CAT 2003 QA - Retake | Previous Year CAT Paper

Previous year paper questions for CAT 2003 QA - Retake

1. CAT 2003 QA - Retake | Geometry - Triangles

A piece of paper is in the shape of a right angled triangle and is cut along a line that is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in the length of the hypotenuse of the triangle. If the area of the original triangle was 34 square inches before the cut, what is the area (in square inches) of the smaller triangle?

A.
16.665
B.
16.565
C.
15.465
D.
14.365

Answer: Option D

Explanation :

Since DE is parallel to AC, ∆ABC is similar to ∆DBE by AAA rule of similarity,

i.e. ΔABC ~ ΔDBE

When two triangles are similar, the ratio of their areas is equal to the ratio of squares of their corresponding sides.

$∴ \frac{A r e a (∆ A B C)}{A r e a (∆ D B E)} = {(\frac{A C}{D E})}^{2} = {(\frac{1}{0.65})}^{2}$

∴ Area (∆DBE) = (0.65)² × Area (∆ABC)

∴ Area (∆DBE) = 0.4225 × 34 = 14.365

Hence, option (d).

2. CAT 2003 QA - Retake | Arithmetic - Ratio, Proportion & Variation

In a coastal village, every year floods destroy exactly half of the huts. After the flood water recedes, twice the number of huts destroyed are rebuilt. The floods occurred consecutively in the last three years namely 2001, 2002 and 2003. If floods are again expected in 2004, the number of huts expected to be destroyed is:

A.
Less than the number of huts existing at the beginning of 2001.
B.
Less than the total number of huts destroyed by floods in 2001 and 2003.
C.
Less than the total number of huts destroyed by floods in 2002 and 2003.
D.
More than the total number of huts built in 2001 and 2002.

3. CAT 2003 QA - Retake | Arithmetic - Ratio, Proportion & Variation

Let a, b, c, d and e be integers such that a = 6b = 12c, and 2b = 9d = 12e. Then which of the following pairs contains a number that is not an integer?

A.
$[\frac{a}{27}, \frac{b}{e}]$
B.
$[\frac{a}{36}, \frac{c}{e}]$
C.
$[\frac{a}{12}, \frac{b d}{18}]$
D.
$[\frac{a}{6}, \frac{c}{d}]$

Answer: Option D

Explanation :

Since a = 6b = 12c and 2b = 9d = 12e

∴ a : b : c = 12 : 2 : 1 and b : d : e = 18 : 4 : 3

∴ a : b : c : d : e = 108 : 18 : 9 : 4 : 3

∴ a = 108k; b = 18k; c = 9k; d = 4k and e = 3k where ‘k’ is an integer

Option 1 is: $(\frac{a}{27}, \frac{b}{e})$ = (4k, 6)

Option 2 is: $(\frac{a}{36}, \frac{c}{e})$ = (3k, 3)

Option 3 is: $(\frac{a}{12}, \frac{b d}{18})$ = (9k, 4k²)

Option 4 is: $(\frac{a}{6}, \frac{c}{d}) = (18 k, \frac{9}{4})$

The 4th option contains $\frac{9}{4}$ , which is not an integer.

Hence, option (d).

4. CAT 2003 QA - Retake | Algebra - Number Theory

If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is:

A.
one
B.
two
C.
three
D.
more than three

5. CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons

A square tin sheet of side 12 inches is converted into a box with open top in the following steps – the sheet is placed horizontally. Then, equal sized squares, each of side x inches, are cut from the four corners of the sheet. Finally, the four resulting sides are bent vertically upwards in the shape of a box. If x is an integer, then what value of x maximizes the volume of the box?

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

Answer: Option D

Explanation :

When the tin sheet is cut across its corners as shown in the figure, the box formed will have a height of x inches and its base will be a square of side (12 – 2x) inches.

Let the volume of the box, V = (12 – 2x)² × x = 4x³ − 48x² + 144x

For V to be maximum, $\frac{d V}{d x}$ shoule be 0.

i.e. 12x² − 96x + 144 = 0

∴ 12(x − 6)(x − 2) = 0

∴ x = 2 or x = 6

However, x cannot be 6 as the length of the side is (12 – 2x).

∴ x = 2

Hence, option (d).

Alternatively,

Since V = (12 – 2x)² × x = [2(6 – x)]² × x = 4x(6 – x)²,

Substituting values of x from 1 to 5, we get V maximum when x = 2 (i.e. V = 128)

6. CAT 2003 QA - Retake | Arithmetic - Time, Speed & Distance | Geometry - Triangles

Two straight roads R₁ and R₂ diverge from a point A at an angle of 120°. Ram starts walking from point A along R₁ at a uniform speed of 3 km/hr. Shyam starts walking at the same time from A along R₂ at a uniform speed of 2 km/hr. They continue walking for 4 hours along their respective roads and reach points B and C on R₁ and R₂, respectively. There is a straight line path connecting B and C. Then Ram returns to point A after walking along the line segments BC and CA. Shyam also returns to A after walking along line segments CB and BA. Their speeds remain unchanged. The time interval (in hours) between Ram’s and Shyam’s return to the point A is:

A.
$\frac{10 \sqrt{19} + 26}{3}$
B.
$\frac{2 \sqrt{19} + 10}{3}$
C.
$\frac{\sqrt{19} + 26}{3}$
D.
$\frac{\sqrt{19} + 10}{3}$

Answer: Option B

Explanation :

