CAT 2021 QA Slot 2 | Previous Year CAT Paper

Answer: 250

Video Explanation

Explanation :

Let total number of black balls be 100b and total number of white balls be 100w.
∴ 100w + 100b = 450
⇒ w + b = 4.5   …(1)

According to the question.
Number of metallic balls:
White metallic = 40% of 100w = 40w
Black metallic = 50% of 100b = 50b

Now, 40w = 50b
⇒ w = 1.25b   …(2)

From (1) and (2)
1.25b + b = 4.5
⇒ b = 2 and w – 2.5

∴ Number of non-metallic white balls = 60% of 100w = 150
and, number of non-metallic black balls = 50% of 100b = 100

∴ Total number of non-metallic balls = 150 + 100 = 250

Hence, 250.

Answer: 4195

Video Explanation

Explanation :

Let the thousands, hundreds, tens and units digits be a, b, c and d.

Given, sum of its digits in the thousands, hundreds and tens places is 14
a + b + c = 14   …(1)
The sum of its digits in the hundreds, tens and units places is 15
b + c + d = 15   …(2)

(2) – (1) ⇒ a = d – 1   …(3)

The tens place digit is 4 more than the units place digit ⇒ c = d + 4   …(4)

For the number to be highest possible d should also be highest possible.

Highest possible of d is 5 (from (4)).
∴ highest possible value of c = 9 and that of a = 4

From (1), b = 14 – a – c = 1

∴ The highest such number is 4195

Hence, 4195.

Answer: Option B

Video Explanation

Explanation :

Let 2, 3, 5 kgs is bought of each variety respectively.

∴ Total quantity bought = 2 + 3 + 5 = 10 kgs

Total cost of tea = 2 × 800 + 3 × 500 + 5 × 300 = 4600.

Profit on total quantity = 50%

∴ Total selling price for 10 kg tea = 4600 × 1.5 = 6900.

Selling price for 1/6^th of 10 kg tea = 10/6 × 700 = 7000/6

⇒ Selling price for remaining 50/6 kg tea = 6900 – 7000/6 = 34400/6

∴ Selling price per kg for the remaining tea = 34400/6 ÷ 50/6 = 688

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

Let the total work to be done = LCM (60, 84) = 420 units.

Efficiency of Anil = 420/60 = 7 units/day
Efficiency of Bimal = 420/84 = 5 units/day

Work done by Anil in 10 + 14 days = 24 × 7 = 168 units

Work done by Bimal in 14 days = 14 × 5 = 70 units

∴ Work done by Charu = 420 – 70 - 168 = 182 units

Fraction of work done by Charu = 182/420 = 91/210

∴ Payment received by Charu = 91/210 × 21,000 = Rs. 9,100

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

Refer the diagram below.

​​​​​​​

ABPD is a parallelogram.

ABPD and ∆BPC have equal base and same height.

∴ Area (ABPD) = 2 × Area (∆BPC)

Also, Area (ABPD) - Area (∆BPC) = 10

⇒ 2 × Area (∆BPC) - Area (∆BPC) = 10

⇒ Area (∆BPC) = 10

∴ Area (ABPD) = 20

⇒ Area (ABCD) = Area (ABPD) + Area (∆BPC) = 20 + 10 = 30

Hence, option (b).

Answer: 1000

Video Explanation

Explanation :

We know that number of distributing n identical objects amongst r different groups = ^n+r-1C_r-1

Distributing balloons
Since each child gets at least 4 balloons, we will first give 4 balloons to each of the 3 children.
Now we have 3 identical balloons to be distributed to 3 children.
Number of ways of doing this = ^3+3-1C_3-1 = 10 ways

Distributing pencils
Since each child gets at least 1 pencil, we will first give 1 pencil to each of the 3 children.
Now we have 3 identical pencils to be distributed to 3 children.
Number of ways of doing this = ^3+3-1C_3-1 = 10 ways

Distributing erasers
We have 3 identical erasers to be distributed to 3 children.
Number of ways of doing this = ^3+3-1C_3-1 = 10 ways

∴ Total number of ways = 10 × 10 × 10 = 1000

Hence, 1000.

