CAT 2006 QA | Previous Year CAT Paper

Answer: Option B

Explanation :

Out of the options, only $\frac{1}{x}$ is negative.

All the others are positive.

$\therefore \frac{1}{x}$ is the smallest.

Hence, option (b).

Answer: Option B

Explanation :

Givne, 2^1/2, 3^1/3, 4^1/4 , 6^1/6  and 12^1/12

Raising all numbers to power 12, we get

(2^1/2)¹², (3^1/3)¹², (4^1/4)¹² , (6^1/6)¹²  and (12^1/12)¹²

(2⁶), (3⁴), (4³) , (6²)  and (12¹)

64, 81, 64, 36, 12

Since 81 is the higest number here, 3^1/3 will be the highest of the original numbers.

Hence, option (b).

Answer: Option A

Explanation :

$\frac{\mathrm{a}}{\mathrm{d}}$ = $\frac{\mathrm{a}}{\mathrm{b}}\times \frac{\mathrm{b}}{\mathrm{c}}\times \frac{\mathrm{c}}{\mathrm{d}}$ = $\frac{1}{3}\times 2\times \frac{1}{2}$ = $\frac{1}{3}$

$\frac{\mathrm{b}}{\mathrm{e}}$ = $\frac{\mathrm{b}}{\mathrm{c}}\times \frac{\mathrm{c}}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{e}}$ = $2\times \frac{1}{2}\times 3$ = 3

$\frac{\mathrm{c}}{\mathrm{f}}$ = $\frac{\mathrm{c}}{\mathrm{d}}\times \frac{\mathrm{d}}{\mathrm{e}}\times \frac{\mathrm{e}}{\mathrm{f}}$ = $\frac{1}{2}\times 3\times \frac{1}{4}$ = $\frac{3}{8}$

∴ $\frac{abc}{def}$ = $\frac{1}{3}\times 3\times \frac{3}{8}$ = $\frac{3}{8}$

Hence, option (a).

Answer: Option E

Explanation :

Let the original length, breadth and height of the room be 3x, 2x and x respectively.

∴ The new length, breadth and height are 6x, x and x/2 respectively.

Area of four walls = (2 × length × height) + (2 × breadth × height)

Original area of four walls = 6x² + 4x² = 10x²

New area of four walls = 6x² + x² = 7x²

∴ Area of wall decreases by [(10x²  − 7x²)/10x²] × 100 = 30%

Hence, option (e).

Answer: Option A

Explanation :

${\mathrm{t}}_{\mathrm{n}}=\frac{\mathrm{n}}{\mathrm{n}+2}$

$\therefore {\mathrm{t}}_3=\frac{3}{3+2}=\frac{3}{5}$

$t_4=\frac{4}{4+2}=\frac{4}{6}$

$t_5=\frac{5}{5+2}=\frac{5}{7}$

...

$t_{51}=\frac{51}{53}$

$t_{52}=\frac{52}{54}$

$t_{53}=\frac{53}{55}$

∴ t₃ × t₄ × t₅ × ... × t₅₃ = $\frac{3}{5}$ × $\frac{4}{6}$ × $\frac{5}{7}$ × ... × $\frac{51}{53}$ × $\frac{52}{54}$ × $\frac{53}{55}$

$=\frac{(3\times 4)}{(54\times 55)}=\frac{2}{495}$

Hence, option (a).

Answer: Option D

Explanation :

Let there be n rows and a students in the first row.

∴ Number of students in the second row = a + 3

∴ Number of students in the third row = a + 6 and so on.

∴ The number of students in each row forms an arithmetic progression with common difference = 3

The total number of students = The sum of all terms in the arithmetic progression

= $\frac{\mathrm{n}[2\mathrm{a}+3(\mathrm{n}-1\left)\right]}{2}$ = 30

Now consider options.

