# CAT 2021 QA Slot 3

**1. CAT 2021 QA Slot 3 | Algebra - Logarithms**

For a real number a, if $\frac{lo{g}_{15}a+lo{g}_{32}a}{lo{g}_{15}a\times lo{g}_{32}a}$ = 4, then a must lie in the range

- A.
4 < a < 5

- B.
2 < a < 3

- C.
a > 5

- D.
3 < a < 4

Answer: Option A

**Explanation** :

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

Converting all logs to base 10.

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

⇒ log_{a}32 + log_{a}15 = 4

⇒ log_{a}(32 × 15) = 4

⇒ a^{4} = 480

This is possible when 4 < a < 5.

Hence, option (a).

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**2. CAT 2021 QA Slot 3 | Geometry - Triangles**

In a triangle ABC, ∠BCA = 50°. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then ∠FDE, in degrees is equal to

- A.
96

- B.
100

- C.
80

- D.
72

Answer: Option C

**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).

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**3. CAT 2021 QA Slot 3 | Algebra - Functions & Graphs**

If f(x) = x^{2} – 7x and g(x) = x + 3, then the minimum value of f(g(x)) – 3x is

- A.
-12

- B.
-16

- C.
-15

- D.
20

Answer: Option B

**Explanation** :

f(x) = x^{2} – 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)) = x^{2} - 4x – 12

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

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

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

Hence, option (b).

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**4. CAT 2021 QA Slot 3 | Arithmetic - Time & Work**

One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the

- A.
10

- B.
12

- C.
12.5

- D.
11.5

Answer: Option A

**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).

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**5. CAT 2021 QA Slot 3 | Algebra - Number Theory**

The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

- A.
160000

- B.
120000

- C.
100000

- D.
90000

Answer: Option B

**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).

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**6. CAT 2021 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is

Answer: 3500

**Explanation** :

Side of the rhombus = ¼ × 40 = 10 cm.

Area of a rhombus = ½ d_{1}d_{2} = 96 [d_{1} and d_{2} are the two diagonals]

⇒ d_{1}d_{2} = 192

We know, (d_{1}/2)^{2} + (d_{2}/2)^{2} = 102

⇒ d_{1}^{2} + d_{2}^{2} = 400

⇒ (d_{1} + d_{2})^{2} - 2d_{1}d_{2} = 400

⇒ (d_{1} + d_{2})^{2} – 2 × 192 = 400

⇒ (d_{1} + d_{2})^{2} = 784

⇒ d_{1} + d_{2} = 28

Hence, the cost of laying electric wore along its two diagonals = 125 × (d_{1} + d_{2}) = 125 × 28 = Rs. 3,500

Hence, 3500.

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**7. CAT 2021 QA Slot 3 | Arithmetic - Time & Work**

Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is

Answer: 3000

**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.

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

One part of a hostel’s monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be

- A.
20000

- B.
20200

- C.
20800

- D.
20500

Answer: Option D

**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).

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**9. CAT 2021 QA Slot 3 | Modern Math - Permutation & Combination**

A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Answer: 50

**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.

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**10. CAT 2021 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

Answer: 8

**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.

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**11. CAT 2021 QA Slot 3 | Algebra - Surds & Indices**

In ‘n’ is a positive integer such that $\left(\sqrt[7]{10}\right)$${\left(\sqrt[7]{10}\right)}^{2}$...${\left(\sqrt[7]{10}\right)}^{n}$ > 999, then the smallest value of n is

Answer: 6

**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.

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**12. CAT 2021 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC is equal to 30° then the area of the parallelogram is sq. cm. is

- A.
25(√3 + √15)/2

- B.
25(√5 + √15)

- C.
25(√3 + √15)

- D.
25(√5 + √15)/2

Answer: Option C

**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 30-60-90 triangle.

⇒ CE^{2} = AC^{2} – AE^{2}

⇒ CE^{2} = 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).

