# CAT 2005 QA | Previous Year Questions

Previous year paper questions for CAT 2005 QA

**1. CAT 2005 QA | Algebra - Number Theory**

If *x* = (16^{3} + 17^{3} + 18^{3} + 19^{3}), then *x* divided by 70 leaves a remainder of

- A.
0

- B.
1

- C.
69

- D.
35

Answer: Option A

**Explanation** :

*x* = 16^{3} + 17^{3} + 18^{3} + 19^{3}

= (16^{3} + 19^{3}) + (17^{3} + 18^{3})

= (16 + 19)(16^{2} − 16 × 19 + 19^{2}) + (17 + 18)(17^{2} − 17 × 18 + 18^{2})

35 × (an odd number) + 35 × (another odd number) = 35 × (an even number)

= 35 × (2k) … (k is a positive integer)

∴ x = 70k

∴ x is divisible by 70.

Remainder when x is divided by 70 = 0

Hence, option (a).

Workspace:

**2. CAT 2005 QA | Arithmetic - Time & Work**

A chemical plant has four tanks (A, B, C, and D), each containing 1000 litres of a chemical. The chemical is being pumped from one tank to another as follows:

From A to B @ 20 litres/minute

From C to A @ 90 litres/minute

From A to D @ 10 litres/minute

From C to D @ 50 litres/minute

From B to C @ 100 litres/minute

From D to B @ 110 litres/minute

Which tank gets emptied first and how long does it take (in minutes) to get empty after pumping starts?

- A.
A, 16.66

- B.
C, 20

- C.
D, 20

- D.
D, 25

Answer: Option C

**Explanation** :

The change in the amount of chemical in each tank after every minute is as follows:

A: −20 – 10 + 90 = 60

B: −100 + 110 + 20 = 30

C: −50 − 90 + 100 = −40

D: −110 + 10 + 50 = −50

Since tank D loses the maximum amount of chemical in a minute, it will be emptied first.

Let n minutes be the time taken by tank D to get empty.

∴ 1000 – 50n = 0

∴ n = 20 minutes

Hence, option (c).

Workspace:

**3. CAT 2005 QA | Geometry - Circles**

Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area in sq. cm of the portion that is common to the two circles is

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

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

- C.
$\frac{\pi}{5}$

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

Answer: Option B

**Explanation** :

Let the two circles with centres P and Q intersect at M and N.

Quadrilateral PMQN is a square.

m ∠MPN = m ∠MQN = 90°

The area common to both the circles = 2(Area of sector P-MN – Area of ∆PMN)

$=2\left[\left(\frac{90}{360}\times \pi \times {1}^{2}\right)-\left(\frac{1}{2}\times {1}^{2}\right)\right]$

$=\frac{\pi}{2}-1$

Hence, option (b).

Workspace:

**4. CAT 2005 QA | Geometry - Mensuration**

A jogging park has two identical circular tracks touching each other, and a rectangular track enclosing the two circles. The edges of the rectangles are tangential to the circles. Two friends, A and B, start jogging simultaneously from the point where one of the circular tracks touches the smaller side of the rectangular track. A jogs along the rectangular track, while B jogs along the two circular tracks in a figure of eight. Approximately, how much faster than A does B have to run, so that they take the same time to return to their starting point?

- A.
3.88%

- B.
4.22%

- C.
4.44%

- D.
4.72%

Answer: Option D

**Explanation** :

Let r be the radius of the circular tracks.

Length and breadth of the rectangular track are 4r and 2r respectively.

Length (perimeter) of the rectangular track = 12r

Length of the two circular tracks (figure of eight) = 4πr

If A and B have to reach their starting points at the same time,

$\frac{12r}{a}=\frac{4\pi r}{b}$

(where a and b are the speeds of A and B respectively)

∴ $\frac{b}{a}=\frac{4\pi}{12}$

⇒ $\frac{(b-a)}{a}=\frac{4\pi -12}{12}$

∴ (b - a) × $\frac{100}{a}$ = 0.047 × 100 = 4.7%

Hence, option (d).

