# CAT 2018 QA Slot 1

**1. CAT 2018 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?

Answer: 15

**Explanation** :

Let distance between X and Y be ‘d’ km and train T travels at speed ‘s’ kmph.

Therefore, speed of train S = $\frac{3}{4}$s kmph

At 4 pm, T must have covered ‘s’ km.

Therefore, distance between the two

= (d – s) km

Time taken for the two to cover this distance = $\frac{d-s}{s+{\displaystyle \frac{3}{4}}s}$ = $\frac{4(d-s)}{7s}$

Distance covered by train T by now = s + $\frac{4(d-s)}{7s}\times s$ = $\frac{3s+4d}{7}$

By the given condition, $\frac{3s+4d}{7}$ = $\frac{3}{5}d$

Solving this, we get d = 15s

To travel from X to Y, train T takes 15s/s = 15 hours

Answer :15

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**2. CAT 2018 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is

Answer: 12

**Explanation** :

Let distance between A and P be ‘x’ units. Therefore, distance between P and B = (3x) units

Let car 1 reaches point P in ‘t’ minutes. Therefore, car 2 takes (t + 60) minutes to reach P.

Speed of cat 1 = x/t and speed of car 2 = 3x/(t + 60)

Hence, $\frac{x}{2t}$ = $\frac{3x}{t+60}$

Solving this, we get

t = 12 minutes

Answer: 12

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**3. CAT 2018 QA Slot 1 | Arithmetic - Simple & Compound Interest**

John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest of the principal amount plus due interest thereon. Then each instalment, in Rs., is

Answer: 121000

**Explanation** :

Let the part of principal amount in the first instalment be Rs. X.

Interest for the first year on Rs. 2,10,0000 = 210000 × 0.1 = 21,000

Therefore, the first instalment = (21000 + X)

Remaining part of the principal = (210000 – X)

Amount due on this principal of (210000 – X) = (210000 – X) × 1.1 = 231000 – 1.1X

Therefore, 21000 + X = 231000 – 1.1X

Solving this, we get X = 1,00,000

∴ Each instalment = 100000 + 21000 = Rs. 1,21,000

Answer: 121000.

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**4. CAT 2018 QA Slot 1 | Arithmetic - Time & Work**

A tank is fitted with pipes, some filling it and the rest draining it. All filling pipes fill at the same rate, and all draining pipes drain at the same rate. The empty tank gets completely filled in 6 hours when 6 filling and 5 draining pipes are on, but this time becomes 60 hours when 5 filling and 6 draining pipes are on. In how many hours will the empty tank get completely filled when one draining and two filling pipes are on?

Answer: 10

**Explanation** :

Let a filling pipe fills the tank at ‘a’ liters per hour and a draining pipe drains at ‘b’ liters per hour.

Therefore,

6(6a – 5b) = 60(5a – 6b)

∴ 6a – 5b = 50a – 60b

∴ 44a = 55b

∴ a = 5b/4

Let the tank gets filled completely in ‘m’ hours when one draining pipe and two filling pipes are on.

∴ m(2a – b) = 6(6a – 5b)

∴ $m\left(2\times \frac{5}{4}b-b\right)$ = $6\left(6\times \frac{5}{4}b-5b\right)$

∴ $m\left(\frac{6b}{4}\right)$ = $6\left(\frac{10b}{4}\right)$

Solving this, we get m = 10

Answer: 10

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**5. CAT 2018 QA Slot 1 | Arithmetic - Average**

A CAT aspirant appears for a certain number of tests. His average score increases by 1 if the first 10 tests are not considered, and decreases by 1 if the last 10 tests are not considered. If his average scores for the first 10 and the last 10 tests are 20 and 30, respectively, then the total number of tests taken by him is

Answer: 60

**Explanation** :

Let there be ‘n’ tests and their average be ‘x’.

