# CAT 1999 QA | Previous Year CAT Paper

**1. CAT 1999 QA | Algebra - Number Theory**

The number of positive integer valued pairs (x, y) satisfying 4x – 17y = 1 and x ≤ 1000 is

- A.
59

- B.
57

- C.
55

- D.
58

Answer: Option A

**Explanation** :

The difference between two integers will be 1, only if one is even and the other one is odd. 4x will always be even, so 17y has to be odd and hence, y has to be odd.

Moreover, the number 17y should be such a number that is 1 less than a multiple of 4. In other words, we have to find all such multiples of 17, which are 1 less than a multiple of 4. The first such multiple is 51.

Now you will find that as the multiples of 17 goes on increasing, the difference between it and its closest higher multiple of 4 is in the following pattern, 0, 3, 2, 1, e.g. 52 – 51 = 1, 68 – 68 = 0, 88 – 85 = 3, 104 – 102 = 2, 120 – 119 = 1,136 – 136 = 0

So the multiples of 17 that we are interested in are 3, 7, 11, 15 .

Now since, x ≤ 1000, 4x ≤ 4000 . The multiple of 17 closest and less than 4000 is 3995 (17 × 235). And incidentally, 3996 is a multiple of 4, i.e. the difference is 4.

This means that in order to find the answer, we need to find the number of terms in the AP formed by 3, 7, 11, 15 … 235, where a = 3, d = 4.

Since, we know that Tn = a + (n – 1)d

⇒ 235 = 3 + (n – 1) × 4

Hence, n = 59.

**Alternate Solution:**

4x – 17y = 1 and x ≤ 1000

so 17y + 1≤ 4000 i.e. y ≤ 235 and moreover every 4th value of y with give value of x.

So number of values $\frac{235}{4}\approx 58$

Hence, total number of terms will be 58 + 1 = 59.

Hence, option (a).

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**2. CAT 1999 QA | Algebra - Number Theory**

Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both defined under the usual decimal number system, if (ab)^{2} = ccb > 300, then the value of b is

- A.
1

- B.
0

- C.
5

- D.
6

Answer: Option A

**Explanation** :

(ab)^{2} = ccb, the greatest possible value of ‘ab’ to be 31.

Since 31^{2} = 961 and since ccb > 300, 300 < ccb < 961, so 18 < ab < 31.

So the possible value of ab which satisfies (ab)^{2} = ccb is 21.

So 21^{2} = 441,

∴ a = 2, b = 1, c = 4.

Hence, option (a).

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**3. CAT 1999 QA | Algebra - Number Theory**

The remainder when 7^{84} is divided by 342 is

- A.
0

- B.
1

- C.
49

- D.
341

Answer: Option B

**Explanation** :

Note: 342 = 7^{3} – 1. On further simplification we get,

$\mathrm{R}\left[\frac{{\left({7}^{3}\right)}^{28}}{342}\right]$ = $\mathrm{R}\left[\frac{{\left(343\right)}^{28}}{342}\right]$ = $\mathrm{R}{\left[\frac{343}{342}\right]}^{28}$ = $\mathrm{R}{\left[\frac{1}{342}\right]}^{28}$ = 1

∴ Remainder = 1,

Hence, option (b).

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**4. CAT 1999 QA | Modern Math - Permutation & Combination**

Ten points are marked on a straight-line and 11 points are marked on another straight-line. How many triangles can be constructed with vertices from among the above points?

- A.
495

- B.
550

- C.
1045

- D.
2475

Answer: Option C

**Explanation** :

The answer is ^{10}C_{2} × 11 + ^{11}C_{2} × 10 = 45 × 11 + 55 x 10 = 1045.

Hence, option (c).

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**5. CAT 1999 QA | Modern Math - Permutation & Combination**

For a scholarship, at the most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is

- A.
3

- B.
4

- C.
6

- D.
5

Answer: Option A

**Explanation** :

At least 1 and at most n are to be selected

⇒ ^{2n+1}C_{1} + ^{2n+1}C_{2} + ... +^{ 2n+1}C_{n} = 63

$\Rightarrow \frac{1}{2}$(2^{2n+1} - 2) = 63

⇒ n = 3

Hence, option (a).

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**6. CAT 1999 QA | Arithmetic - Ratio, Proportion & Variation**

The speed of a railway engine is 42 kmph when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 kmph when 9 compartments are attached, the maximum number of compartments that can be carried by the engine is

- A.
49

- B.
48

- C.
46

- D.
47

Answer: Option B

**Explanation** :

18 ∝ $\sqrt{9}$

42 ∝ $\sqrt{x};$ Here x = number of compartments

$\frac{18}{42}=\frac{\sqrt{9}}{\sqrt{x}}$

Simplifying, x = 49, but this is with reference to maximum speed. Hence number of compartments would be one less in order to run i.e. 48.

Hence, option (b).

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**7. CAT 1999 QA | Arithmetic - Ratio, Proportion & Variation**

Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

- A.
550

- B.
580

- C.
540

- D.
570

Answer: Option A

**Explanation** :

Let x be the fixed cost and y the variable cost 17500 = x + 25y … (i)

30000 = x + 50y … (ii)

Solving the equation (i) and (ii), we get

x = 5000, y = 500

Now if the average expense of 100 boarders be ‘A’.

Then

100 × A = 5000 + 500 × 100

∴ A = 550.

Hence, option (a).

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**8. CAT 1999 QA | Arithmetic - Percentage**

Forty per cent of the employees of a certain company are men, and 75% of the men earn more than Rs. 25,000 per year. If 45% of the company’s employees earn more than Rs. 25,000 per year, what fraction of the women employed by the company earn less than or equal to Rs. 25,000 per year?

- A.
$\frac{2}{11}$

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

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

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

Answer: Option D

**Explanation** :

Men Women

40% 60%

Out of 40% men, 75% earn more than Rs. 25,000.

Hence, 30% of the company (men) earn more than Rs. 25,000.

But, in all 45% of the employees earn more than Rs. 25,000.

Hence, among women 15% earn more than Rs. 25,000 and the remaining (60 – 15)% earn less than or equal to Rs. 25,000.

Therefore, the fraction of women = $\frac{45}{60}=\frac{3}{4}$

Hence, option (d).