Workspace:

**5. CAT 2005 QA | Modern Math - Permutation & Combination**

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls, and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

- A.
200

- B.
216

- C.
235

- D.
256

Answer: Option A

**Explanation** :

Let there be g girls and b boys.

Number of games between two girls = ^{g}C_{2}

∴ g(g – 1)/2 = 45

⇒ g^{2} – g – 90 = 0

⇒ (g – 10)(g + 9) = 0

Since the number of girls cannot be negative, ∴ g = 10

Number of games between two boys = ^{b}C_{2}

∴ b(b – 1)/2 = 190

⇒ b^{2} – b – 380 = 0

⇒ (b + 19)(b – 20) = 0

⇒ b = 20

∴ Number of games in which one player is a boy and the other is a girl = 10 × 20 = 200

Hence, option (a).

Workspace:

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

Ram and Shyam run a race between points A and B, 5 km apart. Ram starts at 9 a.m. from A at a speed of 5 km/hr, reaches B, and returns to A at the same speed. Shyam starts at 9:45 a.m. from A at a speed of 10 km/hr, reaches B and comes back to A at the same speed.

**6. CAT 2005 QA | Arithmetic - Time, Speed & Distance**

At what time do Ram and Shyam first meet each other?

- A.
10:00 a.m.

- B.
10:10 a.m.

- C.
10:20 a.m.

- D.
10:30 a.m.

Answer: Option B

**Explanation** :

The distance between points A and B is 5 km.

Ram starts at 9 a.m. from A at a speed of 5 km/hr. So, he reaches B at 10:00 a.m.

When Ram reaches B (i.e. at 10.00 a.m., or 15 minutes after Shyam started from A), Shyam (running at a speed of 10 km/hr) is 15/60 × 10 = 2.5 km away from A.

Ram meets Shyam $\frac{2.5}{10+5}\times 60$ = 10 minutes after 10:00 a.m., i.e. at 10:10 a.m.

Hence, option (b).

Workspace:

**7. CAT 2005 QA | Arithmetic - Time, Speed & Distance**

At what time does Shyam overtake Ram?

- A.
10:20 a.m.

- B.
10:30 a.m.

- C.
10:40 a.m.

- D.
10:50 a.m..

Answer: Option B

**Explanation** :

Ram reaches B at 10.00 a.m. while Shyam reaches B at 10:15 a.m.

At 10:15 a.m., Ram is (15/60) × 5 = 1.25 km away from B.

Shyam overtakes Ram 1.25/(10 – 5) = 0.25 hrs = 15 minutes after 10:15 a.m., i.e. at 10:30 a.m.

Hence, option (b).

Workspace:

**8. CAT 2005 QA | Algebra - Number Theory**

If R = $\frac{{30}^{65}-{29}^{65}}{{30}^{64}+{29}^{64}}$ then

- A.
0 < R ≤ 0.1

- B.
0.1 < R ≤ 0.5

- C.
0.5 < R ≤ 1.0

- D.
R > 1.0

Answer: Option D

**Explanation** :

R = $\frac{{30}^{65}-{29}^{65}}{{30}^{64}+{29}^{64}}$

∵ *a ^{n}* –

*b*= (

^{n}*a*–

*b*)(

*a*

^{n – 1}+

*a*

^{n – 2}

*b*+

*a*

^{n – 3}

*b*

^{2}+ ... +

*b*

^{n – 1})

∴ R = $\frac{(30-29)[{30}^{64}+({30}^{63}\times 29)+...+{29}^{64}]}{{30}^{64}+{29}^{64}}$

∵ 30^{64} + 30^{63} × 29 + ... + 29^{64} > 30^{64 }+ 29^{64}

∴ R > 1

Hence, option (d).

Workspace:

**9. CAT 2005 QA | Geometry - Circles**

What is the distance in cm between two parallel chords of lengths 32 cm and 24 cm in a circle of radius 20 cm?

- A.
1 or 7

- B.
2 or 14

- C.
3 or 21

- D.
4 or 28

Answer: Option D

**Explanation** :

The two chords AB and CD can be on the same side or the opposite sides of the centre O.

