# CAT 2002 QA

Paper year paper questions for CAT 2002 QA

**1. CAT 2002 QA | Algebra - Number Theory**

On dividing a number by 3, 4 and 7, the remainders are 2, 1 and 4 respectively. If the same number is divided by 84 then the remainder is

- A.
80

- B.
76

- C.
56

- D.
None of these

Answer: Option C

**Explanation** :

Let the given number be N

From the given data, we can conclude that

N = 3a + 2 = 4b + 1 = 7c + 4, where a, b and c are natural numbers.

Thus, 3a = 4b – 1 = 7c + 2

By trial and error, we can see that the first number satisfying the above given equation is 3

All the numbers satisfying this condition are of the form 3 + (l.c.m. of 3 and 4) × k

= 3 + 12k (where k is a natural number)

Thus 12k + 3 = 7c + 2

Hence, 12k + 1 = 7c

By trial and error method, we can see that the smallest number satisfying the above equation is 49.

Thus the smallest required number is 7c + 4 = 49 + 4 = 53

All the numbers satisfying the conditions given in the question will be of the form 53 + (l.c.m of 3, 4 and 7) × m

= 53 + 84m (where m is a natural number)

∴ N when divided by 84 will result in remainder 53.

Hence, option 3.

Workspace:

**2. CAT 2002 QA | Algebra - Number Theory**

There are three pieces of cake weighing 9/2 lbs, 27/4 lbs and 36/5 lbs. Pieces of the cake are equally divided and distributed in such a manner that every guest in the party gets one single piece of cake. Further the weight of the pieces of the cake is as heavy as possible. What is the largest number of guests to whom we can distribute the cake?

- A.
54

- B.
72

- C.
20

- D.
None of these

Answer: Option D

**Explanation** :

The largest possible piece of cake from the three given pieces [9/2, 27/4, 36/5] can be obtained by finding the H.C.F of the three fractions,

∴ H.C.F. $\left[\frac{9}{2},\frac{27}{4},\frac{36}{5}\right]=\frac{\mathrm{H}.\mathrm{C}.\mathrm{F}.\mathrm{of}\mathrm{Numerator}}{\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{Denominator}}=\frac{9}{20}$

∴ The weight of the piece of cake given to each guest 9/20 lbs.

∴ Number of Guests = $\left(\frac{{\displaystyle \frac{9}{2}}}{{\displaystyle \frac{9}{20}}}\right)+\left(\frac{{\displaystyle \frac{27}{4}}}{{\displaystyle \frac{9}{20}}}\right)+\left(\frac{{\displaystyle \frac{36}{5}}}{{\displaystyle \frac{9}{20}}}\right)=$ 10 + 15 + 16 = 41

Hence, option 4.

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**3. CAT 2002 QA | Algebra - Number Theory**

For three real numbers x, y and z, x + y + z = 5, and xy + yz + xz = 3. What is the largest value which x can take?

- A.
$3\sqrt{13}$

- B.
$\sqrt{19}$

- C.
13/3

- D.
$\sqrt{15}$

Answer: Option C

**Explanation** :

x + y + z = 5 …(i)

xy + yz + zx = 3 …(ii)

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

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

∴ x² + y² + z² = 19 …(iii)

When x² is maximum, y² + z² is minimum,

∵ y² + z² ≥ 0 or y² + z² = 0

∴ y = z = 0, This is not possible as it does not satisfy equation (ii)

∴ y ≠ 0 and z ≠ 0

∴ y² + z² will be minimum when y = z

Substituting z = y in (i) and (ii), we get,

x = 5 − 2y …(iii)

xy + y² + xy = 3 …(iv)

Solving (iii) and (iv) for x and y, we get,

y = 3 or 1/3

∴ If y = 3; x = −1 and z = 3

∴ If y = 1/3; x = 13/3 and z = 1/3

∴ The maximum value of x is 13/3.

Hence, option 3.

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**4. CAT 2002 QA | Miscellaneous**

A transport company charges for its vehicles in the following manner:

If the driving is 5 hours or less, the company charges Rs. 60 per hour or Rs. 12 per km (which ever is larger)

If driving is more than 5 hours, the company charges Rs. 50 per hour or Rs. 7.5 per km (which ever is larger)

If Anand drove it for 30 km and paid a total of Rs. 300, then for how many hours does he drive?

- A.
4

- B.
5.5

- C.
7

- D.
6

Answer: Option D

**Explanation** :

The rent of the car = Maxima(number of hours × charge per hour, km travelled × charge per km)

Case 1: Anand drove the Car for 5 hours or less

∴ Rent = Maxima(300, 30 × 12) = 360

Case 2: Anand drove the Car for more than 5 hours

∴ Rent = Maxima(300, 7.5 × 30) = 300

∵ Anand paid Rs. 300 as rent.

∴ Anand drove for 6 hours.

Hence, option 4.

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**5. CAT 2002 QA | Arithmetic - Time, Speed & Distance**

Only a single rail track exists between station A and B on a railway line. One hour after the north bound superfast train N leaves station A for station B, a south passenger train S reaches station A from station B. The speed of the superfast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day N leaves for station B from station A, 20 minutes behind the normal schedule. In order to maintain the schedule both N and S increased their speed. If the superfast train doubles its speed, what should be the ratio (approximately) of the speed of passenger train to that of the superfast train so that passenger train S reaches exactly at the scheduled time at the station A on that day?

- A.
1 : 3

- B.
1 : 4

- C.
1 : 5

- D.
1 : 6

Answer: Option D

**Explanation** :

Let the speed of E be 2x km/hr.

∴ Speed of N = 4x km/hr and speed of S = x km/hr

According to the problem,

Time taken by N to cover AB + Time taken by S to cover BA = 1 hour

$\therefore \frac{d}{4x}+\frac{d}{x}=1$

∴ 5d = 4x or d = $\frac{4x}{5}$

∵ To maintain the schedule (gap of 20 min), N doubles its speed and speed of S becomes y.

$\therefore \frac{{\displaystyle \frac{4x}{5}}}{8x}+\frac{{\displaystyle \frac{4x}{5}}}{y}=\frac{2}{3}$

$\therefore \frac{1}{10}+\frac{4x}{5y}=\frac{2}{3}$

$\therefore \frac{4x}{5y}=\frac{17}{30}$

$\therefore \frac{4x}{y}=\frac{17}{6}$

$\therefore \frac{8x}{y}=\frac{17}{3}or\frac{y}{8x}=\frac{3}{17}\approx \frac{1}{6}$

$\therefore \frac{Speedofpassengertrain}{Speedofsuperfasttrain}\approx \frac{1}{6}$

Hence, option 4.

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**6. CAT 2002 QA | Algebra - Number Theory**

If x^{2} + 5y^{2} + z^{2} = 2y(2x + z), then which of the following statements are necessarily true?

I. x = 2y

II. x = 2z

III. 2x = z

- A.
Only I

- B.
Only II

- C.
Only III

- D.
Only I and II

Answer: Option D

**Explanation** :

x² + 5y² + z² = 4xy + 2yz

∴ x² + 4y² + y² + z² – 4xy – 2yz = 0

∴ (x² + 4y² – 4xy) + (y² – 2yz + z²) = 0

∴ (x – 2y)² + (y – z)² = 0 …(i)

As per equation (i),

∴ (x – 2y)² = 0 and (y – z)² = 0

∴ x = 2y and y = z

Hence, option 4.

