# CAT 2003 QA - Retake

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.

$\therefore \frac{Area(\u2206ABC)}{Area(\u2206DBE)}={\left(\frac{AC}{DE}\right)}^{2}={\left(\frac{1}{0.65}\right)}^{2}$

∴ Area (∆DBE) = (0.65)^{2} × Area (∆ABC)

∴ Area (∆DBE) = 0.4225 × 34 = 14.365

Hence, option 4.

Workspace:

**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.

Answer: Option C

**Explanation** :

Let the number of huts at the beginning of 2001 be n.

From the table, it is clear that only option 3 is satisfied as per the above table.

Hence, option 3

Workspace:

**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.
$\left[\frac{a}{27},\frac{b}{e}\right]$

- B.
$\left[\frac{a}{36},\frac{c}{e}\right]$

- C.
$\left[\frac{a}{12},\frac{bd}{18}\right]$

- D.
$\left[\frac{a}{6},\frac{c}{d}\right]$

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: $\left(\frac{a}{27},\frac{b}{e}\right)$ = (4k, 6)

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

Option 3 is: $\left(\frac{a}{12},\frac{bd}{18}\right)$ = (9k, 4k^{2})

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

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

Hence, option 4.

Workspace:

**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

Answer: Option A

**Explanation** :

a, a + 2, a + 4 are prime numbers.

Now, all prime numbers greater than 3 are of the form 6k ± 1, where k is some natural number.

When a = 6k + 1, then a + 2 = (6k + 1) + 2 = 6k + 3 = 3(2k + 1), which is not prime since it is a multiple of 3

When a = 6k – 1, then a + 4 = (6k – 1) + 4 = 6k + 3 = 3(2k + 1), which is again not prime

So, the only possible values of a that remain are the prime numbers which are less than or equal to 3; i.e. 2 and 3.

When a = 2, a + 2 and a + 4 are obviously not prime numbers

When a = 3, a + 2 = 5 and a + 4 = 7, which satisfies the required condition

Hence, option 1.

Workspace:

**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 – 2*x*)^{2} × *x* = 4*x*^{3} − 48*x*^{2} + 144*x*

For V to be maximum, $\frac{dV}{dx}$ shoule be 0.

i.e. 12*x*^{2} − 96*x* + 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 4.

Alternatively,

Since *V* = (12 – 2*x*)^{2} × *x* = [2(6 – *x*)]^{2} × *x* = 4*x*(6 – *x*)^{2},

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

Workspace:

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

Two straight roads R_{1} and R_{2} diverge from a point A at an angle of 120°. Ram starts walking from point A along R_{1} at a uniform speed of 3 km/hr. Shyam starts walking at the same time from A along R_{2} 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 R_{1} and R_{2}, 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 = $\sqrt{\left({\left(4\sqrt{3}\right)}^{2}+{16}^{2}\right)}$

BC = $\sqrt{304}$

BC = $4\sqrt{19}$

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

Time taken by Shyam to travel = $\frac{BC+BA}{2}=\frac{4\sqrt{19}+12}{2}\phantom{\rule{0ex}{0ex}}$ ...(ii)

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

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

Hence, option 2.

Workspace:

**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

Answer: Option D

**Explanation** :

Option 1: False

∵ (y – x) is equal to an even integer when x = 2 and y = 6

Option 2: False

∵ xy is an even integer whenever x = 2 and y = 6

Option 3: False

∵ (x + y)/x is an even integer when x = 3 and y = 6, 9, 15, 21, 27… and so on. (i.e. Taking any value of x and then taking y as a multiple of that x will work.)

Hence, option 4.

Workspace:

**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

Answer: Option D

**Explanation** :

Here, AC – Air Conditioning, R – Radio and PW – Power Windows

From the given conditions, we have the above Venn diagram.

When we add up all the values in the Venn diagram, we get 23 (15 + 5 + 1 + 2) cars.

∴ 2 (i.e. 25 − 23) cars don’t have any of the three options.

Hence, option 4.

Workspace:

**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** :

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

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 3.

Workspace:

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

If 13*x* + 1 < 2*z*, and *z* + 3 = 5*y*^{2}, 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.

