# CAT 1997 QA | Previous Year Questions

**Direction: Answer the questions based on the following information.**

A certain race is made up of three stretches: A, B and C, each 2 km long, and to be covered by a certain mode of transport. The following table gives these modes of transport for the stretches, and the minimum and maximum possible speeds (in km/hr) over these stretches. The speed over a particular stretch is assumed to be constant. The previous record for the race is 10 min.

**1. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

Anshuman travels at minimum speed by car over A and completes stretch B at the fastest speed. At what speed should he cover stretch C in order to break the previous record?

- A.
Maximum speed for C

- B.
Minimum speed for C

- C.
This is not possible

- D.
None of these

Answer: Option C

**Explanation** :

If he travels at minimum speed over stretch A (i.e. 40 km/hr), the total time taken to cover this stretch $=\left(\frac{2}{40}\right)=\frac{1}{20}$ hr = 3 min.

If he then travels at the fastest speed over stretch B (i.e. 50 km/hr), the total time taken to cover this stretch = $\left(\frac{2}{50}\right)=\frac{1}{25}$ = 2.4 min.

Thus, total time taken over the first two stretches = (3 + 2.4) = 5.4 min.

In order to break the previous record he will have to cover the third stretch in (10 – 5.4) = 4.6 min. To do this he will have to cover the third stretch at $\left(\frac{2}{4.6}\right)$ = 0.434 km per minute or 26.08 km/hr. But the maximum speed over the stretch C is 20 km/hr. Hence, it is not possible for C to break the previous record.

Workspace:

**2. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

Mr Hare completes the first stretch at the minimum speed and takes the same time for stretch B. He takes 50% more time than the previous record to complete the race. What is Mr Hare's speed for the stretch C?

- A.
10.9 km/hr

- B.
13.3 km/hr

- C.
17.1 km/hr

- D.
None of these

Answer: Option B

**Explanation** :

The minimum speed in stretch A is 40 km/hr. If Mr Hare travels the first stretch at this speed, then the time taken by him to cover this stretch = $\left(\frac{2}{40}\right)$ = 3 min. Also he takes 3 min to cover stretch B. And he covers the entire race in (1.5 × 10) = 15 min. This means that he should have taken (15 – 3 – 3) = 9 min to cover stretch C. Hence, his speed over this stretch should be $\left(\frac{2}{9}\right)$ = 0.22 km per minute or 13.3 km/hr.

Workspace:

**3. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

Mr Tortoise completes the race at an average speed of 20 km/hr. His average speed for the first two stretches is four times that for the last stretch. Find the speed over stretch C.

- A.
15 km/hr

- B.
12 km/hr

- C.
10 km/hr

- D.
This is not possible

Answer: Option C

**Explanation** :

Let his average speed over the last stretch be x.

Hence, his average speed for first two stretches = 4x. So the total time taken to cover the three stretches $=\left(\frac{4}{4x}\right)+\left(\frac{2}{x}\right)$

His average speed over the race is 20 km/hr.

Hence, the time taken to complete the race = $\frac{6}{20}$

Hence, we have the equation $\frac{4}{4x}+\frac{2}{x}=\frac{6}{20}$ Solving this equation, we get x = 10 km/hr.

Workspace:

**Direction: Answer the questions based on the following information.**

There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D.

**4. CAT 1997 QA | Arithmetic - Average**

What is the average weight of the students in group D?

- A.
More than the average weight of A

- B.
More than the average weight of C

- C.
Less than the average weight of C

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Although the number of students in group D is more than in any other group, we still cannot say anything about the average weight of this group as nothing is mentioned about the average weights of any of the groups or of the individual students.

Workspace:

**5. CAT 1997 QA | Arithmetic - Average**

If one student from group A is shifted to group B, which of the following will be true?

- A.
The average weight of both groups increases

- B.
The average weight of both the groups decreases

- C.
The average weight of the class remains the same

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Although one student is shifted from group A to group B, the number of students in the class and the total weight of the students remain the same. Therefore, the average weight of the class remains the same.

Workspace:

**6. CAT 1997 QA | Arithmetic - Average**

If all the students of the class have the same weight, then which of the following is false?

- A.
The average weight of all the four groups is the same

- B.
The total weight of A and C is twice the total weight of B

- C.
The average weight of D is greater than the average weight of A

- D.
The average weight of all the groups remains the same even if a number of students are shifted from one group to another

Answer: Option C

**Explanation** :

The total weight of any group will vary according to the number of students in that group. Hence, the total weight of group A and C which has (15 + 25) = 40 will be twice that of students in group B which has 20 students. However, it is clear that if all the students are of same weight, then the average weight of all groups remains same irrespective of how many students are present in each group. Hence, clearly the statement 3 is false

Workspace:

**7. CAT 1997 QA | Arithmetic - Percentage**

A student gets an aggregate of 60% marks in five subjects in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?

- A.
2

- B.
3

- C.
4

- D.
5

Answer: Option C

**Explanation** :

Let his marks be 100, 90, 80, 70 and 60 in the five subjects. Hence, totally he has scored 400 marks. This constitutes only 60% of the total marks. Hence, total marks = $\frac{400}{0.6}=667,$ Since the total marks in each subject is the same, hence maximum marks in each subject will be $\left(\frac{667}{5}\right)$ ≃ 133. Out of this 50% is the passing marks. In other words, to pass in a subject he needs to score 66.5 marks. We can see that only in

one subject he scored less than this, viz. 60. Hence, he passed in 4 subjects.

Workspace:

**8. CAT 1997 QA | Modern Math - Permutation & Combination**

In how many ways can eight directors, the vice chairman and chairman of a firm be seated at a round table, if the chairman has to sit between the vice chairman and a director?

- A.
9! × 2

- B.
2 × 8!

- C.
2 × 7!

- D.
None of these

Answer: Option B

**Explanation** :

If we consider the Chairman and the vice chairman as one set, we can see that this set can fit 8 slots in between the 8 directors. Hence, this can be done in 8! ways. Between themselves, the chairman and the

vice chairman can be arranged in 2 ways. Hence, the required answer = 2 x 8!.

