# CAT 2004 QA | Previous Year CAT Paper

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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

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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

∴ $\frac{\mathrm{BF}}{\mathrm{CS}}=\frac{\mathrm{FD}}{\mathrm{SD}}$

∴ $\frac{1.8}{0.9}=\frac{\mathrm{FD}}{\mathrm{SD}}$

∴ $\frac{\mathrm{FD}}{\mathrm{SD}}$ = 2

∴ FD = 2 × SD

Let SD = FS = x m

ΔAPD ~ ΔBFD

∴ $\frac{\mathrm{AP}}{\mathrm{BF}}=\frac{\mathrm{PD}}{\mathrm{FD}}$

∴ $\frac{6}{1.8}=\frac{2.1+2x}{2x}$

∴ $\frac{20x}{3}=2.1+2x$

∴ $\frac{14x}{3}=2.1$

∴ x = 0.45 m

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

Hence, option (d).

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**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

Answer: Option A

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**Explanation** :

The quantity of milk and water is as shown in the table.

∴ Current proportion of water and milk is 40 : 60 = 2 : 3

Hence, option (a).

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**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

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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).

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**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

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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

^{N}C_{2} – (those sung between adjacent persons) = ^{N}C_{2} – N

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

∴ $\frac{{\mathrm{N}}^{2}-3\mathrm{N}}{2}=14$

∴ (N – 7)(N + 4) = 0

∴ N = 7

Hence, option (b).

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**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

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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 = - $\frac{29d}{2}$

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).

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**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

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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, $\frac{15x}{10}$ = 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).

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**7. CAT 2004 QA | Algebra - Progressions**

On January 1, 2004 two new societies, S_{1}, and S_{2}, 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

Answer: Option A

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**Explanation** :

Members are added on the first day of every month for 6 months before July 2, 2004.

Number of members in S_{1} on July 2, 2004 = n + 6b = 64n (as b = 10.5n)

Number of members in S_{2} on July 2, 2004 = n × r^{6}

By conditions, 64n = n × r^{6}

∴ r = 2

Hence, option (a).

Hence, option (a).

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**8. CAT 2004 QA | Algebra - Inequalities & Modulus**

If f(x) = x^{3} – 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

Answer: Option B

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**Explanation** :

f(x) = x^{3} – 4x + p

∴ f(0) = p and f(1) = p – 3

p – 3 < p

As p and p – 3 are of opposite signs, p – 3 < 0 and p > 0.

∴ p < 3 and p > 0

∴ 0 < p < 3

Hence, option (b).

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**9. CAT 2004 QA | Algebra - Number Theory**

Suppose n is an integer such that the sum of the digits of n is 2, and 10^{10} < n < 10^{11}. The number of different values for n is

- A.
11

- B.
10

- C.
9

- D.
8

Answer: Option A

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**Explanation** :

10^{10} < n < 10^{11}

∴ 10000000000 < n < 100000000000

As the sum of the digits of n = 2, n can be any one out of

11000000000

10100000000

10010000000

...

10000000001 … (10 such numbers)

or n can be 20000000000

∴ There are 11 such values of n.

Hence, option (a).

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**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

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**Explanation** :

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

By property of equal ratios,

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

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

Assuming (a + b + c ≠ 0),

∴ r = $\frac{1}{2}$

If a + b + c = 0, a = –(b + c)

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

Hence, option (c).

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**11. CAT 2004 QA | Algebra - Surds & Indices**

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

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

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**Explanation** :

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

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

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

∴ 2*y*^{2} + 7*y* = 3 + *y*

∴ 2*y*^{2} + 6*y* – 3 = 0

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

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

∴ y = $\frac{\sqrt{15}-3}{2}$

Hence, option (d).

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

Let *f*(*x*) = *ax*^{2} – *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

Answer: Option D

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**Explanation** :

*f*(*x*) = *ax*^{2} – *b*|*x*|

*x*^{2} and |*x*| both are positive. Let *x* ≠ 0.

At x = 0, f(0) = 0

Consider the following cases:

**1.** *a* > 0, *b* > 0

∴ *f*(*x*) > 0, when *ax*^{2} > *b*|*x*|

∴ *f*(*x*) < 0, when *ax*^{2} < *b*|*x*|

*f*(*x*) is neither maximised or minimised when *x* = 0.