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**7. CAT 2002 QA | Geometry - Quadrilaterals & Polygons**

In the following figure, the area of the isosceles right triangle ABE is 7 sq.cm. If EC = 3BE, then the area of rectangle ABCD (in sq. cm.) is

- A.
64

- B.
82

- C.
26

- D.
56

Answer: Option D

**Explanation** :

∵ The area of the ∆ABE = $\frac{1}{2}\times $ AB × BE = 7 cm^{2}

∴ BE × AB = 14

The area of the rectangle ABCD = AB × BC = AB × (BE + EC) = 4 × (BE × AB) = 56 cm^{2}

Hence, option 4.

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**8. CAT 2002 QA | Algebra - Number Theory**

Number S is equal to the square of the sum of the digits of a 2 digit number D. If the difference between S and D is 27, then D is

- A.
32

- B.
54

- C.
64

- D.
52

Answer: Option B

**Explanation** :

Let S be the square of the sum of the digits.

Substituting options,

Option 1, (3 + 2)^{2} – 24 ≠ 27

Option 2, (5 + 4)^{2} − 54 = 27

Hence, option 2.

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**9. CAT 2002 QA | Arithmetic - Average**

A boy finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?

- A.
4

- B.
2

- C.
6

- D.
8

Answer: Option B

**Explanation** :

Let the Arithmetic Mean of the 10 numbers be x and s be the sum of the remaining 9 numbers.

$\frac{x+10a+b}{10}=x$ ...(i)

Interchanging the number ab with ba,

$\frac{x+10b+a}{10}=$ x + 1.8 ... (ii)

Subtracting (i) from (ii), we get,

10b + a − (10a + b) = 9(b − a) = 18

b − a = 2

Hence, option 2.

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**10. CAT 2002 QA | Algebra - Simple Equations**

A string of length 40 metres is divided into three parts of different lengths. The first part is three times the second part, and the last part is 23 metres smaller than the first part. Find the length of the largest part.

- A.
27

- B.
4

- C.
5

- D.
9

Answer: Option A

**Explanation** :

Let l, m and s be the longest, medium and the shortest lengths of the strings.

l = 3m and s = l – 23

l + s + m = 40

$l+\frac{l}{3}+$ l - 23 = 40

$\frac{7l}{3}$ = 63

l = 27

Hence, option 1.

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**11. CAT 2002 QA | Algebra - Number Theory**

For all integers n > 0, 76^{n} – 66^{n} is divisible by

- A.
13

- B.
127

- C.
559

- D.
All of these

Answer: Option D

**Explanation** :

Factorizing the given expression,

(7^{3n} – 6^{3n})(7^{3n} + 6^{3n})

i.e. (7^{n} – 6^{n})(7^{2n} + 6^{2n} + 7^{n}6^{n})(7^{3n} + 6^{3n}) ... (i)

Substitute n = 1 in equation (i),

The three factors are 1, 127 and 559.

Since 559 is divisible by 13, hence 13 will also be a factor.

Hence, option 4.

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**12. CAT 2002 QA | Modern Math - Permutation & Combination**

n_{1}, n_{2}, n_{3} ... n_{10} are 10 numbers such that n_{1} > 0 and the numbers are given in ascending order. How many triplets can be formed using these numbers such that in each triplet, the first number is less than the second number, and the second number is less than the third number?

- A.
109

- B.
27

- C.
36

- D.
None of these

Answer: Option D

**Explanation** :

Let us assume that the 10 numbers are 1 - 10.

When the first number is 1,

2^{nd} number as 2, the last number can be chosen in 8 different ways.

2^{nd} number as 3, the last number can be chosen in 7 different ways, and so on.

When 1 is the first number, the possible number of triplets as per the given condition = 8 + 7 + 6 + … + 1 = 36

When the first number as 2, the possible number of triplets will be 7 + 6 + ... + 1 = 28

∴ The total number of triplets = 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

Hence, option 4.

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**13. CAT 2002 QA | Geometry - Mensuration**

In order to cover less distance, a boy - rather than going along the longer and the shorter lengths of the rectangular path, goes by the diagonal. The boy finds that he saved a distance equal to half the longer side. The ratio of the breadth and length is

- A.
1/2

- B.
2/3

- C.
3/4

- D.
7/15

Answer: Option C

**Explanation** :

Let the breadth of the rectangular field be b and its length be l.

If he had moved along the edge, he would have moved through a distance = l + b = d + l/2

d = b + l/2

Substituting the options, only option 3 gives the values of l, b and d as 4, 3 and 5 which are Pythagorean triplets.

Hence, option 3.

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**14. CAT 2002 QA | Algebra - Quadratic Equations**

The number of roots $\frac{{A}^{2}}{x}+\frac{{B}^{2}}{x-1}=1$ is

- A.
1

- B.
2

- C.
3

- D.
None of these

Answer: Option B

**Explanation** :

$\frac{{A}^{2}}{x}+\frac{{B}^{2}}{x-1}=1(x\ne 1)$

A^{2}(x − 1) + B^{2}(x) = x(x − 1)

A^{2}x − A^{2} + B^{2}x = x^{2} − x

x^{2} − x − A^{2}x − B^{2}x + A^{2} = 0

x^{2} − x[A^{2} + B^{2} + 1] + A^{2} = 0

Since it is a quadratic equation, the number of roots = 2

Now, ∆ = (A^{2} + B^{2} + 1)^{2} − 4A^{2}

= A^{4} + B^{4} + 1 + 2A^{2}B^{2} + 2B^{2} + 2A^{2} − 4A^{2}

= A^{4} + B^{4} + 1 + 2A^{2}B^{2} + 2B^{2 }− 2A^{2}

For roots to be real, A^{4} + B^{4} + 1 + 2A^{2}B^{2} + 2B^{2 }− 2A^{2 }≥ 0

∴ B^{4} + 2B^{2}(A^{2} + 1) + (A^{2} - 1)^{2} ≥ 0

The given equation will have one root if ∆ = 0.

It can be clearly seen that for ∆ = 0, each of the three terms must be equal to 0 since each term is greater than or equal to 0.

∴ A = ±1 and B = 0.

But for these values of A and B, x = 1 which is a contradiction.

So, ∆ ≠ 0

For ∆ > 0, two different roots are obtained.

Therefore, the number of roots of the given equation will be 2.

Hence, option (b).

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**15. CAT 2002 QA | Algebra - Simple Equations**

Mayank, Mirza, Little and Jagbir bought a motorbike for $60. Mayank contributed half of the total amount contributed by others, Mirza contributed one-third of total amount contributed by others, and Little contributed one-fourth of the total amount contributed by others. What was the money paid by Jagbir?

- A.
$12

- B.
$13

- C.
$18

- D.
$20

Answer: Option B

**Explanation** :

Let the money contributed by Mayank, Mirza, Little and Jagbir be a, b, c and d respectively.

a = $\frac{1}{2}$ (b + c + d)

∴ 2a = b + c + d

∴ 2a = 60 - a

∴ a = 20

b = $\frac{1}{3}$ (a + c + d)

∴ 3b = a + c + d

∴ 3b = 60 - b

∴ b = 15

c = $\frac{1}{4}$ (a + b + d)

∴ 4c = a + b + d

∴ 4c = 60 - c

∴ c = 12

Money contributed by Jagbir = 60 − 20 − 15 − 12 = $13

Hence, option 2.

Workspace:

**16. CAT 2002 QA | Algebra - Number Theory**

If U, V, W and m are natural numbers such that U^{m} + V^{m} = W^{m}, then which of the following is true?

