# CAT 2006 QA | Previous Year CAT Paper

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**1. CAT 2006 QA | Algebra - Surds & Indices**

If x = −0.5, then which of the following has the smallest value?

- A.
${2}^{1/x}$

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

- C.
$\frac{1}{{x}^{2}}$

- D.
${2}^{x}$

- E.
$\frac{1}{\sqrt{-x}}$

Answer: Option B

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**Explanation** :

Out of the options, only $\frac{1}{x}$ is negative.

All the others are positive.

$\therefore \frac{1}{x}$ is the smallest.

Hence, option (b).

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**2. CAT 2006 QA | Algebra - Surds & Indices**

Which among 2^{1/2}, 3^{1/3}, 4^{1/4} , 6^{1/6} and 12^{1/12} is the largest ?

- A.
2

^{1/2} - B.
3

^{1/3} - C.
4

^{1/4} - D.
6

^{1/6} - E.
12

^{1/12}

Answer: Option B

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**Explanation** :

Givne, 2^{1/2}, 3^{1/3}, 4^{1/4} , 6^{1/6} and 12^{1/12}

Raising all numbers to power 12, we get

(2^{1/2})^{12}, (3^{1/3})^{12}, (4^{1/4})^{12} , (6^{1/6})^{12} and (12^{1/12})^{12}

(2^{6}), (3^{4}), (4^{3}) , (6^{2}) and (12^{1})

64, 81, 64, 36, 12

Since 81 is the higest number here, 3^{1/3} will be the highest of the original numbers.

Hence, option (b).

Workspace:

**3. CAT 2006 QA | Arithmetic - Ratio, Proportion & Variation**

If $\frac{a}{b}=\frac{1}{3}$, $\frac{b}{c}=2$, $\frac{c}{d}=\frac{1}{2}$, $\frac{d}{e}=3$ and $\frac{e}{f}=\frac{1}{4}$ then what is the value of $\frac{\mathit{a}\mathit{b}\mathit{c}}{\mathit{d}\mathit{e}\mathit{f}}$?

- A.
$\frac{3}{8}$

- B.
$\frac{27}{8}$

- C.
$\frac{3}{4}$

- D.
$\frac{27}{4}$

- E.
$\frac{1}{4}$

Answer: Option A

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**Explanation** :

$\frac{\mathrm{a}}{\mathrm{d}}$ = $\frac{\mathrm{a}}{\mathrm{b}}\times \frac{\mathrm{b}}{\mathrm{c}}\times \frac{\mathrm{c}}{\mathrm{d}}$ = $\frac{1}{3}\times 2\times \frac{1}{2}$ = $\frac{1}{3}$

$\frac{\mathrm{b}}{\mathrm{e}}$ = $\frac{\mathrm{b}}{\mathrm{c}}\times \frac{\mathrm{c}}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{e}}$ = $2\times \frac{1}{2}\times 3$ = 3

$\frac{\mathrm{c}}{\mathrm{f}}$ = $\frac{\mathrm{c}}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{e}}\times \frac{\mathrm{e}}{\mathrm{f}}$ = $\frac{1}{2}\times 3\times \frac{1}{4}$ = $\frac{3}{8}$

∴ $\frac{abc}{def}$ = $\frac{1}{3}\times 3\times \frac{3}{8}$ = $\frac{3}{8}$

Hence, option (a).

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**4. CAT 2006 QA | Geometry - Mensuration**

The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halved while the length is doubled, then the total area of the four walls of the room will

- A.
remain the same.

- B.
decrease by 13.64%.

- C.
decrease by 15%.

- D.
decrease by 18.75%.

- E.
decrease by 30%.

Answer: Option E

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**Explanation** :

Let the original length, breadth and height of the room be 3x, 2x and x respectively.

∴ The new length, breadth and height are 6x, x and x/2 respectively.

Area of four walls = (2 × length × height) + (2 × breadth × height)

Original area of four walls = 6x^{2} + 4x^{2} = 10x^{2}

New area of four walls = 6x^{2} + x^{2} = 7x^{2}

∴ Area of wall decreases by [(10x^{2} − 7x^{2})/10x^{2}] × 100 = 30%

Hence, option (e).

