# CAT 2001 QA | Previous Year Questions

**1. CAT 2001 QA | Arithmetic - Percentage**

A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, the candidate obtained 60% of the total marks. Then the number of papers in which he got more than 50% marks is

- A.
2

- B.
3

- C.
4

- D.
5

Answer: Option C

**Explanation** :

Let the maximum marks for each of the paper = 100

∴ Total maximum marks = 5 × 100 = 500

Let the marks obtained by the student be 6x, 7x, 8x, 9x and 10x in each of the paper.

∴ 6x + 7x + 8x + 9x + 10x = 60% of 500

40x = 300

∴ x = 7.5

∴ The marks obtained in each of the paper will be 45, 52.5, 60, 67.5 and 75.

∴ In 4 papers he got more than 50% marks.

Hence, option (c).

Alternatively

Average of 6, 7, 8, 9 and 10 = $\frac{6+7+8+9+10}{5}=8$

∵ 60% of marks = 8

∴ 50% of marks = $\frac{8}{60}\times 50$

= 6.67

∴ The score in 4 of the subjects is more than 6.67.

Hence, option (c).

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**2. CAT 2001 QA | Geometry - Mensuration**

A square, whose side is 2 metres, has its corners cut away so as to form an octagon with all sides equal. Then the length of each side of the octagon, in metres is

- A.
$\frac{\sqrt{2}}{\sqrt{2}+1}$

- B.
$\frac{2}{\sqrt{2}+1}$

- C.
$\frac{2}{\sqrt{2}-1}$

- D.
$\frac{\sqrt{2}}{\sqrt{2}-1}$

Answer: Option B

**Explanation** :

Let the side of the octagon be x, which is also the hypotenuse of the triangle in the corner of the square.

∴ TYhe other two sides of the triangle = $\frac{x}{\sqrt{2}}$

$\frac{x}{\sqrt{2}}+x+\frac{x}{\sqrt{2}}=2$

$x=\frac{2}{\sqrt{2}+1}$

Hence, option (b).

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**3. CAT 2001 QA | Algebra - Number Theory**

Let x, y and z be distinct integers. x and y are odd and positive, and z is even and positive. Which one of the following statements cannot be true?

- A.
(x − z)

^{2}y is even - B.
(x − z)

^{2}y is odd - C.
(x − z)y is odd

- D.
(x − y)

^{2}z is even

Answer: Option A

**Explanation** :

Assume some suitable values, for example, x = 1, y = 3 and z = 2

Now, substitute in the options and check.

Only option 1 is false.

Hence, option (a).

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**4. CAT 2001 QA | Algebra - Inequalities & Modulus**

If x > 5 and y < −1, then which of the following statements is true?

- A.
(x + 4y) > 1

- B.
x > − 4y

- C.
−4x < 5y

- D.
None of these

Answer: Option D

**Explanation** :

Assume some values for x and y and substitute in the options and check.

For example, (x, y) = (6, −10), (7, −20) etc. None of the options are satisfied by these values.

Hence, option (d).

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**5. CAT 2001 QA | Algebra - Number Theory**

A red light flashes 3 times per minute and a green light flashes 5 times in two minutes at regular intervals. If both lights start flashing at the same time, how many times do they flash together in each hour?

- A.
30

- B.
24

- C.
20

- D.
60

Answer: Option A

**Explanation** :

The flashing time for red light is = 60/3 = 20 sec

The flashing time for green light = 120/5 = 24 sec

So, they will flash together in LCM (20, 24) = 120 sec, i.e. 2 minutes, which is 60/2 = 30 times in an hour

Hence, option (a).

Alternatively,

Red light flashes 3 times in 1 minute i.e. 3 × 60 = 180 times in 1 hour.

Also, green light flashes 5 times in 2 minutes i.e. 5 × 30 = 150 times in 1 hour.

The number of times they flashed together in 1 hour is the HCF of 180 and 150, which is 30.

Hence, option (a).

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**6. CAT 2001 QA | Algebra - Number Theory**

Of 128 boxes of oranges, each box contains at least 120 and at most 144 oranges. The number of boxes containing the same number of oranges is at least

- A.
5

- B.
103

- C.
6

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Suppose we have five different types of fruits − A, B, C, D, E and seven boxes − 1, 2, 3, 4, 5, 6 and 7. We have to fill each box with only one kind of fruits.

To find the least number of boxes that will have same kind of fruit, we try to fill as many boxes as possible with different fruits. So we may have,

1: A

2: B

3: C

4: D

5: E

6: A

7: B

Therefore, there will be at least two boxes with the same fruits.

Applying a similar logic to the question, we try to fill the maximum number of boxes with different possible number of oranges.

Box 1: 120 oranges, Box 2: 121 oranges, …, Box 25: 144 oranges,

Box 26: 120 oranges, Box 27: 121 oranges, …, Box 50: 144 oranges,

Box 51: 120 oranges, Box 52: 121 oranges, …, Box 75: 144 oranges,

Box 76: 120 oranges, Box 77: 121 oranges, …, Box 100: 144 oranges,

Box 101: 120 oranges, Box 102: 121 oranges, …, Box 125: 144 oranges,

Box 126: 120 oranges, Box 127: 121 oranges, Box 128: 122 oranges

We can see that at least 6 boxes will have the same number of fruits.

Hence, option (c).

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**7. CAT 2001 QA | Geometry - Circles**

A certain city has a circular wall around it, and this wall has four gates pointing north, south, east and west. A house stands outside the city, three km north of the north gate, and it can just be seen from a point nine km east of the south gate. What is the diameter of the wall that surrounds the city?

- A.
6 km

- B.
9 km

- C.
12 km

- D.
None of these

Answer: Option B

**Explanation** :

In the figure given below, E is the north gate, A is the south gate and C is the house which can be seen from the point B. AE is the diameter of the wall that surrounds the city.

DB and AB are the tangents to the circle from B.

∴ DB = AB = 9 km

Also, CD^{2} = CE × CA

x2 = 3 × (3 + AE) = 9 + 3AE ...(i)

Now in ΔABC, (x + 9)^{2} = (3 + AE)^{2} + 9^{2}

∴ x^{2} + 9^{2} + 18x = (3 + AE)^{2} + 9^{2}

∴ x^{2} + 18x = (3 + AE)^{2} ...(ii)

Now, start substituting ‘AE’ from the options.

AE = 9, satisfies the equations

Hence, option (b).

Alternatively,

We have 9, AE + 3 and 9 + x, as the sides of a right-angled triangle. The most common Pythagorean triplet is 9, 12, 15. Using this, AE = 9 and x = 6. Substitute these values in the equations and verify.