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**9. CAT 1999 QA | Algebra - Inequalities & Modulus**

If |r − 6| = 11 and |2q − 12| = 8,what is the minimum possible value of $\frac{q}{r}$?

- A.
$-\frac{2}{5}$

- B.
$\frac{2}{17}$

- C.
$\frac{10}{17}$

- D.
None of these

Answer: Option D

**Explanation** :

|r − 6| = 11 ⇒ r − 6 = 11, r = 17

or –(r – 6) = 11, r = –5

|2q – 12| = 8 ⇒ 2q –12 = 8 , q = 10

or 2q – 12 = –8 , q =2

Hence, minimum value of $\frac{q}{r}=\frac{10}{-5}=-2$

Hence, option (d).

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**10. CAT 1999 QA | Algebra - Number Theory**

If n = 1 + x where x is the product of four consecutive positive integers, then which of the following

is/are true?

A. n is odd

B. n is prime

C. n is a perfect square

- A.
A and C only

- B.
A and B only

- C.
A only

- D.
None of these

Answer: Option A

**Explanation** :

Use the method of simulation, viz. take any sample values of x and verify that n is both odd as well as a perfect square.

**Alternate solution:**

Let x = (n – 1) n (n + 1) (n + 2)

= (n^{2} – 1) n(n + 2)

= n^{4 }+ 2n^{3} – n^{2 }– 2n

p = x + 1 = n^{4} + 2n^{3} – n^{2} – 2n + 1

= (n^{2} – n + 1)^{2}

Hence, option (a).

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**11. CAT 1999 QA | Venn Diagram**

In a survey of political preferences, 78% of those asked were in favour of at least one of the proposals: I, II and III.

50% of those asked favoured proposal I, 30% favoured proposal II and 20% favoured proposal III. If 5% of those asked favoured all three of the proposals, what percentage of those asked favoured more than one of the three proposals?

- A.
10

- B.
12

- C.
17

- D.
22

Answer: Option C

**Explanation** :

Let ‘a’ be the percentage of people who favoured exactly one proposal, ‘b’ be the percentage of people who favoured exactly by two proposals and ‘c’ be the percentage of people who favoured exactly three proposals.

a + b + c = 78......(i)

a + 2b + 3c = 100.....(ii)

(ii) – (i) implies b + 2c = 22

Since c = 5, b = 12

Required percentage = b + c = 12 + 5 = 17%.

Hence, option (c).

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**12. CAT 1999 QA | Miscellaneous**

For two positive integers a and b define the function h(a,b) as the greatest common factor (G.C.F) of a, b. Let A be a set of n positive integers. G(A), the G.C.F of the elements of set A is computed by repeatedly using the function h. The minimum number of times h is required to be used to compute G is

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

- B.
(n - 1)

- C.
n

- D.
None of these

Answer: Option B

**Explanation** :

If there are n numbers, the function h has to be performed one time less.

Hence, option (b).

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**13. CAT 1999 QA | Geometry - Quadrilaterals & Polygons**

The figure below shows two concentric circles with centre O. PQRS is a square inscribed in the outer circle. It also circumscribes the inner circle, touching it at points B, C, D and A. What is the ratio of the perimeter of the outer circle to that of polygon ABCD?

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

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

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

- D.
π

Answer: Option C

**Explanation** :

Let the radius of the outer circle be OQ = x.

Hence, perimeter of the circle = 2πx

But OQ = BC = x (diagonals of the square BQCO)

Perimeter of ABCD = 4x

Hence, ratio = $\frac{2\pi x}{4x}=\frac{\pi}{2}.$

Hence, option (c).

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**14. CAT 1999 QA | Miscellaneous**

Three labelled boxes containing red and white cricket balls are all mislabelled. It is known that one of the boxes contains only white balls and another one contains only red balls. The third contains a mixture of red and white balls. You are required to correctly label the boxes with the labels red, white and red and white by picking a sample of one ball from only one box. What is the label on the box you should sample?

- A.
white

- B.
red

- C.
red and white

- D.
Not possible to determine from a sample of one ball

Answer: Option C

**Explanation** :

Test the boxes labelled — Red and White.

Now if the ball is Red, label the box — Red

Now the box which has the label White is either Red or Red and White.

However, it cannot be Red.

Hence, it is Red and White.

The last box is White.

Hence, option (c).

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**15. CAT 1999 QA | Algebra - Number Theory**

If n^{2} = 12345678987654321, what is n?

- A.
12344321

- B.
1335789

- C.
111111111

- D.
11111111

Answer: Option C

**Explanation** :

The square root is 111111111.

Hence, option (c).

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**16. CAT 1999 QA | Arithmetic - Time, Speed & Distance**

Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6.30 a.m. and travels at 50 kmph towards Baroda situated 100 km away. At 7.00 a.m. Howrah-Ahmedabad Express leaves Baroda towards Ahmedabad and travels at 40 kmph. At 7.30 a.m. Mr Shah, the traffic controller at Baroda realizes that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?

- A.
15 min

- B.
20 min

- C.
25 min

- D.
30 min

Answer: Option B

**Explanation** :

At 7.30 a.m., Navjivan Express is at 50 km from A at the same time, Howrah-Ahmedabad Express is at 20 km from B.

Hence, distance between the trains at 7.30 a.m. is 30 km.

Relative speed = 50 + 40 = 90 kmph

Hence, time left = $\frac{30}{90}=\frac{1}{3}$hr = 20 min.

Hence, option (b).

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**17. CAT 1999 QA | Geometry - Quadrilaterals & Polygons**

There is a circle of radius 1 cm. Each member of a sequence of regular polygons S1(n), n = 4, 5, 6, …, where n is the number of sides of the polygon, is circumscribing the circle: and each member of the sequence of regular polygons S2(n), n = 4, 5, 6, … here n is the number of sides of the polygon, is inscribed in the circle. Let L1(n) and L2(n) denote the perimeters of the corresponding

polygons of S1(n) and S2(n), then $\frac{\left\{L1\right(13)+2\pi \}}{L2\left(17\right)}$ is

- A.
greater than $\frac{\pi}{4}$ and less than 1

- B.
greater than 1 and less than 2

- C.
greater than 2

- D.
less than $\frac{\pi}{4}$

Answer: Option C

**Explanation** :

Following rule should be used in this case: The perimeter of any polygon circumscribed about a circle is always greater than the circumference of the circle and the perimeter of any polygon inscribed in a circle is always less than the circumference of the circle.

