CAT 2020 QA Slot 2 questions and solutions

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1. CAT 2020 QA Slot 2 | Algebra - Progressions

Let the m-th and n-th terms of a geometric progression be 3/4 and 12, respectively, where m < n. If the common ratio of the progression is an integer r, then the smallest possible value of r + n - m is

A.
6
B.
2
C.
-4
D.
-2

Answer: Option D

Video Explanation

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Explanation :

If the first term is a and common ratio is r (an integer)

T_m = ¾ = ar^m-1 and

T_n = 12 = ar^n-1

Now, $\frac{T_{n}}{T_{m}} = \frac{12}{3 / 4} = \frac{a r^{n - 1}}{a r^{m - 1}} = r^{n - m}$

∴ r^(n-m) = 16

Since r is an integer, r can be either ± 2, ± 4 or 16.

If r = ± 2, n – m = 4 ⇒ least value of r + n – m = -2 + 4 = 2

If r = ± 4, n – m = 2 ⇒ least value of r + n – m = -4 + 2 = - 2

If r = 16, n – m = 1 ⇒ least value of r + n – m = 16 + 1 = 17

∴ Least possible value of r + n – m = - 2

Hence, option (d).

2. CAT 2020 QA Slot 2 | Algebra - Number Theory

For real x, the maximum possible value of $\frac{x}{\sqrt{1 + x^{4}}}$ is

A.
1/2
B.
1
C.
1/√2
D.
1/√3

Answer: Option C

Video Explanation

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Explanation :

$\frac{x}{\sqrt{1 + x^{4}}}$ = $\frac{1}{\sqrt{\frac{1 + x^{4}}{x^{2}}}}$ = $\frac{1}{\sqrt{\frac{1}{x^{2}} + x^{2}}}$

We know x2 is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e.,

$\frac{1}{x^{2}} + x^{2}$ ≥ 2.

⇒ $\frac{1}{\sqrt{\frac{1}{x^{2}} + x^{2}}}$ will be maximum when 1 $\frac{1}{x^{2}} + x^{2}$ is minimum i.e., 2.

∴ Maximum value of $\frac{1}{\sqrt{\frac{1}{x^{2}} + x^{2}}} = \frac{1}{\sqrt{2}}$

Hence, option (c).

3. CAT 2020 QA Slot 2 | Modern Math - Permutation & Combination

How many 4-digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3?

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4. CAT 2020 QA Slot 2 | Arithmetic - Time & Work

John takes twice as much time as Jack to finish a job. Jack and Jim together take one-thirds of the time to finish the job than John takes working alone. Moreover, in order to finish the job, John takes three days more than that taken by three of them working together. In how many days will Jim finish the job working alone?

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5. CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance

In a car race, car A beats car B by 45 km, car B beats car C by 50 km, and car A beats car C by 90 km. The distance (in km) over which the race has been conducted is

A.
550
B.
450
C.
475
D.
500

Answer: Option B

Video Explanation

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Explanation :

Let their speeds be a, b and c respectively and the length of the race be D.

Car A beats car B by 45 kms

∴ $\frac{a}{b} = \frac{D}{D - 45}$ …(1)

Car B beats car C by 50 kms

∴ $\frac{b}{c} = \frac{D}{D - 50}$ …(2)

Car A beats car C by 90 kms

∴ $\frac{a}{c} = \frac{D}{D - 90}$ …(3)

(1) × (2) = (3)

⇒ $\frac{D}{D - 45} \times \frac{D}{D - 50} = \frac{D}{D - 90}$

⇒ $\frac{D}{(D - 45) (D - 50)} = \frac{1}{D - 90}$

⇒ D² – 95D + 2250 = D² – 90D

⇒ 5D = 2250

⇒ D = 450

Hence, option (b).