After 4 hours,

Distance travelled by Ram, AB = 4 × 3 = 12 km

Distance travelled by Shyam, AC = 4 × 2 = 8 km

Ram’s return path = BC + CA = (x + 8) km

Shyam’s return path = CB + BA = (x + 12) km

Now dropping perpendiculars from C and A to form the right triangle AOC,

In ΔAOC,

OA = AC sin30˚ = 8 × 1/2 = 4

OC = AC sin60° = 8 × $\frac{\sqrt{3}}{2} = 4 \sqrt{3}$

∴ BC = $\sqrt{({(4 \sqrt{3})}^{2} + 16^{2})}$

BC = $\sqrt{304}$

BC = $4 \sqrt{19}$

Time taken by Ram to travel = $\frac{B C + C A}{3} = \frac{4 \sqrt{19} + 8}{3}$ ...(i)

Time taken by Shyam to travel = $\frac{B C + B A}{2} = \frac{4 \sqrt{19} + 12}{2}$ ...(ii)

∴ Time interval between the two reaching point A is (i) subtracted from (ii)

= $\frac{2 \sqrt{19} + 10}{3}$

Hence, option (b).

7. CAT 2003 QA - Retake | Algebra - Number Theory

Let x and y be positive integers such that x is prime and y is composite. Then,

A.
(y – x) cannot be an even integer.
B.
xy cannot be an even integer.
C.
(x + y)/x cannot be an even integer.
D.
None of these

8. CAT 2003 QA - Retake | Algebra - Simple Equations | Venn Diagram

A survey on a sample of 25 new cars being sold at a local auto dealer was conducted to see which of the three popular options- air conditioning, radio and power windows- were already installed. The survey found

15 had air conditioning,
2 had air conditioning and power windows but no radios,
12 had radio,
6 had air conditioning and radio, but no power windows,
11 had power windows,
4 had radio and power windows and
3 had all three options.

What is the number of cars that had none of the options?

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

9. CAT 2003 QA - Retake | Algebra - Number Theory

If n is such that 36 ≤ n ≤ 72, then x = $\frac{n^{2} + 2 \sqrt{n} (n + 4) + 16}{n + 4 \sqrt{n} + 4}$ satisfies

A.
20 < x < 54
B.
23 < x < 58
C.
25 < x < 64
D.
28 < x < 60

Answer: Option C

Explanation :

$\begin{aligned} x & = \frac{n^{2} + 2 \sqrt{n} (n + 4) + 16}{n + 4 \sqrt{n} + 4} \\ = \frac{n^{2} + 2 n \sqrt{n} + 8 \sqrt{n} + 16}{(\sqrt{n} + 2)^{2}} \\ = \frac{n \sqrt{n} (\sqrt{n} + 2) + 8 (\sqrt{n} + 2)}{(\sqrt{n} + 2)^{2}} \\ = \frac{(n \sqrt{n} + 8) (\sqrt{n} + 2)}{(\sqrt{n} + 2)^{2}} \\ ∴ x & = \frac{n \sqrt{n} + 8}{\sqrt{n} + 2} \end{aligned}$

When n = 36,

$x = \frac{36 \times 6 + 8}{6 + 2} = \frac{8 (27 + 1)}{8} = 28$

When n = 72,

$x = \frac{432 \sqrt{2} + 8}{6 \sqrt{2} + 2} = \frac{4 (54 \sqrt{2} + 1)}{3 \sqrt{2} + 1} = \frac{309.42}{5.24} \approx 59$

Only the option 3 includes both these values.

Hence, option (c).

10. CAT 2003 QA - Retake | Algebra - Number Theory

If 13x + 1 < 2z, and z + 3 = 5y², then

A.
x is necessarily less than y.
B.
x is necessarily greater than y.
C.
x is necessarily equal to y.
D.
None of the above is necessarily true.

11. CAT 2003 QA - Retake | Algebra - Number Theory

Let n(>1) be a composite integer such that $\sqrt{n}$ is not an integer.

Consider the following statements:

A: n has a perfect integer - valued divisor which is greater than 1 and less than $\sqrt{n}$ .

B: n has a perfect integer- valued divisor which is greater than $\sqrt{n}$ but less than n.

Then,

A.
Both A and B are false
B.
A is true but B is false
C.
A is false but B is true
D.
Both A and B are true

Answer: Option D

Explanation :

Consider a number n = 10, then $\sqrt{n} \approx 36$ .

A: We have a divisor 2 which is greater than 1 and less than 3.16.

B: We have a divisor 5 which is greater than 3.16 but less than 10.

∴ Both statements A and B are true.

Also, as a rule, any composite number which is not a perfect square has at least one factor less than $\sqrt{n}$ and another factor more than n, such that their product is n.

∴ Both statements are true.

Hence, option (d).

12. CAT 2003 QA - Retake | Algebra - Inequalities & Modulus

If |b| ≥ 1 and x = –|a|b, then which one of the following is necessarily true?

A.
a – xb < 0
B.
a – xb ≥ 0
C.
a – xb > 0
D.
a – xb ≤ 0

Answer the following question based on the information given below.

Consider a cylinder of height h cms and radius r = 2/π cms as shown in the figure (not drawn to scale). A string of a certain length, when wound on its cylindrical surface, starting at point A and ending at point B, gives a maximum of n turns (in other words, the string’s length is the minimum length required to wind n turns.)

13. CAT 2003 QA - Retake | Geometry - Mensuration

What is the vertical spacing in cms between two consecutive turns?

