# CAT 2000 QA

Paper year paper questions for CAT 2000 QA

**1. CAT 2000 QA | Algebra - Number Theory**

Let D be a recurring decimal of the form, D = 0.a_{1}a_{2}a_{1}a_{2}a_{1}a_{2} ......., where digits a_{1} and a_{2} lie between 0 and 9. Further, at most one of them is zero. Then which of the following numbers necessarily produces an integer, when multiplied by D?

- A.
18

- B.
108

- C.
198

- D.
288

Answer: Option C

**Explanation** :

D = 0.a_{1}a_{2}a_{1}a_{2}a_{1}a_{2}.....

∴100D = a_{1}a_{2}∙a_{1}a_{2}.....

100D – D = 99D = a_{1}a_{2}, which is an integer.

∴ If D is multiplied by 99 or any multiple of 99 it will result in an integer.

From the options given, it is clear that 198 (= 99 × 2) will necessarily produce an integer, when multiplied by D.

Hence, option 3.

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**2. CAT 2000 QA | Algebra - Functions & Graphs**

In the above table, for suitably chosen constants a, b and c, which one of the following best describes the relation between y and x?

- A.
y = a + bx

- B.
y = a + bx + cx

^{2} - C.
y = e

^{a + bx} - D.
None of the above

Answer: Option B

**Explanation** :

In the above table, it can be seen that the values are taken at equal intervals. Also, the second differences i.e. values of ∆^{2}y are constant and hence, the function is actually a polynomial of degree 2.

Thus, y = a + bx + cx^{2} best describes the relation between y and x.

Hence, option 2.

Alternatively,

The relation between x and is not a line since $\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}$ is not a constant.

Also, e^{a+bx} will not be a whole number. Hence, by substituting values from the table, it can be checked that the relation is given by y = a + bx + cx^{2}.

Hence, option 2.

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**3. CAT 2000 QA | Algebra - Progressions**

If a_{1} = 1 and a_{n+1} = 2a_{n} + 5, n = 1, 2 ... , then a_{100} is equal to

- A.
(5 × 299 – 6)

- B.
(5 × 299 + 6)

- C.
(6 × 299 + 5)

- D.
(6 × 299 – 5)

Answer: Option D

**Explanation** :

a_{1} = 1 = 6 – 5

a_{2} = 7 = 12 – 5 = 6 × 2 – 5

a_{3 }= 19 = 24 – 5 = 6 × 2^{2} – 5

a_{4} = 43 = 48 – 5 = 6 × 2^{3} – 5

and so on.

Thus, a_{100} = 6 × 2^{99} – 5

Hence, option 4.

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**4. CAT 2000 QA | Algebra - Progressions**

What is the value of the following expression?

$\frac{1}{{2}^{2}-1}$ + $\frac{1}{{4}^{2}-1}$ + $\frac{1}{{6}^{2}-1}$ + ... + $\frac{1}{{20}^{2}-1}$

- A.
$\frac{9}{19}$

- B.
$\frac{10}{19}$

- C.
$\frac{10}{21}$

- D.
$\frac{11}{21}$

Answer: Option C

**Explanation** :

$\frac{1}{{2}^{2}-1}+\frac{1}{{4}^{2}-1}+\frac{1}{{6}^{2}-1}+...+\frac{1}{{20}^{2}-1}$

$=\frac{1}{1\times 3}+\frac{1}{3\times 5}+\frac{1}{5\times 7}+...+\frac{1}{19\times 21}$

$=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)...+\left(\frac{1}{19}-\frac{1}{21}\right)\right]$

$=\frac{1}{2}\left[\left(1-\frac{1}{21}\right)\right]=\frac{1}{2}\left[\frac{20}{21}\right]=\frac{10}{21}$

Hence, option 3.

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**5. CAT 2000 QA | Arithmetic - Ratio, Proportion & Variation**

A truck travelling at 70 kilometres per hour uses 30% more diesel to travel a certain distance than it does when it travels at the speed of 50 kilometres per hour. If the truck can travel 19.5 kilometres on a litre of diesel at 50 kilometres per hour, how far can the truck travel on 10 litres of diesel at a speed of 70 kilometres per hour?

- A.
130

- B.
140

- C.
150

- D.
175

Answer: Option C

**Explanation** :

At 50 km/hr, truck covers 19.5 km in 1 litre diesel.

∴ The truck will use 1.3 litre diesel to cover 19.5 km at 70 km/hr.

∴ At 70 km/hr, in 10 litres of diesel, the truck will cover (19.5 × 10)/1.3 = 150 km

Hence, option 3.

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**6. CAT 2000 QA | Arithmetic - Average**

Consider a sequence of seven consecutive integers. The average of the first five integers is n. The average of all the seven integers is

- A.
n

- B.
n + 1

- C.
K × n, where K is a function of n

- D.
n + (2/7)

Answer: Option B

**Explanation** :

Average of first five integers is n.

∴ The integers are (n – 2), (n – 1), n, (n +1), (n + 2).

∴ The last two numbers are (n + 3), (n + 4).

∴ The average of the seven numbers = (7n + 7)/7 = n + 1

Hence, option 2.

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**7. CAT 2000 QA | Algebra - Inequalities & Modulus**

If x > 2 and y > – 1, Then which of the following statements is necessarily true?

- A.
xy > –2

- B.
–x < 2y

- C.
xy < –2

- D.
–x > 2y

Answer: Option B

**Explanation** :

y > –1

∴ –2y < 2 < x

∴ –x < 2y

Hence, option 2.

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**8. CAT 2000 QA | Modern Math - Permutation & Combination**

One red flag, three white flags and two blue flags are arranged in a line such that,

(A) no two adjacent flags are of the same colour.

(B) the flags at the two ends of the line are of different colours.

In how many different ways can the flags be arranged?

- A.
6

- B.
4

- C.
10

- D.
2

Answer: Option A

**Explanation** :

First arrange one red and two blue flags in 3 ways. (i.e. BBR, BRB, RBB)

Now there are four positions (say 1, 2, 3, 4) to arrange 3 white flags. Since the flags at the ends are of different colours, two white flags can’t be at positions 1 and 4 simultaneously. Thus, the three flags can be arranged at 1, 2, 3 or 2, 3, 4 in 2 ways.

Thus, the six flags can be arranged in 3 × 2 = 6 ways.

Hence, option 1.

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**9. CAT 2000 QA | Algebra - Number Theory**

Let S be the set of integers x such that

(i) 100 < x < 200

(ii) x is odd

(iii) x is divisible by 3 but not by 7

How many elements does S contain?

- A.
16

- B.
12

- C.
11

- D.
13

Answer: Option D

**Explanation** :

100 ≤ x ≤ 200 and x is odd multiple of 3 which is not divisible by 21.

Number of x those are odd multiples of 3 = 16

Number of x those are odd multiples of 21 = 3

∴ Number of elements in S = 16 − 3 = 13

Hence, option 4.

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**10. CAT 2000 QA | Algebra - Number Theory**

Let x, y and z be distinct integers, that are odd and positive. Which one of the following statements cannot be true?

- A.
xyz

^{2}is odd. - B.
(x − y)

^{2}z is even. - C.
(x + y − z)

^{2}(x + y) is even. - D.
(x − y) (y + z) (x + y − z) is odd.

Answer: Option D

**Explanation** :

Option 1:

Product of odd numbers is always odd.

∴ x × y × z × z is always odd

Option 2:

(x − y) is even.