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**13. CAT 2021 QA Slot 3 | Arithmetic - Percentage**

In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be

- A.
78

- B.
84

- C.
80

- D.
86

Answer: Option B

**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).

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**14. CAT 2021 QA Slot 3 | Arithmetic - Simple & Compound Interest**

Bank A offers 6% interest rate per annum compounded half yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is

- A.
2346

- B.
3436

- C.
1436

- D.
2436

Answer: Option D

**Explanation** :

Let Raju invests Rs. A in bank B at r% p.a. for t years.

∴ Raju 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).

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**15. CAT 2021 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement**

If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

- A.
2.5

- B.
3.5

- C.
3

- D.
4

Answer: Option C

**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).

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**16. CAT 2021 QA Slot 3 | Algebra - Inequalities & Modulus**

If 3x + 2|y| + y = 7 and x + |x| + 3y = 1, then x + 2y is

- A.
8/3

- B.
1

- C.
0

- D.
-4/3

Answer: Option C

**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).

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**17. CAT 2021 QA Slot 3 | Algebra - Progressions**

Consider a sequence of real numbers x_{1}, x_{2}, x_{3}, … such that x_{n+1} = x_{n} + n – 1 for all n ≥ 1. If x_{1} = -1 then x_{100}

- A.
4850

- B.
4849

- C.
4950

- D.
4949

Answer: Option A

**Explanation** :

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

x_{2} = x_{1} + 1 – 1 = x_{1} + 0

x_{3} = x_{2} + 2 – 1 = x_{2} + 1 = x_{1} + 0 + 1

x_{4} = x_{3} + 3 – 1 = x_{3} + 2 = x_{1} + 0 + 1 + 2

x_{5} = x_{4} + 4 – 1 = x_{4} + 3 = x_{1} + 0 + 1 + 2 + 3

…

x_{n} = x_{n-1} + 0 + 1 + 2 + … + (n – 2)

⇒ x_{100} = x_{1} + 0 + 1 + 2 + … + 98 = x_{1} + (98 × 99)/2

⇒ x_{100} = -1 + 4851 = 4850

Hence, option (a).

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**18. CAT 2021 QA Slot 3 | Algebra - Quadratic Equations**

A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

Answer: 34

**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

⇒ 10x^{2} + 42x + 36 = 40x^{2}

⇒ 30x^{2} - 42x - 36 = 0

⇒ 5x^{2} – 7x – 6 = 0

⇒ 5x^{2} – 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.

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**19. CAT 2021 QA Slot 3 | Arithmetic - Percentage**

The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is

- A.
68.50

- B.
69.25

- C.
68.75

- D.
68.25

Answer: Option C

**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 popylation 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%.

Hence, option (c).

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**20. CAT 2021 QA Slot 3 | Arithmetic - Average**

The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Answer: 92

**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.

Workspace:

**21. CAT 2021 QA Slot 3 | Algebra - Quadratic Equations**

A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was INR 50 less than that of a large shirt. She paid a total of INR 5000 for the large shirts, and a total of INR 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in INR, is

- A.
175

- B.
200

- C.
150

- D.
225

Answer: Option B

**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

⇒ 4p^{2} - 225p - 5625 = 0

⇒ 4p^{2} - 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).

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**22. CAT 2021 QA Slot 3 | Algebra - Inequalities & Modulus**

The number of distinct pairs of integers (m, n) satisfying |1 + mn| < |m + n| < 5 is

Answer: 12

**Explanation** :

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

Squaring all the expressions

(1 + mn)^{2} < (m + n)^{2} < 25

⇒ 1 + m^{2}n^{2} + 2mn < m^{2} + n^{2} + 2mn

⇒ m^{2} + n^{2} - m^{2}n^{2} – 1 > 0

⇒ (m^{2} – 1) - n^{2}(m^{2} – 1) > 0

⇒ (m^{2} – 1)(1 – n^{2}) > 0

⇒ (m^{2} – 1)(n^{2} – 1) < 0

Case 1: Either m^{2} > 1 and n^{2} < 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 n^{2} > 1 and m^{2} < 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.

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