Concept:

Linear Race

6. CAT 2020 QA Slot 2 | Algebra - Number Theory

If x and y are non-negative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals

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7. CAT 2020 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation

A sum of money is split among Amal, Sunil and Mita so that the ratio of the shares of Amal and Sunil is 3 : 2, while the ratio of the shares of Sunil and Mita is 4 : 5. If the difference between the largest and the smallest of these three shares is Rs 400, then Sunil’s share, in rupees, is

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8. CAT 2020 QA Slot 2 | Geometry - Circles

Let C₁ and C₂ be concentric circles such that the diameter of C₁ is 2 cm longer than that of C₂. If a chord of C₁ has length 6 cm and is a tangent to C₂, then the diameter, in cm, of C₁ is

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9. CAT 2020 QA Slot 2 | Geometry - Circles

Let C be a circle of radius 5 meters having center at O. Let PQ be a chord of C that passes through points A and B where A is located 4 meters north of O and B is located 3 meters east of O. Then, the length of PQ, in meters, is nearest to

A.
8.8
B.
6.6
C.
7.8
D.
7.2

Answer: Option A

Video Explanation

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Explanation :

The following figure can be drawn from the information given in the question.

In right triangle AOB, AB = 5 (Pythagorean triplet)

Also, OM is the altitude = $\frac{3 \times 4}{5}$ = 2.4

In right triangle OMP,

OP² = OM² + MP²

⇒ 5² + (2.4)²+ MP²

⇒ MP = $\sqrt{19.24}$

⇒ PQ = 2MP = 2 × $\sqrt{19.24}$ ≈ 2 × 4.4 = 8.8.

Hence, option (a).

10. CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance

A and B are two points on a straight line. Ram runs from A to B while Rahim runs from B to A. After crossing each other, Ram and Rahim reach their destinations in one minute and four minutes, respectively. If they start at the same time, then the ratio of Ram's speed to Rahim's speed is

A.
√2
B.
2√2
C.
1/2
D.
2

11. CAT 2020 QA Slot 2 | Algebra - Quadratic Equations | Algebra - Inequalities & Modulus

In how many ways can a pair of integers (x , a) be chosen such that x² − 2|x| + |a - 2| = 0?

A.
4
B.
5
C.
6
D.
7

12. CAT 2020 QA Slot 2 | Geometry - Quadrilaterals & Polygons

The sum of the perimeters of an equilateral triangle and a rectangle is 90 cm. The area, T, of the triangle and the area, R, of the rectangle, both in sq cm, satisfy the relationship R = T². If the sides of the rectangle are in the ratio 1 : 3, then the length, in cm, of the longer side of the rectangle, is

A.
27
B.
24
C.
18
D.
21

Answer: Option A

Video Explanation

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Explanation :

Let the side of equilateral triangle be a. Hence, perimeter = 3a.

Let the sides of the rectangle be x and 3x. Hence, perimeter = 8x

Given, 3a + 8x = 90 …(1)

Also given, R = T²

⇒ x × 3x = ${[\frac{\sqrt{3}}{4} a^{2}]}^{2}$

⇒ 16x² = a⁴

⇒ a² = 4x

Substituting x in (1)

⇒ 3a + 2a² = 90

⇒ 2a² + 3a – 90 = 0

⇒ a = -15/2 or 6 (-ve value will be rejected)

∴ x = $\frac{a^{2}}{4}$ = 9

∴ Longer side of the rectangle = 3x = 27.

Hence, option (a).

13. CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance

Two circular tracks T₁ and T₂ of radii 100 m and 20 m, respectively touch at a point A. Starting from A at the same time, Ram and Rahim are walking on track T₁ and track T₂ at speeds 15 km/hr and 5 km/hr respectively. The number of full rounds that Ram will make before he meets Rahim again for the first time is

A.
3
B.
5
C.
4
D.
2

Answer: Option A

Video Explanation

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Explanation :

Time taken by Ram to reach the starting point A = $\frac{2 \times π \times 0.1}{15} = \frac{π}{75}$

Time taken by Rahim to reach the starting point A = $\frac{2 \times π \times 0.02}{5} = \frac{π}{125}$

Both of them will meet only at A since they are running in different circles which meet at A.

Both of them will reach the starting point after LCM $[\frac{π}{75}, \frac{π}{125}]$ = $\frac{π}{25}$

∴ Number of rounds completed by Ram when both of them meet again = $\frac{\frac{π}{25}}{\frac{π}{75}}$ = 3

Hence, (a).

