# CAT 2007 QA | Previous Year CAT Paper

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**1. CAT 2007 QA | Algebra - Progressions**

The price of Darjeeling tea (in rupees per kilogram) is 100 + 0.10n, on the n^{th} day of 2007 (n = 1, 2, ..., 100), and then remains constant. On the other hand, the price of Ooty tea (in rupees per kilogram) is 89 + 0.15n, on the n^{th} day of 2007 (n = 1, 2, ..., 365). On which date in 2007 will the prices of these two varieties of tea be equal?

- A.
May 21

- B.
April 11

- C.
May 20

- D.
April 10

- E.
June 30

Answer: Option C

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**Explanation** :

Note that the price of Darjeeling tea remains constant after the 100^{th} day (n = 100).

If the prices of the two varieties of tea become equal before n = 100, then

100 + 0.1n = 89 + 0.15n

∴ n = 220, which is not possible. (since n is assumed to be less than 100)

∴ The prices of the two varieties will be equal after n = 100,

i.e., when the price of Darjeeling tea = 100 + 0.1 × 100 = 110

∴ 89 + 0.15n = 110

∴ n = 140

2007 is not a leap year. Number of days till 30th April = 31 + 28 + 31 + 30 = 120

The prices of the two varieties will be equal on 20^{th} May.

Hence, option (c).

Workspace:

**2. CAT 2007 QA | Algebra - Quadratic Equations**

A quadratic function f(x) attains a maximum of 3 at x = 1. The value of the function at x = 0 is 1. What is the value of f(x) at x = 10?

- A.
–119

- B.
–159

- C.
–110

- D.
–180

- E.
–105

Answer: Option B

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**Explanation** :

Let f(x) = px^{2} + qx + k, where p, q and k are integers, and p ≠ 0

Given, f(0) = k = 1

∴ f(x) = px^{2} + qx + 1

f(x) will attain maximum value at x = -q/2p

Now f(x) attains maixmum at x = 1

∴ x = 1 = -q/2p

⇒ q = -2p

Maximum value of f(x) = 3 = p(1)^{2} + q(1) + 1

⇒ 3 = p + q + 1

⇒ 3 = p - 2p + 1

⇒ p = -2 and q = 4

∴ f(x) = −2x^{2 }+ 4x + 1

⇒ f(10) = −200 + 40 + 1 = −159

Hence, option (b).

Workspace:

**3. CAT 2007 QA | Geometry - Circles**

Two circles with centres P and Q cut each other at two distinct points A and B. The circles have the same radii and neither P nor Q falls within the intersection of the circles. What is the smallest range that includes all possible values of the angle AQP in degrees?

- A.
Between 0 and 90

- B.
Between 0 and 30

- C.
Between 0 and 60

- D.
Between 0 and 75

- E.
Between 0 and 45

Answer: Option C

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**Explanation** :

P and Q do not lie within the intersection of the two circles.

So they lie on the circumferences or outside the circumferences.

**Case 1**: If they lie on the circumferences, then ΔAPQ forms an equilateral triangle.

So, m ∠AQP = 60°

**Case 2**: From the diagram, if they lie outside the circumferences,

m ∠AQ'P' < 60°

Also, m ∠AQP would be 0° if A, Q and P were collinear.

But as P and Q cut each other in two distinct points, A, Q and P cannot be collinear.

∴ m ∠AQP > 0°

∴ The value, m ∠AQP lies between 0° and 60°

Hence, option (c).

Workspace:

**Answer the next 2 questions based on the information given below.**

Let S be the set of all pairs (i, j) where 1 ≤ i < j ≤ n and n ≥ 4. Any two distinct members of S are called “friends” if they have one constituent of the pairs in common and “enemies” otherwise. For example, if n = 4, then S = {(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}. Here, (1, 2) and (1, 3) are friends, (1, 2) and (2, 3) are also friends, but (1, 4) and (2, 3) are enemies.

**4. CAT 2007 QA | Modern Math - Permutation & Combination**

For general n, how many enemies will each member of S have?

