CAT 2005 QA | Previous Year CAT Paper

Answer: Option A

Video Explanation

Explanation :

x = 16³ + 17³ + 18³ + 19³

= (16³ + 19³) + (17³ + 18³)

= (16 + 19)(16² − 16 × 19 + 19²) + (17 + 18)(17² − 17 × 18 + 18²)

 35 × (an odd number) + 35 × (another odd number) = 35 × (an even number) 

= 35 × (2k) … (k is a positive integer)

∴ x = 70k

∴ x is divisible by 70.

Remainder when x is divided by 70 = 0

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

The change in the amount of chemical in each tank after every minute is as follows:

A: −20 – 10 + 90 = 60

B: −100 + 110 + 20 = 30

C: −50 − 90 + 100 = −40

D: −110 + 10 + 50 = −50
 
Since tank D loses the maximum amount of chemical in a minute, it will be emptied first.

Let n minutes be the time taken by tank D to get empty.

∴ 1000 – 50n = 0

∴ n = 20 minutes

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

Let the two circles with centres P and Q intersect at M and N.

Quadrilateral PMQN is a square.

m ∠MPN = m ∠MQN = 90°

The area common to both the circles = 2(Area of sector P-MN – Area of ∆PMN) 

$=2\left[\left(\frac{90}{360}\times \pi \times 1^2\right)-\left(\frac{1}{2}\times 1^2\right)\right]$

$=\frac{\pi }{2}-1$

Hence, option (b).

Answer: Option D

Video Explanation

Explanation :

​​​​​​​

Let r be the radius of the circular tracks.

Length and breadth of the rectangular track are 4r and 2r respectively.

Length (perimeter) of the rectangular track = 12r

Length of the two circular tracks (figure of eight) = 4πr

If A and B have to reach their starting points at the same time,

 $\frac{12r}{a}=\frac{4\pi r}{b}$
(where a and b are the speeds of A and B respectively)

∴ $\frac{b}{a}=\frac{4\pi }{12}$

⇒ $\frac{(b-a)}{a}=\frac{4\pi -12}{12}$

∴ (b - a) × $\frac{100}{a}$ = 0.047 × 100 = 4.7%

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

Let there be g girls and b boys.

Number of games between two girls = ^gC_{2​​​​​​​}
∴ g(g – 1)/2 = 45 
⇒ g² – g – 90 = 0
⇒ (g – 10)(g + 9) = 0
Since the number of girls cannot be negative, ∴ g = 10

Number of games between two boys = ^bC₂
∴ b(b – 1)/2 = 190
⇒ b² – b – 380 = 0
⇒ (b + 19)(b – 20) = 0
⇒ b = 20

∴ Number of games in which one player is a boy and the other is a girl = 10 × 20 = 200

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

The distance between points A and B is 5 km.

Ram starts at 9 a.m. from A at a speed of 5 km/hr. So, he reaches B at 10:00 a.m.

When Ram reaches B (i.e. at 10.00 a.m., or 15 minutes after Shyam started from A), Shyam (running at a speed of 10 km/hr) is 15/60 × 10 = 2.5 km away from A.

Ram meets Shyam $\frac{2.5}{10+5}\times 60$ = 10 minutes after 10:00 a.m., i.e. at 10:10 a.m. 

Hence, option (b).

Answer: Option B

Video Explanation

Explanation :

Ram reaches B at 10.00 a.m. while Shyam reaches B at 10:15 a.m.

At 10:15 a.m., Ram is (15/60) × 5 = 1.25 km away from B.

Shyam overtakes Ram 1.25/(10 – 5) = 0.25 hrs = 15 minutes after 10:15 a.m., i.e. at 10:30 a.m.

Hence, option (b).