**2.** *a* > 0, *b* < 0

∴ *f*(*x*) = *ax*^{2} + |*b*||*x*| > 0

Thus *f*(*x*) is greater than 0 when *x* ≠ 0.

∴ *f*(*x*) is minimised at x = 0 whenever *a* > 0, *b* < 0.

Hence, option (d).

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**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

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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).

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**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

Answer: Option D

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**Explanation** :

Let f be the number of families.

∴ The number of adults ≤ 2f, children ≥ 3f

Also, adults > boys > girls > families

If f = 2, the minimum possible number of girls and boys is 3 and 4.

Thus, number of children = 3 + 4 = 7 and number of adults ≤ 4

But as adults can be equal to boys, f cannot be 2.

If f = 3, minimum possible number of girls and boys = 4 and 5.

Thus number of children = 9 and number of adults ≤ 6

This is possible. Thus the minimum number of families in the locality = 3.

Hence, option (d).

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**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

Answer: Option C

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**Explanation** :

100 nuts are produced per minute.

∴ Time taken to produce 1000 nuts = 10 minutes.

But the machine needs cleaning for 5 minutes after producing 1000 nuts. Thus effective time taken to produce 1000 nuts = 15 minutes.

Effective time taken to produce 9000 nuts = 9 × 15 – time taken for the last cleaning = 135 – 5 = 130 minutes.

75 bolts are produced per minute.

∴ Time taken to produce 1500 bolts = 20 minutes.

But the machine needs cleaning for 10 minutes after producing 1500 bolts. Thus effective time taken to produce 1500 bolts = 30 minutes.

Effective time taken to produce 9000 bolts = 6 × 30 – time taken for the last cleaning = 180 – 10 = 170 minutes.

Thus time taken to produce 9000 pairs of nuts and bolts = 170 minutes.

Hence, option (c).

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**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

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**Explanation** :

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

x > y

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

∴ $\frac{{x}^{2}}{2}={y}^{2}$

∴ $\frac{{x}^{2}}{2}$ = 4 (∵ y = 2)

∴ x^{2} = 8

∴ x = 2√2

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

Hence, option (b).

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

In the adjoining figure, I and II are circles with centres P and Q respectively. The two circles touch each other and have a common tangent that touches them at points R and S respectively. This common tangent meets the line joining P and Q at O. The diameters of I and II are in the ratio 4 : 3. It is also known that the length of PO is 28 cm.

**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

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**Explanation** :

ΔSQO ~ ΔRPO

⇒ $\frac{SQ}{RP}$ = $\frac{OQ}{OP}$ = $\frac{SO}{OR}$

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

⇒ 42 = 56 – 7x

⇒ x = 2

⇒ PQ = 7 cm, OQ = 21 cm

∴ PQ : OQ = 1 : 3

Hence, option (b).

Workspace:

**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

Answer: Option B

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**Explanation** :

Radius of circle II = 1.5 × 2 = 3 cm.

Hence, option (b).

**Alternately,**

We can solve this question and the previous question together using options as follows:

Consider each of the options in this question.

Option (a): If the radius of circle II is 2, the radius of I is 8/3. PQ is then 14/3 and OQ is 70/3.

∴ PQ : OQ = 1 : 5, which is not there in the options of the previous question.

Option (b): If the radius of circle II is 3, the radius of I is 4. PQ is then 7 and OQ is 21.

∴ PQ : OQ = 1 : 3, which is there in the options of the previous question. This is possible.

Option (c): If the radius of circle II is 4, the radius of I is 16/3. PQ is then 28/3 and OQ is 56/3.

∴ PQ : OQ = 1 : 2, which is not there in the options of the previous question.

Option (d): If the radius of circle II is 5, the radius of I is 20/3. PQ is then 35/3 and OQ is 49/3.

∴ PQ : OQ = 5 : 7, which is not there in the options of the previous question.

∴ The radius of circle II is 3 and PQ : OQ = 1 : 3

Hence, option (b).

Workspace:

**19. CAT 2004 QA | Geometry - Circles**

The length of SO is ________.