A.
$\frac{h}{n}$
B.
$\frac{h}{\sqrt{n}}$
C.
$\frac{h}{n^{2}}$
D.
Cannot be determined with given information

Answer: Option A

Explanation :

The equivalent spacing between two consecutive turns = $\frac{H e i g h t o f t h e c y l i n d e r}{N u m b e r o f t u r n s} = \frac{h}{n}$

Hence, option (a).

14. CAT 2003 QA - Retake | Geometry - Mensuration

The same string, when wound on the exterior four walls of a cube of side n cms, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale).

The length of the string, in cms, is

A.
$\sqrt{2} n$
B.
$\sqrt{17} n$
C.
n
D.
$\sqrt{13} n$

Answer: Option B

Explanation :

If the cube is cut open as shown in the figure below, a rectangle with sides 4n and n is obtained.

The string is the diagonal along this rectangle, with length,

$\sqrt{n^{2} + {(4 n)}^{2}} = n \sqrt{17}$

∴ The length of the string is $n \sqrt{17} .$

Hence, option (b).

15. CAT 2003 QA - Retake | Geometry - Mensuration

In the setup of the previous two questions, how is h related to n?

A.
$h = \sqrt{2} n$
B.
$h = \sqrt{17} n$
C.
h = n
D.
h = $\sqrt{13} n$

Answer: Option C

Explanation :

The string tied on the cylinder and the cube is of the same length $(n \sqrt{17})$ .

Now, we cut the cylinder horizontally such that each turn of the string is now on a separate cylinder. On opening up one of the n cylinders, we get a rectangle whose length is equal to the circumference of the cylinder and whose breadth is equal to h/n. The string will be the diagonal of the rectangle.

Cylinder’s circumference = 2 × π × r = 2 × π × 2/π = 4 units

The string has encircled the cylinder n times, thus the length used in 1 turn

= $\frac{n \sqrt{17}}{n} = \sqrt{17}$

From the above figure,

${(\sqrt{17})}^{2} = {(\frac{h}{n})}^{2} + 4^{2}$

∴ h = n

Hence, option (c).

Answer the following question based on the information given below.

Consider three circular parks of equal size with centres at A₁, A₂ and A₃ respectively. The parks touch each other at the edge as shown in the figure (not drawn to scale). There are three paths formed by the triangles A₁A₂A₃, B₁B₂B₃ and C₁C₂C₃, as shown. Three sprinters A, B, and C begin running from points A₁, B₁ and C₁ respectively. Each sprinter traverses her respective triangular path clockwise and returns to her starting point.

16. CAT 2003 QA - Retake | Geometry - Circles

Let the radius of each circular park be r, and the distances to be traversed by the sprinters A, B and C be a, b and c, respectively. Which of the following is true?

A.
b - a = c - b = 3 $\sqrt{3} r$
B.
b - a = c - b = $\sqrt{3}$ r
C.
b = $\frac{a + c}{2} = 2 (1 + \sqrt{3}) r$
D.
c = 2b - a = (2 + $\sqrt{3}$ )r

Answer: Option A

Explanation :

The radius of each circular park = r

A₁B₁ = A₁E = r

Distance travelled by A = a = 3 × 2r = 6r

∆A₁B₁D is a right angle triangle with ∠A₁B₁D = 30° and ∠B₁A₁D = 60°

∴ B₁D = $\frac{\sqrt{3} r}{2}$

∴B₁B₂ = 2r + 2 × $\frac{r \sqrt{3}}{2} = r (2 + \sqrt{3})$

∴ Distance travelled by B = b = 3r(2 + $\sqrt{3}$ )

Similarly, ∆A₁C₁E is a right angle triangle with ∠A₁C₁E = 30° and ∠C₁A₁E = 60°

∴ C₁E = $r \sqrt{3}$

∴ C₁C₂ = 2r + 2r $\sqrt{3} = 2 r (1 + \sqrt{3})$

∴ Distance travelled by C = c = 6r(1 + $\sqrt{3}$ )

∴ b - a = 3 $\sqrt{3}$ r and c - b = 3 $\sqrt{3}$ r

Hence, option (a).

17. CAT 2003 QA - Retake | Geometry - Circles

Sprinter A traverses distances A₁A₂, A₂A₃, and A₃A₁ at average speeds of 20, 30 and 15 respectively. B traverses her entire path at a uniform speed of $(10 \sqrt{3} + 20) .$ C traverses distances C₁C₂, C₂C₃, and C₃C₁ at average speeds of $\frac{40}{3} (\sqrt{3} + 1), \frac{40}{3} (\sqrt{3} + 1),$ and 120 respectively. All speeds are in the same unit. Where would B and C be respectively when A finishes her sprint?

A.
B₁, C₁
B.
B₃, C₃
C.
B₁, C₃
D.
B₁, Somewhere between C₃, C₁

Answer: Option C

Explanation :

Time taken by A to travel through distance a (A₁A₂ + A₂A₃ + A₃A₁)

$= \frac{2 r}{20} + \frac{2 r}{30} + \frac{2 r}{15} = \frac{3 r}{10}$

Distance travelled by B in the same time,

$= \frac{3 r}{10} \times (10 \sqrt{3} + 20) = 3 r (2 + \sqrt{3}) = b$

∴ B travels a full round in the same time as A. Thus, B will be at B₁.

