CAT 2002 QA questions and solutions

Previous year paper questions for CAT 2002 QA

1. CAT 2002 QA | Algebra - Number Theory

On dividing a number by 3, 4 and 7, the remainders are 2, 1 and 4 respectively. If the same number is divided by 84 then the remainder is

A.
80
B.
76
C.
56
D.
None of these

2. CAT 2002 QA | Algebra - Number Theory

There are three pieces of cake weighing 9/2 lbs, 27/4 lbs and 36/5 lbs. Pieces of the cake are equally divided and distributed in such a manner that every guest in the party gets one single piece of cake. Further the weight of the pieces of the cake is as heavy as possible. What is the largest number of guests to whom we can distribute the cake?

A.
54
B.
72
C.
20
D.
None of these

Answer: Option D

Explanation :

The largest possible piece of cake from the three given pieces [9/2, 27/4, 36/5] can be obtained by finding the H.C.F of the three fractions,

∴ H.C.F. $[\frac{9}{2}, \frac{27}{4}, \frac{36}{5}] = \frac{H . C . F . of Numerator}{L . C . M . of Denominator} = \frac{9}{20}$

∴ The weight of the piece of cake given to each guest 9/20 lbs.

∴ Number of Guests = $(\frac{\frac{9}{2}}{\frac{9}{20}}) + (\frac{\frac{27}{4}}{\frac{9}{20}}) + (\frac{\frac{36}{5}}{\frac{9}{20}}) =$ 10 + 15 + 16 = 41
Hence, option (d).

3. CAT 2002 QA | Algebra - Number Theory

For three real numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?

A.
3√13
B.
√19
C.
13/3
D.
√15

4. CAT 2002 QA | Algebra - Simple Equations

A transport company charges for its vehicles in the following manner:

If the driving is 5 hours or less, the company charges Rs. 60 per hour or Rs. 12 per km (which ever is larger)
If driving is more than 5 hours, the company charges Rs. 50 per hour or Rs. 7.5 per km (which ever is larger)

If Anand drove it for 30 km and paid a total of Rs. 300, then for how many hours does he drive?

A.
4
B.
5.5
C.
7
D.
6

5. CAT 2002 QA | Arithmetic - Time, Speed & Distance

Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day?

A.
1 : 3
B.
1 : 4
C.
1 : 5
D.
1 : 6

Answer: Option D

Explanation :

Let the speed of E be 2x km/hr.

∴ Speed of N = 4x km/hr and speed of S = x km/hr

According to the problem,

Time taken by N to cover AB + Time taken by S to cover BA = 1 hour

$∴ \frac{d}{4 x} + \frac{d}{x} = 1$

∴ 5d = 4x or d = $\frac{4 x}{5}$

∵ To maintain the schedule (gap of 20 min), N doubles its speed and speed of S becomes y.

$∴ \frac{\frac{4 x}{5}}{8 x} + \frac{\frac{4 x}{5}}{y} = \frac{2}{3}$

$∴ \frac{1}{10} + \frac{4 x}{5 y} = \frac{2}{3}$

$∴ \frac{4 x}{5 y} = \frac{17}{30}$

$∴ \frac{4 x}{y} = \frac{17}{6}$

$∴ \frac{8 x}{y} = \frac{17}{3} o r \frac{y}{8 x} = \frac{3}{17} \approx \frac{1}{6}$

$∴ \frac{S p e e d o f p a s s e n g e r t r a i n}{S p e e d o f s u p e r f a s t t r a i n} \approx \frac{1}{6}$

Hence, option (d).

6. CAT 2002 QA | Algebra - Number Theory

If x² + 5y² + z² = 2y(2x + z), then which of the following statements are necessarily true?

I. x = 2y
II. x = 2z
III. 2x = z

A.
Only I
B.
Only II
C.
Only III
D.
Only I and II

7. CAT 2002 QA | Geometry - Quadrilaterals & Polygons

In the following figure, the area of the isosceles right triangle ABE is 7 sq.cm. If EC = 3BE, then the area of rectangle ABCD (in sq. cm.) is

A.
64
B.
82
C.
26
D.
56

Answer: Option D

Explanation :

∵ The area of the ∆ABE = $\frac{1}{2} \times$ AB × BE = 7 cm²

∴ BE × AB = 14

The area of the rectangle ABCD = AB × BC = AB × (BE + EC) = 4 × (BE × AB) = 56 cm²

Hence, option (d).

8. CAT 2002 QA | Algebra - Number Theory

Number S is equal to the square of the sum of the digits of a 2 digit number D. If the difference between S and D is 27, then D is

A.
32
B.
54
C.
64
D.
52

9. CAT 2002 QA | Arithmetic - Average

A boy finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?