Answer: Option D

Video Explanation

Explanation :

R = $\frac{{30}^{65}-{29}^{65}}{{30}^{64}+{29}^{64}}$

∵ aⁿ – bⁿ = (a – b)(a^{n – 1} + a^{n – 2}b + a^{n – 3}b² + ... + b^{n – 1})

∴ R = $\frac{(30-29)[{30}^{64}+({30}^{63}\times 29)+...+{29}^{64}]}{{30}^{64}+{29}^{64}}$

∵ 30⁶⁴ + 30⁶³ × 29 + ... + 29⁶⁴ > 30⁶⁴ + 29⁶⁴

∴ R > 1

Hence, option (d).

Answer: Option D

Video Explanation

Explanation :

The two chords AB and CD can be on the same side or the opposite sides of the centre O.

Let M and N be the midpoints of AB and CD.
∴ MN is the distance between the two chords.

MB = 12 cm and ND = 16 cm

OM and ON are perpendicular to AB and CD respectively.

ON² = OD² - ND² (By Pythagoras theorem)
⇒ ON² = 20² – 16² ​​​​​​​
⇒ ON = 12 cm

Similarly,OM² = OB² - MB² 
⇒ OM = 16 cm 

Case 1: AB and BC are on the same side of the centre.
MN = OM – ON = 4 cm

Case 2: AB and BC are on opposite sides of the centre.
MN = OM + ON = 28 cm

Hence, option (d).

Answer: Option C

Video Explanation

Explanation :

We have, x² = y² and (x − k) ² + y² = 1

Solving the two equations simultaneously, we get,

(x – k) ² + x² = 1   ...(1)

⇒ x² – 2kx + k² + x² = 1

⇒ 2x² – 2kx + (k² – 1) = 0

If this equation has a unique solution for x, then discriminant = 0

∴ 4k² – 8(k² – 1) = 0

⇒ 8 – 4k² = 0

⇒ k² = 2

⇒ k = ± √2

Now, k = - √​​​​​​​2 will give us a negative solution for (1) and hence it is rejected. [x should be positive according to the question.]

∴ k = √2

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

p = (1 × 1!) + (2 × 2!) + (3 × 3!) + (4 × 4!) + … + (10 × 10!)

Now, n × n! = [(n + 1) – 1] × n! = (n + 1)! – n!

∴ p = 2! – 1! + 3! – 2! + 4! – 3! + 5! – 4! +… + 11! – 10!

∴ p = 11! – 1!

∴ p + 2 = 11! + 1

∴ p + 2 when divided by 11! leaves a remainder of 1.

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

The points satisfying the equations x + y < 41, y > 0, x > 0 lie inside the triangle.
 
Integer solutions of x + y < 41 can be found as follows:
 
If x + y = 40, then

(x, y) could be (1, 39), (2, 38), …, (39, 1)  ... (39 solutions)

 If x + y = 39, then

(x, y) could be (1, 38), (2, 37), …, (38, 1)   ... ( 38 solutions)

If x + y = 38, we get 37 solutions and so on till x + y = 2  ... (1 solution)

∴ Total solutions = 1 + 2 + 3 + ... + 39 = 39 × 40/2 = 780 integer solutions to x + y < 41.
 
The number of points with integer coordinates lying inside the circle = 780

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

Let A = 100x + 10y + z (x ≠ 0, x, y, z are single-digit numbers)

∴ B = 100z + 10y + x

∴ B – A = 99(z – x)

As (B – A) is divisible by 7 and 99 is not, (z – x) is divisible by 7.

∴ z and x can have values (8, 1) or (9, 2).
[Since B > A, z > x]

y can have any value from 0 to 9.

A = 1y8 or 2y9

∴ Lowest possible value of A is 108 and the highest possible value of A is 299.

Hence, option (b).

Answer: Option C

Explanation :

a₁ = 1

a_n+1 = 4n + 3a_n – 2

a₂ = 4 + 3(1) – 2 = 5 = 3² – 4

a₃ = 4(2) + 3(5) – 2 = 21 = 3³ – 6

a₄ = 4(3) + 3(21) – 2 = 73 = 3⁴ – 8

∴ a_n = 3ⁿ – 2(n)

∴ a₁₀₀ = 3¹⁰⁰ – 200

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

Let O and E represent odd and even digits respectively.