Since, the circles is of radius 1, its circumference will be 2 π .

Hence, L1(13) > 2π and L2(17) < 2π.

So {L1(13) + 2 π } > 4 π .

Hence, $\frac{\left\{L1\right(13)+2\pi \}}{L2\left(17\right)}$ will be greater than 2.

Hence, option (c).

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**18. CAT 1999 QA | Geometry - Mensuration**

There is a square field of side 500 m long each. It has a compound wall along its perimeter. At onec of its corners, a triangular area of the field is to be cordoned off by erecting a straight-line fence. The compound wall and the fence will form its borders. If the length of the fence is 100 m, what is the maximum area that can be cordoned off?

- A.
2,500 sq.m

- B.
10,000 sq.m

- C.
5,000 sq.m

- D.
20,000 sq.m

Answer: Option A

**Explanation** :

$=\frac{1}{2}\times \frac{100}{\sqrt{2}}\times \frac{100}{\sqrt{2}}=$2,500 sq.m

Area of a Δ is maximum when it is an isosceles Δ.

So perpendicular sides should be of length $\frac{100}{\sqrt{2}}$.

Hence, option (a).

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**Directions: ****Answer the questions based on the following information.**

A young girl Roopa leaves home with x flowers, goes to the bank of a nearby river. On the bank of the river, there are four places of worship, standing in a row. She dips all the x flowers into the river. The number of flowers doubles. Then she enters the first place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the second place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the third place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the fourth place of worship, offers y flowers to the deity. Now she is left with no flowers in hand.

**19. CAT 1999 QA | Algebra - Simple Equations**

If Roopa leaves home with 30 flowers, the number of flowers she offers to each deity is

- A.
.30

- B.
31

- C.
32

- D.
33

Answer: Option C

**Explanation** :

Starting from the fourth place of worship and moving backwards, we find that number of flowers before entering the first place of worship is $\frac{15}{8}y.$

Hence, number of flowers before doubling = $\frac{15}{16}y$

(but this is equal to 30)

Hence, y = 32

Hence, option (c).

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**20. CAT 1999 QA | Algebra - Simple Equations**

The minimum number of flowers that could be offered to each deity is

- A.
0

- B.
15

- C.
16

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Starting from the fourth place of worship and moving backwards, we find that number of flowers before

entering the first place of worship is $\frac{15}{8}y.$

The minimum value of y so that $\frac{15}{16}y$ is a whole number is 16.

Therefore, 16 is the minimum number of flowers that can be offered.

Hence, option (c).

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**21. CAT 1999 QA | Algebra - Simple Equations**

The minimum number of flowers with which Roopa leaves home is

- A.
16

- B.
15

- C.
0

- D.
Cannot be determined

Answer: Option B

**Explanation** :

Starting from the fourth place of worship and moving backwards, we find that number of flowers before

entering the first place of worship is $\frac{15}{8}y.$

For y = 16, the value of $\frac{15}{16}y=15.$

Hence, the minimum number of flowers with which Roopa leaves home is 15.

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**Directions: Answer the questions based on the following information.**

The following table presents the sweetness of different items relative to sucrose, whose sweetness is taken to be 1.00.

**22. CAT 1999 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

What is the minimum amount of sucrose (to the nearest gram) that must be added to one gram of saccharin to make a mixture that will be at least 100 times as sweet as glucose?

- A.
7

- B.
8

- C.
9

- D.
100

Answer: Option C

**Explanation** :

If the mixture is to be made 100 times as sweet as glucose, its sweetness should be 74. The ratio in which saccharin and sucrose be mixed to get the above level of sweetness is given by the following alligation table.

In other words, it means to achieve the given level of sweetness, you need to add 601 g of sucrose to 73 g of saccharin. Hence to 1 g of saccharin, the amount of sucrose to be added is $\frac{601}{73}$ = 8.23.g.

Hence, option (c).

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**23. CAT 1999 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Approximately how many times sweeter than sucrose is a mixture consisting of glucose, sucrose and fructose in the ratio of 1 : 2 : 3?

- A.
1.3

- B.
1.0

- C.
0.6

- D.
2.3

Answer: Option A

**Explanation** :

$\frac{\left[\right(0.74)+(1.000)2+(1.7\left)3\right]}{6}$ = 1.31.

Hence, option (a).

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**Directions : **Answer the questions based on the following information.

A rectangle PRSU, is divided into two smaller rectangles PQTU, and QRST by the line TQ. PQ = 10 cm. QR = 5 cm and RS = 10 cm. Points A, B, F are within rectangle PQTU, and points C, D, E are within the rectangle QRST. The closest pair of points among the pairs (A, C), (A, D), (A, E), (F, C), (F, D), (F, E), (B, C), (B, D), (B, E) are $10\sqrt{3}$ cm apart.

**24. CAT 1999 QA | Geometry - Quadrilaterals & Polygons**

Which of the following statements is necessarily true?

- A.
The closest pair of points among the six given points cannot be (F, C)

- B.
Distance between A and B is greater than that between F and C.

- C.
The closest pair of points among the six given points is (C, D), (D, E), or (C, E).

- D.
None of the above

Answer: Option A

**Explanation** :

The sides of the rectangle PRSU are PR = 15 cm and RS = 10 cm.

∴ Its diagonal = $\sqrt{{15}^{2}+{10}^{2}}=\sqrt{325}\approx 18$ cm.

Given that the minimum distance between any pair of points formed by taking one from A, B and F and the other from C, D and E is $10\sqrt{3}=\sqrt{300}\approx 17$ cm, which is close to the length of the diagonal of PRSU. This implies that each of A, B and F are close to one of the vertices P or U and each of C, D and E are close to one of the vertices S or R.

Note that all the three points in one of the rectangles PQTU and QRST will be close to the same vertex (or else the minimum distance between one of these points and one of the three in the other rectangle will be less than $10\sqrt{3}$ cm). Also, the points (A, B, F) and (C, D, E) have to be diagonally opposite as maximum distance between two vertices on same side is 15 cm.

**Option (a):** The closest two points of the given six points have to be any two out of A, B, F or any two out of C, D, E (since they are closest to the same vertex). Therefore, (F, C) cannot be the closest pair of points as they are diagonally opposite.

