# CAT 2022 QA Slot 3

**1. CAT 2022 QA Slot 3 | Algebra - Inequalities & Modulus**

The minimum possibe value of $\frac{{x}^{2}-6x+10}{3-x}$, for x < 3, is

- A.
-2

- B.
$\frac{1}{2}$

- C.
2

- D.
-$\frac{1}{2}$

Answer: Option C

**Explanation** :

We have $\frac{{\mathrm{x}}^{2}-6\mathrm{x}+10}{3-\mathrm{x}}$

= $\frac{{(\mathrm{x}-3)}^{2}+1}{3-\mathrm{x}}$

= $\frac{{(\mathrm{x}-3)}^{2}}{3-\mathrm{x}}$ + $\frac{1}{3-\mathrm{x}}$

= (3 - x) + $\frac{1}{3-\mathrm{x}}$

Here, since x < 3, 3 – x > 0

Also, we know that sum of a positive number and its reciprocal is always greater than or equal to 2.

⇒ (3 - x) + $\frac{1}{3-\mathrm{x}}$ ≥ 2

Hence, option (c).

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**2. CAT 2022 QA Slot 3 | Geometry - Circles**

In a triangle ABC, AB = AC = 8 cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then, the area in sq. cm, of the overlapping region between the two circles is

- A.
16(π - 1)

- B.
32(π - 1)

- C.
16π

- D.
32π

Answer: Option B

**Explanation** :

Since AB is the diameter of the smaller circle ⇒ ∠ACB = 90°.

In ∆ACB, AC^{2} + BC^{2} = AB^{2}

⇒ AB = √(8^{2} + 8^{2}) = 8√2

∴ Radius of smaller circle is ½ × 8√2 = 4√2

Diameter of larger circle = 8

**Smaller Circle**:

Area of green part = 1/2 × π × (4√2)^{2} = 16π

**Larger Circle**:

Area of blue part = Area of sector CAB – Area of ∆CAB

Area of blue part = 1/4 × π × 8^{2} - 1/2 × 8 × 8 = 16π – 32

∴ Required area = 16π + (16π - 32) = 32(π - 1)

Hence, option (b).

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**3. CAT 2022 QA Slot 3 | Arithmetic - Time & Work**

A group of N people worked on a project. They finished 35% of the project by working 7 hours a day for 10 days. Thereafter, 10 people left the group and the remaining people finished the rest of the project in 14 days by working 10 hours a day. Then the value of N is

- A.
23

- B.
140

- C.
36

- D.
150

Answer: Option B

**Explanation** :

Let the efficiency of each person be ‘e’ units/hour.

Given,

35% work is done by N men in 10 days working for 7 hours/day, while

65% work is done by N-10 men in 14 days working for 10 hours/day

⇒ $\frac{\mathrm{N}\times 10\times 7}{0.35}$= $\frac{(\mathrm{N}-10)\times 14\times 10}{0.65}$

⇒ $\frac{\mathrm{N}\times 7}{7}$= $\frac{(\mathrm{N}-10)\times 14}{13}$

⇒ N = $\frac{(\mathrm{N}-10)\times 14}{13}$

⇒ 13N = 14N – 140

⇒ N = 140

Hence, option (b).

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**4. CAT 2022 QA Slot 3 | Algebra - Surds & Indices**

If ${\left(\sqrt{\frac{7}{5}}\right)}^{3x-y}$ = $\frac{875}{2401}$ and ${\left(\frac{4a}{b}\right)}^{6x-y}$ = ${\left(\frac{2a}{b}\right)}^{y-6x},$ for all non-zero real values of a and b, then the value of x + y is

Answer: 14

**Explanation** :

Givne, ${\left(\sqrt{\frac{7}{5}}\right)}^{3x-y}$ = $\frac{875}{2401}$

⇒ ${\left(\frac{7}{5}\right)}^{\frac{3x-y}{2}}$ = $\frac{125\times 7}{7\times 7\times 7\times 7}$ = ${\left(\frac{5}{7}\right)}^{3}$ = ${\left(\frac{7}{5}\right)}^{-3}$