∴ (x − y)^{2} z is even.

Option 3:

(x + y) is always even.

∴ (x + y − z)^{2} (x + y) is even.

Option 4:

(x − y) is even, (y + z) is even, and (x + y − z) is odd.

∴ (x − y) (y + z) (x + y − z) is even.

Hence, option 4.

Workspace:

**11. CAT 2000 QA | Algebra - Number Theory**

Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply all elements of S. With how many consecutive zeros will the product end?

- A.
1

- B.
4

- C.
5

- D.
10

Answer: Option A

**Explanation** :

There is only one even prime number and one prime number ending with 5 (i.e 5 itself) in set S.

When all elements of set S are multiplied, there will be only 1 zero at the end of the product.

Hence, option 1.

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**12. CAT 2000 QA | Geometry - Triangles**

What is the number of distinct triangles with integral valued sides and perimeter 14?

- A.
6

- B.
5

- C.
4

- D.
3

Answer: Option C

**Explanation** :

Let the sides of the triangle be a, b and c then, a + b > c.

a + b + c = 14

**Case 1**: Largest side is 6

Possible sets of sides are (6, 4, 4), (6, 5, 3) and (6, 6, 2).

**Case 2**: Largest side is 5

Possible set of sides is (5, 5, 4).

∴ 4 triangles are possible with integral sides and perimeter 14.

Hence, option 3.

Workspace:

**13. CAT 2000 QA | Algebra - Number Theory**

Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?

- A.
0

- B.
9

- C.
3

- D.
6

Answer: Option C

**Explanation** :

1421 ≡ 5(mod 12); 1423 ≡ 7(mod 12); 1425 ≡ 9(mod 12)

∴ N ≡ 5(mod 12) × 7(mod 12) × 9(mod 12)

= 315(mod 12) = 3(mod 12)

∴ The remainder is 3.

Hence, option 3.

Alternatively,

N = 1421 × 1423 × 1425

= (1422 – 1)(1422 + 1)(1428 – 3) = (1422^{2} – 12)(1428 – 3)

Both 1422^{2} and 1428 are divisible by 3 as well as 4.

N = (1422^{2} – 1)(1428 – 3)

(1422^{2} – 1) divided by 12 leaves a remainder of −1 and (1428 – 3) divided by 12 leaves a remainder of −3

Effective remainder = −1 × −3 = 3

∴ N when divided by 12 leaves a remainder of 3.

∴ N = X + 3; where X is divisible by 12

Thus, when N = 1421 × 1423 × 1425 is divided by 12, the remainder will be 3.

Hence, option 3.

Workspace:

**14. CAT 2000 QA | Algebra - Number Theory**

The integers 34041 and 32506 when divided by a three-digit integer ‘n’ leave the same remainder. What is ‘n’?

- A.
289

- B.
367

- C.
453

- D.
307

Answer: Option D

**Explanation** :

Let 34041 and 32506 when divided by n leave remainder r.

34041 = nk_{1} + r …(i)

32506 = nk_{2} + r …(ii)

Subtract equation (ii) from equation (i)

n (k_{1} – k_{2}) = 1535

∵ n has to be a factor of 1535.

From the given options, only 307 is a factor of 1535.

Hence, option 4.

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**15. CAT 2000 QA | Algebra - Number Theory**

Each of the numbers x_{1}, x_{2}...., x_{n}, n > 4, is equal to 1 or –1. Suppose,

x_{1}x_{2}x_{3}x_{4} + x_{2}x_{3}x_{4}x_{5} + x_{3}x_{4}x_{5}x_{6} + ... + x_{n–3}x_{n–2}x_{n–1}x_{n} + x_{n–2}x_{n–1}x_{n}x_{1}+ x_{n–1}x_{n}x_{1}x_{2} + x_{n}x_{1}x_{2}x_{3 }= 0, then,

- A.
n is even

- B.
n is odd

- C.
n is an odd multiple of 3

- D.
n is prime

Answer: Option A

**Explanation** :

The terms in the given expression are x_{1}, x_{2}, ... x_{n-1}, x_{n}.

There are n terms in the expression.

The only possible value of each of the terms is either 1 or –1.

For the given expression to be zero there should be even number of terms of which half the terms are –1 and the remaining are 1.

∴ n is even.

Hence, option 1.

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**16. CAT 2000 QA | Arithmetic - Percentage**

The table below shows the age-wise distribution of the population of Reposia. The number of people aged below 35 years is 400 million.

If the ratio of females to males in the ‘below 15 years’ age group is 0.96, then what is the number of females (in millions) in that age group?

- A.
82.8

- B.
90.8

- C.
80.0

- D.
90.0

Answer: Option B

**Explanation** :

Population below 35 years of age = 30 + 17.75 + 17 = 64.75% of the total population = 400 million

∴ 30% of the total population $=\frac{30\times 400}{64.75}$ million ≈ 185 million

The ratio of females to males in the ‘below 15 years’ age group is 0.96.

i.e. if the total population is 196, then there are 96 females.

Approximately, the number of females (in millions) in the ‘below 15 years’ age group

$=\frac{185\times 96}{196}=90.6$

Hence, option 2.

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**17. CAT 2000 QA | Modern Math - Permutation & Combination**

Sam has forgotten his friend’s seven-digit telephone number. He remembers the following: the first three digits are either 635 or 674, the number is odd, and the number nine appears once. If Sam were to use a trial and error process to reach his friend, what is the minimum number of trials he has to make before he can be certain to succeed?

- A.
1000

- B.
2430

- C.
3402

- D.
3006

Answer: Option C

**Explanation** :

**Case i:** The last digit is 9.

∴ The remaining 3 digits can be filled by any of the 9 digits (i.e., 0 to 8) in 9^{3} ways.

∴ Total number of ways = 729

**Case ii:** The last digit is not 9.

∴ The last digit can be filled in 4 ways.

Let any of the other 3 digits be 9. This can happen in 3 ways.

∴ The remaining two digits can be filled in 9^{2}, i.e. 81 ways.

∴ The total number of ways = 9^{2 }× 3 × 4 = 972

∴ Number of trials for last four digits = 729 + 972 = 1701

Also, for the first three digits he has two options.

The minimum number of trials he has to make before he can be certain to succeed = 2 × 1701 = 3402

Hence, option 3.

Workspace:

A, B, C are three numbers. Let

@ (A, B) = average of A and B,

/ (A, B) = product of A and B, and

X (A, B) = the result of dividing A by

**18. CAT 2000 QA | Algebra - Simple Equations**

The sum of A and B is given by

- A.
/(@( A, B), 2)

- B.
X(@(A, B), 2)

- C.
@(/(A, B), 2)

- D.
@(X(A, B), 2)

Answer: Option A

**Explanation** :

Sum of A and B = (Average of A and B) × 2

= /(@(A, B), 2)

∴ Hence, option 1.

Workspace:

**19. CAT 2000 QA | Algebra - Simple Equations**

Average of A, B and C is given by

- A.
@(/(@(/(B, A), 2), C), 3)

- B.
X(@(/(@(B, A), 3), C), 2)

- C.
/(@(X(@(B, A), 2), C), 3)

- D.
/(X(@(/(@(B, A), 2), C), 3), 2)

Answer: Option D

**Explanation** :

We need to find average of A, B and C i.e. (A + B + C)/3

The only operator that can give 3 in the denominator is “X”

**Option 1 **cannot be the answer, as there is no “X”.

**Option 2** also cannot be the answer as X(@(/(@(/(B, A), 2), C), 3), 2) gives 2 in the denominator and not 3.

Similarly option 3 is not the answer.