**Option (b):** It is definitely false as A and B are close to the same vertex, while F and C are close to diagonally opposite vertices.

**Option (c):** It is possible but not necessary that the closest pair of points among the six given points is (C,D), (D, E) or (C, E). The other possibilities are (A, B), (B,F) or (A,F).

Hence, option (a).

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**25. CAT 1999 QA | Geometry - Quadrilaterals & Polygons**

AB > AF > BF ; CD > DE >CE ; and BF = 6$\sqrt{5}$ cm. Which is the closest pair of points among all the six given points?

- A.
B, F

- B.
C, D

- C.
A, B

- D.
None of these

Answer: Option D

**Explanation** :

The maximum possible distance between the pairs (C,D), (D,E) or (C, E) is the length of the diagonal of the rectangle QRST, i.e. $\sqrt{{10}^{2}+{5}^{2}}=5\sqrt{5}$ cm.

Since AB > AF > BF = $6\sqrt{5}$ cm, the closest pair of points of the given six points will be from the set (C, D, E). As CD > DE > CE, so (C, E) will be the pair of closest points.

*Note: There is slight inconsistency regarding the information given in the question. If BF = $6\sqrt{5}$ ≈ 13.4 cm, then A, B and F cannot be close to the same vertex as the length of the diagonal of rectangle PQTU is 14 cm approximately. This in turn will contradict the fact the minimum distance between any point of A, B, F and the other from C, D, E is 10$\sqrt{3}$ cm.

A likely possibility is that the information regarding minimum distance between any point of (A, B, F) and the other from (C, D, E) is specific to previous question only.

Hence, option (d).

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Directions: Answer the questions based on the following information.

In each of the following questions, a pair of graphs F(x) and F1(x) is given. These are composed of straightline segments, shown as solid lines, in the domain x ∈ (−2, 2).

**Choose the answer as**

a. if F1(x) = –F(x)

b. if F1(x) = F(–x)

c. if F1(x) = –F(–x)

d. if none of the above is true

**26. CAT 1999 QA | Algebra - Functions & Graphs**

Answer: 4

**Explanation** :

The graph F(x) represents the function F(x) = | x |, where x is any real number.

The graph of F1(x) represents the function F1(x) = − x, where x is any real number.

None of the given relationships are satisfied by these two functions.

**Alternate solution:**

F1(-2) = 2 = F(−2) and F1(2) = −2. But F(2) = 2.

Hence, option (d).

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**27. CAT 1999 QA | Algebra - Functions & Graphs**

Answer: 2

**Explanation** :

F(x) = $\left\{\begin{array}{l}0\mathrm{when}x\ge 0\\ x\mathrm{when}x0\end{array}\right.$ and F1(x) = $\left\{\begin{array}{l}\begin{array}{ccc}-x& \mathrm{when}& x>0\end{array}\\ \begin{array}{ccc}0& \mathrm{when}& x\le 0\end{array}\end{array}\right.$

Therefore, replacing x by (−x) in above functions, we get

F(-x) = $\left\{\begin{array}{l}0\mathrm{when}x\le 0\\ -x\mathrm{when}x0\end{array}\right.$ and F1(-x) = $\left\{\begin{array}{l}x\mathrm{when}x0\\ 0\mathrm{when}x\ge 0\end{array}\right.$

Clearly, F1(x) = F(−x), hence, option (b) is the correct choice.

**Alternate solution:**

F1(-2) = 0 = F(2) and F1(2) = −2 = F(−2).

So the correct option is (b), i.e. F1(x) = F(−x).

Hence, option (b).

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**28. CAT 1999 QA | Algebra - Functions & Graphs**

Answer: 2

**Explanation** :

F(x) = $\left\{\begin{array}{lll}0& when& x\le 0\\ -x& when& x>0\end{array}\right.$ and F1(x) = $\left\{\begin{array}{lll}x& when& x<0\\ 0& when& x\ge 0\end{array}\right.$

Therefore, replacing x by (−x) in above functions, we get

F(-x) = $\left\{\begin{array}{lll}0& when& x\ge 0\\ x& when& x<0\end{array}\right.$ and F1(x) = $\left\{\begin{array}{lll}-x& when& x>0\\ 0& when& x\le 0\end{array}\right.$

Clearly, F1(x) = F(−x), hence, option (b) is the correct choice.

**Alternate solution:**

F1(-2) = −2 = F(2) and F1(2) =0 = F(−2).

So the correct option is (b), i.e. F1(x) = F(−x)

Hence, option (b).

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**29. CAT 1999 QA | Algebra - Functions & Graphs**

Answer: 3

**Explanation** :

F(x) = $\left\{\begin{array}{lll}1-x& wnen& 0\le x<2\\ 1& when& -2<x<0\end{array}\right.$ and F1(x) = $\left\{\begin{array}{lll}-1-x& when& -2<x<0\\ -1& when& 0\le x<2\end{array}\right.$

Therefore, replacing x by (−x) in above functions, we get

F(-x) = $\left\{\begin{array}{lll}1+x& when& -2<x<0\\ 1& when& 0\le x<2\end{array}\right.$ and F1(-x) = $\left\{\begin{array}{lll}-1+x& when& 0\le x<2\\ -1& when& -2<x<0\end{array}\right.$

∴ -F(-x) = - $\left\{\begin{array}{lll}1+x& when& -2<x<0\\ 1& when& 0\le x<2\end{array}\right.$

= $\left\{\begin{array}{lll}-1-x& when& -2<x<0\\ -1& when& 0\le x-2\end{array}\right.$ = F1(x)

Hence, option (c) is the correct option.

**Alternate solution:**

F1(2) = −1 = F(2) and F1(−2) = 1 = F(−2).

So the correct option is (c), i.e. F1(x) = − F(−x).

Hence, option (c).

Workspace:

**Directions: Answer the questions based on the following information.**

There are blue vessels with known volumes v_{1}, v_{2}..., v_{m}, arranged in ascending order of volume, v_{1} > 0.5 litre, and v_{m} < 1 litre. Each of these is full of water initially. The water from each of these is emptied into a minimum number of empty white vessels, each having volume 1 litre. The water from a blue vessel is not emptied into a white vessel unless the white vessel has enough empty volume to hold all the water of the blue vessel. The number of white vessels required to empty all the blue vessels according to the above rules was n.

