CAT 2021 QA Slot 3 | Previous Year CAT Paper

Answer: Option A

Video Explanation

Explanation :

$\frac{{\log }_{15}a+{\log }_{32}a}{{\log }_{15}a\times {\log }_{32}a}$ = 4

Converting all logs to base 10.

$\frac{{\displaystyle \frac{1}{{\log }_{a}15}}+{\displaystyle \frac{1}{{\log }_{a}32}}}{\frac{1}{{\log }_{a}15}\times \frac{1}{{\log }_{a}32}}$ = 4

⇒ log_a32 + log_a15 = 4

⇒ log_a(32 × 15) = 4

⇒ a⁴ = 480

This is possible when 4 < a < 5.

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

​​​​​​​

In the figure above:

In ∆ABC,
∠A + ∠B + ∠C = 180°
⇒ x + y = 130°

Since DA = DE
⇒ ∠DAE = ∠DEA = x 
⇒ ∠ADE = 180° – 2x

Also, DB = DF
⇒ ∠DBF = ∠DFB = y
⇒ ∠BDC = 180° – 2y

∠ADE + ∠EDF + ∠FDB = 180°
⇒ 180 – 2x + ∠EDF + 180 – 2y = 180
⇒ ∠EDF = 2x + 2y – 180
⇒ ∠EDF = 260° – 180° = 80°

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

f(x) = x² – 7x and g(x) = x + 3

⇒ f(g(x)) – 3x = (g(x))² – 7g(x) – 3x

⇒ f(g(x)) - 3x = (x + 3)² – 7(x + 3) - 3x 

⇒ f(g(x)) - 3x = x^{​​​​​​​2} + 6x + 9 – 7x – 21 – 3x

⇒ f(g(x)) = x² - 4x – 12

f(g(x)) is a quadratic equation. Least value of ax² + bx + c occurs at x = -b/2a

∴ Minimum value of x² - 4x – 12 occurs at x = -(-4)/2 = 2

∴ Minimum value of f(g(x)) = 4 - 8 - 12 = -16.

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

Let the efficiency of Rahul be ‘r’ units per hour and for Gautam be ‘g’ units per day.

Initially, Rahul works for 8 hours and Gautam works for 6 hours.
∴ Work done by them = 8r + 6g   …(1)

If both of them worked starting from 9 AM, they would have completed the work in 7.5 hours.
∴ Work done by them = 7.5r + 7.5g   …(2)

⇒ 8r + 6g = 7.5r + 7.5g
⇒ 0.5r = 1.5g
⇒ r = 3g

∴ Work to be done = 8r + 6g = 8r + 2r = 10r

∴ Time taken by Rahul to complete the work alone = 10r/r = 10 hours.

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

Let the two sides of the rectangle be ‘l’ and ‘b’ ft.

⇒ l × b = 60,000

Cost of fencing = 200 × l + 100 × (l + 2b) = 100 (3l + 2b)

To minimize cost we need to minimize 3l + 2b

Now, we know AP ≥ GP

⇒ $\frac{3l+2b}{2}$ ≥ $\sqrt{3l\times 2b}$
⇒ 3l + 2b ≥ 2 × √$\sqrt{6\times 60,000}$
⇒ 3l + 2b ≥ 1200

∴ Minimum cost of fencing = 100 × 1200 = 1,20,000.

Hence, option (b).

Answer: 3500

Video Explanation

Explanation :

Side of the rhombus = ¼ × 40 = 10 cm.

Area of a rhombus = ½ d₁d₂ = 96 [d₁ and d₂ are the two diagonals]

⇒ d₁d₂ = 192

We know, (d₁/2)² + (d₂/2)² = 10²

⇒ d₁² + d₂² = 400

⇒ (d₁ + d₂)² - 2d₁d₂ = 400

⇒ (d₁ + d₂)² – 2 × 192 = 400

⇒ (d₁ + d₂)² = 784

⇒ d₁ + d₂ = 28

Hence, the cost of laying electric wore along its two diagonals = 125 × (d₁ + d₂) = 125 × 28 = Rs. 3,500

Hence, 3500.

Answer: 3000

Video Explanation

Explanation :

Let the area to be painted = LCM(12, 16) = 48 units.

⇒ Efficiency of Anil = 48/12 = 4 units/day
and Efficiency of Barun = 48/16 = 3 units/day

Work done by Anil in 6 days = 6 × 4 = 24 units
Work done by Anil in 6 days = 6 × 3 = 18 units
∴ Remaining work done by Chandu = 48 – 24 – 18 = 6 units.