Answer: Option D

**Explanation** :

We have,

13*x* + 1 < 2*z* … (i)

*z* + 3 = 5*y*^{2} …(ii)

From (i) and (ii), we get,

13*x* + 1 < 2(5*y*^{2} – 3)

∴ 13*x* + 1 < 10*y*^{2} – 6

∴ *x* < (10*y*^{2} – 7)/13

For *y* = 1, we get *x* < 3/13

∴ *x* < *y*

For *y* = 2, we get *x* < 33/13

∴ *x* could be greater than or less than *y*.

∴ None of the given options are necessarily true.

Hence, option 4.

Workspace:

**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 4.

Workspace:

**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: Option B

**Explanation** :

**Consider the case when b is negative**:

i.e. say *b* = –*k*, where *k* ≥ 1

Then, *x* = –|*a*|*b* = –|*a*| × (–*k*) = |*a*|*k*

∴ *xb* = –|*a*|*k*^{2}

∴ *a* – *xb* = *a* + |*a*|*k*^{2}

Now,

If *a* > 0, then *a* – *xb* = *a* + |*a*|*k*^{2} > 0 since all the terms will be positive

If *a* < 0 (say *a* = –2), then *a* – *xb* = –2 + 2*k*^{2} ≥ 0, since 2*k*^{2} ≥ 2 as *k* ≥ 1

However, if *a* = 0, then *a* – *xb* = 0 + 0 = 0

Hence, when *b* is negative, *a* – *xb* ≥ 0

**Now, consider the case when b is positive**:

i.e. say *b* = +*k*, where *k* ≥ 1

Then, *x* = –|*a*|*b* = –|*a*| × (*k*) = –|*a*|*k*

∴ *xb* = –|*a*|*k*^{2}

This is the same value of *xb* as we got in the previous case. Hence, the same conclusions will hold.

∴ For all cases, *a* − *xb *≥ 0

Hence, option 2.

Workspace:

**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 = $\frac{Heightofthecylinder}{Numberofturns}=\frac{h}{n}$

Hence, option 1.

Workspace:

**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,

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

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

Hence, option 2.

Workspace:

**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 $\left(n\sqrt{17}\right)$.

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,

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

∴ h = n

Hence, option 3.

Workspace:

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

Consider three circular parks of equal size with centres at A_{1}, A_{2} and A_{3} 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 A_{1}A_{2}A_{3}, B_{1}B_{2}B_{3} and C_{1}C_{2}C_{3}, as shown. Three sprinters A, B, and C begin running from points A_{1}, B_{1} and C_{1} 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

A_{1}B_{1} = A_{1}E = *r*

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

∆A_{1}B_{1}D is a right angle triangle with ∠A_{1}B_{1}D = 30° and ∠B_{1}A_{1}D = 60°

∴ B_{1}D = $\frac{\sqrt{3}r}{2}$

∴B_{1}B_{2} = 2r + 2 × $\frac{r\sqrt{3}}{2}=r(2+\sqrt{3})$

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

Similarly, ∆A_{1}C_{1}E is a right angle triangle with ∠A_{1}C_{1}E = 30° and ∠C_{1}A_{1}E = 60°

∴ C_{1}E = $r\sqrt{3}$

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

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

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

Hence, option 1.

Workspace:

**17. CAT 2003 QA - Retake | Geometry - Circles**

Sprinter A traverses distances A_{1}A_{2}, A_{2}A_{3}, and A_{3}A_{1} 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 C_{1}C_{2}, C_{2}C_{3}, and C_{3}C_{1} 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.
B

_{1}, C_{1} - B.
B

_{3}, C_{3} - C.
B

_{1}, C_{3} - D.
B

_{1}, Somewhere between C_{3}, C_{1}

Answer: Option C

**Explanation** :

Time taken by A to travel through distance *a* (A_{1}A_{2} + A_{2}A_{3} + A_{3}A_{1})

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

Distance travelled by B in the same time,

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

∴ B travels a full round in the same time as A. Thus, B will be at B_{1}.

For C, time taken to traverse through C_{1}C_{2}

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

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

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

∴ C will be at C_{3}.

Hence, option 3.

Workspace:

**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 *u*^{2 }: *v*^{2} : *w*^{2} is equal to Area A : Area B : Area C, where Area A, Area B and Area C are the areas of triangles A_{1}A_{2}A_{3}, B_{1}B_{2}B_{3}, and C_{1}C_{2}C_{3} respectively.