- A.
$\mathrm{}$8√3 cm

- B.
10√3 cm

- C.
$\mathrm{}$12√3 cm

- D.
$\mathrm{}$14√3 cm

Answer: Option C

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**Explanation** :

SO^{2} = OQ^{2} – SQ^{2}

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

Hence, option (c).

Workspace:

**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°

Answer: Option D

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**Explanation** :

m∠EBC = 65°

∴ m∠EOC = 130° (∵ angle subtended by an arc at the centre is twice the angle subtended by the same arc on the circumference)

As AC || ED, m∠OCE = m∠DEC

ΔOEC is an isosceles triangle.

∴ m∠OCE = m∠OEC = (180 – 130)/2 = 25°

∴ m∠DEC = 25°

Hence, option (d).

Workspace:

**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

Answer: Option B

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**Explanation** :

Quadrilateral ABCD is an isosceles trapezium. Let AM = DN = x

∵ AD = 8, AO = DO = BO = 4

∴ OM = ON = 4 – x

ΔBMO and ΔBMA are right triangles.

∴ OB^{2} = OM^{2} + MB^{2}

∴ 16 = (4 – *x*)^{2} + *h*^{2} …(i)

Also,

AB^{2} = MB^{2} + AM^{2}

4 = *x*^{2} + *h*^{2} …(ii)

From (i) and (ii), x = 0.5

∴ BC = 2(4 – x)

∴ BC = 7

Hence, option (b).

Workspace:

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

*f*_{1}(*x*) = *x* 0 ≤ *x* ≤ 1

= 1 *x* ≥ 1

= 0 otherwise

*f*_{2}(*x*) = *f*_{1}(–*x*) for all *x*

*f*_{3}(*x*) = –*f*_{2}(*x*) for all *x*

*f*_{4}(*x*) = *f*_{3}(–*x*) for all *x*

**22. CAT 2004 QA | Algebra - Functions & Graphs**

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

*f*_{1}(*x*)*f*_{2}(*x*), *f*_{2}(*x*)*f*_{3}(*x*), *f*_{2}(*x*)*f*_{4}(*x*)?

- A.
0

- B.
1

- C.
2

- D.
3

Answer: Option C

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**Explanation** :

*f*_{1}(*x*) = *x* 0 ≤ *x* ≤ 1

= 1 *x* ≥ 1

= 0 –1 ≤ *x* < 0

= 0 *x* ≤ –1

The table is written after defining the given function as above.

From the table, only *f*_{1}(*x*) × *f*_{2}(*x*) and *f*_{2}(*x*) × *f*_{4}(*x*) are necessarily zero for every x.

Hence, option (c).

Workspace:

**23. CAT 2004 QA | Algebra - Functions & Graphs**

Which of the following is necessarily true?

- A.
*f*_{4}(*x*) =*f*_{1}(*x*) for all*x* - B.
*f*_{1}(*x*) = –*f*_{3}(–*x*) for all*x* - C.
*f*_{2}(–*x*) =*f*_{4}(*x*) for all*x* - D.
*f*_{1}(*x*) +*f*_{3}(*x*) = 0 for all*x*

Answer: Option B

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**Explanation** :

From the table in the previous question,

–*f*_{3}(–*x*) = –*f*_{4}(*x*) = *f*_{1}(*x*).

Hence, option (b).

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

In an examination, there are 100 questions divided into three groups A, B and C such that each group contains at least one question. Each question in group A carries 1 mark, each question in group B carries 2 marks and each question in group C carries 3 marks. It is known that the questions in group A together carry at least 60% of the total marks.

**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

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

b = 23

⇒ a + c = 77

⇒ a = 77 – c

⇒ $\frac{\mathrm{a}}{\mathrm{a}+46+3\mathrm{c}}\ge 0.6$

⇒ 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).

Workspace:

**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

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**Explanation** :

c = 8

∴ a + b = 92 ...(1)

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

⇒ $\frac{2b}{92-b+2b+24}\ge 0.2$

⇒ 2b ≥ 23.2 + 0.2b

⇒ $b\ge \frac{232}{18}$

⇒ b ≥ 13 ...(2)

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

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

⇒ $\frac{a}{a+2(92-a)+24}\ge 0.6$

⇒ 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).