Hence, option (b).

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**8. CAT 2001 QA | Geometry - Quadrilaterals & Polygons**

In the above diagram, ABCD is a rectangle with AE = EF = FB. What is the ratio of the area of the triangle CEF and that of the rectangle?

- A.
$\frac{1}{6}$

- B.
$\frac{1}{8}$

- C.
$\frac{1}{9}$

- D.
None of these

Answer: Option A

**Explanation** :

∵ AB = 3 × EF

$\frac{Areaof(\u2206CEF)}{Areaof(\square ABCD)}=\frac{{\displaystyle \frac{1}{2}}\times EF\times CB}{AB\times CB}$

$\frac{Areaof(\u2206CEF)}{Areaof(\square ABCD)}=\frac{{\displaystyle \frac{1}{2}}\times EF\times CB}{3\times EF\times CB}=\frac{1}{6}$

Hence, option (a).

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**9. CAT 2001 QA | Arithmetic - Time & Work**

A can complete a piece of work in 4 days. B takes double the time taken by A, C takes double that of B, and D takes double that of C to complete the same task. They are paired in groups of two each. One pair takes two thirds the time needed by the second pair to complete the work. Which is the first pair?

- A.
A, B

- B.
A, C

- C.
B, C

- D.
A, D

Answer: Option D

**Explanation** :

If A takes 4 days then, B will take 8 days, C will take 16 days and D will take 32 days.

∴ Time taken by A and B together to complete a work $=\frac{(4\times 8)}{8+4}=\frac{8}{3}$ days

∴ Time taken by A and C together to complete a work $=\frac{(4\times 16)}{16+4}=\frac{16}{5}$ days

∴ Time taken by A and D together to complete a work $=\frac{(4\times 32)}{32+4}=\frac{32}{9}$ days

∴ Time taken by A and C together to complete a work $=\frac{(8\times 16)}{8+16}=\frac{16}{3}$ days

$\therefore \frac{32}{9}=\frac{2}{3}\times \frac{16}{3}$

∴ The pairs are (A, D) and (B, C).

Hence, option (d).

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**10. CAT 2001 QA | Algebra - Number Theory**

In a 4-digit number, the sum of the first two digits is equal to that of the last two digits. The sum of the first and last digits is equal to the third digit. Finally, the sum of the second and fourth digits is twice the sum of the other two digits. What is the third digit of the number?

- A.
5

- B.
8

- C.
1

- D.
4

Answer: Option A

**Explanation** :

Let the four digit number be abcd.

∴ a + b = c + d ...(i)

∴ a + d = c ...(ii)

∴ b + d = 2 × (a + c) ...(iii)

Solving these equations, we get,

b = 2d

d = 4a

c = 5a

∴ c has to be a multiple of 5 and a single digit number.

The only possibility is c = 5

Hence, option (a).

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**11. CAT 2001 QA | Algebra - Progressions**

Two men X and Y started working for a certain company at similar jobs on January 1, 1950. X asked for an initial salary of Rs. 300 with an annual increment of Rs. 30. Y asked for an initial salary of Rs. 200 with a rise of Rs. 15 every six months. Assume that the arrangements remained unaltered till December, 1959. Salary is paid on the last day of the month. What is the total amount paid to them as salary during the period?

- A.
Rs. 93,300

- B.
Rs. 93,200

- C.
Rs. 93,100

- D.
None of these

Answer: Option A

**Explanation** :

X’s salary = [3600] + [3600 + 30 × 12] + [3600 + 30 × 12 × 2] + … + [3600 + 30 × 12 × 9]

= [3600 × 10] + [30 × 12(1 + 2 + 3 + … + 9)]

= Rs. 52,200

Y’s salary = [1200 × 20] + [15 × 6(1 + 2 + 3 + … + 19)]

= Rs. 41,100

∴ Sum of X and Y’s salary = Rs. 52,200 + Rs. 41,100 = Rs. 93,300

Hence, option (a).

Alternatively,

X starts at a salary of 300 and ends at a salary = 300 + 30 × 9 = 570

∴ X's average salary = $=\frac{300+570}{2}=435$

Y starts at a salary of 200 and ends at a salary = 200 + 15 × 19 = 485

∴ Y's average salary = $\frac{200+485}{2}=342.5$

∴ X and Y’s total salary = (435 + 342.5) × 12 × 10 = Rs. 93,300

Hence, option (a).

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**12. CAT 2001 QA | Algebra - Number Theory**

Anita had to do a multiplication. Instead of taking 35 as one of the multipliers, she took 53. As a result, the product went up by 540. What is the new product?

- A.
1050

- B.
540

- C.
1440

- D.
1590

Answer: Option D

**Explanation** :

Let the number to be multiplied be z.

∴ 53z – 35z = 540

∴ z = 30

∴ Required product = 30 × 53 = 1590

Hence, option (d).

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**13. CAT 2001 QA | Arithmetic - Percentage**

A college has raised 75% of the amount it needs for a new building by receiving an average donation of Rs. 600 from the people already solicited. The people already solicited represent 60% of the people the college will ask for donations. If the college is to raise exactly the amount needed for the new building, what should be the average donation from the remaining people to be solicited?

- A.
Rs. 300

- B.
Rs. 250

- C.
Rs. 400

- D.
Rs. 500

Answer: Option A

**Explanation** :

Let the total population be p.

Then the amount already received = 0.6p × 600 = 360p

This is 75% (3/4th) of the amount.

∴ Remaining amount (25%) = (360p)/3 = 120p = 120p

∴ Required contribution per head = (120p)/(0.4p) = Rs. 300

Hence, option (a).

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**14. CAT 2001 QA | Algebra - Inequalities & Modulus**

x and y are real numbers satisfying the conditions 2 < x < 3 and –8 < y < –7. Which of the following expressions will have the least value?

- A.
x

^{2}y - B.
xy

^{2} - C.
5xy

- D.
None of these

Answer: Option D

**Explanation** :

Option 2 can easily be eliminated as it will give positive value, while both option 1 and 3 will be negative.

∵ x^{2} will fall in the range 4 < x^{2} < 9.

∵ 5x will fall between 10 < 5x < 15.

∴ 5x > x^{2}

Also, 5x < 6x < 7x < ... and –8 < y < –7

∴ 5xy > 6xy > 7xy > ...

Hence, option (d).

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**15. CAT 2001 QA | Algebra - Quadratic Equations | Algebra - Inequalities & Modulus**

m is the smallest positive integer such that for any integer n > m, the quantity n^{3} – 7n^{2} + 11n – 5 is positive. What is the value of m?

