# CAT 2008 QA

Paper year paper questions for CAT 2008 QA

**1. CAT 2008 QA | Algebra - Simple Equations**

A shop stores x kg of rice. The first customer buys half this amount plus half a kg of rice. The second customer buys half the remaining amount plus half a kg of rice. Then the third customer also buys half the remaining amount plus half a kg of rice. Thereafter, no rice is left in the shop. Which of the following best describes the value of x?

- A.
2 ≤ x ≤ 6

- B.
5 ≤ x ≤ 8

- C.
9 ≤ x ≤ 12

- D.
11 ≤ x ≤ 14

- E.
13 ≤ x ≤ 18

Answer: Option B

**Explanation** :

The initial quantity of rice is x kg.

The first customer buys half the total rice in the store, and another half kg.

∴ Rice purchased by the first customer = $\frac{x}{2}+\frac{1}{2}=\frac{x+1}{2}$

∴ Remaining rice = $x-\left(\frac{x+1}{2}\right)=\frac{x-1}{2}$

Now, the second customer buys half of this, and another half kg.

∴ Rice purchased by the second customer = $\frac{x-1}{4}+\frac{1}{2}=\frac{x+1}{4}$

∴ Remaining rice = $\frac{x-1}{2}-\frac{x+1}{4}=\frac{x-3}{4}$

Now, the third customer buys half the remaining rice, and another half kg.

∴ Rice purchased by the third customer = $\frac{x-3}{8}+\frac{1}{2}=\frac{x+1}{8}$

Since after this purchase, there is no rice left in the store, we conclude that:

$\frac{x-3}{4}-\frac{x+1}{8}=\frac{x-7}{8}=0$

∴ x = 7

Hence, option (b).

Workspace:

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

Let f(x) = ax^{2} + bx + c, where, a, b and c are certain constants and a ≠ 0. It is known that f(5) = −3f(2) and that 3 is a root of f(x) = 0.

**2. CAT 2008 QA | Algebra - Quadratic Equations | Algebra - Functions & Graphs**

What is the correct root of f(x) = 0?

- A.
-7

- B.
-4

- C.
2

- D.
6

- E.
Cannot be determined

Answer: Option B

**Explanation** :

∵ 3 is a root of f(x) = 0,

∴ 9a + 3b + c = 0 …(i)

Also,

f(5) = −3f(2)

∴ 25a + 5b + c = −3(4a + 2b + c)

∴ 37a + 11b + 4c = 0 …(ii)

On solving equations (i) and (ii), we get,

a – b = 0

∴ a = b

We know that sum of the roots of a quadratic equation (ax^{2} + bx + c = 0) is –b/a

∴ 3 + other root = −1

∴ Other root = −4

Hence, option (b).

Workspace:

**3. CAT 2008 QA | Algebra - Quadratic Equations | Algebra - Functions & Graphs**

What is the value of a + b + c?

- A.
9

- B.
14

- C.
13

- D.
37

- E.
Cannot be determined

Answer: Option E

**Explanation** :

The roots at f(x) = 0 are 3 and −4

∴ The equation can be written as (x – 3)(x + 4) = 0

Or, x^{2} − x + 12 = 0

The co-efficient of x^{2} is 1 here, but all equations which are multiple of this equation will also have same roots.

For example,

10(x^{2} − x + 12) = 0 will also have same roots

∴ (a + b + c) cannot be determined uniquely.

Hence, option (e).

Workspace:

**4. CAT 2008 QA | Algebra - Progressions**

The number of common terms in the two sequences 17, 21, 25, … , 417 and 16, 21, 26, … , 466 is

- A.
78

- B.
19

- C.
20

- D.
77

- E.
22

Answer: Option C

**Explanation** :

The first sequence can be written as 17, 17 + 4, 17 + 8, … , 417 and second sequence can be written as 16, 16 + 5, 16 + 10, … , 466

The common difference for the first sequence is 4 and that for the second sequence is 5 and both the sequences have 21 as the first common term.

∴ Common terms are 21, 21 + L, 21 + 2L, ...