Hence, option 4 must be the answer.

**Option 4:**

/ ( X (@ (/ (@ (B, A), 2), C), 3), 2)

$=/\left(x\left(@\left(/\left(\frac{A+B}{2},2\right),c\right),3\right),2\right)$

=/(X(@((A + B), C), 3), 2)

$=/\left(x\left(\left(\frac{A+B+C}{2}\right),3\right),2\right)=/\left(\frac{A+B+C}{6},2\right)$

$=\frac{2(A+B+C)}{6}=\frac{A+B+C}{3}$

Hence, option 4.

Workspace:

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

For real numbers x, y, let

f(x, y) = Positive square-root of (x + y), if (x + y)^{0.5} is real

= (x + y)^{2}, otherwise

g(x, y) = (x + y)^{2}, if (x + y)^{0.5} is real

= –(x + y), otherwise

**20. CAT 2000 QA | Algebra - Functions & Graphs**

Which of the following expressions yields a positive value for every pair of non-zero real number (x, y)?

- A.
f(x, y) – g(x, y)

- B.
f(x, y) – (g(x, y))

^{2} - C.
g(x, y) – (f(x, y))

^{2} - D.
f(x, y) + g(x, y)

Answer: Option D

**Explanation** :

If (x + y) > 0, (x + y)^{0.5} is real.

∴ For (x + y) > 0;

f(x, y) = Positive square-root of (x + y) > 0

g(x, y) = (x + y)^{2} > 0

Otherwise,(i.e. for (x + y) < 0);

f(x, y) = (x + y)^{2} > 0

g(x, y) = –(x + y) > 0

Now it is clear that f(x, y) + g(x, y) will always be positive.

Hence, option 4.

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**21. CAT 2000 QA | Algebra - Functions & Graphs**

Under which of the following conditions is f(x, y) necessarily greater than g(x, y)?

- A.
Both x and y are less than –1

- B.
Both x and y are positive

- C.
Both x and y are negative

- D.
y > x

Answer: Option A

**Explanation** :

If 1 > (x + y) > 0; f(x, y) > g(x, y) and if (x + y) < –1; f(x, y) > g(x, y).

If both x and y are less than –1, (x + y) will also be less than –1.

Hence, option 1.

Workspace:

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

For three distinct positive real numbers x, y and z, let

f(x, y, z) = min(max(x, y), max(y, z), max(z, x))

g(x, y, z) = max(min(x, y), min(y, z), min(z, x))

h(x, y, z) = max(max(x, y), max(y, z), max(z, x))

j(x, y, z) = min(min(x, y), min(y, z), min(z, x))

m(x, y, z) = max(x, y, z)

n(x, y, z) = min(x, y, z)

**22. CAT 2000 QA | Algebra - Simple Equations**

Which of the following is necessarily greater than 1?

- A.
(h(x, y, z) – f(x, y, z))/j(x, y, z)

- B.
j(x, y, z)/h(x, y, z)

- C.
f(x, y, z)/g(x, y, z)

- D.
(f(x, y, z) + h(x, y, z) – g(x, y, z))/j(x, y, z)

Answer: Option D

**Explanation** :

x, y and z are distinct real numbers.

∴ Without loss of generality, let x < y < z

Then,

f(x, y, z) = y; g(x, y, z) = y; h(x, y, z) = z

j(x, y, z) = x; m(x, y, z) = z; n(x, y, z) = x

Substituting these values in the given options,

Option 1 = (z – y)/x may or may not be greater than 1.

Option 2 = x/z < 1

Option 3 = y/y = 1

Option 4 = (y + z − y)/x > 1

Hence, option 4.

Workspace:

**23. CAT 2000 QA | Algebra - Functions & Graphs**

Which of the following expressions is necessarily equal to 1?

- A.
(f(x, y, z) – m(x, y, z))/(g(x, y, z) – h(x, y, z))

- B.
(m(x, y, z) – f(x, y, z))/(g(x, y, z) – n(x, y, z))

- C.
(j(x, y, z) – g(x, y, z))/h(x, y, z)

- D.
(f(x, y, z) – h(x, y, z))/f(x, y, z)

Answer: Option A

**Explanation** :

Using the values from previous answer,

Option 1 = (y − z)/(y − z)

Option 2 = (z − y)/(y − x)

Option 3 = (x − y)/z

Option 4 = (y − z)/y

It can be easily concluded that the expression in option 1 is necessarily equal to 1.

Hence, option 1.

Workspace:

**24. CAT 2000 QA | Algebra - Functions & Graphs**

Which of the following expressions is indeterminate?

- A.
(f(x, y, z) – h(x, y, z))/(g(x, y, z) – j(x, y, z))

- B.
(f(x, y, z) + h(x, y, z) + g(x, y, z) + j(x, y, z))/(j(x, y, z) + h(x, y, z) – m(x, y, z) – n(x, y, z))

- C.
(g(x, y, z) – j(x, y, z))/(f(x, y, z) – h(x, y, z))

- D.
(h(x, y, z) – f(x, y, z))/(n(x, y, z) – g(x, y, z))

Answer: Option B

**Explanation** :

Using the values found previously,

Option 1 = (y − z)/(y − x)

Option 2 = (y + z + y + x)/(x + z – z − x)

Option 3 = (y − x)/(y – z)

Option 4 = (z − y)/(x – y)

The denominator in the expression in option 2 is zero.

Hence, the expression in option 2 is indeterminate.

Hence, option 2.

Workspace:

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

There are five machines A, B C, D and E situated on a straight line at distances of 10 metres, 20 metres, 30 metres, 40 metres and 50 metres respectively from the origin of the line. A robot is stationed at the origin of the line. The robot serves the machines with raw material whenever a machine becomes idle. All the raw material is located at the origin. The robot is in an idle state at the origin at the beginning of a day. As soon as one or more machines become idle, they send messages to the robot-station and the robot starts and serves all the machines from which it received messages. If a message is received at the station while the robot is away from it, the robot takes notice of the message only when it returns to the station. While moving, it serves the machines in the sequence in which they are encountered, and then returns to the origin. If any messages are pending at the station when it returns, it repeats the process again. Otherwise, it remains idle at the origin till the next message(s) is received.

**25. CAT 2000 QA | Miscellaneous**

Suppose on a certain day, machines A and D have sent the first two messages to the origin at the beginning of the first second, and C has sent a message at the beginning of the 5^{th} second and B at the beginning of the 6^{th} second, and E at the beginning of the 10^{th} second. How much distance in metres has the robot travelled since the beginning of the day, when it notices the message of E? Assume that the speed of movement of the robot is 10 metres per second.

- A.
140

- B.
80

- C.
340

- D.
360

Answer: Option A

**Explanation** :

Machines A and D sent message at the beginning of the first second.

∴ Round 1: Origin to Machine A to Machine D to Origin

Distance covered by the robot = 80 m (40 m + 40 m)

Time taken to return to origin = 8 seconds

At the end of 8th second, messages from C and B are already received,

∴ Round 2: Origin to Machine B to Machine C to Origin

Distance covered by the robot = 60 m (30 m + 30 m)

Time taken to return to origin = 6 seconds

Total Time elapsed = 8 + 6 = 14 seconds

At the end of 14th second, message from E is already there.

The robot has travelled 140 m (80 m + 60 m), when it notices message from E.

Hence, option 1.