Workspace:

**9. CAT 1997 QA | Algebra - Logarithms**

If log_{2} [log_{7} (x^{2} - x + 37)] = 1, then what could be the value of ‘x’?

- A.
3

- B.
5

- C.
4

- D.
None of these

Answer: Option C

**Explanation** :

We know that if log_{a} x = y , then x = a^{y} . So comparing this form with our equation, we can get log_{7} (x^{2} – x + 37) = 2^{1} = 2 and furthermore from this we can say that

(x^{2} – x + 37) = 7^{2} = 49

Thus, we have the equation

x^{2} – x – 12 = 0

The solutions of this equations are,

x = 4 or x = –3.

The value that satisfies the given answer-choices is

x = 4.

Workspace:

**10. CAT 1997 QA | Arithmetic - Profit & Loss**

After allowing a discount of 11.11%, a trader still makes a gain of 14.28%. At how many percentage above the cost price does he mark on his goods?

- A.
28.56%

- B.
35%

- C.
22.22%

- D.
None of these

Answer: Option A

**Explanation** :

Hint: Students please note that the percentages that are given are the basic percentages derived from basic fractions. e.g. 11.11% = $\frac{1}{9}$ and 14.28 = $\frac{1}{7}.$

Hence, you should make use of the most of this kind of knowledge. So let the CP be Re 1. Since he makes a profit of $\frac{1}{7},$ his SP = $\left(1+\frac{1}{7}\right)=\mathrm{Rs}.\frac{8}{7}$.

His marked price should be $\frac{1}{9}$ above this. So if we subtract $\frac{1}{9}$ of MP from the MP, we will get the SP.

So $\left(\mathrm{MP}-\frac{1}{9}\mathrm{MP}\right)=\mathrm{SP}=\frac{8}{7}$

Hence, MP = $\frac{9}{7}$

Therefore, percentage of mark-up on CP = (MP – CP)/CP

$=\left(\frac{9}{7}-1\right)/1=\frac{2}{7}=2\left(\frac{1}{7}\right)$ = 2 × 14.28 = 28.56%

**Alternative method:**

We can use the formula z = x – y – $\frac{xy}{100},$ where

z = Gain percentage

x = Percentage above CP

y = Discount percentage

∴ 14.28% = x – 11.11% – $\frac{11.11x}{100}$

or 14.28 = $\frac{100x-1111-11.11x}{100}$

or 1428 - 1111 = 88.89x

or x = 28.56% (Approximately)

Workspace:

**11. CAT 1997 QA | Algebra - Number Theory**

If n is an integer, how many values of n will give an integral value of $\frac{(16{n}^{2}+7n+6)}{n}$?

- A.
2

- B.
3

- C.
4

- D.
None of these

Answer: Option D

**Explanation** :

The given expression can be written as

$\left(\frac{16{n}^{2}}{n}\right)+\left(\frac{7n}{n}\right)+\left(\frac{6}{n}\right)=16n+7+\left(\frac{6}{n}\right)$

Since n is an integer, the expression (16n + 7) will always be an integer. Hence, for the entire expression to be an integer, the part $\left(\frac{6}{n}\right)$ should also be an integer.

This can be possible only if n is a factor of 6, viz. n = 1, 2, 3, 6, –1, –2, –3 and –6.

Hence, n can have eight values.

Workspace:

**12. CAT 1997 QA | Arithmetic - Profit & Loss**

A dealer buys dry fruits at Rs. 100, Rs. 80 and Rs. 60 per kilogram. He mixes them in the ratio 3 : 4 : 5 by weight, and sells at a profit of 50%. At what price per kilogram does he sell the dry fruit?

- A.
Rs. 80

- B.
Rs. 100

- C.
Rs. 95

- D.
None of these

Answer: Option D

**Explanation** :

Let he mix 3 kg, 4 kg and 5 kg of dry fruits at Rs. 100, Rs. 80 and at Rs. 60 per kilogram respectively. Hence, his effective cost of the dry fruits per kilogram should be the weighted average

$=\left(\frac{3\times 100+4\times 80+5\times 60}{3+4+5}\right)=\frac{920}{12}$

In order to make a 50% profit, he will have to sell it at $\left(\frac{920}{12}\times 1.5\right)=\frac{920}{12}\times \frac{3}{2}=\frac{920}{8}$ = Rs. 115 per kg.

Since none of the answer-choices confirms this, the answer is (d).

Workspace:

**13. CAT 1997 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

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

- A.
2 kg

- B.
2.5 kg

- C.
2.4 kg

- D.
None of these

Answer: Option B

**Explanation** :

20 kg fresh grapes will contain (0.9 × 20) = 18 kg water and 2 kg mass. If the dry grape has to contain 2 kg mass, it should constitute 80% of that. Hence, if 80% of dry grapes corresponds to 2 kg, its total weight will be $\left(\frac{2}{0.8}\right)$ = 2.5 kg

Workspace:

**14. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

An express train travelling at 80 km/hr overtakes a goods train, twice as long and going at 40 km/hr on a parallel track, in 54 s. How long will the express train take to cross a platform of 400 m long?

- A.
36 s

- B.
45 s

- C.
27 s

- D.
None of these

Answer: Option C

**Explanation** :

Effective speed of two trains = (80 – 40) = 40 km/hr. (Since they are moving in the same direction as inferred from the word ‘overtakes’). At this speed in 54 s, they would travel an effective distance of $\frac{(40\times 54)}{3600}$ = 0.6 km or 600 m. This effective distance should be

equal to the sum of the lengths of the two trains. So, if length of the express train is L, length of the goods train will be 2L. Hence, our equation will be L + 2L = 600 or L = 200 m. So, the time taken by this train to cross a platform 400 m long will be $\frac{(200+400)}{\left(80\times {\displaystyle \frac{5}{18}}\right)}=27$ s.

(Note that we have converted the denominator in metres per second. Hence, the factor of $\frac{5}{18}$).

Workspace:

**15. CAT 1997 QA | Arithmetic - Ratio, Proportion & Variation**

A student instead of finding the value of $\frac{7}{8}$ of a number, found the value of $\frac{7}{18}$ of the number. If his answer differed from the actual one by 770, find the number.

