# CAT 2020 QA Slot 1 | Previous Year CAT Paper

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**1. CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to

- A.
75

- B.
82

- C.
87

- D.
78

Answer: Option B

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**Explanation** :

Let the speed and length of train be ‘s’ m/s and ‘t’ m respectively.

Speeds of two persons is 2 kmph and 4 kmph i.e., 5/9 m/s and 10/9 m/s

Train passes the first person in 90 seconds.

∴ $\frac{t}{s-\frac{5}{9}}$ = 90

⇒ t = 90s – 50 …(1)

Train passes the second person in 100 seconds.

∴ $\frac{t}{s-\frac{10}{9}}$ = 100

⇒ t = 100s – 1000/9 …(2)

(1) = (2)

⇒ 90s – 50 = 100s – 1000/9

⇒ s = 55/9 m/s and t = 500 m

∴ Time taken to cross an electric pole = $\frac{500}{55/9}$ = $\frac{4500}{55}$ = $\frac{900}{11}$ = 81.81 ≈ 82 seconds.

Hence, option (b).

**Concept**:

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**2. CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is

- A.
100

- B.
90

- C.
70

- D.
80

Answer: Option B

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**Explanation** :

Let the time taken for them to meet after starting from opposite ends be t mins.

Also, let speeds of Car 2 be b kmph.

Distance AM = 60 × t = b × 20 …(1)

Distance BM = b × t = 60 × 45 …(2)

(1) × (2)

⇒ t^{2} = 900

⇒ t = 30 minutes

From (2) we get,

60 × 30 = b × 20

⇒ b = 90 kmph

Hence, option (b).

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**3. CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Leaving home at the same time, Amal reaches office at 10:15 am if he travels at 8 km/hr, and at 9:40 am if he travels at 15 km/hr. Leaving home at 9:10 am, at what speed, in km/hr, must he travel so as to reach office exactly at 10 am?

- A.
13

- B.
14

- C.
11

- D.
12

Answer: Option D

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**Explanation** :

Let the distance between home and office be d kms.

∴ $\frac{d}{8}-\frac{d}{15}=\frac{35}{60}=\frac{7}{12}$

⇒ d = 10 kms

If he leaves home at 9:10 a.m. and reaches office at 10 a.m., i.e., time taken = 50 minutes = 5/6 hours

∴ Required speed = $\frac{10}{5/6}$ = 12 kmph.

Hence, option (d).

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**4. CAT 2020 QA Slot 1 | Arithmetic - Time, Speed & Distance**

A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to

- A.
58

- B.
61

- C.
67

- D.
50

Answer: Option C

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**Explanation** :

Since the train travelled at 1/3rd of its speed, it will take thrice the usual time.

Hence, if the normal time taken for train is t minutes, now it will take 3t minutes.

∴ 3t – t = 30

⇒ t = 15 minutes.

While coming back train travels at usual speed for 5 minutes and then takes 4 minutes break.

∴ If the train has to come back at scheduled time, it needs to cover the remaining distance in (15 – 5 – 4 =) 6 minutes. Train usually takes 10 minutes to travel this distance.

∴ The time reduced to 6/10 = 3/5th hence speed should become 5/3 times.

∴ % increase = $\frac{\frac{5}{3}-1}{1}$ × 100 = 66.66% ≈ 67%

Hence, option (c).

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**5. CAT 2020 QA Slot 1 | Arithmetic - Profit & Loss**

A person spent Rs. 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is

Answer: 20000

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**Explanation** :

Let the price of desktop be Rs. d, and laptop be Rs. (50,000 - d)

Total profit = 2% of 50,000 = Rs. 1,000

⇒ 1000 = 20% of d - 10% of (50000 - d)

⇒ 100000 = 20d - 500000 + 10d

⇒ 30d = 6,00,000

⇒ d = 20,000

Hence, 20000.

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**6. CAT 2020 QA Slot 1 | Arithmetic - Percentage**

In a group of people, 28% of the members are young while the rest are old. If 65% of the members are literates, and 25% of the literates are young, then the percentage of old people among the illiterates is nearest to

- A.
55

- B.
66

- C.
59

- D.
62

Answer: Option B

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**Explanation** :

According to the information given in the question, we can make the following table.