Answer: Option A

Video Explanation

Explanation :

Given x² – xy – x = 22 and y² – xy + y = 34.

Adding both the equations

⇒ x² + y² – 2xy - (x – y) = 56

⇒ (x - y)² – (x - y) = 56

⇒ (x - y)(x - y - 1) = 56

Let (x – y) = a [a > 0 since x > y]

⇒ a(a - 1) = 56

⇒ a² – a – 56 = 0

⇒ (a – 8)(a + 7) = 0

⇒ a = 8 or -7 (rejected)

∴ a = x - y = 8

Hence, option (a).

Answer: 45

Video Explanation

Explanation :

Initially the container is full of milk. Let the capacity of container be ‘V’ liters.

When 9 liters are drawn, fraction of quantity drawn = $\frac{9}{V}$.

Hence, fraction of milk drawn will also be $\frac{9}{V}$

∴ Fraction of milk remaining will be $\left(1-\frac{9}{V}\right)$

Only water is added.

⇒ Quantity of milk remaining after first replacement = $V\left(1-\frac{9}{V}\right)$

Similarly, after second replacement quantity of milk remaining = $V{\left(1-\frac{9}{V}\right)}^2$

This is $\frac{16}{25}$ ^th  of the total volume.

∴ $V{\left(1-\frac{9}{V}\right)}^2$ = $\frac{16}{25}$V

⇒ 1 – $\frac{9}{V}$ = $\frac{4}{5}$

⇒ V = 45 liters.

Hence, 45. 

Answer: Option B

Video Explanation

Explanation :

Let the lengths of trains A and B be ‘A’ and ‘B’ and their speeds be ‘a’ and ‘b’ respectively.

Time taken for front end of A to cross rear end of B = 46 seconds

⇒ 46 = $\frac{B}{a+b}$

⇒ B = 46a + 46b   …(1)

Total time taken for rear ends of the two train to cross each other = (46 + 69) = 115 seconds

⇒ 115 = $\frac{A+B}{a+b}$

⇒ A + B = 115a + 115b   …(2)

(2) - (1)

⇒ A = 69a + 69b   …(3)

Now, $\frac{A}{B}$ = $\frac{69(a+b)}{46(a+b)}$ = $\frac{69}{46}$ = $\frac{3}{2}$

Hence, option (b). 

Concept:

Trains

Answer: 10

Video Explanation

Explanation :

Let the number of matches played so far be ‘b’ and the goals scored in these ‘n’ matches be ‘g’.

Now if 1 goal is scored in next 10 matches 

⇒ g + 1 = (n + 10) × 0.15

⇒ g = 0.15n + 1.5   …(1)

If 2 goals are scored in next 10 matches 

⇒ g + 2 = (n + 10) × 0.2

⇒ g = 0.2n + 1   …(2)

Solving (1) and (2) we get,

n = 10

Hence, 10.

Answer: 5

Video Explanation

Explanation :

Given, log₂⁡[3 + log₃{4 + log₄(x - 1)}] - 2 = 0

⇒ log₂⁡[3 + log₃{4 + log₄(x - 1)}] = 2

⇒ 3 + log₃{4 + log₄(x - 1)} = 4

⇒ log₃{4 + log₄(x - 1)} = 1

⇒ 4 + log₄(x - 1) = 3

⇒ log₄(x - 1) = -1

⇒ x – 1 = ¼

⇒ x = 5/4

⇒ 4x = 5

Hence, 5.

Answer: Option A

Video Explanation

Explanation :

Let the loss incurred by Raj in first year = P%

∴ Amount remaining after 1st year = 10,000$\left(1-\frac{P}{100}\right)$ > 5,000   …(1)

Now the percentage growth next year = 5P%

∴ Amount after 2 years = 10,000$\left(1-\frac{P}{100}\right)$$\left(1+\frac{5P}{100}\right)$

Overall growth after 2 years is 35%, hence amount after 2 years should be 10,000 × 1.35

⇒ 10,000$\left(1-\frac{P}{100}\right)$$\left(1+\frac{5P}{100}\right)$ = 10,000 × 1.35

⇒ 10000 – 5P² + 400P = 13500

⇒ 5P² - 400P + 3500 = 0

⇒ P² – 80P + 700 = 0

⇒ P = 10% or 70%.