- A.
1584

- B.
2520

- C.
1728

- D.
1656

Answer: Option A

**Explanation** :

This equation is very straightforward. If the number is 'x', then $\frac{7x}{8}-\frac{7x}{18}=770.$ On solving this equation, we get x = 1584. Hint: Students please note that if the difference in $\frac{7}{8}$ and $\frac{7}{18}$ of a number is 770, then the difference in $\frac{1}{8}$ and $\frac{1}{18}$ of the number should be 110. If we express this as an equation, we get $\frac{x}{8}-\frac{x}{18}=110$

or 10x = 110 × 18 × 8

or x = 11 × 18 × 8

You can further proceed from here in two ways: (i) the last digit of the required answer should be (1 × 8 × 8) = 4, (ii) number should be divisible by 11. In both cases, the answer that is obtained from the given choices is 1584.

Workspace:

**16. CAT 1997 QA | Algebra - Number Theory**

P and Q are two positive integers such that PQ = 64. Which of the following cannot be the value of P + Q?

- A.
20

- B.
65

- C.
16

- D.
35

Answer: Option D

**Explanation** :

If we were to express 64 as product of two positive integers, we can get the following combinations: (64 × 1), (32 × 2), (16 × 4), (8 × 8).

Thus, we find that P + Q cannot be 35.

Workspace:

**17. CAT 1997 QA | Arithmetic - Average**

The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.

- A.
55

- B.
60

- C.
62

- D.
Cannot be determined

Answer: Option B

**Explanation** :

Total marks scored by the student in 10 papers = (80 × 10) = 800. If we exclude the papers in which he scored the highest and the lowest marks, then the total marks scored by him in remaining 8 papers = (81 × 8) = 648. Hence, his total in these two papers in which he scored the highest and the lowest marks = (800 – 648) = 152. Since his highest score is 92, his lowest score is (152 – 92) = 60.

Workspace:

**18. CAT 1997 QA | Algebra - Quadratic Equations**

If the roots x_{1} and x_{2} of the quadratic equation x^{2} − 2x + c = 0 also satisfy the equation 7x_{2} – 4x_{1} = 47, then which of the following is true?

- A.
c = -15

- B.
x

_{1}= -5, x_{2}= 3 - C.
x

_{1}= 4.5, x_{2}= -2.5 - D.
None of these

Answer: Option A

**Explanation** :

We know that the sum of the roots = $-\frac{b}{a}$.

Hence, x_{1} + x_{2} = 2. Now we have two equations, viz. x_{1} + x_{2} = 2 ...(i)

and 7x_{2} – 4x_{1} = 47 ...(ii)

Solving these two equations, we get x_{1 }= –3 and x_{2} = 5. Since it does not satisfy options (b) and (c), we will verify it for option (a).

The product of the roots = (–3) × 5 = –15, $\frac{c}{a}$ in our case is c. Hence, c = –15.

**Alternately,**

Put values of x_{1}, x_{2} in equation (ii). Do not match. So put c = –15 in equation (i) to get the roots of equation. After finding the roots of equation (i), check whether they satisfy equation (ii) or not.

The roots (5, –3) satisfy the equation (ii) so answer is (a).

Hence, option (a).

Workspace:

**19. CAT 1997 QA | Geometry - Circles**

The sum of the areas of two circles, which touch each other externally, is 153π. If the sum of their radii is 15, find the ratio of the larger to the smaller radius.

- A.
4

- B.
2

- C.
3

- D.
None of these

Answer: Option A

**Explanation** :

If the radii of two circles are r_{1} and r_{2}, then the two equations can be written πr_{1}^{2} + πr_{2}^{2} = 153π

or (r_{1}^{2} + r_{2}^{2}) = 153 and r_{1} + r_{2} = 15.

Now r_{1}^{2} + r_{2}^{2} = (r_{1} + r_{2})^{2} – 2r_{1}r_{2}

Therefore, 153 = (15)^{2} – 2r_{1}r_{2} or r_{1}r_{2} = 36.

If 36 is to be expressed as the product of two integers, it could be (36 × 1), (18 × 2), (12 × 3), (9 × 4), (6 × 6).

The only two factors that add up to 15 are 12 and 3.

Hence, r_{1} = 12, r_{2} = 3. Therefore, the ratio of larger radius to the smaller one is 12 : 3 = 4.

Workspace:

**20. CAT 1997 QA | Algebra - Number Theory**

If m and n are integers divisible by 5, which of the following is not necessarily true?

- A.
m – n is divisible by 5

- B.
m

^{2}– n^{2}is divisible by 25 - C.
m + n is divisible by 10

- D.
None of these

Answer: Option B

**Explanation** :

The best way to solve this is the method of simulation, e.g. let m = 10 and n = 5. Therefore m – n = 5, which is divisible by 5.

m^{2} – n^{2} = 100 – 25 = 75, divisible by 25. m + n = 10 + 5 = 15 is not divisible by 10.

Hence, the answer is (c).

Note that for the sum of two multiples of 5 to be divisible by 10, either both of them should be odd (i.e. ending in 5) or both of them should be even (i.e. ending in 0).

Workspace:

**21. CAT 1997 QA | Algebra - Surds & Indices**

Which of the following is true?

- A.
${7}^{{3}^{2}}$ = (7

^{3})^{2} - B.
${7}^{{3}^{2}}$ > (7

^{3})^{2} - C.
${7}^{{3}^{2}}$ < (7

^{3})^{2} - D.
None of these

Answer: Option B

**Explanation** :

${7}^{{3}^{2}}$ = 7^{9} and (7^{3})^{2} = 7^{6}. SInce 7^{9} > 7^{6} ⇒ ${7}^{{3}^{2}}$ > (7^{3})^{2}

Workspace:

**Direction: Answer the questions based on the following information.**

A survey of 200 people in a community who watched at least one of the three channels — BBC, CNN and DD — showed that 80% of the people watched DD, 22% watched BBC and 15% watched CNN.

**22. CAT 1997 QA | Venn Diagram**

What is the maximum percentage of people who can watch all the three channels?