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**5. CAT 2006 QA | Algebra - Progressions**

Consider a sequence where the n^{th} term, ${t}_{n}=\frac{n}{n+2},n=1,2,....$

The value of ${t}_{3}\times {t}_{4}\times {t}_{5}\times ...\times {t}_{53}$ equals:

- A.
$\frac{2}{495}$ - B.
$\frac{2}{477}$

- C.
$\frac{12}{55}$

- D.
$\frac{1}{1485}$

- E.
$\frac{1}{2970}$

Answer: Option A

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**Explanation** :

${\mathrm{t}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+2}$

$\therefore {\mathrm{t}}_{3}=\frac{3}{3+2}=\frac{3}{5}$

${t}_{4}=\frac{4}{4+2}=\frac{4}{6}$

${t}_{5}=\frac{5}{5+2}=\frac{5}{7}$

...

${t}_{51}=\frac{51}{53}$

${t}_{52}=\frac{52}{54}$

${t}_{53}=\frac{53}{55}$

∴ t_{3} × t_{4} × t_{5} × ... × t_{53} = $\frac{3}{5}$ × $\frac{4}{6}$ × $\frac{5}{7}$ × ... × $\frac{51}{53}$ × $\frac{52}{54}$ × $\frac{53}{55}$

$=\frac{(3\times 4)}{(54\times 55)}=\frac{2}{495}$

Hence, option (a).

Workspace:

**6. CAT 2006 QA | Algebra - Progressions**

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

- A.
3

- B.
4

- C.
5

- D.
6

- E.
7

Answer: Option D

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**Explanation** :

Let there be n rows and a students in the first row.

∴ Number of students in the second row = a + 3

∴ Number of students in the third row = a + 6 and so on.

∴ The number of students in each row forms an arithmetic progression with common difference = 3

The total number of students = The sum of all terms in the arithmetic progression

= $\frac{\mathrm{n}[2\mathrm{a}+3(\mathrm{n}-1\left)\right]}{2}$ = 30

Now consider options.

Option (a): n = 3

$\frac{3[2a+3(3-1\left)\right]}{2}$ = 630

∴ a = 207

Option (b): n = 4

$\frac{4[2a+3(4-1\left)\right]}{2}$ = 630

∴ a = 153

Option (c): n = 5

$\frac{5[2a+3(5-1\left)\right]}{2}$ = 630

∴ a = 120

Option (d): n = 6

$\frac{6[2a+3(6-1\left)\right]}{2}$ = 630

∴ a = 195/2 = 97.5

Option (e): n = 7

$\frac{7[2a+3(7-1\left)\right]}{2}$ = 630

∴ a = 81

As a is an integer, only n = 6 is not possible.

Hence, option (d).

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**7. CAT 2006 QA | Algebra - Surds & Indices**

What are the values of x and y that satisfy both the equations?

2^{0.7x} . 3^{−1. 25y} = $\frac{8\sqrt{6}}{27}$

4^{0.3x} . 9^{0.2y} = 8 . (81)^{1/5}

- A.
x = 2, y = 5

- B.
x = 2.5, y = 6

- C.
x = 3, y = 5

- D.
x = 3, y = 4

- E.
x = 5, y = 2

Answer: Option E

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**Explanation** :

The fastest way to solve this sum is by substituting the values of x and y from the options.

For x = 5 and y = 2, the first equation becomes

2^{0.7x} × 3^{−1.25y} = 2^{0.7× 5} × 3^{−1.25 × 2}

= 2^{7/2} × 3^{−5/2}

= $\frac{8\sqrt{6}}{27}$

These values of x and y satisfy the second equation also.

Hence, option (e).

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**8. CAT 2006 QA | Algebra - Simple Equations | Algebra - Number Theory**

The number of solutions of the equation 2x + y = 40 where both x and y are positive integers and x ≤ y is:

- A.
7

- B.
13

- C.
14

- D.
18

- E.
20

Answer: Option B

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**Explanation** :

2x + y = 40

∴ y = 40 – 2x

x and y are positive integers and x ≤ y

If x = 1, y = 38

x = 2, y = 36

x = 3, y = 34

.

.