For C, time taken to traverse through C₁C₂

_{$= \frac{2 r (\sqrt{3} + 1)}{\frac{40}{3} (\sqrt{3} + 1)} = \frac{3 r}{20}$}

Time remaining for C = $= (\frac{3}{10} - \frac{3}{20}) r = \frac{3 r}{20}$

Now, distance travelled by C in this time = $\frac{3 r}{20} \times \frac{40}{32} (1 + \sqrt{3}) = 2 r (1 + \sqrt{3}) = C_{2} C_{3}$

∴ C will be at C₃.

Hence, option (c).

18. CAT 2003 QA - Retake | Geometry - Circles

Sprinters A, B and C traverse their respective paths at uniform speeds u, v and w respectively. It is known that u²: v² : w² is equal to Area A : Area B : Area C, where Area A, Area B and Area C are the areas of triangles A₁A₂A₃, B₁B₂B₃, and C₁C₂C₃ respectively.

Where would A and C be when B reaches point B₃?

A.
A₂, C₃
B.
A₃, C₃
C.
A₃, C₂
D.
Somewhere between A₂ and A₃, Somewhere between C₃ and C₁

19. CAT 2003 QA - Retake | Algebra - Progressions

The infinite sum $1 + \frac{4}{7} + \frac{9}{7^{2}} + \frac{16}{7^{3}} + \frac{25}{7^{4}} + . . . .$ equals

A.
$\frac{27}{14}$
B.
$\frac{21}{13}$
C.
$\frac{49}{27}$
D.
$\frac{256}{147}$

Answer: Option C

Explanation :

Let S = $1 + \frac{4}{7} + \frac{9}{7^{2}} + \frac{16}{7^{3}} + \frac{25}{7^{4}} + . . . . .$ (i)

Dividing (i) by 7, we get,

$\frac{S}{7} = \frac{1}{7} + \frac{4}{7^{2}} + \frac{9}{7^{3}} + \frac{16}{7^{4}} + \frac{25}{7^{5}} + . . . .$ (ii)
Subtracting (ii) from (i)

$\frac{6 S}{7} = 1 + \frac{3}{7} + \frac{5}{7^{2}} + \frac{7}{7^{3}} + \frac{9}{7^{4}} + . . . .$ (iii)
Dividing (iii) by 7

$\frac{6 S}{49} = \frac{1}{7} + \frac{3}{7^{2}} + \frac{5}{7^{3}} + \frac{5}{7^{4}} + . . .$ (iv)
Subtracting (iv) from (iii)

$\frac{36 S}{49} = 1 + 2 (\frac{1}{7} + \frac{1}{7^{2}} + \frac{1}{7^{3}} + . . .) = 1 + \frac{2 \times \frac{1}{7}}{1 - \frac{1}{7}} = 1 + \frac{\frac{2}{7}}{\frac{6}{7}} = \frac{4}{3}$

$∴ \frac{36 S}{49} = \frac{4}{3} S = \frac{4}{3} \times \frac{49}{36} = \frac{49}{27}$

Hence, option (c).

20. CAT 2003 QA - Retake | Modern Math - Sets

Consider the sets T_n = {n, n + 1, n + 2, n + 3, n + 4}, where n = 1, 2, 3, … , 96.

How many of these sets contain 6 or any integral multiple thereof (i.e., any one of the numbers 6, 12, 18, ...)?

A.
80
B.
81
C.
82
D.
83

21. CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons

Let ABCDEF be a regular hexagon. What is the ratio of the area of the triangle ACE to that of the hexagon ABCDEF?

A.
$\frac{1}{3}$
B.
$\frac{1}{2}$
C.
$\frac{2}{3}$
D.
$\frac{5}{6}$

Answer: Option B

Explanation :

ABCDEF is a regular hexagon, join centre O with vertices A, C and E.

In quadrilateral AFEO, diagonal EA divides it into two equal area triangles.

i.e. Area (∆AFE) = Area (∆AOE)

Similarly, Area (∆ABC) = Area (∆AOC)

And, Area (∆CDE) = Area (∆COE)

$∴ \frac{A r e a (∆ A C E)}{A r e a (H e x a g o n A B C D E F)} = \frac{3}{6} = \frac{1}{2}$

Hence, option (b).

22. CAT 2003 QA - Retake | Algebra - Quadratic Equations

The number of roots common between the two equations x³+ 3x²+ 4x + 5 = 0 and x³+ 2x²+ 7x + 3 = 0 is:

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

23. CAT 2003 QA - Retake | Algebra - Inequalities & Modulus

A real number x satisfying $1 - \frac{1}{n} < x \leq 3 + \frac{1}{n},$ for every positive integer n, is best described by:

A.
1 < x < 4
B.
1 < x ≤ 3
C.
0 < x ≤ 4
D.
1 ≤ x ≤ 3

Answer: Option C

Explanation :

$(1 - \frac{1}{n}) < x \leq (3 + \frac{1}{n}),$ wher n is a positive integer

If n = 1, then 0 < x ≤ 4

As the value of n increases from 1 to infinity, the range of x approaches (1 < x ≤ 3).

So, the lower limit of x increases from almost 0 (to almost 1), while the higher limit decreases from 4 (to 3).

Thus, the lowest possible limit of x will be > 0 and the highest possible limit will be ≤ 4.

∴ For all range of n, 0 < x ≤ 4

Hence, option (c).