[Here, L = LCM of 4 and 5 = 20]

∴ Common terms are 21, 21 + 20, 21 + 40, ...

The common terms have a common difference of 20 and first term as 21.

∴ Let T_{n} be the last common term, T_{n} = 21 + (n - 1) × 20

Now, Tn should be less than or equal to 417

⇒ T_{n} = 21 + (n - 1) × 20 ≤ 417

⇒ 20n - 20 ≤ 396

⇒ n ≤ 416/20 = 20.8

Hence, the highest possible value of n is 20.

∴ The total number of terms which are common to both the sequences = 20

Hence, option (c).

Workspace:

The figure below shows the plan of a town. The streets are at right angles to each other. A rectangular park (P) is situated inside the town with a diagonal road running through it. There is also a prohibited region (D) in the town.

**5. CAT 2008 QA | Modern Math - Permutation & Combination**

Neelam rides her bicycle from her house at A to her office at B, taking the shortest path. Then the number of possible shortest paths that she can choose is

- A.
60

- B.
75

- C.
45

- D.
90

- E.
72

Answer: Option D

**Explanation** :

We can find the number of shortest possible paths from A to E either by trial and error or by using combinations.

Note that to travel from A to E, we have to take 2 roads to the right and 2 roads downwards (in the diagram) in order that we follow the shortest path. In other words, we have to use 2 + 2 = 4 roads, out of which 2 are towards right and 2 are downwards.

This is equivalent to selecting 2 things (roads towards right) out of 4 things (roads). (The remaining two roads will be downwards.)

The number of ways of doing this is ^{4}C_{2} = 4!/(2!×2!) = 6

∴ From point A to E, there are 6 ways to reach with the minimum distance travelled.

Here E to F is the shortest distance because the third side of a triangle is always less than the sum of the other two sides.

From point F to B, there are ^{6}C_{4} = 6!/(4!×2!) = 15 ways to reach with the minimum distance travelled.

∴ There are 15 × 6 = 90 shortest paths that Neelam can choose.

Hence, option 4.

Workspace:

**6. CAT 2008 QA | Modern Math - Permutation & Combination**

Neelam rides her bicycle from her house at A to her club at C, via B taking the shortest path. Then the number of possible shortest paths that she can choose is

- A.
1170

- B.
630

- C.
792

- D.
1200

- E.
936

Answer: Option A

**Explanation** :

From point A to B, there are 90 paths possible with the minimum distance travelled.

We can travel from B to C in two ways

1. B – K – M – C

To travel from B to K, we have to take 6 roads, out of which one is towards left and 5 are upwards.

There are 6!/(5!×1!) = 6 ways to do this. (Refer to the explanation in the first question of this set.)

2. B – N – C

To travel from B to N, we have to take 7 roads, out of which one is towards left and 6 are upwards.

There are 7!/(6!×1!) = 7 ways to do this.

∴ There are 6 + 7 = 13 ways in which we can travel from B to C.

∴ Overall there are 90 × 13 = 1170 paths possible

Hence, option (a).

Workspace:

**7. CAT 2008 QA | Algebra - Functions & Graphs**

Let f(x) be a function satisfying f(x) × f(y) = f(xy) for all real x, y. Let f(2) = 4, then what is the value of $f\left(\frac{1}{2}\right)?$

- A.
0

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

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

- D.
1

- E.
Cannot be determined

Answer: Option B

**Explanation** :

Given, f(x) × f(y) = f(xy)

Let x = 1 and y = 2

∴ f(1) × f(2) = f(2)

∴ f(1) = 1

Now, let x = 1/2 and y = 2

∴ f(1/2) × f(2) = f(1/2 × 2) = f(1)

∴ f(1/2) × 4 = 1

∴ f(1/2) = 1/4

Hence, option (b).

Workspace:

**8. CAT 2008 QA | Algebra - Progressions**

The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39 times: In each repetition, any two numbers, say a and b, currently on the blackboard are erased and a new number a + b – 1 is written. What will be the number left on the board at the end?