⇒ Payment received by Chandu = 24000/48 × 6 = Rs. 3,000

Alternately, 
Fraction of work done by Anil in 6 days = 6/12 = ½
Fraction of work done by Barun in 6 days = 6/16 = 3/8

⇒ Fraction of work done by Chandu in 6 days = 1 - ½ - 3/8 = 1/8

Since, Chandu completes 1/8^th of the work, he will receive 1/8^th of the payment.

∴ Chandu’s payment = 1/8 × 24,000 = Rs. 3,000

Hence, 3000.

Answer: Option D

Video Explanation

Explanation :

Total cost = Fixed cost + Variable cost

Variable cost = k × b
where, b = number of boarders and k = variable cost per boarder.

Total cost = FC + kb

Total collection = 1600b

Case 1: Total profit = 200 × 50 = 1600 × 50 – (FC + k × 50)   …(1)
Case 2: Total profit = 250 × 75 = 1600 × 75 – (FC + k × 75)   …(2)

(2) - (1), we get
18750 – 10000 = 1600 × 25 - 25k
⇒ k = 1250 and FC = 7,500 

Now, total profit when there are 80 boarders = 1600 × 80 – (7500 + 1250 × 80) = 20,500

Alternately,
Total cost when there are 50 students = (1600 – 200) × 50 = 70,000
Total cost when there are 75 students = (1600 – 250) × 75 = 1,01,250

Due to 25 students cost increases by (101250 - 70000) = 31250
∴ Due to 5 students cost increases by = 31250/5 = 6250

∴ Total cost when there are 80 students = 1,01,250 + 6250 = 1,07,500
Total revenue for 80 students = 1600 × 80 = 1,28,000

∴ Profit = 1,28,000 – 1,07,500 = 20,500.

Hence, option (d).

Answer: 50

Video Explanation

Explanation :

Case 1: Number contains 2, 3, 1, 1
Number of such numbers = 4!/2! = 12

Case 2: Number contains 2, 3, 3, 1
Number of such numbers = 4!/2! = 12

Case 3: Number contains 2, 2, 3, 1
Number of such numbers = 4!/2! = 12

Case 4: Number contains 2, 2, 3, 3
Number of such numbers = 4!/(2!×2!) = 6

Case 5: Number contains 2, 2, 2, 3
Number of such numbers = 4!/3! = 4

Case 6: Number contains 2, 3, 3, 3
Number of such numbers = 4!/3! = 4

∴Total such numbers = 12 + 12 + 12 + 6 + 4 + 4 = 50.

Hence, 50.

Answer: 8

Video Explanation

Explanation :

Let the length of the track be ‘L’ meter and speeds of Mira and Amal be ‘m’ and ‘a’ meters/minute respectively.

When they walk in opposite direction, they meet in 3 minutes.
∴ 3(m + a) = L   
⇒ 3m + 3a = L   …(1)

When they walk in same direction, Amal completes 3 rounds more than Mira. Hence, Amal completes 3L meters more than Mira.
∴ 45a – 45m = 3L   …(2)

(1) × 15 - (2)

⇒ 90m = 12L

⇒ L/m = 7.5 

Hence, it takes Mira 7.5 minutes to complete one round.

Hence, in 60 minutes she will cover = 60/7.5 = 8 rounds.

Hence, 8.

Concept:

Circular Race

Answer: 6

Video Explanation

Explanation :

$\left(\sqrt[7]{10}\right)$${\left(\sqrt[7]{10}\right)}^2$...${\left(\sqrt[7]{10}\right)}^n$ > 999

⇒ 101/7 × 102/7 ×… × 10n/7 > 999

⇒ ${10}^{\frac{1+2+3+...+n}{7}}$ > 999

⇒ ${10}^{\frac{n(n+1)}{14}}$ > 999

This is possible when n(n + 1)/14 ≥ 3.

∴ Least possible positive integral value of n is 6.

Hence, 6.

Answer: Option C

Video Explanation

Explanation :

​​​​​​​

Let AE be a perpendicular from A to DC.