.

x = 12, y =16

x = 13, y = 14

x = 14, y = 12

∴ For x > 13, y ≤ x

∴ There are 13 solutions to the given equation.

Hence, option (b).

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**9. CAT 2006 QA | Venn Diagram**

A survey was conducted of 100 people to find out whether they had read recent issues of Golmal, a monthly magazine. The summarized information regarding readership in 3 months is given below:

Only September: 18; September but not August: 23; September and July: 8; September: 28; July: 48; July and August: 10; None of the three months: 24.

What is the number of surveyed people who have read exactly two consecutive issues (out of the three)?

- A.
7

- B.
9

- C.
12

- D.
14

- E.
17

Answer: Option B

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**Explanation** :

100 – 24 = 76 had read at least one issue.

If x people read all the three issues, then (8 – x) people read only the September and July issues.

23 people read the September issue but not the August issue.

∴ 18 + 8 – x = 23

∴ x = 3

As 28 people read the September issue, [28 – (8 – 3) – 3 – 18] = 2 people read only the August and September issues.

As 10 people read the July and August issues, 10 – 3 = 7 people read only the July and August issues.

∴ The number of people who have read exactly two consecutive issues = 7 + 2 = 9

[**Note**: July and September are not consecutive months, hence we will not consider them.]

Hence, option (b).

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**10. CAT 2006 QA | Algebra - Number Theory**

The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these four numbers?

- A.
21

- B.
25

- C.
41

- D.
67

- E.
73

Answer: Option C

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**Explanation** :

The four consecutive two-digit odd numbers will have (1, 3, 5, 7) or (3, 5, 7, 9) or (5, 7, 9, 1) or (7, 9, 1, 3) or (9, 1, 3, 5) as units digits.

As the sum divided by 10 yields a perfect square, the sum is a multiple of 10.

∴ The units digits have to be (7, 9, 1, 3).

Thus the four numbers will be (10x + 7), (10x + 9), (10x + 11) and (10x + 13),

where 0 < x < 9 (as each of these numbers is a two digit number)

Sum of these numbers = 40x + 40 = 40(x + 1)

Now, 40(x + 1)/10 = 4(x + 1) is a perfect square

As 4 is a perfect square, (x + 1) is some perfect square < 10

If x + 1 = 4, x = 3, and the four numbers are 37, 39, 41 and 43

If x + 1 = 9, x = 8, and the four numbers are 87, 89, 91 and 93

Hence, option (c).

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**11. CAT 2006 QA | Algebra - Functions & Graphs**

The graph of y – x against y + x is as shown below. (All graphs in this question are drawn to scale and the same scale has been used on each axis). Then, which of the options given shows the graph of y against x.

- A.
- B.
- C.
- D.
- E.

Answer: Option D

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**Explanation** :

All the given graphs are drawn to the same scale.

We can see that the line makes an angle which is more than 45° with the horizontal axis.

∴ The slope of the line is greater than 1.

Let the slope be k.

∴ (y – x) = k(y + x) {∵ k > 1}

∴ y – x = ky + kx

$\therefore y=\frac{x(k+1)}{(1-k)}$

$\frac{(k+1)}{(1-k)}$ is negative and $\left|\frac{(k+1)}{(1-k)}\right|>1$

∴ The graph of y against x will be such that when x is positive, y is negative and |x| < |y|, except at (0, 0).

Hence, option (d).

Workspace:

**12. CAT 2006 QA | Algebra - Progressions | Modern Math - Sets**

Consider the set S = {1, 2, 3, …, 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?

- A.
3

- B.
4

- C.
6

- D.
7

- E.
8

Answer: Option D

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**Explanation** :

Let there be n terms (n ≥ 3) in the arithmetic progression having 1 as the first term and 1000 as the last. Let d be the common difference. Then,

1000 = 1 + (n – 1) × d

∴ 999 = (n – 1) × d ... (i)

∴ Factors of 999 are 1, 3, 9, 27, 37, 111, 333 and 999

Substituting in equation (i)

If d = 1, n = 1000

If d = 3, n = 334

If d = 9, n = 112

If d = 27, n = 38

If d = 37, n = 28

If d = 111, n = 10

If d = 333, n = 4

If d = 999, n = 2, which is not possible as n > 2

∴ 7 arithmetic progressions can be formed.