Let M and N be the midpoints of AB and CD.

∴ MN is the distance between the two chords.

MB = 12 cm and ND = 16 cm

OM and ON are perpendicular to AB and CD respectively.

ON^{2} = OD^{2} - ND^{2} (By Pythagoras theorem)

⇒ ON^{2} = 20^{2} – 16^{2}

⇒ ON = 12 cm

Similarly,OM^{2} = OB^{2} - MB^{2}

⇒ OM = 16 cm

**Case 1**: AB and BC are on the same side of the centre.

MN = OM – ON = 4 cm

**Case 2**: AB and BC are on opposite sides of the centre.

MN = OM + ON = 28 cm

Hence, option (d).

Workspace:

**10. CAT 2005 QA | Algebra - Quadratic Equations**

For which value of k does the following pair of equations yield a unique solution for x such that the solution is positive?

*x*^{2} – *y*^{2} = 0

(*x* – *k*)^{ 2} + *y*^{2} = 1

- A.
2

- B.
0

- C.
$\sqrt{2}$

- D.
$-\sqrt{2}$

Answer: Option C

**Explanation** :

We have, *x*^{2} = *y*^{2} and (*x* − *k*)^{ 2} + *y*^{2} = 1

Solving the two equations simultaneously, we get,

(*x* – *k*)^{ 2} + *x*^{2} = 1 ...(1)

⇒ *x*^{2} – 2*kx* + *k*^{2} + *x*^{2} = 1

⇒ 2*x*^{2} – 2*kx* + (*k*^{2} – 1) = 0

If this equation has a unique solution for *x*, then discriminant = 0

∴ 4*k*^{2} – 8(*k*^{2} – 1) = 0

⇒ 8 – 4*k*^{2} = 0

⇒ *k*^{2} = 2

⇒ *k* = ± √2

Now, *k* = - √2 will give us a negative solution for (1) and hence it is rejected. [x should be positive according to the question.]

∴ *k* = √2

Hence, option (c).

Workspace:

**11. CAT 2005 QA | Algebra - Number Theory**

Let n! = 1 × 2 × 3 × ... × n for integer n ≥ 1. If p = 1! + (2 × 2!) + (3 × 3!) + … + (10 × 10!), then p + 2 when divided by 11! leaves a remainder of

- A.
10

- B.
0

- C.
7

- D.
1

Answer: Option D

**Explanation** :

p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!)

Now, n × n! = [(n + 1) – 1] × n! = (n + 1)! – n!

∴ p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +… + 11! – 10!

∴ p = 11! – 1!

∴ p + 2 = 11! + 1

∴ p + 2 when divided by 11! leaves a remainder of 1.

Hence, option (d).

Workspace:

**12. CAT 2005 QA | Geometry - Coordinate Geometry**

Consider a triangle drawn on the X-Y plane with its three vertices at (41, 0), (0, 41) and (0, 0), each vertex being represented by its (X, Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is

- A.
780

- B.
800

- C.
820

- D.
741

Answer: Option A

**Explanation** :

The points satisfying the equations x + y < 41, y > 0, x > 0 lie inside the triangle.

Integer solutions of x + y < 41 can be found as follows:

If x + y = 40, then

(x, y) could be (1, 39), (2, 38), …, (39, 1) ... (39 solutions)

If x + y = 39, then

(x, y) could be (1, 38), (2, 37), …, (38, 1) ... ( 38 solutions)

If x + y = 38, we get 37 solutions and so on till x + y = 2 ... (1 solution)

∴ Total solutions = 1 + 2 + 3 + ... + 39 = 39 × 40/2 = 780 integer solutions to x + y < 41.

The number of points with integer coordinates lying inside the circle = 780

Hence, option (a).

Workspace:

**13. CAT 2005 QA | Algebra - Number Theory**

The digits of a three-digit number A are written in the reverse order to form another three-digit number B. If B > A and B − A is perfectly divisible by 7, then which of the following is necessarily true?