∴ % old illiterates among illiterates = 23.25/35 × 100 = 66.4%

Hence, option (b).

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**7. CAT 2020 QA Slot 1 | Arithmetic - Average**

The mean of all 4-digit even natural numbers of the form ‘aabb’, where a > 0, is

- A.
5050

- B.
4864

- C.
4466

- D.
5544

Answer: Option D

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**Explanation** :

‘aabb’ is an even number hence, b can be either 0 or 2 or 4 or 6 or 8.

∴ ‘aabb’ can be:

1100, 1122, 1144, 1166, 1188

2200, 2222, 2244, 2266, 2288

…

9900, 9999, 9944, 9966, 9988

Adding all these number

= 5 × (1100 + ... + 9900) + 9 × (22 + 44 + 66 + 88)

= (5 × 1100 × 45) + 9 × 22 × 10 = 5500 × 45 + 44 × 45 = 5544 × 45

∴ Average of all such numbers = (5544 × 45)/45 = 5544.

Hence, option (d).

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**8. CAT 2020 QA Slot 1 | Algebra - Simple Equations**

A gentleman decided to treat a few children in the following manner. He gives half of his total stock of toffees and one extra to the first child, and then the half of the remaining stock along with one extra to the second and continues giving away in this fashion. His total stock exhausts after he takes care of 5 children. How many toffees were there in his stock initially?

Answer: 62

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**Explanation** :

**Giving chocolates to 5 ^{th} child**:

Let the man be left with x chocolates after 4

^{th}child.

Now he gives half of x i.e., x/2 chocolates to 5

^{th}child. Hence, he will be left with x/2 chocolates.

The man again gives 1 chocolate to 5th child and is now left with no chocolates.

∴ x/2 – 1 = 0

⇒ x = 2

**Giving chocolates to 4 ^{th} child**:

Now, let the man be left with y chocolates after 3

^{rd}child.

Now he gives half of y i.e., y/2 chocolates to 4

^{th}child. Hence, he will be left with y/2 chocolates.

The man again gives 1 chocolate to 4

^{th}child and is now left with 2 chocolates.

∴ y/2 – 1 = 2

⇒ y = 6

**Giving chocolates to 3 ^{rd} child**:

Now, let the man be left with z chocolates after 2

^{nd}child.

∴ z/2 – 1 = 6

⇒ y = 14

**Giving chocolates to 2 ^{nd} child:**

Now, let the man be left with a chocolates after 1

^{st}child.

∴ a/2 – 1 = 14

⇒ y = 30

**Giving chocolates to 1 ^{st} child:**

Now, let the man have c chocolates in the beginning.

∴ c/2 – 1 = 30

⇒ y = 62

Hence, 62.

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**9. CAT 2020 QA Slot 1 | Arithmetic - Average**

Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option C

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**Explanation** :

Given,

$a+\frac{b+c}{2}=5$

⇒ 2a + b + c = 10 …(1)

Also, $b+\frac{a+c}{2}=7$

⇒ 2b + a + c = 14 …(2)

Solving (1) and (2) we get,

b – a = 4

⇒ b = a + 4

Substituting this in the (1)

⇒ 2a + a + 4 + c = 10

⇒ 3a + c = 6

Given all three as positive integers, only possible value for a is 1. (c cannot be 0)

So, when a = 1, c = 3 and b = 5

∴ a + b = 6.

Hence, option (c).

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**10. CAT 2020 QA Slot 1 | Algebra - Quadratic Equations**

How many distinct positive integer-valued solutions exist to the equation ${\left({x}^{2}-7x+11\right)}^{({x}^{2}-13x+42)}$ = 1?

- A.
2

- B.
8

- C.
4

- D.
6

Answer: Option D

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**Explanation** :

Given, ${\left({x}^{2}-7x+11\right)}^{({x}^{2}-13x+42)}$ = 1?