24. CAT 2003 QA - Retake | Algebra - Logarithms

If $\frac{1}{3} \log_{3} M + 3 \log_{3} N = 1 + \log_{0.008} 5,$ then

A.
$M^{9} = \frac{9}{N}$
B.
$N^{9} = \frac{9}{M}$
C.
$M^{3} = \frac{3}{N}$
D.
$N^{9} = \frac{3}{M}$

Answer: Option B

Explanation :

$\frac{1}{3} \log_{3} M + 3 \log_{3} N + 1 + \log_{0.008} 5$ ...(i)

Consider the R.H.S of (i):

$\begin{aligned} 1 + \log_{0.008} 5 \\ = 1 + \frac{\log 5}{\log 0.008} \\ = 1 + \frac{\log (\frac{10}{2})}{\log (\frac{8}{1000})} \\ = 1 + \frac{\log 10 - \log 2}{\log 8 - \log 1000} = 1 + \frac{\log 10 - \log 2}{3 \log 2 - 3 \log 10} \end{aligned}$

$= 1 + \frac{\log 10 - \log 2}{- 3 (\log 10 - \log 2)} = 1 - \frac{1}{3} = \frac{2}{3}$ ...(ii)

Consider the L.H.S of (i):

$\begin{aligned} \frac{1}{3} \log_{3} M + 3 \log_{3} N \end{aligned}$

$= \log_{3} M^{1 / 3} + \log_{3} N^{3} = \log_{3} (M^{1 / 3} N^{3})$ ....(iii)

Equating equations (ii) and (iii), we get,

$\begin{aligned} \log_{3} (M^{1 / 3} N^{3}) = \frac{2}{3} \\ ∴ M^{1 / 3} N^{3} = 3^{2 / 3} \end{aligned}$

Cubing both sides, we get,

M × N⁹ = 9

$∴ N^{9} = \frac{9}{M}$

Hence, option (b).

25. CAT 2003 QA - Retake | Algebra - Simple Equations

Using only 2, 5, 10, 25 and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 69 paise and Rs 1.01 to three different persons?

A.
19
B.
20
C.
17
D.
18

26. CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons

The length of the circumference of a circle equals the perimeter of a triangle of equal sides, and also the perimeter of a square. The areas covered by the circle, triangle, and square are c, t, and s, respectively. Then,

A.
s > t > c
B.
c > t > s
C.
c > s > t
D.
s > c > t

Answer: Option C

Explanation :

Let radius of the circle be r, a side of the equilateral triangle be a, and a side of the square be x.

The circumference/perimeter of the circle, triangle and square are equal. Hence,

2πr = 3a = 4x = k

$∴ r = \frac{k}{2 π}, a = \frac{k}{3},$ and $x = \frac{k}{4}$

The areas of the circle, triangle and square are c, t, s respectively. Hence,

$\begin{aligned} c = π r^{2} = \frac{π k^{2}}{4 π^{2}} = \frac{k^{2}}{4 π}, \\ t = \frac{\sqrt{3}}{4} a^{2} = \frac{\sqrt{3}}{4} \times \frac{k^{2}}{9} = \frac{k^{2}}{12 \sqrt{3}} \\ s = x^{2} = \frac{k^{2}}{16} \\ ∵ \frac{1}{π} > \frac{1}{4} > \frac{1}{3 \sqrt{3}} \\ ∴ c > s > t \end{aligned}$

Hence, option (c).

27. CAT 2003 QA - Retake | Algebra - Number Theory

What is the remainder when 4⁹⁶ is divided by 6?

A.
0
B.
2
C.
3
D.
4

28. CAT 2003 QA - Retake | Algebra - Simple Equations

If x and y are integers then the equation 5x + 19y = 64 has:

A.
no solution for x < 300 and y < 0
B.
no solution for x > 250 and y > –100
C.
a solution for 250 < x < 300
D.
a solution for –59 < y < –56

Answer the following question based on the information given below.

Two binary operations ⊕ and * are defined over the set {a, e, f, g, h} as per the following tables:

Thus, according to the first table f ⊕ g = a, while according to the second table g * h = f, and so on.

Also, let f² = f * f, g³= g * g * g, and so on.

29. CAT 2003 QA - Retake | Miscellaneous

What is the smallest positive integer n such that gⁿ = e?

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

30. CAT 2003 QA - Retake | Miscellaneous

Upon simplification, f ⊕ [f * {f ⊕ (f * f)}] equals:

A.
e
B.
f
C.
g
D.
h

31. CAT 2003 QA - Retake | Miscellaneous

Upon simplification, (a¹⁰ * (f¹⁰ ⊕ g⁹)} ⊕ e⁸ equals

A.
e
B.
f
C.
g
D.
h

Answer the following question based on the information given below.

A string of three English letters is formed as per the following rules:

The first letter is any vowel.
The second letter is m, n or p.
If the second letter is m, then the third letter is any vowel which is different from the first letter.
If the second letter is n, then the third letter is e or u.
1. If the second letter is p, then the third letter is the same as the first letter.

32. CAT 2003 QA - Retake | Algebra - Simple Equations

How many strings of letters can possibly be formed using the above rules?

A.
40
B.
45
C.
30
D.
35

33. CAT 2003 QA - Retake | Algebra - Simple Equations

How many strings of letters can possibly be formed using the above rules such that the third letter of the string is e?

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

34. CAT 2003 QA - Retake | Modern Math - Permutation & Combination

There are 12 towns grouped into four zones with three towns per zone. It is intended to connect the towns with telephone lines such that every two towns are connected with three direct lines if they belong to the same zone, and with only one direct line otherwise. How many direct telephone lines are required?