By the given conditions,

$\frac{nx-200}{n-10}$ = x + 1

$\frac{nx-300}{n-10}$ = x - 1

∴ $\frac{nx-200}{n-10}$ - 1 = $\frac{nx-300}{n-10}$ + 1

∴ $\frac{nx-200}{n-10}$ - $\frac{nx-300}{n-10}$ = 2

∴ $\frac{100}{n-10}$ = 2

∴ n = 60

Answer: 60

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**6. CAT 2018 QA Slot 1 | Arithmetic - Average**

In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

- A.
27

- B.
28

- C.
26

- D.
25

Answer: Option B

**Explanation** :

Let x is average age of 30 people whose age is 51 years and above. Clearly, x ≥ 51

Let y is average age of 39 people whose age is less than 51 years.

∴ 30x + 39y = 69 × 38

∴ 10x + 13y = 874

x ≥ 51 ⇒ 10x ≥ 510 ⇒ 10x + 13y ≥ 510 + 13y

⇒ 874 ≥ 510 + 13y ⇒ 364 ≥ 13y

⇒ 28 ≥ y

Thus, the largest possible average age of the people whose ages are below 51 years is 28 years.

Hence, option 2.

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**7. CAT 2018 QA Slot 1 | Arithmetic - Percentage**

In an examination, the maximum possible score is N while the pass mark is 45% of N. A candidate obtains 36 marks, but falls short of the pass mark by 68%. Which one of the following is then correct?

- A.
N ≤ 200

- B.
201 ≤ N ≤ 242

- C.
243 ≤ N ≤ 252

- D.
N ≥ 253

Answer: Option C

**Explanation** :

As 36 marks are short of the pass marks by 68%,

36 = (100 – 68)% of 45% of N

∴ 36 = 0.32 × 0.45 × N

∴ N = $\frac{36}{0.32\times 0.45}$ = 250

Hence,option 3.

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**8. CAT 2018 QA Slot 1 | Arithmetic - Profit & Loss**

A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?

- A.
86

- B.
84

- C.
98

- D.
96

Answer: Option D

**Explanation** :

Assume that the wholesaler bought peanut at Rs. X per kg.

∴ Cost price of walnuts = Rs. 3X per kg

The wholesaler sold peanuts at Rs. 1.1X per kg and walnuts at Rs. 3.6X per kg.

Total cost of 8 kg peanuts and 16 kg walnuts for the shopkeeper

= 8 × 1.1X + 16 × 3.6X = 66.4X

Shopkeeper overall profit was 25%

Therefore, revenue = 66.4X × 1.25

He sold (8 – 3 =) 5 kg peanuts and (16 – 5 =) 11 kg walnuts after mixing at Rs. 166 per kg

Therefore, 66.4X × 1.25 = 16 × 166

Solving this, we get

X = 32

Cost price of walnuts for the wholesaler = 3 × 32 = Rs. 96

Hence, option 4.

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**9. CAT 2018 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement | Arithmetic - Profit & Loss**

Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio

- A.
17 : 25

- B.
18 : 25

- C.
21 : 25

- D.
19 : 24

Answer: Option D

**Explanation** :

Let cost of tea A and B be a and b respectively.

If 3 kg of tea A is mixed with 2 kg of tea B, (3a + 2b) × 1.1 = 40× 5 = 200

If 2 kg of tea A is mixed with 3 kg of tea B, (2a + 3b) × 1.05 = 40× 5 = 200

Therefore,

(3a + 2b) × 1.1 = (2a + 3b) × 1.05

∴ 3.3a + 2.2b = 2.1a + 3.15b

∴ a/b = 19/24

Hence, option 4.

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**10. CAT 2018 QA Slot 1 | Arithmetic - Time & Work**

Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any robot) take to finish it?

- A.
36

- B.
32

- C.
45

- D.
40

Answer: Option B

**Explanation** :

Let work-done by a human and a robot in a day are ‘h’ and ‘r’ respectively.

Total work = 30(15h + 5r) = 60(5h + 15r)

∴15h + 5r = 10h + 30r

∴ h = 5r

Let fifteen humans take ‘y’ days to finish the job.

∴ 30(15h + 5r) = y × 15h

∴ 30(15h + h) = y × 15h

∴ y = 32

Hence, option 2.