14. CAT 2020 QA Slot 2 | Algebra - Number Theory

If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of 2601 $(1 + \frac{1}{x}) (1 + \frac{1}{y})$ is

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Answer: 2704

Video Explanation

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Explanation :

Let’s try to calculate the minimum possible value of $(1 + \frac{1}{x}) (1 + \frac{1}{y})$

⇒ $(1 + \frac{1}{x}) (1 + \frac{1}{y})$ = $\frac{1 + x + y + x y}{x y}$ = $\frac{103}{x y}$ + 1

Minimum value of this expression will be when xy is maximum.

Now, when sum of two number is constant their product is maximum when both the number are equal to each other.

∴ x = y = 51

∴ Least possible value of 2601 $(1 + \frac{1}{x}) (1 + \frac{1}{y})$ = 2601 × $\frac{52}{51} \times \frac{52}{51}$ = 2704

Hence, 2704.

15. CAT 2020 QA Slot 2 | Arithmetic - Percentage

In May, John bought the same amount of rice and the same amount of wheat as he had bought in April, but spent ₹ 150 more due to price increase of rice and wheat by 20% and 12%, respectively. If John had spent ₹ 450 on rice in April, then how much did he spend on wheat in May?

A.
Rs. 580
B.
Rs. 560
C.
Rs. 590
D.
Rs. 570

16. CAT 2020 QA Slot 2 | Algebra - Logarithms

The value of⁡ $\log_{a} (\frac{a}{b}) + \log_{b} (\frac{b}{a})$ , for 1 < a ≤ b cannot be equal to

A.
0
B.
-1
C.
1
D.
-0.5

Answer: Option C

Video Explanation

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Explanation :

$\log_{a} (\frac{a}{b}) + \log_{b} (\frac{b}{a})$ = 1 - log_ab + 1 - log_ba = 2 – (log_ab + log_ba)

We know logab is positive and sum of a positive number and its reciprocal is always ≥ 2, i.e.,

∴ log_ab + log_ba ≥ 2

⇒ 2 – (log_ab + log_ba) ≤ 2 – 2

⇒ 2 – (log_ab + log_ba) ≤ 0

∴ 2 – (log_ab + log_ba) cannot take positive value i.e., it cannot be equal to 1 (option (b))

Hence, option (c).

17. CAT 2020 QA Slot 2 | Algebra - Simple Equations

Aron bought some pencils and sharpeners. Spending the same amount of money as Aron, Aditya bought twice as many pencils and 10 less sharpeners. If the cost of one sharpener is ₹ 2 more than the cost of a pencil, then the minimum possible number of pencils bought by Aron and Aditya together is

A.
27
B.
36
C.
33
D.
30

18. CAT 2020 QA Slot 2 | Arithmetic - Simple & Compound Interest

For the same principal amount, the compound interest for two years at 5% per annum exceeds the simple interest for three years at 3% per annum by Rs 1125. Then the principal amount in rupees is

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Answer: 90000

Video Explanation

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Explanation :

Let the Principal be Rs. P.

Simple Interest for 3 years = $\frac{3 \times P \times 3}{100} = \frac{9 P}{100}$ = 0.09P

Compound Interest for 2 years = P × 1.05² – P = 0.1025P

∴ 0.1025P – 0.09P = 1125

⇒ 0.0125P = 1125

⇒ P = 90,000

Hence, 90000.

19. CAT 2020 QA Slot 2 | Algebra - Quadratic Equations

The number of integers that satisfy the equality (x² - 5x + 7)^x+1 = 1 is

A.
5
B.
2
C.
4
D.
3

20. CAT 2020 QA Slot 2 | Venn Diagram

Students in a college have to choose at least two subjects from chemistry, mathematics and physics. The number of students choosing all three subjects is 18, choosing mathematics as one of their subjects is 23 and choosing physics as one of their subjects is 25. The smallest possible number of students who could choose chemistry as one of their subjects is

A.
21
B.
20
C.
19
D.
22

21. CAT 2020 QA Slot 2 | Arithmetic - Time, Speed & Distance

The distance from B to C is thrice that from A to B. Two trains travel from A to C via B. The speed of train 2 is double that of train 1 while traveling from A to B and their speeds are interchanged while traveling from B to C. The ratio of the time taken by train 1 to that taken by train 2 in travelling from A to C is

A.
4 : 1
B.
7 : 5
C.
5 : 7
D.
1 : 4

Answer: Option C

Video Explanation

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Explanation :

Let the distance from A to B = d and that from B to C = 3d.