- A.
n – 3

- B.
$\frac{1}{2}({n}^{2}-3n-2)$

- C.
2n - 7

- D.
$\frac{1}{2}({n}^{2}-5n+6)$

- E.
$\frac{1}{2}({n}^{2}-7n+14)$

Answer: Option D

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**Explanation** :

Enemies of every pair are the pairs formed with all numbers other than the two in the member itself.

∴ If there are n elements then each member has

${}^{(n-2)}{C}_{2}=\frac{(n-2)(n-3)(n-4)!}{2(n-4)!}\phantom{\rule{0ex}{0ex}}=\frac{1}{2}({n}^{2}-5n+6)\mathrm{enemies}$

Hence, option (d).

Workspace:

**5. CAT 2007 QA | Modern Math - Permutation & Combination**

For general n, consider any two members of S that are friends. How many other members of S will be common friends of both these members?

- A.
$\frac{1}{2}({n}^{2}-5n+8)$

- B.
$2n-6$

- C.
$\frac{1}{2}n(n-3)$

- D.
n - 2

- E.
$\frac{1}{2}({n}^{2}-7n+16)$

Answer: Option D

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**Explanation** :

Two members are friends if they have one element in common.

Let the two members be (a, b) and (b, c)

Now, all the members having one constituent as the common element are common friends.

i.e., all members having b as one of the elements will be common friend of the above two members.

There are (n – 3) elements apart from a, b and c and hence (n - 3) such friends.

Also, one pair formed by the uncommon constituents of the two friends is a common friend i.e., (a, c)

∴ There are n – 3 + 1 = n – 2 common friends.

Hence, option (d).

Workspace:

**Answer the next 2 questions based on the information given below.**

Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guarantee maximum returns on her investment. She has three options, each of which can be utilized fully or partially in conjunction with others.

Option A: Invest in a public sector bank. It promises a return of +0.10%

Option B: Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of +5%, while a fall will entail a return of –3%

Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of –2.5%, while a fall will entail a return of +2%

**6. CAT 2007 QA | Arithmetic - Percentage**

The maximum guaranteed return to Shabnam is:

- A.
0.25%

- B.
0.10%

- C.
0.20%

- D.
0.15%

- E.
0.30%

Answer: Option C

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**Explanation** :

Let Shabnam have Rs. 100 to invest. Let Rs. x, Rs. y and Rs. z be invested in option A, B and C respectively.

∴ x + y + z = 100 ... (I)

If there is a rise in the stock market, returns = 0.001x + 0.05y – 0.025z

If there is a fall in the stock market, returns = 0.001x – 0.03y + 0.02z

Now, x, y and z should be such that regardless of whether the market rises or falls, they give the same return, which is the maximum guaranteed return.

∴ 0.001x + 0.05y – 0.025z = 0.001x – 0.03y + 0.02z

∴ y/z = 9/16

Now, consider different possible values of x, y and z. The returns are as follows:

We see that as the values of y and z increase, the returns increase.

∴ The returns are maximum when x = 0%, y = 36% and z = 64% (Note that the values of y and x are multiples of 9 and 16.)

The maximum returns are 0.2%.

Hence, option (c).

Workspace:

**7. CAT 2007 QA | Arithmetic - Percentage**

What strategy will maximize the guaranteed return to Shabnam?

- A.
100% in option A

- B.
36% in option B and 64% in option C

- C.
64% in option B and 36% in option C

- D.
1/3 in each of the three options

- E.
30% in option A, 32% in option B and 38% in option C

Answer: Option B

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**Explanation** :

As shown by the table formulated in the first question, maximum returns are guaranteed by investing 36% in option B and 64% in option C.

Hence, option (b).

Workspace:

**Answer the next 2 questions based on the information given below.**

Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes the schedule of an airline operating non-stop flights between A and B. All the times indicated are local and on the same day.

Assume that planes cruise at the same speed in both directions. However, the effective speed is influenced by a steady wind blowing from east to west at 50 km per hour.

**8. CAT 2007 QA | Arithmetic - Time, Speed & Distance**

What is the time difference between A and B?

- A.
1 hour and 30 minutes

- B.
2 hours

- C.
2 hours and 30 minutes

- D.
1 hour

- E.
Cannot be determined

Answer: Option D

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**Explanation** :

Let the speed of the plane be x kmph.

