# CAT 2021 QA Slot 1 | Previous Year CAT Paper

**1. CAT 2021 QA Slot 1 | Arithmetic - Average**

Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was

Answer: 35

**Explanation** :

Let the number of patients in hospital A = x

∴ The number of patients in hospital B = x + 21

⇒ $\frac{200}{x}$ = $\frac{152}{x+21}$ + 3

⇒ 200x + 4200 = 152x + 3x^{2} + 63x

⇒ 3x^{2} + 15x - 4200 = 0

⇒ x^{2} + 5x - 1400 = 0

⇒ x = 35 or -40 (not possible)

∴ Number of patients in hospital A = 35

Hence, 35.

Workspace:

**2. CAT 2021 QA Slot 1 | Arithmetic - Profit & Loss**

Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is

Answer: 1000

**Explanation** :

Let the total number of pens Amal bought = x

Also, let the wage of the employee = w

∴ Amal’s total cost price = 8x + w

Total selling price of the first 100 pen = 100 × 12 = 1200

**Case 1**: The remaining pen when each is sold at Rs. 11

Total selling price of the remaining pen = (x - 100) × 11

⇒ 1200 + 11(x - 100) = 8x + w + 300

⇒ 3x - 200 = w …(1)

**Case 2**: The remaining pen when each is sold at Rs. 9

Total selling price of the remaining pen = (x - 100) × 9

⇒ 1200 + 9(x - 100) = 8x + w - 300

⇒ x + 600 = w …(2)

Solving (1) and (2) we get

x = 400 and w = 1000.

Hence, 1000.

Workspace:

**3. CAT 2021 QA Slot 1 | Arithmetic - Ratio, Proportion & Variation**

The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is

- A.
7 : 3

- B.
3 : 2

- C.
11 : 3

- D.
11 : 7

Answer: Option C

**Explanation** :

Let the earnings of Neeta, Geeta and Sita be ‘n’, ‘g’ and ‘s’ respectively.

The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days

⇒ 6s = n + g …(1)

The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days

⇒ 2g = s + n

⇒ 2(6s – n) = s + n [From (1)]

⇒ 11s = 3n

⇒ n = 11s/3 …(2)

Substituting n = 11s/3 in (1)

⇒ 6s = 11s/3 + g

⇒ 7s/3 = g …(3)

From (2) and (3)

11s/3 > 7s/3 > s

∴ n > g > s

∴ The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is = 11s/3 : s = 11 : 3

Hence, option (c).

Workspace:

**4. CAT 2021 QA Slot 1 | Algebra - Inequalities & Modulus**

The number of integers n that satisfy the inequalities |n - 60| < |n - 100| < |n - 20| is

- A.
18

- B.
21

- C.
20

- D.
19

Answer: Option D

**Explanation** :

Here the critical points are 20, 60 and 100.

**Case 1**: n ≥ 100

⇒ n - 60 < n - 100

⇒ -60 < -100

This can never be true.

This case is rejected.

**Case 2**: 60 ≤ n < 100

⇒ n - 60 < - n + 100

⇒ 2n < 160

⇒ n < 80

Also,

- n + 100 < n – 20

⇒ 2n > 120

⇒ n > 60

∴ Possible integral values of n are 61, 62, 63, …, 79 i.e., 19 values

**Case 3**: n < 60

⇒ - n + 100 < - n + 20

⇒ 100 < 20

This can never be true.

This case is rejected.

∴ n can take 19 integral values.

Hence, option (d).

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**5. CAT 2021 QA Slot 1 | Algebra - Quadratic Equations**

If r is a constant such that |x^{2} – 4x - 13| = r has exactly three distinct real roots, then the value of r is?

- A.
21

- B.
15

- C.
17

- D.
18

Answer: Option C

**Explanation** :

Given, |x^{2} – 4x - 13| = r

∴ x^{2} – 4x - 13 = ± r

We have two quadratic equations here but only three distinct roots it means one of the quadratic equations will have equal roots.