- A.
100 < A < 299

- B.
106 < A < 305

- C.
112 < A < 311

- D.
118 < A < 317

Answer: Option B

**Explanation** :

Let A = 100x + 10y + z (x ≠ 0, x, y, z are single-digit numbers)

∴ B = 100z + 10y + x

∴ B – A = 99(z – x)

As (B – A) is divisible by 7 and 99 is not, (z – x) is divisible by 7.

∴ z and x can have values (8, 1) or (9, 2).

[Since B > A, z > x]

y can have any value from 0 to 9.

A = 1y8 or 2y9

∴ Lowest possible value of A is 108 and the highest possible value of A is 299.

Hence, option (b).

Workspace:

**14. CAT 2005 QA | Algebra - Progressions**

If a_{1} = 1 and a_{n + 1} – 3a_{n} + 2 = 4n for every positive integer n, then a_{100} equals

- A.
3

^{99}– 200 - B.
3

^{99}+ 200 - C.
3

^{100}– 200 - D.
3

^{100}+ 200

Answer: Option C

**Explanation** :

*a*_{1} = 1

*a*_{n+1} = 4*n* + 3*a _{n}* – 2

*a*_{2} = 4 + 3(1) – 2 = 5 = 3^{2} – 4

*a*_{3} = 4(2) + 3(5) – 2 = 21 = 3^{3} – 6

*a*_{4} = 4(3) + 3(21) – 2 = 73 = 3^{4} – 8

∴ *a _{n}* = 3

^{n}– 2(

*n*)

∴ *a*_{100} = 3^{100 }– 200

Hence, option (c).

Workspace:

**15. CAT 2005 QA | Modern Math - Permutation & Combination**

Let S be the set of five digit numbers formed by the digits 1, 2, 3, 4 and 5, using each digit exactly once such that exactly two odd positions are occupied by odd digits. What is the sum of the digits in the rightmost position of the numbers in S?

- A.
228

- B.
216

- C.
294

- D.
192

Answer: Option B

**Explanation** :

Let O and E represent odd and even digits respectively.

∴ S can have digits of the form:

O _ O _ E or O _ E _ O or E _ O _ O

**Case 1**: O _ O _ E

The first digit can be chosen in 3 ways out of 1, 3 and 5

The third can be chosen in 2 ways.

The fifth digit can be chosen in 2 ways after which the second and fourth digits can be chosen in 2 ways.

∴ There are 3 × 2 × 2 × 2 = 24 ways in which this number can be written.

12 out of these ways will have 2 in the rightmost position and 12 will have 4 in the rightmost position.

∴ The sum of the rightmost digits in Case 1 = (12 × 2) + (12 × 4) = 72

**Case 2**: O _ E _ O

This number can again be written in 24 ways.

8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position.

Thus the sum of the rightmost digits in Case 2 = (8 × 1) + (8 × 3) + (8 × 5) = 72

**Case 3**: E _ O _ O

This number can also be written in 24 ways.

As in Case 2, 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position.

∴ The sum of the rightmost digits in Case 3 = (8 × 1) + (8 × 3) + (8 × 5) = 72

∴ The sum of the digits in the rightmost position of the numbers in S = 72 + 72 + 72 = 216

Hence, option (b).

Workspace:

**16. CAT 2005 QA | Algebra - Number Theory**

The rightmost non-zero digit of the number 30^{2720} is

- A.
1

- B.
3

- C.
7

- D.
9

Answer: Option A

**Explanation** :

30^{2720} = 3^{2720} × 10^{2720}

The rightmost non-zero digit of 30^{2720} will be the digit in the unit’s place of 3^{2720}.

3’s power cycle is 3, 9, 7, 1 and cyclicity is 4.

i.e. if we get a remainder of 1, 2, 3, or 0 when 2720 is divided by 4, then the digit in the units place will be 3, 9, 7, or 1 respectively.

2720 = 680 × 4, i.e. the remainder is 0

∴ The digit in the unit’s place of 3^{2720} is 1.

∴ The rightmost non-zero digit of 30^{2720} is 1.

Hence, option (a).

