# CAT 2021 QA Slot 2 | Previous Year CAT Paper

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**1. CAT 2021 QA Slot 2 | Algebra - Simple Equations**

A box has 450 balls, each either white or black, there being as many metallic white balls as metallic black balls. If 40% of the white balls and 50% of the black balls are metallic, then the number of non-metallic balls in the box is

Answer: 250

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**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.

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**2. CAT 2021 QA Slot 2 | Algebra - Number Theory**

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Answer: 4195

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**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.

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

A person buys tea of three different qualities at ₹ 800, ₹ 500, and ₹ 300 per kg, respectively, and the amounts bought are in the proportion 2 : 3 : 5. She mixes all the tea and sells one-sixth of the mixture at ₹ 700 per kg. The price, in INR per kg, at which she should sell the remaining tea, to make an overall profit of 50%, is

- A.
692

- B.
688

- C.
675

- D.
653

Answer: Option B

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**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).

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**4. CAT 2021 QA Slot 2 | Arithmetic - Time & Work**

Anil can paint a house in 60 days while Bimal can paint it in 84 days. Anil starts painting and after 10 days, Bimal and Charu join him. Together, they complete the painting in 14 more days. If they are paid a total of ₹ 21000 for the job, then the share of Charu, in INR, proportionate to the work done by him, is

- A.
9100

- B.
9000

- C.
9150

- D.
9200

Answer: Option A

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**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).

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**5. CAT 2021 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

The sides AB and CD of a trapezium ABCD are parallel, with AB being the smaller side. P is the midpoint of CD and ABPD is a parallelogram. If the difference between the areas of the parallelogram ABPD and the triangle BPC is 10 sq cm, then the area, in sq cm, of the trapezium ABCD is

- A.
20

- B.
30

- C.
40

- D.
25

Answer: Option B

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**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).

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**6. CAT 2021 QA Slot 2 | Modern Math - Permutation & Combination**

The number of ways of distributing 15 identical balloons, 6 identical pencils and 3 identical erasers among 3 children, such that each child gets at least four balloons and one pencil, is

Answer: 1000

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**Explanation** :

We know that number of distributing n identical objects amongst r different groups = ^{n+r-1}C_{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+3-1}C_{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+3-1}C_{3-1} = 10 ways

Distributing erasers

We have 3 identical erasers to be distributed to 3 children.

Number of ways of doing this = ^{3+3-1}C_{3-1} = 10 ways

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

Hence, 1000.

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**7. CAT 2021 QA Slot 2 | Algebra - Number Theory**

Consider the pair of equations: x^{2} – xy – x = 22 and y^{2} – xy + y = 34. If x > y, then x – y equal.

- A.
8

- B.
7

- C.
6

- D.
4

Answer: Option A

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**Explanation** :

Given x^{2} – xy – x = 22 and y^{2} – xy + y = 34.

Adding both the equations

⇒ x^{2} + y^{2} – 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

⇒ a^{2} – a – 56 = 0

⇒ (a – 8)(a + 7) = 0

⇒ a = 8 or -7 (rejected)

∴ a = x - y = 8

Hence, option (a).

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**8. CAT 2021 QA Slot 2 | Arithmetic - Mixture, Alligation, Removal & Replacement**

From a container filled with milk, 9 litres of milk are drawn and replaced with water. Next, from the same container, 9 litres are drawn and again replaced with water. If the volumes of milk and water in the container are now in the ratio of 16 : 9, then the capacity of the container, in litres, is

Answer: 45

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**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.

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**9. CAT 2021 QA Slot 2 | Arithmetic - Time, Speed & Distance**

Two trains A and B were moving in opposite directions, their speeds being in the ratio 5 : 3. The front end of A crossed the rear end of B 46 seconds after the front ends of the trains had crossed each other. It took another 69 seconds for the rear ends of the trains to cross each other. The ratio of length of train A to that of train B is

- A.
2 : 1

- B.
3 : 2

- C.
2 : 3

- D.
5 : 3

Answer: Option B

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**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**:

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**10. CAT 2021 QA Slot 2 | Arithmetic - Average**

In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

Answer: 10

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**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.