Then the speed from B to A is (x – 50) kmph and that from A to B is (x + 50) kmph.

Note that the plane travels from B to A, halts for 1 hour and travels back to B, all in 12 hrs.

∴ 3000/(x – 50) + 1 + 3000/(x + 50) = 12

Solving this we get x = 550. [you can check options from the 2^{nd} question of this set to figure out the value of x.]

Speed of plane = 550 kmph

Now, the plane takes 3000/500 = 6 hrs to travel from B to A.

It reaches A when the time at B is 8:00 am + 6 hrs = 2:00 p.m.

When plane reaches A, time at B is 2 p.m. while time at A is 3 p.m.

∴ The time difference between A and B is 1 hour.

Hence, option (d).

Workspace:

**9. CAT 2007 QA | Arithmetic - Time, Speed & Distance**

What is the plane’s cruising speed in km per hour?

- A.
700

- B.
550

- C.
600

- D.
500

- E.
Cannot be determined

Answer: Option B

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**Explanation** :

As calculated in the first question, the cruising speed of the plane is 550 kmph.

Hence, option (b).

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**10. CAT 2007 QA | Algebra - Number Theory**

Consider all four digit numbers for which the first two digits are equal and the last two digits are also equal. How many such numbers are perfect squares?

- A.
3

- B.
2

- C.
4

- D.
0

- E.
1

Answer: Option E

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**Explanation** :

Let aabb (a ≠ 0, a and b being single digits) be a perfect square.

⇒ aabb = 1000a + 100a + 10b + b = 1100a + 11b = 11(100a + b)

Also, as aabb is a perfect square, it has to be a multiple of 121.

∴ aabb = 121K, where K is also perfect square.

For K = 4, aabb is a 3 digit number, while for K > 82, K is a 5 digit number.

For 81 ≥ K ≥ 9,

121 × 9 = 1089

121 × 16 = 1936

121 × 25 = 3025

121 × 36 = 4356

121 × 49 = 5929

121 × 64 = 7744

121 × 81 = 9801

∴ There is only one number 7744 of the form aabb, which is a perfect square.

Hence, option (e).

Workspace:

**11. CAT 2007 QA | Modern Math - Permutation & Combination**

In a tournament, there are n teams T_{1} , T_{2} ....., T_{n} with n > 5. Each team consists of k players, k > 3. The following pairs of teams have one player in common:

T_{1} & T_{2} , T_{2} & T_{3} ,......, T_{n − 1} & T_{n} , and T_{n} & T_{1}.

No other pair of teams has any player in common. How many players are participating in the tournament, considering all the n teams together?

- A.
n(k – 1)

- B.
k(n – 1)

- C.
n(k – 2)

- D.
k(k – 2)

- E.
(n – 1)(k – 1)

Answer: Option A

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**Explanation** :

Each team has (k – 2) players to itself and shares 2 players with other two teams.

n pairs of teams have 1 player in common and there are n teams.

Total number of players = n(k – 2) + n

= nk – 2n + n

= nk – n

= n(k – 1)

Hence, option (a).

Workspace:

**Answer the next 2 questions based on the information given below.**

Mr. David manufactures and sells a single product at a fixed price in a niche market. The selling price of each unit is Rs. 30. On the other hand, the cost, in rupees, of producing x units is 240 + bx + cx^{2}, where b and c are some constants. Mr. David noticed that doubling the daily production from 20 to 40 units increases the daily production cost by 66.66%. However, an increase in daily production from 40 to 60 units results in an increase of only 50% in the daily production cost. Assume that demand is unlimited and that Mr. David can sell as much as he can produce. His objective is to maximize the profit.

**12. CAT 2007 QA | Algebra - Quadratic Equations**

How many units should Mr. David produce daily?