**30. CAT 1999 QA | Miscellaneous**

Among the four values given below, which is the least upper bound on e, where e is the total empty volume in the white vessels at the end of the above process?

- A.
mv

_{m} - B.
m(1 - v

_{m}) - C.
mv

_{1} - D.
m(1 - v

_{1})

Answer: Option D

**Explanation** :

Let m = 1. So, option (a) will give the answer as V_{m} and option (c) will give the answer as V_{1}. Both of these cannot be the answers as V_{m} and V_{1} are the amount of volume filled.

Let m = 2. So, option (b) will give the answer as 2 (1 – V_{2}) and option (d) will give the answer as 2(1 – V_{1}). Now consider option (b).

Actual empty volume > 2(1 – V_{2}). Therefore, for this situation m(1 – V_{1}) is the only possible answer.

Hence, option (d).

Workspace:

**31. CAT 1999 QA | Miscellaneous**

Let the number of white vessels needed be n_{1} for the emptying process described above, if the volume of each white vessel is 2 litres. Among the following values, which is the least upper bound on n_{1}?

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

- B.
Smallest integer greater than or equal to $\left(\frac{n}{2}\right)$

- C.
n

- D.
Greatest integer less than or equal to $\left(\frac{n}{2}\right)$

Answer: Option B

**Explanation** :

Let m = 1 and n = 1. Option (a) gives the answer as $\frac{1}{4}$ and option (d) gives the answer as ‘greatest integer less than or equal to $\frac{1}{2}.$ So, both of these cannot be the answer. Option (b) gives the answer as ‘smallest integer greater than or equal to $\frac{1}{2}$ and option (c) gives the answer as 1. But the actual answer can be greater than 1 as the volume of the vessel is 2 l.

Hence, (b) is the answer.

Workspace:

**Directions: **Answer the questions based on the following information.

There are 50 integers a_{1}, a_{2} … a_{50}, not all of them necessarily different. Let the greatest integer of these 50 integers be referred to as G, and the smallest integer be referred to as L. The integers a_{1} through a_{24} form sequence S1, and the rest form sequence S2. Each member of S1 is less than or equal to each member of S2.

**32. CAT 1999 QA | Algebra - Progressions**

All values in S1 are changed in sign, while those in S2 remain unchanged. Which of the following statements is true?

- A.
Every member of S1 is greater than or equal to every member of S2.

- B.
G is in S1.

- C.
If all numbers originally in S1 and S2 had the same sign, then after the change of sign, the largest number of S1 and S2 is in S1.

- D.
None of the above

Answer: Option D

**Explanation** :

The ideal approach is to pick up the options one by one.

**Option (a)** – Let S1 and S2 be two sequences of positive numbers. After change of sign, S1 will consist of negative numbers while S2 remains unchanged.

Clearly, the members of S1 would be less than that of S2.

Hence, option (a) is not correct.

**Option (b) **- Let S1 and S2 be two sequences of positive numbers. After change of sign, S1 will consist of negative numbers while S2 remains unchanged.

Clearly, G would remain in S2 itself. Hence, option (b) is not correct.

**Option (c)** – If S1 and S2 had same sign, say positive initially, then the largest number of S1 and S2 would be in S2. Then after the change of sign, every member of S1 will be negative and therefore, less than every member of S2. This implies that the largest number would remain in S2.

Hence, option (c) is not correct.

Workspace:

**33. CAT 1999 QA | Algebra - Progressions**

Elements of S1 are in ascending order, and those of S2 are in descending order. a_{24} and a_{25} are interchanged. Then which of the following statements is true?

- A.
S1 continues to be in ascending order.

- B.
S2 continues to be in descending order.

- C.
S1 continues to be in ascending order and S2 in descending order.

- D.
None of the above

Answer: Option A

**Explanation** :

The elements of S1 are in the order : a_{1} < a_{2} < a_{3} < a_{4} < . . . < a_{24}

The elements of S2 are in the order: a_{25 }> a_{26} > . . . > a_{49} > a_{50}

Even if a_{24} and a_{25} are interchanged, the elements of S1 continues to be in ascending order. However, nothing can be concluded about the elements of S2.

Workspace:

**34. CAT 1999 QA | Algebra - Progressions**

Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then x cannot be less than

- A.
2

^{10} - B.
the smallest value of S2

- C.
the largest value of S2

- D.
(G – L)

Answer: Option D

**Explanation** :

Since every element of S1 is less than or equal to each member of S2, L will be in S1 and G in S2.

For some i (1≤ i ≤ 24), ai = L and for some j (25 ≤ j ≤ 50), a_{j} = G.

Every other element of S1 is greater than ai and every other member of S2 is less than a_{j}.

Therefore, to make every element of S1 greater than or equal to that of S2, we need to add a minimum of (a_{j} –a_{i}) = G – L.

Workspace:

**Directions: **Answer the questions based on the following information.

Let x and y be real numbers and let

f (x,y) = |x + y| , F(f (x,y)) = −f (x,y)

and G(f (x, y)) = −F(f (x, y))

**35. CAT 1999 QA | Algebra - Functions & Graphs**

Which of the following statements is true?

- A.
F(f(x, y)) ⋅ G(f(x, y)) = –F(f(x, y)) ⋅ G(f(x, y))

- B.
F(f(x, y)) ⋅ G(f(x, y)) > –F(f(x, y)) ⋅ G(f(x, y))

- C.
F(f(x, y)) ⋅ G(f(x, y)) ≠ G(f(x, y)) ⋅ G(f(x, y))

- D.
F(f(x, y)) + G(f(x, y)) + f(x, y) = f(–x, –y)

Answer: Option D

**Explanation** :

f (x,y) = |x + y| − − − This is always positive

F(f (x,y)) = − f (x,y) = −|x + y| − − − This is always negative

G(f (x,y)) = − F(f (x,y)) = − (−|x + y|) = |x + y| − − − This is always positive

F(f(x, y))G(f(x, y)) = -|x + y)^{2}

and G(f(x, y)) G(f(x, y)) = |x + y|^{2}

From the choices, we observe that:

**Option (a):** LHS of the expression is -|x + y)^{2}, which is always non positive. RHS of the expression

is |x + y)^{2}, which is always non negative. The only situation when LHS is equal to RHS is when each is equal to zero. Hence, (a) is not necessarily true. Option (b): The given expression can be written as

-|x + y|^{2} > |x + y|^{2} or 0 > 2|x + y|^{2}

This implies that 0 > |x + y|, which is not true. Hence, (b) is not true.