**Case 1**: x^{2} – 4x - 13 = r has equal roots, i.e., Discriminant = 0

⇒ x^{2} – 4x – 13 - r = 0 had D = 0

⇒ D = 16 – 4(-13 - r) = 0

⇒ 16 + 52 + 4r = 0

⇒ r = - 17

**Case 2**: x^{2} – 4x - 13 = - r has equal roots, i.e., Discriminant = 0

⇒ x^{2} – 4x – 13 + r = 0 had D = 0

⇒ D = 16 – 4(-13 + r) = 0

⇒ 16 + 52 - 4r = 0

⇒ r = 17

Hence, option (c).

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**6. CAT 2021 QA Slot 1 | Arithmetic - Time, Speed & Distance**

Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is

- A.
180

- B.
184

- C.
190

- D.
192

Answer: Option C

**Explanation** :

Let the length of the smaller train be L meters.

Also, let the speeds of faster and slower trains be ‘f’ and ‘s’ m/s.

Given, f - s = 6 × 5/18 = 5/3 m/s

⇒ f = s + 5/3 …(1)

Faster train crosses a lamp post in 12 seconds.

∴ 160/f = 12

⇒ f = 40/3

∴ s = f – 5/3 = 35/3

The two trains cross each other in 14 seconds.

∴ (160 + L)/(f + s) = 14

⇒ L = 14(f + s) – 160

⇒ L = 14 × 25 – 160

⇒ L = 350 – 160 = 190

Hence, option (c).

**Concept**: Trains

Workspace:

**7. CAT 2021 QA Slot 1 | Arithmetic - Time & Work**

Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works everyday. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is

- A.
8

- B.
6

- C.
5

- D.
7

Answer: Option D

**Explanation** :

Let the total work to be done = LCM (15, 12, 20) = 60 units.

Efficiency of Anu = 60/15 = 4 units/day

Efficiency of Vinu = 60/12 = 5 units/day

Efficiency of Manu = 60/20 = 3 units/day

Let us calculate the work done every 2 days.

Vinu works on both the days, hence work done by Vinu = 2 × 5 = 10 units

Anu works on alternate days, hence work done by Anu = 4 units

Manu works on alternate days, hence work done by Manu = 3 units

∴ Total work done in 2 days = 10 + 4 + 3 = 17 units.

Hence total work done in 6 days = 3 × 17 = 51 units.

On 7th day Vinu and Anu will work, hence work done = 5 + 4 = 9 units.

∴ Work done in total 7 days = 51 + 9 = 60 units [Work gets completed now]

∴ The number of days needed to complete the work is 7 days.

Hence, option (d).

Workspace:

**8. CAT 2021 QA Slot 1 | Modern Math - Permutation & Combination**

The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

Answer: 47

**Explanation** :

Out of the seven integers given, 3 and 5 should always be selected.

**Case 1**: Exactly 1 of 7 or 8 is selected.

Exactly one of 7 or 8 can be selected in 2 ways.

Now we have already selected three integers.

For each of the remaining 4 integers, it may be selected or it may not be selected.

∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16

∴ Total number of ways = 2 × 16 = 32

**Case 2**: None of 7 and 8 is selected.

So far we have selected only two integer.

For each of the remaining 4 integers, it may be selected or it may not be selected.

∴ Number of ways of selection for remaining integers = 2 × 2 × 2 × 2 = 16

But this includes one way when no integer is selected.

∴ Number of ways of selecting at least one integer = 16 – 1 = 15

∴ Total number of ways = 32 + 15 = 47.

Hence, 47.

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**9. CAT 2021 QA Slot 1 | Geometry - Mensuration**

If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length in cm of each side of the hexagon is

- A.
4√6

- B.
√6

- C.
6√6

- D.
2√6

Answer: Option D

**Explanation** :

Area of the equilateral triangle = √3/4 × (12)^{2}

Let the length of each side of the regular hexagon be ‘a’.

Area of a regular hexagon with side ‘a’ = 6 × √3/4 × (a)^{2}

Now, 6 × √3/4 × (a)^{2} = √3/4 × (12)^{2}

⇒ a^{2} = 24

⇒ a = 2√6

Hence, option (d).