∆ADE is a 30-60-90 triangle.
⇒ DE = √3/2 × 10 = 5√3
⇒ AE = 10/2 = 5

∆ACE is a right triangle.
⇒ CE² = AC² – AE²
⇒ CE² = 400 – 25 = 375
⇒ CE = 5√15

∴ Area of parallelogram = 2 × Area of triangle ADC
⇒ Area of parallelogram = 2 × (1/2 × CD × AE) = CD × AE
= (DE + EC) × AE
= (5√3 + 5√15) × 5
= 25(√3 + √15)

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

Matches so far:
Total played = 40
Won = 30% of 40 = 12

Let the number of remaining matches be ‘n’.
The team wins 60% i.e., 0.6n matches.

There overall win percentage is 50% i.e., half

⇒ 12 + 0.6n = ½ (40 + n)

⇒ 24 + 1.2n = 40 + n

⇒ n = 80

Now, if the team wins 90% of these 80 remaining matches, they will win 90% of 80 = 72 more matches.

∴ Total matches won = 12 + 72 = 84

Hence, option (b).

Answer: Option D

Video Explanation

Explanation :

Let Raju invests Rs. A in bank B at r% p.a. for t years.
∴ Rupa invests Rs. 10,000 in bank C at 2r% p.a. for 2t years.

Total interest accrued for Raju = Art/100

This is same as interest accrued by investing same amount in bank A for a year.

Amount due after 1 year in bank A = A${\left[1+\frac{3}{100}\right]}^2$ = 1.0609A
Interest accrued = 1.0609A - A = 0.0609A

⇒ 0.0609A = $\frac{Art}{100}$

⇒ rt = 6.09

Now, interest accrued by Rupa = $\frac{10000\times 2r\times 2t}{100}$ = 400 × rt = 400 × 6.09 = 2436.

Hence, option (d).

Answer: Option C

Video Explanation

Explanation :

Let the weight of initial alloy of silver and copper be x kg and it has p% silver.

This alloy is mixed with 3 kg pure silver.

Using alligation:

⇒ $\frac{a}{3}$ = $\frac{100-90}{90-p}$ = $\frac{10}{90-p}$   …(1)

This alloy is not mixed with another alloy of 2kg containing 90% silver.

Using alligation:

⇒ $\frac{a}{2}$ = $\frac{90-84}{84-p}$ = $\frac{6}{84-p}$   …(2)

From (1) and (2), we get

$\frac{30}{90-p}$ = $\frac{12}{84-p}$

⇒ 2520 – 30p = 1080 – 12p

⇒ 18p = 1440

⇒ p = 80%

∴ a = $\frac{30}{90-p}$ = $\frac{30}{90-80}$ = 3 kg.

Hence, option (c).

Answer: Option C

Video Explanation

Explanation :

Case 1: x, y > 0
⇒ 3x + 3y = 7 and 2x + 3y = 1
Solving these two equations we get, 
x = 6 and y = -11/3
This is rejected as y should be positive.

Case 2: x > 0, y < 0
⇒ 3x - y = 7 and 2x + 3y = 1
Solving these two equations we get, 
x = 2 and y = -1
This is accepted.
∴ x + 2y = 2 + 2 × -1 = 0

Case 3: x < 0, y > 0
⇒ 3x + 3y = 7 and 3y = 1
Solving these two equations we get, 
x = 2 and y = 1/3
This is rejected as x should be negative.

Case 4: x, y < 0
⇒ 3x - 3y = 7 and 3y = 1
Solving these two equations we get, 
x = 8/3 and y = 1/3
This is rejected as and y should be negative.

Hence, option (c).

Answer: Option A

Video Explanation

Explanation :

By substituting x = 1 or 2 or 3 and so on, we get

x₂ = x₁ + 1 – 1 = x₁ + 0
x₃ = x₂ + 2 – 1 = x₂ + 1 = x₁ + 0 + 1
x₄ = x₃ + 3 – 1 = x₃ + 2 = x₁ + 0 + 1 + 2
x₅ = x₄ + 4 – 1 = x₄ + 3 = x₁ + 0 + 1 + 2 + 3
…
x_n = x_n-1 +  0 + 1 + 2 + … + (n – 2)

⇒ x₁₀₀ = x₁ + 0 + 1 + 2 + … + 98 = x₁ + (98 × 99)/2

⇒ x₁₀₀ = -1 + 4851 = 4850

Hence, option (a).

Answer: 34

Video Explanation

Explanation :

Let the price of small, medium and large cups be Rs. 2x, 5x and p respectively.