**Option (c): **F(f(x, y)) G(f(x, y)) = - |x + y|^{2}

and G(f(x, y)) G(f(x, y)) = |x + y|^{2}

These two expressions can be equal if |x + y| = 0.

Hence, (c) is not necessarily true.

**Option (d):** F(f (x,y)) + G(f (x,y)) + f (x,y)

=-|x + y| + |x + y| + |x + y| = |x + y|

f (−x,−y) = |(−x) + (−y)| = |−x − y| = |−(x + y)| = |x + y|

Therefore, the two expressions are equal.

Workspace:

**36. CAT 1999 QA | Algebra - Functions & Graphs**

What is the value of f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2))))?

- A.
3

- B.
2

- C.
1

- D.
0

Answer: Option C

**Explanation** :

f (x,y) = |x + y| This is always positive

F(f (x,y)) = − f (x,y) = −|x + y| This is always negative

G(f (x,y)) = − F(f (x,y)) = − (−|x + y|) = |x + y| This is always positive.

ƒ(G(ƒ(1, 0)), ƒ(F (ƒ(1,2)), G(ƒ(1, 2))))

= ƒ(G(ƒ(1, 0)), ƒ(3, − 3))

= ƒ(G(ƒ(1, 0)), 0)

= ƒ(−1, 0) = 1.

Workspace:

**37. CAT 1999 QA | Algebra - Functions & Graphs**

Which of the following expressions yields x^{2} as its result?

- A.
F(f (x, − x))⋅G(f (x, − x))

- B.
F(f (x, x))⋅G(f (x, x))⋅ 4

- C.
-F(f(x, x)) ∙ G(f(x, x)) ÷ log

_{2}16 - D.
f (x, x)⋅ f (x, x)

Answer: Option C

**Explanation** :

f (x,y) = |x + y| − − − This is always positive

F(f (x,y)) = − f (x,y) = −|x + y| − − − This is always negative

G(f (x,y)) = − F(f (x,y)) = − (−|x + y|) = |x + y| − − − This is always positive.

The option (c) yields x^{2}.

-F(f(x, x)) ∙ G(f(x, x)) ÷ log_{2} 16

= -(-2x ∙ 2x) ÷ log_{2} 16

= $\frac{4{x}^{2}}{{\mathrm{log}}_{2}{2}^{4}}={x}^{2}$

Workspace:

**Directions: **Answer the questions based on the following information.

A robot moves on a graph sheet with X and Y-axis. The robot is moved by feeding it with a sequence of instructions. The different instructions that can be used in moving it, and their meanings are:

**38. CAT 1999 QA | Geometry - Coordinate Geometry**

The robot reaches point (6, 6) when a sequence of three instructions is executed, the first of which is a GOTO(x,y) instruction, the second is WALKX(2) and the third is WALKY(4). What are the value of x and y?

- A.
2, 4

- B.
0, 0

- C.
4, 2

- D.
2, 2

Answer: Option C

**Explanation** :

The final point is (6, 6). The previous point is (6, 2) and the one before is (4, 2).

Workspace:

**39. CAT 1999 QA | Geometry - Coordinate Geometry**

The robot is initially at (x, y), x > 0 and y < 0. The minimum number of instructions needed to be executed to bring it to the origin (0, 0) if you are prohibited from using the GOTO instruction is

- A.
2

- B.
1

- C.
x + y

- D.
0

Answer: Option A

**Explanation** :

Two instructions are needed, one parallel to the X-axis and the other parallel to the Y-axis. i.e. WALKX(–x) and WALKY (–y)

Workspace:

**Directions: **Answer the questions based on the following information.

A road network (shown in the figure below) connects cities A, B, C and D. All road segments are straight lines. D is the mid-point on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long.

Ms X and Mr Y leave A at 8.00 a.m., take different routes to city C and reach at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 kmph. Y takes the direct route AC and travels at 45 kmph on segment AD. Y’s speed on segment DC is 55 kmph.

**40. CAT 1999 QA | Arithmetic - Time, Speed & Distance**

What is the average speed of Y?

- A.
47.5 kmph

- B.
49.5 kmph

- C.
50 kmph

- D.
52 kmph

Answer: Option C

**Explanation** :

Since AD = DC, the distance travelled is same for the two stretches.

Hence, the average speed is given by $\frac{2ab}{(a+b)}$ = $\frac{(2\times 45\times 55)}{(45+55)}=$ 49.5 kmph

Workspace:

**41. CAT 1999 QA | Arithmetic - Time, Speed & Distance**

The total distance travelled by Y during the journey is approximately

- A.
105 km

- B.
150 km

- C.
130 km

- D.
Cannot be determined

Answer: Option A

**Explanation** :

Now, since X and Y reach C at the same time, $\frac{100+BC}{64.875}=\frac{AC}{49.5},$ Hence, you need to take help from options. i.e.

If AC = 105, BC^{2} = AC^{2} – AB^{2} = 105^{2} – 100^{2} = 1025 or BC = 32

So, $\frac{100+BC}{AC}=\frac{132}{105}=1.25$

Workspace:

**42. CAT 1999 QA | Arithmetic - Time, Speed & Distance**

What is the length of the road segment BD?

- A.
50 km

- B.
52.5 km

- C.
55 km

- D.
Cannot be determined

Answer: Option B

**Explanation** :

In this triangle, AD = DC = BD = $\frac{105}{2}$ = 52.5 km.

(Note: The best way to solve this question is to treat questions 104 and 105 together. You would figure out that the answer to question 105 has to be half of the answer to question 104, and the only pairs of answer choices that fit into this condition are 105 and 52.5)

Workspace:

**Directions:** Answer the questions based on the following information.

Rajiv reaches city B from city A in 4 hours, driving at speed of 35 kmph for the first two hour and at 45 kmph for the next two hours. Aditi follows the same route, but drives at three different speeds: 30, 40 and 50 kmph, covering an equal distance in each speed segment. The two cars are similar with petrol consumption characteristics (km per litre) shown in the figure below.