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**11. CAT 2018 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

Raju and Lalitha originally had marbles in the ratio 4 : 9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5 : 6. What fraction of her original number of marbles was given by Lalitha to Raju?

- A.
6/19

- B.
1/5

- C.
7/33

- D.
1/4

Answer: Option C

**Explanation** :

Let Raju and Lalitha originally had (4x) and (9x) marbles. Later Lalitha gave (y) marbles to Raju.

Therefore, $\frac{4x+y}{9x-y}$ = $\frac{5}{6}$

Solving this, we get

y = 21x/11

The required fraction = $\frac{{\displaystyle \frac{21}{11}}x}{9x}$ = $\frac{7}{33}$

Hence, option 3.

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**12. CAT 2018 QA Slot 1 | Arithmetic - Time & Work**

When they work alone, B needs 25% more time to finish a job than A does. They two finish the job in 13 days in the following manner: A works alone till half the job is done, then A and B work together for four days, and finally B works alone to complete the remaining 5% of the job. In how many days can B alone finish the entire job?

- A.
18

- B.
22

- C.
16

- D.
20

Answer: Option D

**Explanation** :

Ratio of time taken by A and B to complete a certain task = 4 : 5.

∴ Ratio of their efficiencies = 5 : 4.

Now, A completes 50% of the work alone and B completes 5% of the work alone. That means, A and B together complete 45% of the work together in 4 days.

Out of this 45%, work done by A and B individually will be in the ratio of their efficiencies.

∴ Work done by A B in 4 days = 5/9th of 45% = 20%.

⇒ B completes 20% of the work in 4 days.

Thus, B alone can finish the entire job in 20 days.

Hence, option 4.

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**13. CAT 2018 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement | Arithmetic - Profit & Loss**

A trader sells 10 litres of a mixture of paints A and B, where the amount of B in the mixture does not exceed that of A. The cost of paint A per litre is Rs. 8 more than that of paint B. If the trader sells the entire mixture for Rs. 264 and makes a profit of 10%, then the highest possible cost of paint B, in Rs. per litre, is

- A.
20

- B.
26

- C.
16

- D.
22

Answer: Option A

**Explanation** :

Let the 10 litres of mixture has ‘Y’ litres of A and (10 – Y) litres of B. Let cost of paint B be Rs. X and that of A be Rs. (X + 8).

We know that, Y ≥ (10 – Y) ⇒ Y ≥ 5

The trader makes 10% profit by selling this mixture at Rs. 264.

∴ Cost price of the mixture = $\frac{264}{1.1}$ = Rs. 240

∴ (X + 8) × Y + (10 – Y) × X = 240

∴ 10X + 8Y = 240

∴ X = 24 – 0.8Y

For maximum value of X, we need to consider minimum value of Y.

∴ X = 24 – (0.8 × 5) = Rs. 20

Hence, option 1.

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**14. CAT 2018 QA Slot 1 | Arithmetic - Time, Speed & Distance**

The distance from A to B is 60 km. Partha and Narayan start from A at the same time and move towards B. Partha takes four hours more than Narayan to reach B. Moreover, Partha reaches the mid-point of A and B two hours before Narayan reaches B. The speed of Partha, in km per hour, is

- A.
4

- B.
3

- C.
6

- D.
5

Answer: Option D

**Explanation** :

Let the time taken by Narayan to cover 60 km be ‘t’ hours.

Therefore, Partha takes (t + 4) hours to cover 60 km.

Parth takes (t – 2) hours to cover half the distance.

So, to cover remaining distance i.e., 30 km, he takes (t + 4) – (t – 2) = 6 hours.

Hence, Partha's speed = 30/5 = 6 kmph

Hence, option 4.