- A.
m < Min(U, V, W)

- B.
m > Max(U, V, W)

- C.
m < Max(U, V, W)

- D.
None of these

Answer: Option C

**Explanation** :

Taking m = 1, we have,

U^{1} + V^{1} = W^{1}

The least values of U and V are 1 and 1.

∴ Least value of W = 2

Also, taking m = 2, we have,

U^{2} + V^{2} = W^{2}

The least values of U and V are 3 and 4 (not necessarily in that order).

∴ Least value of W = 5

∴ m is always less than Max(U, V, W).

Hence, option 3.

Workspace:

**17. CAT 2002 QA | Algebra - Functions & Graphs**

If $f\left(x\right)=\mathrm{log}\left(\frac{1+x}{1-x}\right),$ then f(x) + f(y) =

- A.
f(x + y)

- B.
f(1 + xy)

- C.
(x + y) f(1 + xy)

- D.
$f\left(\frac{x+y}{1+xy}\right)$

Answer: Option D

**Explanation** :

$f\left(x\right)+f\left(y\right)=\mathrm{log}\left(\frac{1+x}{1-x}\right)+\mathrm{log}\left(\frac{1+y}{1-y}\right)\phantom{\rule{0ex}{0ex}}=\mathrm{log}\frac{\left(1+x\right)(1+y)}{\left(1-x\right)(1-y)}\phantom{\rule{0ex}{0ex}}=\mathrm{log}\frac{1+xy+x+y}{1+xy-(x+y)}\phantom{\rule{0ex}{0ex}}=\mathrm{log}\frac{1+{\displaystyle \frac{x+y}{1+xy}}}{1-{\displaystyle \frac{x+y}{1+xy}}}\phantom{\rule{0ex}{0ex}}=f\left(\frac{x+y}{1+xy}\right)$

Hence, option 4.

Workspace:

**18. CAT 2002 QA | Algebra - Progressions**

On a straight road XY, 100 metres in length, 5 stones are kept beginning from the end X. The distance between two adjacent stones is 2 metres. A man is asked to collect the stones one at a time and put at the end Y. What is the distance covered by him?

- A.
460 metres

- B.
540 metres

- C.
860 metres

- D.
920 metres

Answer: Option C

**Explanation** :

The first stone is kept at 'X' and the subsequent stones are equidistant from each other, i.e. 2 metres.

∴ The total distance covered by the man to keep all the stones at Y = 100 + 2 × (98 + 96 + 94 + 92) = 100 + 2 × (380) = 860 metres

Hence, option 3.

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**19. CAT 2002 QA | Geometry - Triangles**

The internal bisector of an angle A in a triangle ABC meets the side BC at point D. AB = 4, AC = 3 and ∠A = 60°. Then what is the length of the bisector AD?

- A.
$\frac{12\sqrt{3}}{7}$

- B.
$\frac{12\sqrt{13}}{7}$

- C.
$\frac{4\sqrt{13}}{7}$

- D.
$\frac{4\sqrt{3}}{7}$

Answer: Option A

**Explanation** :

Area of ∆ADC = $\frac{1}{2}$ AD × 3 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ AD × 4 × sin 30°

Area of ∆ADC = $\frac{1}{2}$ × 3 × 4 × sin 60°

$\frac{1}{2}$ AD × 3 × sin 30° + $\frac{1}{2}$ × AD × 4 × sin 30° = $\frac{1}{2}$ × 3 × 4 × sin 60°

AD$\left[3\times \frac{1}{2}+4\times \frac{1}{2}\right]=12\times \frac{\sqrt{3}}{2}$

AD$\frac{7}{2}=\frac{12\sqrt{3}}{2}$

∴ AD = $\frac{12\sqrt{3}}{7}$

Hence, option 1.

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**20. CAT 2002 QA | Geometry - Circles**

In the figure given below, find the distance PQ.

- A.
7 m

- B.
4.5 m

- C.
10.5 m

- D.
6 m

Answer: Option A

**Explanation** :

BC^{2} = AB^{2} + AC^{2} = 15^{2} + 20^{2} = 625

BC = 25

Let BD = x, so DC = 25 − x

In ΔABD,

BD^{2} + AD^{2} = AB^{2}

x^{2} + AD^{2} = 225 ... (i)

In ΔADC,

AD^{2} + (25 − x)^{2} = 400

∴ AD^{2} + 625 − 50x + x^{2} = 400

∴ (AD^{2} + x^{2}) + 625 − 50x = 400 ... (ii)

Substituting (i) in (ii),

225 + 625 − 50x = 400

∴ x = 9

BD = 9 and DC = 16

We use the formula A = r × S, where A is the area of a triangle, r is the inradius and S is the semi-perimeter to find the radii of the two circles.

The radius of the circle inscribed in ∆ABD = 3 m

The radius of the circle inscribed in ∆ADC = 4 m

∴ PQ = radius of ∆ABC + radius of ∆ADC = 7 m

Hence, option 1.

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**21. CAT 2002 QA | Algebra - Number Theory**

The remainder when 2^{256} is divided by 17 is

- A.
7

- B.
13

- C.
11

- D.
1

Answer: Option D

**Explanation** :

2^{4} = −1 (mod 17)

∴ (24)^{64} = [−1 (mod 17)]^{64} = (−1)^{64} = 1

Hence, option 4.

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**22. CAT 2002 QA | Algebra - Progressions**

Let S = 2x + 5x^{2} + 9x^{3} + 14x^{4} + 20x^{5} ... infinity (x < 1)

The coefficient of nth term = $\frac{n(n+3)}{2}.$ The sum is

- A.
$\frac{x(2-x)}{{(1-x)}^{3}}$

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

- C.
$\frac{x(2-x)}{{(1-x)}^{2}}$

- D.
None of these

Answer: Option A

**Explanation** :

S = 2x + 5x^{2} + 9x^{3} + 14x^{4} + 20x^{5} + ... …(i)

Multiplying both sides by x, we get

xS = 2x^{2} + 5x^{3} + 9x^{4} + 14x^{5} + 20x^{6} + ... ...(ii)

Subtracting (ii) from (i),

(1 – x)S = 2x + 3x^{2} + 4x^{3} + 5x^{4} + 6x^{5} + ... ...(iii)

Again multiplying both the sides by x, we get

x(1 – x)S = 2x^{2} + 3x^{3} + 4x^{4} + 5x^{5} + 6x^{6} + ... ...(iv)

Subtracting (iv) from (iii),

(1 – x)S – x(1 – x)S = 2x + x^{2} + x^{3} + x^{4} + x^{5} + ...

(1 – x)^{2}S = 2x + x^{2} + x^{3} + x^{4} + x^{5} +...

(1 – x)^{2}S = x + (x + x^{2} + x^{3} + ... + ∞)

(1 - x)^{2}S = x + $\frac{x}{1-x}$ ∵ S_{∞} = $\frac{a}{1-r}$ for r < 1

∴ (1 - x)^{2}S = $\frac{x-{x}^{2}+x}{1-x}$

∴ S = $\frac{x(2-x)}{{(1-x)}^{3}}$

Hence, option 1.

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**23. CAT 2002 QA | Geometry - Circles**

There is a common chord of 2 circles with radius 15 and 20. The distance between the two centres is 25. The length of the chord is

- A.
48

- B.
24

- C.
36

- D.
28

Answer: Option B

**Explanation** :

In ∆ABC,

AB^{2} + AC^{2} = BC^{2}

∴∠BAC = 90°

Let BP = x

∴ PC = 25 − x

202 = x^{2} + AP^{2} ...(i)

In ∆APC,

225 = AP^{2} + (25 – x)^{2}

225 = AP^{2} + 625 + x^{2} – 50x ...(ii)

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

225 = 625 + 400 – 50x

∴ 50x = 800

∴ x = 16

AP^{2} = 20^{2} – 16^{2} = 12^{2}

∴ AP = 12

∴ Length of the chord = 2 × AP = 24

Hence, option 2.

