CAT 2004 QA questions and solutions

Previous year paper questions for CAT 2004 QA

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1. CAT 2004 QA | Geometry - Trigonometry

A father and his son are waiting at a bus stop in the evening. There is a lamp post behind them. The lamp post, the father and his son stand on the same straight line. The father observes that the shadows of his head and his son's head are incident at the same point on the ground. If the heights of the lamp post, the father and his son are 6 metres, 1.8 metres and 0.9 metres respectively, and the father is standing 2.1 metres away from the post, then how far (in metres) is the son standing from his father?

A.
0.9
B.
0.75
C.
0.6
D.
0.45

Answer: Option D

Video Explanation

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Explanation :

Let PA, FB and SC denote the lamp post, the father and the son respectively and let D denote the point where the shadows of the father’s and son’s heads are incident.

ΔBFD ~ ΔCSD

∴ $\frac{BF}{CS} = \frac{FD}{SD}$

∴ $\frac{1.8}{0.9} = \frac{FD}{SD}$

∴ $\frac{FD}{SD}$ = 2

∴ FD = 2 × SD

Let SD = FS = x m

ΔAPD ~ ΔBFD

∴ $\frac{AP}{BF} = \frac{PD}{FD}$

∴ $\frac{6}{1.8} = \frac{2.1 + 2 x}{2 x}$

∴ $\frac{20 x}{3} = 2.1 + 2 x$

∴ $\frac{14 x}{3} = 2.1$

∴ x = 0.45 m

∴ The son is standing 0.45 m away from the father.

Hence, option (d).

2. CAT 2004 QA | Arithmetic - Mixture, Alligation, Removal & Replacement

A milkman mixes 20 litres of water with 80 litres of milk. After selling one-fourth of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk?

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

3. CAT 2004 QA | Arithmetic - Time, Speed & Distance

Karan and Arjun run a 100 metre race, where Karan beats Arjun by 10 metres. To do a favour to Arjun, Karan starts 10 metres behind the starting line in a second 100 metre race. They both run at their earlier speeds. Which of the following is true in connection with the second race?

A.
Karan and Arjun reach the finishing line simultaneously.
B.
Arjun beats Karan by 1 metre.
C.
Arjun beats Karan by 11 metres.
D.
Karan beats Arjun by 1 metre.

Answer: Option D

Video Explanation

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Explanation :

In the first race when Karan runs 100 m, Arjun runs 90 m.

∴ Ratio of Karan’s and Arjun’s speed = 10 : 9

In the second race, when Karan runs 110 m, Arjun runs $\frac{110 \times 9}{10}$ = 99 m

∴ Karan beats Arjun by 1 m.

Hence, option (d).

4. CAT 2004 QA | Modern Math - Permutation & Combination

N persons stand on the circumference of a circle at distinct points. Each possible pair of persons, not standing next to each other, sings a two-minute song one pair after the other. If the total time taken for singing is 28 minutes, what is N?

A.
5
B.
7
C.
9
D.
None of the above

Answer: Option B

Video Explanation

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Explanation :

Total time taken for singing = 28 minutes

∴ Total songs sung = 14

As there are N persons, the total number of songs sung

^NC₂ – (those sung between adjacent persons) = ^NC₂ – N

∴ $\frac{N!}{(N - 2)! \times 2!}$ - N = 14

∴ $\frac{N^{2} - 3 N}{2} = 14$

∴ (N – 7)(N + 4) = 0

∴ N = 7

Hence, option (b).

5. CAT 2004 QA | Algebra - Progressions

If the sum of the first 11 terms of an arithmetic progression equals that of the first 19 terms, then what is the sum of the first 30 terms?

A.
0
B.
-1
C.
1
D.
Not unique

Answer: Option A

Video Explanation

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Explanation :

Let a and d be the first term and the common difference of the AP.

Sum of the first n terms of this A.P. = $\frac{n}{2}$ [2a + (n - 1)d]

∴ The sum of the first 30 terms = 15 × (2a + 29d) …(i)

By conditions,

$\frac{11}{2}$ × (2a + 10d) = $\frac{19}{2}$ × (2a + 18d)

∴ 11a + 55d = 19a + 171d

∴ 8a = –116d

∴ a = - $\frac{29 d}{2}$

From (i),

∴ The sum of the first 30 terms = 15 × (–29d + 29d)

∴ The sum of the first 30 terms = 0

Hence, option (a).

Alternatively,

If the sum of the first p terms of an A.P. is equal to the sum of the first q terms of the A.P. such that p and q are different, then the sum of the first (p + q) terms of the A.P. is zero.