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**15. CAT 2018 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is

- A.
2 : 5

- B.
4 : 9

- C.
3 : 8

- D.
1 : 3

Answer: Option D

**Explanation** :

y > x

Side of smaller square = $\sqrt{{x}^{2}+{y}^{2}}$

By the given condition,

x^{2} + y^{2} = 62.5% of (x + y)^{2}

x^{2} + y^{2} = 5/8 (x + y)^{2}

8x^{2} + 8y^{2} = 5x^{2} + 5y^{2} + 10xy

3x^{2} + 3y^{2} - 10xy = 0

3${\left(\frac{x}{y}\right)}^{2}$ -10$\left(\frac{x}{y}\right)$ + 3 = 0

$\left(\frac{x}{y}\right)$ = $\frac{10\pm \sqrt{{10}^{2}-4\times 3\times 3}}{6}$ = $\frac{10\pm 8}{6}$ = 9/3 of 1/3

As y > x, x/y < 1

∴ x : y = 1 : 3

Hence, option 4.

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**16. CAT 2018 QA Slot 1 | Geometry - Circles**

In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is

- A.
√12

- B.
√13

- C.
√11

- D.
√14

Answer: Option B

**Explanation** :

OP⊥ PQ and OQ⊥ AB

Let OP = x cm

OC = OB

∴ Radius = $\sqrt{P{C}^{2}+Q{O}^{2}}$ = $\sqrt{Q{B}^{2}+Q{O}^{2}}$

∴ $\sqrt{{3}^{2}+{x}^{2}}$ = $\sqrt{{2}^{2}+{(x+1)}^{2}}$

∴ x = 2

∴ Radius = $\sqrt{{9}^{2}+{2}^{2}}$ = √13

Hence, option 2.

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**17. CAT 2018 QA Slot 1 | Geometry - Circles**

In a circle with center O and radius 1 cm, an arc AB makes an angle 60 degrees at O. Let R be the region bounded by the radii OA, OB and the arc AB. If C and D are two points on OA and OB, respectively, such that OC = OD and the area of triangle OCD is half that of R, then the length of OC, in cm, is

- A.
(π/6)

^{1/2} - B.
(π/3√3)

^{1/2} - C.
(π/4√3)

^{1/2} - D.
(π/4)

^{1/2}

Answer: Option B

**Explanation** :

Let OC = OD = x

△OCD is an equilateral triangle of side x.

A(△OCD) = $\frac{\sqrt{3}}{4}{x}^{2}$

A(sector O-AB) = $\frac{\mathrm{\pi}{\left(1\right)}^{2}}{6}$ = $\frac{\mathrm{\pi}}{6}$

By the given condition,

$\frac{\sqrt{3}}{4}{x}^{2}$ = $\frac{1}{2}\times \frac{\mathrm{\pi}}{6}$

∴ x^{2} = $\frac{\mathrm{\pi}}{3\sqrt{3}}$

∴ x = ${\left(\frac{\mathrm{\pi}}{3\sqrt{3}}\right)}^{1/2}$

Hence, option 2.

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**18. CAT 2018 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

In a parallelogram ABCD of area 72 sq cm, the sides CD and AD have lengths 9 cm and 16 cm, respectively. Let P be a point on CD such that AP is perpendicular to CD. Then the area, in sq cm, of triangle APD is

- A.
12√3

- B.
24√3

- C.
18√3

- D.
32√3

Answer: Option D

**Explanation** :

Area of parallelogram = base × height

∴ A(□ABCD) = CD × AP

∴ 72 = 9 × AP

∴ AP = 8

Consider ∆APD.

PD = $\sqrt{{16}^{2}+{8}^{2}}$ = 8√3

A(APD) = 1/2 × 8√3 × 8 = 32√3 sq.cm.

Hence, option 4.

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**19. CAT 2018 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?

- A.
24, 12

- B.
24, 10

- C.
25, 10

- D.
25, 9

Answer: Option B

**Explanation** :

As □ABCD is a rectangle, AC = DB = 2 × 13 = 26 cm

AB^{2} + BC^{2} = 26^{2} = 676

(24^{2} + 10^{2}) = 676

Hence, option 2.