Where would A and C be when B reaches point B_{3}?

- A.
A

_{2}, C_{3} - B.
A

_{3}, C_{3} - C.
A

_{3}, C_{2} - D.
Somewhere between A

_{2}and A_{3}, Somewhere between C_{3}and C_{1}

Answer: Option B

**Explanation** :

Given that *u*^{2} : *v*^{2} : *w*^{2} = Area A : Area B : Area C …(i)

∵ The area of an equilateral triangle is proportional to the square of its side,

∴ Area A : Area B : Area C = (A_{1}A_{2})^{2} : (B_{1}B_{2})^{2} : (C_{1}C_{2})^{2} …(ii)

From (i) and (ii),

*u* : *v* : *w* = A_{1}A_{2} : B_{1}B_{2} : C_{1}C_{2}

∴ 2*u* : 2*v* : 2*w* = 2(A_{1}A_{2}) : 2(B_{1}B_{2}) : 2(C_{1}C_{2}) = (A_{1} to A_{3}) : (B_{1} to B_{3})_{ }: (C_{1} to C_{3})

Thus, C and A would be at C_{3} and A_{3} respectively when B reaches B_{3}.

Hence, option 2.

Workspace:

**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)

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

Dividing (iii) by 7

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

Subtracting (iv) from (iii)

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

$\therefore \frac{36S}{49}=\frac{4}{3}\phantom{\rule{0ex}{0ex}}S=\frac{4}{3}\times \frac{49}{36}=\frac{49}{27}$

Hence, option 3.

Workspace:

**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

Answer: Option A

**Explanation** :

*n *= 1, 2, 3, … , 96 and *T _{n}* = {

*n*,

*n*+ 1,

*n*+ 2,

*n*+ 3,

*n*+ 4}

*n* could be either 6*k* or 6*k + *1 or 6*k +* 2 or 6*k +* 3 or 6*k +* 4 or 6*k +* 5.

When *n* = 6*k*, then set *T _{n}* will definitely contain a multiple of 6 as it contains

*n*

When *n* = 6*k* + 5, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 1 = 6

*k*+ 6

When *n* = 6*k* + 4, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 2 = 6

*k*+ 6

When *n* = 6*k* + 3, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 3 = 6

*k*+ 6

When *n* = 6*k* + 2, then set *T _{n}* will contain a multiple of 6 as it contains

*n*+ 4 = 6

*k*+ 6

However, for every *n* = 6*k* + 1, the set *T _{n}* will not contain any multiple of 6There will be 16 such sets for

*k*= 0 to 15, for which

*T*will not contain a multiple of 6

_{n}∴ (96 – 16) = 80 sets contain multiples of 6

Hence, option 1.

Workspace:

**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)

$\therefore \frac{Area(\u2206ACE)}{Area(HexagonABCDEF)}=\frac{3}{6}=\frac{1}{2}$

Hence, option 2.

Workspace:

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

The number of roots common between the two equations *x*^{3 }+ 3*x*^{2 }+ 4*x* + 5 = 0 and *x*^{3 }+ 2*x*^{2 }+ 7*x* + 3 = 0 is:

- A.
0

- B.
1

- C.
2

- D.
3

Answer: Option A

**Explanation** :

*x*^{3 }+ 3*x*^{2} + 4*x* + 5 = 0 ... (i)

*x*^{3} + 2*x*^{2} + 7*x* + 3 = 0 ... (ii)

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

*x*^{2 }– 3*x* + 2 = 0

∴ (*x* – 1)(*x* – 2) = 0

∴ *x* = 1 or *x* = 2

Since both *x* = 1 and *x* = 2, do not satisfy either (i) or (ii), there exists no common roots for equations (i) and (ii).

Hence, option 1.

Workspace:

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

A real number x satisfying $1-\frac{1}{n}<x\le 3+\frac{1}{n},$ 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** :

$\left(1-\frac{1}{n}\right)<x\le \left(3+\frac{1}{n}\right),$ 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 3.