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**11. CAT 2021 QA Slot 2 | Algebra - Logarithms**

log_{2 }[3 + log_{3} {4 + log_{4} (x - 1)}] - 2 = 0, then 4x equals

Answer: 5

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**Explanation** :

Given, log_{2}[3 + log_{3}{4 + log_{4}(x - 1)}] - 2 = 0

⇒ log_{2}[3 + log_{3}{4 + log_{4}(x - 1)}] = 2

⇒ 3 + log_{3}{4 + log_{4}(x - 1)} = 4

⇒ log_{3}{4 + log_{4}(x - 1)} = 1

⇒ 4 + log_{4}(x - 1) = 3

⇒ log_{4}(x - 1) = -1

⇒ x – 1 = ¼

⇒ x = 5/4

⇒ 4x = 5

Hence, 5.

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**12. CAT 2021 QA Slot 2 | Arithmetic - Percentage**

Raj invested ₹ 10000 in a fund. At the end of first year, he incurred a loss but his balance was more than ₹ 5000. This balance, when invested for another year, grew and the percentage of growth in the second year was five times the percentage of loss in the first year. If the gain of Raj from the initial investment over the two year period is 35%, then the percentage of loss in the first year is

- A.
10

- B.
15

- C.
5

- D.
70

Answer: Option A

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**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 – 5P^{2} + 400P = 13500

⇒ 5P^{2} - 400P + 3500 = 0

⇒ P^{2} – 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).

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**13. CAT 2021 QA Slot 2 | Algebra - Progressions**

For a sequence of real numbers x_{1}, x_{2}, …, x_{n}, if x_{1} - x_{2} + x_{3} - … + (-1)^{(n+1)} x_{n} = n^{2 }+ 2n for all natural numbers n, then the sum x_{49} + x_{50} equals.

- A.
-2

- B.
2

- C.
-200

- D.
200

Answer: Option A

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**Explanation** :

Given, x_{1} - x_{2 }+ x_{3 }- … + (-1)^{(n+1)} x_{n }= n^{2 }+ 2n

Put n = 1, ⇒ x_{1} = 1^{2} + 2 × 1 = 3

Put n = 2, ⇒ x_{1} - x_{2} = 2^{2} + 2 × 2 = 8 ⇒ x_{2} = 3 – 8 = -5

Put n = 3, ⇒ x_{1} - x_{2} + x_{3} = 3^{2} + 2 × 3 = 15 ⇒ x_{3} = 8 – 15 = -7

…

⇒ x_{n} = (-1)^{n+1} × (2n+1)

∴ x_{49} = 99 and x_{50} = -101

⇒ x_{49} + x_{50} = 99 – 101 = -2

Hence, option (a).

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**14. CAT 2021 QA Slot 2 | Arithmetic - Ratio, Proportion & Variation**

Anil, Bobby and Chintu jointly invest in a business and agree to share the overall profit in proportion to their investments. Anil’s share of investment is 70%. His share of profit decreases by ₹ 420 if the overall profit goes down from 18% to 15%. Chintu’s share of profit increases by ₹ 80 if the overall profit goes up from 15% to 17%. The amount, in INR, invested by Bobby is

- A.
2000

- B.
2400

- C.
2200

- D.
1800

Answer: Option A

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**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).

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**15. CAT 2021 QA Slot 2 | Algebra - Surds & Indices**

For all possible integers n satisfying 2.25 ≤ 2 + 2^{n+2} ≤ 202, the number of integer values of 3 + 3^{n+1} is

Answer: 7

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**Explanation** :

Given, 2.25 ≤ 2 + 2^{n+2} ≤ 202

⇒ 0.25 ≤ 2^{n+2} ≤ 200

Case 1: 0.25 ≤ 2^{n+2}

∴ n + 2 ≥ -2

⇒ n ≥ -4

Case 2: 2^{n+2} ≤ 200

This is true when n ≤ 5

∴ -4 ≤ n ≤ 5

Now, 3 + 3^{n+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.