Workspace:

**26. CAT 2000 QA | Miscellaneous**

Suppose there is a second station with raw material for the robot at the other extreme of the line which is 60 metres from the origin, that is, 10 metres from E. After finishing the services in a trip, the robot returns to the nearest station. If both stations are equidistant, it chooses the origin as the station to return to. Assuming that both stations receive the messages sent by the machines and that all the other data remains the same, what would be the answer to the above question?

- A.
120

- B.
140

- C.
340

- D.
70

Answer: Option A

**Explanation** :

Machines A and D sent message at the beginning of the first second.

∴ Round 1: Origin 1 to Machine A to Machine D to Origin 2

Distance covered by the robot = 10 m + 30 m + 20 m = 60 m

Time taken for round 1 = 6 seconds

At the end of 6th second, messages from C and B are already received,

∴ Round 2: Origin 2 to Machine C to Machine B to Origin 1

Distance covered by the robot = 30 m + 10 m + 20 m = 60 m

Time taken to return to origin = 6 seconds

Total Time elapsed = 6 + 6 = 12 seconds

At the end of 12th second, message from E is already there.

The robot has travelled 120 m (60 m + 60 m), when it notices message from E.

Hence, option 1.

Workspace:

Given below are three graphs made up of straight-line segments shown as thick lines. In each case choose the answer as

1. if f(x) = 3 f(–x);

2. if f(x) = –f(–x);

3. if f(x) = f(–x); and

4. if 3 f(x) = 6 f(–x), for x ≥ 0.

**27. CAT 2000 QA | Algebra - Functions & Graphs**

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

This is a graph of straight line that is parallel to x-axis passing through (0, 1)

∴ for every x, f(x) = 1

∴ f(x) = f(−x)

Hence, option 3.

Workspace:

**28. CAT 2000 QA | Algebra - Functions & Graphs**

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

As per the given figure,

*f*(1) = 2**, ** *f*(−1) = 1

∴ (1), (2) and (3) cannot be correct choices.

*f*(1) = 2*f*(−1)

i.e., 3 *f*(1) = 6 *f*(−1)

∴ 3*f*(*x*) = 6 *f*(−*x*)

Hence, option 4.

Workspace:

**29. CAT 2000 QA | Algebra - Functions & Graphs**

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

In the given figure, *f*(3) = −1, *f*(−3) = 1

i.e.* f*(3) = −*f*(−3)

∴ (1), (3) and (4) cannot be correct choices.

If we check with remaining values we conclude that *f*(*x*) = −*f*(−*x*)

Hence, option 2.

Workspace:

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

There are three bottles of water, A, B, C, whose capacities are 5 litres, 3 litres, and 2 litres respectively. For transferring water from one bottle to another and to drain out the bottles, there exists a piping system. The flow thorough these pipes is computer controlled. The computer that controls the flow through these pipes can be fed with three types of instructions, as explained below.

Initially, A is full with water, and B and C are empty.

**30. CAT 2000 QA | Arithmetic - Time & Work**

After executing a sequence of three instructions, bottle A contains one litre of water. The first and the third of these instructions are shown below

First instruction FILL (C, A)

Third instruction FILL (C, A)

Then which of the following statements about the instructions is true?

- A.
The second instruction is FILL (B, A)

- B.
The second instruction is EMPTY (C, B)

- C.
The second instruction transfers water from B to C

- D.
The second instruction involves using the water in bottle A.

Answer: Option B

**Explanation** :

Instruction 1: FILL (C, A)

2 litres of water is transferred from A to C

∴ A has 3 litres of water left and C has 2 litres of water.

Instruction 3: FILL (C, A)

2 litres of water is transferred from A to C

∴ A has 1 litres of water left and C has 2 litres of water.

Since the third instruction again transfers water from A to C, C should be empty before the third operation. Thus, instruction 2 should be either “EMPTY (C, B)” or “DRAIN (C)”.

Hence, option 2.

Workspace:

**31. CAT 2000 QA | Arithmetic - Time & Work**

Consider the same sequence of three instructions and the same initial state mentioned above. Three more instructions are added at the end of the above sequence to have A contain 4 litres of water. In this total sequence of six instructions, the fourth one is DRAIN (A). This is the only DRAIN instruction in the entire sequence. At the end of the execution of the above sequence, how much water (in litres) is contained in C?

- A.
One

- B.
Two

- C.
Zero

- D.
None of these

Answer: Option C

**Explanation** :

After third instruction A has 1 litre, B has 2 litres and C has 2 litres of water.

Instruction 4: DRAIN A

Instruction 5: EMPTY (B, A)

Instruction 6: EMPTY (C, A)

After 6^{th} instruction, C has no water as all 2 litres have been transferred to A.

Hence, option 3.

Workspace:

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

For a real number x, let

f(x) = 1/(1 + x), if x is non-negative

= 1+ x, if x is negative

f n(x) = f(f n – 1(x)), n = 2, 3, ....

**32. CAT 2000 QA | Algebra - Functions & Graphs**

What is the value of the product, f(2)f^{2}(2)f^{3}(2)f^{4}(2)f^{5}(2)?

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

- B.
3

- C.
$\frac{1}{18}$

- D.
None of these

Answer: Option C

**Explanation** :

When x = 2, f(x) = 1/(1 + x)

${f}^{3}\left(2\right)=\frac{1}{1+2}=\frac{1}{3},{f}^{2}\left(2\right)=f\left(f\right(2\left)\right)=f\left(\frac{1}{3}\right)=f\left(\frac{1}{3}\right)=\frac{1}{1+{\displaystyle \frac{1}{3}}}=\frac{3}{4}$

${f}^{3}\left(2\right)=f\left(\frac{3}{4}\right)=\frac{1}{1+{\displaystyle \frac{3}{4}}}=\frac{4}{7}$

${f}^{4}\left(2\right)=f\left(\frac{4}{7}\right)=\frac{1}{1+{\displaystyle \frac{4}{7}}}=\frac{7}{11}$

${f}^{5}\left(2\right)=f\left(\frac{7}{11}\right)=\frac{1}{1+{\displaystyle \frac{7}{11}}}=\frac{11}{18}$

∴ f(2)f^{2}(2)f^{3}(2)f^{4}(2)f^{5}(2)=$\frac{1}{3}\times \frac{3}{4}\times \frac{4}{7}\times \frac{7}{11}\times \frac{11}{18}=\frac{1}{18}$

Hence, option 3.

Hence, option 3.

Workspace:

**33. CAT 2000 QA | Algebra - Functions & Graphs**

r is an integer > 2. Then, what is the value of f ^{r – 1}(–r) + f ^{r}(–r)+ f^{ r + 1} (–r)?

- A.
-1

- B.
0

- C.
1

- D.
None of these

Answer: Option B

**Explanation** :

f(–r) = 1 + (–r) = 1 – r < 0 (∵ r ≥ 2)

f^{2}(–r) = f(f(–r) = f(1 – r) = 1 + (1 – r) = 2 – r

f^{2}(–r) = 2 – r = 0, if r = 2

< 0, if r > 2

Similarly,

f^{3}(–r) = 3 – r = 0, if r = 3

< 0, if r > 3

and so on.

i.e. f^{r–1}(–r) < 0 and fr(–r) = 0

∴ f^{r+1}(–r) = 1/(1+0) = 1

∴ f^{r – 1} (–r) + f^{r} (–r) + f ^{r + 1} (–r) = –1 + 0 + 1 = 0.

Hence, option 2.