- A.
130

- B.
100

- C.
70

- D.
150

- E.
Cannot be determined

Answer: Option B

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**Explanation** :

The cost function C(x) = 240 + bx + cx^{2}

C(20) = 240 + 20b + 400c

C(40) = 240 + 40b + 1600c

C(60) = 240 + 60b + 3600c

By conditions,

2/3 × C(20) = C(40) – C(20)

⇒ C(40) = 5/3 × C(20)

⇒ 240 + 40b + 1600c = 400 + 100b/3 + 2000c/3

⇒ 20b/3 + 2800c/3 = 160

⇒ 20b + 2800c = 480 ... (1)

Also,

½ × C(40) = C(60) – C(40)

⇒ 3/2 × C(40) = C(60)

⇒ 360 + 60b + 2400c = 240 + 60b + 3600c

⇒ c = 1/10

⇒ b = 10 ... (from 1)

Profit P(x) for x units is 30x – C(x)

⇒ P(x) = 30x – 240 – 10x – x^{2}/10 = – x^{2}/10 +20x – 240

This is a quadratic expression whose maximum value will occur at x = -(20/2 × -1/10) = 100

∴ x = 100

Hence, option (b).

Workspace:

**13. CAT 2007 QA | Algebra - Quadratic Equations**

What is the maximum daily profit, in rupees, that Mr. David can realize from his business?

- A.
620

- B.
920

- C.
840

- D.
760

- E.
Cannot be determined

Answer: Option D

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**Explanation** :

Following from the first question, at x = 100 the profit is maximum.

At that level of production

P(100) = –240 + 20 (100) – (100)^{2}/10 = –240 + 2000 – 1000 = 760

Hence, option (d).

Workspace:

**Answer the next 2 questions based on the information given below.**

Let a_{1} = p and b_{1} = q, where p and q are positive quantities.

Define:

a_{n} = pb_{n−1} b_{n} = qb_{n−1}, for even n > 1 and

a_{n} = pa_{n − 1} b_{n} = qa_{n − 1}, for odd n > 1.

**14. CAT 2007 QA | Algebra - Progressions**

Which of the following best describes ${a}_{n}+{b}_{n}$ for even n?

- A.
$q{\left(pq\right)}^{\frac{1}{2}n-1}{(p+q)}^{\frac{1}{2}n}$

- B.
$q{\left(pq\right)}^{\frac{1}{2}n-1}(p+q)$

- C.
$q{p}^{\frac{1}{2}n-1}(p+q)$

- D.
${q}^{\frac{1}{2}n}(p+q)$

- E.
${q}^{\frac{1}{2}n}{(p+q)}^{\frac{1}{2}n}$

Answer: Option B

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**Explanation** :

We have the following for different values of n

∴ For even n (say, n = 4),

${a}_{n}+{b}_{n}={p}^{2}{q}^{2}+p{q}^{3}=p{q}^{2}(p+q)$

Now consider the given options, for n = 4

Option (a) gives pq² (p + q)²

Option (b) gives pq²(p + q)

Option (c) gives pq(p + q)

Option (d) gives q²(p + q)

Option (e) gives q²(p + q)²

Hence, option (b).

Workspace:

**15. CAT 2007 QA | Algebra - Progressions**

If p = 1/3 and q = 2/3, then what is the smallest odd n such that a_{n} + b_{n} < 0.01?

- A.
7

- B.
13

- C.
11

- D.
9

- E.
15

Answer: Option D

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**Explanation** :

For odd n, a^{n} + b^{n} = p^{(n + 1)/2} q^{(n – 1)/2} + q^{(n + 1)/2} p^{(n – 1)/2} = p^{(n – 1)/2} q^{(n – 1)/2} (p + q)

Here, p = 1/3, q = 2/3

∴ p + q = 1

∴ a_{n} + b_{n }= p^{(n – 1)/2} q^{(n – 1)/2} = (2/9)^{(n – 1)/2}

Now considering options starting from the lowest,

For n = 7, a_{n} + b_{n} = 8/729 ≈ 1/91 > 1/100

For n = 9, a_{n} + b_{n} = 16/6561 ≈ 1/410 < 1/100

Hence, option (d).

Workspace:

**16. CAT 2007 QA | Arithmetic - Average | Data Sufficiency**

**Each question is followed by two statements A and B. Answer each question using the following instructions.**

Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.

Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.

Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone.

Mark (4) if the question cannot be answered on the basis of the two statements.