A.
72
B.
90
C.
96
D.
144

35. CAT 2003 QA - Retake | Geometry - Triangles

In the figure (not drawn to scale) given below, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD is parallel to CP. In ∆ARC, ∠ARC = 90°, and in ΔPQS, ∠PSQ = 90°. The length of QS is 6 cm. What is the ratio AP : PD?

A.
10 : 3
B.
2 : 1
C.
7 : 3
D.
8 : 3

Answer: Option C

Explanation :

In ∆ABC, PQ is parallel to AC,

BP : AP = BQ : QC = 3 : 4

In ∆PBC, QD is parallel to CP,

BD : DP = BQ : QC = 3 : 4

The given figure may be depicted as follows:

Let AP = 4x and PB = 3x

Since DB : PD = 3 : 4,

∴ PD = $\frac{4}{7} \times 3 x = \frac{12 x}{7}$

∴ AP : PD = 4x : $\frac{12 x}{7}$ = 7 : 3.

Hence, option (c).

36. CAT 2003 QA - Retake | Geometry - Trigonometry

A car is being driven, in a straight line and at a uniform speed, towards the base of a vertical tower. The top of the tower is observed from the car and, in the process, it takes 10 minutes for the angle of elevation to change from 45° to 60°. After how much more time will this car reach the base of the tower?

A.
$5 (\sqrt{3} + 1)$
B.
$6 (\sqrt{3} + 1)$
C.
$7 (\sqrt{3} - 1)$
D.
$8 (\sqrt{3} - 2)$

Answer: Option A

Explanation :

Let x be the distance from the later position of the car and the tower (i.e. when the angle of elevation was 60°).

Since the triangle formed (i.e. ∆ABD) is a 30°-60°-90° triangle, we have,

height of the tower, h = x $\sqrt{3}$

Now, since the triangle formed by the initial position of the car (i.e. ∆ABC) is an isosceles triangle, AB = BC

i.e. BC = $x \sqrt{3}$

∴ DC = x $\sqrt{3}$ - x = x( $\sqrt{3}$ - 1)

Time taken to travel distance DC is 10 minutes, thus,

Speed $s = \frac{x (\sqrt{3} - 1)}{10}$

Time taken to travel distance x = $\frac{x}{\frac{x (\sqrt{3} - 1)}{10}} = \frac{10}{\sqrt{3} - 1} = 5 (\sqrt{3} + 1)$

Hence, option (a).

37. CAT 2003 QA - Retake | Geometry - Triangles

In the figure (not drawn to scale) given below, if AD = CD = BC, and ∠BCE = 96°, how much is ∠DBC?

A.
32°
B.
84°
C.
64°
D.
Cannot be determined

38. CAT 2003 QA - Retake | Algebra - Quadratic Equations

If both a and b belong to the set {1, 2, 3, 4}, then the number of equations of the form ax² + bx + 1 = 0 having real roots is:

A.
10
B.
7
C.
6
D.
12

39. CAT 2003 QA - Retake | Algebra - Number Theory

If three positive real numbers x, y, z satisfy y – x = z – y and xyz = 4, then what is the minimum possible value of y?

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

Answer: Option B

Explanation :

y − x = z − y

2y = x + z ... (i)

xyz = 4 ... (ii)

It is known that Arithmetic Mean (A.M.) is greater than or equal to Geometric Mean (G.M.)

i.e. A.M. ≥ G.M.

Hence, $\frac{x + y + z}{3}$ ≥ (xyz)^1/3 .....(iii)

From (i), (ii) and (iii), we get $\frac{3 y}{3}$ ≥ 4^1/3

∴ y ≥ 2^2/3

∴ The minimum value of y is 2^2/3

Hence, option (b).

40. CAT 2003 QA - Retake | Modern Math - Permutation & Combination

An intelligence agency forms a code of two distinct digits selected from 0, 1, 2, ... , 9 such that the first digit of the code is non-zero. The code, handwritten on a slip, can however potentially create confusion, when read upside down – for example, the code 91 may appear as 16. How many codes are there for which no such confusion can arise?

A.
80
B.
78
C.
71
D.
69

41. CAT 2003 QA - Retake | Geometry - Mensuration

Consider two different cloth-cutting processes. In the first one, n circular cloth pieces are cut from a square cloth piece of side a in the following steps: the original square of side a is divided into n smaller squares, not necessarily of the same size; then a circle of maximum possible area is cut from each of the smaller squares. In the second process, only one circle of maximum possible area is cut from the square of side a and the process ends there. The cloth pieces remaining after cutting the circles are scrapped in both the processes. The ratio of the total area of scrap cloth generated in the former to that in the latter is:

A.
1 : 1
B.
$\sqrt{2} : 1$
C.
$\frac{n (4 - π)}{4 n - π}$
D.
$\frac{4 n - π}{n (4 - π)}$

Answer: Option A

Explanation :

Let the areas of the n squares formed from the original square be:

A₁, A₂, A₃, A₄, A₅, ... , A_n

Also, let A₁ + A₂ + A₃ + A₄ + A₅ + ... + A_n = A ... (i)

Where, A will be the Area of the original square.