Workspace:

**24. CAT 2003 QA - Retake | Algebra - Logarithms**

If $\frac{1}{3}{\mathrm{log}}_{3}M+3{\mathrm{log}}_{3}N=1+{\mathrm{log}}_{0.008}5,$ 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** :

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

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

$\begin{array}{rl}& 1+{\mathrm{log}}_{0.008}5\\ & =1+\frac{\mathrm{log}5}{\mathrm{log}0.008}\\ & =1+\frac{\mathrm{log}\left(\frac{10}{2}\right)}{\mathrm{log}\left(\frac{8}{1000}\right)}\\ & =1+\frac{\mathrm{log}10-\mathrm{log}2}{\mathrm{log}8-\mathrm{log}1000}=1+\frac{\mathrm{log}10-\mathrm{log}2}{3\mathrm{log}2-3\mathrm{log}10}\end{array}$

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

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

$\begin{array}{rl}& \frac{1}{3}{\mathrm{log}}_{3}M+3{\mathrm{log}}_{3}N\end{array}$

$={\mathrm{log}}_{3}{M}^{1/3}+{\mathrm{log}}_{3}{N}^{3}={\mathrm{log}}_{3}\left({M}^{1/3}{N}^{3}\right)$ ....(iii)

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

$\begin{array}{rl}& {\mathrm{log}}_{3}\left({M}^{1/3}{N}^{3}\right)=\frac{2}{3}\\ & \therefore {M}^{1/3}{N}^{3}={3}^{2/3}\end{array}$

Cubing both sides, we get,

*M* × *N*^{9} = 9

$\therefore {N}^{9}=\frac{9}{M}$

Hence, option 2.

Workspace:

**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

Answer: Option A

**Explanation** :

**Case1: **To pay 78 paise using minimum number of coins:

78 = 1(50) + 2(10) + 4(2), i.e. a total of 7 coins were used

**Case 2: **To pay 69 paise using minimum number of coins:

69 = 1(50) + 1(10) + 1(5) + 2(2), i.e. a total of 5 coins were used

**Case 3: **To pay 101 paise using minimum number of coins:

101 = 1(50) + 1(25) + 2(10) + 3(2), i.e. a total of 7 coins were used

∴ Minimum coins used were 7 + 5 + 7 = 19 coins

Hence, option 1.

Workspace:

**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

$\therefore r=\frac{k}{2\pi},a=\frac{k}{3},$ and $x=\frac{k}{4}$

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

$\begin{array}{rl}& c=\pi {r}^{2}=\frac{\pi {k}^{2}}{4{\pi}^{2}}=\frac{{k}^{2}}{4\pi},\\ & t=\frac{\sqrt{3}}{4}{a}^{2}=\frac{\sqrt{3}}{4}\times \frac{{k}^{2}}{9}=\frac{{k}^{2}}{12\sqrt{3}}\\ & s={x}^{2}=\frac{{k}^{2}}{16}\\ & \because \frac{1}{\pi}>\frac{1}{4}>\frac{1}{3\sqrt{3}}\\ & \therefore c>s>t\end{array}$

Hence, option 3.

Workspace:

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

What is the remainder when 4^{96} is divided by 6?

- A.
0

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

4^{1} mod 6 = 4

4^{2 }mod 6 = 4

4^{3} mod 6 = 4

4^{4} mod 6 = 4

That is, 4* ^{n}* (

*n*being any positive integer) when divided by 6, will always give the remainder 4.

Hence, option 4.

Workspace:

**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: Option C

**Explanation** :

5x + 19y = 64 where x, y ∈ I

This means that the values of x have an interval of 19 between each other and the values of y will have an interval of 5 between each other.

Now, there are 2 possible cases; y could either be positive or negative:

**Case 1:**

When y = 1, then x = 9

When y = 6, then x = −10

When y = 11, then x = −29 and so on

You will notice that the values of y are in intervals of 5 and that of x are in intervals of 19.

Generally speaking, when y is positive, we will get integral values of x when y’s unit’s digit is either 1 or 6.

**Case 2:**

When y = −4, then x = 28

When y = −9, then x = 47

Again, the values of y are in intervals of 5 and that of x are in intervals of 19.

That is, when y is negative, we will get integral values of x when y’s unit’s digit is either 4 or 9.

Now, let’s evaluate the options:

**Option 1: **“no solution for x < 300 and y < 0” is False.

∵ According to Case 2, we should get integral values of x when y is −4, −9 or −14 and so on.

**Option 2:** “no solution for x > 250 and y > –100” is False.

According to Case 2, we should get integral values of x when y is −99, −94, −74 or −69 etc.