- A.
820

- B.
821

- C.
781

- D.
819

- E.
780

Answer: Option C

**Explanation** :

Initial sum of the terms of the sequence 1, 2, 3, ...., 40 = $\frac{40\times 41}{2}$ = 820

After erasing two numbers a and b, and replacing with (a + b − 1), the new sum of the terms of the sequence = 820 − 1

Similarly, after every operation, the sum of the terms of the sequence reduces by 1.

∴ The last number left (i.e. final sum) = 820 − 39 = 781

Hence, option (c).

Workspace:

**9. CAT 2008 QA | Algebra - Number Theory**

Suppose, the seed of any positive integer n is defined as follows:

seed(n) = n, if n < 10

= seed(s(n)), otherwise,

where s(n) indicates the sum of digits of n. For example, seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc.

How many positive integers n, such that n < 500, will have seed(n) = 9?

- A.
39

- B.
72

- C.
81

- D.
108

- E.
55

Answer: Option E

**Explanation** :

Sum of the digits of multiples of 9 is always 9.

∴ Seed of any number will be 9 if and only if it is a multiple of 9.

There are 55 multiples of 9 which are less than 500 (as 500/9 = 55.555)

∴ There are 55 positive integers which will have seed = 9

Hence, option (e).

Workspace:

**10. CAT 2008 QA | Geometry - Triangles**

In a triangle ABC, the lengths of the sides AB and AC equal 17.5 cm and 9 cm respectively. Let D be a point on the line segment BC such that AD is perpendicular to BC. If AD = 3 cm, then what is the radius (in cm) of the circle circumscribing the triangle ABC?

- A.
17.05

- B.
27.85

- C.
22.45

- D.
32.25

- E.
26.25

Answer: Option E

**Explanation** :

We know that the area (A) of triangle (ABC) is related to the circum radius (R) and sides of the triangle as follows:

R = $\frac{AB\times BC\times AC}{4A}$

Where, Area (A) = ½ × AD × BC

∴ R = $\frac{\mathrm{AB}\times \mathrm{BC}\times \mathrm{AC}}{4\times {\displaystyle \frac{1}{2}}\times \mathrm{AD}\times \mathrm{BC}}$ = $\frac{\mathrm{AB}\times \mathrm{AC}}{2\mathrm{AD}}$ = $\frac{17.5\times 9}{2\times 3}$ = 26.25 cm

Hence, option (e).

Workspace:

**11. CAT 2008 QA | Algebra - Number Theory**

What are the last two digits of 7^{2008}?

- A.
21

- B.
61

- C.
01

- D.
41

- E.
81

Answer: Option C

**Explanation** :

7^{1} = 07

7^{2} = 49

7^{3} = 343

7^{4} = 2,401

7^{5} = 16,807

7^{6} = 1,16,649

7^{7} = 8,23,543

7^{8} = 57,64,801

As we can see, for every 4^{th} power of 7, the last two digits are 01. Since 2008 is divisible by 4, we can conclude that last two digits of 7^{2008} are 01.

Hence, option (c).

Workspace:

**12. CAT 2008 QA | Algebra - Quadratic Equations**

If the roots of the equation x^{3}− ax^{2} + bx – c = 0 are three consecutive integers, then what is the smallest possible value of b?

- A.
$\frac{-1}{\sqrt{3}}$

- B.
-1

- C.
0

- D.
1

- E.
$\frac{1}{\sqrt{3}}$

Answer: Option B

**Explanation** :

If p, q and r are the roots of a cubic equation ax^{3} + bx^{2} + cx + d = 0,

then pq + pr + qr = c/a

If a is 1, then pq + qr + pr = c

Comparing the equation ax^{3} + bx^{2} + cx + d = 0 with the equation in the question x^{3}− ax^{2} + bx – c = 0, we get

pq + qr + pr = b

Let the three roots of the given equation be (n – 1), n and (n + 1).