P cannot be 70% since amount remaining after 1st year has to be greater than 5000 [from (1)]

∴ P = 10%

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

Given, x₁ - x₂ + x₃ - … + (-1)⁽ⁿ⁺¹⁾ x_n = n² + 2n

Put n = 1, ⇒ x₁ = 1² + 2 × 1 = 3

Put n = 2, ⇒ x₁ - x₂ = 2² + 2 × 2 = 8 ⇒ x₂ = 3 – 8 = -5

Put n = 3, ⇒ x₁ - x₂ + x₃ = 3² + 2 × 3 = 15 ⇒ x₃ = 8 – 15 = -7
…
⇒ x_n = (-1)ⁿ⁺¹ × (2n+1)

∴ x₄₉ = 99 and x₅₀ = -101

⇒ x₄₉ + x₅₀ = 99 – 101 = -2

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

Let the total investment done by all three of them is I.

When total profit decreases by 3% (from 18% to 15%) Anil’s share decreases by 420.

Decrease in total profit = 3% of I.

Decrease in Anil’s profit will be 70% of decrease in total profit.

⇒ 420 = 70% of 3% of I

⇒ 420 = 70/100 × 3/100 × I

⇒ I = 20,000

Now, when total profit increases by 2% (from 15% to 17%) Charu’s share increases by 80.

Increase in total profit = 2% of I.

Increase in Charu’s profit = C% of I [C% = percentage on investment contribution by Charu]

⇒ 80 = C% of 2% of I

⇒ 80 = C/100 × 2/100 × 20000

⇒ C = 20%

∴ Bobby’s percentage share of investment = 100 – 70 – 20 = 10%

⇒ Investment done by Bobby = 10% of 20,000 = Rs. 2,000

Hence, option (a).

Answer: 7

Video Explanation

Explanation :

Given, 2.25 ≤ 2 + 2ⁿ⁺² ≤ 202

⇒ 0.25 ≤ 2ⁿ⁺² ≤ 200

Case 1: 0.25 ≤ 2ⁿ⁺²
∴ n + 2 ≥ -2
⇒ n ≥ -4

Case 2: 2ⁿ⁺² ≤ 200
This is true when n ≤ 5

∴ -4 ≤ n ≤ 5

Now, 3 + 3ⁿ⁺¹ should be an integer

This is possible when n = -1, 0, 1, 2, 3, 4 and 5

∴ n can take 7 values.

Hence, 7.

Answer: Option D

Video Explanation

Explanation :

Case 1: 3x ≥ 40 ⇒ x ≥ 13.33

⇒ 3x – 20 + 3x – 40 = 20

⇒ 6x = 80

⇒ x = 13.33

Case 2: 20 ≤ 3x < 40 ⇒ 6.67 ≤ x < 13.33

⇒ 3x – 20 - 3x + 40 = 20

⇒ 20 = 20

This is always true.

Case 3: 3x ≤ 20

⇒ - 3x + 20 - 3x + 40 = 20

⇒ -6x = -40

⇒ x = 6.67

Combining solution from all the three cases, we get

6.67 ≤ x ≤ 13.33

Only option (d) is a subset of this range.

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

Since coefficients of the quadratic are rational numbers, hence the roots will be conjugate of each other.

If 2 + √3 is one of the roots, the other root will be 2 - √3.

∴ Sum of the roots = -(-b)/a = (2 + √3) + (2 - √3)
⇒ b/a = 4
⇒ b = 4a

Also, product of the roots = c/a = (2 + √3) × (2 + √3)
⇒ c/a = 4 – 3
⇒ c = a

Now, b = c³

⇒ 4a = a³

⇒ a² = 4

⇒ a = ±2

∴ |a| = 2

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

Let the filling and emptying capacity of A and B be ‘a’ and ‘b’ units/hour respectively.