This is possible when either (x² - 7x + 11) = 1 or

(x² - 13x + 42) = 0 or

(x² - 7x + 11) = -1 and (x² - 13x + 42) = even number.

**Case 1** : x² - 7x + 11 = 1

⇒ x² - 7x + 10 = 0

⇒ (x - 5)(x - 2) = 0

⇒ x = 5 or 2.

**Case 2** : (x² - 13x + 42) = 0

⇒ (x - 6)(x - 7) = 0

⇒ x = 6 or 7.

**Case 3** : x² - 7x + 11 = -1 and (x² - 13x + 42) = even

⇒ x² - 7x + 12 = 0

⇒ (x - 3)(x - 4) = 0

⇒ x = 3 or 4.

Now, (x² - 13x + 42) is even for all values of x.

∴ We have a total of 6 possible values of x i.e. 5 or 2, 6 or 7 and 3 or 4.

Hence, option (d).

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**11. CAT 2020 QA Slot 1 | Algebra - Surds & Indices**

The number of real-valued solutions of the equation 2^{x} + 2^{-x} = 2 – (x – 2)² is

- A.
0

- B.
Infinite

- C.
2

- D.
1

Answer: Option A

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**Explanation** :

2^{x} + 2^{-x} is of the form y + 1/y

Minimum value the expression can take is 2.

Hence, 2^{x} + 2^{-x} ≥ 2

Now, 2 - (x - 2)^{2}

We know, (x - 2)^{2} ≥ 0

∴ 2 - (x - 2)^{2} ≤ 2 (From 2 we are subtracting a non-negative number)

Maximum value this expression can have is 2.

The only possibility is both sides are = 2

LHS = 2^{x} + 2^{-x} = 0

This is possible only when x = 0.

When x = 0, RHS = 2 - (x - 2)^{2} = 2 - 2^{2} = - 2

Hence, at x = 0, LHS ≠ RHS.

∴ There is no solution possible.

Hence, option (a).

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**12. CAT 2020 QA Slot 1 | Algebra - Number Theory**

How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Answer: 21

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**Explanation** :

Let the digits of the 3-digit number be p, q, & r.

∴ 2 < p × q × r < 7

⇒ p × q × r can take the values 3, 4, 5, or 6.

Let's start with prime numbers 3 & 5.

Since they are prime, they can't be split, and hence if one of p, q or r is 3, the remaining two should be 1.

So, the possible combinations are

If p × q × r = 3

The number can be {113, 131, 311}

If p × q × r = 5

The number can be {115, 151, 511}

If p × q × r = 4

4 can be written as 1 × 2 × 2 or 1 × 1 × 4.

Therefore, the possible combinations of p, q, r are {122, 212, 221, 114, 141, 411}

If p × q × r = 6

6 can be written as 1 × 3 × 2 or 1 × 1 × 6.

Therefore, the possible combinations of p, q, r are {123, 132, 213, 231, 312, 321, 116, 161, 611}

Therefore, the total number of possibilities are 3 + 3 + 6 + 9 = 21.

Hence, 21.

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**13. CAT 2020 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A solution, of volume 40 litres, has dye and water in the proportion 2 : 3. Water is added to the solution to change this proportion to 2 : 5. If one-fourths of this diluted solution is taken out, how many litres of dye must be added to the remaining solution to bring the proportion back to 2 : 3?

Answer: 8

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**Explanation** :

40 litres solution had dye and water in the ratio 2 : 3

∴ dye = 16 l and water = 24 l.

After adding x liters of water, now proportion is 2 : 5

⇒ $\frac{16}{24+x}=\frac{2}{5}$

⇒ x = 16 liters

Dye = 16 l and water = 40 l now in the solution.

Now, 1/4^{th} is removed from the solution, hence dye left = ¾ × 16 = 12 l and water left = ¾ × 40 = 30 l

Now, After adding y liters of dye, proportion becomes 2 : 3.

⇒ $\frac{12+y}{30}=\frac{2}{3}$

⇒ y = 8 liters

Hence, 8.