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**10. CAT 2021 QA Slot 1 | Algebra - Logarithms**

If 5 – ${\mathrm{log}}_{10}\left(\sqrt{1+x}\right)$ + 4${\mathrm{log}}_{10}\left(\sqrt{1-x}\right)$ = ${\mathrm{log}}_{10}\left(\frac{1}{\sqrt{1-{x}^{2}}}\right)$, then 100x equals

Answer: 99

**Explanation** :

Given,

5 – ${\mathrm{log}}_{10}\left(\sqrt{1+x}\right)$ + 4${\mathrm{log}}_{10}\left(\sqrt{1-x}\right)$ = ${\mathrm{log}}_{10}\left(\frac{1}{\sqrt{1-{x}^{2}}}\right)$

⇒ 5 – ${\mathrm{log}}_{10}{\left(1+x\right)}^{1/2}$ + ${\mathrm{log}}_{10}{\left(1-x\right)}^{4/2}$ = ${\mathrm{log}}_{10}{\left(1-{x}^{2}\right)}^{-1/2}$

⇒ 5 = ${\mathrm{log}}_{10}{\left(1+x\right)}^{1/2}$ - ${\mathrm{log}}_{10}{\left(1-x\right)}^{2}$ - ${\mathrm{log}}_{10}{\left(1-{x}^{2}\right)}^{1/2}$

⇒ 5 = ${\mathrm{log}}_{10}\left[\frac{{\left(1+x\right)}^{1/2}}{{(1-x)}^{2}\times {\left\{\right(1+x\left)\right(1-x\left)\right\}}^{1/2}}\right]$

⇒ 5 = ${\mathrm{log}}_{10}{(1-x)}^{-5/2}$

⇒ 5 = -5/2 × log_{10}(1 - x)

⇒ -2 = log_{10}(1 - x)

⇒ 1 – x = 10^{-2} = 1/100

⇒ 100 – 100x = 1

⇒ 100x = 99

Hence, 99.

Workspace:

**11. CAT 2021 QA Slot 1 | Algebra - Simple Equations**

A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

- A.
13

- B.
12

- C.
11

- D.
10

Answer: Option A

**Explanation** :

Let the cost of each apple be ‘a’, each orange be ‘o’ and each mango be ‘m’.

∴ 2a + 4o + 6m = a + 4o + 8m = 8o + 7m

⇒ 2a + 4o + 6m = 8o + 7m

⇒ 4o + m = 2a …(1)

Also, a + 4o + 8m = 8o + 7m

⇒ 4o – m = a …(2)

Solving (1) and (2) we get

8o = 3a and m = a/2

∴ Value of each basked in terms of mangoes

= 2a + 4o + 6m

= 4m + 3a/2 + 6m

= 10m + 3/2 × 2m

= 10m + 3m

= 13m

∴ Value of each basket is same as value of 13 mangoes.

Hence, option (a).

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**12. CAT 2021 QA Slot 1 | Arithmetic - Mixture, Alligation, Removal & Replacement**

The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is

Answer: 200

**Explanation** :

Let x cc is removed from first bottle and replaced with solution from second bottle.

Using alligation rule

⇒ (800-x)/x = 4/12 = 1/3

⇒ 2400 – 3x = x

⇒ x = 600 cc.

∴ Solution left in second bottle = 800 – 600 = 200 cc.

Hence, 200.

Workspace:

**13. CAT 2021 QA Slot 1 | Arithmetic - Average**

Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

- A.
20

- B.
16

- C.
18

- D.
26

Answer: Option C

**Explanation** :

Let the amounts spent by the family each month be LCM (10, 20, 25, 50) = Rs. 100 for the first 3 months and then Rs. 50 for the next two months.

Amount of onion bought during month 1 = 100/10 = 10 kgs

Amount of onion bought during month 2 = 100/20 = 5 kgs

Amount of onion bought during month 3 = 100/25 = 4 kgs

Amount of onion bought during month 4 = 50/25 = 2 kgs

Amount of onion bought during month 5 = 50/50 = 1 kgs

∴ Total amount of onion bought = 10 + 5 + 4 + 2 + 1 = 22 kgs

Total amount spend on onions = 100 + 100 + 100 + 50 + 50 = Rs. 400.

