CAT 2008 QA | Previous Year CAT Paper

Answer: Option B

Video Explanation

Explanation :

The initial quantity of rice is x kg.

The first customer buys half the total rice in the store, and another half kg.

∴ Rice purchased by the first customer = $\frac{x}{2}+\frac{1}{2}=\frac{x+1}{2}$
∴ Remaining rice = $x-\left(\frac{x+1}{2}\right)=\frac{x-1}{2}$

Now, the second customer buys half of this, and another half kg.

∴ Rice purchased by the second customer = $\frac{x-1}{4}+\frac{1}{2}=\frac{x+1}{4}$
∴ Remaining rice = $\frac{x-1}{2}-\frac{x+1}{4}=\frac{x-3}{4}$

Now, the third customer buys half the remaining rice, and another half kg.

∴ Rice purchased by the third customer = $\frac{x-3}{8}+\frac{1}{2}=\frac{x+1}{8}$

Since after this purchase, there is no rice left in the store, we conclude that:

$\frac{x-3}{4}-\frac{x+1}{8}=\frac{x-7}{8}=0$

∴ x = 7

Alternately,
If the shopkeeper had x kg rice before a customer comes, the amount of rice left after every time a customer buys half the amount of rice + half a kg more = $\frac{\mathrm{x}}{2}-\frac{1}{2}$

For 3^rd customer.
Now, $\frac{\mathrm{x}}{2}-\frac{1}{2}$ = 0 [No rice is left after third customer]
⇒ x = 1 kg.

Now, the shopkeeper had 1 kg rice when 3^rd customer came, hence after 2^nd customer left he had 1 kg of rice.

For 2^nd customer.
$\frac{\mathrm{x}}{2}-\frac{1}{2}$ = 1
⇒ x = 3 kg.

Now, the shopkeeper had 3 kg rice when 2^nd customer came, hence after 1^st customer left he had 3 kg of rice.

For 1^st customer.
$\frac{\mathrm{x}}{2}-\frac{1}{2}$ = 3
⇒ x = 7 kg.

Now, the shopkeeper had 7 kg rice when 1^st customer came.

Hence, option (b).

Answer: Option B

Video Explanation

Explanation :

∵ 3 is a root of f(x) = 0,

∴ 9a + 3b + c = 0           …(i)

Also,

f(5) = −3f(2)

∴ 25a + 5b + c = −3(4a + 2b + c)

∴ 37a + 11b + 4c = 0       …(ii)

On solving equations (i) and (ii), we get,

a – b = 0

∴ a = b

We know that sum of the roots of a quadratic equation (ax² + bx + c = 0) is –b/a

∴ 3 + other root = −1

∴ Other root = −4

Hence, option (b).

Answer: Option E

Video Explanation

Explanation :

The roots at f(x) = 0 are 3 and −4

∴ The equation can be written as (x – 3)(x + 4) = 0

Or, x² − x + 12 = 0

The co-efficient of x² is 1 here, but all equations which are multiple of this equation will also have same roots.

For example,

 10(x² − x + 12) = 0 will also have same roots

∴ (a + b + c) cannot be determined uniquely.

Hence, option (e).

Answer: Option C

Video Explanation

Explanation :

The first sequence can be written as 17, 17 + 4, 17 + 8, … , 417 and second sequence can be written as 16, 16 + 5, 16 + 10, … , 466

The common difference for the first sequence is 4 and that for the second sequence is 5 and both the sequences have 21 as the first common term.

∴ Common terms are 21, 21 + L, 21 + 2L, ...

[Here, L = LCM of 4 and 5 = 20]

∴ Common terms are 21, 21 + 20, 21 + 40, ...

The common terms have a common difference of 20 and first term as 21.

∴ Let T_n be the last common term, T_n = 21 + (n - 1) × 20

Now, T_n should be less than or equal to 417 

⇒ T_n = 21 + (n - 1) × 20 ≤ 417

⇒ 20n - 20 ≤ 396

⇒ n ≤ 416/20 = 20.8

Hence, the highest possible value of n is 20.

∴ The total number of terms which are common to both the sequences = 20

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

We can find the number of shortest possible paths from A to E either by trial and error or by using combinations.

Note that to travel from A to E, we have to take 2 roads to the right and 2 roads downwards (in the diagram) in order that we follow the shortest path. In other words, we have to use 2 + 2 = 4 roads, out of which 2 are towards right and 2 are downwards.