- A.
4

- B.
5

- C.
8

- D.
None of these

Answer: Option B

**Explanation** :

n^{3} − 7n^{2} + 11n − 5

For n = 1, the expression reduces to zero

∴ (n − 1) is a factor.

n^{3} − 7n^{2} + 11n − 5 = n^{3 }− n^{2} − 6n^{2} + 6n + 5n − 5

= n^{2}(n − 1) − 6n(n − 1) + 5(n − 1)

= (n − 1)(n^{2} − 6n + 5)

= (n − 1)(n − 1)(n − 5)

= (n − 1)^{2}(n − 5)

Since, (n − 1)^{2} is always positive.

∴ The expression is positive only when n > 5.

∴ n = 6 and m = 5

Hence, option (b).

Alternatively,

Check using options. If we take m = 4 then n = 5, then we do not get a positive value of the given equation. Therefore, check the next higher value.

n = 6 and m = 5 gives the positive value.

Hence, option (b).

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**16. CAT 2001 QA | Geometry - Trigonometry**

A ladder leans against a vertical wall. The top of the ladder is 8 m above the ground. When the bottom of the ladder is moved 2 m farther away from the wall, the top of the ladder rests against the foot of the wall. What is the length of the ladder?

- A.
10 m

- B.
15 m

- C.
20 m

- D.
17 m

Answer: Option D

**Explanation** :

Let the foot of the ladder be x metres away from the foot of the wall.

Now, the length of the ladder will be = x + 2

∴ (x + 2)^{2} = x^{2} + 8

On solving, we get, x = 15

∴ Length of ladder = 15 + 2 = 17 m

Hence, option (d).

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**17. CAT 2001 QA | Arithmetic - Ratio, Proportion & Variation**

Three friends, returning from a movie, stopped to eat at a restaurant. After dinner, they paid their bill and noticed a bowl of mints at the front counter. Sita took 1/3 of the mints, but returned four because she had a momentary pang of guilt. Fatima then took 1/4 of what was left but returned three for similar reasons. Eswari then took half of the remainder but threw two back into the bowl. The bowl had only 17 mints left when the raid was over. How many mints were originally in the bowl?

- A.
38

- B.
31

- C.
41

- D.
None of these

Answer: Option D

**Explanation** :

Sita takes 1/3rd of the total mints which implies that the total number of mints in the bowl should be a multiple of 3. None of the options is a multiple of 3.

Hence, option (d).

Alternatively

This problem can be best solved by working backwards.

Number of mints before Eswari = (17 − 2) × 2 = 30

∴ Number of mints before a Fatima = $\frac{(30-3)\times 4}{3}=36$

∴ Number of mints before Sita = $\frac{(36-4)\times 3}{2}=48$

∴ Total number of mints in the bowl = 48

Hence, option (d).

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**18. CAT 2001 QA | Algebra - Number System**

In a number system the product of 44 and 11 is 1034. The number 3111 of this system, when converted to the decimal number system, becomes

- A.
406

- B.
1086

- C.
216

- D.
691

Answer: Option A

**Explanation** :

Let the base be b.

∴ (4b + 4)(b + 1) = b^{3} + 3b + 4

∴ b^{3} − 4b^{2} − 5b = 0

Solving we get, b = 0, –1, 5

∵ Base cannot be zero or negative, so base is 5.

∴ (3111)_{5} = 3 × 125 + 25 + 5 + 1 = 406

Hence, option (a).

Alternatively,

Let the required base be x.

We know that the answer of 44 × 11 in the required base is 1034.

∴ As 4 occurs in the product, we can say that the base is greater than 4.

Also, 44 × 11 = 484 in base 10.

As 1034 > 484, the base is lesser than 10.

So we can represent the multiplication as follows:

44 × 11 = 44(x + 1) = 44x + 44 = 1034

4 + 4 → 3 or 13 or 23 …

As the base is less than 10, 4 + 4, which is 8 in base 10 cannot be expressed as 3 in the required base.

∴ 4 + 4 → 13 or 23…

4 + 4 → 13

∴ 8 → 13

∴ 8 → x + 3

∴ x = 5

If 4 + 4 → 23

∴ 8 → 23

∴ 8 → x + 13

∴ x = –5, which is not possible.

∴ (3111)_{5} = 1 × 5^{0} + 1 × 5^{1} + 1 × 5^{2} + 3 × 5^{3} = (406)_{10}

Hence, option (a).

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**19. CAT 2001 QA | Arithmetic - Time, Speed & Distance**

At his usual rowing rate, Rahul can travel 12 miles downstream in a certain river in six hours less than it takes him to travel the same distance upstream. But if he could double his usual rowing rate for this 24 miles round trip, the downstream 12 miles would then take only one hour less than the upstream 12 miles. What is the speed of the current in miles per hour?

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

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

- C.
$\frac{5}{3}$

- D.
$\frac{8}{3}$

Answer: Option D

**Explanation** :

Let the speed of Rahul in still water be p and speed of current be q miles per hour.

$\therefore \frac{12}{(p+q)}=\left\{\frac{12}{(p-q)}-6\right\}$ ...(i)

$\therefore \frac{12}{(2p+q)}=\left\{\frac{12}{(2p-q)}-1\right\}$ ...(ii)

12_{p} − 12_{q} = 12_{p }+ 12_{q} − 6(p − q)(p + q)

On solving, we get,

4q = p^{2} − q^{2} … (iii)

24p – 12q = 24p + 12q – (4p^{2} − q^{2})

On solving we get,

4p^{2} − q^{2} = 24q … (iv)

Solving (iii) and (iv), we get,

24q = 9q^{2}

$q=\frac{8}{3}$

Hence, option (d).

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**20. CAT 2001 QA | Algebra - Simple Equations**

Every ten years the Indian government counts all the people living in the country. Suppose that the director of the census has reported the following data on two neighbouring villages Chota hazri and Mota hazri:

Chota hazri has 4,522 fewer males than Mota hazri.

Mota hazri has 4,020 more females than males.

Chota hazri has twice as many females as males.

Chota hazri has 2,910 fewer females than Mota hazri.

What is the total number of males in Chota hazri?