Now, when y = −74, x = 294

∴ A solution exists.

**Option 3:** “a solution for 250 < x < 300” is True.

∵ y = −74, x = 294 is a possible solution

**Option 4:** “a solution for –59 < y < –56” is False.

∵ From Case 2, when y is negative, we will get integral values of x only when y’s unit’s digit is either 4 or 9.

Hence, option 3.

Workspace:

**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 f^{2} = f * f, g^{3 }= g * g * g, and so on.

**29. CAT 2003 QA - Retake | Miscellaneous**

What is the smallest positive integer *n* such that g* ^{n}* = e?

- A.
4

- B.
5

- C.
2

- D.
3

Answer: Option A

**Explanation** :

From the second table,

g^{2} = g * g = h

g^{3} = h * g = f

g^{4} = f * g = e

Hence, option 1.

Workspace:

**30. CAT 2003 QA - Retake | Miscellaneous**

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

- A.
e

- B.
f

- C.
g

- D.
h

Answer: Option D

**Explanation** :

f ⊕ [f * {f ⊕ (f * f)}]

Using the simplification rule, start from the innermost bracket.

f * f = h

f ⊕ e = f

f ⊕ f = h

Hence, option 4.

Workspace:

**31. CAT 2003 QA - Retake | Miscellaneous**

Upon simplification, (a^{10} * (f^{10} ⊕ g^{9})} ⊕ e^{8} equals

- A.
e

- B.
f

- C.
g

- D.
h

Answer: Option A

**Explanation** :

a10 = a ...(∵ a * a = a)

f^{10} = (f^{2})^{5} = h^{5} = h * (h^{2})^{2} = h * e^{2} = h * e = h

g^{9} = g * (g^{2})^{4 }= g * h^{4} = g * e = g

e^{8} = e .....(∵ e^{2} = e)

∴ {a^{10} * (f^{10} ⊕ g^{9})} ⊕ e^{8} = {a * (h ⊕ g)} ⊕ e = {a * f} ⊕ e = a ⊕ e = e

Workspace:

**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.
- 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

Answer: Option D

**Explanation** :

**Case 1:** When the 2^{nd} letter is m:

The 1^{st} letter can be any of the 5 vowels.

The 3^{rd} letter will be any of the 4 remaining vowels (i.e. different from the 1^{st} one).

Number of possible 3 letter combinations = 5 × 4 = 20

**Case 2:** When the 2^{nd} letter is n:

The 1^{st} letter can be any of the 5 vowels.

The 3^{rd} letter will be either e or u.

Number of possible 3 letter combinations = 5 × 2 = 10

**Case 3:** When the 2^{nd} letter is p:

The 1^{st} letter can be any of the 5 vowels.

The 3^{rd} letter will be the same as the 1^{st} letter.

Number of possible 3 letter combinations = 5 × 1 = 5

∴ Total number of possible 3 letter combinations = 20 + 10 + 5 = 35

Hence, option 4.

Workspace:

**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

Answer: Option C

**Explanation** :

**Case 1:** The 2^{nd} letter is m and the 3^{rd} letter is e:

The 1^{st} letter may be any of the 4 remaining vowels (i.e. different from e)

Number of possible 3 letter combinations = 4

**Case 2:** The 2^{nd} letter is n and the 3^{rd} letter is e:

The 1^{st} letter may be any of the 5 vowels.

Number of possible 3 letter combinations = 5

**Case 3:** The 2^{nd} letter is p and the 3^{rd} letter is e:

The 1^{st} letter will be the same as the 3^{rd} letter.

Number of possible 3 letter combinations = 1 (i.e. ‘epe’)

∴ Total number of possible 3 letter combinations = 4 + 5 + 1 = 10

Hence, option 3.

Workspace:

**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

Answer: Option B

**Explanation** :

**Case 1: Lines within Zones**

Within every zone there will be 3 + 3 + 3 = 9 lines

∴ Total number of phone lines within each zone for all 4 zones together = 9 × 4 = 36

**Case 2: Lines connecting different Zones**

Let Zone 1 have towns A, B and C.

A will be connected to 9 towns of the other 3 zones, each through a single direct line.

Similarly, B and C will also be connected to 9 different towns.