Case 1: A is opened at 2 pm and B at 3 pm
Total work done till 10 pm = 8a - 7b

Case 2: A is opened at 2 pm and B at 4 pm
Total work done till 6 pm = 4a - 2b

Since work done is same in both cases, we have

8a – 7b = 4a – 2b

⇒ 4a = 5b

Now time taken by A alone to fill the tank = (Total work)/a = (4a – 2b)/a = (5b – 2b)/(5b/4) = 12/5 hours = 144 minutes.

Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

Let the three integers x, y and z be (a - d), a, (a + d) respectively.

[(a – d), a and (a + d) are all positive integers]

Since y – x > 2, hence d > 2.

Given, xyz = 5(x + y + z) 

⇒ (a – d) × a × (a + d) = 5 × 3a

⇒ (a – d)(a + d) = 15

Here (a – d) and (a + d) are positive integers

∴ We need to write 15 as product of 2 positive integers. This can be done in two ways, 1 × 15 or 3 × 5

Hence, (a, d) is either (8, 7) or (4, 1).

Since d > 0 hence, (4, 1) is rejected.

∴ (a, d) = (8, 7)

∴ z – x = (a + d) – (a - d) = 2d = 14

Hence, option (c).

Answer: 30

Video Explanation

Explanation :

Consider the figure below.

​​​​​​​

Consider ∆AED and ∆BED.
Height of both triangles is same, hence ratio of area will be same as ratio of their base.
∴ Area(∆AED)/Area(∆BED) = AD/BD = 2/1

Area (∆BED) = 8/2 = 4

∴ Area (∆ABE) = 8 + 4 = 12

Now, consider ∆AED and ∆BED.
Height of both triangles is same, hence ratio of area will be same as ratio of their base.
∴ Area(∆ABE)/Area(∆CBE) = AE/CE = 2/3

Area (∆CBE) = 3/2 × 12 = 18

∴ Area (∆ABC) = 12 + 18 = 30 sq. cm.

Hence, 30.

Answer: Option B

Video Explanation

Explanation :

Consider the diagram below.

​​​​​​​

In a rhombus diagonals perpendicularly bisect each other.

∴ Let OA = OC = x and OB = OD = y

In ∆AOB ⇒ x² + y² = 5²  …(1)

Also, area of the rhombus = ½ × 2x × 2y = 12

⇒ 2xy = 12   …(2)

(1) + (2)
⇒ x² + y² + 2xy = 25 + 12 = 37
⇒ (x + y)² = 37
⇒ x + y = √37   …(3)

(1) - (2)
⇒ x² + y² - 2xy = 25 - 12 = 13
⇒ (x - y)² = 13
⇒ x - y = √13   …(4)

Solving (3) and (4) we get,

2x = √37 + √13 and
2y = √37 - √13

∴ The longer diagonal = 2x = √37 + √13

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

Given, f(x) = $\frac{x^2+2x+4}{2x^2+4x+9}$

⇒ f(x) = $\frac{1}{2}\left[\frac{2x^2+4x+8}{2x^2+4x+9}\right]$

⇒ f(x) = $\frac{1}{2}\left[\frac{2x^2+4x+9-1}{2x^2+4x+9}\right]$

⇒ f(x) = $\frac{1}{2}\left[1-\frac{1}{2x^2+4x+9}\right]$

f(x) will be minimum when (2x² + 4x + 9) is minimum.

Now, 2x² + 4x + 9 will be minimum when x = -(4)/2 × 2 = -1

∴ Minimum value of 2x² + 4x + 9 = 2(-1)² + 4(-1) + 9 = 7

∴ Minimum value of f(x) = $\frac{1}{2}\left[1-\frac{1}{7}\right]$ = $\frac{3}{7}$

f(x) will be maximum when (2x² + 4x + 9) is maximum.

Maximum value of (2x² + 4x + 9) will be ∞.

∴ Maximum value of f(x) = $\frac{1}{2}\left[1-\frac{1}{\infty }\right]$ = $\frac{1}{2}$

∴ Range of f(x) = $\left[\frac{3}{7},\frac{1}{2}\right)$

Upper value of ½ is in open bracket as value of (2x² + 4x + 9) will never actually be ∞.
 
Hence, option (a).