⇒ 3x - y = - 6

⇒ y = 3x + 6 ...(1)

Also, ${\left(\frac{4a}{b}\right)}^{6x-y}$ = ${\left(\frac{2a}{b}\right)}^{y-6x}$ = ${\left(\frac{b}{2a}\right)}^{6x-y}$

⇒ ${\left(\frac{4\mathrm{a}}{\mathrm{b}}\times \frac{2\mathrm{a}}{\mathrm{b}}\right)}^{6\mathrm{x}-\mathrm{y}}$ = 1

⇒ ${\left(\frac{8{\mathrm{a}}^{2}}{{\mathrm{b}}^{2}}\right)}^{6\mathrm{x}-\mathrm{y}}$ = 1

This is true for all values of a and b. This is only possible when 6x - y = 0

⇒ y = 6x ...(2)

Solving (1) and (2), we get x = 2 and y = 12.

⇒ x + y = 2 + 12 = 14.

Hence, 14.

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**5. CAT 2022 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Moody takes 30 seconds to finish riding an escalator if he walks on it at his normal speed in the same direction. He takes 20 seconds to finish riding the escalator if he walks at twice his normal speed in the same direction. If Moody decides to stand still on the escalator, then the time, in seconds, needed to finish riding the escalator is

Answer: 60

**Explanation** :

Let speed of Moody and escalator be ‘m’ and ‘e’ steps/sec and total number of steps in the escalator be ‘N’.

We will assume that Moody is going in the same direction as the escalator. If that is not the case value of e will come out to be negative.

Moody takes 30 seconds at normal speed ⇒ N = 30m + 30 ×e …(1)

Moody takes 20 seconds at twice the speed ⇒ N = 20 × 2m + 20 ×e …(2)

4 × (1) – 3 × (2)

N = 60e

⇒ N/e = 60

∴ If someone was standing on the escalator, he/she will take 60 seconds to climb the escalator.

Hence, 60.

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**6. CAT 2022 QA Slot 3 | Arithmetic - Average**

In an examination, the average marks of students in sections A and B are 32 and 60, respectively. The number of students in section A is 10 less than that in section B. If the average marks of all the students across both the sections combined is an integer, then the difference between the maximum and minimum possible number of students in section A is

Answer: 63

**Explanation** :

Let the number of students in section B is x and that in A is (x – 10).

⇒ $\frac{(\mathrm{x}-10)\times 32+\mathrm{x}\times 60}{2\mathrm{x}-10}$ = integer

⇒ $\frac{92\mathrm{x}-320}{2\mathrm{x}-10}$ = integer

⇒ $\frac{46\mathrm{x}-160}{\mathrm{x}-5}$ = integer

⇒ $\frac{46(\mathrm{x}-5)+70}{\mathrm{x}-5}$ = integer

⇒ (x - 5) + $\frac{70}{\mathrm{x}-5}$ = integer

⇒ $\frac{70}{\mathrm{x}-5}$ should be an integer

∴ (x – 5) should be a factor of 70 while x > 10

⇒ Highest possible value of x = 75 while lowest possible value is 12

∴ Difference between highest and lowest values of x = 75 – 12 = 63.

Hence, 63.

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**7. CAT 2022 QA Slot 3 | Arithmetic - Time, Speed & Distance**

Two cars travel from different locations at constant speeds. To meet each other after starting at the same time, they take 1.5 hours if they travel towards each other, but 10.5 hours if they travel in the same direction. If the speed of the slower car is 60 km/hr, then the distance travelled, in km, by the slower car when it meets the other car while traveling towards each other, is

- A.
90

- B.
120

- C.
100

- D.
150

Answer: Option A

**Explanation** :

When travelling towards each other, the two cars take 1.5 hours to meet.

⇒ The distance travelled by slower car in this case till they meet = 1.5 × 60 = 90 kms.

Hence, option (a).