Hence, option (d).

Workspace:

**Answer the next 2 questions based on the information given below:**

A punching machine is used to punch a circular hole of diameter two units from a square sheet of aluminium of width 2 units, as shown below. The hole is punched such that the circular hole touches one corner P of the square sheet and the diameter of the hole originating at P is in line with a diagonal of the square.

**13. CAT 2006 QA | Geometry - Mensuration**

The proportion of the sheet area that remains after punching is:

- A.
$\frac{\pi +2}{8}$

- B.
$\frac{6-\pi}{8}$

- C.
$\frac{4-\pi}{4}$

- D.
$\frac{\pi -2}{4}$

- E.
$\frac{14-3\pi}{6}$

Answer: Option B

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**Explanation** :

Let PQRS be the square sheet and let the hole have centre O.

As P lies on the circumference of the circle and as m ∠APC = 90°, AC is a diameter.

∵ BP is a diameter, m ∠PAB = m ∠BCP = 90°.

∵ BP = AC, ABCP is a square.

∴ m ∠POC = 90° and OP = OC = 1 unit

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆ OPC)

$=2\times \left[\left(\frac{\pi \times {1}^{2}}{4}\right)-\left(\frac{1}{2}\times {1}^{2}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{\pi -2}{2}\mathrm{sq}.\mathrm{units}$

Area of part of hole on sheet = Area of hole − Area of part of the circle falling outside the square sheet

$=\pi -\left(\frac{\pi -2}{2}\right)=\frac{\pi +2}{2}\mathrm{sq}.\mathrm{units}$

Part of square remaining after punching = Area of square − Area of part of hole on sheet

$=4-\left(\frac{\pi +2}{2}\right)=\frac{6-\pi}{2}\mathrm{sq}.\mathrm{units}$

∴ Proportion of sheet area that remains after punching $=\frac{\left({\displaystyle \frac{6-\pi}{2}}\right)}{4}=\frac{6-\pi}{8}$

Hence, option (b).

Workspace:

**14. CAT 2006 QA | Geometry - Mensuration**

Find the area of the part of the circle (round punch) falling outside the square sheet.

- A.
$\frac{\pi}{4}$

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

- C.
$\frac{\pi -1}{4}$

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

- E.
$\frac{\pi -2}{4}$

Answer: Option D

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**Explanation** :

We have calculated this value while solving the previous question.

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆OPC)

$=\frac{\pi -2}{2}\mathrm{sq}.\mathrm{units}$

Hence, option (d).

Workspace:

**15. CAT 2006 QA | Algebra - Quadratic Equations | Algebra - Surds & Indices**

What values of x satisfy ${x}^{\frac{2}{3}}+{x}^{\frac{1}{3}}-2\le 0$?

- A.
–8 ≤ x ≤ 1

- B.
–1 ≤ x ≤ 8

- C.
1 < x < 8

- D.
1 ≤ x ≤ 8

- E.
–8 ≤ x ≤ 8

Answer: Option A

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**Explanation** :

${x}^{\frac{2}{3}}+{x}^{\frac{1}{3}}-2\le 0$ ...(1)

Put y = x^{1/3}

Then equation (1) becomes y^{2} + y – 2 ≤ 0

(y + 2)(y – 1) ≤ 0

–2 ≤ y ≤ 1

$-2\le {x}^{\frac{1}{3}}\le 1$

–8 ≤ x ≤ 1

Hence, option (a).

Workspace:

**16. CAT 2006 QA | Algebra - Functions & Graphs**

Let f(x) = max (2x + 1, 3 − 4x), where x is any real number. Then the minimum possible value of f(x) is:

- A.
1/3

- B.
1/2

- C.
2/3

- D.
4/3

- E.
5/3

Answer: Option E

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**Explanation** :

Let the two lines represent the equations y = 2x + 1 and y = 3 – 4x

f(x) = max{2x + 1, 4 - 3x} is represented by green and blue lines.

f(x) is greater than 5/3, when x < 1/3 (green part) or x > 1/3 (blue part).

∴ f(x) is minimum at x = 1/3 and this value is 5/3.

Hence, option (e).