Workspace:

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

Sixteen teams have been invited to participate in the ABC Gold Cup cricket tournament. The tournament is conducted in two stages. In the first stage, the teams are divided into two groups. Each group consists of eight teams, with each team playing every other team in its group exactly once. At the end of the first stage, the top four teams from each group advance to the second stage while the rest are eliminated. The second stage comprises of several rounds. A round involves one match for each team. The winner of a match in a round advances to the next round, while the loser is eliminated. The team that remains undefeated in the second stage is declared the winner and claims the Gold Cup.

The tournament rules are such that each match results in a winner and a loser with no possibility of a tie. In the first stage, a team earns one point for each win and no points for a loss. At the end of the first stage teams in each group are ranked on the basis of total points to determine the qualifiers advancing to the next stage. Ties are resolved by a series of complex tie-breaking rules so that exactly four teams from each group advance to the next stage.

**34. CAT 2000 QA | Algebra - Simple Equations**

to be updated

What is the total number of matches played in the tournament?

- A.
28

- B.
55

- C.
63

- D.
35

Answer: Option C

**Explanation** :

There are 8 teams in each group.

Stage 1:

Within a group each team played with every other team.

∴ Total number of matches = 7 + 6 + 5 + 4 + 3 + 2 + 1 = (7 × 8) / 2 = 28 matches in one group

∴ Total number of matches in both the groups = 56 matches

Stage 2:

Number of matches = 4 matches (Round 1) + 2 matches (Round 2) + 1 Match (Round 3) = 7

∴ Total number of matches played in the tournament = 63

Hence, option 3.

Workspace:

**35. CAT 2000 QA | Algebra - Simple Equations**

The minimum number of wins needed for a team in the first stage to guarantee its advancement to the next stage is

- A.
5

- B.
6

- C.
7

- D.
4

Answer: Option B

**Explanation** :

28 matches are played between 8 teams.

Consider option 1:

There is a possibility that 5 teams win exactly 5 matches each (5 × 5 = 25 matches played), one team wins 2 matches, one team wins exactly one match and one team lose all.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F; B wins matches against C, D, E, F, G; C wins matches against D, E, F, G, H; G wins matches against A, D, E, F, H and H wins matches against A, B, D, E, F. Also E wins match against D. F wins matches against D and E. D loses all the matches}

Thus, for a team, there is no guarantee of its advancement to the next stage if it wins 5 matches.

But if a team wins six matches, there is no possibility that 5 teams will win 6 matches each. (As there are only 28 matches) Hence, the team will definitely advance to the next stage.

Hence, option 2.

Workspace:

**36. CAT 2000 QA | Algebra - Simple Equations**

What is the highest number of wins for a team in the first stage in spite of which it would be eliminated at the end of first stage?

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

We need to find, maximum value of x, such that despite winning x matches, a team gets definitely eliminated at the end of the first stage.

If x = 4; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D(i.e. wins 2 matches);; C wins matches against D, E(i.e. wins 2 matches); D wins match against E; E wins match against B; F wins matches against B, C, D, E(i.e. wins 4 matches); G loose only one match against A (i.e. wins 6 matches); and H loose only one match against G (i.e. wins 6 matches).}

If x = 3; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D; C wins matches against D, E; D wins match against E and F(i.e. wins 2 matches); E wins match against B; F wins matches against B, C, E(i.e. wins 3 matches); G loose only one match against A (i.e. wins 6 matches); and H loose only one match against G (i.e. wins 6 matches).}

If x = 2; the team can reach in second stage.

{For example: Let A, B, C, D, E, F, G, and H are the eight teams. A wins matches against B, C, D, E, F, G(i.e. 6 matches); B wins matches against C, D; C wins matches against D, E; D wins match against E and F(i.e. wins 2 matches); E wins match against B and F(i.e. wins 2 matches); F wins matches against B, C,(i.e. wins 2 matches); G loose only one match against A (i.e. wins 6 matches); and H loose only one match against G (i.e. wins 6 matches).}

Thus, for x = 2, 3 and 4; we definitely cannot say that the team will not reach the second stage.

We can say that the value of x must be 1, because only four teams reach the second stage and it is not possible that more than two teams win exactly one match each.

Hence, option 1.

Workspace:

**37. CAT 2000 QA | Algebra - Simple Equations**

What is the number of rounds in the second stage of the tournament?

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

When the second stage starts, the first round will have 4 matches between 8 teams. From this round, only 4 teams will advance to the second round. So, second round will have 2 matches. In the third round, two teams will play 1 match and after that the winner will be declared.

Thus, the number of rounds in the second stage of the tournament = 3

Hence, option 3.

Workspace:

**38. CAT 2000 QA | Algebra - Simple Equations**

Which of the following statements is true?

- A.
The winner will have more wins than any other team in the tournament.

- B.
At the end of the first stage, no team eliminated from the tournament will have more wins than any of the teams qualifying for the second stage.

- C.
It is the possible that the winner will have the same number of wins in the entire tournament as a team eliminated at the end of the first stage.

- D.
The number of teams with exactly one win in the second stage of the tournament is 4.

Answer: Option C

**Explanation** :

We have seen that even if a team wins 2 matches, it can be in the second stage of the tournament. Consider Team A wins 2 matches and enters the second round and wins the tournament. Then its total score will be 5 points. While there can be a team having more than 6 points but not the winner of the tournament. Thus, option 1 may not be true.

We have seen that even if a team wins 5 matches is can be eliminated and a team winning only two matches can enter second round. Thus, option 2 may not be true.

Consider a team in group I eliminated and the score is 5. A team in group II advances to the second stage with score 2. If this team (i.e. the team with score 2) is a winner then the score will be 2 + 3 = 5. Thus, option 3 is true.

Eight teams enter second stage. Of these, 4 teams advances to the second round scoring 1 point each. In the second round, only two teams win and their score in the second round so far will be 2. Thus, only two teams will have score 1. Thus, option 4 is not true.

Hence, option 3.

Workspace:

**39. CAT 2000 QA | Algebra - Number Theory**

Let N = 55^{3 }+ 17^{3} – 72^{3}. N is divisible by

- A.
both 7 and 13

- B.
both 3 and 13

- C.
both 17 and 7

- D.
both 3 and 17

Answer: Option D

**Explanation** :

N = 55^{3} + 17^{3} − 72^{3} = 55^{3} + 17^{3} +(−72^{3})

It is known that if a + b + c = 0 then a^{3 }+ b^{3} + c^{3} = 3abc.

In the given expression, 55 + 17 + (−72) = 0

∴ N = 3 × 55 × 17 × (−72)

∴ N is divisible by both 3 and 17.

Hence, option 4

Workspace:

**40. CAT 2000 QA | Algebra - Inequalities & Modulus**

If x^{2} + y^{2} = 0.1 and |x – y| = 0.2, then |x| + |y| is equal to

- A.
0.3

- B.
0.4

- C.
0.2

- D.
0.6

Answer: Option B

**Explanation** :

x^{2} + y^{2} = 0.1 …(i)

|x – y| = 0.2

Squaring both the sides, we get,

x^{2} + y^{2} – 2xy = 0.04 …(ii)

From (i) and (ii),

2xy = 0.1 – 0.04 = 0.06

Thus, x and y both are positive or both are negative.

∴ 2xy = 2|x||y|= 0.06

(|x| + |y|)^{2} = x^{2} + y^{2} + 2|x||y| = 0.1 + 0.06 = 0.16

|x| + |y| > 0

∴ |x| + |y| = 0.4

Hence, option 2.