- A.
12.5%

- B.
8.5%

- C.
15%

- D.
Data insufficient

Answer: Option C

**Explanation** :

Here, a + b + c + d + e + f + g = 200 ...(i)

80% of the people watch DD implies

c + d + f + g = 160 ...(ii)

22% of the people watch BBC implies

a + d + e + g = 44 ...(iii)

15% of the people watch CNN implies

b + e + f + g = 30 ...(iv)

(ii) + (iii) + (iv) gives

a + b + c + 2(d + e + f) + 3g = 234

Subtracting (i) from this equation,

d + e + f + 2g = 34 ...(v)

To maximize g, in equation (v), we put d = e = f = 0

∴ Maximum value of g = $\frac{34}{2}$ = 17

∴ Required percentage = $\frac{17}{200}$ × 100 = 8.5%

Workspace:

**23. CAT 1997 QA | Venn Diagram**

If 5% of people watched DD and CNN, 10% watched DD and BBC, then what percentage of people watched BBC and CNN only?

- A.
2%

- B.
5%

- C.
8.5%

- D.
Cannot be determined

Answer: Option A

**Explanation** :

Here, a + b + c + d + e + f + g = 200 ...(i)

80% of the people watch DD implies

c + d + f + g = 160 ...(ii)

22% of the people watch BBC implies

a + d + e + g = 44 ...(iii)

15% of the people watch CNN implies

b + e + f + g = 30 ...(iv)

(ii) + (iii) + (iv) gives

a + b + c + 2(d + e + f) + 3g = 234

Subtracting (i) from this equation,

d + e + f + 2g = 34 ...(v)

5% of people watch DD and CNN implies

f + g = 10 ...(vi)

10% of people watch DD and BBC implies

d + g = 20 ...(vii)

(v) – (vi) – (vii) gives

e = 4

∴ Required percentage = $\frac{4}{200}$ × 100 = 2%

Workspace:

**24. CAT 1997 QA | Venn Diagram**

Referring to the previous question, what percentage of people watched all the three channels?

- A.
3.5%

- B.
0%

- C.
8.5%

- D.
Cannot be determined

Answer: Option D

**Explanation** :

Here, a + b + c + d + e + f + g = 200 ...(i)

80% of the people watch DD implies

c + d + f + g = 160 ...(ii)

22% of the people watch BBC implies

a + d + e + g = 44 ...(iii)

15% of the people watch CNN implies

b + e + f + g = 30 ...(iv)

(ii) + (iii) + (iv) gives

a + b + c + 2(d + e + f) + 3g = 234

Subtracting (i) from this equation,

d + e + f + 2g = 34 ...(v)

From equation (v), we have

(d + 4 + f) + 2g = 34

⇒ (d + f) + 2g = 30

Since we cannot find the values of d and f, the value of g cannot be ascertained.

Workspace:

**25. CAT 1997 QA | Arithmetic - Percentage**

A man earns x% on the first Rs. 2,000 and y% on the rest of his income. If he earns Rs. 700 from income of Rs. 4,000 and Rs. 900 from if his income is Rs. 5,000, find x%.

- A.
20%

- B.
15%

- C.
25%

- D.
None of these

Answer: Option B

**Explanation** :

The two equations can be written

2000 $\left(\frac{x}{100}\right)$ + 2000 $\left(\frac{y}{100}\right)$ = 700 and 2000 $\left(\frac{x}{100}\right)$ + 3000 $\left(\frac{y}{100}\right)=900$

The equations can be simplified to x + y = 35 and 2x + 3y = 90. Solving these two equations simultaneously, we get x = 15%.

Workspace:

**26. CAT 1997 QA | Geometry - Circles**

AB is the diameter of the given circle, while points C and D lie on the circumference as shown. If AB is 15 cm, AC is 12 cm and BD is 9 cm, find the area of the quadrilateral ACBD.

- A.
54π sq.cm

- B.
216π sq.cm

- C.
162π sq.cm

- D.
None of these

Answer: Option D

**Explanation** :

Since AB is the diameter of the circle, ∠ACB would be right angle. In this triangle, we know AB = 15 and AC = 12. So we can find BC. Hence, BC = 9. Since BC = BD, AD = AC (similar triangles).

Hence, area of ΔACB = Area of Δ ABD $=\frac{1}{2}$ × AC × CB = $\frac{1}{2}$ × 12 × 9 = 54 cm^{2}

So the area of quadrilateral ACBD = 2 × 54 = 108 sq. cm.

Workspace:

**27. CAT 1997 QA | Algebra - Number Theory**

P, Q and R are three consecutive odd numbers in ascending order. If the value of three times P is 3 less than two times R, find the value of R.

- A.
5

- B.
7

- C.
9

- D.
11

Answer: Option C

**Explanation** :

As P, Q and R are consecutive odd numbers, Q = P + 2 and R = P + 4. Now 3P = 2(P + 4) – 3. On solving this equation, we get P = 5.

Therefore, R = 5 + 4 = 9

Workspace:

**Direction: Answer the questions based on the following information.**

For these questions the following functions have been defined.

la(x, y, z) = min(x + y, y + z)

le(x, y, z) = max (x − y, y − z)

ma(x, y, z) = $\frac{1}{2}$ [le(x, y, z) + la(x, y, z)]

**28. CAT 1997 QA | Algebra - Functions & Graphs**

Given that x > y > z > 0. Which of the following is necessarily true?

- A.
la(x, y, z) < le(x, y, z)

- B.
ma(x, y, z) < la(x, y, z)

- C.
ma(x, y, z) < le(x, y, z)

- D.
None of these

Answer: Option B

**Explanation** :

This question can be done by assuming some values for x, y and z, e.g. let x = 4, y = 3 and z = 1. Thus, la(4, 3, 1) = min(7, 4) = 4; le(x, y, z) = max(1, 2) = 2; ma(x, y, z) = $\frac{1}{2}$ (4 + 2) = 3. Hence, we can see that the only answer-choice that satisfies the relationship is ma(x, y, z) < la(x, y, x).

Workspace:

**29. CAT 1997 QA | Algebra - Functions & Graphs**

What is the value of ma(10, 4, le(la(10, 5, 3), 5, 3))?