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**20. CAT 2018 QA Slot 1 | Geometry - Mensuration**

A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With π = 22/7, the volume, in cubic ft, of the remaining part of the cone is

Answer: 198

**Explanation** :

AB = 3 ft, AC = 12 ft and CD = 4 ft ⇒ BE = 1 ft

The required volumn of the cone = $\frac{1}{3}\mathrm{\pi}\times {4}^{2}\times 12$ - $\frac{1}{3}\mathrm{\pi}\times {1}^{2}\times 3$ = $\frac{1}{3}\mathrm{\pi}\times (192-3)$ = $\frac{1}{3}\times \frac{22}{7}\times 189$ = 198 cubic ft.

Hence, 198.

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**21. CAT 2018 QA Slot 1 | Algebra - Surds & Indices**

Given that x^{2018}y^{2017} = 1/2 and x^{2016}y^{2019} = 8, the value of x^{2} + y^{3} is

- A.
37/4

- B.
33/4

- C.
35/4

- D.
31/4

Answer: Option B

**Explanation** :

x^{2018}y^{2017} = 1/2 ...(1)

x^{2016}y^{2019} = 8 ...(2)

Dividing the 2 given equations, we get

$\frac{{x}^{2018}{y}^{2017}}{{x}^{2016}{y}^{2019}}$ = $\frac{{\displaystyle \frac{1}{2}}}{4}=\frac{1}{16}$

x^{2} = y^{2}/16

From Eq (1)

x^{2(1009)}y^{2017} = 1/2

∴ ${\left(\frac{{y}^{2}}{16}\right)}^{1009}{y}^{2017}$ = 1/2

∴ y^{4035} = 1/2 × 16^{1009}

∴ y^{4035} = 1/2 × 2^{4 × }^{1009} = 2^{4035}

∴ y = 2

Again, from equation 1

x^{2018}y^{2017} = 1/2

∴ x^{2018}2^{2017} = 1/2

∴ x^{2018} = 1/2^{2018}

∴ x = 1/2

Now, x^{2} + y^{3} = 1/4 + 8 = 33/4

Hence, option 2.

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**22. CAT 2018 QA Slot 1 | Algebra - Logarithms**

If x is a positive quantity such that 2^{x} = ${3}^{{\mathrm{log}}_{5}\left(2\right)}$ , then x is equal to

- A.
1 + ${\mathrm{log}}_{3}\left(\frac{5}{3}\right)$

- B.
log

_{5}9 - C.
log

_{5}8 - D.
1 + ${\mathrm{log}}_{5}\left(\frac{3}{5}\right)$

Answer: Option D

**Explanation** :

2^{x} = ${3}^{{\mathrm{log}}_{5}\left(2\right)}$

Taking log on both the sides,

x log2 = (log_{5}2)(log3)

∴ x = $\frac{\left({\mathrm{log}}_{5}2\right)\left(\mathrm{log}3\right)}{\left(\mathrm{log}2\right)}$ = $\frac{\left(\mathrm{log}2\right)\left(\mathrm{log}3\right)}{\left(\mathrm{log}5\right)\left(\mathrm{log}2\right)}$ = $\frac{\left(\mathrm{log}3\right)}{\left(\mathrm{log}5\right)}$ = log_{5}3

[1]: 1 + ${\mathrm{log}}_{3}\left(\frac{5}{3}\right)$ = 1 + log_{3}5 - log_{3}3 = log_{3}5

[2]: log_{5}9 = 2(log_{5}3)

[3]: log_{5}8 = 3(log_{5}2)

[4]: 1 + ${\mathrm{log}}_{5}\left(\frac{3}{5}\right)$ = 1 + log_{5}3 - log_{5}5 = log_{5}3

Hence, option 4.