Workspace:

**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

Answer: Option C

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**Explanation** :

Let πr = x.

∴ The circumference of the circular path = 2x.

∴ The sprinter covers 2x metres in one round. The time taken to complete each round is as under:

Thus, we see that the ratio of time taken for the second round to that for the 1st round = 16.

Similarly, ratio of time taken for the third round to that for the 2nd round = 16 and so on.

Hence, option (c).

Workspace:

**27. CAT 2004 QA | Algebra - Progressions**

Consider the sequence of numbers *a*_{1}, *a*_{2}, *a*_{3}, ... to infinity where *a*_{1} = 81.33 and *a*_{2} = –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

Answer: Option C

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**Explanation** :

*a*_{1} = 81.33

*a*_{2} = –19

*a*_{3} = *a*_{2} - *a*_{1} = –100.33

*a*_{4} = *a*_{3} - *a*_{2} = –81.33

*a*_{5} = *a*_{4} - *a*_{3} = 19

*a*_{6} = *a*_{5} - *a*_{4} = 100.33

*a*_{7} = *a*_{2} - *a*_{5} = 81.33

*a*_{8} = *a*_{7} - *a*_{6} = –19

We can see that the sequence repeats itself after every 6 terms.

Sum of the first 6 terms of the sequence = 0

Thus, the sum of the first 6000 terms of this sequence = 0

The sum of the 6001^{st} and 6002^{nd} terms = 81.33 – 19 = 62.33

Hence, option (c).

Workspace:

**28. CAT 2004 QA | Algebra - Number Theory**

The remainder, when (15^{23} + 23^{23}) is divided by 19, is:

- A.
4

- B.
15

- C.
0

- D.
18

Answer: Option C

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**Explanation** :

(a^{n} + b^{n}) is always divisible by (a + b) if n is odd

∴ (15^{23} + 23^{23}) is divisible by (15 + 23) ie, 38.

⇒ 15^{23} + 23^{23} = 38x, which is divisible by 19.

⇒ Remainder when (15^{23} + 23^{23}) is divided by 19 is 0.

Hence, option (c).

Workspace:

**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

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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).

Workspace:

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

Let C be a circle with centre P_{0} and AB be a diameter of C. Suppose P_{1} is the mid-point of the line segment P_{0}B, P_{2} is the mid-point of the line segment P_{1}B and so on. Let C_{1}, C_{2}, C_{3}, ... be circles with diameters P_{0}P_{1}, P_{1}P_{2}, P_{2}P_{3} ... respectively. Suppose the circles C_{1},C_{2}, C_{3}, ... 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

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**Explanation** :

Let the radius of C be r.

Then P_{0}P_{1} = $\frac{r}{2}$

P_{1}P_{2} = $\frac{r}{4}$

P_{2}P_{3} = $\frac{r}{8}$

and so on.

Thus the areas of circles with diameters P_{0}P_{1}, P_{1}P_{2}, P_{2}P_{3}, … are

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

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

∴ Shaded area = $\frac{\pi {r}^{2}}{12}$

Unshaded area = πr^{2} - $\frac{\pi {r}^{2}}{12}=\frac{11\pi {r}^{2}}{12}$

∴ Required ratio = 11 : 12

Hence, option (d).

Workspace:

**31. CAT 2004 QA | Algebra - Logarithms**

Let u = (log_{2 }*x*)^{2} – 6 log_{2} *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

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**Explanation** :

*u* = (log_{2}*x*)^{2} – 6log_{2}*x* + 12 ...(1)

*x ^{u}* = 256

Taking log on both sides,

*u*log_{2}*x* = log_{2}256 = 8

Let log_{2}*x* = *m*

∴ u × m = 8

Substituting in (1),

$\frac{8}{m}$ = m^{2} - 6m + 12

⇒ *m*^{3} – 6*m*^{2} + 12*m* – 8 = 0

⇒ (*m* – 2)^{3} = 0

⇒ m = 2

⇒ log_{2}*x* = 2

⇒ x = 4

∴ The given equation has exactly one solution for x.