∴ As the sum of the first 11 and the sum of the first 19 terms of the A.P. is equal, the sum of the first (11 + 19) = 30 terms of the A.P. is zero.

Hence, option (a).

6. CAT 2004 QA | Arithmetic - Time, Speed & Distance

If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

A.
11 km/hr
B.
12 km/hr
C.
13 km/hr
D.
14 km/hr

Answer: Option B

Video Explanation

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Explanation :

Let x hrs be the time required at 15kmph to reach the place at 11:00 am
∴ Distance that the man has to cycle = 15x km

Now, $\frac{15 x}{10}$ = x + 2
∴ 15x = 10x + 20
∴ 5x = 20
∴ x = 4 hrs

∴ Distance to be travelled = 60 km

∴ To reach the place at noon, the man should cycle for x + 1 = 5 hrs.

Let the new speed be y.

∴ 5 = $\frac{60}{y}$

∴ y = 12 km/hr

Hence, option (b).

7. CAT 2004 QA | Algebra - Progressions

On January 1, 2004 two new societies, S₁, and S₂, are formed, each with n members. On the first day of each subsequent month, S1 adds b members while S2 multiplies its current number of members by a constant factor r. Both the societies have the same number of members on July 2, 2004. If b = 10.5n, what is the value of r?

A.
2.0
B.
1.9
C.
1.8
D.
1.7

8. CAT 2004 QA | Algebra - Inequalities & Modulus

If f(x) = x³ – 4x + p, and f(0) and f(1) are of opposite signs, then which of the following is necessarily true?

A.
–1 < p < 2
B.
0 < p < 3
C.
–2 < p < 1
D.
–3 < p < 0

9. CAT 2004 QA | Algebra - Number Theory

Suppose n is an integer such that the sum of the digits of n is 2, and 10¹⁰ < n < 10¹¹. The number of different values for n is

A.
11
B.
10
C.
9
D.
8

10. CAT 2004 QA | Arithmetic - Ratio, Proportion & Variation

If $\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{a + b} = r,$ then r cannot take any value except _________.

A.
$\frac{1}{2}$
B.
-1
C.
$\frac{1}{2}$ or -1
D.
$- \frac{1}{2}$ or -1

Answer: Option C

Video Explanation

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Explanation :

Given, $\frac{a}{b + c} = \frac{b}{c + a} = \frac{c}{c + a} = r$

By property of equal ratios,

$\frac{a + b + c}{b + c + c + a + a + b}$ = r

∴ $\frac{a + b + c}{2 (a + b + c)}$ = r

Assuming (a + b + c ≠ 0),

∴ r = $\frac{1}{2}$

If a + b + c = 0, a = –(b + c)

Then from the original ratio, r = $\frac{a}{b + c}$ = $\frac{- (b + c)}{(b + c)}$ = -1

Hence, option (c).

11. CAT 2004 QA | Algebra - Surds & Indices

Let $y = \frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + . . . .}}}}$

What is the value of y?

A.
$\frac{\sqrt{13} + 3}{2}$
B.
$\frac{\sqrt{13} - 3}{2}$
C.
$\frac{\sqrt{15} + 3}{2}$
D.
$\frac{\sqrt{15} - 3}{2}$

Answer: Option D

Video Explanation

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Explanation :

y = $\frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + . . . .}}}}$

y = $\frac{1}{2 + \frac{1}{3 + y}}$

y = $\frac{3 + y}{2 y + 7}$

∴ 2y² + 7y = 3 + y

∴ 2y² + 6y – 3 = 0

∴ y = $\frac{- 6 \pm \sqrt{36 - 4 (- 3) (2)}}{4}$ = $\frac{\pm \sqrt{15} - 3}{2}$

As all the terms in y are positive, y is positive.

∴ y = $\frac{\sqrt{15} - 3}{2}$

Hence, option (d).

12. CAT 2004 QA | Algebra - Functions & Graphs | Algebra - Inequalities & Modulus

Let f(x) = ax² – b|x|, where a and b are constants. Then at x = 0, f(x) is

A.
maximized whenever a > 0, b > 0
B.
maximized whenever a > 0, b < 0
C.
minimized whenever a > 0, b > 0
D.
minimized whenever a > 0, b < 0

13. CAT 2004 QA | Arithmetic - Time, Speed & Distance

Two boats, travelling at 5 and 10 kms per hour, head directly towards each other. They begin at a distance of 20 kms from each other. How far apart are they (in kms) one minute before they collide?

A.
$\frac{1}{12}$
B.
$\frac{1}{6}$
C.
$\frac{1}{4}$
D.
$\frac{1}{3}$

Answer: Option C

Video Explanation

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Explanation :

The distance between the boats one minute before they collide will be same as the total distance they travel in last one minute.