A.
4
B.
2
C.
6
D.
8

Answer: Option B

Explanation :

Let the Arithmetic Mean of the 10 numbers be x and s be the sum of the remaining 9 numbers.

$\frac{x + 10 a + b}{10} = x$ ...(i)

Interchanging the number ab with ba,

$\frac{x + 10 b + a}{10} =$ x + 1.8 ... (ii)

Subtracting (i) from (ii), we get,

10b + a − (10a + b) = 9(b − a) = 18

b − a = 2

Hence, option (b).

10. CAT 2002 QA | Algebra - Simple Equations

A string of length 40 metres is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 metres smaller than the first part. Find the length of the largest part.

A.
27
B.
4
C.
5
D.
9

Answer: Option A

Explanation :

Let l, m and s be the longest, medium and the shortest lengths of the strings.

l = 3m and s = l – 23

l + s + m = 40

$l + \frac{l}{3} +$ l - 23 = 40

$\frac{7 l}{3}$ = 63

l = 27

Hence, option (a).

11. CAT 2002 QA | Algebra - Number Theory

For all integers n > 0, 7⁶ⁿ – 6⁶ⁿ is divisible by

A.
13
B.
127
C.
559
D.
All of these

12. CAT 2002 QA | Modern Math - Permutation & Combination

n₁, n₂, n₃ ... n₁₀ are 10 numbers such that n₁ > 0 and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number?

A.
109
B.
27
C.
36
D.
None of these

13. CAT 2002 QA | Geometry - Mensuration

In order to cover less distance, a boy - rather than going along the longer and the shorter lengths of the rectangular path, goes by the diagonal. The boy finds that he saved a distance equal to half the longer side. The ratio of the breadth and length is

A.
1/2
B.
2/3
C.
3/4
D.
7/15

14. CAT 2002 QA | Algebra - Quadratic Equations

The number of roots $\frac{A^{2}}{x} + \frac{B^{2}}{x - 1} = 1$ is

A.
1
B.
2
C.
3
D.
None of these

Answer: Option B

Explanation :

$\frac{A^{2}}{x} + \frac{B^{2}}{x - 1} = 1 (x \neq 1)$

A²(x − 1) + B²(x) = x(x − 1)
A²x − A² + B²x = x² − x
x² − x − A²x − B²x + A² = 0
x² − x[A² + B² + 1] + A² = 0

Since it is a quadratic equation, the number of roots = 2
Now, ∆ = (A² + B² + 1)² − 4A²
= A⁴ + B⁴ + 1 + 2A²B² + 2B² + 2A² − 4A²
= A⁴ + B⁴ + 1 + 2A²B² + 2B²− 2A²

For roots to be real, A⁴ + B⁴ + 1 + 2A²B² + 2B²− 2A²≥ 0
∴ B⁴ + 2B²(A² + 1) + (A² - 1)² ≥ 0

The given equation will have one root if ∆ = 0.
It can be clearly seen that for ∆ = 0, each of the three terms must be equal to 0 since each term is greater than or equal to 0.
∴ A = ±1 and B = 0.
But for these values of A and B, x = 1 which is a contradiction.

So, ∆ ≠ 0

For ∆ > 0, two different roots are obtained.
Therefore, the number of roots of the given equation will be 2.

Hence, option (b).

15. CAT 2002 QA | Algebra - Simple Equations

Mayank, Mirza, Little and Jagbir bought a motorbike for $60. Mayank contributed half of the total amount contributed by others, Mirza contributed one-third of total amount contributed by others, and Little contributed one-fourth of the total amount contributed by others. What was the money paid by Jagbir?

A.
$12
B.
$13
C.
$18
D.
$20

Answer: Option B

Explanation :

Let the money contributed by Mayank, Mirza, Little and Jagbir be a, b, c and d respectively.

a = $\frac{1}{2}$ (b + c + d)

∴ 2a = b + c + d

∴ 2a = 60 - a

∴ a = 20

b = $\frac{1}{3}$ (a + c + d)

∴ 3b = a + c + d

∴ 3b = 60 - b

∴ b = 15

c = $\frac{1}{4}$ (a + b + d)

∴ 4c = a + b + d

∴ 4c = 60 - c

∴ c = 12

Money contributed by Jagbir = 60 − 20 − 15 − 12 = $13

Hence, option (b).

16. CAT 2002 QA | Algebra - Number Theory

If U, V, W and m are natural numbers such that U^m + V^m = W^m, then which of the following is true?