Hence, option (b).

Workspace:

**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

Answer: Option A

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**Explanation** :

The first stripe can be coloured in 4 ways. Then the second can be coloured in 3 ways. The third, fourth, fifth and sixth can be coloured thereafter in 3 ways each.

∴ The total number of ways in which the flag can be coloured

= 4 × 3 × 3 × 3 × 3 × 3

= 12 × 81 ways.

Hence, option (a).

Workspace:

**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

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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).

Workspace:

**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

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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).

Workspace:

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

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

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

**Choose 2 **if the question can be answered by using either of the statements alone.

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

**Choose 4 **if the question cannot be answered on the basis of the two statements.

Ravi spent less than Rs.75 to buy one kilogram each of potato, onion, and gourd. Which one of the three vegetables bought was the costliest?

A. 2 kg potato and 1 kg gourd cost less than 1 kg potato and 2 kg gourd.

B. 1 kg potato and 2 kg onion together cost the same as 1 kg onion and 2 kg gourd.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

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**Explanation** :

Let one kilogram of potatoes, onions and gourd cost Rs. p, Rs. o and Rs. g respectively. Then p + o + g < 75.

**Consider statement A:** 2p + g < p + 2g

∴ p < g

But, nothing is known about o.

∴ Statement A is insufficient.

**Consider statement B: **p + 2o = o + 2g

∴ o + p = 2g

∴ The cost of 1 kg of gourd is the average of the costs of 1 kg potatoes and 1 kg onions.

∴ g is neither the costliest nor the cheapest.

But the costliest vegetable cannot be found.

∴ Statement B is insufficient.

**Considering both statements together:** p < g and g is the average of p and o

∴ o > g

∴ Onions are the costliest.

Hence, option (c).

Workspace:

**36. CAT 2004 QA | Modern Math - Permutation & Combination | Data Sufficiency**

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

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

**Choose 2 **if the question can be answered by using either of the statements alone.

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

**Choose 4 **if the question cannot be answered on the basis of the two statements.

Tarak is standing 2 steps to the left of a red mark and 3 steps to the right of a blue mark. He tosses a coin. If it comes up heads, he moves one step to the right; otherwise he moves one step to the left. He keeps doing this until he reaches one of the two marks, and then he stops. At which mark does he stop?

A. He stops after 21 coin tosses.

B. He obtains three more tails than heads.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

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**Explanation** :

Let Tarak get h heads and t tails.

Let Tarak be at 0 on the number line.

Then the blue mark is at –3 and the red mark is at 2.

**Consider statement A:** h + t = 21 and

h – t = 2 (if he stops at the red mark) or

h – t = –3 (if he stops at the blue mark)

Solving, 2h = 23 (however, this is not possible)

or 2h = 18

∴ h = 9 and t = 12

∴ Tarak stops at the blue mark.

∴ Statement A is sufficient.

**Consider statement B:** h = t – 3 and h – t = 2 or h – t = –3

∴ He stops at the blue mark.

∴ Statement B is also sufficient.

Hence, option (b).

Workspace:

**37. CAT 2004 QA | Algebra - Simple Equations | Data Sufficiency**

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

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

**Choose 2 **if the question can be answered by using either of the statements alone.

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

**Choose 4 **if the question cannot be answered on the basis of the two statements.

Nandini paid for an article using currency notes of denominations Re. l, Rs. 2, Rs. 5, and Rs. 10 using at least one note of each denomination. The total number of five and ten rupee notes used was one more than the total number of one and two rupee notes used. What was the price of the article?

A. Nandini used a total of 13 currency notes.

B. The price of the article was a multiple of Rs. 10.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

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**Explanation** :

Let Nandini use a, b, c and d notes of denominations 1, 2, 5 and 10 respectively.

Then c + d = a + b + 1 and price of the article = a + 2b + 5c + 10d

**Consider statement A:** a + b + c + d = 13

∴ 2(a + b) + 1 = 13

∴ a + b = 6, c + d = 7

Price of the article cannot be found from this data.

∴ Statement A is insufficient.

**Consider statement B:** a + 2b + 5c + 10d = 10k

As a, b, c, d ≥ 1, 10k ≥ 18

∴ k ≥ 2

But, statement B is also insufficient.

