# CAT 2020 QA Slot 2 | Previous Year CAT Paper

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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

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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).

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**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

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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).

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**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?

Answer: 315

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**Explanation** :

Out of 4 digits we already have 7 and 3. Now we have to select 2 more digits.

**Case 1**: zero is not selected.

We can select 2 more digits out of 7 in ^{7}C_{2} = 21 ways.

These 4 digits can be arranged in 4! = 24 ways

∴ Total number of ways = 21 × 24 = 504 ways.

Now, in half of these numbers 7 would be before 3 and in other half 3 would be before 7.

∴ 7 comes before 3 in 504/2 = 252 ways.

**Case 2:** zero is selected.

We can select 1 more digit out of 7 in ^{7}C_{1} = 7 ways.

These 4 digits can be arranged in 3 × 3! = 18 ways

∴ Total number of ways = 7 × 18 = 126 ways.

Now, in half of these numbers 7 would be before 3 and in other half 3 would be before 7.

∴ 7 comes before 3 in 126/2 = 63 ways.

∴ Total numbers in which 7 comes before 3 = 252 + 63 = 315.

Hence, 315.

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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?

Answer: 4

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**Explanation** :

Let John’s efficiency be 1 unit/day.

∴ Jack’s efficiency is 2 units/day.

Jack and Jim’s efficiency is thrice John’s efficienc.

∴ 2 + e_{Jim} = 3 × 1

⇒ e_{Jim} = 1

John takes three days more than that taken by three of them working together

Let the time taken by all three together is d.

∴ Total work to be done = 4 × d = 1 × (d + 3)

⇒ d = 1

∴ John finished the work in d + 3 = 4 days.

Since, John and Jim have same efficiency, Jim will also finish the work in 4 days.

Hence, 4.

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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

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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)

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

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

⇒ D^{2} – 95D + 2250 = D^{2} – 90D

⇒ 5D = 2250

⇒ D = 450

Hence, option (b).

**Concept**:

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**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

Answer: 23

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**Explanation** :

Given,

x = z – 9 and

y = z – 1

Also, x + y < z + 5

⇒ (z – 9) + (z – 1) < z + 5

⇒ z < 15

∴ Highest integral value of z is 14.

Now, we have to calculate maximum value of 2x + y

= 2(z - 9) + (z - 1)

= 3z – 19

= 3 × 14 – 19 = 23

Hence, 23.

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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

Answer: 800

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**Explanation** :

Let the amount received by them be a, s and m respectively.

Given, a : s = 3 : 2 and s : m = 4 : 5

∴ a : s : m = 6 : 4 : 5

⇒ Amal’s share = 6x and Sunil’s share = 4x.

Given, difference between the largest and the smallest of these three shares is Rs 400.

⇒ 6x – 4x = 400

⇒ x = 200.

∴ Sunil’s share = 4x = Rs. 800.

Hence, 800.

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**8. CAT 2020 QA Slot 2 | Geometry - Circles**

Let C_{1} and C_{2} be concentric circles such that the diameter of C_{1} is 2 cm longer than that of C_{2}. If a chord of C_{1} has length 6 cm and is a tangent to C_{2}, then the diameter, in cm, of C_{1} is

Answer: 10

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**Explanation** :

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

Perpendicular OA will bisect the chord, hence AB = 3.

Let the radius of smaller circle be r, hence radius of bigger circle will be r + 1.

In right triangle OAB,

OB^{2} = OA^{2} + AB^{2}

⇒ (r + 1)^{2} = r^{2} + 3^{2}

∴ r = 4

Diameter of C_{1} (the bigger circle) = 2(r + 1) = 10 cm.

Hence, 10.

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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

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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,

OP^{2} = OM^{2} + MP^{2}

⇒ 5^{2} + (2.4)^{2 }+ MP^{2}

⇒ MP = $\sqrt{19.24}$

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

Hence, option (a).

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**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

Answer: Option D

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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 Ram and Rahim be a and b respectively.