This is equivalent to selecting 2 things (roads towards right) out of 4 things (roads). (The remaining two roads will be downwards.)

The number of ways of doing this is ⁴C₂ = 4!/(2!×2!) = 6

∴ From point A to E, there are 6 ways to reach with the minimum distance travelled.

Here E to F is the shortest distance because the third side of a triangle is always less than the sum of the other two sides.

From point F to B, there are ⁶C₄ = 6!/(4!×2!) = 15 ways to reach with the minimum distance travelled.

∴ There are 15 × 6 = 90 shortest paths that Neelam can choose.

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

From point A to B, there are 90 paths possible with the minimum distance travelled.

We can travel from B to C in two ways

1. B – K – M – C

To travel from B to K, we have to take 6 roads, out of which one is towards left and 5 are upwards.

There are 6!/(5!×1!) = 6 ways to do this. (Refer to the explanation in the first question of this set.)

2. B – N – C

To travel from B to N, we have to take 7 roads, out of which one is towards left and 6 are upwards.

There are 7!/(6!×1!) = 7 ways to do this.

∴ There are 6 + 7 = 13 ways in which we can travel from B to C.

∴ Overall there are 90 × 13 = 1170 paths possible

Hence, option (a).

Answer: Option B

Video Explanation

Explanation :

Given, f(x) × f(y) = f(xy)

Let x = 1 and y = 2

∴ f(1) × f(2) = f(2)
∴ f(1) = 1

Now, let x = 1/2 and y = 2

∴ f(1/2) × f(2) = f(1/2 × 2) = f(1)

∴ f(1/2) × 4 = 1

∴ f(1/2) = 1/4

Hence, option (b).

Answer: Option C

Video Explanation

Explanation :

Initial sum of the terms of the sequence 1, 2, 3, ...., 40 = $\frac{40\times 41}{2}$ = 820.

When two numbers (a + b) are replaced by (a + b - 1), the total decreases by 1.

After erasing two numbers a and b, and replacing with (a + b − 1), the new sum of the terms of the sequence = 820 − 1

Similarly, after every operation, the sum of the terms of the sequence reduces by 1.

There are a total of 40 numbers and in each step 2 numbers are replaced by 1 number i.e., the number of numbers reduces by 1. For 40 numbers to reduce to just 1 number, this process must be repeated a total of 39 times.

 ∴ The last number left (i.e. final sum) = 820 − 39 = 781

Hence, option (c).

Answer: Option E

Video Explanation

Explanation :

Sum of the digits of multiples of 9 is always 9.

∴ Seed of any number will be 9 if and only if it is a multiple of 9.

There are 55 multiples of 9 which are less than 500 (as 500/9 = 55.555)

∴ There are 55 positive integers which will have seed = 9

Hence, option (e).

Answer: Option E

Video Explanation

Explanation :

We know that the area (A) of triangle (ABC) is related to the circum radius (R) and sides of the triangle as follows:

R = $\frac{AB\times BC\times AC}{4A}$

Where, Area (A) = ½ × AD × BC

∴ R = $\frac{\mathrm{AB}\times \mathrm{BC}\times \mathrm{AC}}{4\times {\displaystyle \frac{1}{2}}\times \mathrm{AD}\times \mathrm{BC}}$ = $\frac{\mathrm{AB}\times \mathrm{AC}}{2\mathrm{AD}}$ = $\frac{17.5\times 9}{2\times 3}$ = 26.25 cm

Hence, option (e).

Answer: Option C

Video Explanation

Explanation :

Last 2 digits of

7¹ = 07
7² = last 2 digits of (07 × 7) = 49
7³ = last 2 digits of (49 × 7) = 43
7⁴ = last 2 digits of (43 × 7) = 01
7⁵ = last 2 digits of (01 × 7) = 07
7⁶ = last 2 digits of (07 × 7) = 49
7⁷ = last 2 digits of (49 × 7) = 43
7⁸ = last 2 digits of (43 × 7) = 01

As we can see, for every 4^th power of 7, the last two digits are 01.

Since 2008 is divisible by 4, we can conclude that last two digits of 7²⁰⁰⁸ are 01.

Hence, option (c).

Answer: Option B

Video Explanation

Explanation :

If p, q and r are the roots of a cubic equation ax³ + bx² + cx + d = 0,

then pq + pr + qr = c/a

If a is 1, then pq + qr + pr = c

Comparing the equation ax³ + bx² + cx + d = 0 with the equation in the question x³− ax² + bx – c = 0, we get

pq + qr + pr = b

Let the three roots of the given equation be (n – 1), n and (n + 1).