- A.
11264

- B.
14174

- C.
5632

- D.
10154

Answer: Option C

**Explanation** :

Let a = no. of males in Chota hazri, b = no. of males in Mota hazri, x = no. of females in Chota hazri and y = no. of females in Mota hazri

∴ a + 4522 = b ...(i)

∴ y = b + 4020 ...(ii)

∴ x = 2a ...(iii)

∴ x = y – 2910 ...(iv)

Using (iv), (ii) and (iii), we have,

2a = b + 4020 – 2910

Substituting from (i), we have,

2a = a + 4522 + 4020 – 2910

a = 5632

Hence, option (c).

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**21. CAT 2001 QA | Arithmetic - Average**

Three math classes; X, Y, and Z, take an algebra test.

The average score in class X is 83.

The average score in class Y is 76.

The average score in class Z is 85.

The average score of all students in classes X and Y together is 79.

The average score of all students in classes Y and Z together is 81.

What is the average for all three classes?

- A.
81

- B.
81.5

- C.
82

- D.
84.5

Answer: Option B

**Explanation** :

Let there be x number of students in class X, y number of students in class Y and z number of students in class Z.

∴ 83x + 76y = 79(x + y)

∴ 4x = 3y

Similarly,

76y + 85z = 81(y + z)

∴ 4z = 5y

∴ 20x = 15y = 12z

∴ x : y : z = 3 : 4 : 5

∴ The average of all the three classes

$=\frac{83\times 3+76\times 4+85\times 5}{3+4+5}=81.5$

Hence, option (b).

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**22. CAT 2001 QA | Geometry - Mensuration**

Two sides of a plot measure 32 metres and 24 metres and the angle between them is a perfect right angle. The other two sides measure 25 metres each and the other three are not right angles.

What is the area of the plot?

- A.
768

- B.
534

- C.
696.5

- D.
684

Answer: Option D

**Explanation** :

$AC=\sqrt{{24}^{2}+{23}^{2}}=40$

Let us draw DE, which is perpendicular to AC.

Since ΔADC is isosceles, therefore DE will be its median and altitude both.

∴ DE = $\sqrt{{25}^{2}-{20}^{2}}=15$

∴ Required area = $\left(\frac{1}{2}\times 24\times 32\right)+\left(\frac{1}{2}\times 40\times 15\right)$ = 384 + 300 = 684

Hence, option (d).

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**23. CAT 2001 QA | Algebra - Progressions**

All the page numbers from a book are added, beginning at page 1. However, one page number was mistakenly added twice. The sum obtained was 1000. Which page number was added twice?

- A.
44

- B.
45

- C.
10

- D.
12

Answer: Option C

**Explanation** :

First we have to find the total number of pages in the book. For this we need to find the sum of the first few natural numbers, such that we reach to a number just short of 1000.

By trial and error, we can find that,

Sum of the first 44 natural numbers = $=\frac{44\left(45\right)}{2}$ = 90

∵ 990 + 10 = 1000

∴ The page number added twice was 10.

Hence, option (c).

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**24. CAT 2001 QA | Arithmetic - Time, Speed & Distance**

Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constant speed. Shyama takes three steps for every two of Vyom's steps. Shyama gets to the top of the escalator after having taken 25 steps. While Vyom (because his slower pace lets the escalator do a little more of the work) takes only 20 steps to reach the top. If the escalator were turned off, how many steps would they have to take to walk up?

- A.
40

- B.
50

- C.
60

- D.
80

Answer: Option B

**Explanation** :

Assume Shyama takes 3 steps and Vyom takes 2 steps in 6 seconds and let the escalator moves up by x steps per second.

∴Shyama takes 6/3 = 2 seconds for a step and Vyom takes 6/2 = 3 seconds for a step

∴ Shyama took 25 × 2 = 50 seconds to go up

∴ Height of the stairway = (25 + 50x) steps

Vyom took 20 × 3 = 60 seconds to go up

Similarly, in Vyom’s case, the height of the stairway = (20 + 60x) steps

∴ 20 + 60x = 25 + 50x

∴ x = 1/2

∴ If the escalator was turned off, they would have to take (20 + 60 × 1/2) = 50 steps

Hence, option (b).

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**25. CAT 2001 QA | Algebra - Simple Equations**

At a certain fast food restaurant, Brian can buy 3 burgers, 7 shakes, and one order of fries for Rs. 120 exactly. At the same place it would cost Rs. 164.5 for 4 burgers, 10 shakes, and one order of fries. How much would it cost for an ordinary meal of one burger, one shake, and one order of fries?

- A.
Rs. 31

- B.
Rs. 41

- C.
Rs. 21

- D.
Cannot be determined

Answer: Option A

**Explanation** :

Let the cost of one burger, one shake and one order of fries be b, s and f respectively.

∴ 3b + 7s +1f = 120 ...(i)

∴ 4b + 10s + 1f = 164.5 ...(ii)

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

1b + 3s = 44.5

Multiplying by 2, we get,

2b + 6s = 89 ...(iii)

Subtracting equation (iii) from equation (i), we get,

b + s + f = 31

Hence, option (a).

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**26. CAT 2001 QA | Algebra - Inequalities & Modulus**

If a, b, c and d are four positive real numbers such that abcd = 1, what is the minimum value of (1 + a)(1 + b)(1 + c)(1 + d)?

- A.
4

- B.
1

- C.
16

- D.
18

Answer: Option C

**Explanation** :

This is possible when, a = b = c = d = 1

∵ a, b, c, d are positive real numbers

∴ (1 + a)(1 + b)(1 + c)(1 + d) = 2 × 2 × 2 × 2 = 16

Hence, option (c).

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**27. CAT 2001 QA | Arithmetic - Time & Work**

There's a lot of work in preparing a birthday dinner. Even after the turkey is in oven, there are still the potatoes and gravy, yams, salad, and cranberries, not to mention setting the table. Three friends, Asit, Arnold, and Afzal, work together to get all of these chores done. The time it takes them to do the work together is six hours less than Asit would have taken working alone, one hour less than Arnold would have taken, and half the time Afzal would have taken working alone.

How long did it take them to do these chores working together?

- A.
20 minutes

- B.
30 minutes

- C.
40 minutes

- D.
50 minutes

Answer: Option C

**Explanation** :

Suppose Asit, Arnold and Afzal finish the work together in x hours.

Then Asit, Arnold and Afzal will take x + 6, x + 1 and 2x hour respectively.

$\therefore \frac{1}{x+6}+\frac{1}{x+1}+\frac{1}{2x}=\frac{1}{x}\phantom{\rule{0ex}{0ex}}\therefore \frac{1}{x+6}+\frac{1}{x+1}=\frac{1}{2x}\phantom{\rule{0ex}{0ex}}\frac{x+1+x+6}{(x+6)(x+1)}=\frac{1}{2x}\phantom{\rule{0ex}{0ex}}\frac{2x(2x+7)}{{x}^{2}+7x+6}=1$

3x^{2} + 7x – 6 = 0

(x + 3)(3x − 2) = 0

$\therefore x=\frac{2}{3}$ or -3

∵ Number of hours cannot be negative.