Workspace:

**41. CAT 2000 QA | Geometry - Quadrilaterals & Polygons**

ABCD is a rhombus with the diagonals AC and BD intersecting at the origin on the x-y plane. The equation of the straight line AD is x + y = 1. What is the equation of BC?

- A.
x + y = –1

- B.
x – y = –1

- C.
x + y = 1

- D.
None of the above

Answer: Option A

**Explanation** :

Equation of line AD is x + y = 1.

∴ AD intersects x-axis at (1, 0) and y-axis at (0, 1).

ABCD is a rhombus, with diagonals AC and BD intersecting at origin.

The four vertices are (1, 0), (0, 1), (−1, 0) and (0, −1).

Since AD pass through (1, 0) and (0, 1), BC passed through (−1, 0) and (0, −1).

Equation of BC $\frac{x}{-1}+\frac{y}{-1}=1$

i.e. x + y = −1

Hence, option 1.

Workspace:

**42. CAT 2000 QA | Geometry - Circles**

Consider a circle with unit radius. There are 7 adjacent sectors, S_{1}, S_{2}, S_{3},...., S_{7} in the circle such that their total area is (1/8)th of the area of the circle. Further, the area of the jth sector is twice that of the (j –1)th sector, for j = 2, ..., 7. What the angle, in radians, subtended by the arc of S_{1} at the centre of the circle?

- A.
π/508

- B.
π/2040

- C.
π/1016

- D.
π/1524

Answer: Option A

**Explanation** :

Let the area of sector S_{1} be x units.

The area of the sectors S_{2}, S_{3}, S_{4}, S_{5}, S_{6}, S_{7} will be 2x, 4x, 8x, 16x, 32x and 64x

∴ The total area of 7 sectors = 127x units = (1/8) × total area of circle = (1/8) π

∴ 127x = π/8 units

A circle subtends an angle of 2π at the centre.

Hence, (1/8)th of the circle will subtend an angle of π/4 at the centre.

i.e. area of the seven sectors i.e. 127x will subtend an angle of π/4 at the centre.

Sector S_{1}, whose area is x will subtend an angle of π/(127 × 4) at the centre.

∴The required angle = π/508 radians

Hence, option 1.

Workspace:

**43. CAT 2000 QA | Miscellaneous**

There is a vertical stack of books marked 1, 2, and 3 on Table-A, with 1 at the bottom and 3 on top. These are to be placed vertically on Table-B with 1 at the bottom and 2 on the top, by making a series of moves from one table to the other. During a move, the topmost book, or the topmost two books, or all the three, can be moved from one of the tables to the other. If there are any books on the other table, the stack being transferred should be placed on top of the existing books, without changing the order of books in the stack that is being moved in that move. If there are no books on the other table, the stack is simply placed on the other table without disturbing the order of books in it. What is the minimum number of moves in which the above task can be accomplished?

- A.
One

- B.
Two

- C.
Three

- D.
Four

Answer: Option D

**Explanation** :

In four moves the task can be accomplished.

One of the ways is given below.

Initial state: Table A: 3(on top) 2(in middle) 1(bottom)

Step 1: Move book 3 from table A to table B

Step 2: Move book 2 from table A to table B

Step 3: Move books 2(on top) and 3(below book 2) from table B to table A on book 1(bottommost)

Step 4: Move books 2, 3 and 1 from table A to table B.

Hence, option 4

Workspace:

**44. CAT 2000 QA | Geometry - Coordinate Geometry**

The area bounded by the three curves |x + y| = 1, |x| = 1, and |y| = 1, is equal to

- A.
4

- B.
3

- C.
2

- D.
1

Answer: Option B

**Explanation** :

We need to find area bounded by |x + y| = 1, |x| = 1, and |y| = 1

i.e. all points P(x, y) in the plane such that −1≤ x + y ≤ 1, −1≤ x ≤ 1, and −1≤ y ≤ 1

−1≤ x + y will be the origin side of line x + y = −1

x + y ≤ 1 will be the origin side of line x + y = 1

−1≤ x ≤ 1 will be the area between the lines x = −1 and x = 1.

−1≤ y ≤ 1 will be the area between the lines y = −1 and y = 1.

Tracing these curves, we get the area shown in the graph below

∴ Shaded Area = Area of 3 squares of side 1 unit = 3 × (1)2 = 3

Hence, option 2.

Workspace:

**45. CAT 2000 QA | Algebra - Quadratic Equations**

If the equation x^{3} – ax^{2} + bx – a = 0 has three real roots, then it must be the case that,

- A.
b = 1

- B.
b ≠ 1

- C.
a = 1

- D.
a ≠ 1

Answer: Option B

**Explanation** :

x^{3} – ax^{2} + bx – a = x^{3} + bx – ax^{2} – a = x(x^{2} + b) – a(x^{2} + 1)

∴ x(x^{2} + b) – a(x^{2} + 1) = 0

If b = 1 then x(x^{2} + 1) – a(x^{2} + 1) = (x – a)(x^{2} + 1) = 0

∴ (x – a) = 0 or (x^{2} + 1) = 0

But then (x^{2} + 1) = 0 does not have real roots.

∴ b ≠ 1

Hence, option 2.

Workspace:

**46. CAT 2000 QA | Geometry - Triangles**

If a, b, c are the sides of a triangle, and a^{2} + b^{2} + c^{2} = bc + ca + ab, then the triangle is

- A.
equilateral

- B.
isosceles

- C.
right angled

- D.
obtuse angled

Answer: Option A

**Explanation** :

a, b, c are the sides of a triangle.

It is given that a^{2} + b^{2} + c^{2} = bc + ca + ab

⇒ a^{2} + b^{2} + c^{2} - bc - ca - ab = 0

⇒ 2a^{2} + 2b^{2} + 2c^{2} - 2bc - 2ca - 2ab = 0

⇒ (a^{2} -2ab + b^{2}) + (b^{2} - 2bc + c^{2}) + (c^{2} - 2ca + a^{2}) = 0

⇒ (a - b)^{2} + (b - c)^{2} + (c - a)^{2} = 0

This is possible only if a = b = c

i.e., the triangle is equilateral.

Hence, option 1.

Workspace:

**47. CAT 2000 QA | Geometry - Triangles**

In the figure above, AB = BC = CD = DE = EF = FG = GA. Then ∠DAE is approximately

- A.
15°

- B.
20°

- C.
30°

- D.
25°

Answer: Option D

**Explanation** :

Consider the given figure,

Let ∠DAE = x

In ΔABC, AB = BC

∴ m∠ACB = m∠BAC = x

∴ Being an exterior angle of ΔABC, m∠CBF = 2x

In ΔAGF, FG = GA

∴ m∠AFG = m∠GAF = x

∴ Being an exterior angle of ΔAGF, m∠CGF = 2x

In ΔBCD, BC = CD

∴ m∠CBD = m∠CDB = 2x

∴ Being an exterior angle of ΔACD, m∠DCE = 3x

∵ DE = CD, in ΔDCE, m∠CED = m∠ECD = 3x

In ΔGFE, EF = FG

∴ m∠FGE = m∠FEG = 2x

∴ Being an exterior angle of ΔAFE, m∠EFD = 3x

∵ DE = EF, in ΔDFE, m∠EFD = m∠EDF = 3x

In ΔADE,

x + 3x + 3x = 7x = 180°

∴ x ≡ 25°

∴ m∠DAE ≡ 25°

Hence, option 4.