Workspace:

**17. CAT 2005 QA | Geometry - Circles**

Four points A, B, C and D lie on a straight line in the X-Y plane, such that AB = BC = CD, and the length of AB is 1 metre. An ant at A wants to reach a sugar particle at D. But there are insect repellents kept at points B and C. The ant would not go within one metre of any insect repellent. The minimum distance in metres the ant must traverse to reach the sugar particle is

- A.
$3\sqrt{2}$

- B.
1 + π

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

- D.
5

Answer: Option B

**Explanation** :

The ant will not go into the circles with centers B and C and radius = 1 m

The minimum distance that the ant has to traverse = the distance of the path A-H-G-D

HG = 1 m

AH = GD = $\frac{1}{4}$ × Circumference of circle = $\frac{\pi}{2}$

AH + GD = π m

∴ The ant must traverse (1 + π)m.

Hence, option (b).

Workspace:

**18. CAT 2005 QA | Algebra - Logarithms**

If x ≥ y and y > 1, then the value of the expression

${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ can never be

- A.
-1

- B.
-0.5

- C.
0

- D.
1

Answer: Option D

**Explanation** :

${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ = log_{x} x - log_{x} y + log_{y} y - log_{y} x

⇒ ${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ = 1 - log_{x} y + 1 - log_{y} x

⇒ ${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ = 2 - log_{x} y - log_{y} x

⇒ ${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ = 2 - (log_{x} y + log_{y} x)

As x ≥ y and y > 1,

log_{y} x ≥ 0

Now, (log_{x} y + log_{y} x) = ${\mathrm{log}}_{x}\left(y\right)+\frac{1}{{\mathrm{log}}_{x}\left(y\right)}$ [This is sum of a positive number and its reciprocal]

Now, sum of a positive number and its reciprocal is always greater than or equal to 2.

∴ (log_{x} y + log_{y} x) = ${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ ≥ 2

⇒ ${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ = 2 - (log_{x} y + log_{y} x) ≤ 0

∴ ${\mathrm{log}}_{x}\left(\frac{x}{y}\right)+{\mathrm{log}}_{y}\left(\frac{y}{x}\right)$ ≠ 1

Hence, option (d).

Workspace:

**19. CAT 2005 QA | Algebra - Number Theory**

For a positive integer n, let *P _{n}* denote the product of the digits of

*n*, and

*S*denote the sum of the digits of

_{n}*n*. The number of integers between 10 and 1000 for which

*P*+

_{n}*S*=

_{n}*n*is

- A.
81

- B.
16

- C.
18

- D.
9

Answer: Option D

**Explanation** :

n can be a 2-digit or a 3-digit number.

**Case 1: **Let n be a 2 digit number.

Let n = 10x + y, where x and y are non-negative integers,

P_{n} = xy and S_{n} = x + y

Now, P_{n} + S_{n} = n

∴ xy + x + y = 10x + y

⇒ xy = 9x or y = 9

There are 9 two-digit numbers (19, 29, 29, … ,99) for which y = 9

**Case (2)**: Let n be a 3-digit number.

Let n = 100x + 10y + z, where x, y and z are non-negative integers,

P_{n} = xyz and S_{n} = x + y + z

Now, P_{n} + S_{n} = n

∴ xyz + x + y + z = 100x + 10y + z

⇒ xyz = 99x + 9y

⇒ z = 99/y + 9/x

From the above expression, 0 < x, y ≤ 9

But, we cannot find any value of x and y, for which z is a single-digit number. z will be minimum when x and y are both 9, but even then its value is 12.

∴ There are no 3-digit numbers which satisfy *P _{n}* +

*S*=

_{n}*n*

Hence, option (d).

Workspace:

**20. CAT 2005 QA | Geometry - Mensuration**

Rectangular tiles each of size 70 cm by 30 cm must be laid horizontally on a rectangular floor of size 110 cm by 130 cm, such that the tiles do not overlap. A tile can be placed in any orientation so long as its edges are parallel to the edges of the floor. No tile should overshoot any edge of the floor. The maximum number of tiles that can be accommodated on the floor is

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option C

**Explanation** :

This problem can be solved by trying different ways of placing the tiles on the floor. The maximum number of tiles that can be accommodated is 6, as shown in the figure.