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**8. CAT 2022 QA Slot 3 | Modern Math - Permutation & Combination**

The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421, including itself, is

- A.
2222

- B.
2592

- C.
2442

- D.
3333

Answer: Option A

**Explanation** :

8. Let us consider the sum of unit’s digit of all such numbers.

__ __ __ __ x __

If 1 occupies the units digit, number of such numbers = 3! = 6

Sum of all the units digit of such numbers = 6 × 1 = 6

If 2 occupies the units digit, number of such numbers = 3!/2! = 3

Sum of all the units digit of such numbers = 3 × 2 = 6

If 4 occupies the units digit, number of such numbers = 3!/2! = 3

Sum of all the units digit of such numbers = 3 × 4 = 12

∴ Sum of unit’s digits of all possible numbers = 6 + 6 + 12 = 24

Similarly,

Sum of ten’s digits of all possible numbers = 24 × 10 = 240

Sum of hundred’s digits of all possible numbers = 24 × 100 = 2400

Sum of thousand’s digits of all possible numbers = 24 × 1000 = 24000

⇒ Sum of all such numbers = 24 + 240 + 2400 + 24000 = 26664

Also, number of such numbers = 4!/2! = 12

∴ Average of all such numbers = 2664/12 = 2222

Hence, option (a).

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**9. CAT 2022 QA Slot 3 | Algebra - Quadratic Equations**

Suppose k is any integer such that the equation 2x^{2} + kx + 5 = 0 has no real roots and the equation x^{2} + (k - 5)x + 1 = 0 has two distinct real roots for x. Then, the number of possible values of k is

- A.
8

- B.
7

- C.
9

- D.
13

Answer: Option C

**Explanation** :

2x^{2} + kx + 5 = 0 has no real roots ⇒ D < 0

⇒ k^{2} – 4 × 2 × 5 < 0

⇒ k^{2} < 40

⇒ -√40 < k < √40

∴ Possible integral values of k are -6, -5, -4, …, 0, …4, 5, 6 …(1)

Also, x^{2} + (k - 5)x + 1 = 0 has two distinct roots ⇒ D > 0

⇒ (k - 5)^{2} – 4 × 1 × 1 > 0

⇒ k^{2} + 25 – 10k – 4 > 0

⇒ k^{2} – 10k + 21 > 0

⇒ (k - 7)(k - 3) > 0

⇒ k ∈ (-∞, 3) ∪ (7, ∞) …(2)

The integral value of k satisfying both (1) and (2) are

-6, -5, -4, -3, -2, -1, 0, 1, 2 i.e., 9 values.

Hence, option (c).

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**10. CAT 2022 QA Slot 3 | Geometry - Quadrilaterals & Polygons**

The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number of possible lengths of the fourth side is

- A.
5

- B.
4

- C.
3

- D.
6

Answer: Option A

**Explanation** :

In a quadrilateral, the longest side is less than the sum of other three sides and greater than the least of the difference of any 2 of the other three sides.

Let the fourthe sides of the quadrilateral be x.

⇒ 2 - 1 < x < 1 + 2 + 4

⇒ 1 < x < 7

∴ x can be 2, 3, 4, 5 or 6

Hence, x can take 5 integral values.

Hence, option (a).

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**11. CAT 2022 QA Slot 3 | Geometry - Triangles**

Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

- A.
6

- B.
8

- C.
12

- D.
4

Answer: Option C

**Explanation** :

Initially both ships are at a distance of 24 kms from the port (A).

Let the distance travelled by faster ship be F kms when slower ship travels 8 kms.

Now, ∆ABC is a right triangle when ∠BAC = 60°

⇒ AC = ½ × AB = 8 kms.

∴ 24 – F = 8

⇒ F = 16 kms

Ratio of speeds of faster and slower ships = 16 : 8 = 2 : 1

Now, when the faster ships reaches port it would have travelled 24 kms and the slower ship would have travelled 12 kms.

∴ Slower ship will be 24 – 12 = 12 kms from the port.

Hence, option (c).