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**23. CAT 2018 QA Slot 1 | Algebra - Logarithms**

If log_{12}81 = p, then $3\left(\frac{4-p}{4+p}\right)$ is equal to

- A.
log

_{2}8 - B.
log

_{6}16 - C.
log

_{6}8 - D.
log

_{4}16

Answer: Option C

**Explanation** :

log_{12}81 = p

⇒ p = 4log_{12}3

Now, $\frac{4-p}{4+p}=\frac{4-4{\mathrm{log}}_{12}3}{4+4{\mathrm{log}}_{12}3}$

= $\frac{1-{\mathrm{log}}_{12}3}{1+{\mathrm{log}}_{12}3}$

= $\frac{{\mathrm{log}}_{12}12-{\mathrm{log}}_{12}3}{{\mathrm{log}}_{12}12+{\mathrm{log}}_{12}3}=\frac{{\mathrm{log}}_{12}12/3}{{\mathrm{log}}_{12}12\times 3}=\frac{{\mathrm{log}}_{12}4}{{\mathrm{log}}_{12}36}$

= $\frac{2{\mathrm{log}}_{12}2}{2{\mathrm{log}}_{12}6}$ = ${\mathrm{log}}_{6}2$

Hence, $3\left(\frac{4-p}{4+p}\right)=3{\mathrm{log}}_{6}2={\mathrm{log}}_{6}8$

Alternately,

log_{12}81 = p ⇒ 81 = 12^{p}

⇒ 3^{(4 − p)} = 2^{2p}

Taking log on both the sides,

(4 – p) (log 3) = (2p) (log 2)

∴ $\frac{\mathrm{log}3}{\mathrm{log}2}$ = $\frac{2p}{4-p}$

∴ $\frac{\mathrm{log}3+\mathrm{log}2}{\mathrm{log}2}$ = $\frac{2p+(4-p)}{4-p}$ …(by adding 1 both sides)

∴ $\frac{\mathrm{log}6}{\mathrm{log}2}$ = $\frac{4+p}{4-p}$

∴ $\frac{4-p}{4+p}$ = $\frac{\mathrm{log}2}{\mathrm{log}6}$ = log_{6}2

∴ $3\left(\frac{4-p}{4+p}\right)$ = 3log_{6}2 = log_{6}8

Hence, option 3.

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**24. CAT 2018 QA Slot 1 | Algebra - Logarithms**

If log_{2}(5 + log_{3} a) = 3 and log_{5}(4a + 12 + log_{2} b) = 3, then a + b is equal to

- A.
40

- B.
67

- C.
59

- D.
32

Answer: Option C

**Explanation** :

log_{2}(5 + log_{3}a) = 3

∴ (5 + log_{3}a) = 2^{3} = 8

∴ log_{3}a = 3

∴ a = 3^{3} = 27

4a + 12 + log_{2}b = (4 × 27) + 12 + log_{2}b

= 120 + log_{2}b

log_{5}(4a + 12 + log_{2}b) = log_{5}(120 + log_{2}b) = 3

∴ (120 + log_{2}b) = 5^{3} = 125

∴ log_{2}b = 5

∴ b = 2^{5} = 32

Therefore, a + b = 27 + 32 = 59

Hence, option 3.

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**25. CAT 2018 QA Slot 1 | Algebra - Quadratic Equations**

If u^{2} + (u−2v−1)^{2} = −4v(u + v), then what is the value of u + 3v?

- A.
1/2

- B.
-1/4

- C.
0

- D.
1/4

Answer: Option B

**Explanation** :

u^{2} + (u – 2v – 1)^{2} = −4v(u + v)

u^{2} + (u – 2v – 1)^{2} + 4v(u + v) = 0

(u – 2v – 1)^{2} + u^{2} + 4uv + 4v^{2} = 0

(u – 2v – 1)^{2} + (u + 2v)^{2} = 0

(u – 2v – 1)^{2} & (u + 2v)^{2 }are both non-negative terms. Sum of two negative terms will be zero only when both the terms are equal to zero.

∴ u – 2v – 1 = 0 and u + 2v = 0

Solving the two equations, we get

u = 1/2 and v = -1/4

u + 3v = 1/2 - 3/4 = -1/4

Hence, option 2.