Hence, option (c).

Workspace:

**21. CAT 2005 QA | Algebra - Functions & Graphs**

In the X-Y plane, the area of the region bounded by the graph of |x + y| + |x – y| = 4 is

- A.
8

- B.
12

- C.
16

- D.
20

Answer: Option C

**Explanation** :

|x + y| + |x – y| = 4 … (i)

Consider the case when x and y are both positive. This is the area of quadrant I.

In this case, two cases are possible.

Case I: x > y

In this case, expression (i) becomes:

x + y + x – y = 4

∴ 2x = 4

∴ x = 2

Case II: x < y

In this case, expression (i) becomes:

x + y + y – x = 4

∴ 2y = 4

∴ y = 2

∴ The area for the first quadrant is as shown in the figure.

Extending the same logic to other quadrants, we get the area as shown in the above diagram.

∴ Its area = 4 × 4 = 16 sq. units

Hence, option (c).

Workspace:

**22. CAT 2005 QA | Geometry - Circles**

In the following figure, the diameter of the circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE : EB = 1 : 2, and DF is perpendicular to MN such that NL : LM = 1 : 2. The length of DH in cm is

- A.
$2\sqrt{2}-1$

- B.
$\frac{(2\sqrt{2}-1)}{2}$

- C.
$\frac{(3\sqrt{2}-1)}{2}$

- D.
$\frac{(2\sqrt{2}-1)}{3}$

Answer: Option B

**Explanation** :

AO = OD = 1.5 cm

AE + EB = 3 cm and AE : EB = 1 : 2

∴ AE = 1 cm and EB = 2 cm

OE = AO – AE = 1.5 – 1 = 0.5 cm

Similarly, NL = 1 cm, LM = 2 cm, and OL = 0.5 cm

OEHL is a square as all its angles are right angles and OE = OL

∴ EH = HL = 0.5 cm

In ∆ODL, OD^{2} = OL^{2} + DL^{2}

1.5² = 0.5² + (0.5 + DH)²

2.25 = 0.25 + 0.25 + DH + DH²

DH² + DH – 1.75 = 0

DH = $\frac{-1\pm \sqrt{1-4(-1.75)}}{2}$ = $\frac{2\sqrt{2}-1}{2}$ (DH > 0)

Hence, option (b).

Workspace:

**23. CAT 2005 QA | Geometry - Triangles**

Consider the triangle ABC shown in the following figure where BC = 12 cm, DB = 9 cm, CD = 6 cm and ∠BCD = ∠BAC. What is the ratio of the perimeter of the triangle ADC to that of the triangle BDC?

- A.
7/9

- B.
8/9

- C.
6/9

- D.
5/9

Answer: Option A

**Explanation** :

m ∠BCD = m ∠BAC and B is common to triangles ABC and CBD.

∆ABC is similar to ∆CBD.

AB/CB = BC/BD = AC/CD

AB/12 = 12/9 = AC/6

AB = 16 cm and AC = 8 cm

AD = AB – BD = 16 – 9 = 7 cm

∴ Perimeter of ∆ADC = 7 + 6 + 8 = 21 cm

∴ Perimeter of ∆BDC = 9 + 6 + 12 = 27 cm

∴ Required ratio = 21/27 = 7/9

Hence, option (a).

Workspace:

**24. CAT 2005 QA | Geometry - Circles**

P, Q, S, and R are points on the circumference of a circle of radius r, such that PQR is an equilateral triangle and PS is a diameter of the circle. What is the perimeter of the quadrilateral PQSR?

- A.
2r(1 + $\sqrt{3}$)

- B.
2r(2 + $\sqrt{3}$)

- C.
r(1 + $\sqrt{5}$)

- D.
2r + $\sqrt{3}$

Answer: Option A

**Explanation** :

∆ PQR is an equilateral triangle and PS is the diameter.