Now, if A = a²(where a is a side of the square), then the area of the largest circle which can be drawn in it will have an area of π(a/2)² = π/4 × a² = π/4 × A

∴ Area of the maximum circle which can be cut from a square of area A is $\frac{π A}{4}$

Case 1: When the cloth is cut using the 2^nd process

The area of the scrap material will be:

Area of Square - Area of the single maximum area Circle = A - $(\frac{π A}{4}) = \frac{A}{4} (4 - π)$

Case 2: When the cloth is cut using the 1^st process

The sum of the areas of the maximum circles that can be cut out from the n squares

$= \frac{π}{4} A_{1} + \frac{π}{4} A_{2} + \frac{π}{4} A_{3} + \dots + \frac{π}{4} A_{n} = \frac{π}{4} (A_{1} + A_{2} + A_{3} + \dots + A_{n}) = \frac{π}{4} A$

Also, the sum of the areas of the n squares = Area of the original square = A

∴ Area of the scrap material will be:

$A - \frac{π}{4} A = \frac{A}{4} (4 - π)$

From Cases 1 and 2, it is clear that the ratio of scrap left in the 1st process to the 2nd process is 1 : 1.

Hence, option (a).

42. CAT 2003 QA - Retake | Geometry - Circles

In the figure below (not drawn to scale), rectangle ABCD is inscribed in the circle with centre at O. The length of side AB is greater than that of side BC.

The ratio of the area of the circle to the area of the recrangle ABCD is π : $\sqrt{3}$ .

The line segment DE intersects AB at E such that ∠ODC = ∠ADE. What is the ratio AE : AD?

A.
1 : $\sqrt{3}$
B.
1 : $\sqrt{2}$
C.
1 : 2 $\sqrt{3}$
D.
1 : 2

Answer: Option A

Explanation :

Let r be the radius of the circle and let a and b be the length and breadth of the rectangle ABCD respectively.

$\frac{π r^{2}}{a b} = \frac{π}{\sqrt{3}}$

$\sqrt{3} r^{2} = a b$ ...(i)

In ΔDBC,

$\tan θ = \frac{B C}{D C} = \frac{b}{a}$ ...(ii)

In ΔDAE,

$\tan θ = \frac{A E}{A D} = \frac{A E}{b}$ ...(iii)

From (ii) and (iii),

$\frac{A E}{A D} = \frac{b}{a}$

In ΔDBC,

$\begin{aligned} 4 r^{2} = a^{2} + b^{2} \\ 4 r^{2} = a^{2} + \frac{3 r^{4}}{a^{2}} \\ a^{4} - 4 r^{2} a^{2} + 3 r^{4} = 0 \\ a^{4} - 3 r^{2} a^{2} - r^{2} a^{2} + 3 r^{4} = 0 \\ a^{2} (a^{2} - 3 r^{2}) - r^{2} (a^{2} - 3 r^{2}) = 0 \\ a^{2} = r^{2} and a^{2} = 3 r^{2} \\ ∴ a = r and a = \sqrt{3} r \end{aligned}$

When a = ±r, then b = ± $\sqrt{3}$ r; when a = ± $\sqrt{3} r,$ then b = ±r

∴ The required ratio is either 1 : $\sqrt{3}$ or $\sqrt{3}$ : 1.

However, only the former is present in the options.

Hence, option (a).

43. CAT 2003 QA - Retake | Algebra - Logarithms

If log₁₀ x - log₁₀ $\sqrt{x}$ = 2 log_x 10, then a possible value of x is given by:

A.
10
B.
$\frac{1}{100}$
C.
$\frac{1}{1000}$
D.
None of these

Answer: Option B

Explanation :

log₁₀ x - log₁₀ $\sqrt{x}$ = 2 log_x 10 ....(i)

Consider L.H.S of (i):

log₁₀ x - $\frac{1}{2}$ log₁₀ x = $\frac{1}{2}$ log₁₀ x

Equating the L.H.S. and the R.H.S., we get,

log₁₀ x = 4 log_x 10

∴ log₁₀ x = 4 log_x 10⁴

Now, let log₁₀ x = log_x 10⁴ = y

∴ 10^y = x and x^y = 10⁴

Solving these two equations, we get,

y = ±2

∴ x = 10² or x = 10^–2

As x = 100 is not in the options,

∴ x = 10⁻² = 1/100

Hence, option (b).

44. CAT 2003 QA - Retake | Algebra - Number Theory

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?

A.
666
B.
676
C.
683
D.
777

Answer: Option B

Explanation :

Any number that gives a remainder of 3 when divided by 7 will be of the form 7k + 3.

Since we only need two-digit numbers, k will range from 1 to 13 {where 7(1) + 3 = 10 and 7(13) + 3 = 94}

Sum of all these numbers = $\sum_{k = 1}^{13} 7 k + 3$

= 13 × 3 + 7(1 + 2 + ... + 13)

= 39 + $\frac{7 \times 13 \times 14}{2}$

= 39 + 637 = 676

Hence, option (b).