⇒ 2x × 5x × p = 800   …(1)

Also, (2x + 6) × (5x + 6) × p = 3200   …(2)

(2) = (1) × 4

⇒ (2x + 6) × (5x + 6) × p = 2x × 5x × p × 4

⇒ 10x² + 42x + 36 = 40x²

⇒ 30x² - 42x - 36 = 0

⇒ 5x² – 7x – 6 = 0

⇒ 5x² – 10x + 3x – 6 = 0

⇒ (5x + 3)(x - 2) = 0

⇒ x = 2

∴ Price of small cup = 2x = 4
Price of medium cup = 5x = 10
Price of large cup = 800/(4 × 10) = 20

⇒ Sum of the prices of three cups = 4 + 10 + 20 = 34.

Hence, 34.

Answer: Option C

Video Explanation

Explanation :

Let the male and female population be ‘m’ and ‘f’ in 1970. 

Using alligation

$\frac{m}{f}$ = $\frac{25-20}{40-25}$ = $\frac{1}{3}$

∴ f = 3m

Total population in 1970 = f + m = 4m

⇒ Number of females in 1980 = 1.2f = 3.6m and number of males in 1980 = 1.4m

Population in 1990
Number of females = 1.25 × 3.6m = 4.5m
Number of males = ½ × 4.5m = 2.25m
Total population = 4.5m + 2.25m = 6.75m

⇒ Total population increases from 4m in 1970 to 6.75m in 1990.

∴ Percentage increase = 2.75m/4m×100 = 68.75%.

Alternately,
Let the number of males and females in 1970 be 'm' and 'f' respectively.

∴ In 1980, number of males = 1.4m, females = 1.2f and total population = 1.25(m + f)
⇒ 1.4m + 1.2f = 1.25(m + f)
⇒ 0.15m = 0.05f
⇒ m : f = 1 : 3

∴ In 1990, number of females = 1.25 × 1.2f = 1.5f, number of males = 1/2 × 1.5f = 0.75f 
∴ Total population in 1990 = 1.5f + 0.75f = 2.25f

Also, total population in 1970 = m + f = f + f/3 = 4f/3

⇒ Required percentage = $\left(\frac{2.25f-4f/3}{4f/3}\right)$ × 100% = $\frac{11}{16}$ × 100% = 68.75%

Hence, option (c).

Answer: 92

Video Explanation

Explanation :

Let the score of 5 toppers be x each.

To maximize the score of toppers we need to minimize the scores of the remaining 20 students.

∴ Score of remaining 25 students will be 30, 31, 32, … , 49.

⇒ 25 × 50 = 30 + 31 + … + 49 + 5x

⇒ 1250 = 790 + 5x

⇒ x = 92

Hence, 92.

Answer: Option B

Video Explanation

Explanation :

Let the price of a smaller shirt be Rs. p. Hence, price of a larger shirt will be Rs. (p + 50).

⇒ $\frac{5000}{p+50}$ + $\frac{1800}{p}$ = 64

⇒ 6800p + 90000 = 64p² + 3200p

⇒ 4p² - 225p - 5625 = 0

⇒ 4p² - 300p + 75p - 5625 = 0

⇒ (4p + 75)(p - 75) = 0

⇒ p = 75 or -75/4 (rejected)

⇒ Price of a smaller and a larger shirt = 75 + (75 + 50) = 200

Hence, option (b).

Answer: 12

Video Explanation

Explanation :

|1 + mn| < |m + n| < 5

Squaring all the expressions

(1 + mn)² < (m + n)² < 25

⇒ 1 + m²n² + 2mn < m² + n² + 2mn
⇒ m² + n² - m²n² – 1 > 0
⇒ (m² – 1) - n²(m² – 1) > 0
⇒ (m² – 1)(1 – n²) > 0
⇒ (m² – 1)(n² – 1) < 0

Case 1: Either m² > 1 and n² < 1
⇒ n = 0 and since |1 + mn| < |m + n| < 5
⇒ 1 < |m| < 5
⇒ m = ±2, ±3 or ±4
∴ 6 possible pairs of (m, n)

Case 2: Either n² > 1 and m² < 1
⇒ m = 0 and since |1 + mn| < |m + n| < 5
⇒ 1 < |n| < 5
⇒ n = ±2, ±3 or ±4
∴ 6 possible pairs of (m, n)

∴ Total 6 + 6 = 12 possible pairs of (m, n)

Hence, 12.