∴ m ∠PQS = m ∠PRS = 90° (angles subtended in a semi-circle)

and m ∠PQM = m ∠PRM = m ∠QPR = 60° (each angle in an equilateral triangle = 60°)

PS bisects ∠QPS as it is the median of ∆PQR.

m ∠PMQ = m ∠PMR = 90°

∴ m ∠QPS = m ∠RPS = 30°

∴ m ∠PSQ = m ∠PSR = 60°

Radius = r

∴ PS = 2r

As ∆ PQS, ∆ PQM, ∆ MQS are 30°-60°-90° triangles,

QS = r, PQ = √3r

Similarly, RS = r and PR = √3r

∴ Perimeter of quadrilateral PQRS = 2r + 2√3r = 2r(1 + √3)

Hence, option (a).

Workspace:

**25. CAT 2005 QA | Algebra - Number Theory | Modern Math - Sets**

Let S be a set of positive integers such that every element n of S satisfies the conditions

a. 1000 ≤ n ≤ 1200

b. every digit in n is odd

Then how many elements of S are divisible by 3?

- A.
9

- B.
10

- C.
11

- D.
12

Answer: Option A

**Explanation** :

n will be of the form 11ab, where a and b are odd numbers.

We are looking for all n’s divisible by 3.

∴ 1 + 1 + a + b = 3 or 6 or 9 or 12 or 15 or 18

∴ a + b = 1 or 4 or 7 or 10 or 13 or 16

∴ a + b = 1 or 7 or 13 is not possible as the sum of two odd numbers cannot be odd.

∴ (a, b) = (1, 3), (3, 1), (1, 9), (3, 7), (5, 5), (7, 3), (9, 1), (7, 9), (9, 7)

∴ 9 elements of S are divisible by 3.

Hence, option (a).

Workspace:

**26. CAT 2005 QA | Algebra - Surds & Indices**

Let x = $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-....toinfinity.}}}}$ Then x equals

- A.
3

- B.
$\left(\frac{\sqrt{13}-1}{2}\right)$

- C.
$\left(\frac{\sqrt{13}+1}{2}\right)$

- D.
$\sqrt{13}$

Answer: Option C

**Explanation** :

x = $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-....toinfinity}}}}$

∴ x^{2} - 4 = $\sqrt{4-\sqrt{4+\sqrt{4-....toinfinity}}}$

∴ (x^{2} - 4)^{2} - 4 = - $\sqrt{4+\sqrt{4-...toinfinity}}$

∴ (x^{2} - 4)^{2} - 4 = - x

Now, substituting options, we find that only option (c) satisfies the above equation.

Hence, option (c).

Workspace:

**27. CAT 2005 QA | Algebra - Functions & Graphs**

Let g(x) be a function such that g(x + 1) + g(x – 1) = g(x) for every real x. Then for what value of p is the relation g(x + p) = g(x) necessarily true for every real x?

- A.
5

- B.
3

- C.
2

- D.
6

Answer: Option D

**Explanation** :

g(x + 1) + g(x – 1) = g(x)

∴ g(x + 1) = g(x) – g(x – 1)

Now, let g(x − 1) = a and g(x) = b

∴ g(x + 1) = b – a

∴ g(x + 2) = b – a – b = –a

∴ g(x + 3) = −a – b + a = –b

∴ g(x + 4) = −b + a = a – b

∴ g(x + 5) = a – b + b = a = g(x – 1)

∴ g(x + 6) = a – a + b = b = g(x)

and so on.

Thus we observe that the values of g(x + 6) and g(x) are always equal.

Hence, option (d).

Workspace:

**28. CAT 2005 QA | Algebra - Simple Equations**

A telecom service provider engages male and female operators for answering 1000 calls per day. A male operator can handle 40 calls per day whereas a female operator can handle 50 calls per day. The male and the female operators get a fixed wage of Rs. 250 and Rs. 300 per day respectively. In addition, a male operator gets Rs. 15 per call he answers and a female operator gets Rs. 10 per call she answers. To minimize the total cost, how many male operators should the service provider employ assuming he has to employ more than 7 of the 12 female operators available for the job?