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**26. CAT 2018 QA Slot 1 | Algebra - Functions & Graphs**

If f(x + 2) = f(x) + f(x + 1) for all positive integers x, and f(11) = 91, f(15) = 617, then f(10) equals

Answer: 54

**Explanation** :

f(15) = f(14) + f(13)

= {f(13) + f(12)} + f(13)

= 2 × f(13) + f(12)

= 2{f(12) + f(11)} + f(12)

= 3 × f(12) + 2 × f(11)

= 3{f(11) + f(10)} + 2 × f(11)

= 5 × f(11) + 2 × f(10)

∴ 617 = 5 × 91 + 3 × f(10)

Solving this we get, f(10) = 54

Answer: 54

Workspace:

**27. CAT 2018 QA Slot 1 | Algebra - Number Theory**

The number of integers x such that 0.25 < 2^{x} < 200, and 2x + 2 is perfectly divisible by either 3 or 4, is

Answer: 5

**Explanation** :

For x = (−2), 2^{x} = 0.25 and x = 7 is the largest value that x can take for 2^{x} ≤ 200

For x = (−2) and (−1), (2^{x} + 2) is not an integer.

For x = 0, (2^{x} + 2) = 3 i.e., divisible by 3.

For x = 1, (2^{x} + 2) = 4 i.e., divisible by 4.

For x = 2, (2^{x} + 2) = 6 i.e., divisible by 3.

For x = 3, (2^{x} + 2) = 10 i.e., neither divisible by 3 nor by 4.

For x = 4, (2^{x} + 2) = 18 i.e., divisible by 3.

For x = 5, (2^{x} + 2) = 34 i.e., neither divisible by 3 nor by 4.

For x = 6, (2^{x} + 2) = 66 i.e., divisible by 3.

For x = 7, (2^{x} + 2) = 130 i.e., neither divisible by 3 nor by 4.

Thus, for x = 0, 1, 2, 4 and 6, 2^{x} + 2 is perfectly divisible by either 3 or 4.

Hence, 5.

Workspace:

**28. CAT 2018 QA Slot 1 | Algebra - Number Theory | Algebra - Progressions**

While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is

Answer: 40

**Explanation** :

Let X and Y are the two real numbers.

X × Y × 73 − X × Y × 37 = 720

X × Y × (73 – 37) = 720

X × Y × 36 = 720

X × Y = 20

AM(X^{2}, Y^{2}) ≥ GM(X^{2}, Y^{2})

∴ $\frac{{X}^{2}+{Y}^{2}}{2}$ ≥ $\sqrt{{X}^{2}\times {Y}^{2}}$

∴ (X^{2} + Y^{2}) ≥ 40

Answer : 40

Alternately,

We know, X × Y = 40

Also, (X - Y)^{2} ≥ 0 [Square of any number is always greater than or equal to zero]

∴ X^{2} + Y^{2} - 2XY ≥ 0

∴ X^{2} + Y^{2} ≥ 2XY

∴ X^{2} + Y^{2} ≥ 40

Workspace:

**29. CAT 2018 QA Slot 1 | Modern Math - Permutation & Combination**

How many numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, 9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?

Answer: 502

**Explanation** :

As the digits appear in ascending order, there is only one way in which selected group of integers can be arranged.

Thus for two digit number, we need to select two digits.

This can be done in ^{9}C_{2} = 36 ways

As ^{n}C_{m} =^{9}C_{(n−m)} ; ^{9}C_{2} = ^{9}C_{7} = Number of ways in which number can have 2 or 7 digits = ^{9}C_{2} = ^{9}C_{7} = 36 ways

Number of ways in which number can have 3 or 6 digits = ^{9}C_{3} = 84 ways

Number of ways in which number can have 4 or 5 digits = ^{9}C_{4} = 126 ways

Number of ways in which number can have 8 digits = ^{9}C_{8} = 8 ways

Number of ways in which number can have 9 digits = 1 ways

1 + 8 + 2(36) + 2(84) + 2(126) = 502 numbers have 2 or more distinct digits such that the digits appear in ascending order.

Answer: 502

Workspace:

**30. CAT 2018 QA Slot 1 | Algebra - Functions & Graphs**

Let f(x) = min{2x^{2},52 − 5x}, where x is any positive real number. Then the maximum possible value of f(x) is

Answer: 32

**Explanation** :

For positive value of x, (2x^{2}) is increasing.