∴ The number of direct lines from Zone 1 to other towns (in Zones 2, 3 and 4) = 3 × 9 = 27

For Zone 2, we must only count the number of lines to Zones 3 and 4 (lines between Zone 1 and Zone 2 have already been considered.)

∴ The number of lines from the 3 towns of Zone 2 to other towns (in Zones 3 and 4) = 3 × 6 = 18

Now, for Zone 3, we should only count the number of lines to Zone 4 (lines to Zones 1 and 2 have already been considered.)

∴ The number of direct lines from the 3 towns of Zone 3 to the 3 towns in Zone 4 = 3 × 3 = 9

∴ Total number of direct telephone lines = 36 + 27 + 18 + 9 = 90

Hence, option 2.

Workspace:

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

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

Hence, option 3.

Workspace:

**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.
$5\left(\sqrt{3}+1\right)$

- B.
$6\left(\sqrt{3}+1\right)$

- C.
$7\left(\sqrt{3}-1\right)$

- D.
$8\left(\sqrt{3}-2\right)$

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 = $\frac{x}{{\displaystyle \frac{x(\sqrt{3}-1)}{10}}}=\frac{10}{\sqrt{3}-1}=5(\sqrt{3}+1)$

Hence, option 1.

Workspace:

**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

Answer: Option C

**Explanation** :

∵ AD = DC,

∴ ∠DAC = ∠DCA = x

∴ ∠CDB = ∠DAC + ∠DCA = 2x

∵ DC = BC

∴ ∠CDB = ∠CBD = 2x

Also,

x + y = 180° − 96° = 84° ... (i)

y + 4x = 180° ... (ii)

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

x = 32°

∴ ∠DBC = 2x = 64°

Hence, option 3.

Workspace:

**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 *ax*^{2} + *bx* + 1 = 0 having real roots is:

- A.
10

- B.
7

- C.
6

- D.
12

Answer: Option B

**Explanation** :

*ax*^{2} + *bx* + 1 = 0 … (i)

For equation (i) to have real roots, *b*^{2} − 4*a* ≥ 0

i.e. *a* ≤ *b*^{2}/4 … (ii)

For *b* = 4: to satisfy (ii), *a* = 1, 2, 3, 4

∴ 4 equations are possible.

For *b* = 3: to satisfy (ii), *a* = 1, 2

∴ 2 equations are possible.

For *b* = 2: to satisfy (ii), *a* = 1

∴ 1 equation is possible.

Thus, total number of possible equations = 7

Hence, option 2.

Workspace:

**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*

2*y* = *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 $\frac{3y}{3}$ ≥ 4^{1/3}

∴ y ≥ 2^{2/3}

∴ The minimum value of *y* is 2^{2/3}

Hence, option 2.

Workspace:

**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

Answer: Option C

**Explanation** :

From the digits 0, 1, 2, … , 9, there are 4 digits that can create confusion are 1, 6, 8 and 9

Numbers with 0 in unit’s place cannot be counted because reversing them will give an invalid code.

**Codes for which confusion can arise:**

Possible number of digits in ten’s place = 4

Possible number of digits in unit’s place, such that the 2 digits are distinct = 3

Total number of ways = 4 × 3 = 12

However, the numbers 69 and 96 do not create confusion when written upside down.

∴ Total number of ways for codes for which confusion can arise = 12 − 2 = 10

**Total codes possible with digits from 0 to 9:**

Possible number of digits in ten’s place = 9 (0 cannot be used)

Possible number of digits in one’s place, such that the 2 digits are distinct = 9

Total number of all possible codes = 9 × 9 = 81

∴ Number of codes with no confusion = 81 – 10 = 71

Hence, option 3.

Workspace:

**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.
$\frac{n(4-\pi )}{4n-\pi}$

- D.
$\frac{4n-\pi}{n(4-\pi )}$

Answer: Option A

**Explanation** :

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

*A*_{1}, *A*_{2}, *A*_{3}, *A*_{4}, *A*_{5}, ... , *A _{n}*

Also, let *A*_{1} + *A*_{2} + *A*_{3} + *A*_{4} + *A*_{5} + ... + *A _{n}* =

*A*... (i)

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

Now, if *A* = *a*^{2 }(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 × *a*^{2} = π/4 × *A*

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

**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 - $\left(\frac{\pi A}{4}\right)=\frac{A}{4}(4-\pi )$

**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

$=\frac{\pi}{4}{A}_{1}+\frac{\pi}{4}{A}_{2}+\frac{\pi}{4}{A}_{3}+\cdots +\frac{\pi}{4}{A}_{n}=\frac{\pi}{4}({A}_{1}+{A}_{2}+{A}_{3}+\cdots +{A}_{n})=\frac{\pi}{4}A$

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

∴ Area of the scrap material will be:

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

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 1.