Workspace:

**24. CAT 2002 QA | Algebra - Simple Equations**

A man received a cheque. The amount in Rs. has been transposed for paise and vice versa. After spending Rs. 5 and 42 paise, he discovered he now had exactly 6 times the value of the correct cheque amount. What amount he should have received?

- A.
Rs. 5.30

- B.
Rs. 6.44

- C.
Rs. 60.44

- D.
Rs. 16.44

Answer: Option B

**Explanation** :

Let the man has received a cheque of x rupees and y paise.

∴ The amount on cheque = (100x + y) …(i)

The amount actually received by him = 100y + x

After spending Rs. 5 and 42 paise, the remaining amount = (100y + x – 542) …(ii)

But, (100y + x – 542) = 6 × (100x + y) …(iii)

Substituting the values from the given options, x = 6 and y = 44

Hence, option 2.

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**25. CAT 2002 QA | Algebra - Functions & Graphs**

Suppose, for any real number x, [x] denotes the greatest integer less than or equal to x. Let L(x, y) = [x] + [y] + [x + y] and R(x, y) = [2x] + [2y]. Then it’s impossible to find any two positive real numbers x and y for which of the following?

- A.
L(x, y) = R(x, y)

- B.
L(x, y) ≠ R(x, y)

- C.
L(x, y) < R(x, y)

- D.
L(x, y) > R(x, y)

Answer: Option D

**Explanation** :

Assume the values for x and y.

Let x = 1.1 and y = 2.1

∴ L(x, y) = 6 and R(x, y) = 6

Let x = 1.1 and y = 2.5

L(x, y) = 6 and R(x, y) = 7

Thus, options (1), (2) and (3) are eliminated.

Hence, option 4.

Workspace:

Sum of first n natural numbers = S(n)

Sum given by student = 575

S(10) = $\frac{10\times 11}{2}=$ 55

S(20) = $\frac{20\times 21}{2}=$ 210

S(30) = $\frac{30\times 31}{2}=$ 465

S(40) = $\frac{40\times 41}{2}=$ 820

∴ The student stopped counting somewhere between 30 and 40.

Consider S(35) = $\frac{36\times 35}{2}=$ 630

The student stopped somewhere before 35.

∴ S(31) = 496, S(32) = 528, S(33) = 561 and S(34) = 595

But the student gave 575 as the sum, so the student missed on the number 20.

Hence, option 4.

**26. CAT 2002 QA | Algebra - Progressions**

A student finds the sum 1 + 2 + 3 + ... as his patience runs out. He found the sum as 575. When the teacher declared the result wrong, the student realized that he missed a number. What was the number the student missed?

- A.
16

- B.
18

- C.
14

- D.
20

Answer: Option D

**Explanation** :

Sum of first n natural numbers = S(n)

Sum given by student = 575

S(10) = $\frac{10\times 11}{2}$ = 55

S(20) = $\frac{20\times 21}{2}$= 210

S(30) = $\frac{30\times 31}{2}$= 465

S(40) = $\frac{40\times 41}{2}$= 820

∴ The student stopped counting somewhere between 30 and 40.

Consider S(35) = $\frac{36\times 35}{2}$= 630

The student stopped somewhere before 35.

∴ S(31) = 496, S(32) = 528, S(33) = 561 and S(34) = 595

But the student gave 575 as the sum, so the student missed on the number 20.

Hence, option 4.

Workspace:

**27. CAT 2002 QA | Algebra - Simple Equations**

A thief was stealing diamonds from a jewellery store. On his way out, he encountered three guards, each was given half of the existing diamonds and two over it by the thief. In the end, he was left with one diamond. How many did the thief steal?

- A.
40

- B.
36

- C.
42

- D.
38

Answer: Option B

**Explanation** :

Assume that the thief had stolen x diamonds.

The 1st watchman got = x/2 + 2

Diamonds left = x − (x/2 + 2) = x/2 – 2

The 2nd watchman got = (x/2 − 2)/2 + 2 = x/4 + 1

Diamonds left = (x/2 − 2) − (x/4 + 1) = x/4 − 3

The 3rd watchman got = (x/4 − 3)/2 + 2 = x/8 + 1/2

Diamonds left = (x/4 − 3) − (x/8 + 1/2) = x/8 − 7/2 = 1

∴ x/8 = 9/2

∴ x = 36

Hence, option 2.

Alternatively,

One may proceed by substituting the options that are even.

Workspace:

**28. CAT 2002 QA | Arithmetic - Average**

Three friends went for a picnic. First brought five apples and the second brought three. The third friend however brought only Rs. 8. What is the share of the first friend?

- A.
8

- B.
7

- C.
1

- D.
None of these

Answer: Option B

**Explanation** :

There are 8 apples available for 3 friends. So, share of apples of each person = 8/3

The 1st person gives out 5 − 8/3 apples = 7/3 apples to 3rd person

The 2nd person gives out 3 − 8/3 apples = 1/3 apples to 3rd person

∴ 3rd person will distribute Rs. 8 in the ratio 7 : 1 to 1st and 2nd person respectively and thus Rs. 7 and Rs. 1 are given to them respectively.

∴ Share of 1st friend is Rs. 7.

Hence, option 2.

Workspace:

**29. CAT 2002 QA | Algebra - Simple Equations**

Amar went for a holiday to his friend's place. They together either went for yoga in the morning or played tennis in the evening but not both. 14 mornings and 24 evenings, they both stayed home and they both went out together for 22 days. How many days did Amar stay at his friend's place?

- A.
20

- B.
16

- C.
30

- D.
40

Answer: Option C

**Explanation** :

Let y be the number of days that they went for Yoga, t be the number of days that they played Tennis and f be the number of days that they were Free.

On 14 mornings they did not do anything, then they either played tennis in the evening or did nothing.

∴ 14 = t + f ...(i)

Similarly, 24 = y + f ...(ii)

Also, 22 = t + y …(iii)

Adding (i) and (ii) and substituting in (iii), we get,

2f = 16

∴ f = 8 days

∴ Total number of days Amar stayed = 22 + f = 22 + 8 = 30 days

Hence, option 3.

Workspace:

**30. CAT 2002 QA | Geometry - Coordinate Geometry**

The area of the triangle with the vertices (a, a), (a + 1, a) and (a, a + 2) is

- A.
a

^{3} - B.
1

- C.
0

- D.
None of these

Answer: Option B

**Explanation** :

Area (∆) = $\left|\begin{array}{ccc}{x}_{1}& {y}_{1}& 1\\ {x}_{2}& {y}_{2}& 2\\ {x}_{3}& {y}_{3}& 3\end{array}\right|$

where *x*_{1} = *y*_{1} = *a*,* x*_{2} = *a* + 1, *y*_{2} = *a*, *x*_{3} = *a* and *y*_{3} = *a* + 2

∴ Area (∆) = $\frac{1}{2}\left|\begin{array}{ccc}a& a& 1\\ a+1& a& 1\\ a& a+12& 1\end{array}\right|$

∴ Area (∆) = $\frac{1}{2}${a[a(1) - (a + 2)(1)] - a[(a + 1)(1) - a(1)] + 1[(a + 1)(a + 2) - a^{2}]} = $\frac{1}{2}$ {- 2a - a + 3a + 2} = 1

Hence, option 2.

