆SXR)

Also, A(∆RTX) = A(∆TSX) + A(∆SXR)

Now, triangles PSX and TSX lie within the same parallel lines – SX and PT – and hence, have the same area.

∴ A(∆PSX) = A(∆TSX)

∴ A(∆PSR) = A(∆RTX) = 2.9 sq.cm

Hence, option (a).

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**5. IIFT 2018 QA | Algebra - Functions & Graphs**

Given P(*x,y*) = *x*^{2} + *xy* + *y*^{2}; Q(*x,y*) = *x*^{2} – *xy* +* y*^{2}. Find the value of P(7, Q(9,4))

- A.
4169

- B.
4197

- C.
4089

- D.
4127

Answer: Option B

**Explanation** :

Q(9, 4) = (9)^{2} – (9)(4) + (4)^{2}

= 81 – 36 + 16 = 61

∴ P(7,Q(9,4)) = P(7,61)

= (7)^{2} + (7)(61) + (61)^{2}

= 49 + 427 + 3721 = 4197

Hence, option (b).

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**6. IIFT 2018 QA | Geometry - Mensuration**

In the given figure, PA = QB and PRQ is the arc of the circle, centre of which is O such that angle POQ = 90⁰. If AB = 25√2 cm and the perpendicular distance of AB from centre O is 30cm. Find the area if the shaded region?

- A.
625√2 sq.cm

- B.
$625\left(\frac{1}{2}+\frac{\pi}{4}\right)$

- C.
$750\sqrt{2}-625\left(\frac{1}{2}+\frac{\pi}{4}\right)$

- D.
None

Answer: Option C

**Explanation** :

PA = QB and angles A and B are right angles.

Hence, PABQ is a rectangle.

∴ Required area

= A(rectangle PABQ) – A(segment O-PRQ)

= A(rectangle PABQ) – [A(sector O-PRQ) – A(∆OPQ)]

= A(rectangle PABQ) – A(sector O-PRQ) + A(∆OPQ)

Now, PQ = AB = 25√2 cm and ∠POQ = 90°

∴ Radius of sector = PO = OQ = PQ/√2 = 25

Now, A(∆OPQ) = (1/2) × OP × OQ

= (1/2) × 25 × 25 = 625/2 sq.cm

A(sector O-PRQ) = (90/360) × π × (OP)^{2}

= (π/4) × (25)^{2} = (625π/4)

Now, height of ∆OPQ = 25/√2

Distance from O to AB is 30 cm

∴ PA = QB = 30 – (25/√2) cm

∴ A(rectangle PABQ) = PA × AB

= [30 – (25/√2)] × (25√2) = 750√2 – 625 sq.cm

∴ Required area = $(750\sqrt{2}-625)$ - $\frac{625\mathrm{\pi}}{4}$ + $\frac{625}{2}$

= 750√2 - 625/2 - 625π/4

= 750√2 - 625$\left(\frac{1}{2}+\frac{\mathrm{\pi}}{4}\right)$

Hence, option (c).

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**7. IIFT 2018 QA | Algebra - Quadratic Equations**

The roots of quadratic equation *y*^{2 }– 8*y* + 14 = 0 are α and β. Find the value of (1 + α + β^{2})( 1 + β + α^{2})

- A.
419

- B.
431

- C.
485 + 3√22

- D.
453 + √22

Answer: Option B

**Explanation** :

The roots of the equation y^{2} – 8y + 14 = 0 are α and β.

∴ Sum of roots = α + β = −(−8)/1 = 8 and Product of roots = αβ = (14)/1 = 14

∴ (1 + α + β^{2}) (1 + β + α^{2}) = 1 + α + β^{2} + β + αβ + β^{3} + α^{2} + α^{3} + α^{2}β^{2}

= 1 + (α + β) + (αβ) + (αβ)^{2} + (α^{2} + β^{2}) + (α^{3} + β^{3})

= 1 + (α + β) + (αβ) + (αβ)^{2} + (α + β)^{2} – (2αβ) + (α + β)^{3} – (3αβ)(α + β)

= 1 + 8 + 14 + (14)2 + (8)^{2} – 2(14) + (8)^{3 }– 3(14)(8)

= 23 + 196 + 64 – 28 + 512 – 336 = 431

Hence, option (b).

Workspace:

**8. IIFT 2018 QA | Algebra - Logarithms**

$\frac{1}{{\mathrm{log}}_{x}yz+1}+\frac{1}{{\mathrm{log}}_{y}xz+1}+\frac{1}{{\mathrm{log}}_{z}xy+1}=?$

- A.
0

- B.
1

- C.
xyz

- D.
$\frac{\mathrm{log}xyz}{\mathrm{log}xyz+1}$

Answer: Option B

**Explanation** :

Since there is no condition imposed on x, y and z, assume *x* = *y* = *z* = 10

Consider log* _{x}yz* + 1.

When *x* = *y* = *z* = 10; log* _{x}yz* + 1

= log_{10}(10 × 10) + 1

= log_{10}(10)^{2} + 1 = 2 log_{10}(10) + 1

= 2(1) + 1 = 3

Similarly, log* _{y}xz* + 1 = log

*+ 1 = 3*

_{z}xy∴ Required value = (1/3) + (1/3) + (1/3) = 1

Hence, option (b).

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**9. IIFT 2018 QA | Arithmetic - Time & Work**

Ram, Ravi and Ratan can alone finish an assignment in 9 days, 12 days and 15 days respectively. They decide to complete a work by working in turns. Ram works alone on Monday, Ravi does the work alone on Tuesday, followed by Ratan working alone on Wednesday & so on. What proportion of the complete work is done by Ravi?

- A.
2/9

- B.
12/47

- C.
1/3

- D.
4/9

Answer: Option C

**Explanation** :

Let the total work be 180 units (LCM of 9, 12 and 15) and let Ram, Ravi and Ratan do a, b and c units of work per day.

∴ a = 180/9 = 20; b = 180/12 = 15 and c

= 180/15 = 12

Ram works alone on Monday, Ravi works alone on Tuesday and Ratan works alone on Wednesday.

∴ Work done in three days = a + b + c

= 20 + 15 + 12 = 47

∴ Work done in nine days = 3(47)

= 141 units

On the 10th day, Ram does 20 units, thereby taking the completed total work to 141 + 20 = 161 units.

On the 11th day, Ravi does 15 units, hence, taking the completed total work to 176 units.

The work gets completed on the 12th day.

Hence, Ravi has done work equivalent to four days when the work is completed.

∴ Ravi’s total work = 4(15) = 60 units

∴ Required proportion = 60/180 = 1/3

Hence, option (c).

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**10. IIFT 2018 QA | Geometry - Mensuration**

Let S_{1} be a square of side 4 cm. Circle C_{1} circumscribes the square S_{1} such that all its corners are on C_{1}. Another square S_{2} circumscribes the circle C_{1}. Circle C_{2} circumscribes the square S_{2}, and square S_{3} circumscribes circle C_{2}, & so on. If A_{N} is the area between the square S_{N} and the circle C_{N}, where N is the natural number, then the ratio of sum of all A_{N} to A_{1} is

- A.
1

- B.
$\frac{\pi}{2}-1$

- C.
Can’t be determined

- D.
None of the above

Answer: Option C

**Explanation** :

Here, the area of each square and each circle will keep on increasing. Hence, each A_{n} will keep on increasing.

Hence, if the value of n is not known, the ratio will not be found.

Here, the value of n is unknown. Hence, the value of the ratio cannot be determined.

Hence, option (c).

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**11. IIFT 2018 QA | Geometry - Circles**

Joseph diametrically crosses a semi-circular playground and takes 48 seconds less than if he crosses the playground along the semi-circular path. If he walks 50 metres in one minute, the diameter of playground is

- A.
54 metres

- B.
70 metres

- C.
85 metres

- D.
35 metres

Answer: Option B

**Explanation** :

Since Joseph takes less time while crossing the path diametrically rather than in a semi-circular manner, the distance travelled along the diameter is less.

Let the radius of the path be r m.

∴ Difference in distance = πr – 2r

= (π – 2)r m

Also, difference in distance = difference in time × constant speed

∴ (π – 2)r = 48 × (50/60)

∴ 1.14r = 40 i.e. r = 40/(8/7) = 35 m

∴ Diameter = 2r = 70 m

Hence, option (b).

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**12. IIFT 2018 QA | Algebra - Simple Equations**

Garima had only Rs. 200, Rs. 500 and Rs. 2000 notes in her wallet. She goes to shoppers Stop, purchases some dresses and gives half of her Rs. 2000 notes & in turn receives same number of Rs. 200 notes. She then goes to a restaurant and gives all her Rs. 500 notes and receives thirty Rs. 2000 notes, which increases the number of Rs. 2000 notes she had by 75%. If now she has fifty Rs. 200 notes, what were the original number of Rs. 2000 and Rs. 200 notes she had at the start?

- A.
60, 10

- B.
60, 15

- C.
80, 10

- D.
80, 15

Answer: Option C

**Explanation** :

Let Garima initially have (2n) notes of Rs. 2,000.

After the first transaction (in which she gives away half her notes), Garima has n notes of Rs. 2,000.

Now, Garima gets 30 notes of Rs. 2,000 which increases her number of Rs. 2,000 notes by three-fourth.

∴ n + 30 = n + (3n/4)

∴ (3n/4) = 30 i.e. n = 40

∴ Initial number of Rs. 2,000 notes = 2n = 80

This eliminates options 1 and 2.

Let Garima initially have x notes of Rs. 200.

She then gets n notes of Rs. 200, thereby taking the total notes of Rs. 200 to 50.

∴ x + n = 50

∴ x + 40 = 50 i.e. x = 10

∴ Initial number of Rs. 200 notes = x = 10

Hence, option (c).

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**13. IIFT 2018 QA | Geometry - Mensuration**

A metallic solid is made up of a solid cylindrical base with a solid cone on its top. The radius of the base of the cone is 5cm. and the ratio of the height of the cylinder and the cone is 3:2. A cylindrical hole is drilled through the solid with height equal to 2/3rd of the height of solid. What should be the radius (in cm) of the hole so that the volume of the hole is 1/3rd of the volume of the metallic solid after drilling?

- A.
$\sqrt{\frac{45}{8}}$

- B.
$\sqrt{\frac{35}{8}}$

- C.
$\sqrt{\frac{65}{8}}$

- D.
$\sqrt{\frac{55}{8}}$

Answer: Option D

**Explanation** :

Volume of original solid = volume of cylinder + volume of cone

Radius of original cylinder = radius of cone = 5 cm

Let height of the original cylinder be 3x cm. Hence, height of cone = 2x cm

Volume of solid = $\pi \left({5}^{2}\right)\left(3x\right)+\frac{\pi}{3}\left({5}^{2}\right)\left(2x\right)$

$=25\pi \times \frac{11x}{3}=\frac{275\pi x}{3}$

Let the radius of the cylindrical hole be r cm.

Also, height of cylindrical hole = (2/3)(5x) = (10x/3) cm

∴ Volume of hole = $\pi \left({r}^{2}\right)\frac{10x}{3}$

Hence, as per the given condition:

$\frac{\pi \left({r}^{2}\right){\displaystyle \frac{10x}{3}}}{{\displaystyle \frac{275x}{3}}-\pi \left({r}^{2}\right){\displaystyle \frac{10x}{3}}}=\frac{1}{3}$

$\therefore \frac{10{r}^{2}}{275-10{r}^{2}}=\frac{1}{3}$

∴ 275 – 10*r*^{2} = 30*r*^{2}

∴ 40*r*^{2} = 245

$\therefore {r}^{2}=\frac{245}{40}=\frac{55}{8}$

$\therefore r=\sqrt{\left(\frac{55}{8}\right)}$

Hence, option (d).

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**14. IIFT 2018 QA | Arithmetic - Time & Work**

Nitin installed an overhead tank on the roof of his newly constructed house. Three taps are connected to the tank: 2 taps A and B to fill the tank and one tap C to empty it. Tap A alone can fill the tank in 12 hours, while tap B alone takes one and a half times more time than tap A to fill the tank completely. Tap C alone can empty a completely filled tank in 36 hours. Yesterday, to fill the tank, Nitin first opened tap A, and then after 2 hours opened tap B also. However after 6 hours he realised that tap C was open from the very beginning. He quickly closes tap C. What will be the total time required to fill the tank?

- A.
8 hours 48 minutes

- B.
8 hours 30 minutes

- C.
9 hours 12 minutes

- D.
9 hours 36 minutes

Answer: Option C

**Explanation** :

Tab B takes 1.5 times more than tap A to fill the tank.

A takes 12 hours to fill the tank alone.

Hence, time taken by tap B alone = 1.5 × 12 = 18 hours

Let the total capacity of the tank be 180 litres.

Let taps A and B fill a and b litres per hour and let tap C empty c litres per hour.

∴ a = 180/12 = 15; b = 180/18 = 10 and c = 180/36 = 5

For the first two hours, only A and C are working; for the next four hours, all three of A, B and C are working and for the next n hours, only A and B are working.

∴ 2(a – c) + 4(a + b – c) + n(a + b) = 180

∴ 2(15 – 5) + 4(15 + 10 – 5) + n(15 + 10) = 180

∴ 2(10) + 4(20) + 25n = 180

∴ 25n = 180 – 20 – 80 = 80

∴ n = 80/25 = 3.2 hours

∴ Total time = 6 + 3.2 = 9.2 hours

i.e. 9 hours and 12 minutes

Hence, option (c).