Option (a): n = 3

$\frac{3[2a+3(3-1\left)\right]}{2}$ = 630

∴ a = 207

Option (b): n = 4

$\frac{4[2a+3(4-1\left)\right]}{2}$ = 630

∴ a = 153

Option (c): n = 5

$\frac{5[2a+3(5-1\left)\right]}{2}$ = 630

∴ a = 120

Option (d): n = 6

$\frac{6[2a+3(6-1\left)\right]}{2}$ = 630

∴ a = 195/2 = 97.5

Option (e): n = 7

$\frac{7[2a+3(7-1\left)\right]}{2}$ = 630

∴ a = 81

As a is an integer, only n = 6 is not possible.

Hence, option (d).

Answer: Option E

Explanation :

The fastest way to solve this sum is by substituting the values of x and y from the options.

For x = 5 and y = 2, the first equation becomes

2^0.7x × 3^−1.25y  = 2^{0.7× 5} × 3^{−1.25 × 2}

              = 2^7/2 × 3^−5/2

             = $\frac{8\sqrt{6}}{27}$

These values of x and y satisfy the second equation also.

Hence, option (e).

Answer: Option B

Explanation :

2x + y = 40

∴ y = 40 – 2x

x and y are positive integers and x ≤ y

If x = 1, y = 38

x = 2, y = 36

x = 3, y = 34
.
.
.
x = 12, y =16

x = 13, y = 14

x = 14, y = 12

∴ For x > 13, y ≤ x

∴ There are 13 solutions to the given equation.

Hence, option (b).

Answer: Option B

Explanation :

100 – 24 = 76 had read at least one issue.

If x people read all the three issues, then (8 – x) people read only the September and July issues.

23 people read the September issue but not the August issue.

∴ 18 + 8 – x = 23

∴ x = 3

As 28 people read the September issue, [28 – (8 – 3) – 3 – 18] = 2 people read only the August and September issues.

As 10 people read the July and August issues, 10 – 3 = 7 people read only the July and August issues.

∴ The number of people who have read exactly two consecutive issues = 7 + 2 = 9

[Note: July and September are not consecutive months, hence we will not consider them.]

Hence, option (b).

Answer: Option C

Explanation :

The four consecutive two-digit odd numbers will have (1, 3, 5, 7) or (3, 5, 7, 9) or (5, 7, 9, 1) or (7, 9, 1, 3) or (9, 1, 3, 5) as units digits.

As the sum divided by 10 yields a perfect square, the sum is a multiple of 10.

∴ The units digits have to be (7, 9, 1, 3).

Thus the four numbers will be (10x + 7), (10x + 9), (10x + 11) and (10x + 13),

where 0 < x < 9 (as each of these numbers is a two digit number)

Sum of these numbers = 40x + 40 = 40(x + 1)

Now, 40(x + 1)/10 = 4(x + 1) is a perfect square

As 4 is a perfect square, (x + 1) is some perfect square < 10

If x + 1 = 4, x = 3, and the four numbers are 37, 39, 41 and 43

If x + 1 = 9, x = 8, and the four numbers are 87, 89, 91 and 93

Hence, option (c).

Answer: Option D

Explanation :

All the given graphs are drawn to the same scale.

We can see that the line makes an angle which is more than 45° with the horizontal axis.

∴ The slope of the line is greater than 1.

Let the slope be k.

∴ (y – x) = k(y + x)     {∵ k > 1}

∴ y – x = ky + kx

$\therefore y=\frac{x(k+1)}{(1-k)}$

$\frac{(k+1)}{(1-k)}$ is negative and $\left|\frac{(k+1)}{(1-k)}\right|>1$

∴ The graph of y against x will be such that when x is positive, y is negative and |x| < |y|, except at (0, 0).

Hence, option (d).