∴ Average expense for onion per kg for these 5 months = 400/22 = 18.18 ≈ Rs. 18/kg.

Hence, option (c).

Workspace:

**14. CAT 2021 QA Slot 1 | Geometry - Quadrilaterals & Polygons**

Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm, is

- A.
√14

- B.
√12

- C.
√13

- D.
√15

Answer: Option C

**Explanation** :

Side of the regular hexagon = 2 cm.

Consider the figure below.

Consider the isosceles ∆ATF

TU is the altitude from T to AF.

We know, in a regular hexagon the distance between any two parallel sides = √3 × side.

∴ TU = √3 × 2 = 2√3 cm.

Since ATF is an isosceles triangle, U will be the mid-point of AF

∴ AU = 1 cm.

In ∆ATU

AT^{2} = TU^{2} + AU^{2}

⇒ AT^{2} = (2√3)^{2} + 1^{2}

⇒ AT^{2} = 13

⇒ AT = √13

Hence, option (c).

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**15. CAT 2021 QA Slot 1 | Geometry - Circles**

A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches then the area of triangle in square inches is.

Answer: 120

**Explanation** :

Here, BM = BN = 4 (Radius of the circle)

BC = 10 cm, hence, MC = CO = 10 – 4 = 6 cm.

Let AN = x, hence, AO = x

In right ∆ABC,

AB^{2} + BC^{2} = AC^{2}

⇒ (x + 4)^{2} + 10^{2} = (x + 6)^{2}

⇒ x^{2} + 16 + 8x + 100 = x^{2}+ 12x + 36

⇒ 4x = 80

⇒ x = 20

∴ Area of triangle ABC = ½ × AB × BC = ½ × (20 + 4) × 10 = 120 sq. inches.

Hence, 120.

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**16. CAT 2021 QA Slot 1 | Algebra - Functions & Graphs**

f(x) = $\frac{{x}^{2}+2x-15}{{x}^{2}-7x-18}$ is negative if and only if

- A.
X < -5 or 3 < x < 9

- B.
-2 < x < 3 or x > 9

- C.
-5 < x < -2 or 3 < x < 9

- D.
x < -5 or -2 < x < 3

Answer: Option C

**Explanation** :

Given, f(x) = $\frac{{x}^{2}+2x-15}{{x}^{2}-7x-18}$ < 0

⇒ $\frac{(x+5)(x-3)}{(x-9)(x+2)}$ < 0

Here, the critical points are -5, -2, 3 and 9.

∴ f(x) will be positive when -5 < x < -2 or 3 < x < 9.

Hence, option (c).

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**17. CAT 2021 QA Slot 1 | Arithmetic - Time & Work**

Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is

Answer: 32

**Explanation** :

Let the total work to be done = LCM (12, 16, 24) = 48 units.

Let the efficiency of Amar, Akbar and Anthony be ‘x’, ‘y’ and ‘z’ units/month

⇒ x + y = 48/12 = 4 …(2)

⇒ y + z = 48/16 = 3 …(1)

⇒ z + x = 48/24 = 2 …(3)

Adding all these equations

⇒ x + y + z = 9/2 = 4.5 …(4)

Solving these four equations we get,

x = 1.5, y = 2.5 and z = 0.5

∴ The person who is neither slowest not fastest is Amar with efficiency of 1.5 units/month.

∴ Time required by Amar to complete the task alone = 48/1.5 = 32 months.

Hence, 32.

Workspace:

**18. CAT 2021 QA Slot 1 | Arithmetic - Percentage**

Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to

- A.
127

- B.
135

- C.
144

- D.
124

Answer: Option A

**Explanation** :

Let the weight and price of small box is ‘w’ and ‘p’ respectively.

∴ Selling price per gram for small box = p/w

Now, selling price per gram for large box = 0.88p/w

Also, selling price for the whole box = 2p.