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**12. CAT 2022 QA Slot 3 | Algebra - Inequalities & Modulus**

If c = $\frac{16x}{y}$ + $\frac{49y}{x}$ for some non-zero real numbers x and y, then c cannot take the value

- A.
-60

- B.
-70

- C.
60

- D.
-50

Answer: Option D

**Explanation** :

Given, c = $\frac{16x}{y}$ + $\frac{49y}{x}$

Let $\frac{\mathrm{x}}{\mathrm{y}}$ = a

⇒ c = 16a + 49/a

**Now, for two positive numbers, AM ≥ GM**

If a > 0

⇒ $\frac{16\mathrm{a}+{\displaystyle \frac{49}{\mathrm{a}}}}{2}$ ≥ $\sqrt{16\mathrm{a}\times \frac{49}{\mathrm{a}}}$

⇒ $\frac{\mathrm{c}}{2}$ ≥ 28

⇒ c ≥ 56

∴ option (c) is rejected.

**Now, for two negative numbers, AM ≤ - GM**

If a < 0

⇒ $\frac{16\mathrm{a}+{\displaystyle \frac{49}{\mathrm{a}}}}{2}$ ≤ - $\sqrt{16\mathrm{a}\times \frac{49}{\mathrm{a}}}$

⇒ $\frac{\mathrm{c}}{2}$ ≤ - 28

⇒ c ≤ - 56

∴ option (a) and (b) are rejected.

∴ c cannot be equal to -50

Hence, option (d).

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**13. CAT 2022 QA Slot 3 | Algebra - Progressions**

The average of all 3-digit terms in the arithmetic progression 38, 55, 72, ..., is

Answer: 548

**Explanation** :

38, 55, 72, … forms an AP whose first term is 38 and common difference is 17.

∴ T_{n} = 38 + (n - 1) × 17

To find the average we need to find the highest and lowest 3-digit numbers of this sequence.

**Lowest**: 38 + (n - 1) × 17 > 99

⇒ 17n - 17 > 61

⇒ 17n > 78

⇒ n > 4

∴ Least possible value of n = 5

⇒ Least such number = 38 + 4 × 17 = 106

**Highest**: 38 + (n - 1) × 17 < 999

⇒ 17n - 17 < 961

⇒ 17n < 978

⇒ n < 56.5

∴ Highest possible value of n = 56

⇒ Highest such number = 38 + 56 × 17 = 990

∴ The average of the sequence (AP) is same as the average of lowest and highest terms = $\frac{106+990}{2}$ = 548

Hence, 548.

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**14. CAT 2022 QA Slot 3 | Algebra - Number Theory**

A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

Answer: 150

**Explanation** :

The number of students will be of the form = LCM(9, 10, 12, 25) × k + 4 = 900k + 4

The number of students is less than 500 and also completely divisible by 11. This is possible when k = 2 and hence the number the students = 1804.

Maximum number of groups of 12 students that can be formed = Quotient of [1804 / 12] = 150

Hence, 150.

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**15. CAT 2022 QA Slot 3 | Arithmetic - Average**

Consider six distinct natural numbers such that the average of the two smallest numbers is 14, and the average of the two largest numbers is 28. Then, the maximum possible value of the average of these six numbers is

- A.
23

- B.
22.5

- C.
24

- D.
23.5

Answer: Option B

**Explanation** :

The average of lowest two number is 14, hence their sum = 28

The average of highest two number is 28, hence their sum = 56

To maximize the average of these six numbers, we need to maximize the middle two numbers. But it is also given that all the numbers are distinct.

Since all numbers are distinct out of the two highest numbers, one must be greater than 28 (a) and the other less than 28 (b).

To maximize the middle two numbers we need to maximize ‘b’. The maximum value ‘b’ can take is 27.

∴ Maximum value of the two middle numbers can be 25 and 26.

⇒ Maximum sum of the six numbers = 28 + 25 + 26 + 56 = 135

⇒ Maximum average of the six numbers = 135/6 = 22.5

Hence, option (b).