**43. CAT 1999 QA | Algebra - Simple Equations**

The quantity of petrol consumed by Aditi for the journey is

- A.
8.3 I

- B.
8.6 I

- C.
8.9 I

- D.
9.2 I

Answer: Option C

**Explanation** :

Distance between A and B = (35 × 2) + (45 × 2) = 160 km.

Distance covered by Aditi in each speed segment $=\frac{160}{3}$

Hence, total petrol consumed

$=\left(\frac{160}{3}\times \frac{1}{16}\right)+\left(\frac{160}{3}\times \frac{1}{24}\right)+\left(\frac{160}{3}\times \frac{1}{16}\right)=8.9l$

Workspace:

**44. CAT 1999 QA | Algebra - Simple Equations**

Zoheb would like to drive Aditi’s car over the same route from A to B and minimize the petrol consumption for the trip. What is the quantity of petrol required by him?

- A.
6.67 l

- B.
7 l

- C.
6.33 l

- D.
6.0 l

Answer: Option A

**Explanation** :

For minimum petrol consumption, Zoheb should drive at 40 kmph, petrol consumption = $\frac{160}{24}$ = 6.67 I.

Workspace:

**Directions: **Answer the questions based on the following information.

Recently, Ghosh Babu spent his winter vacation on Kyakya Island. During the vacation, he visited the local casino where he came across a new card game. Two players, using a normal deck of 52 playing cards, play this game. One player is called the ‘dealer’ and the other is called the ‘player’. First, the player picks a card at random from the deck. This is called the base card. The amount in rupees equal to the face value of the base card is called the base amount. The face values of ace, king, queen and jack are ten. For other cards the face value is the number on the card. Once the ‘player’ picks a card from the deck, the ‘dealer’ pays him the base amount. Then the ‘dealer’ picks a card from the deck and this card is called the top card. If the top card is of the same suit as the base card, the ‘player’ pays twice the base amount to the ‘dealer’. If the top card is of the same colour as the base card (but not the same suit), then the ‘player’ pays the base amount to the ‘dealer’. If the top card happens to be of a different colour than the base card, the ‘dealer’ pays the base amount to the ‘player’.

Ghosh Babu played the game four times. First time he picked eight of clubs and the ‘dealer’ picked queen of clubs. Second time, he picked ten of hearts and the ‘dealer’ picked two of spades. Next time, Ghosh Babu picked six of diamonds and the ‘dealer’ picked ace of hearts. Lastly, he picked eight of spades and the ‘dealer’ picked jack of spades. Answer the following questions based on these four games.

**45. CAT 1999 QA | Algebra - Simple Equations**

If Ghosh Babu stopped playing the game when his gain would be maximized, the gain in Rs. would have been

- A.
12

- B.
20

- C.
16

- D.
4

Answer: Option A

**Explanation** :

Hence, we see that the maximum gain is Rs. 12

Workspace:

**46. CAT 1999 QA | Algebra - Simple Equations**

The initial money Ghosh Babu had (before the beginning of the game sessions) was Rs. X. At no point did he have to borrow any money. What is the minimum possible value of X?

- A.
16

- B.
8

- C.
100

- D.
24

Answer: Option B

**Explanation** :

Since the maximum negative that Ghosh Babu goes into is –8, he should begin with at least Rs. 8, so that he does not have to borrow any money at any point.

Workspace:

**47. CAT 1999 QA | Algebra - Simple Equations**

If the final amount of money that Ghosh Babu had with him was Rs. 100, what was the initial amount he had with him?

- A.
120

- B.
8

- C.
4

- D.
96

Answer: Option D

**Explanation** :

From the above table it is evident that in four games, Ghosh Babu makes a profit of Rs. 4. Hence, if the final amount left with Ghosh Babu is Rs. 100, the initial amount that he had would be Rs. 96.

Workspace:

**48. CAT 1999 QA | Arithmetic - Average | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

The average weight of students in a class is 50 kg. What is the number of students in the class?

I. The heaviest and the lightest members of the class weigh 60 kg and 40 kg respectively.

II. Exclusion of the heaviest and the lightest members from the class does not change the average weight of the students.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement I gives the weight of the heaviest and lightest members of the class but no indication as to the number

of students in the class or the total weight of the students is there. The second statement is also inconclusive, making our answer choice as (d).

Workspace:

**49. CAT 1999 QA | Geometry - Mensuration | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

A small storage tank is spherical in shape. What is the storage volume of the tank?

I. The wall thickness of the tank is 1 cm.

II. When an empty spherical tank is immersed in a large tank filled with water, 20 l of water overflow from the large tank.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Statement I gives the thickness of the wall which is of no use to find the volume of the tank since we do not know the radius of the sphere.

Statement II gives us the answer as the volume of water displaced is equal to the volume of the immersed tank (from Archimedes’ principle)

So to find the exact storage volume of the tank both the statements are needed.

Workspace:

**50. CAT 1999 QA | Geometry - Quadrilaterals & Polygons | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

Mr X starts walking northwards along the boundary of a field from point A on the boundary, and after walking for 150 m reaches B, and then walks westwards, again along the boundary, for another 100 m when he reaches C. What is the maximum distance between any pair of points on the boundary of the field?

I. The field is rectangular in shape.

II. The field is a polygon, with C as one of its vertices and A as the mid-point of a side.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Statement I by itself does not solve the problem but it does tell us about the shape of the field. However, it fails to give information about the points A, B and C as to whether they be at the end of the field, etc. This data is given by the second statement, from which it is known that

The polygon has the length = 150 × 2 = 300 m and the breadth = 100 m and also that it is a rectangle (from A). Thus, the maximum distance is the diagonal length of the rectangle.

Workspace:

**51. CAT 1999 QA | Geometry - Trigonometry | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

A line graph on a graph sheet shows the revenue for each year from 1990 through 1998 by points and joins the successive points by straight-line segments. The point for revenue of 1990 is labelled A, that for 1991 as B, and that for 1992 as C. What is the ratio of growth in revenue between 1991-92 and 1990-91?

I. The angle between AB and X-axis when measured with a protractor is 40°, and the angle between CB and X-axis is 80°.

II. The scale of Y-axis is 1 cm = Rs. 100

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Ratio of revenues = $\frac{RQ}{QP}$

Since in a line graph, the years are uniformly spaced

$\Rightarrow \frac{RQ}{QP}=\frac{\mathrm{tan}80\xb0}{\mathrm{tan}40\xb0}$

So the ratio can be determined from statement I alone.

Statement II is immaterial because we intend to find the ratio and not absolute figures.

Workspace:

**52. CAT 1999 QA | Geometry - Circles | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

There is a circle with centre C at the origin and radius r cm. Two tangents are drawn from an external

point D at a distance d cm from the centre. What are the angles between each tangent and the X-axis.

I. The coordinates of D are given.

II. The X-axis bisects one of the tangents.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

r and d are given

From statement I, when co-ordinates of D are given,

only one pair of tangents can be drawn onto the given circle from D. So angle made by x-axis for each can be found out.

Hence, statement I alone is sufficient.

Consider statement II. Let the x-axis bisect the tangent QD, i.e. QA = AD.

Here QA = $\frac{1}{2}$QD = $\frac{1}{2}\sqrt{{d}^{2}-{r}^{2}}$. So using trigonometric ratios in right ΔCQA, we can determine ∠CAQ. Therefore, ∠DAB is equal to ∠CAQ CAQ (vertically opposite angles).

Consider the other tangent DP. Let it intersect x-axis at point L.

∠CDQ can be determined using trigonometric ratios (as two of the sides are given in right ΔCDQ). Also, ∠CDQ is equal to ∠CDP (since the two right ΔCQD and ΔCPD are congruent). Drop a perpendicular DB on x-axis. In right ΔDBL, we can find ∠BDL = 180o − (∠ADB + ∠CDQ + ∠CDP)

= 180° − 2∠CDQ − ∠ADB . Applying angle sum property of triangle, we can determine ∠DLB.

Hence, statement II alone is sufficient.

Workspace:

**53. CAT 1999 QA | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

Find a pair of real numbers x and y that satisfy the following two equations simultaneously. It is known that the values of a, b, c, d, e and f are non-zero.

ax + by = c

dx + ey = f

I. a = kd and b = ke, c = kf, k ≠ 0

II. a = b = 1, d = e = 2, f ≠ 2c

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement I when used to solve the sum gives us the same equation as the second substituted in to the first equation.

kdx + key = kf

∴ k(dx + ey) = kf

as k ≠ 0, dx + ey = f which is same as second equation.

So it is of no use as we get infinite solutions and not a unique one.

Statement II gives us the following equations.

x + y = c

2x + 2y = f.

These are two linear equations in x and y, such that

$\frac{1}{2}=\frac{1}{2}\ne \frac{c}{f}$

As $\frac{c}{f}\ne \frac{1}{2}$ (Given), the system will have no solution.

As the data given in both the statements is inconsistent, the question cannot be answered.

Workspace:

**54. CAT 1999 QA | Algebra - Number Theory | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

Three professors A, B and C are separately given three sets of numbers to add. They were expected to find the answers to 1 + 1, 1 + 1 + 2, and 1 + 1 respectively. Their respective answers were 3, 3 and 2. How many of the professors are mathematicians?

I. A mathematician can never add two numbers correctly, but can always add three numbers correctly.

II. When a mathematician makes a mistake in a sum, the error is +1 or –1.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

**Statement I:**

As C added up two numbers correctly, he is not a mathematician. However, from the given information, it is not necessary that any person who adds up two numbers incorrectly is a athematician.

Therefore, A or B may or may not be mathematicians. Hence, statement I alone is not sufficient.

**Statement II:**

If a mathematician makes a mistake in a sum, the error is +1 or -1. But it doesn't implies that if a

person makes an error of +1 or-1, he is a mathematician.

Hence, statement II alone is not sufficient.

Even on combining the two statements, we cannot conclude anything concrete.

Workspace:

**55. CAT 1999 QA | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

How many students among A, B, C and D have passed the examination?

I. The following is a true statement: A and B passed the examination.

II. The following is a false statement: At least one among C and D has passed the examination.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

From I, we know A and B passed the examination.

From II, we know the condition that among C and D at least one passed (or both passed) is false.

Therefore, it is obvious that both C and D have failed.

Thus, both statements are necessary to find the answer.

Workspace:

**56. CAT 1999 QA | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

What is the distance x between two cities A and B in integral number of kilometres?

I. x satisfies the equation ${\mathrm{log}}_{2}x=\sqrt{x}$

II. x ≤ 10 km

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

**Statement I:** Given that x satisfies the equation,

${\mathrm{log}}_{2}x=\sqrt{x}\phantom{\rule{0ex}{0ex}}\therefore x={2}^{\sqrt{x}}$

This equation is satisfied by the values x = 4 and 16.

Hence, statement I alone is not sufficient.

**Statement II: **Nothing concrete can be concluded from the fact that x ≤ 10 km.

Hence, statement II alone is not sufficient.

**Combining statements I and II, **we get a unique value of x = 4 km.

Workspace:

**57. CAT 1999 QA | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

Mr Mendel grew 100 flowering plants from black seeds and white seeds, each seed giving rise to one plant. A plant gives flowers of only one colour. From a black seed comes a plant giving red or blue flowers. From a white seed comes a plant giving red or white flowers. How many black seeds were used by Mr Mendel?

I. The number of plants with white flowers was 10.

II. The number of plants with red flowers was 70.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement I gives us the number of white flowers. But we know that a white seed gives both red or white flowers. Thus, proving statement II, gives the number of red flowers. But both black and white seeds give red flowers, again providing no solutions.

Workspace:

**58. CAT 1999 QA | Algebra - Logarithms | Data Sufficiency**

**Directions: **Each question is followed by two statements I and II. Mark:

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

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

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

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

What is the distance x between two cities A and B in integral number of kilometres?

I. x satisfies the equation ${\mathrm{log}}_{2}x=\sqrt{x}$

II. x ≤ 10 km

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

**Statement I:** Given that x satisfies the equation,

${\mathrm{log}}_{2}x=\sqrt{x}\phantom{\rule{0ex}{0ex}}\therefore x={2}^{\sqrt{x}}$

This equation is satisfied by the values x = 4 and 16.

Hence, statement I alone is not sufficient.

**Statement II: **Nothing concrete can be concluded from the fact that x ≤ 10 km.

Hence, statement II alone is not sufficient.

**Combining statements I and II, **we get a unique value of x = 4 km.

Workspace:

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