The boats together travel 10 + 5 = 15 km in 60 minutes.

∴ In one minute they can travel $\frac{1}{4}$ km.

∴ They are $\frac{1}{4}$ km part, one minute before they collide.

Hence, option (c).

14. CAT 2004 QA | Miscellaneous

Each family in a locality has at most two adults, and no family has fewer than 3 children. Considering all the families together, there are more adults than boys, more boys than girls, and more girls than families. Then the minimum possible number of families in the locality is:

A.
4
B.
5
C.
2
D.
3

15. CAT 2004 QA | Arithmetic - Time & Work

In Nuts And Bolts factory, one machine produces only nuts at the rate of 100 nuts per minute and needs to be cleaned for 5 minutes after production of every 1000 nuts. Another machine produces only bolts at the rate of 75 bolts per minute and needs to be cleaned for 10 minutes after production of every 1500 bolts. If both the machines start production at the same time, what is the minimum duration required for producing 9000 pairs of nuts and bolts?

A.
130 minutes
B.
135 minutes
C.
170 minutes
D.
180 minutes

16. CAT 2004 QA | Geometry - Mensuration

A rectangular sheet of paper, when halved by folding it at the mid-point of its longer side, results in a rectangle, whose longer and shorter sides are in the same proportion as the longer and shorter sides of the original rectangle. If the shorter side of the original rectangle is 2, what is the area of the smaller rectangle?

A.
4√2
B.
2√2
C.
√2
D.
None of the above

Answer: Option B

Video Explanation

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Explanation :

Let the length and breadth of the rectangle be x and y respectively.

x > y

By conditions, $\frac{x}{y} = \frac{y}{\frac{x}{2}}$

∴ $\frac{x^{2}}{2} = y^{2}$

∴ $\frac{x^{2}}{2}$ = 4 (∵ y = 2)

∴ x² = 8

∴ x = 2√2

∴ Area of the smaller rectangle = 2√2 sq.units.

Hence, option (b).

17. CAT 2004 QA | Geometry - Circles

What is the ratio of the length of PQ to that of QO?

A.
1 : 4
B.
1 : 3
C.
3 : 8
D.
3 : 4

Answer: Option B

Video Explanation

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Explanation :

ΔSQO ~ ΔRPO

⇒ $\frac{S Q}{R P}$ = $\frac{O Q}{O P}$ = $\frac{S O}{O R}$

⇒ $\frac{1.5}{2}$ = $\frac{28 - 3.5 x}{28}$

⇒ 42 = 56 – 7x

⇒ x = 2

⇒ PQ = 7 cm, OQ = 21 cm

∴ PQ : OQ = 1 : 3

Hence, option (b).

18. CAT 2004 QA | Geometry - Circles

What is the radius of the circle II?

A.
2 cm
B.
3 cm
C.
4 cm
D.
5 cm

19. CAT 2004 QA | Geometry - Circles

The length of SO is ________.

A.
8√3 cm
B.
10√3 cm
C.
12√3 cm
D.
14√3 cm

Answer: Option C

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Explanation :

SO² = OQ² – SQ²

∴ SO = $\sqrt{21^{2} - 3^{2}}$ = 12√3 cm

Hence, option (c).

20. CAT 2004 QA | Algebra - Simple Equations | Geometry - Circles

In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If ∠CBE = 65°, then what is the value of ∠DEC?

A.
35°
B.
55°
C.
45°
D.
25°

21. CAT 2004 QA | Geometry - Circles

On a semicircle with diameter AD, chord BC is parallel to the diameter. Further, each of the chords AB and CD has length 2, while AD has length 8. What is the length of BC?

A.
7.5
B.
7
C.
7.75
D.
None of the above

22. CAT 2004 QA | Algebra - Functions & Graphs

How many of the following products are necessarily zero for every x

f₁(x)f₂(x), f₂(x)f₃(x), f₂(x)f₄(x)?

A.
0
B.
1
C.
2
D.
3

23. CAT 2004 QA | Algebra - Functions & Graphs

Which of the following is necessarily true?

A.
f₄(x) = f₁(x) for all x
B.
f₁(x) = –f₃(–x) for all x
C.
f₂(–x) = f₄(x) for all x
D.
f₁(x) + f₃(x) = 0 for all x

24. CAT 2004 QA | Arithmetic - Percentage

If group B contains 23 questions, then how many questions are there in group C?

A.
1
B.
2
C.
3
D.
Cannot be determined

Answer: Option A

Video Explanation

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Explanation :

Let there be a, b and c questions in groups A, B and C respectively.