A.
m < Min(U, V, W)
B.
m > Max(U, V, W)
C.
m < Max(U, V, W)
D.
None of these

17. CAT 2002 QA | Algebra - Functions & Graphs

If $f (x) = \log (\frac{1 + x}{1 - x}),$ then f(x) + f(y) =

A.
f(x + y)
B.
f(1 + xy)
C.
(x + y) f(1 + xy)
D.
$f (\frac{x + y}{1 + x y})$

Answer: Option D

Explanation :

$f (x) + f (y) = \log (\frac{1 + x}{1 - x}) + \log (\frac{1 + y}{1 - y}) = \log \frac{(1 + x) (1 + y)}{(1 - x) (1 - y)} = \log \frac{1 + x y + x + y}{1 + x y - (x + y)} = \log \frac{1 + \frac{x + y}{1 + x y}}{1 - \frac{x + y}{1 + x y}} = f (\frac{x + y}{1 + x y})$

Hence, option (d).

18. CAT 2002 QA | Algebra - Progressions

On a straight road XY, 100 metres in length, 5 stones are kept beginning from the end X. The distance between two adjacent stones is 2 metres. A man is asked to collect the stones one at a time and put at the end Y. What is the distance covered by him?

A.
460 metres
B.
540 metres
C.
860 metres
D.
920 metres

19. CAT 2002 QA | Geometry - Triangles

The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC = 3 and ∠A = 60°. Then what is the length of the bisector AD?

A.
$\frac{12 \sqrt{3}}{7}$
B.
$\frac{12 \sqrt{13}}{7}$
C.
$\frac{4 \sqrt{13}}{7}$
D.
$\frac{4 \sqrt{3}}{7}$

Answer: Option A

Explanation :

Area of ∆ADC = $\frac{1}{2}$ AD × 3 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ AD × 4 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ × 3 × 4 × sin 60°

$\frac{1}{2}$ AD × 3 × sin 30° + $\frac{1}{2}$ × AD × 4 × sin 30° = $\frac{1}{2}$ × 3 × 4 × sin 60°

AD $[3 \times \frac{1}{2} + 4 \times \frac{1}{2}] = 12 \times \frac{\sqrt{3}}{2}$

AD $\frac{7}{2} = \frac{12 \sqrt{3}}{2}$

∴ AD = $\frac{12 \sqrt{3}}{7}$

Hence, option (a).

20. CAT 2002 QA | Geometry - Circles

In the figure given below, find the distance PQ.

A.
7 m
B.
4.5 m
C.
10.5 m
D.
6 m

21. CAT 2002 QA | Algebra - Number Theory

The remainder when 2²⁵⁶ is divided by 17 is

A.
7
B.
13
C.
11
D.
1

22. CAT 2002 QA | Algebra - Progressions

Let S = 2x + 5x² + 9x³ + 14x⁴ + 20x⁵ ... infinity (x < 1)

The coefficient of nth term = $\frac{n (n + 3)}{2} .$ The sum is

A.
$\frac{x (2 - x)}{{(1 - x)}^{3}}$
B.
$\frac{(2 - x)}{{(1 - x)}^{3}}$
C.
$\frac{x (2 - x)}{{(1 - x)}^{2}}$
D.
None of these

Answer: Option A

Explanation :

S = 2x + 5x² + 9x³ + 14x⁴ + 20x⁵ + ... …(i)

Multiplying both sides by x, we get

xS = 2x² + 5x³ + 9x⁴ + 14x⁵ + 20x⁶ + ... ...(ii)

Subtracting (ii) from (i),

(1 – x)S = 2x + 3x² + 4x³ + 5x⁴ + 6x⁵ + ... ...(iii)

Again multiplying both the sides by x, we get

x(1 – x)S = 2x² + 3x³ + 4x⁴ + 5x⁵ + 6x⁶ + ... ...(iv)

Subtracting (iv) from (iii),

(1 – x)S – x(1 – x)S = 2x + x² + x³ + x⁴ + x⁵ + ...

(1 – x)²S = 2x + x² + x³ + x⁴ + x⁵ +...

(1 – x)²S = x + (x + x² + x³ + ... + ∞)

(1 - x)²S = x + $\frac{x}{1 - x}$ ∵ S_∞ = $\frac{a}{1 - r}$ for r < 1

∴ (1 - x)²S = $\frac{x - x^{2} + x}{1 - x}$

∴ S = $\frac{x (2 - x)}{{(1 - x)}^{3}}$

Hence, option (a).