- A.
7

- B.
6.5

- C.
8

- D.
7.5

Answer: Option B

**Explanation** :

ma(10, 4, le(la(10, 5, 3), 5, 3))

= ma(10,4, le(min(15,8),5,3))

= ma (10, 4, le (8, 5, 3))

= ma(10, 4, max (3,2))

= ma (10, 4, 3)

= $\frac{1}{2}$ [le(10, 4, 3) + la(10, 4, 3)]

= $\frac{1}{2}$[max(6,1) + min(14, 7)]

= $\frac{1}{2}$(6 + 7) = 6.5

Workspace:

**30. CAT 1997 QA | Algebra - Functions & Graphs**

For x = 15, y = 10 and z = 9 , find the value of le(x, min(y, x − z), le (9, 8, ma(x, y, z))).

- A.
5

- B.
12

- C.
9

- D.
4

Answer: Option C

**Explanation** :

le (15,min (10, 6),le (9,8,ma (15,10,9)))

Now ma(15,10, 9) = $\frac{1}{2}$ [le(15, 10, 9) + la(15, 10, 9)]

= $\frac{1}{2}$ [max(5, 1) + min(25,19)]

= $\frac{1}{2}$ (5 + 19) = 12

Hence, our original expression would now be

le(15,min(10,6),le(9,8,12))

= le(15, 6,max(1, − 4))

= le(15, 6,1) = max(9, 5) = 9

Workspace:

**31. CAT 1997 QA | Algebra - Number Theory**

ABC is a three-digit number in which A > 0. The value of ABC is equal to the sum of the factorials of its three digits. What is the value of B?

- A.
9

- B.
7

- C.
4

- D.
2

Answer: Option C

**Explanation** :

The value of ABC, three-digit number is should satisfy the condition 100A + 10B + C = A! + B! + C! ....(i)

The maximum value of three digit number is 999 and minimum is 100

We observe that 7! = 5040 > 999

So the 3-digits of the number must be

6 as 6! = 720

and/or 5 as 5! = 120

and/or 4 as 4! = 24

and/or 3 as 3! = 6

and/or 2 as 2! = 4

and/or 1 as 1! = 1

If we consider 6 at the hundred’s place digit we see that condition (1) is not satisfied as 600 < 720 (6!) So we conclude that 6 cannot occupy any position in the number.

If we place ‘5’ at the hundred’s place then the number should lie between the range of 500 and 600. Considering the RHS of equation (1) by putting A + B = C = 5 we get the sum as 360 which is less than 500. Similarly, putting 4, 3, 2 at the hundred’s place does not satisfies the given condition (1).

Only 1 can be placed at hundred place and 5 should be one of the digit at other two position in order to make it a three digit number.

Thus, only combination we satisfies the given condition (1) is (1, 4, 5) i.e. 145 = 1! + 4! + 5! = 145.

Workspace:

**32. CAT 1997 QA | Geometry - Quadrilaterals & Polygons**

The adjoining figure shows a set of concentric squares. If the diagonal of the innermost square is 2 units, and if the distance between the corresponding corners of any two successive squares is 1 unit, find the difference between the areas of the eighth and the seventh squares, counting from the innermost square.

- A.
10√2 sq. units

- B.
30 sq. units

- C.
35√2 sq. units

- D.
None of these

Answer: Option B

**Explanation** :

The diagonal of the innermost square is 2 units. The diagonal of every successive square would increase by 2 units (since corners are one unit apart). So the diagonal of the 7th square = 14 units and that of the 8th square = 16 units. Areas of the 7th square = $\frac{1}{2}$ (14)^{2} and that of 8th square = $\frac{1}{2}$(16)^{2}, and 128 respectively. Hence, the difference in their areas = (128 – 98) = 30 sq.units.

Workspace:

**33. CAT 1997 QA | Algebra - Surds & Indices**

A, B and C are defined as follows.

A = 2.000004 ÷ [(2.000004)^{2} + (4.000008)]

B = 3.000003 ÷ [(3.000003)^{2} + (9.000009)]

C = 4.000002 ÷ [(4.000002)^{2} + (8.000004)]

Which of the following is true about the values of the above three expressions?

- A.
All of them lie between 0.18 and 0.2

- B.
A is twice of C

- C.
C is the smallest

- D.
B is the smallest

Answer: Option D

**Explanation** :

$A=\frac{2.000004}{[{(2.000004)}^{2}+2(2.000004\left)\right]}$

$=\frac{2.000004}{2.000004\left[\right(2.000004)2]}$

= $\frac{1}{\left[\right(2.000004)2]}$

$=\frac{1}{4.000004}$

= $\frac{1}{4}=0.25$ (Approximately)

$B=\frac{3.000003}{[{(3.000003)}^{3}+3(3.000003\left)\right]}$

$=\frac{3.000003}{3.000003\left[\right(3.000003)3]}$

$=\frac{1}{\left[\right(3.000003)3]}=\frac{1}{6.000003}$

$C=\frac{4.000002}{[{(4.000002)}^{2}+2(4.000002\left)\right]}$

$=\frac{4.000002}{4.000002\left[\right(4.000002)2]}$

$=\frac{1}{\left[\right(4.000002)2]}=\frac{1}{6.000002}$

Looking at the answer choices, we can see that the only (d) satisfies the relationship, viz. B is the smallest.

Workspace:

**34. CAT 1997 QA | Arithmetic - Ratio, Proportion & Variation**

The value of each of a set of coins varies as the square of its diameter, if its thickness remains constant, and it varies as the thickness, if the diameter remains constant. If the diameter of two coins are in the ratio 4 : 3, what should be the ratio of their thickness if the value of the first is four times that of the second?

- A.
16 : 9

- B.
9 : 4

- C.
9 : 16

- D.
4 : 9

Answer: Option B

**Explanation** :

Let D_{1}, T_{1} and D_{2}, T_{2} denote the diameters and the thickness of the two coins respectively. If V_{1} and V_{2 }are the values of the two coins.