Then, a + b + c =100

Total marks = a + 2b + 3c

Also, $\frac{a}{a + 2 b + 3 c} \geq \frac{60}{100}$

b = 23

⇒ a + c = 77

⇒ a = 77 – c

⇒ $\frac{a}{a + 46 + 3 c} \geq 0.6$

⇒ a ≥ 0.6a + 27.6 + 1.8c

⇒ 4a ≥ 276 + 18c

⇒ a ≥ 69 + 4.5c

⇒ a – 4.5c ≥ 69

⇒ 77 – c – 4.5c ≥ 69

⇒ 77 ≥ 69 + 5.5c

⇒ c ≤ 8/5.5

⇒ c = 1

Hence, option (a).

Alternatively,
Evaluating options,

If C has 1 question then B and A have 23 and 76 questions respectively.
The total number of marks = 125
⇒ Group A has 60.8% marks.
⇒ Option (a) is possible.

If C has 2 questions then B and A have 23 and 75 questions respectively.
The total number of marks = 127
⇒ Group A has 59.05% marks.
⇒ Option (b) is not possible.

If C has 3 questions then B and A have 23 and 74 questions respectively.
The total number of marks = 129
⇒ Group A has 57.36% marks.
⇒ Option (c) is not possible.

Hence, option (a).

25. CAT 2004 QA | Arithmetic - Percentage

If group C contains 8 questions and group B carries at least 20% of the total marks, which of the following best describes the number of questions in group B?

A.
11 or 12
B.
12 or 13
C.
13 or 14
D.
14 or 15

Answer: Option C

Video Explanation

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Explanation :

c = 8
∴ a + b = 92 ...(1)

Also, $\frac{2 b}{a + 2 b + 3 c} \geq 0.2$

⇒ $\frac{2 b}{92 - b + 2 b + 24} \geq 0.2$

⇒ 2b ≥ 23.2 + 0.2b

⇒ $b \geq \frac{232}{18}$

⇒ b ≥ 13 ...(2)

Also, as group A carries at least 60% of the total marks,

$\frac{a}{a + 2 b + 3 c} \geq 0.6$

⇒ $\frac{a}{a + 2 (92 - a) + 24} \geq 0.6$

⇒ aa ≥ 0.6a + 110.4 - 1.2a + 14.4

⇒ 1.6a ≥ 124.8

⇒ a ≥ 78 ...(3)

From (1), (2) and (3)
⇒ b can be 13 or 14.

Hence, option (c).

Alternately,
Consider options.

For c = 8 and b = 11, a = 81. The total marks = 127.
Group B has 17.32% marks, which is not possible.

For c = 8 and b = 12, a = 80. The total marks = 128.
Group B has 18.75% marks, which is not possible.

For c = 8 and b = 13, a = 79. The total marks = 129.
Group B has 20.15% marks and group A has 61.24% marks, which is possible.

For c = 8 and b = 14, a = 78. The total marks = 130.
Group B has 21.53% marks and group A has 60% marks, which is possible.

For c = 8 and b = 15, a = 77. The total marks = 131.
Group B has 22.9% marks and group A has 58.77 marks, which is not possible.

Hence, option (c).

26. CAT 2004 QA | Arithmetic - Time, Speed & Distance

A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute) is πr during the first 30 seconds, πr/2 during next one minute, πr/4 during next 2 minutes, πr/8 during next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for the previous round?

A.
4
B.
8
C.
16
D.
32

27. CAT 2004 QA | Algebra - Progressions

Consider the sequence of numbers a₁, a₂, a₃, ... to infinity where a₁ = 81.33 and a₂ = –19 and a_j = a_j_–1– a_j_–2 for j ≥ 3. What is the sum of the first 6002 terms of this sequence?

A.
-100.33
B.
-30.00
C.
62.33
D.
119.33

28. CAT 2004 QA | Algebra - Number Theory

The remainder, when (15²³ + 23²³) is divided by 19, is:

A.
4
B.
15
C.
0
D.
18

29. CAT 2004 QA | Modern Math - Permutation & Combination

In the adjoining figure, the lines represent one-way roads allowing travel only northwards or only westwards. Along how many distinct routes can a car reach point B from point A?

A.
15
B.
56
C.
120
D.
336

Answer: Option B

Video Explanation

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Explanation :

To reach point B from A, the car has to take a northward route thrice and a westward route five times. Thus we can visualise a set of 8 routes out of which 3 will be northward and 5 westward.

The number of ways in which the car can reach point B from A is a permutation of 8 routes, 5 (W) and 3 (N) of which are repeated.