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**16. CAT 2022 QA Slot 3 | Algebra - Functions & Graphs**

Let r be a real number and f(x) = $\left\{\begin{array}{cc}2\mathrm{x}-\mathrm{r}& \mathrm{if}\mathrm{x}\ge \mathrm{r}\\ \mathrm{r}& \mathrm{if}\mathrm{x}\mathrm{r}\end{array}.\right.$Then, the equation f(x) = f(f(x)) holds for all real values of x where

- A.
x ≠ r

- B.
x ≥ r

- C.
x > r

- D.
x ≤ r

Answer: Option D

**Explanation** :

**Case 1**: x < r

⇒ f(x) = r

⇒ f(f(x)) = f(f(r)) = 2r – r = r

∴ f(x) = f(f(x))

**Case 2**: x = r

⇒ f(x) = 2r – r = r

⇒ f(f(x)) = f(f(r)) = 2r – r = r

∴ f(x) = f(f(x))

**Case 3**: x > r

⇒ f(x) = 2x – r > r

⇒ f(f(x)) = f(f(2x - r)) = 2(2x - r) – r = 4x - 3r

∴ f(x) ≠ f(f(x))

⇒ f(x) = f(f(x) when x ≤ r

Hence, option (d).

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**17. CAT 2022 QA Slot 3 | Geometry - Triangles**

Suppose the medians BD and CE of a triangle ABC intersect at a point O. If area of triangle ABC is 108 sq. cm., then, the area of the triangle EOD, in sq. cm., is

Answer: 9

**Explanation** :

BD, CE and AF are medians of the triangle ABC.

We know centroid (O) divides the triangle in 6 smaller triangles of equal area.

⇒ Area(∆EOB) = Area(∆BOF) = Area(∆FOC) = Area(∆COD) = 1/6 × 108 = 18.

∆AED ~ ∆ABC

⇒ $\frac{\mathrm{Area}(\u2206\mathrm{AED})}{\mathrm{Area}(\u2206\mathrm{ABC})}$ = ${\left(\frac{\mathrm{AE}}{\mathrm{AB}}\right)}^{2}$ = $\frac{1}{4}$

⇒ Area(EBCD) = ¾ × 108 = 81

Now,

Area(∆EOD) = Area(EBCD) – [Area(∆EOB) + Area(∆BOF) + Area(∆FOC) + Area(∆COD)]

= 81 – 72 = 9

Hence, 9.

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**18. CAT 2022 QA Slot 3 | Algebra - Quadratic Equations**

If (3 + 2$\sqrt{2}$) is a root of the equation ax^{2} + bx + c = 0, and (4 + 2$\sqrt{3}$) is a root of the equation ay^{2} + my + n = 0, where a, b, c, m and n are integers, then the value of $\left(\frac{b}{m}+\frac{c-2b}{n}\right)$ is

- A.
1

- B.
0

- C.
4

- D.
3

Answer: Option C

**Explanation** :

(3 + 2√2) is a root of the equation ax2 + bx + c = 0 whose coefficients are integers

⇒ Since coefficients are rational the other root will be (3 - 2√2)

∴ Sum of roots = 6 = -b/a

⇒ b = -6a …(1)

∴ product of roots = 1 = c/a

⇒ c = a …(2)

(4 + 2√3) is a root of the equation ay2 + my + n = 0 whose coefficients are integers

⇒ Since coefficients are rational the other root will be (4 - 2√3)

∴ Sum of roots = 8 = -m/a

⇒ m = -8a …(3)

∴ product of roots = 4 = n/a

⇒ n = 4a …(4)

Now, $\left(\frac{\mathrm{b}}{\mathrm{m}}+\frac{\mathrm{c}-2\mathrm{b}}{\mathrm{n}}\right)$

= $\left(\frac{-6\mathrm{a}}{-8\mathrm{a}}+\frac{\mathrm{a}+12\mathrm{a}}{4\mathrm{a}}\right)$

= $\frac{3}{4}+\frac{13}{4}$ = 4

Hence, option (c).