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**16. CAT 2021 QA Slot 2 | Algebra - Inequalities & Modulus**

For a real number x the condition |3x - 20| + |3x - 40| = 20 necessarily holds if

- A.
6 < x < 11

- B.
9 < x < 14

- C.
10 < x < 15

- D.
7 < x < 12

Answer: Option D

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**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).

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**17. CAT 2021 QA Slot 2 | Algebra - Quadratic Equations**

Suppose one of the roots of the equation ax^{2} – bx + c = 0 is 2 + √3, where a, b and c are rational numbers and a ≠ 0. If b = c^{3} then |a| equals

- A.
2

- B.
4

- C.
3

- D.
1

Answer: Option A

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**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 = c^{3}

⇒ 4a = a^{3}

⇒ a^{2} = 4

⇒ a = ±2

∴ |a| = 2

Hence, option (a).

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**18. CAT 2021 QA Slot 2 | Arithmetic - Time & Work**

Two pipes A and B are attached to an empty water tank. Pipe A fills the tank while pipe B drains it. If pipe A is opened at 2 pm and pipe B is opened at 3 pm, then the tank becomes full at 10 pm. Instead, if pipe A is opened at 2 pm and pipe B is opened at 4 pm, then the tank becomes full at 6 pm. If pipe B is not opened at all, then the time, in minutes, taken to fill the tank is:

- A.
140

- B.
120

- C.
144

- D.
264

Answer: Option C

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**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).

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**19. CAT 2021 QA Slot 2 | Algebra - Progressions**

Three positive integers x, y and z are in arithmetic progression. If y − x > 2 and xyz = 5(x + y + z), then z − x equals

- A.
12

- B.
8

- C.
14

- D.
10

Answer: Option C

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**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).

Workspace:

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

Let D and E be points on sides AB and AC, respectively, of a triangle ABC, such that AD : BD = 2 : 1 and AE : CE = 2 : 3. If the area of the triangle ADE is 8 sq cm, then the area of the triangle ABC, in sq cm, is

Answer: 30

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**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.

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**21. CAT 2021 QA Slot 2 | Geometry - Quadrilaterals & Polygons**

If a rhombus has area 12 sq cm and side length 5 cm, then the length, in cm, of its longer diagonal is

- A.
√13 + √12

- B.
√37 + √13

- C.
(√13 + √12)/2

- D.
(√37 + √13)/2

Answer: Option B

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**Explanation** :

Consider the diagram below.

In a rhombus diagonals perpendicularly bisect each other.

∴ Let OA = OC = x and OB = OD = y

In ∆AOB ⇒ x^{2} + y^{2} = 5^{2} …(1)

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

⇒ 2xy = 12 …(2)

(1) + (2)

⇒ x^{2} + y^{2} + 2xy = 25 + 12 = 37

⇒ (x + y)^{2} = 37

⇒ x + y = √37 …(3)

(1) - (2)

⇒ x^{2} + y^{2} - 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).

Workspace:

**22. CAT 2021 QA Slot 2 | Algebra - Quadratic Equations**

For all real values of x, the range of the function f(x) = $\frac{{x}^{2}+2x+4}{2{x}^{2}+4x+9}$ is

- A.
$\left[\frac{3}{7},\frac{1}{2}\right)$

- B.
$\left[\frac{3}{7},\frac{8}{9}\right)$

- C.
$\left[\frac{4}{9},\frac{1}{2}\right]$

- D.
$\left(\frac{3}{7},\frac{1}{2}\right)$

Answer: Option A

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**Explanation** :

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

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

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

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

f(x) will be minimum when (2x^{2} + 4x + 9) is minimum.

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

∴ Minimum value of 2x^{2} + 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 (2x^{2} + 4x + 9) is maximum.

Maximum value of (2x^{2} + 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 (2x^{2} + 4x + 9) will never actually be ∞.

Hence, option (a).

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