23. CAT 2002 QA | Geometry - Circles

There is a common chord of 2 circles with radius 15 and 20. The distance between the two centres is 25. The length of the chord is

A.
48
B.
24
C.
36
D.
28

24. CAT 2002 QA | Algebra - Simple Equations

A man received a cheque. The amount in Rs. has been transposed for paise and vice versa. After spending Rs. 5 and 42 paise, he discovered he now had exactly 6 times the value of the correct cheque amount. What amount he should have received?

A.
Rs. 5.30
B.
Rs. 6.44
C.
Rs. 60.44
D.
Rs. 16.44

25. CAT 2002 QA | Algebra - Functions & Graphs

Suppose, for any real number x, [x] denotes the greatest integer less than or equal to x. Let L(x, y) = [x] + [y] + [x + y] and R(x, y) = [2x] + [2y]. Then it’s impossible to find any two positive real numbers x and y for which of the following?

A.
L(x, y) = R(x, y)
B.
L(x, y) ≠ R(x, y)
C.
L(x, y) < R(x, y)
D.
L(x, y) > R(x, y)

26. CAT 2002 QA | Algebra - Progressions

A student finds the sum 1 + 2 + 3 + ... as his patience runs out. He found the sum as 575. When the teacher declared the result wrong, the student realized that he missed a number. What was the number the student missed?

A.
16
B.
18
C.
14
D.
20

Answer: Option D

Explanation :

Sum of first n natural numbers = S(n)

Sum given by student = 575

S(10) = $\frac{10 \times 11}{2}$ = 55

S(20) = $\frac{20 \times 21}{2}$ = 210

S(30) = $\frac{30 \times 31}{2}$ = 465

S(40) = $\frac{40 \times 41}{2}$ = 820

∴ The student stopped counting somewhere between 30 and 40.

Consider S(35) = $\frac{36 \times 35}{2}$ = 630

The student stopped somewhere before 35.

∴ S(31) = 496, S(32) = 528, S(33) = 561 and S(34) = 595

But the student gave 575 as the sum, so the student missed on the number 20.

Hence, option (d).

27. CAT 2002 QA | Algebra - Simple Equations

A thief was stealing diamonds from a jewellery store. On his way out, he encountered three guards, each was given half of the existing diamonds and two over it by the thief. In the end, he was left with one diamond. How many did the thief steal?

A.
40
B.
36
C.
42
D.
38

28. CAT 2002 QA | Arithmetic - Average

Three friends went for a picnic. First brought five apples and the second brought three. The third friend however brought only Rs. 8. What is the share of the first friend?

A.
8
B.
7
C.
1
D.
None of these

29. CAT 2002 QA | Algebra - Simple Equations

Amar went for a holiday to his friend's place. They together either went for yoga in the morning or played tennis in the evening but not both. 14 mornings and 24 evenings, they both stayed home and they both went out together for 22 days. How many days did Amar stay at his friend's place?

A.
20
B.
16
C.
30
D.
40

30. CAT 2002 QA | Geometry - Coordinate Geometry

The area of the triangle with the vertices (a, a), (a + 1, a) and (a, a + 2) is

A.
a³
B.
1
C.
0
D.
None of these

Answer: Option B

Explanation :

Area (∆) = $|\begin{matrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 2 \\ x_{3} & y_{3} & 3 \end{matrix}|$

where x₁ = y₁ = a, x₂ = a + 1, y₂ = a, x₃ = a and y₃ = a + 2

∴ Area (∆) = $\frac{1}{2} |\begin{matrix} a & a & 1 \\ a + 1 & a & 1 \\ a & a + 12 & 1 \end{matrix}|$

∴ Area (∆) = $\frac{1}{2}$ {a[a(1) - (a + 2)(1)] - a[(a + 1)(1) - a(1)] + 1[(a + 1)(a + 2) - a²]} = $\frac{1}{2}$ {- 2a - a + 3a + 2} = 1

Hence, option (b).

31. CAT 2002 QA | Arithmetic - Time, Speed & Distance

On a 20 km tunnel connecting two cities A and B there are three gutters. The distance between gutter 1 and 2 is half the distance between gutter 2 and 3. The distance from city A to its nearest gutter, gutter 1 is equal to the distance of city B from gutter 3. On a particular day the hospital in city A receives information that an accident has happened at the third gutter. The victim can be saved only if an operation is started within 40 minutes. An ambulance started from city A at 30 km/hr and crossed the first gutter after 5 minutes. If the driver had doubled the speed after that, what is the maximum amount of time the doctor would get to attend the patient at the hospital? Assume 1 minute is elapsed for taking the patient into and out of the ambulance.

