CAT 2018 QA Slot 1 questions and solutions

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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?

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Answer: 15

Video Explanation

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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 = $\frac{d - s}{s + \frac{3}{4} s}$ = $\frac{4 (d - s)}{7 s}$

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

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

Solving this, we get d = 15s

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

Hence, 15.

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

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Answer: 12

Video Explanation

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Explanation :

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

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

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

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

Solving this, we get

t = 12 minutes.

Hence, 12.

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

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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?

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Answer: 10

Video Explanation

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Explanation :

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

Work done by 6 filling and 5 emptying pipes in 6 hours = 6(6a - 5b)
Work done by 5 filling and 6 emptying pipes in 60 hours = 60(5a - 6b)

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.
Work done by 2 filling and 1 emptying pipes in m hours = m(2a - b)

∴ m(2a – b) = 6(6a – 5b)

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

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

Solving this, we get m = 10

Hence, 10.

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

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Answer: 60

Video Explanation

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Explanation :

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

By the given conditions,

When first 10 tests are not considered:
⇒ $\frac{n x - 200}{n - 10}$ = x + 1

⇒ x = $\frac{n x - 200}{n - 10}$ - 1 ...(1)

When last 10 tests are not considered:
⇒ $\frac{n x - 300}{n - 10}$ = x - 1

⇒ x = $\frac{n x - 300}{n - 10}$ + 1 ...(2)

Equating (1) and (2)
∴ $\frac{n x - 200}{n - 10}$ - 1 = $\frac{n x - 300}{n - 10}$ + 1

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

∴ $\frac{100}{n - 10}$ = 2

∴ n = 60

Hence, 60.

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

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

Video Explanation

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Explanation :

Passing marks = 45% of N

As 36 marks are short of the pass marks by 68%, it means the cadidate got (100 - 68 = ) 32% of the passing marks.

⇒ 36 = 32% of 45% of N

∴ 36 = 0.32 × 0.45 × N

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

Hence, option (c).

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

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

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

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

Video Explanation

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Explanation :

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

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

⇒ 6(4x + y) = 5(9x - y)
⇒ 24x + 6y = 45x - 5y
⇒ 11y = 21x
∴ y = 21x/11

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

Hence, option (c).

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

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

Video Explanation

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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 (a).

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

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

Video Explanation

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Explanation :

y > x

Side of smaller square = $\sqrt{x^{2} + y^{2}}$
∴ Area of smaller square = (x² + y²)

Side of bigger square = (x + y)
∴ Area of bigger square = (x + y)²

By the given condition,

⇒ x² + y² = 62.5% of (x + y)²

⇒ x² + y² = 5/8 (x + y)²

⇒ 8x² + 8y² = 5x² + 5y² + 10xy

⇒ 3x² + 3y² - 10xy = 0

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

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

As y > x, x/y < 1

∴ x : y = 1 : 3

Hence, option (d).

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

Video Explanation

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Explanation :

OP⊥ PQ and OQ⊥ AB

In a circle perpendicular to the chord bisects the chord.
∴ CP = PD = 3 and AQ = QB = 2

Let OP = x cm

OC = OB

In ∆OPC and ∆OQB
∴ Radius = $\sqrt{P C^{2} + P O^{2}}$ = $\sqrt{Q B^{2} + Q O^{2}}$

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

∴ x = 2

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

Hence, option (b).

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

Video Explanation

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Explanation :

Let OC = OD = x

△OCD is an equilateral triangle of side x. [All angles are 60 degree.]

A(△OCD) = $\frac{\sqrt{3}}{4} x^{2}$

A(sector O-AB) = $\frac{60}{360} \times π {(1)}^{2}$ = $\frac{π}{6}$

By the given condition,

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

∴ x² = $\frac{π}{3 \sqrt{3}}$

∴ x = ${(\frac{π}{3 \sqrt{3}})}^{1 / 2}$

Hence, option (b).

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

Video Explanation

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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 (d).

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

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

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Answer: 198

Video Explanation

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Explanation :

∆ABE ~ ∆ACD

⇒ $\frac{AB}{AC}$ = $\frac{BE}{CD}$

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

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

Hence, 198.

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

Given that x²⁰¹⁸y²⁰¹⁷ = 1/2 and x²⁰¹⁶y²⁰¹⁹ = 8, the value of x² + y³ is

A.
37/4
B.
33/4
C.
35/4
D.
31/4

Answer: Option B

Video Explanation

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Explanation :

x²⁰¹⁸y²⁰¹⁷ = 1/2 ...(1)

x²⁰¹⁶y²⁰¹⁹ = 8 ...(2)

Dividing the 2 given equations, we get

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

x² = y²/16

From Eq (1)

x^2(1009)y²⁰¹⁷ = 1/2

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

∴ y⁴⁰³⁵ = 1/2 × 16¹⁰⁰⁹

∴ y⁴⁰³⁵ = 1/2 × 2^{4 ×}¹⁰⁰⁹ = 2⁴⁰³⁵

∴ y = 2

Again, from equation 1

x²⁰¹⁸y²⁰¹⁷ = 1/2

∴ x²⁰¹⁸2²⁰¹⁷ = 1/2

∴ x²⁰¹⁸ = 1/2²⁰¹⁸

∴ x = ± 1/2

Now, x² + y³ = 1/4 + 8 = 33/4

Hence, option (b).