The average weight of a class of 100 students is 45 kg. The class consists of two sections, I and II, each with 50 students. The average weight, W_{I}, of Section I is smaller than the average weight, W_{II}, of Section II. If the heaviest student, say Deepak, of Section II is moved to Section I, and the lightest student, say Poonam, of Section I is moved to Section II, then the average weights of the two sections are switched, i.e., the average weight of Section I becomes W_{II} and that of Section II becomes W_{I}. What is the weight of Poonam?

A. W_{II} – W_{I} = 1.0

B. Moving Deepak from Section II to I (without any move from I to II) makes the average weights of the two sections equal.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

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**Explanation** :

Let the weights of Deepak and Poonam be d and p respectively.

(50W_{II} + 50W_{I})/100 = 45

∴ W_{II} + W_{I} = 90 ...(i)

50W_{I} + d – p = 50W_{II}

50W_{II} – d + p = 50W_{I}

∴ 50(W_{II} – W_{I}) = d – p ...(ii)

From Statement A, W_{II} – W_{I} = 1 ...(iii)

From (i), (ii) and (iii)

W_{I} and W_{II} can be found. Also, d – p = 50 ...(iv)

However this information does not give us the value of p. Statement A is insufficient to answer the question.

From Statement B,

W_{I} = W_{II} = (Sum_{I} + d)/51 = (Sum_{II} – d)/49

∴ 49(Sum_{I}) + 49d = 51(Sum_{II}) – 51d

∴ 100d = 51(50W_{II}) – 49(50W_{I})

∴ 2d = 51W_{II} – 49W_{I} ...(v)

This alone cannot help us find the value of p. Statement B is insufficient to answer the question.

Considering both statements together, we have values of W_{I} and W_{II}, which can be substituted in (v) to find d, which can be used to find p using (iv).

Hence, option (c).

Workspace:

**17. CAT 2007 QA | Algebra - Number Theory**

**Each question is followed by two statements A and B. Answer each question using the following instructions.**

Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.

Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.

Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone.

Mark (4) if the question cannot be answered on the basis of the two statements.

Consider integers x, y and z. What is the minimum possible value of x^{2} + y^{2} + z^{2} ?

A. x + y + z = 89

B. Among x, y, z two are equal.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option A

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**Explanation** :

Statement A:

x + y + z = 89

x^{2} + y^{2} + z^{2} will be minimum when x = y = z = 89/3

But 89/3 is a non-integer. ∴ We consider integer values of x, y, z which are as close as possible to 89/3.

We get two cases

1. x, y, z = 30, 30, 29

x^{2} + y^{2} + z^{2} = 2641

2. x, y, z = 31, 29, 29

x^{2} + y^{2} + z^{2} = 2643

Minimum possible value of x^{2} + y^{2} + z^{2} is 2641. Thus statement A is sufficient to get the answer. Though statement B states a fact related to the minimum value, it is not necessary to arrive at the minimum value.

Hence, option (a).

Workspace:

**18. CAT 2007 QA | Geometry - Quadrilaterals & Polygons**

**Each question is followed by two statements A and B. Answer each question using the following instructions.**

Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.

Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.

Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone.

Mark (4) if the question cannot be answered on the basis of the two statements.

Rahim plans to draw a square JKLM with a point O on the side JK but is not successful. Why is Rahim unable to draw the square?

A. The length of OM is twice that of OL.

B. The length of OM is 4 cm.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option A

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**Explanation** :

Let p be the side of square JKLM.

From Statement A,

OM = 2 × OL

OM is maximum when it is the diagonal of the square and has length $\sqrt{2}p$

When OM is maximum, $OM=\sqrt{2}\times OL$

∴ OM ≠ 2 × OL if O lies on JK.

∴ Rahim is unable to draw the square.

Statement B offers no additional or relevant information.

Hence, option (a).

Workspace:

**19. CAT 2007 QA | Geometry - Mensuration**

Mark (1) if the question can be answered by using statement A alone but not by using statement B alone.

Mark (2) if the question can be answered by using statement B alone but not by using statement A alone.

Mark (3) if the question can be answered by using both the statements together but not by using either of the statements alone.

Mark (4) if the question cannot be answered on the basis of the two statements.