$\therefore x=\frac{2}{3}$ hr = 40 minutes

Hence, option (c).

Workspace:

**28. CAT 2001 QA | Geometry - Triangles**

Euclid has a triangle in mind, Its longest side has length 20 and another of its sides has length 10. Its area is 80. What is the exact length of its third side?

- A.
$\sqrt{260}$

- B.
$\sqrt{250}$

- C.
$\sqrt{240}$

- D.
$\sqrt{270}$

Answer: Option A

**Explanation** :

Let the perpendicular on the longest side from the other vertices be h.

$\therefore \frac{1}{2}\times 20\times h$ = 80

∴ h = 8

The perpendicular has two triangles on its two sides. On its left, there is one with a hypotenuse of 10. If the two sides are 10 and 8, the third one must be 6.

∴ The base of the other triangle is 20 − 6 = 14

The two sides being 8 and 14, the hypotenuse must be $\sqrt{{14}^{2}+{8}^{2}}=\sqrt{260}$

Hence, option (a).

Alternatively,

Let the third side of the triangle be x.

$S=\frac{(20+10+x)}{2}=\frac{(30+x)}{2}$

Area of triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Area of triangle = 80 = $\sqrt{\frac{30+x}{2}\times \frac{x-10}{2}\times \frac{x+10}{2}\times \frac{30-x}{2}}$

$\therefore {80}^{2}=\frac{({30}^{2}-{x}^{2})\times ({x}^{2}-100)}{4\times 4}$

$(900-{x}^{2})({x}^{2}-100)$ = 6400 × 16 = 102400

Using options, we get, x = $\sqrt{260}$

(900 − 260) × 160 = 102400

Hence, option (a).

Workspace:

**29. CAT 2001 QA | Algebra - Progressions**

For a Fibonacci sequence, from the third term onwards, each term in the sequence is the sum of the previous two terms in that sequence. If the difference in squares of seventh and sixth terms of this sequence is 517, what is the tenth term of this sequence?

- A.
147

- B.
76

- C.
123

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Let the nth term of the series be F_{n}.

∵ (F_{7})^{2} – (F_{6})^{2} = 517

∴ (F_{7} + F_{6})(F_{7} – F_{6}) = 517

∴ (F_{7} + F_{6}) (F_{7} – F_{6}) = 11 × 47

∴ F_{8} × (F_{6} + F_{5} – F_{6}) = 11 × 47

∴ F_{8} × F_{5} = 11 × 47

∴ F_{8} = 47 and F_{5} = 11

∴F_{8} = F_{7} + F_{6} = 2F_{6} + F_{5} = 3F_{5} + 2F_{4}

∴ F_{4} = 7

Now the series F4 onwards is: 7, 11, 18, 29, 47, 76, 123

∴ The 10th term will be 123.

Hence, option (c).

Workspace:

**30. CAT 2001 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

Fresh grapes contain 90% water by weight while dry grapes contain 20% water by weight. What is the weight of dry grapes available from 20 kg of fresh grapes?

- A.
2 kg

- B.
2.4 kg

- C.
2.5 kg

- D.
None of these

Answer: Option C

**Explanation** :

Total weight of fresh grapes = 20 kg

Weight of grape mass =

In dried grapes, water is 20%.

∴ Grape mass is 80%.

∴ Total weight of dried grapes =

Hence, option (c).

Workspace:

**31. CAT 2001 QA | Arithmetic - Time, Speed & Distance**

A train X departs from station A at 11.00 a.m. for station B, which is 180 km away. Another train Y departs from station B at 11.00 a.m. for station A. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 minutes at station C, which is 60 km away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to point where the trains cross other?

- A.
112

- B.
118

- C.
120

- D.
None of these

Answer: Option A

**Explanation** :

Let us assume that they meet after t hours.

∴ 70 × t + 50$\left(t-\frac{1}{4}\right)$ = 180

120t = 192.5

$t=\left(\frac{192.5}{120}\right)$

∴ The trains should meet at = $\left(\frac{192.5}{120}\right)hrs\times 70\frac{km}{hrs}\cong $ 112 km

Hence, option (a).

Workspace:

**32. CAT 2001 QA | Arithmetic - Average**

A set of consecutive positive integers beginning with 1 is written on the blackboard. A student came along and erased one number.

The average of the remaining numbers is $35\frac{7}{17}$.

What was the number erased?

- A.
7

- B.
8

- C.
9

- D.
None of these

Answer: Option A

**Explanation** :

Let there were n consecutive integers starting with 1 in the original set.

∴ The original average was = $\frac{n\times (n+1)}{2}\times \frac{1}{n}=\frac{n+1}{2}$

∴ Average = 35, if n = 69

∴ Average = 35.5, if n = 70

However, as the new average has 17 in the denominator, we can say that the number of numbers in the new set (n − 1) is 68.

∴ n = 69

∴ Sum of numbers from 1 to 69 = $\frac{69\times 70}{2}$ = 2415

$\because 35\frac{7}{17}=\frac{602}{17}=\frac{2408}{68}$

∴ 68 numbers that remained on the blackboard added up to 2408.

∴ The number that was erased was = 2415 − 2408 = 7

Hence, option (a).

Workspace:

**33. CAT 2001 QA | Geometry - Triangles**

In ∆DEF shown below, points A, B, and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If ∠D = 40°, then what is ∠ACB in degrees?

- A.
140

- B.
70

- C.
100

- D.
None of these

Answer: Option C

**Explanation** :

At point C,

(180° − 2x) + z + (180° − 2y) = 180°

z = 2x + 2y − 180° = 2(x + y) − 180° ...(i)

Also, in ΔDEF, x + y + 40° = 180°

x + y = 140°

Substituting in equation (i), we get,

z = 2 × 140 − 180 = 100°

Hence, option (c).

Workspace:

**34. CAT 2001 QA | Arithmetic - Percentage**

The owner of an art shop conducts his business in the following manner: Every once in a while he raises his prices by X%, then a while later he reduces all the new prices by X%. After one such up-down cycle, the price of a painting decreased by Rs. 441. After a second up-down cycle the painting was sold for Rs. 1,944.81. What was the original price of the painting?