Workspace:

**31. CAT 2002 QA | Arithmetic - Time, Speed & Distance**

On a 20 km tunnel connecting two cities A and B there are three gutters. The distance between gutter 1 and 2 is half the distance between gutter 2 and 3. The distance from city A to its nearest gutter, gutter 1 is equal to the distance of city B from gutter 3. On a particular day the hospital in city A receives information that an accident has happened at the third gutter. The victim can be saved only if an operation is started within 40 minutes. An ambulance started from city A at 30 km/hr and crossed the first gutter after 5 minutes. If the driver had doubled the speed after that, what is the maximum amount of time the doctor would get to attend the patient at the hospital? Assume 1 minute is elapsed for taking the patient into and out of the ambulance.

- A.
4 minutes

- B.
2.5 minutes

- C.
1.5 minutes

- D.
Patient died before reaching the hospital

Answer: Option C

**Explanation** :

As per the description,

AB = 20 km

AG_{1} = BG_{3}

2G_{1}G_{2} = G_{2}G_{3}

∴ x = 2.5 km

y + 2y = 20 − 2x

∴ y = 5 km

∴ Time to cover A to G_{3} = $\frac{1}{12}+\frac{15}{60}$ = 20 min

∴ While coming back his speed is 60 km/hr.

∴ Time taken to cover the distance from G_{3} to A (i.e. 17.5 km) = 17.5 minutes

∴ Required time = 20 + 17.5 + 1 = 38.5 minutes

∴ The doctor will have 1.5 minutes to attend the patient.

Hence, option 3.

Workspace:

**32. CAT 2002 QA | Geometry - Mensuration**

Neeraj has a rectangular field of size 20 × 40 sq.m. He has to mow the field with a mowing machine of width 1 m. If he mows the field from the extremes to the centre, then the number of rounds taken by him to mow half of the field will be

- A.
3.5

- B.
3.8

- C.
3

- D.
4

Answer: Option B

**Explanation** :

The movement of the mower may be considered as shown in the figure.

Total area of the plot = 20 × 40 = 800 sq.m.

After 1st mowing, area left = (20 – 2) × (40 – 2) = 684 sq.m.

After 2nd mowing, area left = (20 – 4) × (40 – 4) = 576 sq.m.

After 3rd mowing, area left = (20 – 6) × (40 – 6) = 476 sq.m.

After 4th mowing, area left = (20 – 8) × (40 – 8) = 384 sq.m.

∴ Half the area is mowed between 3 to 4 rounds.

∵ 476 − 384 = 92 sq.m.

$3+\left[\frac{(476-400)}{92}\right]=3+\left(\frac{76}{92}\right)=3.8$

Hence, option 2.

Workspace:

**33. CAT 2002 QA | Geometry - Mensuration**

On the corners of a square field of side 14 metres, 4 horses are tethered in such a way the adjacent horses just reach each other. There is a circular pond of area 20 sq.m. in the centre of the square. What is the area left ungrazed?

- A.
154 sq. m

- B.
22 sq. m

- C.
120 sq. m

- D.
None of these

Answer: Option B

**Explanation** :

Side of square field = 14 m

∴ Area of square field = 196 sq.m.

Area of the circle = 20 sq.m.

Area grazed by horses = $4\times \frac{90\xb0}{360\xb0}$ × π × 7^{2}

= 22 × 7

= 154 sq.m.

∴ Area capable of being grazed = 196 – 20 = 176 sq.m.

∴ Area left ungrazed = 176 – 154 = 22 sq.m.

Hence, option 2.

Workspace:

**34. CAT 2002 QA | Modern Math - Permutation & Combination**

How many numbers between 0 and one million can be formed using 0, 7 and 8?

- A.
486

- B.
1086

- C.
728

- D.
None of these

Answer: Option C

**Explanation** :

Numbers possible with 0, 7, 8:

Possible single-digit numbers = 2

Possible 2-digit numbers = 2 × 3 = 6

Possible 3-digit numbers = 2 × 3 × 3 = 18

Possible 4-digit numbers = 2 × 3^{3} = 54

Possible 5-digit numbers = 2 × 3^{4} = 162

Possible 6-digit numbers = 2 × 3^{5} = 486

Total possible numbers = 486 + 162 + 54 + 18 + 6 + 2 = 728

Hence, option 3.

Workspace:

**35. CAT 2002 QA | Modern Math - Permutation & Combination**

In how many ways, we can choose a black and a white square on a chess board such that the two are not in the same row or column?

- A.
32

- B.
96

- C.
24

- D.
None of these

Answer: Option D

**Explanation** :

The number of ways of selecting 1 black square is 32.

∵ 8 white squares in the corresponding row and column cannot be selected.

∴ The number of ways of selecting the white square is 24.

∴ Total number of required ways = 32 × 24 = 768

Hence, option 4.

Workspace:

**36. CAT 2002 QA | Algebra - Number Theory**

A rich merchant had collected many gold coins. He did not want anybody to know about them. One day, his wife asked. "How many gold coins do we have?" After pausing a moment, he replied, "Well! If I divide the coins into two unequal numbers, then 48 times the difference of the numbers is equal to the difference of their squares. The wife looked puzzled. Can you help the merchant's wife by finding out how many gold coins the merchant has?

- A.
48

- B.
96

- C.
32

- D.
36

Answer: Option A

**Explanation** :

Let the two unequal numbers be x and y.

∴ 48(x – y) = x^{2} – y^{2}

∴ 48(x – y) = (x + y) (x – y)

∴ (x + y) = 48

∴ Total number of coins is 48.

Hence, option 1.

Workspace:

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

In the diagram below, ∠ABC = 90° = ∠DCH = ∠DOE = ∠EHK = ∠FKL = ∠GLM = ∠LMN, AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN

**37. CAT 2002 QA | Geometry - Mensuration**

The magnitude of ∠FGO =

- A.
30°

- B.
45°

- C.
60°

- D.
None of these

Answer: Option D

**Explanation** :

Given that AB = BC = 2CH = 2CD = EH = EK = 2HK = 4KL = 2LM = MN

And EO = FP

Also,

2CD = EH

EO = FP = CD

∴ KL = PG = $\frac{CD}{2}$

FP = CD : PG = $\frac{CD}{2}$ : ∠FPG = 90°

∵ The angle are proportionate to the sides opposite to the angles.

∴ ∠FGO = ∠FGP = tan^{–1 }2

Hence, option 4.

Workspace:

**38. CAT 2002 QA | Geometry - Mensuration**

The ratio of the areas of the two quadrangles ABCD and DEFG is

- A.
1 : 2

- B.
2 : 1

- C.
12 : 7

- D.
None of these

Answer: Option C

**Explanation** :

Area of trapezium ABCD = $\frac{1}{2}$ BC (CD + AB)

Also BC = 2CD, AB = BC = 2CD

Area of trapezium ABCD = $\frac{1}{2}$ × CD (CD + 2DC) = 3CD^{2}

Area of trapezium DEFG = $\frac{1}{2}$ EO(EF + DG)

EO = CD

EF = CD

DG = CH + HK + KL = CD + CD + $\frac{CD}{2}=\frac{5}{2CD}$

Area of trapezium DEFG = $\frac{1}{2}\times CD\left(CD+\frac{5}{2}CD\right)=\frac{1}{2}\times CD\left(\frac{7CD}{2}\right)=\frac{7C{D}^{2}}{4}...\left(ii\right)$

Ratio of ABCD : DEFG = 12 : 7

Hence, option 3.