∴ (n – 1)n + n(n + 1) + (n – 1)(n + 1) = b

∴ n^{2} – n + n^{2 }+ n + n^{2} – 1 = b

∴ 3n^{2} – 1 = b

∵ n^{2} ≥ 0, minimum value of b occurs at n = 0

∴ Minimum value of b = –1

Hence, option (b).

Workspace:

**13. CAT 2008 QA | Geometry - Triangles**

Consider obtuse-angled triangles with sides 8 cm, 15 cm and x cm. If x is an integer, then how many such triangles exist?

- A.
5

- B.
21

- C.
10

- D.
15

- E.
14

Answer: Option C

**Explanation** :

We know that for an obtuse triangle of sides a, b and c (where c is the largest side),

a^{2} + b^{2} < c^{2}

We also know that for a triangle, a + b > c

Let the third side be x.

These present us with two limiting cases.

**Case 1**: Let 8 cm and 15 cm be the shorter sides. The value of the largest side (x) must be greater than

$\sqrt{{8}^{2}+{15}^{2}}=17cm$

Also, x < 8 + 15 = 23.

The possible integer values of x are 18, 19, 20, 21 and 22 cm.

We cannot consider values from 23 onwards because 8 + 15 = 23 and this violates the second condition.

**Case 2:** Let 8 and x be the shorter sides and 15 cm is the largest side.

The value of the remaining side (x) must be less than

$\sqrt{{15}^{2}-{8}^{2}}=12.69cm$

Also, x > 15 - 8 = 7

The possible integer values are 12, 11, 10, 9 and 8 cm.

We cannot consider values less than 8 because 7 + 8 = 15 and this violates the second condition.

Thus, we have 10 possible values for x.

Hence, option (c).

Workspace:

**14. CAT 2008 QA | Modern Math - Permutation & Combination**

How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4, if repetition of digits is allowed?

- A.
499

- B.
500

- C.
375

- D.
376

- E.
501

Answer: Option D

**Explanation** :

The minimum number that can be formed is 1000 and the maximum number that can be formed is 4000.

As 4000 is the only number in which the first digit is 4, first let us calculate the numbers less than 4000 and then we will add 1 to it.

∴ First digit can be 1, 2 or 3.

Remaining 3 digits can be any of the 5 digits.

∴ Total numbers that can be formed, which are less than 4000 = 3 × 5 × 5 × 5 = 375

∴ Total numbers that satisfy the given condition = 375 + 1 = 376

Hence, option (d).

Workspace:

**15. CAT 2008 QA | Modern Math - Permutation & Combination**

What is the number of distinct terms in the expansion of (a + b + c)^{20}?

- A.
231

- B.
253

- C.
242

- D.
210

- E.
228

Answer: Option A

**Explanation** :

Consider (a + b + c)^{20}

∵ The degree of the expression is 20, the degree of each term of the expression after expansion will be 20.

∴ We have to divide 20 into three parts which can be done by using the distribution rule

^{n+r-1}C_{r-1}

Where,

n is number of things to be distributed.

r is number of parts into which the things are to be distributed.

∴ To divide 20 into 3 parts we have,

^{20+3-1}C_{3-1} = ^{22}C_{2} = $\frac{22\times 21}{2\times 1}$ = 11 × 21 = 231

**Alternatively**,

This can be solved without using much knowledge of permutations and combinations as follows,

(a + b + c)^{1} = a + b + c [i.e. 3 terms = (1 + 2) terms]

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2ab + 2bc + 2ac [i.e. 6 terms = (1 + 2 + 3) terms]

(a + b + c)^{3} = a^{3} + b^{3} + c^{3} + 6abc + 3ab^{2} + 3ac^{2} + 3a^{2}b + 3bc^{2} + 3a^{2}c + 3b^{2}c [i.e. 10 terms = (1 + 2 + 3 + 4) terms]

Similarly,

(a + b + c)^{n} will have (1 + 2 + 3 + … + (n + 1)) terms

∴ (a + b + c)^{20} will have (1 + 2 + 3 + … + 21) = 231 terms

Hence, option (a).