$\frac{{V}_{1}}{{V}_{2}}=\left(\frac{{{D}_{1}}^{2}{T}_{1}}{{{D}_{2}}^{2}{T}_{2}}\right)={\left(\frac{{D}_{1}}{{D}_{2}}\right)}^{2}\left(\frac{{T}_{1}}{{T}_{2}}\right)$

Therefore, $\frac{4}{1}={\left(\frac{4}{3}\right)}^{2}\left(\frac{{T}_{1}}{{T}_{2}}\right)\Rightarrow \left(\frac{{T}_{1}}{{T}_{2}}\right)=\frac{9}{4}$

Workspace:

**35. CAT 1997 QA | Geometry - Triangles**

In Δ ABC, points P, Q and R are the mid-points of sides AB, BC and CA respectively. If area of Δ ABC is 20 sq. units, find the area of Δ PQR.

- A.
10 sq.units

- B.
5√3 sq. units

- C.
5 sq. units

- D.
None of these

Answer: Option C

**Explanation** :

In a triangle, the line joining the mid-points of any two sides is half the length of its third side. Hence, every side of ΔPQR would be half the sides of ΔABC. Hence, area of ΔPQR would be $\frac{1}{4}$ the area of ΔABC

$=\frac{1}{4}$ × 20 = 5 sq.units

Workspace:

**36. CAT 1997 QA | Geometry - Quadrilaterals & Polygons**

In a rectangle, the difference between the sum of the adjacent sides and the diagonal is half the length of the longer side. What is the ratio of the shorter to the longer side?

- A.
√3 : 2

- B.
1 : √3

- C.
2 : 5

- D.
3 : 4

Answer: Option D

**Explanation** :

Let L and B denote the length and the breadth of the rectangle. So the diagonal will be $\sqrt{({L}^{2}+{B}^{2})}$. Hence, from the condition given, (L + B) – $\sqrt{({L}^{2}+{B}^{2})}=\frac{1}{2}L$

$\Rightarrow \sqrt{({L}^{2}+{B}^{2})}=\frac{L}{2}+B$

Squaring both sides, we get

(L^{2}+ B^{2}) = ${\left(\frac{L}{2}+B\right)}^{2}$

⇒ L^{2} = $\frac{{L}^{2}}{4}+LB$

⇒ $\frac{3{L}^{2}}{4}=LB$

⇒ $\frac{B}{L}=\frac{3}{4}$

**Shortcut:**

First write the relation (L + B) – $\sqrt{({L}^{2}+{B}^{2})}=\frac{1}{2}L.$

or $\frac{L}{2}+B=\sqrt{({L}^{2}+{B}^{2})}$

Put the values of options. Option (d) satisfies. So the answer is (d).

Workspace:

**Direction: Answer the questions based on the following information.**

The Weirdo Holiday Resort follows a particular system of holidays for its employees. People are given holidays on the days where the first letter of the day of the week is the same as the first letter of their names. All employees work at the same rate.

**37. CAT 1997 QA | Arithmetic - Time & Work**

Raja starts working on February 25, 1996, and finishes the job on March 2, 1996. How much time would T and J take to finish the same job if both start on the same day as Raja?

- A.
4 days

- B.
5 days

- C.
Either (a) or (b)

- D.
Cannot be determined

Answer: Option C

**Explanation** :

As there is no day in the week whose first letter is R, it can be concluded that Raja does not have any holidays. Since 1996 is a leap year, we can figure out that Raja has totally worked for 7 days. Let his rate of doing the job be one unit per day. So he would complete 7 units of work in a week. J's situation is similar to Raja and does not have any holiday during the week. T will have two holidays in a week (Tuesday and Thursday).

Since the rate of working for all the three of them is the same, the working pattern of J and T would be as

follows.

We can see that depending on which day is February 25, 1996, to complete 7 units, they would either take 4 days or 5 days. Hence, the answer is (c).

Workspace:

**38. CAT 1997 QA | Arithmetic - Time & Work**

Starting on February 25, 1996, if Raja had finished his job on April 2, 1996, when would T and S together likely to have completed the job, had they started on the same day as Raja?

- A.
March 15, 1996

- B.
March 14, 1996

- C.
March 22, 1996

- D.
Data insufficient

Answer: Option C

**Explanation** :

Now Raja has worked for (5 days in February + 31 days in March + 2 days in April) = 38 days. Let us assume his rate to be the same as in the previous question, viz. one unit a day. Hence, he completes 38 units totally. In a week, T takes holiday on Tuesday and Thursday, while S takes holiday on Saturday and Sunday. We can see that their working pattern would be as follows.

So in a week they work together 10 units work. Thus, in three weeks, they would complete 30 units work. It

can be found out that February 25 is Sunday. So the remaining 8 units of work can be completed only on Friday, i.e. March 22.

Workspace:

**Direction: **Answer the questions based on the following information.

Boston is 4 hr ahead of Frankfurt and 2 hr behind India. X leaves Frankfurt at 6 p.m. on Friday and reaches Boston the next day. After waiting there for 2 hr, he leaves exactly at noon and reaches India at 1 a.m. On his return journey, he takes the same route as before, but halts at Boston for 1hr less than his previous halt there. He then proceeds to Frankfurt.

**39. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

If his journey, including stoppage, is covered at an average speed of 180 mph, what is the distance between Frankfurt and India?

- A.
3,600 miles

- B.
4,500 miles

- C.
5,580 miles

- D.
Daat insufficient

Answer: Option B

**Explanation** :

Let us convert all the time to same time zone, viz. Boston. So X left Frankfurt at 6 p.m. on Friday (Frankfurt time) or 10 p.m. on Friday (Boston time). X reached Boston at 10 a.m. on Saturday (Boston time). In other words, X has taken 12 hr in all to go from Frankfurt to Boston. After 2 hr wait, X leaves at 12 noon (Boston time). Now X reaches India at 1 a.m. on Sunday (Indian time) or 11 a.m. on Saturday (Boston time). Thus, X takes 11 hr in all to go from Boston to India.

Overall, X has travelled for 25 hr (including stoppages) at an average speed of 180 miles per hour. Hence, the

distance between Frankfurt and India is (25 × 180) = 4500 miles.

Workspace:

**40. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

If X had started the return journey from India at 2.55 a.m. on the same day that he reached there, after how much time would he reach Frankfurt?