As value of x increases, value of (52 – 5x) decreases.

Since, these 2 functions are continuously increasing and decreasing respectively, the max value of f(x) will occur when both these functions are equal.

i.e. 2x^{2} = 52 − 5x, for x = 4 and −6.5

At x = 4, function attains maximum value for positive real value of x.

∴ at x = 4, f(x) = 32

Answer: 32

Workspace:

**31. CAT 2018 QA Slot 1 | Venn Diagram**

Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is

Answer: 52

**Explanation** :

Let ‘a’ and ‘c’ denote the number of students studying only H and only E respectively while ‘b’ denote the number of students studying H and P (but not E) and ‘d’ denote the number of students studying E and P (but not H).

a + b + c + d + 30 = 74 ⇒ a + b + c + d = 44

We know that a + b = c + d

∴ a + b = c + d = 22

∴ the number of students studying in

H = 22 + 30 = 52

Answer: 52

Workspace:

**32. CAT 2018 QA Slot 1 | Geometry - Triangles**

Given an equilateral triangle T_{1} with side 24 cm, a second triangle T_{2} is formed by joining the midpoints of the sides of T_{1}. Then a third triangle T_{3} is formed by joining the midpoints of the sides of T_{2}. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T_{1}, T_{2}, T_{3},... will be

- A.
248√3

- B.
164√3

- C.
188√3

- D.
192√3

Answer: Option D

**Explanation** :

Consider ∆ABC (i.e. T_{1}) and ∆DEF (i.e., T_{2}).

D and E are midpoints of AB and AC respectively. Therefore, BC = 2 × DE

Side of T_{2} = 1/2 × Side of T_{1}

Area of T_{1} = A(T_{1}) = √3/4 × 24^{2}

Similarly, A(T_{2}) = √3/4 × 12^{2}

A(T_{3}) = √3/4 × 6^{2} ... and so on

Sum of areas of infinitely many T_{i}’s

= √3/4 × 24^{2 }+ √3/4 × 12^{2 }+ √3/4 × 6^{2} + ...

= √3/4 (24^{2 }+ 12^{2 }+ 6^{2} + ...)

Here, (24^{2 }+ 12^{2 }+ 6^{2} + ...) is an infinite series with r = 1/4

Hence, = √3/4 (24^{2 }+ 12^{2 }+ 6^{2} + ...) = $\frac{\sqrt{3}}{4}\left(\frac{{24}^{2}}{\left(1-{\displaystyle \frac{1}{4}}\right)}\right)$ = 192√3

Hence, option 4.

Workspace:

**33. CAT 2018 QA Slot 1 | Venn Diagram**

If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be

- A.
26

- B.
96

- C.
93

- D.
23

Answer: Option C

**Explanation** :

The minimum number of students who like only burger = 134 – 105 = 29

We can eliminate options [1] and [4].

The maximum number of students who like only burger = 200 – 105 = 95

We can eliminate option [2].

Thus, it can be concluded that the number of students who like only burger can possibly be 93.

Hence, option 3.

Workspace:

**34. CAT 2018 QA Slot 1 | Algebra - Progressions**

Let x, y, z be three positive real numbers in a geometric progression such that x < y < z. If 5x, 16y, and 12z are in an arithmetic progression then the common ratio of the geometric progression is

- A.
1/6

- B.
5/2

- C.
3/6

- D.
3/2

Answer: Option B

**Explanation** :

Let the common ratio be ‘k’.

⇒ x = y/k and z = yk

As 5x = 5y/k, 16y and 12z = 12yk are in arithmetic progression,

2 × 16y = 5y/k + 12yk

∴ 32 = 5/k + 12k

∴ 12k^{2} – 32k + 5 = 0

∴ (2k – 5)(6k – 1) = 0

∴ k = 5/2 or 1/6

As x < y < z, k has to be greater than 1.

∴ k = 5/2

Hence, option 2.

Workspace:

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