Answer: Option D

Explanation :

Let there be n terms (n ≥ 3) in the arithmetic progression having 1 as the first term and 1000 as the last. Let d be the common difference. Then,

1000 = 1 + (n – 1) × d

∴ 999 = (n – 1) × d                        ... (i)

∴ Factors of 999 are 1, 3, 9, 27, 37, 111, 333 and 999

Substituting in equation (i)

If d = 1, n = 1000

If d = 3, n = 334

If d = 9, n = 112

If d = 27, n = 38

If d = 37, n = 28

If d = 111, n = 10

If d = 333, n = 4

If d = 999, n = 2, which is not possible as n > 2

∴ 7 arithmetic progressions can be formed.

Hence, option (d).

Answer: Option B

Explanation :

Let PQRS be the square sheet and let the hole have centre O.

As P lies on the circumference of the circle and as m ∠APC = 90°, AC is a diameter.

∵ BP is a diameter, m ∠PAB = m ∠BCP = 90°.

∵ BP = AC, ABCP is a square.

∴ m ∠POC = 90° and OP = OC = 1 unit

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆ OPC)

$$\begin{gathered} =2\times \left[\left(\frac{\pi \times 1^2}{4}\right)-\left(\frac{1}{2}\times 1^2\right)\right] \\ =\frac{\pi -2}{2} \mathrm{sq}.\mathrm{units} \end{gathered}$$

Area of part of hole on sheet = Area of hole − Area of part of the circle falling outside the square sheet

$=\pi -\left(\frac{\pi -2}{2}\right)=\frac{\pi +2}{2} \mathrm{sq}.\mathrm{units}$

Part of square remaining after punching = Area of square − Area of part of hole on sheet

$=4-\left(\frac{\pi +2}{2}\right)=\frac{6-\pi }{2} \mathrm{sq}.\mathrm{units}$

∴ Proportion of sheet area that remains after punching $=\frac{\left({\displaystyle \frac{6-\pi }{2}}\right)}{4}=\frac{6-\pi }{8}$

Hence, option (b).

Answer: Option D

Explanation :

We have calculated this value while solving the previous question.

The area of part of the circle falling outside the square sheet

= 2 × (Area of sector OPC – Area of ∆OPC)

$=\frac{\pi -2}{2} \mathrm{sq}.\mathrm{units}$

Hence, option (d).

Answer: Option A

Explanation :

$x^{\frac{2}{3}}+x^{\frac{1}{3}}-2\le 0$   ...(1)

Put y = x^1/3

Then equation (1) becomes y² + y – 2 ≤ 0

(y + 2)(y – 1) ≤ 0

–2 ≤ y ≤ 1

$-2\le x^{\frac{1}{3}}\le 1$

–8 ≤ x ≤ 1

Hence, option (a).

Answer: Option E

Explanation :

Let the two lines represent the equations y = 2x + 1 and y = 3 – 4x

f(x) = max{2x + 1, 4 - 3x} is represented by green and blue lines.

f(x) is greater than 5/3, when x < 1/3 (green part) or x > 1/3 (blue part).

∴ f(x) is minimum at x = 1/3 and this value is 5/3.

Hence, option (e).

Note: The minimum value of the function f(x) = max (ax + b, cx + d) occurs when ax + b = cx + d (Only if one line has a positive slope and the other one has a negative slope)

Answer: Option D

Explanation :

Let f kg be the free luggage allowance and let Raja and Praja have r kg and p kg excess luggage respectively.

Let x be the fixed rate per kg for excess luggage.

∴ 2f + r + p = 60   ... (i)

rx = 1200   ... (ii)

px = 2400   ... (iii)

(60 – f)x = 5400   ... (iv)

From (ii) and (iii),

p = 2r    ... (v)

Substituting in (i),

2f + 3r = 60

∴ f = 30 – 3r/2    ... (vi)

Substituting in (iv),

(60 – 30 + 3r/2)x = 5400

∴ 30x + 3rx/2 = 5400

From (ii),

rx = 1200

∴30x = 3600

∴ x = 120

∴ r = 10, p = 20 and f = 15

∴ Weight of Praja's luggage = p + f = 35 kg

Hence, option (d).