∴ (n – 1)n + n(n + 1) + (n – 1)(n + 1) = b

∴ n² – n + n² + n + n² – 1 = b

∴ 3n² – 1 = b

∵ n² ≥ 0, minimum value of b occurs at n = 0

∴ Minimum value of b = –1

Hence, option (b).

Answer: Option C

Video Explanation

Explanation :

We know that for an obtuse triangle of sides a, b and c (where c is the largest side),

a² + b² < c²

We also know that for a triangle, a + b > c

Let the third side be x.

These present us with two limiting cases.

Case 1: Let 8 cm and 15 cm be the shorter sides. The value of the largest side (x) must be greater than

$\sqrt{8^2+{15}^2}=17 cm$

Also, x < 8 + 15 = 23.

The possible integer values of x are 18, 19, 20, 21 and 22 cm.

We cannot consider values from 23 onwards because 8 + 15 = 23 and this violates the second condition.

Case 2: Let 8 and x be the shorter sides and 15 cm is the largest side.

The value of the remaining side (x) must be less than

$\sqrt{{15}^2-8^2}=12.69 cm$

Also, x > 15 - 8 = 7

The possible integer values are 12, 11, 10, 9 and 8 cm.

We cannot consider values less than 8 because 7 + 8 = 15 and this violates the second condition.

Thus, we have 10 possible values for x.

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

The minimum number that can be formed is 1000, hence the number is a 4-digit number,

The maximum number that can be formed is 4000.

As 4000 is the only number in which the first digit is 4, first let us calculate the numbers less than 4000 and then we will add 1 to it.

∴ First digit can be 1, 2 or 3.

Remaining 3 digits can be any of the 5 digits.

∴ Total numbers that can be formed, which are less than 4000 = 3 × 5 × 5 × 5 = 375

∴ Total numbers that satisfy the given condition = 375 + 1 = 376

Hence, option (d).

Answer: Option A

Video Explanation

Explanation :

Consider (a + b + c)²⁰

Each term in the expansion of (a + b + c)²⁰ can be written as ²⁰C_r × a^x × b^y × c^z, where (x + y + z) = 20

Number of distinct terms in the expansion will be same as the number of distinct pairs of x, y and z.

∴ We have to divide 20 into three parts which can be done by using the distribution rule = ^n+r-1C_r-1

Where,

n is number of things to be distributed.

r is number of parts into which the things are to be distributed.

∴ To divide 20 into 3 parts we have,

^20+3-1C_3-1 = ²²C₂ = $\frac{22\times 21}{2\times 1}$ = 11 × 21 = 231

Alternatively,
This can be solved without using much knowledge of permutations and combinations as follows,

(a + b + c)¹ = a + b + c [i.e. 3 terms = (1 + 2) terms]

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac [i.e. 6 terms = (1 + 2 + 3) terms]

(a + b + c)³ = a³ + b³ + c³ + 6abc + 3ab² + 3ac² + 3a²b + 3bc² + 3a²c + 3b²c [i.e. 10 terms = (1 + 2 + 3 + 4) terms]

Similarly,

(a + b + c)ⁿ will have (1 + 2 + 3 + … + (n + 1)) terms

∴ (a + b + c)²⁰ will have (1 + 2 + 3 + … + 21) = 231 terms

Hence, option (a).

Answer: Option E

Video Explanation

Explanation :

Let the length of the sides of the square be 2s.

Consider ∆BQF,

BF = s

In 30° - 60° - 90° triangle,

QF = $\frac{s}{\sqrt{3}}$

∴ Area of ∆BQF = $\frac{1}{2}$ × s × $\frac{s}{\sqrt{3}}$

Area of ABQCDP = Area of square ABCD – 4 × Area of ∆BQF $=4s^2-4\left(\frac{1}{2}\times s\times \frac{s}{\sqrt{3}}\right)$

∴ Required ratio = $\frac{4s^2-4\left({\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{s}{\sqrt{3}}}\right)}{4\left({\displaystyle \frac{1}{2}}\times s\times {\displaystyle \frac{s}{\sqrt{3}}}\right)}$ = $2\sqrt{3}-1$

Hence, option (e).

Answer: Option A

Video Explanation

Explanation :

Let the three numbers be (a – 2), (a – 1) and a.

∴ (a – 2) + (a – 1)² + a³ = p²

Where p is the sum of the three integers.