- A.
Rs. 2,756.25

- B.
Rs. 2,256.25

- C.
Rs. 2,500

- D.
Rs. 2,000

Answer: Option A

**Explanation** :

As the price decreases after the first cycle, it has to decrease after the second cycle too. Also the decrease in the second cycle will be less than 441 as the original price for the second cycle is less than the original price for the first cycle.

∴ Price after First Cycle – 1944.81 < 441

Now we consider options.

So, options 2 and 4 are eliminated.

As the percentage change in the price in the first and second cycles is equal,

$\frac{Originalprice}{Decreaseinpriceafterthefirstcycle}$ should be equal to

$\frac{Priceafterfirstcycle}{Decreaseinpriceafterthesecondcycle}$

Only option 1 satisfies this.

Hence, option (a).

Workspace:

**35. CAT 2001 QA | Arithmetic - Time, Speed & Distance**

Three runners A, B and C run a race, with runner A finishing 12 metres ahead of runner B and 18 metres ahead of runner C, while runner B finishes 8 metres ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race?

- A.
36 metres

- B.
48 metres

- C.
60 metres

- D.
72 metres

Answer: Option B

**Explanation** :

Let the length of track is d metres.

$\therefore \frac{A}{B}=\frac{d}{d-12}\phantom{\rule{0ex}{0ex}}\frac{A}{C}=\frac{d}{d-18}\phantom{\rule{0ex}{0ex}}\frac{B}{C}=\frac{d}{d-8}$

Solving we get, d = 48 m

Hence, option (b).

Alternatively,

When A finished, the gap between B and C was 6 m. After B ran another 12 m, the gap is increased by another 2 m.

∴ A gap of 2 m is created when B runs 12 m.

∴ A gap of 8 m is created when B runs = 12 × 4 = 48 m

Workspace:

**36. CAT 2001 QA | Algebra - Inequalities & Modulus**

Let x, y be two positive numbers such that x + y = 1. Then, the minimum value of ${\left(x+\frac{1}{x}\right)}^{2}+{\left(y+\frac{1}{y}\right)}^{2}$ is ______.

- A.
12

- B.
20

- C.
12.5

- D.
13.3

Answer: Option C

**Explanation** :

The sum of two numbers, when their product is known, is minimum when they are equal.

The product of two numbers, when their sum is known, is maximum when they are equal.

${\left(x+\frac{1}{x}\right)}^{2}+{\left(y+\frac{1}{y}\right)}^{2}$

= (x^{2 }+ y^{2}) + $\frac{{x}^{2}+{y}^{2}}{{x}^{2}{y}^{2}}+4$

This expression is minimum when (x^{2} + y^{2}) is minimum and x^{2}y^{2} is maximum.

We have x + y = 1

Squaring,

x^{2} + 2xy + y^{2} = 1

∴ x^{2} + y^{2} = 1 – 2xy

x^{2} + y^{2} is minimum when xy is maximum.

xy is maximum when x = y

∴ For minimum value, both x and y have to be equal.

∴ x = y = $\frac{1}{2}$

$\because {\left(x+\frac{1}{x}\right)}^{2}+{\left(y+\frac{1}{y}\right)}^{2}={\left(2+\frac{1}{2}\right)}^{2}+{\left(2+\frac{1}{2}\right)}^{2}$

= (2.5)^{2} + (2.5)^{2}

= 12.5

Hence, option (c).

Workspace:

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

The batting average (BA) of a test batsman is computed from runs scored and innings played-completed innings and incomplete innings (not out) in the following manner:

r_{1} = number of runs scored in completed innings; n_{1} = number of completed innings

r_{2} = number of runs scored in incomplete innings; n_{2} = number of incomplete innings

BA = $\frac{{r}_{1}+{r}_{2}}{{n}_{1}}$

To better assess batsman's accomplishments, the ICC is considering two other measures MBA_{1} and MBA_{2} defined as follows:

MBA_{1} = $\frac{{r}_{1}}{{n}_{1}}+\frac{{n}_{2}}{{n}_{1}}\times max\left[0,\left(\frac{{r}_{2}}{{n}_{2}}-\frac{{r}_{1}}{{n}_{1}}\right)\right]$

MBA_{2} = $\frac{{r}_{1}+{r}_{2}}{{n}_{1}+{n}_{2}}$

**37. CAT 2001 QA | Arithmetic - Average**

Based on the information provided which of the following is true?

- A.
MBA

_{1}≤ BA ≤ MBA_{2} - B.
BA ≤ MBA

_{2}≤ MBA_{1} - C.
MBA

_{2}≤ BA ≤ MBA_{1} - D.
None of these

Answer: Option D

**Explanation** :

BA = $\frac{{r}_{1}}{{n}_{1}}+\frac{{r}_{2}}{{n}_{1}}$

MBA_{1} = $\frac{{r}_{1}}{{n}_{1}}+\frac{{n}_{2}}{{n}_{1}}\times \frac{{r}_{2}{n}_{2}-{r}_{1}{n}_{2}}{{n}_{1}{n}_{2}}=\frac{{r}_{1}}{{n}_{1}}+\frac{{r}_{2}}{{n}_{1}}-\frac{{r}_{1}{n}_{2}}{{n}_{1}}$

Whatever may be the values, MBA_{1} < BA

Similarly, we can observe that MBA_{2} ≤ BA

Hence, option (d).

Workspace:

**38. CAT 2001 QA | Arithmetic - Average**

An experienced cricketer with no incomplete innings has a BA of 50. The next time he bats, the innings is incomplete and he scores 45 runs. It can be inferred that:

- A.
BA and MBA

_{1}will both increase

- B.
BA will increase and MBA

_{2}will decrease - C.
BA will increase and not enough data is available to assess change in MBA

_{1 }and MBA_{2} - D.
None of these

Answer: Option B

**Explanation** :

As 45 is less than his current average of 50, MBA_{2} goes down. But with any incomplete innings, BA increases.

Hence, option (b).

Workspace:

**39. CAT 2001 QA | Geometry - Quadrilaterals & Polygons**

Based on the figure below, what is the value of x, if y = 10?

- A.
0

- B.
11

- C.
12

- D.
None of these

Answer: Option B

**Explanation** :

The third side of the triangle with sides 10 and x - 3 is $\sqrt{100-{(x-3)}^{2}}$

∴ In the large triangle,

(x - 3)^{2} + (x + 4)^{2} = ${\left[x+\sqrt{100-{(x-3)}^{2}}\right]}^{2}$

Only x = 11 satisfies this equation

Hence, option (b).