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**14. CAT 2020 QA Slot 1 | Geometry - Mensuration**

On a rectangular metal sheet of area 135 sq in, a circle is painted such that the circle touches two opposite sides. If the area of the sheet left unpainted is two-thirds of the painted area then the perimeter of the rectangle in inches is

- A.
$3\sqrt{\mathrm{\pi}}\left(\frac{5}{2}+\frac{6}{\mathrm{\pi}}\right)$

- B.
$4\sqrt{\mathrm{\pi}}\left(3+\frac{9}{\mathrm{\pi}}\right)$

- C.
$3\sqrt{\mathrm{\pi}}\left(5+\frac{12}{\mathrm{\pi}}\right)$

- D.
$5\sqrt{\mathrm{\pi}}\left(3+\frac{9}{\mathrm{\pi}}\right)$

Answer: Option C

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**Explanation** :

Area of the sheet left unpainted is two-thirds of painted area i.e., area of the circles will be three-fifth of the total area.

∴ Area of the circle = 3/5 × 135 = 54 sq. in.

Let the radius of the circle be r and l be the length of the rectangle. Width of the rectangle will be 2r.

⇒ πr^{2} = 81

⇒ r = $\frac{9}{\sqrt{\pi}}$

Area of rectangle = 135 = 2r × l

⇒ l = $\frac{135}{2r}$ = $\frac{135}{2\times \frac{9}{\sqrt{\pi}}}$ = $\frac{15\sqrt{\pi}}{2}$

∴ Perimeter of the rectangle = 2(l + r) = $2\left(\frac{15\sqrt{\pi}}{2}+\frac{18}{\sqrt{\pi}}\right)$ = $3\sqrt{\pi}\left(5+\frac{12}{\pi}\right)$

Hence, option (c).

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**15. CAT 2020 QA Slot 1 | Algebra - Functions & Graphs**

Among 100 students, x_{1} have birthdays in January, x_{2} have birthdays in February, and so on. If x_{0} = max(x_{1}, x_{2}, …., x_{12}), then the smallest possible value of x_{0} is

- A.
10

- B.
8

- C.
12

- D.
9

Answer: Option D

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**Explanation** :

Given, x_{0} = max(x_{1}, x_{2}, …., x_{12})

x_{0} will be minimum when we distribute students all 100 students as equally among the 12 months as possible.

This can be done in the following way: 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9 (Adding these 12 we will get 100)

∴ x_{0} = max(8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9).

⇒ Minimum value of x_{0} = 9

Hence, option (d).

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**16. CAT 2020 QA Slot 1 | Geometry - Mensuration**

A solid right circular cone of height 27 cm is cut into two pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in volume of the two pieces is 225 cc, the volume, in cc, of the original cone is

- A.
256

- B.
232

- C.
264

- D.
243

Answer: Option D

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**Explanation** :

When a cone of height ‘H’ is cut at a distance of ‘h’ from the top

⇒ $\frac{Volumeofsmallercone}{Volumeoforiginalcone}$ = ${\left(\frac{h}{H}\right)}^{3}$

∴ $\frac{Volumeofsmallercone}{Volumeoforiginalcone}$ = ${\left(\frac{9}{27}\right)}^{3}$ = $\frac{1}{27}$

⇒ If volume of the smaller cone = x, the volume of original cone is 27x, and volume of the frustum = 26x

⇒ 26x – x = 225

⇒ x = 9

∴ Volume of the original cone = 27x = 27 × 9 = 243.

Hence, option (d).

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**17. CAT 2020 QA Slot 1 | Algebra - Surds & Indices**

If x = (4096)^{7+4√3}, then which of the following equals 64?

- A.
$\frac{{\mathrm{x}}^{\frac{7}{2}}}{{\mathrm{x}}^{\frac{4}{\sqrt{3}}}}$

- B.
$\frac{{\mathrm{x}}^{7}}{{\mathrm{x}}^{2\sqrt{3}}}$

- C.
$\frac{{\mathrm{x}}^{\frac{7}{2}}}{{\mathrm{x}}^{2\sqrt{3}}\mathrm{}}$

- D.
$\frac{{\mathrm{x}}^{7}}{{\mathrm{x}}^{4\sqrt{3}}}$

Answer: Option C

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**Explanation** :

Given,

x = (4096)^{7+4√3}

⇒ x = (64^{2})^{7+4√3}

⇒ x = (64)^{14+8√3}

⇒ ${x}^{\frac{1}{14+8\sqrt{3}}}$ = 64

Now, let’s rationalize $\frac{1}{14+8\sqrt{3}}$,

⇒ $\frac{1}{14+8\sqrt{3}}$ = $\frac{1}{14+8\sqrt{3}}\times \frac{14-8\sqrt{3}}{14-8\sqrt{3}}$ = $\frac{14-8\sqrt{3}}{196-192}$ = $\frac{7-4\sqrt{3}}{2}$

⇒ ${x}^{\frac{1}{14+8\sqrt{3}}}$ = ${x}^{\frac{7-4\sqrt{3}}{2}}$= 64

⇒ ${x}^{\frac{7}{2}}\times {x}^{\frac{-4\sqrt{3}}{2}}$ = 64

⇒ 64 = $\frac{{x}^{\frac{7}{2}}}{{x}^{2\sqrt{3}}}$

Hence, option (c).

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**18. CAT 2020 QA Slot 1 | Algebra - Quadratic Equations**

The number of distinct real roots of the equation ${\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^{2}-3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+2$ = 0 equals

Answer: 1

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**Explanation** :

Let x + 1/x = y

⇒ y^{2} – 3y + 2 = 0

⇒ (y – 1)(y – 2) = 0

⇒ y = 1 or 2

We know, sum of a number and its reciprocal are either ≤ 2 or ≥ 2.

∴ y = 2

⇒ x + 1/x = 2

This is only possible when x = 1 hence, only one real root.

Hence, 1.

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**19. CAT 2020 QA Slot 1 | Algebra - Number Theory**

If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

- A.
49

- B.
56

- C.
46

- D.
59

Answer: Option C

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**Explanation** :

Given ab = 432, bc = 96 and c < 9.

To get the minimum value for a + b + c, the two bigger numbers should be as close as possible.

So possible values are

a = 36, b = 12, c = 8 ⇒ Sum = 46

a = 27, b = 16, c = 6 ⇒ Sum = 49

a = 18, b = 24, c = 4 ⇒ Sum = 46

a = 9, b = 48, c = 2 ⇒ Sum = 59

Least possible value = 46

Hence, option (c).

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**20. CAT 2020 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement**

An alloy is prepared by mixing three metals A, B and C in the proportion 3 : 4 : 7 by volume. Weights of the same volume of the metals A, B and C are in the ratio 5 : 2 : 6. In 130 kg of the alloy, the weight, in kg, of the metal C is

- A.
70

- B.
84

- C.
96

- D.
48

Answer: Option B

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**Explanation** :

Let 1 liter of A, B and C weight 5kg, 2kg and 6 kg respectively.

Now, let’s suppose 3, 4 and 7 liters of A, B and C are mixed.

∴ Total weight of 14 liters of this solution = 3 × 5 + 4 × 2 + 7 × 6 = 65 kgs.

⇒ 65 kgs of the solution contains 7 × 6 = 42 kgs of C

∴ 130 kgs of the solution contains 2 × 42 = 84 kgs of C

Hence, option (b).

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**21. CAT 2020 QA Slot 1 | Arithmetic - Simple & Compound Interest**

Veeru invested Rs. 10,000 at 5% simple annual interest, and exactly after two years, Joy invested Rs. 8,000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

Answer: 12

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**Explanation** :

Let the amounts become equal in t years.

Veeru’s investment after t years = 10000$\left(1+\frac{5\times t}{100}\right)$

Joy’s investment after t years = 8000$\left(1+\frac{10\times (t-2)}{100}\right)$

∴ 10000$\left(1+\frac{5\times t}{100}\right)$ = 8000$\left(1+\frac{10\times (t-2)}{100}\right)$

⇒ 5$\left(1+\frac{5\times t}{100}\right)$ = 4$\left(1+\frac{10\times (t-2)}{100}\right)$

⇒ 500 + 25t = 400 + 40t - 80

⇒ t = 12

Hence, 12.

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**22. CAT 2020 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of circle to the area of rhombus is

- A.
5π/18

- B.
2π/15

- C.
3π/25

- D.
6π/25

Answer: Option D

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**Explanation** :

The following diagram can be drawn from the given information.

⇒ Area of the Rhombus = $\frac{1}{2}\times 12\times 16$ = 96

Radius of the circle is same as the height of ∆ABC

In a right triangle, altitude from right angle = $\frac{Productofshortersides}{hypotenuese}$ = $\frac{\left(6\times 8\right)}{10}$ = $\frac{24}{5}$

⇒ Area of circle = $\pi {\left(\frac{24}{5}\right)}^{2}$

∴ Area of Circle : Area of Rhombus = $\pi {\left(\frac{24}{5}\right)}^{2}:96$ = 6π : 25.

Hence, option (d).

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**23. CAT 2020 QA Slot 1 | Algebra - Logarithms**

If y is a negative number such that ${2}^{{\mathrm{y}}^{2}{\mathrm{log}}_{3}5}={5}^{{\mathrm{log}}_{2}3}$, then y equals

- A.
log₂(1/5)

- B.
–log₂ (1/3)

- C.
–log₂ (1/5)

- D.
log₂ (1/3)

Answer: Option D

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**Explanation** :

Given, ${2}^{{\mathrm{y}}^{2}{\mathrm{log}}_{3}5}={5}^{{\mathrm{log}}_{2}3}$,

Taking log on both sides (Choosing the base to be 3)

⇒ y^{2} × log_{3}5 × log_{3}2 = log_{2}3 × log_{3}5

⇒ y^{2} × log_{3}2 = log_{2}3

⇒ y^{2} = (log23)^{2}

⇒ y = - log_{2}3 (∵ y is a negative number)

⇒ y = log_{2}(1/3)

Hence, option (d).

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**24. CAT 2020 QA Slot 1 | Algebra - Functions & Graphs**

If f(5 + x) = f(5 - x) for every real x, and f(x) = 0 has four distinct real roots, then the sum of these roots is

- A.
20

- B.
40

- C.
0

- D.
10

Answer: Option 20

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**Explanation** :

Given f(5 + x) = f(5 – x)

Assuming one of the roots is (5 + α).

⇒ f(5 + α) = 0

Now, we know, f(5 + α) = f(5 - α) = 0

⇒ If one of the roots is 5 + α, the other root will be 5 – α.

∴ If (5 + α) and (5 + β) are two of the roots then (5 – α) and (5 – β) will be the other two roots.

⇒ Sum of the roots = (5 + α) + (5 + β) + (5 – α) + (5 – β) = 20

Hence, 20.

Workspace:

**25. CAT 2020 QA Slot 1 | Algebra - Logarithms**

If log₄ 5 = (log₄ y)(log₆ √5), then y equals

Answer: 36

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**Explanation** :

Given, log_{4}5 = log_{4}y × log_{6}√5

⇒ log_{4}5 × log_{y}4 = log_{6}√5

⇒ log_{y}5 = ½ × log_{6}5

⇒ log_{y}5 = log_{36}5

⇒ y = 36.

Hence, 36.

Workspace:

**26. CAT 2020 QA Slot 1 | Algebra - Functions & Graphs**

The area of the region satisfying the inequalities |x| - y ≤ 1, y ≥ 0 and y ≤ 1 is

Answer: 3

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**Explanation** :

Given,

|x| - y ≤ 1

⇒ y ≥ |x| - 1

y ≥ 0 and y ≤ 1.

The following diagram can be drawn from the given information.

The required area is highlighted in orange i.e., a trapezium whose parallel sides are 4 units and 2 units and height is 1 unit.

∴ Area of the trapezium = 1/2 × (2 + 4) × 1 = 3 square units.

Hence, 3.

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