Workspace:

**39. CAT 2002 QA | Algebra - Simple Equations**

If X_{n} = (-1)^{n}X_{n-1} and X_{0} = x, then

- A.
X

_{n}is positive for n = even - B.
X

_{n}is negative for n = even - C.
X

_{n}is positive for n = odd - D.
None of these

Answer: Option D

**Explanation** :

It is given to us that X_{0} = x

∴ X_{1} = −x

X_{2} = −x

X_{3} = x

X_{4} = x

There is no trend for odd or even X_{n}.

∴ No concrete statement can be made.

Hence, option 4.

Workspace:

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

There are 11 alphabets A, H, I, M, O, T, U, V, W, X, Y. They are called symmetrical alphabets. The remaining alphabets are known as asymmetrical alphabets.

**40. CAT 2002 QA | Modern Math - Permutation & Combination**

How many four-lettered passwords can be formed by using symmetrical letters only? (repetitions not allowed)

- A.
1086

- B.
255

- C.
7920

- D.
None of these

Answer: Option C

**Explanation** :

4 out of the 11 symmetrical alphabets need to be selected,

Also the order of alphabets needs to be accounted for (as it is a password)

∴ ^{11}P_{4} = $\frac{11!}{7!}$ = 11 × 10 × 9 × 8 = 720 × 11 = 7920

Hence, option 3.

Workspace:

**41. CAT 2002 QA | Modern Math - Permutation & Combination**

How many three-lettered words can be formed such that at least one symmetrical letter is there?

- A.
12870

- B.
18330

- C.
16420

- D.
None of these

Answer: Option D

**Explanation** :

Number of asymmetrical letters = 15

Total number possible of words = 26 × 26 × 26 = 17576

Number of words without any symmetrical letters = 15 × 15 × 15 = 3375

Number of words with at least one symmetrical letter = 17576 − 3375 = 14201

Hence, option 4.

Workspace:

**42. CAT 2002 QA | Arithmetic - Time & Work**

It takes 6 technicians a total of 10 hours to build a new server from Direct Computer, with each working at the same rate. If six technicians start to build the server at 11 a.m. and one technician per hour is added beginning at 5 p.m., at what time will the server be complete?

- A.
6:40 p.m.

- B.
7:00 p.m.

- C.
7:20 p.m.

- D.
8:00 p.m.

Answer: Option D

**Explanation** :

6 technicians → 10 hours to 1 job

∴ 1 technician → 60 hours to 1 job

∴ In 6 hrs (11 a.m. - 5 p.m.)6 technicians → ${\frac{3}{5}}^{th}$ of the work.

In the hour after 5p.m.,

7 technicians → ${\frac{7}{60}}^{th}$ of the job.

In the nect hour, 9 technicians will complete ${\frac{9}{60}}^{th}$ of the job.

∴ In 3 hours after 5:00 p.m. → $\frac{24}{60}={\frac{2}{5}}^{th}$ of the job is complete.

∴ At the end of 9^{th} hour i.e. 8:00 p.m. the work is complete.

Hence, option 4.

Workspace:

**43. CAT 2002 QA | Algebra - Quadratic Equations**

Davji shop sells samosas in boxes of different sizes. The samosas are priced at Rs. 2 per samosa upto 200 samosas. For every additional 20 samosas, the price of the whole lot goes down by 10 paisa per samosa. What should be the maximum size of the box that would maximize the revenue?

- A.
240

- B.
300

- C.
400

- D.
None of these

Answer: Option B

**Explanation** :

Consider that N extra packets of 20 samosas each are bought.

∴ Total cost, C = (200 + 20N) × (2 – 0.1N)

= 400 − 20N + 40N − 2N^{2}

= 400 + 20N − 2N^{2}

Maxima (C) = 20 − 4N = 0

∴ N = 5

∴ Revenue will be maximum at N = 5

∴ Total number of samosas = 200 + 20(5) = 300

Hence, option 2.

Workspace:

**44. CAT 2002 QA | Arithmetic - Time & Work**

Three small pumps and one large pump are filling a tank. Each of the three small pumps works at 2/3rd the rate of the large pump. If all 4 pumps work at the same time, then they should fill the tank in what fraction of time that it would have taken the large pump alone?

- A.
$\frac{4}{7}$

- B.
$\frac{1}{3}$

- C.
$\frac{2}{3}$

- D.
$\frac{3}{4}$

Answer: Option B

**Explanation** :

Work done by the large pump = 3 units

∴ Work done by the 3 small pumps = 2 × 3 = 6 units

∴ Work done by all 4 pumps together = 9 units

$\therefore \frac{\mathrm{Rate}\mathrm{of}\mathrm{work}\mathrm{of}\mathrm{the}\mathrm{large}\mathrm{pump}}{\mathrm{Rate}\mathrm{of}\mathrm{work}\mathrm{of}\mathrm{all}4\mathrm{pumps}}=\frac{3}{9}=\frac{1}{3}$

Rate of work is inversely proportional to time taken for completing the work.

$\therefore \frac{\mathrm{TIme}\mathrm{taken}\mathrm{by}\mathrm{the}\mathrm{large}\mathrm{pump}}{\mathrm{Time}\mathrm{taken}\mathrm{by}\mathrm{all}4\mathrm{pumps}}=\frac{3}{1}$

Hence, option 2.

Workspace:

**45. CAT 2002 QA | Algebra - Surds & Indices**

If pqr = 1 then $\frac{1}{1+p+{q}^{-1}}+\frac{1}{1+q+{r}^{-1}}+\frac{1}{1+r+{p}^{-1}}$ is equivalent to

- A.
p + q + r

- B.
$\frac{1}{p+q+r}$

- C.
1

- D.
p

^{-1}+ q^{-1}+ r^{-1}

Answer: Option C

**Explanation** :

∵ pqr = 1

Case 1:

Let p = 2/3, q = 3/2 and r = 1

Substituting the values of p, q, r, we get,

$\frac{1}{1+p+{q}^{-1}}+\frac{1}{1+q+{r}^{-1}}+\frac{1}{1+r+{p}^{-1}}=\frac{3}{7}+\frac{2}{7}+\frac{2}{7}=1$

Case 2:

Let p = 1, q = 1 and r = 1

Substituting the values of p, q, r, we get,

$\frac{1}{1+p+{q}^{-1}}+\frac{1}{1+q+{r}^{-1}}+\frac{1}{1+r+{p}^{-1}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1$

Hence, option 3.

Workspace:

**46. CAT 2002 QA | Arithmetic - Time, Speed & Distance**

There is a tunnel connecting city A and B. There is a CAT which is standing at 3/8 the length of the tunnel from A. It listens a whistle of the train and starts running towards the entrance where, the train and the CAT meet. In another case, the CAT started running towards the exit and the train again met the CAT at the exit. What is the ratio of their speeds?

- A.
4 : 1

- B.
1 : 2

- C.
8 : 1

- D.
None of these

Answer: Option A

**Explanation** :

Solution:

Let the speed of the train be St and the speed of the CAT be S_{c}.

Case 1:

The CAT runs towards the train.

$\frac{d}{{S}_{t}}=\frac{3x}{8{S}_{c}}$

$\therefore \frac{3x}{8}\times \frac{{S}_{t}}{{S}_{c}}=d$ ...(i)

Case 2:

The CAT runs away from the train.