ABC Corporation is required to maintain at least 400 Kilolitres of water at all times in its factory, in order to meet safety and regulatory requirements. ABC is considering the suitability of a spherical tank with uniform wall thickness for the purpose. The outer diameter of the tank is 10 meters. Is the tank capacity adequate to meet ABC’s requirements?

A. The inner diameter of the tank is at least 8 meters.

B. The tank weighs 30,000 kg when empty, and is made of a material with density of 3 gm/cc.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option B

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**Explanation** :

Let the inner radius be r meter. Capacity of tank = (1 m^{3} = 1 kilolitre)

From statement A, since r ≥ 4m

∴ Capacity of tank > 256 m^{3}

Since the capacity needed is more than 256 m^{3} statement A is insufficient.

From statement B,

Volume of the material of tank = mass/density = 30000kg/(3 gm/cc) = 10,000,000 cm^{3} = 10 m^{3}

Hence the inner volume of tank = Outer volume – Volume of material of tank

Therefore, we can say that the tank capacity is adequate.

Hence, option (b).

Workspace:

**20. CAT 2007 QA | Algebra - Simple Equations**

Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?

- A.
17

- B.
16

- C.
18

- D.
15

- E.
19

Answer: Option C

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**Explanation** :

Let number of 50, 10 and 1 miso notes used is x, y and z respectively.

⇒ 50x + 10y + z = 107

Now, since x, y and z are whole numbers x can take only three values i.e., 0, 1 and 2.

**Case 1**: x = 0

⇒ 10y + z = 107

⇒ y can take values from 0 till 10 i.e., 11 values

∴ 11 ways to pay bill of 107 misos.

**Case 2**: x = 1

⇒ 10y + z = 57

⇒ y can take values from 0 till 5 i.e., 6 values

∴ 6 ways to pay bill of 107 misos.

**Case 3**: x = 2

⇒ 10y + z = 7

⇒ y can take only 1 value i.e., 0

∴ 1 way to pay bill of 107 misos.

⇒ Total ways the bill can be paid = 11 + 6 + 1 = 18 ways,

The bill can be paid in 18 ways as shown in the above table.

Hence, option (c).

Workspace:

**21. CAT 2007 QA | Algebra - Number Theory**

How many pairs of positive integers m, n satisfy 1/m + 4/n = 1/12 where n is an odd integer less than 60?

- A.
6

- B.
4

- C.
7

- D.
5

- E.
3

Answer: Option E

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**Explanation** :

1/m + 4/n = 1/12

∴ 1/m = 1/12 – 4/n

∴ m = 12n/(n – 48)

As, m is a positive integer, n should be greater than 48 and moreover since n is a positive odd integer lesser than 60, n can take values 49, 51, 53, 55, 57 and 59.

If n = 49, 51, 57 then m is a positive integer.

If n = 53, 55, 59 then m is not an integer.

∴ 3 pairs of values of m and n satisfy the given equation.

Hence, option (e).

Workspace:

**22. CAT 2007 QA | Algebra - Simple Equations | Algebra - Number Theory**

A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja, giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise, Shailaja noticed that she was left with exactly three times as much as the amount on the cheque. Which of the following is a valid statement about the cheque amount?

- A.
Over Rupees 13 but less than Rupees 14

- B.
Over Rupees 7 but less than Rupees 8

- C.
Over Rupees 22 but less than Rupees 23

- D.
Over Rupees 18 but less than Rupees 19

- E.
Over Rupees 4 but less than Rupees 5

Answer: Option D

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**Explanation** :

Let the amount on Shailaja’s cheque be Rs. x and paise y = (100x + y) paise (x and y are positive integers)

The teller gives her (100y + x) paise.

Now, 100y + x – 50 = 3(100x + y)

∴ 97y – 299x = 50

∴ y = (50 + 299x)/97 = [50 + 8x + 291x]/97 = [(50 + 8x)/97] + 3x

Now as y is an integer, (50 + 8x) has to be a multiple of 97 with x, y ≤ 99

50 + 8x = 97k (k is an integer)

∴ k = 2, 10, 18…

∴ x = 18, 115, 212…

∴ x = 18 is the only possible value.

This implies that y = 5

∴ The amount on Shailaja’s cheque is over Rs. 18 but less than Rs. 19.

Hence, option (d).