Distance AM = a × t = b × 4 …(1)

Distance BM = b × t = a × 1 …(2)

(1) × (2)

⇒ t^{2} = 4

⇒ t = 2

From (1) we get,

a × 2 = b × 4

⇒ a : b = 4 : 2 = 2 : 1

Hence, option (d).

**Concept**:

Time to Reach Other End After Meeting

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**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 x^{2} − 2|x| + |a - 2| = 0?

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option D

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**Explanation** :

Given, x^{2} − 2|x| + |a - 2| = 0

This can be written as

|x|^{2} − 2|x| + 1 – 1 + |a - 2| = 0

⇒ (|x| - 1)^{2} + |a - 2| - 1 = 0

Now, ((|x| - 1)^{2} ≥ 0 and |a - 2| - 1 will be an integer since a is an integer.

∴ |a - 2| - 1 can take only take non-positive values i.e., 0 or -1.

**Case 1**: |a - 2| - 1 = 0 and (|x| - 1) = 0

⇒ |a – 2| = 1 and |x| = 1

⇒ a = 1 or 3 and x = ± 1

∴ 4 possible combinations of (x, a)

**Case 2**: |a - 2| - 1 = -1 and (|x| - 1) = ±1

|a - 2| = 0 and |x| = 0 or 2

⇒ a = 2 and x = 0 or ± 2

∴ 3 possible combinations of (x, a)

∴ Total 7 possible combination of (x, a) are there.

Hence, option (d).

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**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

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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 = T^{2}

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

⇒ 16x^{2} = a^{4}

⇒ a^{2} = 4x

Substituting x in (1)

⇒ 3a + 2a^{2} = 90

⇒ 2a^{2} + 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).

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

Two circular tracks T_{1} and T_{2} 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 T_{1} and track T_{2} 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

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**Explanation** :

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

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

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$\left[\frac{\pi}{75},\frac{\pi}{125}\right]$ = $\frac{\pi}{25}$

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

Hence, (a).

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**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$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ is

Answer: 2704

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**Explanation** :

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

⇒ $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ = $\frac{1+x+y+xy}{xy}$ = $\frac{103}{xy}$ + 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$\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)$ = 2601 × $\frac{52}{51}\times \frac{52}{51}$ = 2704

Hence, 2704.

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**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

Answer: Option B

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**Explanation** :

John spent Rs. 450 on rice in April hence he will spend 450 × 1.2 = 540 on rice in May.

He spends Rs. 90 extra on rice.

He also spends a total of Rs. 150 more in May compared to April

⇒ He spends 150 – 90 = Rs. 60 more on wheat in May compared to April.

∴ His expenditure on wheat in April on wheat = 60/0.12 = Rs. 500.

⇒ His expenditure on wheat in May = 500 + 60 = Rs. 560.

Hence, option (b).

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**16. CAT 2020 QA Slot 2 | Algebra - Logarithms**

The value of ${\mathrm{log}}_{a}\left(\frac{a}{b}\right)+{\mathrm{log}}_{b}\left(\frac{b}{a}\right)$, for 1 < a ≤ b cannot be equal to

- A.
0

- B.
-1

- C.
1

- D.
-0.5

Answer: Option C

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**Explanation** :

${\mathrm{log}}_{a}\left(\frac{a}{b}\right)+{\mathrm{log}}_{b}\left(\frac{b}{a}\right)$ = 1 - log_{a}b + 1 - log_{b}a = 2 – (log_{a}b + log_{b}a)

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

∴ log_{a}b + log_{b}a ≥ 2

⇒ 2 – (log_{a}b + log_{b}a) ≤ 2 – 2

⇒ 2 – (log_{a}b + log_{b}a) ≤ 0

∴ 2 – (log_{a}b + log_{b}a) cannot take positive value i.e., it cannot be equal to 1 (option (b))

Hence, option (c).

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**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

Answer: Option C

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**Explanation** :

Let the price of a pencil = x and that of a sharpener = x + 2.

Suppose Aron bought p pencils and s sharpeners. Aditya buys 2x pencils and s – 10 sharpeners.

∴ px + s(x + 2) = 2px + (s – 10)(x + 2)

⇒ px + sx + 2s = 2px + sx + 2s – 10x – 20

⇒ 20 = px – 10x

⇒ x(p – 10) = 20

Here, the least value of p can be 11.