$\frac{d+x}{{S}_{t}}=\frac{5x}{8{S}_{c}}$

$\therefore d=\frac{5x}{8}\times \frac{{S}_{t}}{{S}_{C}}-x$ ...(ii)

Equating (i) and (ii),

$\frac{3x}{8}\times \frac{{S}_{t}}{{S}_{c}}=\frac{5x}{8}\times \frac{{S}_{t}}{{S}_{c}}-x\phantom{\rule{0ex}{0ex}}\therefore \frac{3}{8}x\times \frac{{S}_{t}}{{S}_{c}}=\frac{5x}{8}\times \frac{{S}_{t}}{{S}_{c}}-x\phantom{\rule{0ex}{0ex}}\therefore \frac{3}{8}\frac{{S}_{t}}{{S}_{c}}=\frac{5}{8}\frac{{S}_{t}}{{S}_{c}}-1$

∴ S_{t} : S_{c} = 4 : 1

Hence, option 1.

Workspace:

**47. CAT 2002 QA | Algebra - Number Theory**

In a book store, the words of the glowsign board "MODERN BOOK STORES" are individually flashed after 5/2, 17/4 and 41/8 seconds respectively. Each word is put off after a second. What is the least time after which full name of the book store can be read?

- A.
73.5 seconds

- B.
79.4 seconds

- C.
68.2 seconds

- D.
None of these

Answer: Option A

**Explanation** :

The words MODERN, BOOK and STORES appear after 5/2, 17/4 and 41/8 seconds respectively and are put off after 1 second.

So, if all the words flashed together at t = 0 seconds, then

MODERN will flash after (5/2 + 1) = 7/2 seconds

BOOK will flash after (17/4 + 1) = 21/4 seconds

STORES will flash after (41/8 + 1) = 49/8 seconds.

So, all the words will flash together again after LCM(7/2, 21/4, 49/8) seconds.

∴ L.C.M. $\left(\frac{7}{2},\frac{21}{4},\frac{49}{8}\right)=\frac{\mathrm{L}.\mathrm{C}.\mathrm{M}.\mathrm{of}\mathrm{Numerators}}{\mathrm{H}.\mathrm{C}.\mathrm{F}.\mathrm{of}\mathrm{Denominators}}=\frac{147}{2}=$ 73.5 seconds

Hence, option 1.

Workspace:

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

A boy is supposed to put a mango into a basket if ordered 1, an orange if ordered 2 and an apple if ordered 3. He took out 1 mango and 1 orange if ordered 4. He was given the following sequence of orders.

12332142314223314113234

**48. CAT 2002 QA | Miscellaneous**

At the end of the sequence, what will be the number of oranges in the basket?

- A.
2

- B.
3

- C.
4

- D.
6

Answer: Option A

**Explanation** :

The sequence,

1 2 3 3 2 1 4 2 3 1 4 2 2 3 3 1 4 1 1 3 2 3 4

Number of oranges put in = Number of times 2 was ordered = 6

Number of oranges taken out = Number of times 4 was ordered = 4

Number of oranges at the end of the sequence = 6 – 4 = 2

Hence, option 1.

Workspace:

**49. CAT 2002 QA | Miscellaneous**

At the end of the sequence, what will be the total number of fruits in the basket?

- A.
10

- B.
11

- C.
13

- D.
17

Answer: Option B

**Explanation** :

Number of fruits put in the basket = Number of times of orders 1, 2 or 3 = 19

Number of fruits taken out = 2 fruits taken out for every order for 4 = 2 × 4 = 8

Number of fruits in the basket after the sequence = 19 – 8 = 11

Hence, option 2.

Workspace:

**50. CAT 2002 QA | Data Sufficiency**

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

**Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3)**if the question can be solved by using both the statements together and not by any one of them.

**Answer (4)**if the question cannot be solved with the help of the given data and more data is required.

In a hockey match, the Indian team was behind by 2 goals with 5 minutes remaining. Did they win the match?

A. Deepak Thakur, the Indian striker scored 3 goals in the last 5 minutes of the match.

B. Korea scored a total of 3 goals in the match.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

From statement A alone, it is not clear whether Korea too scored in the last 5 minutes or not.

From statement B alone, it is not clear whether India scored in the last 5 minutes or not.

Using both the statements, there are two cases possible:

Case 1:

In this case, the match ended in a draw.

Case 2:

In this case, India wins the match.

∴ Even after combining both the statements, we cannot say whether India won the match or not.

Hence, option 4.

Workspace:

**51. CAT 2002 QA | Algebra - Number Theory | Data Sufficiency**

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

**Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3)**if the question can be solved by using both the statements together and not by any one of them.

**Answer (4)**if the question cannot be solved with the help of the given data and more data is required.

Four students were added to a dance class. Would the teacher be able to divide her students evenly into a dance team (or teams) of 8?

A. If 12 students were added, then the teacher could put everyone in teams of 8 without any left overs.

B. The number of students in the class is currently not divisible by 8.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Consider statement A:

Let x be the total number of students present in the class.

∵ (x + 12) − (x + 4) = 8

∴ If (x + 12) is divisible by 8, then (x + 4) too is divisible by 8.

Statement A alone is sufficient to answer the question.

From statement B alone, we cannot conclude anything.

Hence, option 1.

Workspace:

**52. CAT 2002 QA | Algebra - Number Theory | Data Sufficiency**

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

**Answer (1) if the question can be solved by any one of the statements, but not the other one.
Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3)**if the question can be solved by using both the statements together and not by any one of them.

**Answer (4)**if the question cannot be solved with the help of the given data and more data is required.

Is x = y?

A. (x + y) $\left(\frac{1}{x}+\frac{1}{y}\right)$ = 4

B. (x − 50)^{2} = (y − 50)^{2}

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

**Consider statement A:**

$(x+y)\frac{(x+y)}{xy}=4$

∴ (x + y)^{2} = 4xy

∴ x^{2} + y^{2} + 2xy − 4xy = 0

∴ (x − y)^{2} = 0

∴ x = y

∴ Statement A alone is sufficient.

**Consider statement B:**

(x – 50)^{2} = (y – 50)^{2}

∴ Either (x − 50) = (y − 50) or (x − 50) = −(y − 50)

∴ Either x = y or x + y = 100, which gives infinite values for x and y.

∴ Statement B alone is not sufficient.

Hence, option 1.

Workspace:

**53. CAT 2002 QA | Arithmetic - Profit & Loss | Data Sufficiency**

Answer (2)

**Answer (3) **if the question can be solved by using both the statements together and not by any one of them.

**Answer (4) **if the question cannot be solved with the help of the given data and more data is required.

A dress was initially listed at a price that would have fetched the store a profit of 20% on the wholesale cost. What was the wholesale cost of the dress?

A. After reducing the listed price by 10% the dress was sold for a net profit of 10 dollars.

B. The dress was sold for 50 dollars.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Let 100x be the wholesale cost of the dress.

∴ List price of the dress = 120x

**Consider statement A:**

Selling Price = 0.9 × List Price = 0.9 × 120x = 108x

Now, Selling Price – Cost Price = Profit

∴ (108x) − (100x) = 10

∴ x = 10/8

∴ Wholesale Cost = 100x = Rs. 125

∴ Statement A alone is sufficient.

**Consider statement B:**

This gives the selling price of the dress but it is not mentioned whether any discount is provided on the list price or not.

∴ Statement B alone is not sufficient.

Hence, option 1.