Workspace:

**16. CAT 2008 QA | Geometry - Quadrilaterals & Polygons**

Consider a square ABCD with midpoints E, F, G, H of AB, BC, CD and DA respectively. Let L denote the line passing through F and H. Consider points P and Q, on L and inside ABCD, such that the angles APD and BQC both equal 120°. What is the ratio of the area of ABQCDP to the remaining area inside ABCD?

- A.
$\frac{4\sqrt{2}}{3}$

- B.
$2+\sqrt{3}$

- C.
$\frac{10-3\sqrt{3}}{9}$

- D.
$1+\frac{1}{\sqrt{3}}$

- E.
$2\sqrt{3}-1$

Answer: Option E

**Explanation** :

Let the length of the sides of the square be 2s.

Consider ∆BQF,

BF = s

In 30° - 60° - 90° triangle,

QF = $\frac{s}{\sqrt{3}}$

∴ Area of ∆BQF = $\frac{1}{2}$ × s × $\frac{s}{\sqrt{3}}$

Area of ABQCDP = Area of square ABCD – 4 × Area of ∆BQF $=4{s}^{2}-4\left(\frac{1}{2}\times s\times \frac{s}{\sqrt{3}}\right)$

∴ Required ratio = $\frac{4{s}^{2}-4\left({\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{s}{\sqrt{3}}}\right)}{4\left({\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{s}{\sqrt{3}}}\right)}$ = $2\sqrt{3}-1$

Hence, option (e).

Workspace:

**17. CAT 2008 QA | Algebra - Number Theory | Algebra - Quadratic Equations**

Three consecutive positive integers are raised to the first, second and third powers respectively and then added. The sum so obtained is a perfect square whose square root equals the total of the three original integers. Which of the following best describes the minimum, say m, of these three integers?

- A.
1 ≤ m ≤ 3

- B.
4 ≤ m ≤ 6

- C.
7 ≤ m ≤ 9

- D.
10 ≤ m ≤ 12

- E.
13 ≤ m ≤ 15

Answer: Option A

**Explanation** :

Let the three numbers be (a – 2), (a – 1) and a.

∴ (a – 2) + (a – 1)^{2} + a^{3 }= p^{2}

Where p is the sum of the three integers.

Now, a – 2 + a^{2} – 2a + 1 + a^{3 }= p^{2}

∴ a^{3} + a^{2} – a – 1 = p^{2}

∴ a^{2}(a + 1) – 1(a + 1) = p^{2}

∴ (a^{2} – 1)(a + 1) = p^{2}

∴ (a + 1)^{2}(a – 1) = p^{2}

For the above condition to be satisfied, (a – 1) must be a perfect square.

The smallest possible value for a - 1 is 4, since (a – 2) cannot be zero, giving us (a − 2) = 3

The minimum of the three is therefore 3.

Hence, option (a).

Workspace:

**18. CAT 2008 QA | Algebra - Progressions**

Find the sum $\sqrt{1+\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}}+\sqrt{1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}}+.....+\sqrt{1+\frac{1}{{2007}^{2}}+\frac{1}{{2008}^{2}}}$

- A.
$2008-\frac{1}{2008}$

- B.
$2007-\frac{1}{2007}$

- C.
$2007-\frac{1}{2008}$

- D.
$2008-\frac{1}{2007}$

- E.
$2008-\frac{1}{2009}$

Answer: Option A

**Explanation** :

Consider only first term,

$\sqrt{1+\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}}$ = $\sqrt{1+1+\frac{1}{4}}$ = $\sqrt{\frac{9}{4}}$ = $\frac{3}{2}$ = $2-\frac{1}{2}$

Now consider fisrt two terms,

$\sqrt{1+\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}}$ + $\sqrt{1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}}$ = $\frac{3}{2}$ + $\sqrt{1+\frac{1}{4}+\frac{1}{9}}$ = $\frac{3}{2}$ + $\sqrt{\frac{49}{36}}$ = $\frac{3}{2}+\frac{7}{6}$ = $\frac{8}{3}$ = $3-\frac{1}{3}$

Similarly,

$\sqrt{1+\frac{1}{{1}^{2}}+\frac{1}{{2}^{2}}}$ + $\sqrt{1+\frac{1}{{2}^{2}}+\frac{1}{{3}^{2}}}$ + ... + $\sqrt{1+\frac{1}{{2007}^{2}}+\frac{1}{{2008}^{2}}}$ = $2008-\frac{1}{2008}$

Hence, option (a).