- A.
24 hr

- B.
25 hr

- C.
26 hr

- D.
Data insufficient

Answer: Option A

**Explanation** :

Let us convert all the time to same time zone, viz. Boston. So X left Frankfurt at 6 p.m. on Friday (Frankfurt time) or 10 p.m. on Friday (Boston time). X reached Boston at 10 a.m. on Saturday (Boston time). In other words, X has taken 12 hr in all to go from Frankfurt to Boston. After 2 hr wait, X leaves at 12 noon (Boston time). Now X reaches India at 1 a.m. on Sunday (Indian time) or 11 a.m. on Saturday (Boston time). Thus, X takes 11 hr in all to go from Boston to India.

On the return journey, X halts at Boston for one hour less than his previous halt there. Therefore, X takes 24 hr for his return journey.

Workspace:

**41. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

What is X's average speed for the entire journey (to and fro)?

- A.
176 mph

- B.
180 mph

- C.
165 mph

- D.
Data insufficient

Answer: Option A

**Explanation** :

Let us convert all the time to same time zone, viz. Boston. So X left Frankfurt at 6 p.m. on Friday (Frankfurt time) or 10 p.m. on Friday (Boston time). X reached Boston at 10 a.m. on Saturday (Boston time). In other words, X has taken 12 hr in all to go from Frankfurt to Boston. After 2 hr wait, X leaves at 12 noon (Boston time). Now X reaches India at 1 a.m. on Sunday (Indian time) or 11 a.m. on Saturday (Boston time). Thus, X takes 11 hr in all to go from Boston to India.

Since distance between Frankfurt and India is 4,500 miles, overall distance travelled by him (to and fro) = 9000 miles. And he has taken (25 + 24 + $1\frac{11}{12}$*)

= $50\frac{11}{12}$ hr in all to cover this distance.

*Note: $1\frac{11}{12}$ hr has been accounted for the halt that he had in India (from 1 a.m. to 2.55 a.m.). Hence, his average speed for the entire journey = $\frac{9000}{\left(50{\displaystyle \frac{11}{12}}\right)}$ i.e. 176.75 mph.

Workspace:

**42. CAT 1997 QA | Geometry - Circles**

In the adjoining figure, points A, B, C and D lie on the circle. AD = 24 and BC = 12. What is the ratio of the area of Δ CBE to that of Δ ADE?

- A.
1 : 4

- B.
1 : 2

- C.
1 : 3

- D.
Data insufficient

Answer: Option A

**Explanation** :

In ΔBEC and ΔAED

∠CBE = ∠CDE (angles in the same segment of a circle are equal)

Similarly ∠BCE = ∠EAD (angles in the same segment of a circle are equal)

∠BEC = ∠AED (Vertical angles are equal)

∴ By AAA similarity

ΔCEB ~ ΔAED

We know that the ratio of the areas of two similar triangles is equal to the ratio of the squares of the

corresponding sides

∴ $\frac{area(\u2206BEC)}{area(\u2206AED)}={\left(\frac{BC}{DA}\right)}^{2}={\left(\frac{12}{24}\right)}^{2}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{4}$

Workspace:

**43. CAT 1997 QA | Geometry - Quadrilaterals & Polygons**

In the given figure, EADF is a rectangle and ABC is a triangle whose vertices lie on the sides of EADF and AE = 22, BE = 6, CF = 16 and BF = 2. Find the length of the line joining the mid-points of the sides AB and BC.

- A.
$4\sqrt{2}$

- B.
5

- C.
3.5

- D.
None of these

Answer: Option B

**Explanation** :

We know that length of the line joining the mid-points of two sides of a triangle is half the length of third side. Hence, the required length is half the length of side AC. Since EADF is rectangle, EF = AD = 8.

CD = (22 – 16) = 6.

So in the right-angled ΔADC, AD = 8 and CD = 6.

Therefore, AC = 10. Hence, length of the line joining the mid-points of AB and BC = $\frac{1}{2}$(10) = 5.

Workspace:

**Direction: Answer the question based on the following information.**

A thief, after committing the burglary, started fleeing at 12 noon, at a speed of 60 km/hr. He was then chased by a policeman X. X started the chase, 15 min after the thief had started, at a speed of 65 km/hr.

**44. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

At what time did X catch the thief?

- A.
3.30 p.m.

- B.
3 p.m.

- C.
3.15 p.m.

- D.
None of these

Answer: Option C

**Explanation** :

Since the policeman started 15 min late, in this time the thief would have already covered $\left(\frac{60}{4}\right)$ = 15 km. To

catch the thief, the policeman will have to make up for this distance of 15 km. Every hour the policeman is travelling (65 – 60) = 5 km more than the thief. Hence, to make up the distance of 15 km, he would take 3 hr.

Since policeman started at 12.15 p.m., he would catch the thief at 3.15 p.m.

Workspace:

**45. CAT 1997 QA | Arithmetic - Time, Speed & Distance**

If another policeman had started the same chase along with X, but at a speed of 60 km/hr, then how far behind was he when X caught the thief?

- A.
18.75 km

- B.
15 km

- C.
21 km

- D.
37.5 km

Answer: Option B

**Explanation** :

Every hour the second policeman covers (65 – 60) = 5 km less than the first one. Since the first policeman catches the thief in 3 hr, in this time the second policeman will be (3 × 5) = 15 km behind.

Workspace:

**46. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

1. if the question can be answered with the help of one statement alone.

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**What is the value of a ^{3} + b^{3} ?**

I. a^{2 }+ b^{2} = 22

II. ab = 3

Answer: 4

**Explanation** :

a^{3} + b^{3} = (a + b)(a^{2} + b^{2} – ab). Combining statements I and II, we get the value of (a + b) = √28 or –√28. Since we do not have the unique value of (a + b), we cannot get the unique answer a^{3} + b^{3}.

Workspace:

**47. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

1. if the question can be answered with the help of one statement alone.

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**Is the number completely divisible by 99?**

I. The number is divisible by 9 and 11 simultaneously.

II. If the digits of the number are reversed, the number is divisible by 9 and 11.

Answer: 2

**Explanation** :

11 and 9 are coprimes of 99, and hence the number divisible by 99 must be divisible by 9 and 11. Certainly statement I alone is sufficient to answer the question. Statement II says when the digits of the number are reversed, the new number formed is divisible by 9 and 11. The best way to handle this particular case is by simulation. Let us select any number which is divisible by 9 and 11. Let us select 1386 which is divisible by 9 and 11. Hence, the original number will be 6831 which, in turn, is also divisible by 99. Hence, statement II also is independently sufficient to answer the question. Since both the statements are independently sufficient to answer the question, the answer is (b).