∴ S can have digits of the form:

O _ O _ E or O _ E _ O or E _ O _ O 

Case 1: O _ O _ E
The first digit can be chosen in 3 ways out of 1, 3 and 5
The third can be chosen in 2 ways.
The fifth digit can be chosen in 2 ways after which the second and fourth digits can be chosen in 2 ways.
 ∴ There are 3 × 2 × 2 × 2 = 24 ways in which this number can be written.
12 out of these ways will have 2 in the rightmost position and 12 will have 4 in the rightmost position.
∴ The sum of the rightmost digits in Case 1 = (12 × 2) + (12 × 4) = 72 

Case 2: O _ E _ O
This number can again be written in 24 ways.
8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position. 
Thus the sum of the rightmost digits in Case 2 = (8 × 1) + (8 × 3) + (8 × 5) = 72 

Case 3: E _ O _ O
This number can also be written in 24 ways.
As in Case 2, 8 out of these ways will have 1 in the rightmost position, 8 will have 3 in the rightmost position and 8 will have 5 in the rightmost position.
∴ The sum of the rightmost digits in Case 3 = (8 × 1) + (8 × 3) + (8 × 5) = 72 

∴ The sum of the digits in the rightmost position of the numbers in S = 72 + 72 + 72 = 216

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

30²⁷²⁰ = 3²⁷²⁰ × 10²⁷²⁰

The rightmost non-zero digit of 30²⁷²⁰ will be the digit in the unit’s place of 3²⁷²⁰.

3’s power cycle is 3, 9, 7, 1 and cyclicity is 4. 

i.e. if we get a remainder of 1, 2, 3, or 0 when 2720 is divided by 4, then the digit in the units place will be 3, 9, 7, or 1 respectively.

2720 = 680 × 4, i.e. the remainder is 0 

∴ The digit in the unit’s place of 3²⁷²⁰ is 1. 

∴ The rightmost non-zero digit of 30²⁷²⁰ is 1.

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

The ant will not go into the circles with centers B and C and radius = 1 m

The minimum distance that the ant has to traverse = the distance of the path A-H-G-D

HG = 1 m

AH = GD = $\frac{1}{4}$ × Circumference of circle = $\frac{\pi }{2}$

AH + GD = π m

∴ The ant must traverse (1 + π)m.

Hence, option (b).

Answer: Option D

Video Explanation

Explanation :

${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = log_x x - log_x y + log_y y - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 1 - log_x y + 1 - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - log_x y - log_y x

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - (log_x y + log_y x)

As x ≥ y and y > 1,
log_y x ≥ 0
Now, (log_x y + log_y x) = ${\log }_x\left(y\right)+\frac{1}{{\log }_x\left(y\right)}$ [This is sum of a positive number and its reciprocal]

Now, sum of a positive number and its reciprocal is always greater than or equal to 2.

∴ (log_x y + log_y x) = ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ ≥ 2

⇒ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ = 2 - (log_x y + log_y x) ≤ 0

∴ ${\log }_x\left(\frac{x}{y}\right)+{\log }_y\left(\frac{y}{x}\right)$ ≠ 1

Hence, option (d).

Answer: Option D

Explanation :

n can be a 2-digit or a 3-digit number.

Case 1: Let n be a 2 digit number.
Let n = 10x + y, where x and y are non-negative integers,
P_n = xy and S_n = x + y
Now, P_n + S_n = n
∴ xy + x + y = 10x + y
⇒ xy = 9x or y = 9
There are 9 two-digit numbers (19, 29, 29, … ,99) for which y = 9

Case (2): Let n be a 3-digit number.
Let n = 100x + 10y + z, where x, y and z are non-negative integers, 
P_n = xyz and S_n = x + y + z
Now, P_n + S_n = n
∴ xyz + x + y + z = 100x + 10y + z
⇒ xyz = 99x + 9y
⇒ z = 99/y + 9/x
From the above expression, 0 < x, y ≤ 9
But, we cannot find any value of x and y, for which z is a single-digit number. z will be minimum when x and y are both 9, but even then its value is 12.
∴ There are no 3-digit numbers which satisfy P_n + S_n = n

Hence, option (d).