22. CAT 2018 QA Slot 1 | Algebra - Logarithms

If x is a positive quantity such that 2^x = $3^{\log_{5} (2)}$ , then x is equal to

A.
1 + $\log_{3} (\frac{5}{3})$
B.
log₅9
C.
log₅8
D.
1 + $\log_{5} (\frac{3}{5})$

Answer: Option D

Video Explanation

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Explanation :

Given, 2^x = $3^{\log_{5} (2)}$

⇒ 2^x = $2^{\log_{5} (3)}$

⇒ x = log₅3

Option (a): 1 + $\log_{3} (\frac{5}{3})$ = 1 + log₃5 - log₃3 = log₃5

Option (b): log₅9 = 2(log₅3)

Option (c): log₅8 = 3(log₅2)

Option (d): 1 + $\log_{5} (\frac{3}{5})$ = 1 + log₅3 - log₅5 = log₅3

Hence, option (d).

23. CAT 2018 QA Slot 1 | Algebra - Logarithms

If log₁₂81 = p, then $3 (\frac{4 - p}{4 + p})$ is equal to

A.
log₂8
B.
log₆16
C.
log₆8
D.
log₄16

Answer: Option C

Video Explanation

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Explanation :

log₁₂81 = p

⇒ p = 4log₁₂3

Now, $\frac{4 - p}{4 + p} = \frac{4 - 4 \log_{12} 3}{4 + 4 \log_{12} 3}$

= $\frac{1 - \log_{12} 3}{1 + \log_{12} 3}$

= $\frac{\log_{12} 12 - \log_{12} 3}{\log_{12} 12 + \log_{12} 3} = \frac{\log_{12} 12 / 3}{\log_{12} 12 \times 3} = \frac{\log_{12} 4}{\log_{12} 36}$

= $\frac{2 \log_{12} 2}{2 \log_{12} 6}$ = $\log_{6} 2$

Hence, $3 (\frac{4 - p}{4 + p}) = 3 \log_{6} 2 = \log_{6} 8$

Alternately,
log₁₂81 = p
⇒ 81 = 12^p
⇒ 3⁴ = 3^p × 2^2p
⇒ 3^{(4 − p)} = 2^2p

Taking log on both the sides,

(4 – p) (log 3) = (2p) (log 2)

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

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

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

∴ $\frac{4 - p}{4 + p}$ = $\frac{\log 2}{\log 6}$ = log₆2

∴ $3 (\frac{4 - p}{4 + p})$ = 3log₆2 = log₆8

Hence, option (c).

24. CAT 2018 QA Slot 1 | Algebra - Logarithms

If log₂(5 + log₃ a) = 3 and log₅(4a + 12 + log₂ b) = 3, then a + b is equal to

A.
40
B.
67
C.
59
D.
32

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

If u² + (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

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

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

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

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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

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Answer: 40

Video Explanation

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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(X², Y²) ≥ GM(X², Y²)

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

∴ (X² + Y²) ≥ 40

Alternately,
We know, X × Y = 40

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

∴ X² + Y² - 2XY ≥ 0

∴ X² + Y² ≥ 2XY

∴ X² + Y² ≥ 40

Hence, 40.

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?

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

Let f(x) = min {2x², 52 − 5x}, where x is any positive real number. Then the maximum possible value of f(x) is

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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

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32. CAT 2018 QA Slot 1 | Geometry - Triangles

Given an equilateral triangle T₁ with side 24 cm, a second triangle T₂ is formed by joining the midpoints of the sides of T₁. Then a third triangle T₃ is formed by joining the midpoints of the sides of T₂. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T₁, T₂, T₃,... will be

A.
248√3
B.
164√3
C.
188√3
D.
192√3

Answer: Option D

Video Explanation

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Explanation :

Consider ∆ABC (i.e. T₁) and ∆DEF (i.e., T₂).

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

Side of T₂ = 1/2 × Side of T₁

Area of T₁ = A(T₁) = √3/4 × 24²

Similarly, A(T₂) = √3/4 × 12²

A(T₃) = √3/4 × 6² ... and so on

Sum of areas of infinitely many T_i’s

= √3/4 × 24²+ √3/4 × 12²+ √3/4 × 6² + ...

= √3/4 (24²+ 12²+ 6² + ...)

Here, (24²+ 12²+ 6² + ...) is an infinite series with r = 1/4

Hence, = √3/4 (24²+ 12²+ 6² + ...) = $\frac{\sqrt{3}}{4} (\frac{24^{2}}{(1 - \frac{1}{4})})$ = 192√3

Hence, option (d).

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

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

CAT 2018 QA Slot 1 | Previous Year CAT Paper

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