Workspace:

**48. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

1. if the question can be answered with the help of one statement alone.

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**A person is walking from Mali to Pali, which lies to its north-east. What is the distance between Mali and Pali?**

I. When the person has covered $\frac{1}{3}$ the distance, he is 3 km east and 1 km north of Mali.

II. When the person has covered $\frac{2}{3}$ the distance, he is 6 km east and 2 km north of Mali.

Answer: 2

**Explanation** :

The diagram will be as shown. M is Mali and P is Pali. Consider statement I. When the person covers $\frac{1}{3}$ distance, he is 3 km east and 1 km north of Mali. Based on this statement alone, we can easily find out where

he will be on line MP by using Pythagoras’ theorem. Once we find this distance, we can easily get distance MP by multiplying it by 3. Similarly, based on statement II alone also, we can find distance MP. Hence, both the statements are independently sufficient to answer the question. Hence, the answer is (b).

Workspace:

**49. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**What is the value of x and y?**

I. 3x + 2y = 45

II. 10.5x + 7y = 157.5

Answer: 4

**Explanation** :

Students! be careful. Generally, as we see two unknowns (i.e. x and y in this case) and two equations, we tempt to mark the answer as c, i.e. combining two statements, we can easily find the values of x and y. But have a look at the equations 3x + 2y = 45 and 10.5x + 7y = 157.5. Multiplying 1st equation by 3.5, we get 2nd equation. Hence, these are not really two different equations. Hence, data is insufficient to answer the question. In general, remember the following rule. If we have two equations Ax + By = k_{1} and Cx + Dy = k_{2}, and A × D = B × C, then the equations cannot be solved.

Workspace:

**50. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**Three friends P, Q and R are wearing hats, either black or white. Each person can see the hats of the other two persons. What is the colour of P's hat?**

I. P says that he can see one black hat and one white hat.

II. Q says that he can see one white hat and one black hat.

Answer: 4

**Explanation** :

P says he can see one black and one white hat. So either Q is wearing white and R is wearing black, or Q is wearing black and R is wearing white. Q also makes same statement. Still we cannot say the colour of the hat which P is wearing.

Workspace:

**51. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**What is the speed of the car?**

I. The speed of a car is 10 (km/hr) more than that of a motorcycle.

II. The motorcycle takes 2 hr more than the car to cover 100 km.

Answer: 3

**Explanation** :

Let the speed of the motorcycle be x km/hr. Therefore, speed of the car will be (x + 10) km/hr.

From statement II, we can form the following equation.

$\left(\frac{100}{(x+10)}\right)=\left(\left(\frac{100}{x}\right)+2\right)$

After solving this equation, we can get the speed of the car. Hence, this question can be answered by combining both the statements.

Workspace:

**52. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**What is the ratio of the volume of the given right circular cone to the one obtained from it?**

I. The smaller cone is obtained by passing a plane parallel to the base and dividing the original height in the ratio 1 : 2.

II. The height and the base of the new cone are one-third those of the original cone.

Answer: 2

**Explanation** :

Let V_{1} be the original volume and r_{1} and h_{1 }be the radius and height of the cone respectively.

V_{1} = $\left(\frac{1}{3}\right)$ × π × (r_{1})^{2} × h_{1}. Consider statement I. If the cone is cut parallel to base and dividing the height in

the ratio 1 : 2, then r_{2} = $\left(\frac{1}{2}\right)\times {r}_{1}$ and h_{2} = $\left(\frac{1}{2}\right)$ × h_{1}, where r_{2} and h_{2} are the radius and height of the new

cone respectively. If V_{2} is the volume of new cone, then V2 = $\left(\frac{1}{3}\right)$ × π(r_{2})^{2 }× h_{2}

$=\left(\frac{1}{3}\right)\times \pi \times {\left(\frac{1}{2}\times {r}_{1}\right)}^{2}\times \left(\left(\frac{1}{2}\right){h}_{1}\right)=\left(\frac{1}{8}\right)\times {V}_{1}$

Hence, statement I alone is sufficient to answer the question (as we get the ratio as 1 : 8).

Similarly, based on statement II alone, we can find the ratio (which will be 1 : 27).

Workspace:

**53. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**What is the area bounded by the two lines and the coordinate axes in the first quadrant?**

I. The lines intersect at a point which also lies on the lines 3x – 4y = 1 and 7x – 8y = 5.

II. The lines are perpendicular, and one of them intersects the Y-axis at an intercept of 4.

Answer: 3

**Explanation** :

If we solve the two given equations, we get the point of intersection as (3, 2). Let A = (3, 2). The lines of our interest (let it be L_{1} and L_{2}) also pass through A. One of the lines passes through (0, 4). Let L_{1} passes through (0, 4), but it also passes through (3, 2). Hence, we can find the slope of L_{2} (which is equal to – $\frac{2}{3}$). Hence, slope of L_{2} will be $\frac{3}{2}$ since L_{1} and L_{2} are perpendicular.

Hence, equations of L_{1} and L_{2} can be obtained by using slope point form. (Students! we need not really find out the equations.) After getting both the equations, we can find the area bounded by L_{1} and L_{2 }and coordinate axes.

Workspace:

**54. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**What is the cost price of the chair?**

I. The chair and the table are sold at profits of 15% and 20% respectively.

II. If the cost price of the chair is increased by 10% and that of the table is increased by 20%, the profit reduces by Rs. 20.

Answer: 4

**Explanation** :

Let A and B be the CP of the chair and the table respectively. So 1.15A + 1.2B = SP.

Hence, profit = 0.15A + 0.2B. Now consider statement II, CP = 1.1A + 1.2B. As per new CP, now profit will be

SP – CP = (1.15A + 1.2B) – (1.1A + 1.2B) = 0.05A + 0xB = 0.05A. Combining both statements, we get the

equation as

0.05A = 0.15A + 0.2B – 20. Still we cannot find the answer.

Workspace:

**55. CAT 1997 QA | