# Arithmetic - Ratio, Proportion & Variation - Previous Year CAT/MBA Questions

You can practice all previous year OMET questions from the topic Arithmetic - Ratio, Proportion & Variation. This will help you understand the type of questions asked in OMET. It would be best if you clear your concepts before you practice previous year OMET questions.

**XAT 2020 QADI | Arithmetic - Ratio, Proportion & Variation**

A box contains 6 cricket balls, 5 tennis balls and 4 rubber balls. Of these, some balls are defective. The proportion of defective cricket balls is more than the proportion of defective tennis balls but less than the proportion of defective rubber balls. Moreover, the overall proportion of defective balls is twice the proportion of defective tennis balls. What BEST can be said about the number of defective rubber balls in the box?

- A.
It is exactly 3

- B.
It is either 3 or 4

- C.
It is exactly 2

- D.
It is either 2 or 3

- E.
It is either 0 or 1

Answer: Option A

**Explanation** :

Let the number of defective cricket, tennis and rubber balls = c, t and r respectively.

Given, r/4 > c/6 > t/5 ...(1)

and (c + t + r)/15 = 2 × t/5

⇒ c + r = 5t

Also, maximum value of c + r can be 5 + 4 = 9

⇒ The value of t can only be 1.

⇒ c + r = 5 …(2)

The only possible value to satisfy (1) and (2) is r = 3 and c = 2.

Hence, option (a).

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**XAT 2020 QADI | Arithmetic - Ratio, Proportion & Variation**

X, Y and Z start a web-based venture together. X invests Rs. 2.5 lakhs, Y invests Rs. 3.5 lakhs, and Z invests Rs. 4 lakhs. In the first year, the venture makes a profit of Rs. 2 lakhs. A part of the profit is shared between Y and Z in the ratio of 2 : 3, and the remaining profit is divided among X, Y and Z in the ratio of their initial investments.

The amount that Z receives is four times the amount that X receives. How much amount does Y receive?

- A.
Rs. 80,200

- B.
Rs. 75,000

- C.
Rs. 93,750

- D.
Rs. 74,250

- E.
Rs. 1,02,500

Answer: Option B

**Explanation** :

Let Rs. 5P is divided between Y and Z in the ratio of 2 : 3.

∴ Y received 2P and Z received 3P.

Remaining profit = 2,00,000 – 5P

This will be dived amongst X, Y and Z in the ratio 2.5 : 3.5 : 4 = 5 : 7 : 8.

∴ X receives 5/20 × (2,00,000 – 5P) and Z receives 8/20 × (2,00,000 – 5P)

Total amount received by X = 5/20 × (2,00,000 – 5P)

Total amount received by Z = 3P + 8/20 × (2,00,000 – 5P)

According to the question:

3P + 8/20 × (2,00,000 – 5P) = 4(5/20 × (2,00,000 – 5P))

⇒ 3P + 80,000 – 2P = 2,00,000 – 5P

⇒ 6P = 1,20,000

⇒ P = 20,000

∴ Y receives = 2P + 7/20 × (2,00,000 – 5P) = 40,000 + 7/20 × 1,00,000 = 75,000

Hence, option (b).

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**XAT 2019 QADI | Arithmetic - Ratio, Proportion & Variation**

Two numbers a and b are inversely proportional to each other. If a increases by 100%, then b decreases by

- A.
200%

- B.
50%

- C.
100%

- D.
80%

- E.
150%

Answer: Option B

**Explanation** :

Given that ‘a’ and ‘b’ are inversely proportional

∴ a ∝ 1/b

⇒ a × b = constant

Hence, if a is doubled (i.e., increased by 100%) b needs to become half.

∴ % change in b = (1/2 - 1) × 100% = 50%.

Hence, option (b).

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**XAT 2018 QADI | Arithmetic - Ratio, Proportion & Variation**

The number of boys in a school was 30 more than the number of girls. Subsequently, a few more girls joined the same school. Consequently, the ratio of boys and girls became 3:5. Find the minimum number of girls, who joined subsequently.

- A.
31

- B.
51

- C.
52

- D.
55

- E.
Solution not possible

Answer: Option C

**Explanation** :

Let the number of girls in the school initially be ‘g’.

∴ Number of boys = g + 30.

If ‘x’ new girls who joined the school.

⇒ (g + 30)/(g + x) = 3/5

∴ 5g + 150 = 3g + 3x

∴ x = 2g/3 + 50

For x to be minimum g should be minimum. Also, g should be a multiple of 3 for x to be an integer.

⇒ Minimum value of g = 3

∴ The minimum value of x = 52.

Hence, option (c).

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**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 2.

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**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 3.

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**IIFT 2017 QA | Arithmetic - Ratio, Proportion & Variation**

Suntex Company plans to manufacture a new product line of Razor next year and sell it at a price of Rs. 12 per unit. The variable costs per unit in each production run is estimated to be 50% of the selling price, and the fixed costs for each production run is estimated to be Rs. 50,400. Based om their estimated costs how many units of the new product will company Suntex need to manufacture and sell in order for their revenue to be equal to their total costs for each production run?

- A.
5400

- B.
4200

- C.
8400

- D.
2100

Answer: Option C

**Explanation** :

Let n units be sold.

Variable cost per unit = 0.5 × 12 = Rs. 6 per unit

∴ Total cost for one production run = Rs. (50400 + 6n)

Total revenue by selling n units = Rs. (12n)

When cost = revenue; 50400 + 6n = 12n

∴ 6n = 50400 i.e. n = 8400

Hence, option 3.

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**IIFT 2015 QA | Arithmetic - Ratio, Proportion & Variation**

Sailesh is working as a sales executive with a reputed FMCG Company in Hyderabad. As per the Company’s policy, Sailesh gets a commission of 6% on all sales upto Rs.1,00,000 and 5% on all sales in excess of this amount. If Sailesh remits Rs. 2,65,000 to the FMCG company after deducting his commission, his total slaes were worth:

- A.
Rs. 1,20,000

- B.
Rs. 2,90,526

- C.
Rs. 2,21,054

- D.
Rs. 2,80,000

Answer: Option D

**Explanation** :

Sailesh earns Rs. 6,000 as commission from first Rs. 1,00,000.

Let his total sales = X

∴ Total commission = 6000 + 0.05 (X – 100000) = 0.05X + 1000

∴ X – 0.05X + 1000 = 265000

Solving, we get X = 280000

Hence, option 4.

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**Read the following information and Tables and answer the questions.**

BHUBANESWAR, CHENNAI, KANYAKUMARI, KOCHI, MUMBAI and VIZAG are 6 major Indian cities. For some reason people use only a certain mode of transport between a pair of cities. The modes of transport are provided in Table 1, while in Table 2 the distances between different pairs of cities are given. Table 3 provides the speed of the mode of transport and the cost associated with each of them.

**IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

For which of the following options, travel time is the least?

- A.
MUMBAI – KANYAKUMARI

- B.
BHUBANESWAR - CHENNAI

- C.
CHENNAI - KOCHI

- D.
MUMBAI - CHENNAI

Answer: Option A

**Explanation** :

Time taken to travel = $\frac{\mathrm{Distance}}{\mathrm{Speed}\mathrm{of}\mathrm{available}\mathrm{mode}\mathrm{of}\mathrm{transport}}$

Considering the options, we have the travel times for (Mumbai – Kanyakumari, Bhubaneshwar – Chennai, Chennai – Kochi, Mumbai – Chennai) as,

$\frac{950}{40}$, $\frac{950}{30}$, $\frac{901}{30}$, $\frac{1000}{30}$

It can be observed that $\left(\frac{950}{40}\right)$ has the least value.

Hence, option 1.

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**IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

Mr. Ranjith lives in MUMBAI and wants to travel to KOCHI. However, the train services are on halt due to laying of track for bullet trains across the country. In this scenario, which of the following is the least cost route to reach KOCHI?

- A.
MUMBAI - BHUBANESWAR – KOCHI

- B.
MUMBAI – CHENNAI - KOCHI

- C.
MUMBAI – KANYAKUMARI – KOCHI

- D.
MUMBAI - VIZAG - KOCHI

Answer: Option B

**Explanation** :

Cost of travel = Distance × Cost per KM of available mode of transport

Considering the cost of travel in each option.

Option 1: (701 × 2) + (798 × 5)

≈ 1400 + 4000 = 5400

Option 2: (1000 × 1.5) + (901 × 1.5)

≈ 1900 × 1.5 = 2850

Option 3: (950 × 2) + (1100 × 2.5)

= 1900 + 2750 > 2850

Option 4: (500 × 5) + (600 × 5) > 2850

Hence, option 2.

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**IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

A school in Chennai is planning for an excursion tour for its students. They want to show them KANYAKUMARI, VIZAG, and BHUBANESWAR, not necessarily in the same order. What is the minimum travel cost (in RS.) the school should charge from each of the student for the entire tour?

- A.
Rs. 4300

- B.
Rs. 5000

- C.
Rs. 7500

- D.
Rs. 6800

Answer: Option A

**Explanation** :

Observe that on a per-km basis, ship is the cheapest and airplane is the costliest. Hence, between the given cities, try to select the order of travel such that most of the journey is done by ship, followed by bus, then train and so on.

The overall route to be followed is: Chennai – city 1 – city 2 - city 3 - Chennai.

Chennai is connected to Kanyakumari and Vizag by train and to Bhubaneshwar by ship.

Among the three given cities, Kanyakumari is also the farthest from Chennai.

Hence, Chennai to Kanyakumari is always costlier than Chennai to the other two cities.

Hence, those two cities should be the first and last to be visited (in any order, due to the symmetry of the journey) while Kanyakumari should be visited between the two, as shown below.

Chennai → Vizag/Bhubaneshwar → Kanyakumari → Bhubaneshwar/ Vizag → Chennai

Now, find individual costs:

Chennai – Vizag: (300 × 2.5) = 750

Chennai – Bhubaneshwar: (950 × 1.5) = 1425

Vizag – Kanyakumari: (250 ×1.5) = 375

Kanyakumari – Bhubaneshwar: (700 ×2.5) = 1750

The total cost is Rs. 4300/-.

Hence, option 1.

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**IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

Which of the following cities can be reached from BHUBANESWAR in least time?

- A.
CHENNAI

- B.
KANYAKUMARI

- C.
MUMBAI

- D.
VIZAG

Answer: Option C

**Explanation** :

Considering the options, we have the travel times as,

$\frac{950}{40}$, $\frac{700}{25}$, $\frac{701}{40}$, $\frac{1002}{25}$

It can be observed that $\left(\frac{701}{40}\right)$ has the least value.

Hence, option 3.

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**IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

What is the least cost way to reach to VIZAG from KOCHI?

- A.
Take a flight from KOCHI to VIZAG

- B.
Take a ship from KOCHI to CHENNAI and then take a train to VIZAG

- C.
Take a train from KOCHI to KANYKUMARI and then take a ship to VIZAG

- D.
Take a train from KOCHI to MUMBAI and then take a flight to VIZAG

Answer: Option B

**Explanation** :

Considering the cost in each option.

Option 1: (600 × 5) = 3000

Option 2: (901 × 1.5) + (300 × 2.5) ≈ 2100

Option 3: (1100 × 2.5) + (250 × 1.5) > 2100

Option 4: (300 × 2.5) + (500 × 5) > 2100

Hence, option 2.

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**IIFT 2014 QA | Arithmetic - Ratio, Proportion & Variation**

Ravindra and Rekha got married 10 years ago, their ages were in the ratio of 5 : 4. Today Ravindra’s age is one sixth more than Rekha’s age. After marriage, they had 6 children including a triplet and twins. The age of the triplets, twins and the sixth child is in the ratio of 3 : 2 : 1. What is the largest possible value of the present total age of the family?

- A.
79

- B.
93

- C.
101

- D.
107

Answer: Option D

**Explanation** :

Let the present age of Ravindra and Rekha be R_{v} and R_{e} respectively.

$\frac{{R}_{v}-10}{{R}_{e}-10}$ = $\frac{5}{4}$

R_{v} = (1 + 1/6)R_{e}

Solving these equations, we get R_{v} = 35 and R_{e} = 30

Maximum age of a child = 9 years (10 year after marriage)

To maximize the age of family,

Age of triplets = 9 years, twins = 6 years and sixth child = 3 years

Age of the family = 35 + 30 + (9 × 3) + (6 × 2) + 3 = 107

Hence, option 4.

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**XAT 2012 QA | Arithmetic - Ratio, Proportion & Variation**

Ram and Shyam form a partnership (with Shyam as working partner) and start a business by investing Rs. 4000 and Rs. 6000 respectively. The conditions of partnership were as follows:

- In case of profits till Rs. 200,000 per annum, profits would be shared in the radio of the invested capital.
- Profits from Rs. 200,001 till Rs. 400,000 Shyam would take 20% out of the profit, before the division of remaining profits, which will then be based on ratio of invested capital.
- Profits in excess of Rs. 400,000, Shyam would take 35% out of the profits beyond Rs. 400,000, before the division of remaining profits, which will then be based on ratio of invested capital.

If Shyam’s share in a particular year was Rs. 367000, which option indicates the total business profit (in Rs.) for that year?

- A.
520,000

- B.
530,000

- C.
540,000

- D.
550,000

- E.
None of the above

Answer: Option D

**Explanation** :

For first Rs. 200000, Shyam gets, 6000/(4000 + 6000) × 100 = 60% of the profit.

For next Rs. 200000, he gets 20% + plus 60% of the remaining profit.

i.e. 20% + 80 × 0.6% = 68%

Similarly, for a profit margin greater than Rs. 400000, he will get, 35% + 65 × 0.6% = 74% of the profits beyond Rs. 400000

Now, for a profit of first Rs. 400000, Shyam will receive 200000 × (68 + 60)/100 = 256000

But Shyam earns a total profit of 367000.

Let total profit earned by them be Rs. 400000 + x.

Hence, Shyam received 367000 – 256000 = Rs. 110000 from Rs. x profit.

i.e. Rs. 110000 is 74% of x.

Hence, x = 110000/0.74 = 150000

Hence, total profit earned by them = 400000 + 150000 = Rs. 550000

Hence, option 4.

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**IIFT 2012 QA | Arithmetic - Ratio, Proportion & Variation**

A sum of Rs.1400 is divided amongst A, B, C and D such that A's share : B's share = B's share : C's share = C's share : D's share = $\frac{3}{4}$

How much is C’s share?

- A.
Rs.72

- B.
Rs.288

- C.
Rs.216

- D.
Rs.384

Answer: Option D

**Explanation** :

A : B = 3 : 4, B : C = 3 : 4, C : D = 3 : 4

∴ A : D = ${\left(\frac{3}{4}\right)}^{3}$ = $\frac{27}{64}$

Let D = 64

Then, C = $\frac{3}{4}$ × 64 = 48

B = $\frac{3}{4}$ × 48 = 36

A = $\frac{3}{4}$ × 36 = 27

∴ A : B : C : D = 27 : 36 : 48 : 64

∴ C's share = $\left\{\frac{48}{27+36+48+64}\right\}$ × 1400 = Rs. 384

Hence, option 4.

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**IIFT 2011 QA | Arithmetic - Ratio, Proportion & Variation**

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, $\frac{l}{3}$ = $\frac{v}{2}$ + 5 ...(ii)

Solving i and ii, we get,

l = 90, and v = 50

Hence, required difference = 70 – 50 = 20

Hence, option 3.

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**IIFT 2011 QA | Arithmetic - Ratio, Proportion & Variation**

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.

∴ $\frac{a}{3}$ = $\frac{b}{2}$

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 4.

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**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)

Answer: Option B

**Explanation** :

Let p, m and b be the number of pineapple, mango and black-forest pastries respectively.

∴ p + m + b = 23 … (i)

Each pastry cost as many rupees as there were pastries of that kind.

∴ p^{2} + m^{2} + b^{2} = 211 … (ii)

Substituting options in (i) and (ii), we find that only option 2 satisfies both the equations

Hence, option 2.

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**IIFT 2009 QA | Arithmetic - Ratio, Proportion & Variation**

Kartik’s mother asked him to get the vegetables, milk and butter from the market and gave him the money in the denomination of 1 Rupee, 2 Rupee and 5 Rupee coins. Kartik first goes to the grocery shop to buy vegetables. At the grocery shop he gives half of his 5 Rupee coins and in return receives the same number of 1 Rupee coins. Next he goes to a dairy shop to buy milk and butter and gives all 2 Rupee coins and in return gets thirty 5 Rupee coins which increases the number of 5 Rupee coins to 75% more than the original number. If the number of 1 Rupee coins now is 50, the number of 1 Rupee and 5 Rupee coins originally were:

- A.
10, 60

- B.
10, 70

- C.
10, 80

- D.
None of the above

Answer: Option D

**Explanation** :

Let Kartik initially have x, y and z, 1 Rupee, 2 Rupee and 5 Rupee coins respectively.

At the grocery shop, the number of 5 Rupee coins becomes z/2 and the number of 1 rupee coins becomes x + z/2

At the dairy, the number of 5 Rupee coins changes to z/2 + 30

By conditions given in the question,

z/2 + 30 = 1.75z

∴ z = 24

Also, x + z/2 = 50

∴ x = 38

Hence, option 4.

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**IIFT 2009 QA | Arithmetic - Ratio, Proportion & Variation**

Find the ratio of shaded area to unshaded area.

- A.
$\frac{1}{5}$($\sqrt{21}$ - 2)

- B.
$\frac{1}{5}$(3$\sqrt{7}$ - 2)

- C.
$\frac{1}{5}$(3$\sqrt{7}$ - 2$\sqrt{3}$)

- D.
None of the above

Answer: Option D

**Explanation** :

Area of shaded pentagon = $\frac{5}{4}$$\left(\sqrt{1+\frac{2}{\sqrt{5}}}\right){a}^{2}$

Area of 5, unshaded equilateral triangles = 5 × $\frac{\sqrt{3}}{4}{a}^{2}$

∴ Ratio of shaded area to unshaded area = $\frac{{\displaystyle \frac{5}{4}}\left(\sqrt{1+{\displaystyle \frac{2}{\sqrt{5}}}}\right){a}^{2}}{5\times {\displaystyle \frac{\sqrt{3}}{4}}{a}^{2}}$ = $\sqrt{\frac{1}{3}+\frac{2}{3\sqrt{5}}}$

None of the option is right.

Hence, option 4.

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