Answer: Option B

Explanation :

As calculated in the solution to the previous question, f = 15 kg

Hence, option (b).

Answer: Option C

Explanation :

Arun has travelled 60 km when Barun starts.

Barun overtakes Arun in 60/(40 – 30) = 6 hrs

In this time, Barun travels 6 × 40 = 240 km from the starting point.

Kiranmala overtakes Arun at the same point.

Kiranmala takes 240/60 = 4 hrs to reach there.

Arun takes 240/30 = 8 hrs to reach there.

∴ Kiranmala starts 8 – 4 = 4 hrs after Arun.

Hence, option (c).

Answer: Option B

Explanation :

Let 10x + y be a two digit number, where x and y are positive single digit integers and x > 0.

Its reverse = 10y + x

Now, |10y + x – 10x – y| = 18

 ∴ 9 |y – x| = 18

∴ |y – x| = 2

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

∴ Other than 13, there are 6 such numbers.

Hence, option (b).

Answer: Option B

Explanation :

Let CB = x cm

In a right triangle if DC is the altitude from right vertex, DC² = AC × BC
∴ 6² = 2 × x
⇒ ​​​​​​​x = 18

⇒ AB = 2 + 18 = 20

∴ Diameter of the semicircle = 20

⇒ Radius of the semicircle = 10

∴ Area = 1/2 × π × 10² = 50π sq. cm.

Hence, option (b).

Answer: Option A

Explanation :

Task 2 can be assigned in 2 ways (either to person 3 or 4).

Task 1 can then be assigned in 3 ways (persons 4 or 3, 5 and 6)

The remaining 4 tasks can be assigned to the remaining 4 persons in 4! = 24 ways

∴ The assignment can be done in 24 × 2 × 3 = 144 ways

Hence, option (a).

Answer: Option E

Explanation :

Consider options. As the number of employees is prime we can add the numerator and denominator of ratios directly to find the number of employees.

Option (a): Number of employees = 101 + 88 = 189
Number of employees = 189, which is not a prime number.
∴ Option (a) is eliminated.

Option (b): Number of employees = 87 + 100 = 187
Number of employees = 187, which is not a prime number.
∴ Option (b) is eliminated.

Option (c): Number of employees = 110 + 111 = 221
Number of employees = 221, which is not a prime number.
∴ Option (c) is eliminated.

Option (d): Number of employees = 85 + 98 = 183
Number of employees = 183, which is not a prime number.
∴ Option (d) is eliminated.

Option (e): Number of employees = 97 + 84 = 181
Number of employees = 181, which is a prime number.
∴ The ratio of employees = 97 : 84

Hence, option (e).

Answer: Option E

Explanation :

If log_yx = a × log_zy = b × log_xz = ab

⇒ log_yx = ab   ...(1)

⇒ a × log_zy = ab ⇒ log_zy = b   ...(2)

⇒ b × log_xz = ab ⇒ log_xz = a   ...(3)

From (1), (2) and (3), we get
log_yx = log_xz × log_zy

∴  $\frac{\log x}{\log y}$ = $\frac{\log z}{\log x}\times \frac{\log y}{\log z}$ = $\frac{\log y}{\log x}$

∴ (log x)² = (log y)²

∴ log x = ± log y

∴ log x = log y or log x = - log y

∴  x = y or x = $\frac{1}{y}$

∴ ab = log_y x = 1 or -1

Only option (e) does not satisfy this.

Hence, option (e).

Answer: Option E

Explanation :

BP = PC = BC

m ∠BPC = m ∠PCB = m ∠PBC = 60°

Also, PC = CD = BP = AB

∆ ABP and ∆ PCD are isosceles triangles.

m ∠ABP = m ∠PCD = 90 – 60 = 30°

∴ m ∠APB = m ∠DPC = (180 – 30)/2 = 75°

∴ m ∠APD = 360 – (m ∠APB + m ∠DPC + m ∠BPC) = 360 – (75 + 75 + 60) = 150°

Hence, option (e).