Workspace:

**54. CAT 2002 QA | Arithmetic - Average | Data Sufficiency**

Answer (2)

**Answer (3) **if the question can be solved by using both the statements together and not by any one of them.

**Answer (4) **if the question cannot be solved with the help of the given data and more data is required.

Is 500 the average (arithmetic mean) score of the GMAT?

A. Half of the people who take GMAT score above 500 and half of the people score below 500.

B. The highest GMAT score is 800 and the lowest score is 200.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

**Consider statement A:**

This statement alone does not give sufficient information as we do not know how much above/below 500 these students scored.

For example, it is possible that one student scored 800, one scored 750, the third scored 450, while the last scored 200.

Then, average = (800 + 750 + 450 + 200)/4 = 550 ≠ 500

**Consider statement B:**

Using this statement alone, we cannot say whether 500 is the average score of GMAT. The previous example can be used here as well.

Combining both statements together:

Even now, we cannot determine the average. (Again, the above example can be used.)

Hence, option 4.

Workspace:

**55. CAT 2002 QA | Algebra - Number Theory | Data Sufficiency**

Answer (2)

**Answer (3) **if the question can be solved by using both the statements together and not by any one of them.

**Answer (4) **if the question cannot be solved with the help of the given data and more data is required.

Is |x − 2| < 1?

A. |x| > 1

B. |x − 1| < 2

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

|x − 2| < 1

**Consider statement A:**

|x| > 1

x < −1 or x > 1

For x = 1.5, |x − 2| < 1 is true.

For x = 4, |x − 2| < 1 is false.

∴ Statement A alone is not sufficient.

**Consider statement B:**

|x − 1| < 2

−2 < x − 1 < 2

−1 < x < 3

For −1 < x < 1, |x − 2| < 1 is false.

For 1 < x < 3, |x − 2| < 1 is true.

∴ Statement B alone is not sufficient.

**Consider both the statements:**

We have, 1 < x < 3

For this range, |x − 2| < 1 is true.

∴ Both the statements combined together are sufficient to answer the question.

Hence, option 3.

Workspace:

**56. CAT 2002 QA | Venn Diagram | Data Sufficiency**

**Answer (1) **if the question can be solved by any one of the statements, but not the other one.

**Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3) **if the question can be solved by using both the statements together and not by any one of them.

**Answer (4) **if the question cannot be solved with the help of the given data and more data is required.

Members in a club either speak French or Russian or both. Find the number of members in a club who speak only French.

A. There are 300 members in the club and the number of members who speak both French and Russian is 196.

B. The number of members who speak only Russian is 58.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Let F represent the set of members speaking French and R represent the set of members speaking Russian.

Also, let x represent the number of members speaking only French.

**Consider statement A:**

Here, total number of members = 300

∴ Number of members who speak only Russian = 300 – 196 – x = 104 – x

However, we cannot find the value of x.

∴ Statement A alone is not sufficient.

**Consider statement B:**

We are given that the number of members speaking Russian = 58

∴ Statement B alone is not sufficient.

**Consider both statements together:**

We have, 104 – x = 58

∴ Number of members speaking only French, x = 104 – 58 = 46

∴ The question can be answered using both statements together.

Hence, option 3.

Workspace:

**57. CAT 2002 QA | Arithmetic - Ratio, Proportion & Variation | Data Sufficiency**

**Answer (1) **if the question can be solved by any one of the statements, but not the other one.

**Answer (2) **if the question can be solved by using either of the two statements.

**Answer (3) **if the question can be solved by using both the statements together and not by any one of them.

**Answer (4) **if the question cannot be solved with the help of the given data and more data is required.

A sum of Rs. 38,500 was divided among Jagdish, Punit and Girish. Who received the minimum amount?

A. Jagdish received 2/9 of what Punit and Girish together received.

B. Punit received 3/11 of what Jagdish and Girish together received.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

**Consider statement A:**

Jagdish : (Punit + Girish) = 2 : 9

∴ Jagdish's share = $\frac{2}{11}$ × Total = 18.18% of the total

However, we don’t know the percentage distribution between Punit and Girish.

∴ Statement A alone is not sufficient.

**Consider statement B:**

Punit : (Jagdish + Girish) = 3 : 11

∴ Punit's share = $\frac{3}{14}$ × Total = 21.4% of the total

However, we do not know the percentage distribution between Jagdish and Girish.

∴ Statement B alone is not sufficient.

**Consider both statements together:**

Jagdish = 18.18% of the total amount

Punit = 21.4% of the total amount

Girish = 100 − 18.18 − 21.4 = 60.42% of the total amount

∴ Jagdish received the minimum amount.

Hence, option 3.

Workspace:

**58. CAT 2002 QA | Data Sufficiency**

**Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer**

For what reason Purohit did not get the offer of employment?

Statement:

I. Purohit passed the interview.

II. Purohit's friend passed the medical test who passed the interview along with Purohit.

III. Purohit's father did not want him to take the job.

IV. Purohit has another employment offer from another company.

V. Purohit did not clear the mandatory medical test.

- A.
III and IV only

- B.
III, IV and V only

- C.
I, III and IV only

- D.
V only

Answer: Option D

**Explanation** :

Non-clearance of the mandatory medical test is the reason for not getting the offer of employment.

Hence, option 4.

Workspace:

**59. CAT 2002 QA | Data Sufficiency**

**Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer**

What were the possible reasons due to which DESCO incurred losses for the last two years?

Statement:

I. The company's shares are not registered in the stock exchange.

II. The company does not export its products.

III. The company has an inefficient labour force.

IV. The price of its product has fallen in the last two years due to competitive market.

V. Entry of similar foreign goods at a cheaper rate.

- A.
Only III, IV and V

- B.
Only II, III and IV

- C.
Only IV and V

- D.
Only I and II

Answer: Option A

**Explanation** :

Inefficient labour forces, fall in product price, and entry of similar foreign goods at lower rates are the possible reasons for incurring losses by DESCO.

Hence, option 1.

Workspace:

**60. CAT 2002 QA | Data Sufficiency**

**Each question given below is followed by five statements numbered I, II, III, IV and V. The answer choice given below each question consists of one or more statements. You have to choose the choice which gives more relevant / useful information in answering the question correctly. Read all the statements together with the question and choose your answer**

On which day of the week did Sunil get his letter of promotion?

Statement:

I. Sunil purchased a new shirt on Friday

II. Sunil was given a party that Saturday.

III. Sunil was given the letter of promotion on the day before he purchased the shirt.

IV. Tuesday being his birthday, Sunil gave a party to all his friends.

V. Sunil's friend was promoted on Friday.

- A.
I and II only

- B.
II, III and IV only

- C.
I and III only

- D.
II, III and V only

Answer: Option C

**Explanation** :

Sunil purchased a new shirt on Friday and he got the letter of promotion one day before.

∴ Sunil got his letter of promotion on Thursday.

Hence, option 3.

Workspace:

**61. CAT 2002 QA | Data Sufficiency**

Who among A, B, C, D and E is the heaviest?

Statement:

I. B and C are heavier than A and D.

II. C is heavier than D.

III. C is heavier than A and lighter than B.

IV. E is heavier than B

V. D is lighter than E.

- A.
I, III and IV only

- B.
II, III and V only

- C.
III and IV only

- D.
I, III and V only

Answer: Option A

**Explanation** :

From statement (I), we get,

B > A, B > D, C > A and C > D … (i)

From statement (III), we get,

B > C > A ... (ii)

From statement (IV), we get,

E > B … (iii)

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

E > B > C > A, D

∴ E is the heaviest.

Hence, option 1.

Workspace:

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