A.
4 minutes
B.
2.5 minutes
C.
1.5 minutes
D.
Patient died before reaching the hospital

Answer: Option C

Explanation :

As per the description,

AB = 20 km

AG₁ = BG₃

2G₁G₂ = G₂G₃

∴ x = 2.5 km

y + 2y = 20 − 2x

∴ y = 5 km

∴ Time to cover A to G₃ = $\frac{1}{12} + \frac{15}{60}$ = 20 min

∴ While coming back his speed is 60 km/hr.

∴ Time taken to cover the distance from G₃ to A (i.e. 17.5 km) = 17.5 minutes

∴ Required time = 20 + 17.5 + 1 = 38.5 minutes

∴ The doctor will have 1.5 minutes to attend the patient.

Hence, option (c).

32. CAT 2002 QA | Geometry - Mensuration

Neeraj has a rectangular field of size 20 × 40 sq.m. He has to mow the field with a mowing machine of width 1 m. If he mows the field from the extremes to the centre, then the number of rounds taken by him to mow half of the field will be

A.
3.5
B.
3.8
C.
3
D.
4

Answer: Option B

Explanation :

The movement of the mower may be considered as shown in the figure.

Total area of the plot = 20 × 40 = 800 sq.m.

After 1st mowing, area left = (20 – 2) × (40 – 2) = 684 sq.m.

After 2nd mowing, area left = (20 – 4) × (40 – 4) = 576 sq.m.

After 3rd mowing, area left = (20 – 6) × (40 – 6) = 476 sq.m.

After 4th mowing, area left = (20 – 8) × (40 – 8) = 384 sq.m.

∴ Half the area is mowed between 3 to 4 rounds.

∵ 476 − 384 = 92 sq.m.

$3 + [\frac{(476 - 400)}{92}] = 3 + (\frac{76}{92}) = 3.8$

Hence, option (b).

33. CAT 2002 QA | Geometry - Mensuration

On the corners of a square field of side 14 metres, 4 horses are tethered in such a way the adjacent horses just reach each other. There is a circular pond of area 20 sq.m. in the centre of the square. What is the area left ungrazed?

A.
154 sq. m
B.
22 sq. m
C.
120 sq. m
D.
None of these

Answer: Option B

Explanation :

Side of square field = 14 m

∴ Area of square field = 196 sq.m.

Area of the circle = 20 sq.m.

Area grazed by horses = $4 \times \frac{90 °}{360 °}$ × π × 7²

= 22 × 7

= 154 sq.m.

∴ Area capable of being grazed = 196 – 20 = 176 sq.m.

∴ Area left ungrazed = 176 – 154 = 22 sq.m.

Hence, option (b).

34. CAT 2002 QA | Modern Math - Permutation & Combination

How many numbers between 0 and one million can be formed using 0, 7 and 8?

A.
486
B.
1086
C.
728
D.
None of these

35. CAT 2002 QA | Modern Math - Permutation & Combination

In how many ways, we can choose a black and a white square on a chess board such that the two are not in the same row or column?

A.
32
B.
96
C.
24
D.
None of these

36. CAT 2002 QA | Algebra - Number Theory

A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife asked. "How many gold coins do we have?" After pausing a moment, he replied, "Well! If I divide the coins into two unequal numbers, then 48 times the difference of the numbers is equal to the difference of their squares. The wife looked puzzled. Can you help the merchant's wife by finding out how many gold coins the merchant has?

A.
48
B.
96
C.
32
D.
36

37. CAT 2002 QA | Geometry - Mensuration

The magnitude of ∠FGO =

A.
30°
B.
45°
C.
60°
D.
None of these

Answer: Option D

Explanation :

Given that AB = BC = 2CH = 2CD = EH = EK = 2HK = 4KL = 2LM = MN

And EO = FP

Also,

2CD = EH

EO = FP = CD

∴ KL = PG = $\frac{C D}{2}$

FP = CD : PG = $\frac{C D}{2}$ : ∠FPG = 90°

∵ The angle are proportionate to the sides opposite to the angles.

∴ ∠FGO = ∠FGP = tan^–12

Hence, option (d).

38. CAT 2002 QA | Geometry - Mensuration

The ratio of the areas of the two quadrangles ABCD and DEFG is

A.
1 : 2
B.
2 : 1
C.
12 : 7
D.
None of these

Answer: Option C

Explanation :

Area of trapezium ABCD = $\frac{1}{2}$ BC (CD + AB)
Also BC = 2CD, AB = BC = 2CD

Area of trapezium ABCD = $\frac{1}{2}$ × CD (CD + 2DC) = 3CD²

Area of trapezium DEFG = $\frac{1}{2}$ EO(EF + DG)

EO = CD

EF = CD

DG = CH + HK + KL = CD + CD + $\frac{C D}{2} = \frac{5}{2 C D}$

Area of trapezium DEFG = $\frac{1}{2} \times C D (C D + \frac{5}{2} C D) = \frac{1}{2} \times C D (\frac{7 C D}{2}) = \frac{7 C D^{2}}{4} . . . (i i)$

Ratio of ABCD : DEFG = 12 : 7

Hence, option (c).

39. CAT 2002 QA | Algebra - Simple Equations

If X_n = (-1)ⁿX_n-1 and X₀ = x, then

A.
X_n is positive for n = even
B.
X_n is negative for n = even
C.
X_n is positive for n = odd
D.
None of these

40. CAT 2002 QA | Modern Math - Permutation & Combination

How many four-lettered passwords can be formed by using symmetrical letters only? (repetitions not allowed)

A.
1086
B.
255
C.
7920
D.
None of these

Answer: Option C

Explanation :

4 out of the 11 symmetrical alphabets need to be selected,

Also the order of alphabets needs to be accounted for (as it is a password)

∴ ¹¹P₄ = $\frac{11!}{7!}$ = 11 × 10 × 9 × 8 = 720 × 11 = 7920

Hence, option (c).

41. CAT 2002 QA | Modern Math - Permutation & Combination

How many three-lettered words can be formed such that at least one symmetrical letter is there?

A.
12870
B.
18330
C.
16420
D.
None of these

42. CAT 2002 QA | Arithmetic - Time & Work

It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 a.m. and one technician per hour is added beginning at 5 p.m., at what time will the server be complete?

A.
6:40 p.m.
B.
7:00 p.m.
C.
7:20 p.m.
D.
8:00 p.m.

Answer: Option D

Explanation :

6 technicians → 10 hours to 1 job

∴ 1 technician → 60 hours to 1 job

∴ In 6 hrs (11 a.m. - 5 p.m.)6 technicians → ${\frac{3}{5}}^{t h}$ of the work.

In the hour after 5p.m.,

7 technicians → ${\frac{7}{60}}^{t h}$ of the job.

In the nect hour, 9 technicians will complete ${\frac{9}{60}}^{t h}$ of the job.

∴ In 3 hours after 5:00 p.m. → $\frac{24}{60} = {\frac{2}{5}}^{t h}$ of the job is complete.

∴ At the end of 9^th hour i.e. 8:00 p.m. the work is complete.

Hence, option (d).

43. CAT 2002 QA | Algebra - Quadratic Equations

Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per samosa upto 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?

A.
240
B.
300
C.
400
D.
None of these

44. CAT 2002 QA | Arithmetic - Time & Work

Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone?

A.
$\frac{4}{7}$
B.
$\frac{1}{3}$
C.
$\frac{2}{3}$
D.
$\frac{3}{4}$

Answer: Option B

Explanation :

Work done by the large pump = 3 units

∴ Work done by the 3 small pumps = 2 × 3 = 6 units

∴ Work done by all 4 pumps together = 9 units

$∴ \frac{Rate of work of the large pump}{Rate of work of all 4 pumps} = \frac{3}{9} = \frac{1}{3}$

Rate of work is inversely proportional to time taken for completing the work.

$∴ \frac{TIme taken by the large pump}{Time taken by all 4 pumps} = \frac{3}{1}$

Hence, option (b).

45. CAT 2002 QA | Algebra - Surds & Indices

If pqr = 1 then $\frac{1}{1 + p + q^{- 1}} + \frac{1}{1 + q + r^{- 1}} + \frac{1}{1 + r + p^{- 1}}$ is equivalent to

A.
p + q + r
B.
$\frac{1}{p + q + r}$
C.
1
D.
p^-1 + q^-1 + r^-1

Answer: Option C

Explanation :

∵ pqr = 1

Case 1:

Let p = 2/3, q = 3/2 and r = 1

Substituting the values of p, q, r, we get,

$\frac{1}{1 + p + q^{- 1}} + \frac{1}{1 + q + r^{- 1}} + \frac{1}{1 + r + p^{- 1}} = \frac{3}{7} + \frac{2}{7} + \frac{2}{7} = 1$

Case 2:

Let p = 1, q = 1 and r = 1

Substituting the values of p, q, r, we get,

$\frac{1}{1 + p + q^{- 1}} + \frac{1}{1 + q + r^{- 1}} + \frac{1}{1 + r + p^{- 1}} = \frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1$

Hence, option (c).

46. CAT 2002 QA | Arithmetic - Time, Speed & Distance

There is a tunnel connecting city A and B. There is a CAT which is standing at 3/8 the length of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the CAT meet. In another case, the CAT started running towards the exit and the train again met the CAT at the exit. What is the ratio of their speeds?

A.
4 : 1
B.
1 : 2
C.
8 : 1
D.
None of these

Answer: Option A

Explanation :

Solution:

Let the speed of the train be St and the speed of the CAT be S_c.

Case 1:

The CAT runs towards the train.

$\frac{d}{S_{t}} = \frac{3 x}{8 S_{c}}$

$∴ \frac{3 x}{8} \times \frac{S_{t}}{S_{c}} = d$ ...(i)

Case 2:

The CAT runs away from the train.

$\frac{d + x}{S_{t}} = \frac{5 x}{8 S_{c}}$

$∴ d = \frac{5 x}{8} \times \frac{S_{t}}{S_{C}} - x$ ...(ii)

Equating (i) and (ii),

$\frac{3 x}{8} \times \frac{S_{t}}{S_{c}} = \frac{5 x}{8} \times \frac{S_{t}}{S_{c}} - x ∴ \frac{3}{8} x \times \frac{S_{t}}{S_{c}} = \frac{5 x}{8} \times \frac{S_{t}}{S_{c}} - x ∴ \frac{3}{8} \frac{S_{t}}{S_{c}} = \frac{5}{8} \frac{S_{t}}{S_{c}} - 1$

∴ S_t : S_c = 4 : 1

Hence, option (a).

47. CAT 2002 QA | Algebra - Number Theory

In a book store, the words of the glowsign board "MODERN BOOK STORES" are individually flashed after 5/2, 17/4 and 41/8 seconds respectively. Each word is put off after a second. What is the least time after which full name of the book store can be read?

A.
73.5 seconds
B.
79.4 seconds
C.
68.2 seconds
D.
None of these

Answer: Option A

Explanation :

The words MODERN, BOOK and STORES appear after 5/2, 17/4 and 41/8 seconds respectively and are put off after 1 second.

So, if all the words flashed together at t = 0 seconds, then

MODERN will flash after (5/2 + 1) = 7/2 seconds

BOOK will flash after (17/4 + 1) = 21/4 seconds

STORES will flash after (41/8 + 1) = 49/8 seconds.

So, all the words will flash together again after LCM(7/2, 21/4, 49/8) seconds.

∴ L.C.M. $(\frac{7}{2}, \frac{21}{4}, \frac{49}{8}) = \frac{L . C . M . of Numerators}{H . C . F . of Denominators} = \frac{147}{2} =$ 73.5 seconds

Hence, option (a).

48. CAT 2002 QA | Miscellaneous

At the end of the sequence, what will be the number of oranges in the basket?

A.
2
B.
3
C.
4
D.
6

49. CAT 2002 QA | Miscellaneous

At the end of the sequence, what will be the total number of fruits in the basket?

A.
10
B.
11
C.
13
D.
17

50. CAT 2002 QA | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

In a hockey match, the Indian team was behind by 2 goals with 5 minutes remaining. Did they win the match?

A. Deepak Thakur, the Indian striker scored 3 goals in the last 5 minutes of the match.
B. Korea scored a total of 3 goals in the match.

A.
1
B.
2
C.
3
D.
4

51. CAT 2002 QA | Algebra - Number Theory | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

Four students were added to a dance class. Would the teacher be able to divide her students evenly into a dance team (or teams) of 8?

A. If 12 students were added, then the teacher could put everyone in teams of 8 without any left overs.
B. The number of students in the class is currently not divisible by 8.

A.
1
B.
2
C.
3
D.
4

52. CAT 2002 QA | Algebra - Number Theory | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

Is x = y?

A. (x + y) $(\frac{1}{x} + \frac{1}{y})$ = 4

B. (x − 50)² = (y − 50)²

A.
1
B.
2
C.
3
D.
4

Answer: Option A

Explanation :

Consider statement A:

$(x + y) \frac{(x + y)}{x y} = 4$
∴ (x + y)² = 4xy

∴ x² + y² + 2xy − 4xy = 0

∴ (x − y)² = 0

∴ x = y

∴ Statement A alone is sufficient.

Consider statement B:

(x – 50)² = (y – 50)²

∴ Either (x − 50) = (y − 50) or (x − 50) = −(y − 50)

∴ Either x = y or x + y = 100, which gives infinite values for x and y.

∴ Statement B alone is not sufficient.

Hence, option (a).

53. CAT 2002 QA | Arithmetic - Profit & Loss | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

A dress was initially listed at a price that would have fetched the store a profit of 20% on the wholesale cost. What was the wholesale cost of the dress?

A. After reducing the listed price by 10% the dress was sold for a net profit of 10 dollars.
B. The dress was sold for 50 dollars.

A.
1
B.
2
C.
3
D.
4

54. CAT 2002 QA | Arithmetic - Average | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

Is 500 the average (arithmetic mean) score of the GMAT?

A. Half of the people who take GMAT score above 500 and half of the people score below 500.
B. The highest GMAT score is 800 and the lowest score is 200.

A.
1
B.
2
C.
3
D.
4

55. CAT 2002 QA | Algebra - Number Theory | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

Is |x − 2| < 1?

A. |x| > 1
B. |x − 1| < 2

A.
1
B.
2
C.
3
D.
4

56. CAT 2002 QA | Venn Diagram | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

Members in a club either speak French or Russian or both. Find the number of members in a club who speak only French.

A. There are 300 members in the club and the number of members who speak both French and Russian is 196.
B. The number of members who speak only Russian is 58.

A.
1
B.
2
C.
3
D.
4

57. CAT 2002 QA | Arithmetic - Ratio, Proportion & Variation | Data Sufficiency

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

Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) if the question can be solved by using either of the two statements.
Answer (3) if the question can be solved by using both the statements together and not by any one of them.
Answer (4) if the question cannot be solved with the help of the given data and more data is required.

A sum of Rs. 38,500 was divided among Jagdish, Punit and Girish. Who received the minimum amount?

A. Jagdish received 2/9 of what Punit and Girish together received.
B. Punit received 3/11 of what Jagdish and Girish together received.

A.
1
B.
2
C.
3
D.
4

Answer: Option C

Explanation :

Consider statement A:

Jagdish : (Punit + Girish) = 2 : 9

∴ Jagdish's share = $\frac{2}{11}$ × Total = 18.18% of the total

However, we don’t know the percentage distribution between Punit and Girish.

∴ Statement A alone is not sufficient.

Consider statement B:

Punit : (Jagdish + Girish) = 3 : 11

∴ Punit's share = $\frac{3}{14}$ × Total = 21.4% of the total

However, we do not know the percentage distribution between Jagdish and Girish.

∴ Statement B alone is not sufficient.

Consider both statements together:

Jagdish = 18.18% of the total amount

Punit = 21.4% of the total amount

Girish = 100 − 18.18 − 21.4 = 60.42% of the total amount

∴ Jagdish received the minimum amount.

Hence, option (c).

58. CAT 2002 QA | Data Sufficiency

Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer

For what reason Purohit did not get the offer of employment?

Statement:

I. Purohit passed the interview.
II. Purohit's friend passed the medical test who passed the interview along with Purohit.
III. Purohit's father did not want him to take the job.
IV. Purohit has another employment offer from another company.
V. Purohit did not clear the mandatory medical test.

A.
III and IV only
B.
III, IV and V only
C.
I, III and IV only
D.
V only

59. CAT 2002 QA | Data Sufficiency

Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer

What were the possible reasons due to which DESCO incurred losses for the last two years?

Statement:

I. The company's shares are not registered in the stock exchange.
II. The company does not export its products.
III. The company has an inefficient labour force.
IV. The price of its product has fallen in the last two years due to competitive market.
V. Entry of similar foreign goods at a cheaper rate.

A.
Only III, IV and V
B.
Only II, III and IV
C.
Only IV and V
D.
Only I and II

60. CAT 2002 QA | Data Sufficiency

Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer

On which day of the week did Sunil get his letter of promotion?

Statement:

I. Sunil purchased a new shirt on Friday
II. Sunil was given a party that Saturday.
III. Sunil was given the letter of promotion on the day before he purchased the shirt.
IV. Tuesday being his birthday, Sunil gave a party to all his friends.
V. Sunil's friend was promoted on Friday.

A.
I and II only
B.
II, III and IV only
C.
I and III only
D.
II, III and V only

61. CAT 2002 QA | Data Sufficiency

Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer

Who among A, B, C, D and E is the heaviest?

Statement:

I. B and C are heavier than A and D.
II. C is heavier than D.
III. C is heavier than A and lighter than B.
IV. E is heavier than B
V. D is lighter than E.

A.
I, III and IV only
B.
II, III and V only
C.
III and IV only
D.
I, III and V only

CAT 2002 QA | Previous Year CAT Paper

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