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**15. IIFT 2018 QA | Geometry - Trigonometry**

At the foot of the mountain, the angle of elevation of the summit at the top of the mountain is 45⁰. After ascending 100 metres, at a slope of 30⁰ up the mountain towards the summit, the angle of elevation of the summit is 60⁰. Find the height of the summit.

- A.
$50\left(\sqrt{3}+1\right)$ metres

- B.
$50\left(\sqrt{5}+1\right)$ metres

- C.
$50\left(\sqrt{3}+2\right)$ metres

- D.
$50\sqrt{3}$ metres

Answer: Option A

**Explanation** :

Let height of the mountain = x

From the figure, AD = distance travelled by man = 100 m

Also, DE = AD × sin 30° = 100 × 0.5 = 50 m

Also, AE = AD × cos 30° = 100 × (√3/2) = 50√3 m

Now, ∠CAB = 45°

Hence, ABC is a 45-45-90 triangle and AB = BC = x

Also, EB = AB – AE = (x − 50√3) m

∴ DF = EB = (x − 50√3) m

Similarly, DE = FB = 50 m

∴ CF = BC – FB = (x – 50) m

Now, in ∆CDF, tan 60° = CF/DF

∴ √3 = (x – 50)/(x − 50√3)

∴ √3x – 50(3) = x − 50

∴ (√3 – 1)x = 150 – 50 = 100

∴ x = 100/((√3 – 1) = 50(√3 + 1) [Rationalising]

Hence, option (a).

Workspace:

**16. IIFT 2018 QA | Arithmetic - Ratio, Proportion & Variation**

Land Cruiser Prado, the latest SUV from Toyota Motors, consumes diesel at the rate of $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}$ per km, when travelling at the speed of x km/hr. The diesel costs Rs. 65 per litre and the driver is paid Rs. 50 per hour. Find the steady speed that will minimize the total cost of a 1000 km trip?

- A.
33 km/hr

- B.
36 km/hr

- C.
39 km/hr

- D.
52 km/hr

Answer: Option B

**Explanation** :

Total cost incurred = total cost incurred due to petrol + total cost incurred due to driver charges

= (total litres consumed × cost per litre) + (total driver hours × cost per hour)

Total petrol consumed in 1000 km = 1000 × $\frac{1}{400}\left\{\frac{1000}{x}+x\right\}=2.5\left\{\frac{1000}{x}+x\right\}$

∴ Total petrol cost = $2.5\left\{\frac{1000}{x}+x\right\}\times 65$

$=162.5\times \left\{\frac{1000}{x}+x\right\}$

Time taken to travel 1000 km = 1000/x

∴ Total driver cost = (1000/x) × 50 = Rs. (50000/x)

∴ Total cost = 162.5 × $\left\{\frac{1000}{x}+x\right\}+\left(\frac{50000}{x}\right)$

$=\left(\frac{1000}{x}\right)(162.5+50)+162.5x$

$=\frac{212500}{x}+162.5x$

You can now substitute options and verify that the least cost is for x = 36.

Hence, option (b).

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**17. IIFT 2018 QA | Algebra - Simple Equations**

In a survey on the viewership of the TV channels, 73% of those surveyed viewed at least one of the three Channels: Star Plus, Sab TV, and Sony. 38% of those surveyed viewed Star Plus, 39% viewed Sony, and 23% viewed Sab TV. 11% of all those surveyed viewed all the three channels. What percentage of those surveyed, viewed more than one of the three TV channels?

- A.
16

- B.
38

- C.
27

- D.
Data Inadequate

Answer: Option A

**Explanation** :

Let there be 100 people surveyed in all.

Let a, b, c correspond to people who watched only Star Plus, only Sony and only SAB TV.

Let d, e, f correspond to people who watched only Star Plus and Sony, only Sony and SAB TV and only SAB TV and Star Plus.

Let g correspond to people who watched all three channels.

73% people i.e. 73 people watched atleast one channel.

∴ a + b + c + d + e + f + g = 73 … (i)

Number of people watching Star Plus, Sony and SAB TV is 38%, 39% and 23% respectively i.e. 38, 39 and 23.

∴ a + d + f + g = 38 … (ii)

b + d + e + g = 39 … (iii)

c + e + f + g = 23 … (iv)

Adding (i), (ii) and (iii):

(a + b + c) + 2(d + e + f) + 3g = 39 + 38 + 23 = 100 … (v)

Solving (i) and (v), we get:

(d + e + f) + 2g = 100 – 73 = 27

Number of people surveyed who watch more than one channel = (d + e + f) + g

Also, number of people who watch all channels = g = 11

∴ (d + e + f) + g + 11 = 27

∴ (d + e + f) + g = 16

∴ Required % = (16/100) × 100 = 16%

Hence, option (a).