Answer: Option C

Video Explanation

Explanation :

This problem can be solved by trying different ways of placing the tiles on the floor. The maximum number of tiles that can be accommodated is 6, as shown in the figure.

Hence, option (c).

Answer: Option C

Explanation :

|x + y| + |x – y| = 4         … (i)

Consider the case when x and y are both positive. This is the area of quadrant I.

In this case, two cases are possible.

Case I: x > y

In this case, expression (i) becomes:

x + y + x – y = 4

∴ 2x = 4

∴ x = 2

Case II: x < y

In this case, expression (i) becomes:

x + y + y – x = 4

∴ 2y = 4

∴ y = 2

∴ The area for the first quadrant is as shown in the figure.

Extending the same logic to other quadrants, we get the area as shown in the above diagram.

∴ Its area = 4 × 4 = 16 sq. units

Hence, option (c).

Answer: Option B

Explanation :

AO = OD = 1.5 cm

 AE + EB = 3 cm and AE : EB = 1 : 2

∴ AE = 1 cm and EB = 2 cm

OE = AO – AE = 1.5 – 1 = 0.5 cm

Similarly, NL = 1 cm, LM = 2 cm, and OL = 0.5 cm

OEHL is a square as all its angles are right angles and OE = OL

∴ EH = HL = 0.5 cm

In ∆ODL, OD² = OL² + DL²

1.5² = 0.5² + (0.5 + DH)²

2.25 = 0.25 + 0.25 + DH + DH²

DH² + DH – 1.75 = 0

DH = $\frac{-1\pm \sqrt{1-4(-1.75)}}{2}$ = $\frac{2\sqrt{2}-1}{2}$ (DH > 0)

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

m ∠BCD = m ∠BAC and B is common to triangles ABC and CBD.

∆ABC is similar to ∆CBD.

AB/CB = BC/BD = AC/CD

AB/12 = 12/9 = AC/6

AB = 16 cm and AC = 8 cm

AD = AB – BD = 16 – 9 = 7 cm

∴ Perimeter of ∆ADC = 7 + 6 + 8 = 21 cm

∴ Perimeter of ∆BDC = 9 + 6 + 12 = 27 cm

∴ Required ratio = 21/27 = 7/9

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

∆ PQR is an equilateral triangle and PS is the diameter.

∴ m ∠PQS = m ∠PRS = 90°   (angles subtended in a semi-circle)

and m ∠PQM = m ∠PRM = m ∠QPR = 60°   (each angle in an equilateral triangle = 60°)

PS bisects ∠QPS as it is the median of ∆PQR.

m ∠PMQ = m ∠PMR = 90°

∴ m ∠QPS = m ∠RPS = 30°

∴ m ∠PSQ = m ∠PSR = 60°

Radius = r

∴ PS = 2r

As ∆ PQS, ∆ PQM, ∆ MQS are 30°-60°-90° triangles,

QS = r, PQ = √3r

Similarly, RS = r and PR = √3r

∴ Perimeter of quadrilateral PQRS = 2r + 2√3r = 2r(1 + √3)

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

n will be of the form 11ab, where a and b are odd numbers.

We are looking for all n’s divisible by 3.

∴ 1 + 1 + a + b = 3 or 6 or 9 or 12 or 15 or 18

∴ a + b = 1 or 4 or 7 or 10 or 13 or 16

∴ a + b = 1 or 7 or 13 is not possible as the sum of two odd numbers cannot be odd.

∴ (a, b) = (1, 3), (3, 1), (1, 9), (3, 7), (5, 5), (7, 3), (9, 1), (7, 9), (9, 7)

∴ 9 elements of S are divisible by 3.

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

x = $\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4-....to infinity}}}}$

∴ x² - 4 = $\sqrt{4-\sqrt{4+\sqrt{4-.... to infinity}}}$

∴ (x² - 4)² - 4 = - $\sqrt{4+\sqrt{4-...to infinity}}$

∴ (x² - 4)² - 4 = - x

Now, substituting options, we find that only option (c) satisfies the above equation.