**Note**: The minimum value of the function f(x) = max (ax + b, cx + d) occurs when ax + b = cx + d (Only if one line has a positive slope and the other one has a negative slope)

Workspace:

**Answer the following question based on the information given below.**

An airline has a certain free luggage allowance and charges for excess luggage at a fixed rate per kg. Two passengers, Raja and Praja have 60 kg of luggage between them, and are charged Rs. 1200 and Rs. 2400 respectively for excess luggage. Had the entire luggage belonged to one of them, the excess luggage charge would have been Rs. 5400.

**17. CAT 2006 QA | Arithmetic - Ratio, Proportion & Variation**

What is the weight of Praja‘s luggage?

- A.
20 kg

- B.
25 kg

- C.
30 kg

- D.
35 kg

- E.
40 kg

Answer: Option D

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**Explanation** :

Let f kg be the free luggage allowance and let Raja and Praja have r kg and p kg excess luggage respectively.

Let x be the fixed rate per kg for excess luggage.

∴ 2f + r + p = 60 ... (i)

rx = 1200 ... (ii)

px = 2400 ... (iii)

(60 – f)x = 5400 ... (iv)

From (ii) and (iii),

p = 2r ... (v)

Substituting in (i),

2f + 3r = 60

∴ f = 30 – 3r/2 ... (vi)

Substituting in (iv),

(60 – 30 + 3r/2)x = 5400

∴ 30x + 3rx/2 = 5400

From (ii),

rx = 1200

∴30x = 3600

∴ x = 120

∴ r = 10, p = 20 and f = 15

∴ Weight of Praja's luggage = p + f = 35 kg

Hence, option (d).

Workspace:

**18. CAT 2006 QA | Arithmetic - Ratio, Proportion & Variation**

What is the free luggage allowance?

- A.
10 kg

- B.
15 kg

- C.
20 kg

- D.
25 kg

- E.
30 kg

Answer: Option B

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**Explanation** :

As calculated in the solution to the previous question, f = 15 kg

Hence, option (b).

Workspace:

**19. CAT 2006 QA | Arithmetic - Time, Speed & Distance**

Arun, Barun and Kiranmala start from the same place and travel in the same direction at speeds of 30 km/hr, 40 km/hr and 60 km/hr respectively. Barun starts two hours after Arun. If Barun and Kiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?

- A.
3

- B.
3.5

- C.
4

- D.
4.5

- E.
5

Answer: Option C

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**Explanation** :

Arun has travelled 60 km when Barun starts.

Barun overtakes Arun in 60/(40 – 30) = 6 hrs

In this time, Barun travels 6 × 40 = 240 km from the starting point.

Kiranmala overtakes Arun at the same point.

Kiranmala takes 240/60 = 4 hrs to reach there.

Arun takes 240/30 = 8 hrs to reach there.

∴ Kiranmala starts 8 – 4 = 4 hrs after Arun.

Hence, option (c).

Workspace:

**20. CAT 2006 QA | Algebra - Simple Equations | Algebra - Number Theory**

When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?

- A.
5

- B.
6

- C.
7

- D.
8

- E.
10

Answer: Option B

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**Explanation** :

Let 10x + y be a two digit number, where x and y are positive single digit integers and x > 0.

Its reverse = 10y + x

Now, |10y + x – 10x – y| = 18

∴ 9 |y – x| = 18

∴ |y – x| = 2

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

∴ Other than 13, there are 6 such numbers.

Hence, option (b).

Workspace:

**21. CAT 2006 QA | Geometry - Circles**

A semicircle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn meeting the circumference of the semicircle at D. Given that AC = 2 cm and CD = 6 cm, the area of the semicircle (in sq. cm.) will be:

- A.
32π

- B.
50π

- C.
40.5π

- D.
81π

- E.
undeterminable

Answer: Option B

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**Explanation** :

Let CB = x cm

In a right triangle if DC is the altitude from right vertex, DC^{2} = AC × BC

∴ 6^{2} = 2 × x

⇒ x = 18

⇒ AB = 2 + 18 = 20

∴ Diameter of the semicircle = 20

⇒ Radius of the semicircle = 10

∴ Area = 1/2 × π × 10^{2} = 50π sq. cm.