⇒ Weight of the large box = $\frac{2p}{0.88p/w}$ = $\frac{2w}{0.88}$ = $\frac{25w}{11}$

∴ Required percentage = $\frac{{\displaystyle \frac{25}{11}}w-w}{w}$ × 100 = 14/11 × 100 ≈ 127%

Hence, option (a).

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**19. CAT 2021 QA Slot 1 | Algebra - Simple Equations**

How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Answer: 70

**Explanation** :

Let the required three-digit number be ‘abc’.

According to the question:

‘abc’ + 198 = ‘cba’

⇒ 100a + 10b + c + 198 = 100c + 10b + a

⇒ 99c – 99a = 198

⇒ c – a = 2

∴ Value of (c, a) can be (9, 7) or (8, 6) or (7, 5) or (6, 4) or (5, 3) or (4, 2) or (3, 1) i.e., total 7 values.

For each of these 7 values of c and a, b can taken any of the 10 values from 0 to 9.

∴ Total such possible numbers = 7 × 10 = 70.

Hence, 70.

Workspace:

**20. CAT 2021 QA Slot 1 | Arithmetic - Simple & Compound Interest**

Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to

- A.
934.65

- B.
926.84

- C.
931.72

- D.
929.48

Answer: Option C

**Explanation** :

Let the interest accrued during first year = I and rate of interest be r%

We know in case of compound interest, interest for each year increases by r% every year.

∴ Interest for 2^{nd} year = I × $\left(1+\frac{r}{100}\right)$ = 806.25 …(1)

Interest for 3^{rd} year = I × ${\left(1+\frac{r}{100}\right)}^{2}$ = 866.72 …(2)

(2) ÷ (1)

⇒ $\left(1+\frac{r}{100}\right)$ = $\frac{866.72}{806.25}$ = 1 - $\frac{60.47}{806.25}$

⇒ r = 7.5%

∴ Interest for 4^{th} year = Interest for 3^{rd} year × (1+r/100)

= 866.72 × $\left(1+\frac{7.5}{100}\right)$ = 931.72

Hence, option (c).

Workspace:

**21. CAT 2021 QA Slot 1 | Algebra - Progressions**

If x_{0} = 1, x_{1} = 2 and x_{n+2} = $\frac{1+{x}_{n+1}}{{x}_{n}}$, n = 0, 1, 2, 3, …, then x_{2021} is equal to

- A.
1

- B.
3

- C.
4

- D.
2

Answer: Option D

**Explanation** :

Given, x_{n+2} = $\frac{1+{x}_{n+1}}{{x}_{n}}$

x_{0} = 1, x_{1} = 2

Substituting n = 0 we get

x_{2} = $\frac{1+2}{1}$ = 3

Similarly,

x_{3} = 2

x_{4} = 1

x_{5} = 1

x_{6} = 2

x_{7} = 3

x_{8} = 2

x_{9} = 1

… and so on.

Here, we can see that the value of xn repeats after every 5 terms.

∴ x_{2021} = x_{2020+1} = x_{1} = 2

Hence, option (d).

Workspace:

**22. CAT 2021 QA Slot 1 | Algebra - Progressions**

The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15^{th} group is equal to

- A.
4941

- B.
6119

- C.
7471

- D.
6090

Answer: Option B

**Explanation** :

Here,

1^{st} group has 1 integer

2^{nd} group has 3 integers

3^{rd} group has 5 integers

∴ nth groups will have (2n – 1) integers.

Total inters used till

1^{st} group = 1 = 1^{2}

2^{nd} group = 1 + 3 = 2^{2}

3^{rd} group = 1 + 3 + 5 = 9 = 3^{2}

∴ Total integers used till n^{th} group = n^{2}

∴ Sum of integers in 15^{th} group = Sum of all integers used till 15^{th} group - Sum of all integers used till14^{th} group

= (1 + 2 + 3 + … + 15^{2}) - (1 + 2 + 3 + … + 14^{2})

= $\frac{225\times 226}{2}$ - $\frac{196\times 197}{2}$

= 25425 – 19306 = 6119

Hence, option (b).

Workspace:

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