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

g(x + 1) + g(x – 1) = g(x)

∴ g(x + 1) = g(x) – g(x – 1)

Now, let g(x − 1) = a and g(x) = b

∴ g(x + 1) = b – a

∴ g(x + 2) = b – a – b = –a

∴ g(x + 3) = −a – b + a = –b

∴ g(x + 4) = −b + a = a – b

∴ g(x + 5) = a – b + b = a = g(x – 1)

∴ g(x + 6) = a – a + b = b = g(x)

and so on.

Thus we observe that the values of g(x + 6) and g(x) are always equal.

Hence, option (d).

Answer: Option D

Video Explanation

Explanation :

Let us calculate the cost per call for a male and a female.

Male: A male can make 40 calls/day.
Total cost = 250 + 15 × 40 = 850
∴ Cost/call for a male = 850/40 = Rs. 21.25

Female: A female can make 50 calls/day.
Total cost = 300 + 10 × 50 = 800
∴ Cost/call for a female = 800/50 = Rs. 16

Hence, it could be cost effective to employ as many females as possible.

We have maximum 12 females who can make 12 × 50 = 600 calls/day.

Remaining (1000 - 600 =) 400 calls should be made by males.
∴ Number of males to be employed = 400/40 = 10 males.

Alternately,
Let x females and y males be employed.

As the total number of calls to be answered = 1000 and males and females can handle 40 and 50 calls respectively everyday

50x + 40y = 1000
40y = 1000 – 50x
∴ y = 25 – x – x/4

As 7 < x ≤ 12, x can be 8 or 12.
If x = 8, y = 15 and if x = 12, y = 10

The total cost of employing x females and y males
= 300x + 250y + (50 × 10 × x) + (40 × 15 × y)
= 800x + 850y 

If x = 8 and y = 15, cost = Rs. 19,150
​​​​​​​If x = 12 and y = 10, cost = Rs. 18,100

Thus, cost is minimized when the number of male operators is 10.

Hence, option (d).

Answer: Option C

Video Explanation

Explanation :

Let E₁, E₂ and E₃ be the three Englishmen and F₁, F₂ and F₃ be the three Frenchmen.

Let E₃ be the only Englishman knowing French.

Now, Let A ↔ B denote a phone call between A and B, where they both tell each other their secrets.The following phone calls will ensure that all six persons know all the six secrets.

1.  E₁ ↔ E₃
2.  E₂ ↔ E₃ (Now E₃ knows all the secrets with the Englishmen)

3.  F₁ ↔ F₃
4.  F₂ ↔ F₃ (Now F₃ knows all the secrets with the Frenchmen)

5.  F₃ ↔ E₃ (Now F₃ and E₃ know all the secrets)

6.  E₃ ↔ E₂
7.  E₃ ↔ E₁ (Now E₁ and E₂ know all the secrets)

8.  F₃ ↔ F₂
9.  F₃ ↔ F₁ (Now F₁ and F₂ know all the secrets)

Thus, a minimum of 9 calls are needed to pass all the secrets to all the six persons.

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

Let each square tile have side = 1 unit

Let the length of the rectangular floor be m units and the breadth be n units.

Number of red tiles = (m – 2)(n – 2)

Number of white tiles = mn − (m – 2)(n – 2)

Now, (m – 2)(n – 2) = mn − (m – 2)(n – 2)
⇒ 2(mn – 2m – 2n + 4) − mn = 0
⇒ mn – 4m – 4n + 8 = 0
⇒ n(m – 4) = 4m − 8 
⇒ n = 4(m – 2)/(m – 4)

 Now, consider the options.

Option (a): If m = 10, n = 32/6, which is not possible as n is an integer

Option (b): If m = 12, n = 40/8 = 5, which is possible

Option (c): If m = 14, n = 48/10, which is not possible as n is an integer

Option (d): If m = 16, n = 56/12, which is not possible as n is an integer

Hence, option (b).