∴ Aron and Aditya together bought 3p pencils and least value of 3p = 3 × 11 = 33.

Hence, option (c).

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**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

Answer: 90000

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**Explanation** :

Let the Principal be Rs. P.

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

Compound Interest for 2 years = P × 1.05^{2} – P = 0.1025P

∴ 0.1025P – 0.09P = 1125

⇒ 0.0125P = 1125

⇒ P = 90,000

Hence, 90000.

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**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

Answer: Option D

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**Explanation** :

For (x² - 5x + 7)^{x+1} = 1, either

**Case 1**: x + 1 = 0

⇒ x = -1

**Case 2**: (x² - 5x + 7) = 1

⇒ x² - 5x + 6 = 0

⇒ x = 2 or 3

**Case 3**: (x² - 5x + 7) = -1 and x + 1 is even

⇒ x² - 5x + 8 = 0

No integral value of x is possible in this case.

∴ Possible values of x are -1, 2 and 3 i.e., 3 values.

Hence, option (d).

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**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

Answer: Option B

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**Explanation** :

The following Venn diagram can be drawn from the given information.

Since students must choose at least 2 subjects, there will be no one who chose a single subject.

Now, maximum number of students who choose only Math and Physics can be 5. (∵ 23 students chose Math)

Hence, for Physics we still have 25 – 18 – 5 = 2 students left.

These 2 students will have to chose Chemistry along with Physics.

∴ Least number of students who can choose Chemistry = 2 + 18 = 20.

Hence, option (b).

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**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

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**Explanation** :

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

Speed of train T_{1} = s from A to B and 2s from B to C.

Total time taken by T_{1} = $\frac{d}{s}+\frac{3d}{2s}=\frac{5d}{2s}$

Speed of train T_{2} = 2s from A to B and s from B to C.

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

∴ Ratio of time taken by T_{1} and T_{2} = $\frac{5d}{2s}:\frac{7d}{2s}$ = 5 : 7

Hence, option (c).

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**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

Answer: 17

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**Explanation** :

Given, 2x + 5y = 99

⇒ x = (99 - 5y)/2

For x to be an integer, y has to be odd.

Possible value of x and y

** x y**

97 -19

92 -17

…

17 13

∴ Total 17 possible values.

Hence, 17.

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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.
S

^{2}/√3 - B.
√3S

^{2}/2 - C.
S

^{2}/2√3 - D.
2S

^{2}/√3

Answer: Option A

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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 = $\frac{2s}{\sqrt{3}}$

∴ 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).

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**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

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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).

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**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

Answer: Option A

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**Explanation** :

Let the cost price be c and marked price be m. Total selling price for 12 toys = Rs. 12c.

Anil sells 8 toys at 20% discount. Hence, total selling price for 8 toys = 8 × 0.8m = 6.4m

He then sells 4 toys at further 25% discount. Hence, total selling price for 8 toys = 4 × 0.75 × 0.8m = 2.4m

Given, 6.4m + 2.4m = 2112

⇒ m = 240

Now, he earns an overall profit of 10%

∴ 12c × 1.1 = 2112

⇒ c = 160

∴ With no discounts his profit % = (240 - 160)/160 × 100 = 50%

Hence, option (a).

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**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

Answer: Option C

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**Explanation** :

Given, g(x) = f(x + 1) – f(x – 1).

f(x) = x² + ax + b

∴ g(x) = (x + 1)² + a(x + 1) + b – [(x - 1)² + a(x - 1) + b]

⇒ g(x) = x2 + 2x + 1 + ax + a + b - x2 + 2x - 1 - ax + a – b

⇒ g(x) = 4x + 2a

Given, g(20) = 72

⇒ 4 × 20 + 2a = 72

⇒ a = -4

Also given, f(x) ≥ 0

⇒ x² - 4x + b ≥ 0

This is possible when discriminant is less than or equal to 0.

⇒ 16 – 4b ≤ 0

⇒ b ≥ 4

∴ Least possible value of b is 4.

Hence, option (c).

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