Workspace:

**23. CAT 2007 QA | Arithmetic - Average | Modern Math - Sets**

Consider the set S = {2, 3, 4, ..., 2n + 1}, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X – Y?

- A.
0

- B.
1

- C.
n/2

- D.
n + 1/2n

- E.
2008

Answer: Option B

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**Explanation** :

Y = (2 + 4 + 6 + 8 + … + 2n)/n

Average of numbers in AP is same as average of first and the last terms.

⇒ Y = (2 + 2n)/2 = 1 + n

X = (3 + 5 + 7 + 9 + … + (2n + 1))/n

Average of numbers in AP is same as average of first and the last terms.

⇒ X = (3 + 2n+1)/2 = 2 + n

∴ X – Y = 2 + n - (1 + n) = 1

**Note**: The information that 'n is a positive integer larger than 2007' does not affect the answer in any way.

Hence, option (b).

Workspace:

**24. CAT 2007 QA | Arithmetic - Average**

Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average age of this eight member joint family is nearest to:

- A.
23 years

- B.
22 years

- C.
21 years

- D.
25 years

- E.
24 years

Answer: Option E

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**Explanation** :

The sum of the ages of the members of the family ten years ago = 231

∴ The sum of the ages of 8 (including the newborn) members of the family seven years ago = 231 + (3 × 8) – 60 = 195

∴ The sum of the ages of 8 (including the newborn) members of the family four years ago = 195 + (3 × 8) – 60 = 159

∴ The sum of the ages of 8 members of the family now = 159 + (4 × 8) = 191

∴ Required average = 191/8 = 23.875 ≈ 24

Hence, option (e).

Workspace:

**25. CAT 2007 QA | Algebra - Functions & Graphs**

A function f(x) satisfies f(1) = 3600, and f(1) + f(2) + ... + f(n) = n²f(n), for all positive integers n > 1. What is the value of f(9)?

- A.
80

- B.
240

- C.
200

- D.
100

- E.
120

Answer: Option A

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**Explanation** :

f(1) + f(2) + f (3) + … + f(n −1) + f(n) = n^{2}f(n) ... (i)

Similarly, f(1) + f(2) + f (3) + … + f(n − 1) = (n − 1)^{2} f(n −1) ... (ii)

⇒ f(n) = n^{2} f(n) – (n – 1)^{2}f(n − 1) ... (i) – (ii)

⇒ (n^{2} – 1)f(n) = (n – 1)^{2}f(n – 1)

⇒ f(n) = $\frac{{(\mathrm{n}-1)}^{2}\mathrm{f}(\mathrm{n}-1)}{({\mathrm{n}}^{2}-1)}$

⇒ f(n) = $\frac{(\mathrm{n}-1)\mathrm{f}(\mathrm{n}-1)}{(\mathrm{n}+1)}$

Now, putting

n = 2 we get f(2) = $\frac{1}{3}\times \mathrm{f}\left(1\right)$

n = 3 we get f(2) = $\frac{2}{4}\mathrm{f}\left(2\right)$ = $\frac{2}{4}\times \frac{1}{3}\times \mathrm{f}\left(1\right)$

...

n = 9 we get f(9) = $\frac{8}{10}\mathrm{f}\left(8\right)$ = $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}\times \frac{5}{7}\times \frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times \mathrm{f}\left(1\right)$

∴ f(9) = $\frac{8}{10}\times \frac{7}{9}\times \frac{6}{8}\times \frac{5}{7}\times \frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\times \frac{1}{3}\times 3600$

= $\frac{2}{10\times 9}\times 3600$ = 80

Hence, option (a).

Workspace:

**26. CAT 2007 QA | Miscellaneous | Data Sufficiency**

**Each question is followed by two statements, I and II. Answer each question using the following instructions:**

**Mark (1)** if the question can be answered by using statement I alone but not by using statement II alone.

**Mark (2)** if the question can be answered by using statement II alone but not by using statement I alone.

**Mark (3)** if the question can be answered by using either of the statements alone.

**Mark (4)** if the question can be answered by using both the statements together but not by either of the statements alone.

**Mark (5)** if the question cannot be answered by using any of the statements.

In a football match, at the half-time, Mahindra and Mahindra Club was trailing by three goals. Did it win the match?