- A.
15

- B.
14

- C.
12

- D.
10

Answer: Option D

**Explanation** :

Let us calculate the cost per call for a male and a female.

**Male**: A male can make 40 calls/day.

Total cost = 250 + 15 × 40 = 850

∴ Cost/call for a male = 850/40 = Rs. 21.25

**Female**: A female can make 50 calls/day.

Total cost = 300 + 10 × 50 = 800

∴ Cost/call for a female = 800/50 = Rs. 16

Hence, it could be cost effective to employ as many females as possible.

We have maximum 12 females who can make 12 × 50 = 600 calls/day.

Remaining (1000 - 600 =) 400 calls should be made by males.

∴ Number of males to be employed = 400/40 = 10 males.

**Alternately**,

Let x females and y males be employed.

As the total number of calls to be answered = 1000 and males and females can handle 40 and 50 calls respectively everyday

50x + 40y = 1000

40y = 1000 – 50x

∴ y = 25 – x – x/4

As 7 < x ≤ 12, x can be 8 or 12.

If x = 8, y = 15 and if x = 12, y = 10

The total cost of employing x females and y males

= 300x + 250y + (50 × 10 × x) + (40 × 15 × y)

= 800x + 850y

If x = 8 and y = 15, cost = Rs. 19,150

If x = 12 and y = 10, cost = Rs. 18,100

Thus, cost is minimized when the number of male operators is 10.

Hence, option (d).

Workspace:

**29. CAT 2005 QA | Miscellaneous**

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

- A.
5

- B.
10

- C.
9

- D.
15

Answer: Option C

**Explanation** :

Let E_{1}, E_{2} and E_{3} be the three Englishmen and F_{1}, F_{2} and F_{3} be the three Frenchmen.

Let E_{3} be the only Englishman knowing French.

Now, Let A ↔ B denote a phone call between A and B, where they both tell each other their secrets.The following phone calls will ensure that all six persons know all the six secrets.

1. E_{1} ↔ E_{3}

2. E_{2} ↔ E_{3} (Now E_{3} knows all the secrets with the Englishmen)

3. F_{1} ↔ F_{3}

4. F_{2} ↔ F_{3} (Now F_{3} knows all the secrets with the Frenchmen)

5. F_{3} ↔ E_{3} (Now F_{3} and E_{3} know all the secrets)

6. E_{3} ↔ E_{2}

7. E_{3} ↔ E_{1} (Now E_{1} and E_{2} know all the secrets)

8. F_{3} ↔ F_{2}

9. F_{3} ↔ F_{1 }(Now F_{1} and F_{2} know all the secrets)

Thus, a minimum of 9 calls are needed to pass all the secrets to all the six persons.

Hence, option (c).

Workspace:

**30. CAT 2005 QA | Geometry - Mensuration**

A rectangular floor is fully covered with square tiles of identical size. The tiles on the edges are white and the tiles in the interior are red. The number of white tiles is the same as the number of red tiles. A possible value of the number of tiles along one edge of the floor is

- A.
10

- B.
12

- C.
14

- D.
16

Answer: Option B

**Explanation** :

Let each square tile have side = 1 unit

Let the length of the rectangular floor be m units and the breadth be n units.

Number of red tiles = (m – 2)(n – 2)

Number of white tiles = mn − (m – 2)(n – 2)

Now, (m – 2)(n – 2) = mn − (m – 2)(n – 2)

⇒ 2(mn – 2m – 2n + 4) − mn = 0

⇒ mn – 4m – 4n + 8 = 0

⇒ n(m – 4) = 4m − 8

⇒ n = 4(m – 2)/(m – 4)

Now, consider the options.

Option (a): If m = 10, n = 32/6, which is not possible as n is an integer

Option (b): If m = 12, n = 40/8 = 5, which is possible

Option (c): If m = 14, n = 48/10, which is not possible as n is an integer

Option (d): If m = 16, n = 56/12, which is not possible as n is an integer

Hence, option (b).

Workspace:

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