Workspace:

**19. CAT 2008 QA | Geometry - Circles**

Two circles, both of radii 1 cm, intersect such that the circumference of each one passes through the centre of the circle of the other. What is the area (in sq cm) of the intersecting region?

- A.
$\frac{\pi}{3}-\frac{\sqrt{3}}{4}$

- B.
$\frac{2\pi}{3}+\frac{\sqrt{3}}{2}$

- C.
$\frac{4\pi}{3}-\frac{\sqrt{3}}{2}$

- D.
$\frac{4\pi}{3}+\frac{\sqrt{3}}{2}$

- E.
$\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$

Answer: Option E

**Explanation** :

Let O and P be the centres of the circles.

OR = OP = PR = 1cm

∴ ∆PRO is an equilateral triangle.

∴ m ∠ROP = 60°

∴ m ∠ROS = 120°

Now, area of the intersecting region = 2(area of sector O-RPS – area of ∆ORS)

= 2(area of sector O-RPS – area of ∆PRO) [area of ∆PRO = area of ∆ORS]

Area of sector O - RPS = $\frac{120}{360}\left(\pi \right)$ = $\frac{\pi}{3}$

Area of ∆PRO = $\frac{\sqrt{3}}{4}\left({1}^{2}\right)$

∴ Area of the intersecting region = $2\left(\frac{\pi}{3}\right)$ - $2\left(\frac{\sqrt{3}}{4}\right)$ = $\frac{2\pi}{3}-\frac{\sqrt{3}}{2}$

Hence, option (e).

Workspace:

**20. CAT 2008 QA | Arithmetic - Time, Speed & Distance**

Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arriving there from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B, located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that C is located between west and northwest of B, with BC at 60° to AB. Also, C is located between south and southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and still catch the train is closest to

- A.
6:15 am

- B.
6:30 am

- C.
6:45 am

- D.
7:00 am

- E.
7:15 am

Answer: Option B

**Explanation** :

The angles of the triangle formed by A, B and C tell us that ABC is a right-angle triangle, with right-angle at vertex C, 30° at vertex A and 60° at vertex B.

Since AB = 500 km, in 30°-60°-90° triangle ABC, we get,

AC = 250$\sqrt{3}$ km and BC = 250 km

The train, travels at 50 km/hr. It will travel from B to C (i.e. 250 km) in 5 hours. Since it leaves at 8:00 a.m., it will reach C at 1:00 p.m.

Now, Rahim must be at C latest by 12:45 p.m. (15 minutes before the train)

Travelling at 70 km/hr, he will take approximately 6.2 hours to travel from A to C. Therefore, he must leave at least by

12.75 – 6.2 = 6.55 hours after mid-night.

This is a little after 6:30 a.m. If he leaves by 6:45 a.m., he will not make it to point C 15 minutes before the train arrives.

Hence, option (b).

Workspace:

**21. CAT 2008 QA | Geometry - Mensuration**

Consider a right circular cone of base radius 4 cm and height 10 cm. A cylinder is to be placed inside the cone with one of the flat surface resting on the base of the cone. Find the largest possible total surface area (in sq. cm) of the cylinder.

- A.
$\frac{100\pi}{3}$

- B.
$\frac{80\pi}{3}$

- C.
$\frac{120\pi}{7}$

- D.
$\frac{130\pi}{9}$

- E.
$\frac{110\pi}{7}$

Answer: Option A

**Explanation** :

As shown in the figure, ABC is the cross section of a cone of height 10 cm and radius of base 4 cm.