Now, a – 2 + a² – 2a + 1 + a³ = p²

∴ a³ + a² – a – 1 = p²

∴ a²(a + 1) – 1(a + 1) = p²

∴ (a² – 1)(a + 1) = p²

∴ (a + 1)²(a – 1) = p²

For the above condition to be satisfied, (a – 1) must be a perfect square.

The smallest possible value for a - 1 is 4, since (a – 2) cannot be zero, giving us (a − 2) = 3

The minimum of the three is therefore 3.

Hence, option (a).

Answer: Option A

Video Explanation

Explanation :

Consider only first term,

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}$ = $\sqrt{1+1+\frac{1}{4}}$ = $\sqrt{\frac{9}{4}}$ = $\frac{3}{2}$ = $2-\frac{1}{2}$

Now consider fisrt two terms,

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}$ + $\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}$ = $\frac{3}{2}$ + $\sqrt{1+\frac{1}{4}+\frac{1}{9}}$ = $\frac{3}{2}$ + $\sqrt{\frac{49}{36}}$ = $\frac{3}{2}+\frac{7}{6}$ = $\frac{8}{3}$ = $3-\frac{1}{3}$

Similarly,

$\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}$ + $\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}$ + ... + $\sqrt{1+\frac{1}{{2007}^2}+\frac{1}{{2008}^2}}$ = $2008-\frac{1}{2008}$

Hence, option (a).

Answer: Option E

Video Explanation

Explanation :

Let O and P be the centres of the circles.

OR = OP = PR = 1cm

∴ ∆PRO is an equilateral triangle.

∴ m ∠ROP = 60°

∴ m ∠ROS = 120°

Now, area of the intersecting region = 2(area of sector O-RPS – area of ∆ORS)
                                                         = 2(area of sector O-RPS – area of ∆PRO) [area of ∆PRO = area of ∆ORS]

Area of sector O - RPS = $\frac{120}{360}\left(\pi \right)$ = $\frac{\pi }{3}$

Area of ∆PRO = $\frac{\sqrt{3}}{4}\left(1^2\right)$     

∴ Area of the intersecting region = $2\left(\frac{\pi }{3}\right)$ - $2\left(\frac{\sqrt{3}}{4}\right)$ = $\frac{2\pi }{3}-\frac{\sqrt{3}}{2}$

Hence, option (e).

Answer: Option B

Video Explanation

Explanation :

The angles of the triangle formed by A, B and C tell us that ABC is a right-angle triangle, with right-angle at vertex C, 30° at vertex A and 60° at vertex B.

Since AB = 500 km, in 30°-60°-90° triangle ABC, we get,

AC = 250$\sqrt{3}$ km and BC = 250 km

The train, travels at 50 km/hr. It will travel from B to C (i.e. 250 km) in 5 hours. Since it leaves at 8:00 a.m., it will reach C at 1:00 p.m.

Now, Rahim must be at C latest by 12:45 p.m. (15 minutes before the train)

Travelling at 70 km/hr, he will take approximately 6.2 hours to travel from A to C. Therefore, he must leave at least by

12.75 – 6.2 = 6.55 hours after mid-night.

This is a little after 6:30 a.m. If he leaves by 6:45 a.m., he will not make it to point C 15 minutes before the train arrives.

Hence, option (b).

Answer: Option A

Video Explanation

Explanation :

As shown in the figure, ABC is the cross section of a cone of height 10 cm and radius of base 4 cm.

PQRS is the cross section of a cylinder which needs to be fitted inside the cone such that one of the flat faces of the cylinder (represented by RS) coincides with the base of the cone (represented by BC).

∵ ∆ABD and ∆PBR are similar triangles.

∴ $\frac{10}{4}=\frac{h}{4-r}$

∴ h = 10 - 2.5 × r

Total surface area of the cylinder = S = 2πr × (h + r) = 2πr × (10 - 2.5 × r + r) cm²

⇒ S = πr(20 - 3r) cm²   ...(1)

Now consider two terms 3r and (20 - 3r)

Now, $\frac{3\mathrm{r}+(20-3\mathrm{r})}{2}$ ≥ $\sqrt{3r\times (20-3r)}$ [AM ≥ GM]

⇒ 10 ≥ $\sqrt{3r\times (20-3r)}$

⇒ 3r(20 - 3r) ≤ 100

∴ r(20 - 3r) ≤ 100/3

From (1), we have

⇒ S = πr(20 - 3r) ≤ π × 100/3

Hence, option (a).