I. In the second-half Mahindra and Mahindra Club scored four goals.

II. The opponent scored four goals in the match.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option E

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**Explanation** :

From Statement I, the MM club scored 4 goals in the second half. The number of goals scored by the opponent is not known. So the winner cannot be determined. Statement I is insufficient.

From Statement II, the opponent scored 4 goals in the match, but we do not know the number of goals that the MM club scored. Statement II is insufficient.

Considering both the statements we have following.

Thus, MM club could have won the match or could have tied. The question cannot be answered.

Hence, option (e).

Workspace:

**27. CAT 2007 QA | Data Sufficiency**

**Each question is followed by two statements, I and II. Answer each question using the following instructions:**

**Mark (1)** if the question can be answered by using statement I alone but not by using statement II alone.

**Mark (2)** if the question can be answered by using statement II alone but not by using statement I alone.

**Mark (3)** if the question can be answered by using either of the statements alone.

**Mark (4)** if the question can be answered by using both the statements together but not by either of the statements alone.

**Mark (5)** if the question cannot be answered by using any of the statements.

In a particular school, sixty students were athletes. Ten among them were also among the top academic performers. How many top academic performers were in the school?

I. Sixty per cent of the top academic performers were not athletes.

II. All the top academic performers were not necessarily athletes.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option A

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**Explanation** :

From Statement I, 40% of the top academic performers were athletes.

∴ If there are x top academic performers, 10 = 0.4x

∴ x = 25

Statement I is sufficient.

Statement II does not give any useful information.

Hence, option (a).

Workspace:

**28. CAT 2007 QA | Data Sufficiency**

**Each question is followed by two statements, I and II. Answer each question using the following instructions:**

**Mark (1)** if the question can be answered by using statement I alone but not by using statement II alone.

**Mark (2)** if the question can be answered by using statement II alone but not by using statement I alone.

**Mark (3)** if the question can be answered by using either of the statements alone.

**Mark (4)** if the question can be answered by using both the statements together but not by either of the statements alone.

**Mark (5)** if the question cannot be answered by using any of the statements.

Five students Atul, Bala, Chetan, Dev and Ernesto were the only ones who participated in a quiz contest. They were ranked based on their scores in the contest. Dev got a higher rank as compared to Ernesto, while Bala got a higher rank as compared to Chetan. Chetan’s rank was lower than the median. Who among the five got the highest rank?

I. Atul was the last rank holder.

II. Bala was not among the top two rank holders.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option D

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**Explanation** :

Chetan’s rank = 4 or 5

Now, Bala < Chetan and Dev < Ernesto.

From Statement I,

The highest rank holder cannot be determined. Statement I is insufficient.

Statement II is also insufficient.

Considering both statements together (refer to the table), Case 2 holds. Dev got the highest rank.

Hence, option (d).

Workspace:

**29. CAT 2007 QA | Arithmetic - Percentage | Data Sufficiency**

**Mark (1)** if the question can be answered by using statement I alone but not by using statement II alone.

**Mark (2)** if the question can be answered by using statement II alone but not by using statement I alone.

**Mark (3)** if the question can be answered by using either of the statements alone.

**Mark (4)** if the question can be answered by using both the statements together but not by either of the statements alone.

**Mark (5)** if the question cannot be answered by using any of the statements.

Thirty percent of the employees of a call centre are males. Ten per cent of the female employees have an engineering background. What is the percentage of male employees with engineering background?

I. Twenty five per cent of the employees have engineering background.

II. Number of male employees having an engineering background is 20% more than the number of female employees having an engineering background.

- A.
1

- B.
2

- C.
3

- D.
4

- E.
5

Answer: Option C

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**Explanation** :

Let there be 100x employees. So, 30x are male and 70x are female.

∴ 7x female employees have an engineering background.

From statement I, 25x employees have an engineering background.

∴ 18x male employees have an engineering background.

Required percentage = 18x × 100/ 30x

Statement I is sufficient.

From Statement II, Number of male employees having an engineering background = 1.2 × 7x

Required percentage = 1.2 × 7x × 100/30x

Hence, option (c).

Workspace:

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