45. CAT 2003 QA - Retake | Geometry - Circles

In the figure given below (not drawn to scale), A, B and C are three points on a circle with centre O. The chord BA is extended to a point T such that CT becomes a tangent to the circle at point C. If ∠ATC = 30° and ∠ACT = 50°, then the angle ∠BOA is:

A.
100°
B.
150°
C.
80°
D.
Cannot be determined

46. CAT 2003 QA - Retake | Algebra - Logarithms

What is the sum of n terms in the series

$\log m + \log \frac{m^{2}}{n} + \log \frac{m^{3}}{n^{2}} + \log \frac{m^{4}}{n^{3}} + \dots + \log \frac{m^{n}}{n^{n - 1}} ?$

A.
$\log {[\frac{n^{n - 1}}{m^{(n + 1)}}]}^{\frac{n}{2}}$
B.
$\log {[\frac{m^{m}}{n^{n}}]}^{\frac{n}{2}}$
C.
$\log {[\frac{m^{(1 - n)}}{n^{(1 - m)}}]}^{\frac{n}{2}}$
D.
$\log {[\frac{m^{(n + 1)}}{n^{(n - 1)}}]}^{\frac{n}{2}}$

Answer: Option D

Explanation :

$\begin{aligned} \log m + \log (\frac{m^{2}}{n}) + \log (\frac{m^{3}}{n^{2}}) + \log (\frac{m^{4}}{n^{3}}) + \dots + \log (\frac{m^{n}}{n^{n - 1}}) \\ = \log [\frac{m \times m^{2} \times m^{3} \times \dots \times m^{n}}{1 \times n \times n^{2} \times \dots \times n^{n - 1}}] \\ = \log [\frac{m^{1 + 2 + 3 + 4 + \dots + n}}{n^{0 + 1 + 2 + 3 + \dots + (n - 1)}}] \\ = \log [\frac{m^{\frac{n}{2} (n + 1)}}{n^{\frac{n}{2} (n - 1)}}] \\ = \log {[\frac{m^{(n + 1)}}{n^{(n - 1)}}]}^{\frac{n}{2}} \end{aligned}$

Hence, option (d).

47. CAT 2003 QA - Retake | Geometry - Quadrilaterals & Polygons

Let S₁be a square of side a. Another square S₂ is formed by joining the mid-points of the sides of S₁. The same process is applied to S₂ to form yet another square S₃, and so on. If A₁, A₂, A₃... are the areas and P₁, P₂, P₃... are the perimeters of S₁, S₂, S₃... respectively,

then the ratio $\frac{P_{1} + P_{2} + P_{3} + . . . . .}{A_{1} + A_{2} + A_{3} + . . .}$ equals:

A.
$\frac{2 (1 + \sqrt{2})}{a}$
B.
$\frac{2 (2 - \sqrt{2})}{a}$
C.
$\frac{2 (2 + \sqrt{2})}{a}$
D.
$\frac{2 (1 + 2 \sqrt{2})}{a}$

Answer: Option C

Explanation :

Area and perimeter of square S₁is a², 4a respectively.

Now, the side of S₂ will be $\frac{a \sqrt{2}}{2} = \frac{a}{\sqrt{2}}$

Thus, the area and perimeter of S₂ = $\frac{a^{2}}{2}, \frac{4 a}{\sqrt{2}}$

The side of S₃ will be $\frac{a}{2 \sqrt{2}} \times \sqrt{2} = \frac{a}{2}$

Thus, area and perimeter of S₃ = $\frac{a^{2}}{4}, \frac{4 a}{{(\sqrt{2})}^{2}}$

Similarly, the area and perimeter of S₄ = $\frac{a^{2}}{8}, \frac{4 a}{{(\sqrt{2})}^{3}}$

∴ The required ratio

$\begin{aligned} = \frac{4 a + \frac{4 a}{\sqrt{2}} + \frac{4 a}{(\sqrt{2})^{2}} + \frac{4 a}{(\sqrt{2})^{3}} + \dots}{a^{2} + \frac{a^{2}}{2} + \frac{a^{2}}{4} + \frac{a^{2}}{8} + \dots} \\ = \frac{4 a (1 + \frac{1}{\sqrt{2}} + \frac{1}{(\sqrt{2})^{2}} + \frac{1}{(\sqrt{2})^{3}} \dots)}{a^{2} (1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} \dots)} \\ = \frac{4 a (\frac{1}{1 - \frac{1}{\sqrt{2}}})}{a^{2} (\frac{1}{1 - \frac{1}{2}})} = \frac{4 a (\frac{\sqrt{2}}{\sqrt{2} - 1})}{a^{2} \times 2} \\ = \frac{4 a \times \sqrt{2} (\sqrt{2} + 1)}{2 a^{2}} = \frac{2 (2 + \sqrt{2})}{a} \end{aligned}$

Hence, option (c).

Answer the following question based on the information given below.

The seven basic symbols in a certain numeral system and their respective values are as follows:

I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, and M = 1000

In general, the symbols in the numeral system are read from left to right, starting with the symbol representing the largest value; the same symbol cannot occur continuously more than three times; the value of the numeral is the sum of the values of the symbols.

For example, XXVII = 10 + 10 + 5 + 1 + 1 = 27.

An exception to the left-to-right reading occurs when a symbol is followed immediately by a symbol of greater value; then, the smaller value is subtracted from the larger.

For example, XLVI = (50 – 10) + 5 + 1 = 46.

48. CAT 2003 QA - Retake | Miscellaneous

The value of the numeral MDCCLXXXVII is:

A.
1687
B.
1787
C.
1887
D.
1987

49. CAT 2003 QA - Retake | Miscellaneous

The value of the numeral MCMXCIX is

A.
1999
B.
1899
C.
1989
D.
1889

50. CAT 2003 QA - Retake | Miscellaneous

Which of the following can represent the numeral for 1995?

a. MCMLXXV
b. MCMXCV
c. MVD
d. MVM

A.
Only (a) and (b)
B.
Only (c) and (d)
C.
Only (b) and (d)
D.
Only (d)

CAT 2003 QA - Retake | Previous Year CAT Paper

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