Workspace:

**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.

$\frac{\pi {r}^{2}}{ab}=\frac{\pi}{\sqrt{3}}$

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

In ΔDBC,

$\mathrm{tan}\theta =\frac{BC}{DC}=\frac{b}{a}$ ...(ii)

In ΔDAE,

$\mathrm{tan}\theta =\frac{AE}{AD}=\frac{AE}{b}$ ...(iii)

From (ii) and (iii),

$\frac{AE}{AD}=\frac{b}{a}$

In ΔDBC,

$\begin{array}{rl}& 4{r}^{2}={a}^{2}+{b}^{2}\\ & 4{r}^{2}={a}^{2}+\frac{3{r}^{4}}{{a}^{2}}\\ & {a}^{4}-4{r}^{2}{a}^{2}+3{r}^{4}=0\\ & {a}^{4}-3{r}^{2}{a}^{2}-{r}^{2}{a}^{2}+3{r}^{4}=0\\ & {a}^{2}({a}^{2}-3{r}^{2})-{r}^{2}({a}^{2}-3{r}^{2})=0\\ & {a}^{2}={r}^{2}\text{and}{a}^{2}=3{r}^{2}\\ & \therefore a=r\text{and}a=\sqrt{3}r\end{array}$

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 1.

Workspace:

**43. CAT 2003 QA - Retake | Algebra - Logarithms**

If log_{10} x - log_{10} $\sqrt{x}$ = 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** :

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

Consider L.H.S of (i):

log_{10} x - $\frac{1}{2}$ log_{10} x = $\frac{1}{2}$ log_{10} x

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

log_{10} x = 4 log_{x} 10

∴ log_{10} x = 4 log_{x} 10^{4}

Now, let log_{10} *x* = log* _{x}* 10

^{4}=

*y*

∴ 10* ^{y}* =

*x*and

*x*= 10

^{y}^{4}

Solving these two equations, we get,

*y* = ±2

∴ *x* = 10^{2} or *x* = 10^{–2}

As *x* = 100 is not in the options,

∴* x* = 10^{−2} = 1/100

Hence, option 2.

Workspace:

**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 = $\sum _{k=1}^{13}7k+3$

= 13 × 3 + 7(1 + 2 + ... + 13)

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

= 39 + 637 = 676

Hence, option 2.

Workspace:

**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

Answer: Option A

**Explanation** :

In ∆ACT,

∵ ∠ACT = 50° and ∠ATC = 30°,

∴ ∠CAT = 100°

Applying the Alternate Segment theorem,

∠ABC = 50°

Since ∠CAT is the external angle of ∆ABC, the sum of ∠ABC and ∠BCA is 100°,

∴ ∠BCA = 50°

∴ ∠BOA = 100°

Hence, option 1.

Workspace:

**46. CAT 2003 QA - Retake | Algebra - Logarithms**

What is the sum of n terms in the series

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

- A.
$\mathrm{log}{\left[\frac{{n}^{n-1}}{{m}^{(n+1)}}\right]}^{\frac{n}{2}}$

- B.
$\mathrm{log}{\left[\frac{{m}^{m}}{{n}^{n}}\right]}^{\frac{n}{2}}$

- C.
$\mathrm{log}{\left[\frac{{m}^{(1-n)}}{{n}^{(1-m)}}\right]}^{\frac{n}{2}}$

- D.
$\mathrm{log}{\left[\frac{{m}^{(n+1)}}{{n}^{(n-1)}}\right]}^{\frac{n}{2}}$

Answer: Option D

**Explanation** :

$\begin{array}{rl}& \mathrm{log}m+\mathrm{log}\left(\frac{{m}^{2}}{n}\right)+\mathrm{log}\left(\frac{{m}^{3}}{{n}^{2}}\right)+\mathrm{log}\left(\frac{{m}^{4}}{{n}^{3}}\right)+\dots +\mathrm{log}\left(\frac{{m}^{n}}{{n}^{n-1}}\right)\\ & =\mathrm{log}\left[\frac{m\times {m}^{2}\times {m}^{3}\times \dots \times {m}^{n}}{1\times n\times {n}^{2}\times \dots \times {n}^{n-1}}\right]\\ & =\mathrm{log}\left[\frac{{m}^{1+2+3+4+\dots +n}}{{n}^{0+1+2+3+\dots +(n-1)}}\right]\\ & =\mathrm{log}\left[\frac{{m}^{\frac{n}{2}(n+1)}}{{n}^{\frac{n}{2}(n-1)}}\right]\\ & =\mathrm{log}{\left[\frac{{m}^{(n+1)}}{{n}^{(n-1)}}\right]}^{\frac{n}{2}}\end{array}$

Hence, option 4.

Workspace:

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

Let S_{1 }be a square of side *a*. Another square S_{2} is formed by joining the mid-points of the sides of S_{1}. The same process is applied to S_{2} to form yet another square S_{3}, and so on. If *A*_{1}, *A*_{2}, *A*_{3}... are the areas and *P*_{1}, *P*_{2}, *P*_{3}... are the perimeters of S_{1}, S_{2}, S_{3}... 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 S_{1 }is *a*^{2}, 4*a* respectively.

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

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

The side of S_{3} will be $\frac{a}{2\sqrt{2}}\times \sqrt{2}=\frac{a}{2}$

Thus, area and perimeter of S_{3} = $\frac{{a}^{2}}{4},\frac{4a}{{\left(\sqrt{2}\right)}^{2}}$

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

∴ The required ratio

$\begin{array}{rl}& =\frac{4a+\frac{4a}{\sqrt{2}}+\frac{4a}{(\sqrt{2}{)}^{2}}+\frac{4a}{(\sqrt{2}{)}^{3}}+\cdots}{{a}^{2}+\frac{{a}^{2}}{2}+\frac{{a}^{2}}{4}+\frac{{a}^{2}}{8}+\cdots}\\ & =\frac{4a\left(1+\frac{1}{\sqrt{2}}+\frac{1}{(\sqrt{2}{)}^{2}}+\frac{1}{(\sqrt{2}{)}^{3}}\dots \right)}{{a}^{2}\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}\dots \right)}\\ & =\frac{4a\left(\frac{1}{1-\frac{1}{\sqrt{2}}}\right)}{{a}^{2}\left(\frac{1}{1-\frac{1}{2}}\right)}=\frac{4a\left(\frac{\sqrt{2}}{\sqrt{2}-1}\right)}{{a}^{2}\times 2}\\ & =\frac{4a\times \sqrt{2}(\sqrt{2}+1)}{2{a}^{2}}=\frac{2(2+\sqrt{2})}{a}\end{array}$

Hence, option 3.

Workspace:

**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

Answer: Option B

**Explanation** :

The expression MDCCLXXXVII is expanded as,

1000 + 500 + 100 + 100 + 50 + 10 + 10 + 10 + 5 + 1 + 1

= 1787

Hence, option 2.

Workspace:

**49. CAT 2003 QA - Retake | Miscellaneous**

The value of the numeral MCMXCIX is

- A.
1999

- B.
1899

- C.
1989

- D.
1889

Answer: Option A

**Explanation** :

The expression MCMXCIX is expanded as,

1000 + (1000 – 100) + (100 – 10) + (10 – 1)

= 1000 + 900 + 90 + 9

= 1999

Hence, option 1.

Workspace:

**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)

Answer: Option C

**Explanation** :

Option (a): MCMLXXV

= 1000 + (1000 – 100) + 50 + 10 + 10 + 5

= 1000 + 900 + 50 + 25 = 1975

Option (b): MCMXCV

= 1000 + (1000 – 100) + (100 – 10) + 5

= 1000 + 900 + 90 + 5 = 1995

Option (c): MVD

= 1000 + (500 – 5) = 1000 + 495 = 1495

Option (d): MVM

= 1000 + (1000 – 5) = 1000 + 995 = 1995

Options (b) and (d) give 1995.

Hence, option 3.

Workspace:

## Feedback

**Help us build a Free and Comprehensive CAT/MBA Preparation portal by providing
us your valuable feedback about Apti4All and how it can be improved.**