PQRS is the cross section of a cylinder which needs to be fitted inside the cone such that one of the flat faces of the cylinder (represented by RS) coincides with the base of the cone (represented by BC).

∵ ∆ABD and ∆PBR are similar triangles.

∴ $\frac{10}{4}=\frac{h}{4-r}$

∴ h = 10 - 2.5 × r

Total surface area of the cylinder = S = 2πr × (h + r) = 2πr × (10 - 2.5 × r + r) cm^{2}

∴ S = 2π × (10r - 1.5 × r^{2}) cm^{2}

Differentiating ‘S’ with respect to ‘r’, we get,

∴ $\frac{dS}{dr}$ = 2π × (10 - 3r)

For maximum surface area, $\frac{dS}{dr}$ must be equated to zero which gives r = $\frac{10}{3}$ and h = $\frac{5}{3}$ cm

$\therefore $ Maximum totalsurface area of the cylinder = 2π × $\frac{10}{3}$ × $\left(\frac{5}{3}+\frac{10}{3}\right)$ = $\frac{100\pi}{3}$ cm^{2}

Hence, option (a).

Workspace:

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

Five horses, Red, White, Grey, Black and Spotted participated in a race. As per the rules of the race, the persons betting on the winning horse get four times the bet amount and those betting on the horse that came in second get thrice the bet amount. Moreover, the bet amount is returned to those betting on the horse that came in third, and the rest lose the bet amount. Raju bets Rs. 3000, Rs. 2000 Rs. 1000 on Red, White and Black horses respectively and ends up with no profit and no loss.

**22. CAT 2008 QA | Miscellaneous**

Suppose, in addition, it is known that Grey came in fourth. Then which of the following cannot be true?

- A.
Spotted came in first

- B.
Red finished last

- C.
White came in second

- D.
Black came in second

- E.
There was one horse between Black and White

Answer: Option C

**Explanation** :

We solve this question by options.

If we consider option (c) to be true, then White finishes second and one of the Red or Black horses will come in the first or third positions.

With White at the second position, the amount Raju receives at the end of the race will be at least Rs. 6000, and from Red or Black he will earn some money.

Therefore, the total money Raju receives will be more than Rs. 6000. Since his investment at the start of the race was only Rs. 6000, his profit could never be zero.

∴ Option (c) cannot be true.

Hence, option (c).

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**23. CAT 2008 QA | Miscellaneous**

Which of the following cannot be true?

- A.
At least two horses finished before Spotted

- B.
Red finished last

- C.
There were three horses between Black and Spotted

- D.
There were three horses between White and Red

- E.
Grey came in second

Answer: Option D

**Explanation** :

We solve this question by options.

If we consider option 4 to be true, then either the White or Red horse will finish first. It means that the amount Raju receives at the end of the race will be at least Rs. 8000 or Rs. 12000 (depending on which of the two horses finish first).

However, his investment at the start of the race was only Rs. 6000. So, his profit could never be zero; in the worst scenario he will at least make Rs. 2000.

∴ Option (d) cannot be true.

Hence, option (d).

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**Each question is followed by two statements, A and B. Answer each question using the following instructions:**

**Mark option (1) if the question can be answered by using statement A alone but not by using statement B alone.
Mark option (2) if the question can be answered by using statement B alone but not by using statement A alone.
Mark option (3) if the question can be answered by using either statement alone.
Mark option (4) if the question can be answered by using both the statements together but not by either of the statements alone.
Mark option (5) if the question cannot be answered on the basis of the two statements.**

In a single elimination tournament, any player is eliminated with a single loss. The tournament is played in multiple rounds subject to the following rules:

a. If the number of players, say n, in any round is even, then the players are grouped in to n/2 pairs. The players in each pair play a match against each other and the winner moves on to the next round.

b. If the number of players, say n, in any round is odd, then one of them is given a bye, that is, he automatically moves on to the next round. The remaining (n − 1) players are grouped into (n − 1)/2 pairs. The players in each pair play a match against each other and the winner moves on to the next round. No player gets more than one bye in the entire tournament.