Hence, option (b).

Workspace:

**22. CAT 2006 QA | Modern Math - Permutation & Combination**

There are 6 tasks and 6 persons. Task 1 cannot be assigned either to person 1 or to person 2; task 2 must be assigned to either person 3 or person 4. Every person is to be assigned one task. In how many ways can the assignment be done?

- A.
144

- B.
180

- C.
192

- D.
360

- E.
716

Answer: Option A

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**Explanation** :

Task 2 can be assigned in 2 ways (either to person 3 or 4).

Task 1 can then be assigned in 3 ways (persons 4 or 3, 5 and 6)

The remaining 4 tasks can be assigned to the remaining 4 persons in 4! = 24 ways

∴ The assignment can be done in 24 × 2 × 3 = 144 ways

Hence, option (a).

Workspace:

**23. CAT 2006 QA | Algebra - Number Theory**

The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio of the number of employees who are graduates and above, to that of employees who are not, can possibly be:

- A.
101 : 88

- B.
87 : 100

- C.
110 : 111

- D.
85 : 98

- E.
97 : 84

Answer: Option E

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**Explanation** :

Consider options. As the number of employees is prime we can add the numerator and denominator of ratios directly to find the number of employees.

Option (a): Number of employees = 101 + 88 = 189

Number of employees = 189, which is not a prime number.

∴ Option (a) is eliminated.

Option (b): Number of employees = 87 + 100 = 187

Number of employees = 187, which is not a prime number.

∴ Option (b) is eliminated.

Option (c): Number of employees = 110 + 111 = 221

Number of employees = 221, which is not a prime number.

∴ Option (c) is eliminated.

Option (d): Number of employees = 85 + 98 = 183

Number of employees = 183, which is not a prime number.

∴ Option (d) is eliminated.

Option (e): Number of employees = 97 + 84 = 181

Number of employees = 181, which is a prime number.

∴ The ratio of employees = 97 : 84

Hence, option (e).

Workspace:

**24. CAT 2006 QA | Algebra - Logarithms**

If log_{y}x = a × log_{z}y = b × log_{x}z = ab, then which of the following pairs of values for (a, b) is not possible?

- A.
$-2,\frac{1}{2}$

- B.
$1,1$

- C.
0.4, 2.5

- D.
$\pi ,\frac{1}{\pi}$

- E.
2, 2

Answer: Option E

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**Explanation** :

If log_{y}x = a × log_{z}y = b × log_{x}z = ab

⇒ log_{y}x = ab ...(1)

⇒ a × log_{z}y = ab ⇒ log_{z}y = b ...(2)

⇒ b × log_{x}z = ab ⇒ log_{x}z = a ...(3)

From (1), (2) and (3), we get

log_{y}x = log_{x}z × log_{z}y

∴ $\frac{\mathrm{log}x}{\mathrm{log}y}$ = $\frac{\mathrm{log}z}{\mathrm{log}x}\times \frac{\mathrm{log}y}{\mathrm{log}z}$ = $\frac{\mathrm{log}y}{\mathrm{log}x}$

∴ (log x)^{2} = (log y)^{2}

∴ log x = ± log y

∴ log x = log y or log x = - log y

∴ x = y or x = $\frac{1}{y}$

∴ ab = log_{y} x = 1 or -1

Only option (e) does not satisfy this.

Hence, option (e).

Workspace:

**25. CAT 2006 QA | Geometry - Quadrilaterals & Polygons**

An equilateral triangle BPC is drawn inside a square ABCD. What is the value of the angle APD in degrees?

- A.
75

- B.
90

- C.
120

- D.
135

- E.
150

Answer: Option E

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**Explanation** :

BP = PC = BC

m ∠BPC = m ∠PCB = m ∠PBC = 60°

Also, PC = CD = BP = AB

∆ ABP and ∆ PCD are isosceles triangles.

m ∠ABP = m ∠PCD = 90 – 60 = 30°

∴ m ∠APB = m ∠DPC = (180 – 30)/2 = 75°

∴ m ∠APD = 360 – (m ∠APB + m ∠DPC + m ∠BPC) = 360 – (75 + 75 + 60) = 150°

Hence, option (e).

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