**Considering both statements together:** We know that a + b = 6, c + d = 7 and Price = 10k

When c = 2 and d = 5, Price = a + 2b + 60

When c = 4 and d = 3, Price = a + 2b + 50

Even if we assume that there is only one value of (a, b) such that a + 2b is a multiple of 10, we will still get at least 2 values for the price of the article.

Hence, data is insufficient.

Hence, option (d).

Workspace:

**38. CAT 2004 QA | Data Sufficiency**

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

**Choose 2 **if the question can be answered by using either of the statements alone.

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

**Choose 4 **if the question cannot be answered on the basis of the two statements.

Four candidates for an award obtain distinct scores in a test. Each of the four casts a vote to choose the winner of the award. The candidate who gets the largest number of votes wins the award. In case of a tie in the voting process, the candidate with the highest score wins the award. Who wins the award?

A. The candidates with the top three scores each vote for the top scorer amongst the other three.

B. The candidate with the lowest score votes for the player with the second highest score.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

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**Explanation** :

Let A, B, C and D get the highest to lowest scores in that order.

**Consider statement A:** A votes for B, B votes for A and C votes for A.

If D votes for A, A wins.

If D votes for B, there is a tie between A and B and A wins as he has the highest score.

If D votes for C/D, A wins.

∴ Statement A alone is sufficient.

**Consider statement B: **D votes for B. This statement alone is not sufficient.

Hence, option (a).

Workspace:

**39. CAT 2004 QA | Data Sufficiency**

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

**Choose 2 **if the question can be answered by using either of the statements alone.

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

**Choose 4 **if the question cannot be answered on the basis of the two statements.

In a class of 30 students, Rashmi secured the third rank among the girls, while her brother Kumar studying in the same class secured the sixth rank in the whole class. Between the two, who had a better overall rank?

A. Kumar was among the top 25% of the boys merit list in the class in which 60% were boys.

B. There were three boys among the top five rank holders, and three girls among the top ten rank holders.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

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**Explanation** :

**Consider statement A:** There were 60% of 30 = 18 boys in the class.

Kumar’s rank among boys ≤ |0.25 × 18| or 4

If Kumar’s rank among boys = 4, there are 2 girls in the first 6 rankers.

∴ Rashmi’s rank is seven.

If Kumar’s rank among boys is 3, 2 or 1, there are 3, 4 or 5 girls in the first 6 rankers.

∴ Rashmi’s rank is three.

∴ Statement A is insufficient.

**Consider statement B: **There were three boys in the top 5 and Kumar was 6th.

∴ There were 4 boys and 2 girls in the top 6.

∴ Rashmi was not in the top six.

∴ Kumar had a better overall rank.

Statement B is sufficient.

Hence, option (a).

Workspace:

**40. CAT 2004 QA | Arithmetic - Percentage | Data Sufficiency**

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

**Choose 2 **if the question can be answered by using either of the statements alone.

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

**Choose 4 **if the question cannot be answered on the basis of the two statements.

Zakib spends 30% of his income on his children’s education, 20% on recreation and 10% on healthcare. The corresponding percentages for Supriyo are 40%, 25%, and 13%. Who spends more on children’s education?

A. Zakib spends more on recreation than Supriyo.

B. Supriyo spends more on healthcare than Zakib.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

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**Explanation** :

Let Zakib’s and Supriyo’s incomes be z and s respectively.

**Consider statement A:**

0.2z > 0.25s

$\therefore \frac{z}{s}<1.25$

∴ z > s

Zakib and Supriyo spend 0.3z and 0.4s on children’s education.

$\frac{0.3z}{0.4s}>0.3\times \frac{1.25}{0.4}\phantom{\rule{0ex}{0ex}}\frac{0.3z}{0.4s}>0.9375$

∴ Statement A alone is not sufficient.

**Consider statement B:**

0.13s > 0.1z

$\therefore \frac{z}{s}<1.3$

$\frac{0.3z}{0.4s}<0.975$

∴ Supriyo spends more than Zakib on children’s education.

∴ Statement B alone is sufficient.

Hence, option (a).

Workspace:

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