Speed of train T₁ = s from A to B and 2s from B to C.

Total time taken by T₁ = $\frac{d}{s} + \frac{3 d}{2 s} = \frac{5 d}{2 s}$

Speed of train T₂ = 2s from A to B and s from B to C.

Total time taken by T₂ = $\frac{d}{2 s} + \frac{3 d}{s} = \frac{7 d}{2 s}$

∴ Ratio of time taken by T₁ and T₂ = $\frac{5 d}{2 s} : \frac{7 d}{2 s}$ = 5 : 7

Hence, option (c).

22. CAT 2020 QA Slot 2 | Algebra - Number Theory

The number of pairs of integer (x, y) satisfying x ≥ y ≥ - 20 and 2x + 5y = 99 is

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23. CAT 2020 QA Slot 2 | Geometry - Triangles

From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is

A.
S²/√3
B.
√3S²/2
C.
S²/2√3
D.
2S²/√3

Answer: Option A

Video Explanation

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Explanation :

The figure can be drawn as below.

Sum of areas of three smaller ∆s = Area of ∆ABC

⇒ $\frac{1}{2} \times h_{1} \times a$ + $\frac{1}{2} \times h_{2} \times a$ + $\frac{1}{2} \times h_{3} \times a$ = $\frac{\sqrt{3}}{4} a^{2}$

⇒ $\frac{1}{2} \times s \times a$ = $\frac{\sqrt{3}}{4} a^{2}$

⇒ a = $\frac{2 s}{\sqrt{3}}$

∴ Area of ∆ABC = $\frac{\sqrt{3}}{4} a^{2}$ = $\frac{\sqrt{3}}{4} \times \frac{4}{3} \times s^{2}$ = $\frac{s^{2}}{\sqrt{3}}$

Hence, option (a).

24. CAT 2020 QA Slot 2 | Arithmetic - Average

In a group of 10 students, the mean of the lowest 9 scores is 42 while the mean of the highest 9 scores is 47. For the entire group of 10 students, the maximum possible mean exceeds the minimum possible mean by

A.
4
B.
5
C.
3
D.
6

Answer: Option A

Video Explanation

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Explanation :

Let the highest and lowest score be h and l respectively and total score of remaining 8 students be x.

The mean of the lowest 9 scores is 42

⇒ x + l = 9 × 42 = 378 …(1)

The mean of the highest 9 scores is 47

⇒ x + h = 9 × 47 = 423 …(2)

(2) – (1)

⇒ h – l = 423 – 378 = 45

Case 1: Least possible average is when we minimize the highest marks. The least highest marks can be 47.

∴ Lowest score = 47 – 45 = 2

∴ Least average = $\frac{9 \times 47 + 2}{10}$ = 42.5

Case 2: Highest possible average is when we maximize the lowest marks. The highest lowest marks can be 42.

∴ Highest score = 42 + 45 = 87

∴ Highest average = $\frac{9 \times 42 + 87}{10}$ = 46.5

∴ The required difference = 46.5 – 42.5 = 4

Hence, option (a).

25. CAT 2020 QA Slot 2 | Arithmetic - Profit & Loss

Anil buys 12 toys and labels each with the same selling price. He sells 8 toys initially at 20% discount on the labeled price. Then he sells the remaining 4 toys at an additional 25% discount on the discounted price. Thus, he gets a total of Rs 2112, and makes a 10% profit. With no discounts, his percentage of profit would have been

A.
50
B.
54
C.
60
D.
55

26. CAT 2020 QA Slot 2 | Algebra - Functions & Graphs

Let f(x) = x² + ax + b and g(x) = f(x + 1) – f(x – 1). If f(x) ≥ 0 for all real x, and g(20) = 72, then the smallest possible value of b is

A.
16
B.
1
C.
4
D.
0

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