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**19. CAT 2022 QA Slot 3 | Arithmetic - Time & Work**

Bob can finish a job in 40 days, if he works alone. Alex is twice as fast as Bob and thrice as fast as Cole in the same job. Suppose Alex and Bob work together on the first day, Bob and Cole work together on the second day, Cole and Alex work together on the third day, and then, they continue the work by repeating this three-day roster, with Alex and Bob working together on the fourth day, and so on. Then, the total number of days Alex would have worked when the job gets finished, is

Answer: 11

**Explanation** :

Bob takes 40 days to finish the work. Alex takes is twice as fast as Bob and hence will take half the time taken by Bob i.e., 20 days. Alex is also thrice as fast as Cole, hence Cole will take thrice the time taken by Alex, i.e., 60 days.

Time take by

Alex = 20 days

Bob = 40 days

Cole = 60 days

Let the total work to be done = 120 units.

∴ Efficiency of

Alex = 6 units/day

Bob = 3 units/day

Cole = 2 units/day

Work done/cycle = 22 units.

∴ Work done in 5 cycles = 5 × 22 = 110 units

Work left after 5 cycles (15 days) = 120 – 110 = 10 units.

On 16th day Alex and Bob will together complete 9 units of work while remaining 1 unit of work will be completed by Bob and Cole on 17th day.

∴ Alex worked for 10 days in 5 complete cycles + on 16th day i.e., total 11 days.

Hence, 11.

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**20. CAT 2022 QA Slot 3 | Arithmetic - Simple & Compound Interest**

Nitu has an initial capital of Rs. 20,000. Out of this, she invests Rs. 8,000 at 5.5% in bank A, Rs. 5,000 at 5.6% in bank B and the remaining amount at x% in bank C, each rate being simple interest per annum. Her combined annual interest income from these investments is equal to 5% of the initial capital. If she had invested her entire initial capital in bank C alone, then her annual interest income, in rupees, would have been

- A.
800

- B.
700

- C.
900

- D.
1000

Answer: Option A

**Explanation** :

Rs. 8000 is invested at 5.5% earning yearly interest of 8000 × 5.5% = Rs. 440

Rs. 5000 is invested at 5.6% earning yearly interest of 5000 × 5.6% = Rs. 280

Rs. 7000 is invested at x% earning yearly interest of 7000 × x% = Rs. 70x

Overall, 20,000 is invested which earns 5% yearly interest = 5% of 20000 = 1000

⇒ 440 + 280 + 70x = 1000

⇒ 70x = 1000 – 720 = 280

⇒ x = 4%

∴ If she had invested her entire initial capital in bank C alone, then her annual interest income = 4% of 20000 = Rs. 800

Hence, option (a).

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**21. CAT 2022 QA Slot 3 | Arithmetic - Mixture, Alligation, Removal & Replacement**

A glass contains 500 cc of milk and a cup contains 500 cc of water. From the glass, 150 cc of milk is transferred to the cup and mixed thoroughly. Next, 150 cc of this mixture is transferred from the cup to the glass. Now, the amount of water in the glass and the amount of milk in the cup are in the ratio

- A.
10 : 13

- B.
10 : 3

- C.
3 : 10

- D.
1 : 1

Answer: Option D

**Explanation** :

When two containers contain equal amount of milk and water respectively and then equal amounts are transferred from 1^{st} to 2^{nd} and then from 2^{nd} to 1^{st}.

The amount of milk in 1^{st} becomes equal to amount of water in 2nd while

amount of water in 1^{st} becomes equal to amount of milk in 2^{nd}.

Hence, option (d).

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**22. CAT 2022 QA Slot 3 | Algebra - Simple Equations**

A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is

Answer: 12

**Explanation** :

Let the number of 100, 250 and 500 notes be a, b and c respectively.

⇒ a + b + c = 100 …(1)

⇒ 100a + 250b + 500c = 15250

⇒ 2a + 5b + 10c = 305 …(2)

(2) – 2 × (1)

⇒ 3b + 8c = 105

⇒ c = $\frac{105-3\mathrm{b}}{8}$ = $\frac{104+1-3\mathrm{b}}{8}$ = 13 + $\frac{1-3\mathrm{b}}{8}$

c will be maximum when b is minimum. The least value b can take such that c is integer is 3.

∴ c = 13 + $\frac{1-9}{8}$ = 12

Hence, 12.

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