Workspace:

**48. CAT 2000 QA | Arithmetic - Average**

A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kg. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kg. What is the weight, in kg, of the heaviest box?

- A.
60

- B.
62

- C.
64

- D.
Cannot be determined

Answer: Option B

**Explanation** :

Let a, b, c, d and e be the weights, in kg ,of the five boxes with the shipping clerk, where,

a < b < c < d < e.

110 = a + b < a + c < ….. < c + e < d + e = 121

i.e. a + c = 112 and c + e = 120

Each box is weighed 4 times.

∴ 4a + 4b + 4c + 4d + 4e = 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1156

∴ a + b + c + d + e = 289

Now it is clear that a + b = 110 and d + e = 121

∴ 110 + c + 121 = 289

∴ c = 58

Substituting this value in c + e = 120

∴ e = 62

Hence, option 2.

Workspace:

**49. CAT 2000 QA | Modern Math - Permutation & Combination**

There are three cities A, B and C, each of these cities is connected with the other two cities by at least one direct road. If a traveller wants to go from one city (origin) to another city (destination), she can do so either by traversing a road connecting the two cities directly, or by traversing two roads, the first connecting the origin to the third city and the second connecting the third city to the destination. In all there are 33 routes from A to B (including those via C). Similarly there are 23 routes from B to C (including those via A). How many roads are there from A to C directly?

- A.
6

- B.
3

- C.
5

- D.
10

Answer: Option A

**Explanation** :

Let there be,

x roads connecting A and B directly,

y roads connecting B and C directly and

z roads connecting C and A directly

∴ Total number of routes connecting A and B: x + yz = 33 …(i)

∴ Total number of routes connecting B and C: y + xz = 23 …(ii)

Subtracting (ii) from (i)

(x − y) + z(y − x) = 10

−1(y − x) + z(y − x) = 10

(y − x)(z − 1) = 10

(y − x)(z − 1) = 5 × 2 …(iii)

From the options, the possible values for z are 3 and 6.

Consider z = 3

∴ y – x = 5 …(iv)

From equations (i), (ii) and (iv) we get the values as x = 4.5 and y = 9.5 which is not possible

Consider z = 6

∴ y – x = 2 …(v)

Solving (i), (ii) and (v), we get y = 5, x = 3

Thus, there are 6 direct roads between A and C.

Hence, option 1.

Workspace:

**50. CAT 2000 QA | Algebra - Functions & Graphs | Modern Math - Sets**

The set of all positive integers is the union of two disjoint subsets

{f(1), f(2) ....f(n),......} and {g(1), g(2),......,g(n),......}, where

f (1) < f(2) <...< f(n) ....., and g(1) < g(2) <...< g(n) ......., and

g(n) = f(f(n)) + 1 for all n ≥ 1.

What is the value of g(1)?

- A.
Zero

- B.
Two

- C.
One

- D.
Cannot be determined

Answer: Option B

**Explanation** :

The functions f(n) and g(n) are disjoint sets and union of these two sets is the set of all positive integers.

∵ g(n) = f(f(n)) + 1 for all n ≥ 1

and f (1) < f(2) <...< f(n) ....., and g(1) < g(2) <...< g(n) .......,

∴ f(1) = 1 or 2

If f(1) = 1

g(1) = f(f(1)) + 1

∴ g(1) = f (1) + 1 = 1 + 1 = 2

If f(1) = 2

g(1) = f(f(1)) + 1

∴ g(1) = f (2) + 1

∴ g(1) is greater than f(1), i.e. it is greater than 2.

But the set of all positive integers is the union of these two disjoint sets.

∴ This set has to include 1 which is not possible in this case as f(1) is 2 and g(1) will be greater than f(1).

∴ f(1) = 1 and g(1) = 2

Hence, option 2.

Workspace:

**51. CAT 2000 QA | Modern Math - Permutation & Combination**

ABCDEFGH is a regular octagon. A and E are opposite vertices of the octagon. A frog starts jumping from vertex to vertex, beginning from A. From any vertex of the octagon except E, it may jump to either of the two adjacent vertices. When it reaches E, the frog stops and stays there. Let an be the number of distinct paths of exactly n jumps ending in E. Then what is the value of a_{2n – 1}?

- A.
Zero

- B.
Four

- C.
2n – 1

- D.
Cannot be determined

Answer: Option A

**Explanation** :

The frog has to jump at least four times to reach E.

i.e. a_{4} = 2

(i.e. (i) A-B, B-C,C-D and D-E; (ii) A-H, H-G, G-F and F-E).

If the frog keeps jumping on left (right) hand side vertices it will not take more than four jumps.

If it jumps on right vertex and then keeps on jumping left vertices then it will take 6 jumps to reach E.

∴ We will not find any path such that the frog takes 5 jumps and reaches E.

∴ a_{5} = 0

Similarly, we find that only for even values of n the frog can reach E.

2n − 1 is an odd number

∴ No route is possible with odd number of jumps.

Hence, option 1.

Workspace:

**52. CAT 2000 QA | Algebra - Functions & Graphs**

For all non-negative integers x and y, f(x, y) is defined as below

f(0, y) = y + 1

f(x + 1, 0) = f(x, 1)

f(x + 1,y + 1) = f(x, f(x + 1, y))

Then, what is the value of f(1, 2)?

- A.
Two

- B.
Four

- C.
Three

- D.
Cannot be determined

Answer: Option B

**Explanation** :

f (1, 2) = f(0 + 1, 1 + 1) = f(0, f(0 + 1, 1)) = f(0, f (1, 1))

f (1, 1) = f(0, f(1, 0))

f(1, 0) = f(0, 1) = 2

∴ f(1, 1) = f(0, 2) = 2 + 1 = 3

f (1, 2) = f(0 + 1, 1 + 1) = f(0, f(0 + 1, 1)) = f(0, f (1, 1)) = f(0, 3)

f(0, 3) = 3 + 1 = 4

Hence, option 2.

Workspace:

**53. CAT 2000 QA | Algebra - Number System**

Convert the number 1982 from base 10 to base 12. The result is

- A.
1182

- B.
1912

- C.
1192

- D.
1292

Answer: Option C

**Explanation** :

To convert 1982 from base 10 to 12:

Divide 1982 by 12, write the quotient in the column below 1982 and the remainder in the next column. Repeat the process till the quotient is 0.

After that, write the remainders in the reverse order i.e. upward direction.

1982_{10} = 1192_{12}

Hence, option 3.

Workspace:

**54. CAT 2000 QA | Geometry - Mensuration**

Two full tanks, one shaped like a cylinder and the other like a cone, contain jet fuel. The cylindrical tank holds 500 litres more than the conical tank. After 200 litres of fuel has been pumped out from each tank the cylindrical tank contains twice the amount of fuel in the conical tank. How many litres of fuel did the cylindrical tank have when it was full?

- A.
700

- B.
1000

- C.
1100

- D.
1200

Answer: Option D

**Explanation** :

Let the volume of the cylindrical tank be x.

∴ Volume of the conical tank = x − 500

When 200 litres of fuel is pumped out of each tank, the amount in the cylindrical tank is twice that in the cone.

∴ (x − 200) = 2(x − 500 − 200)

Solving this equation, we get x = 1200

∴ The cylindrical tank had 1200 litres of jet fuel when it was full.

Hence, option 4.