Answer: Option C

Video Explanation

Explanation :

We solve this question by options.

If we consider option (c) to be true, then White finishes second and one of the Red or Black horses will come in the first or third positions. 

With White at the second position, the amount Raju receives at the end of the race will be at least Rs. 6000, and from Red or Black he will earn some money. 

Therefore, the total money Raju receives will be more than Rs. 6000. Since his investment at the start of the race was only Rs. 6000, his profit could never be zero.

∴ Option (c) cannot be true.

Hence, option (c).

Answer: Option D

Video Explanation

Explanation :

We solve this question by options.

If we consider option 4 to be true, then either the White or Red horse will finish first. It means that the amount Raju receives at the end of the race will be at least Rs. 8000 or Rs. 12000 (depending on which of the two horses finish first). 

However, his investment at the start of the race was only Rs. 6000. So, his profit could never be zero; in the worst scenario he will at least make Rs. 2000.

∴ Option (d) cannot be true.

Hence, option (d).

Answer: Option D

Explanation :

From statement (A) alone:

The entry list for the tournament consists of 83 players.

In round 1, 1 of the 83 players gets a bye and directly moves on to the next round.

∴ 42 players move on to round 2.

Similarly, 21 players move on to round 3, 11 players move on to round 4, 6 players move on to round 5, 3 players move on to round 6, 2 players move on to round 7.

The winner of the tournament would have played one match in each of the rounds; i.e. a total of 7 matches, provided he doesn’t get a bye.

However, we are not told whether or not the champion received a bye at some point in the tournament.

∴ We cannot answer the question on the basis of statement (A) alone.

From statement (B) alone:

The champion received one bye.

From this statement, we cannot find the number of matches played by the champion.

∴ We cannot answer the question on the basis of statement (B) alone.

From both the statements (A) and (B) together:

The champion must have played 7 matches if he did not receive any bye.

But it is given that the champion has got one bye in the tournament.

∴ He must have played only 6 matches.

∴ We can answer the question using both the statements (A) and (B) together.

Hence, option (d).

Answer: Option D

Explanation :

From statement (A) alone:

Exactly 1 player received a bye in the entire tournament. We get many values of n between 65 and 128 that satisfy this condition.

For example, n can have the value 124 in round 1, to follow the pattern, [124-62-31-16-8-4-2-1].

Also, n can have the value 127 in round 1, to follow the pattern, [127-64-32-16-8-4-2-1].

∴ We cannot answer the question on the basis of statement (A) alone.

From statement (B) alone:

One player received a bye while moving on to the fourth round from the third round.

Here also, we get multiple values of n.

For example, n can have the value 124 in round 1, where 1 player received a bye while moving from round 3 to round 4 following the pattern, [124-62-31-16-8-4-2-1].

Also, n can have the value 122 in round 1, where 1 player received a bye while moving from round 3 to round 4 following the pattern, [122-61-31-16-8-4-2-1].

∴ We cannot answer the question on the basis of statement (B) alone.

From statements (A) and (B) together:

n can only have the value 124 in round 1, where exactly 1 player received a bye while moving from round 3 to round 4 following the pattern [124-62-31-16-8-4-2-1].

∴ We can answer the question using both the statements (A) and (B) together.

Hence, option (d).

Note: An analysis of how 124 was arrived at when using both conditions together:

Let the number of players in the first round be n. Since only one player gets a bye, and that too when moving from the third to the fourth round, hence we have the following conditions:

1. There will be n players in the first round, where n is even.

2. There will be n/2 players in the second round, where n/2 is even.

3. There will be n/4 players in the third round, where n/4 is odd.

4. There will be $\frac{{\displaystyle \frac{n}{4}}+1}{2}=\frac{n+4}{8}$ players in the fourth round, where $\frac{n+4}{8}$ should be even.

5. All numbers of players in the subsequent rounds should also be even.

From condition 3, we can conclude that:

$\frac{n+4}{8}=2k,$ where k is an integer

Hence, n = 16k – 4; so, within the given range, n could be 76 or 92 or 108 or 124.

Writing the pattern for each of the above possible values of n, we have:

76: [76-38-19-10-5-3-2-1]

92: [92-46-23-12-6-3-2-1]

108: [108-54-27-14-7-4-2-1]

124: [124-62-31-16-8-4-2-1]

We see that only 124 satisfies condition 5.

Hence, option (d).