Thus, number of ways = $\frac{8!}{3! \times 5!}$ = 56

Hence, option (b).

30. CAT 2004 QA | Geometry - Circles | Algebra - Progressions

Let C be a circle with centre P₀ and AB be a diameter of C. Suppose P₁ is the mid-point of the line segment P₀B, P₂ is the mid-point of the line segment P₁B and so on. Let C₁, C₂, C₃, ... be circles with diameters P₀P₁, P₁P₂, P₂P₃ ... respectively. Suppose the circles C₁,C₂, C₃, ... are all shaded. The ratio of the area of the unshaded portion of C to that of the original circle C is:

A.
8 : 9
B.
9 : 10
C.
10 : 11
D.
11 : 12

Answer: Option D

Video Explanation

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Explanation :

Let the radius of C be r.

Then P₀P₁ = $\frac{r}{2}$

P₁P₂ = $\frac{r}{4}$

P₂P₃ = $\frac{r}{8}$

and so on.

Thus the areas of circles with diameters P₀P₁, P₁P₂, P₂P₃, … are

$\frac{π r^{2}}{16}, \frac{π r^{2}}{64}, \frac{π r^{2}}{256}, \frac{π r^{2}}{1024} . . . .$

The sum of these areas = $\frac{π r^{2}}{16} \times \frac{1}{1 - \frac{1}{4}} = \frac{π r^{2}}{12}$

∴ Shaded area = $\frac{π r^{2}}{12}$

Unshaded area = πr² - $\frac{π r^{2}}{12} = \frac{11 π r^{2}}{12}$

∴ Required ratio = 11 : 12

Hence, option (d).

31. CAT 2004 QA | Algebra - Logarithms

Let u = (log₂x)² – 6 log₂ x + 12 where x is a real number. Then the equation x^u = 256, has

A.
no solution for x
B.
exactly one solution for x
C.
exactly two distinct solutions for x
D.
exactly three distinct solutions for x

Answer: Option B

Video Explanation

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Explanation :

u = (log₂x)² – 6log₂x + 12 ...(1)

x^u = 256

Taking log on both sides,

ulog₂x = log₂256 = 8

Let log₂x = m

∴ u × m = 8

Substituting in (1),

$\frac{8}{m}$ = m² - 6m + 12

⇒ m³ – 6m² + 12m – 8 = 0
⇒ (m – 2)³ = 0
⇒ m = 2
⇒ log₂x = 2
⇒ x = 4

∴ The given equation has exactly one solution for x.

Hence, option (b).

32. CAT 2004 QA | Modern Math - Permutation & Combination

A new flag is to be designed with six vertical stripes using some or all of the colours yellow, green, blue and red. Then, the number of ways this can be done such that no two adjacent stripes have the same colour is:

A.
12 × 81
B.
16 × 192
C.
20 × 125
D.
24 × 216

33. CAT 2004 QA | Geometry - Mensuration

If the lengths of diagonals DF, AG and CE of the cube shown in the adjoining figure are equal to the three sides of a triangle, then the radius of the circle circumscribing that triangle will be _______.

A.
equal to the side of the cube
B.
$\sqrt{3}$ times the side of the cube
C.
$\frac{1}{\sqrt{3}}$ times the side of the cube
D.
impossible to find from the given information

Answer: Option A

Video Explanation

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Explanation :

Let the side of the cube be a units.

DF, AG and CE are body diagonals each of length a√3 units.

∴ Circumradius of the equilateral triangle = $\frac{a \sqrt{3}}{\sqrt{3}}$ = a units

∴ Circumradius = side of cube

Hence, option (a).

34. CAT 2004 QA | Geometry - Circles

A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle?

A.
$3 - 2 \sqrt{2}$
B.
$4 - 2 \sqrt{2}$
C.
$7 - 4 \sqrt{2}$
D.
$6 - 4 \sqrt{2}$

Answer: Option D

Video Explanation

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Explanation :

AB = AK = 2

Quadrilateral ABCD is a square.
∴ AC = 2√2

⇒ KC = AC - AK = 2√2 - 2

Let the centre of the smaller circle be O and let its radius be x.

∴ OC = x√2 and KC = KO + OC = x + x√2

⇒ x + x√2 = 2√2 - 2

⇒ x = $\frac{2 \sqrt{2} - 2}{1 + \sqrt{2}}$

⇒ x = $\frac{(2 \sqrt{2} - 2)}{(1 + \sqrt{2})} \times \frac{(\sqrt{2} - 1)}{(\sqrt{2} - 1)}$ = 6 - 4√2

Hence, option (d).

35. CAT 2004 QA | Algebra - Simple Equations | Data Sufficiency