Workspace:

**55. CAT 2000 QA | Geometry - Mensuration**

A farmer has decided to build a wire fence along one straight side of his property. For this, he planned to place several fence-posts at six metre intervals, with posts fixed at both ends of the side. After he bought the posts and wire, he found that the number of posts he had bought was five less than required. However, he discovered that the number of posts he had bought would be just sufficient if he spaced them eight metres apart. What is the length of the side of his property and how many posts did he buy?

- A.
100 metres, 15

- B.
100 metres, 16

- C.
120 metres, 15

- D.
120 metres, 16

Answer: Option D

**Explanation** :

Let the number of posts to be bought be n.

The length of the side of the property, when posts are at 6 meters distance = 6(n − 1) …(i)

The length of the side of the property, when posts are at 8 meters distance = 8(n − 5− 1) …(ii)

From (i) and (ii),

∴ 6(n − 1) = 8(n − 5− 1)

Solving this, we get n = 21

∴ Length of property = 6(21 − 1) = 120

∴ Number of posts the farmer bought = 21 − 5 = 16

Hence, option 4.

Workspace:

**56. CAT 2000 QA | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Consider three real numbers, X, Y and Z. Is Z the smallest of these numbers?

- X is greater than at least one of Y and Z.
- Y is greater than at least one of X and Z.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

From statement A, there are four possibilities.

X > Y > Z

X > Z > Y

Z > X > Y

Y > X > Z

∴ This statement alone is not sufficient.

From statement B, there are four possibilities.

Y > X > Z

Y > Z > X

X > Y > Z

Z > Y > X

∴ This statement alone is not sufficient.

Using both statements together we get,

Either X > Y > Z or Y > X > Z.

In both the cases Z is smallest.

Hence, option 3.

Workspace:

**57. CAT 2000 QA | Algebra - Number Theory | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Let X be a real number. Is the modulus of X necessarily less than 3?

- X(X + 3) < 0
- X(X – 3) > 0

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

From statement A

X (X + 3) < 0

∴ –3 < X < 0

∴ |X| < 3

∴ Statement A alone is sufficient.

From statement B

X(X – 3) > 0

X < 0 or X > 3

This statement does not give a unique answer.

Hence, option 1.

Workspace:

**58. CAT 2000 QA | Venn Diagram | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

How many people are watching TV programme P?

- Number of people watching TV programme Q is 1000 and number of people watching both the programmes, P and Q, is 100.
- Number of people watching either P or Q or both is 1500.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

From Statement A

Q = 1000

P ∩ Q = 100

This doesn’t gives any information regarding how many people are watching TV programme P

From Statement B

P ∪ Q = 1500

This statement alone is not sufficient.

Let’s combine the information in both the statements.

∴ P ∪ Q = P + Q − P ∩ Q

∴ 1500 = P + 1000 – 100

∴ P = 600

Hence, option 3.

Alternatively

By combining both the statements together we can arrive at the above Venn diagram which gives the number of people watching programme P.

Hence, option 3.

Workspace:

**59. CAT 2000 QA | Geometry - Circles | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Triangle PQR has angle PRQ equal to 90 degrees. What is the value of PR + RQ?

- Diameter of the inscribed circle of the triangle PQR is equal to 10 cm.
- Diameter of the circumscribed circle of the triangle PQR is equal to 26 cm.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

From statement A

PF = PR – FR = PR – OD = PR – 5

QD = QR – DR = QR – OF = QR – 5

∵ PE = PF and QE = QD

∴ PE = PR – 5 and QE = QR – 5

∴ PQ = PE + QE = PR – 5 + QR – 5

∴ PR + QR = PQ + 10

This statement alone is not sufficient to find PR + RQ.

From Statement B

Diameter of the circumscribing circle = hypotenuse of the triangle

∴ PQ = 26

∵ PR^{2} + RQ^{2 }= PQ^{2}

But this does not give the value of PR + RQ.

∴ This statement alone is not sufficient.

Let’s combine both the statements.

∴ PR + QR = PQ + 10 = 26 + 10 = 36

Hence, option 3.

Workspace:

**60. CAT 2000 QA | Arithmetic - Profit & Loss | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Harshad bought shares of a company on a certain day, and sold them the next day. While buying and selling he had to pay to the broker one percent of the transaction value of the shares as brokerage. What was the profit earned by him per rupee spent on buying the shares?

- The sales price per share was 1.05 times that of its purchase price.
- The number of shares purchased was 100.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

From Statement A

The cost of buying the shares for Harshad is

CP + 0.01 CP = 1.01 CP

The cost of selling the shares for Harshad is

SP – 0.01 SP = 0.99 SP

∴ Profit = Cost of selling – Cost of buying

= 0.99 SP – 1.01 CP

∵ SP = 1.05 CP

Profit = (0.99 × 1.05 CP) – 1.01 CP

= 0.0295 CP

∴ Profit earned per rupee spent on buying the shares is Rs 0.0295.

From the number of shares given in the statement B, we cannot conclude anything.

Hence, option 1.

Workspace:

**61. CAT 2000 QA | Geometry - Coordinate Geometry | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

There are two straight lines in the x-y plane with equations ax + by = c , dx + ey = f. Do the two straight lines intersect?

- a, b, c, d, e and f are distinct real numbers.
- c and f are non-zero.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement A implies that a, b, c, d, e, f are distinct real numbers.

But if a/d = b/e = c/f then the lines may be parallel and might not intersect at all.

∴ Statement A alone is not sufficient.

Statement B implies that the equations are not homogenous equations.

∴ Statement B is also not sufficient.

The condition for intersecting lines is a/d ≠ b/e

Even after combining both the statements the above condition is not clear.

∴ We cannot be sure whether the lines are intersecting or parallel.

Hence, option 4.

Workspace:

**62. CAT 2000 QA | Geometry - Circles | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

O is the centre of two concentric circles. ae is a chord of the outer circle and it intersects the inner circle at point; b and d. c is a point on the chord in between b and d. What is the value of ac/ce?

- bc/cd = 1
- A third circle intersects the inner circle at b and d and the point c is on the line joining the centres of the third circle and the inner circle.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

From statement A

bc = cd

If c is the midpoint of bd it would also be midpoint of ae because circles are concentric.

∴ ac = ce

∴ The question can be answered using statement A alone.

From Statement B

If c is the point on the line joining the two centres, it has to bisect the chord bd.

∴ c will also bisect the chord ae as the circles are concentric.

∴ ac = ce

∴ The question can be answered using statement B alone also.

Hence, option 2.

Workspace:

**63. CAT 2000 QA | Arithmetic - Time, Speed & Distance | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Ghosh Babu has decided to take a non-stop flight from Mumbai to No-man’s-land in South America. He is scheduled to leave Mumbai at 5 am, Indian Standard Time on December 10, 2000. What is the local time at No-man’s-land when he reaches there?

- The average speed of the plane is 700 kilometres per hour.
- The flight distance is 10,500 kilometres.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

By combining both the statements we can get the duration of the flight. But for the arrival time we should have the information regarding the time zone difference of Mumbai and No man’s land.

Hence, option 4.

Workspace:

**64. CAT 2000 QA | Algebra - Number Theory | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

What are the ages of two individuals, X and Y?

- The age difference between them is 6 years.
- The product of their ages is divisible by 6.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

From statement A

X – Y = 6

From statement B

XY is divisible by 6.

∵ There are many possible values for example (12, 6), (18, 12), (24, 18) ...

∴ The question cannot be answered even by combining both the statements.

Hence, option 4.

Workspace:

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