Thus, if n is even, then n/2 players move on to the next round while if n is odd, then (n + 1)/2 players move on to the next round. The process is continued till the final round, which obviously is played between two players. The winner in the final round is the champion of the tournament.

**24. CAT 2008 QA | Data Sufficiency**

What is the number of matches played by the champion?

**A:** The entry list for the tournament consists of 83 players.

**B:** The champion received one bye.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option D

**Explanation** :

From statement (A) alone:

The entry list for the tournament consists of 83 players.

In round 1, 1 of the 83 players gets a bye and directly moves on to the next round.

∴ 42 players move on to round 2.

Similarly, 21 players move on to round 3, 11 players move on to round 4, 6 players move on to round 5, 3 players move on to round 6, 2 players move on to round 7.

The winner of the tournament would have played one match in each of the rounds; i.e. a total of 7 matches, provided he doesn’t get a bye.

However, we are not told whether or not the champion received a bye at some point in the tournament.

∴ We cannot answer the question on the basis of statement (A) alone.

From statement (B) alone:

The champion received one bye.

From this statement, we cannot find the number of matches played by the champion.

∴ We cannot answer the question on the basis of statement (B) alone.

From both the statements (A) and (B) together:

The champion must have played 7 matches if he did not receive any bye.

But it is given that the champion has got one bye in the tournament. ∴ He must have played only 6 matches.

∴ We can answer the question using both the statements (A) and (B) together.

Hence, option (d).

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**25. CAT 2008 QA | Data Sufficiency**

If the number of players, say n, in the first round was between 65 and 128, then what is the exact value of n?

A. Exactly one player received a bye in the entire tournament.

B. One player received a bye while moving on to the fourth round from the third round

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option D

**Explanation** :

From statement (A) alone:

Exactly 1 player received a bye in the entire tournament. We get many values of n between 65 and 128 that satisfy this condition.

For example, n can have the value 124 in round 1, to follow the pattern, [124-62-31-16-8-4-2-1].

Also, n can have the value 127 in round 1, to follow the pattern, [127-64-32-16-8-4-2-1].

∴ We cannot answer the question on the basis of statement (A) alone.

From statement (B) alone:

One player received a bye while moving on to the fourth round from the third round.

Here also, we get multiple values of n.

For example, n can have the value 124 in round 1, where 1 player received a bye while moving from round 3 to round 4 following the pattern, [124-62-31-16-8-4-2-1].

Also, n can have the value 122 in round 1, where 1 player received a bye while moving from round 3 to round 4 following the pattern, [122-61-31-16-8-4-2-1].

∴ We cannot answer the question on the basis of statement (B) alone.

From statements (A) and (B) together:

n can only have the value 124 in round 1, where exactly 1 player received a bye while moving from round 3 to round 4 following the pattern [124-62-31-16-8-4-2-1].

∴ We can answer the question using both the statements (A) and (B) together.

Hence, option 4.

Note: An analysis of how 124 was arrived at when using both conditions together:

Let the number of players in the first round be n. Since only one player gets a bye, and that too when moving from the third to the fourth round, hence we have the following conditions:

1. There will be n players in the first round, where n is even.

2. There will be n/2 players in the second round, where n/2 is even.

3. There will be n/4 players in the third round, where n/4 is odd.

4. There will be $\frac{{\displaystyle \frac{n}{4}}+1}{2}=\frac{n+4}{8}$ players in the fourth round, where $\frac{n+4}{8}$ should be even.

5. All numbers of players in the subsequent rounds should also be even.

From condition 3, we can conclude that:

$\frac{n+4}{8}=2k,$ where k is an integer

Hence, n = 16k – 4; so, within the given range, n could be 76 or 92 or 108 or 124.

Writing the pattern for each of the above possible values of n, we have:

76: [76-38-19-10-5-3-2-1]

92: [92-46-23-12-6-3-2-1]

108: [108-54-27-14-7-4-2-1]

124: [124-62-31-16-8-4-2-1]

We see that only 124 satisfies condition 5.

Hence, option (d).

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