Alternatively,

x – 3 and x + 4 form part of a Pythagorean triplet. x = 12 and x = 10 does not give a Pythagorean triplet. With x = 11, we have 8, 15, 17.

This means that the large triangle has a hypotenuse of 17.

Therefore, one side of the smallest triangle is 17 – x = 17 – 11 = 6

As 6, 8, and 10 form a Pythagorean triplet, x is definitely equal to 11.

Workspace:

**40. CAT 2001 QA | Geometry - Mensuration**

A rectangular pool 20 metres wide and 60 metres long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 square metres, how wide, in metres, is the walkway?

- A.
43

- B.
4.3

- C.
3

- D.
3.5

Answer: Option C

**Explanation** :

Suppose the width of the walkway is x metres.

∴ Area of walkway = 516 = (60 + 2x) × (20 + 2x) – 60 × 20

516 = 1200 + 120x + 40x + 4x^{2} – 1200

516 = 4x^{2} + 160x

x^{2} + 40x – 129 = 0

(x + 43)(x – 3) = 0

x = 3 or x = − 43

∵ Negative value of width is not possible.

∴ x = 3

Hence, option (c).

Workspace:

**41. CAT 2001 QA | Algebra - Number Theory**

Let b be a positive integer and a = b^{2} – b. If b ≥ 4, then a^{2} – 2a is divisible by

- A.
15

- B.
20

- C.
24

- D.
None of these

Answer: Option C

**Explanation** :

a^{2} – 2a = a(a – 2)

Substituting a = b^{2} – b, we get,

a^{2} – 2a = (b^{2} – b)[(b^{2} – b) – 2]

= b(b – 1)(b^{2} – b – 2)

= b(b – 1)(b – 2)( b + 1)

= (b – 2)(b – 1)b(b + 1)

This is a product of 4 consecutive positive integers which will definitely have a multiple of 2, 3 and 4.

∴ This must be divisible by 2 × 3 × 4 = 24

Hence, option (c).

Workspace:

**42. CAT 2001 QA | Algebra - Number Theory**

Ashish is given Rs. 158 in one rupee denominations. He has been asked to allocate them into a number of bags such that any amount required between Re. 1 and Rs. 158 can be given by handing out a certain number of bags without opening them. What is the minimum number of bags required?

- A.
11

- B.
12

- C.
13

- D.
None of these

Answer: Option D

**Explanation** :

To express any number with the least number of bags of coins, we need these bags of coins to consist of the number of coins as 2^{0}, 2^{1}, 2^{2}, 2^{3}, …, and the remaining coins should be put in one bag.

∴ The bags would consist of 1, 2, 4, 8, 16, 32 and 64 coins.

This takes care of 127 coins.

There are 31 coins remaining.

Now, any number, greater than 127 and less than 158, can be expressed as,

n = 158 − k ...(where k is a positive integer less than 31)

As, the present 7 bags are sufficient to count 31 different coins, any number greater than 127 can be counted even if we place remaining 31 coins in a single bag.

∴ Total number of bags = 7 + 1 = 8

Hence, option (d).

Workspace:

**43. CAT 2001 QA | Algebra - Number Theory**

In some code, letters, a, b, c, d and e represent numbers 2, 4, 5, 6 and 10. However, we don't know which letter represent which number. Consider the following relationships:

i. a + c = e

ii. b – d = d

iii. e + a = b

- A.
b = 4, d = 2

- B.
a = 4, e = 6

- C.
b = 6, e = 2

- D.
a = 4, c = 6

Answer: Option B

**Explanation** :

a + c = e ... (i)

b – d = d

b = 2d ... (ii)

e + a = b ... (iii)

From (ii), we get,

Either b = 4 and d = 2 or b = 10 and d = 5

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

2a + c = b ... (iv)

∴ b = 4 can easily be eliminated because there is no value of b that will satisfy equation (iv).

∴ b = 10 and d = 5

a = 4 or 6 and e = 6 or 4, but c = 2

Using equation (i), we have,

a = 4 and e = 6

Hence, option (b).

Workspace:

**44. CAT 2001 QA | Algebra - Quadratic Equations**

Ujakar and Keshab attempted to solve a quadratic equation. Ujakar made a mistake in writing down the constant term. He ended up with the roots (4, 3). Keshab made a mistake in writing down the coefficient of x. He got the root as (3, 2). What will be the exact roots of the original quadratic equation?

- A.
(6, 1)

- B.
(–3, –4)

- C.
(4, 3)

- D.
(–4, –3)

Answer: Option A

**Explanation** :

In Ujakar’s case, the constant term is wrong.

∴ Product of roots will be wrong but the sum of roots will be correct.

∴ Correct Sum of the roots = 4 + 3 = 7

In Keshab’s case, coefficient of x is wrong.

∴ Sum of roots will be wrong but the product of roots will be correct.

∴ Correct product of roots = 3 × 2 = 6

∴ Correct quadratic equation is x^{2} – 7x + 6 = 0

(x – 6)(x – 1) = 0

∴ x = 6 or x = 1

Hence, option (a).

Workspace:

**45. CAT 2001 QA | Algebra - Simple Equations**

A change making machine contains 1 rupee, 2 rupee and 5 rupee coins. The total number of coins is 300. The amount is Rs. 960. If the number of 1 rupee coins and the number of 2 rupee coins are interchanged, the value comes down by Rs. 40. The total number of 5 rupee coins is

- A.
100

- B.
140

- C.
60

- D.
150

Answer: Option B

**Explanation** :

Let the number of Re. 1, Re. 2 and Re. 5 coins be x, y and z respectively.

x + y + z = 300 ...(i)

x + 2y + 5z = 960 ...(ii)

2x + y + 5z = 920 ...(iii)

Adding equation (ii) and equation (iii), we get,

3x + 3y + 10z = 1880 ...(iv)

Multiply equation (i) by 3, we get,

3x + 3y + 3z = 900 ...(v)

Subtracting equation (v) from equation (iv), we get,

7z = 980

∴ z = 140

Hence, option (b).

Workspace:

**46. CAT 2001 QA | Modern Math - Permutation & Combination**

The figure below shows the network connecting cities A, B, C, D, E and F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F?

- A.
9

- B.
10

- C.
11

- D.
None of these

Answer: Option B

**Explanation** :

Numbers of possible paths from A to F are ADEF, ABCF, ABF, ADCF, ADCEF, ABDEF, ABEF, ABCEF, ABDCF and ABDCEF.

There are10 ways in all.

Hence, option (b).

Workspace:

**47. CAT 2001 QA | Modern Math - Permutation & Combination**

Let n be the number of different 5 digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?