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**18. IIFT 2018 QA | Arithmetic - Ratio, Proportion & Variation**

A physical therapist of Russian football team knows that the team will play 40% of its matches on artificial turf, this season. Because of his vast experience, he knows that a football player’s chances of incurring a knee injury is 50% higher if he is playing on artificial turf instead of grass. If the player's chances of a knee injury is 0.42, what is the probability that a football player with knee injury, incurred the injury while playing on grass?

- A.
0.28

- B.
0.336

- C.
0.5

- D.
None of the above

Answer: Option C

**Explanation** :

P(playing on artificial turf) = 0.4

∴ P(playing on grass) = 1 – P(playing on artificial turf) = 1 – 0.4 = 0.6

P(injury/artificial turf) = 0.42

Also, P(injury/artificial turf) = 1.5 × P(injury/grass)

∴ P(injury/grass) = 0.42/1.5 = 0.28

Now, P(injury) = P(injury/grass) × P(playing on grass) + P(injury/artificial turf) × P(playing on artificial turf)

= (0.28 × 0.6) + (0.42 × 0.4)

= 0.168 + 0.168 = 0.336

Also, P(injury/grass) × P(playing on grass) = P(grass/injury) × P(injury)

∴ P(grass/injury) = 0.168/0.336 = 0.5

Hence, option (c).

Workspace:

**19. IIFT 2018 QA | Algebra - Simple Equations**

The square root of $1+{x}^{2}+\sqrt{1+{x}^{2}+{x}^{4}}$1

- A.
$\frac{1}{\sqrt{2}}\left[\sqrt{1+x+{x}^{2}}+\sqrt{1-x+{x}^{2}}\right]$

- B.
$\frac{1}{\sqrt{2}}\left[\sqrt{1+x+{x}^{2}}-\sqrt{1-x+{x}^{2}}\right]$

- C.
$\frac{1}{\sqrt{2}}\left[\sqrt{(1+{x}^{2}+{x}^{4}+{x}^{8})}\right]$1

- D.
None of the above

Answer: Option A

**Explanation** :

Instead of solving algebraically, solve the question by first substituting x = 1. You will get one original value and three values in the options. Square each value from the options and match it with the original value of the expression.

Original expression:$1+{x}^{2}+\sqrt{(1+{x}^{2}+{x}^{4})}$

$=1+{\left(1\right)}^{2}+\sqrt{[1+{\left(1\right)}^{2}+{\left(1\right)}^{4}]}$

$=1+1+\sqrt{3}=2+\sqrt{3}$

Option 1:

$\frac{1}{\sqrt{2}}\left[\sqrt{1+x+{x}^{2}}+\sqrt{1-x+{x}^{2}}\right]$

$=\frac{1}{\sqrt{2}}\left[\sqrt{1+1+{1}^{2}}+\sqrt{1-1+{1}^{2}}\right]$

$=\frac{\sqrt{3}+1}{\sqrt{2}}$

Now, square this value.

$=\frac{3+1+2\sqrt{3}}{2}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}$

Hence, option (a).

Workspace:

**20. IIFT 2018 QA | Algebra - Simple Equations**

${\mathrm{log}}_{2}x{\mathrm{log}}_{\frac{x}{64}}2={\mathrm{log}}_{\frac{x}{16}}2;\; then\; x\; =\; ?$

- A.
2

- B.
4

- C.
12

- D.
16

Answer: Option B

**Explanation** :

log_{2}*x* × log(*x*/64)^{2} = log(*x*/16)^{2}

$\therefore \frac{\mathrm{log}x}{\mathrm{log}2}\times \frac{\mathrm{log}2}{\mathrm{log}\left({\displaystyle \frac{x}{64}}\right)}=\frac{\mathrm{log}2}{\mathrm{log}\left({\displaystyle \frac{x}{16}}\right)}$

It is clear that x has to be a power of 2 to solve the equation i.e. x ≠ 12.

Hence, option (c) is eliminated.

If x = 16, then the base of the RHS (i.e. x/16) becomes 1, which is not possible for a logarithm i.e. x ≠ 16. Hence, option (d) is eliminated.

If x = 2, then you get (x/64) = (x/16); which is not possible.

Hence, x = 4

Hence, option (b).

Workspace:

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