- A.
144

- B.
168

- C.
192

- D.
None of these

Answer: Option C

**Explanation** :

For a number to be divisible by 4, the last two digits should be divisible by 4.

There are eight such two digit combinations that can be formed from the given digits so that no digits are repeated.

The last two digit combinations are 12, 16, 24, 32, 36, 52, 56 and 64.

∴ There are 8 ways to fill the last two places.

∴ Number of ways in which the first three places can be filled = 4 × 3 × 2 = 24

∴ Total number of ways = 24 × 8 = 192

Hence, option (c).

Workspace:

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

The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the graph below

**48. CAT 2001 QA | Arithmetic - Time, Speed & Distance**

Manasa makes the 200 km trip from Mumbai to Pune at a steady speed of 60 km/hour. What is the amount of petrol consumed for the journey?

- A.
12.5 litres

- B.
13.33 litres

- C.
16 litres

- D.
19.75 litres

Answer: Option B

**Explanation** :

Time for the trip = 200/60 = 10/3 hours

∴ Required fuel = (10/3) × 4 = 40/3 = 13.33 litres

Hence, option (b).

Workspace:

**49. CAT 2001 QA | Arithmetic - Time, Speed & Distance**

Manasa would like to minimize the fuel consumption for the trip by driving at the appropriate speed. How should she change the speed?

- A.
Increase the speed

- B.
Decrease the speed

- C.
Maintain the speed at 60 km/hour

- D.
Cannot be determined

Answer: Option B

**Explanation** :

If the speed is 40 km/hour, then fuel consumption is given by,

(200/40) × 2.5 = 12.5 litres (< 13.33 litres)

∴ For reducing the fuel consumption, she should reduce the speed from 60 km/hour.

Hence, option (b).

Workspace:

**50. CAT 2001 QA | Algebra - Number Theory | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

What are the values of m and n?

- n is an even integer, m is an odd integer, and m is greater than n.
- Product of m and n is 30.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Statement A doesn’t provide any specific values of m and n.

∴ Statement A alone is not sufficient to answer the question.

Using statement B alone:

m × n = 30 = 1 × 30 = 2 × 15 = 3 × 10 = 5 × 6

We cannot say what the values of m and n are.

∴ Statement B alone is also not sufficient to answer the question.

If we combine both the statements A and B together, then we have only one possibility m = 15 and n = 2

∴ We have answered the question using both the statements A and B together.

Hence, option (c).

Note: Negative integers are not considered as Even or Odd.

Workspace:

**51. CAT 2001 QA | Arithmetic - Percentage | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Is Country X's GDP higher than country Y's GDP?

- GDPs of the countries X and Y have grown over the past five years at compounded annual rate of 5% and 6% respectively.
- Five years ago, GDP of country X was higher than that of country Y.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement A does not give us the actual values of their GDPs 5 years ago.

Using statement B alone:

We know that GDP of country X is greater than that of Y but we do not know by how much it is greater.

Even after combining both the statements the question cannot be answered.

Hence, option (d).

Workspace:

**52. CAT 2001 QA | Algebra - Number Theory | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

What is the value of X?

- X and Y are unequal even integers, less than 10, and X/Y is an odd integer.
- X and Y are even integers, each less than 10, and product of X and Y is 12.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Using statement A alone:

∵ Both have to be even numbers, the only possibility is X = 6 and Y = 2

∴ Statement A alone is sufficient to answer the question.

Using statement B alone:

12 = 1 × 12 = 2 × 6 = 3 × 4

∵ Both are even integers and less than 10, the possible set of values of X and Y are (2, 6) and (6, 2).

∴ X is either 2 or 6.

∴ No unique solution can be found using statement B alone.

Hence, option (a).

Note: Negative integers are not considered as Even or Odd.

Workspace:

**53. CAT 2001 QA | Miscellaneous | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

On a given day a boat ferried 1500 passengers across the river in twelve hours. How many round trips did it make?

- The boat can carry two hundred passengers at any time.
- It takes 40 minutes each way and 20 minutes of waiting time at each terminal.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Statement A only gives the maximum capacity of the boat.

∴ The number of trips cannot be calculated.

∴ Statement A alone is not sufficient to answer the question.

Using statement B alone:

Time required to make one round trip = 40 + 40 + 40 = 120 minutes = 2 hours

∴ The total number of trips made = 12/2 = 6

∴ Statement B alone is sufficient to answer the question.

Hence, option (a).

Workspace:

**54. CAT 2001 QA | Arithmetic - Time & Work | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

What will be the time for downloading software?

- Transfer rate is 6 Kilobytes per second.
- The size of the software is 4.5 megabytes.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

Statement A alone is not sufficient because it gives only the rate at which data is transferred but it does not give the size of the software.

Statement B alone is not sufficient because it gives only the size of the software but it does not give the rate of data transfer.

Combining both the statements A and B together, we can get the time to download.

∵ Time required for data transfer = (Size of the software)/(Rate of data transfer)

Hence, option (c).

Workspace:

**55. CAT 2001 QA | Geometry - Circles | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

A square is inscribed in a circle. What is the difference between the area of the circle and that of the square?

- The diameter of the circle is 25 cm.
- The side of the square is 25 cm.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option B

**Explanation** :

Using statement A alone:

∵ Diameter of the circle = Diagonal of the square

∴ Area of the square can be computed and thus the required difference.

∴ Statement A alone is sufficient to answer the question.

Using statement B alone:

∵ Diagonal of the square = × side = Diameter of the circle

∴ Area of the circle can be computed and thus the required difference.

∴ Statement B alone is also sufficient to answer the question.

∴ We can answer the question using either statement alone.

Hence, option (b).

Workspace:

**56. CAT 2001 QA | Algebra - Simple Equations | Data Sufficiency**

**Choose 1**; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.

**Choose 2**; if the question can be answered by using either statement alone.

**Choose 3**; if the question can be answered by using both statements together, but cannot be answered using either statement alone.

**Choose 4**; if the question cannot be answered even by using both statements together.

Two friends, Ram and Gopal, bought apples from a wholesale dealer. How many apples did they buy?

- Ram bought one-half the number of apples that Gopal bought.
- The wholesale dealer had a stock of 500 apples.

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option D

**Explanation** :

Statement A simply gives the ratio of apples between Ram and Gopal.

∴ Statement A is not alone sufficient to answer the question.

Statement B doesn’t provide any information related to number of apples bought by Ram and Gopal.

Even after combining both the statements A and B together, the question cannot be answered.

Hence, option (d).

Workspace:

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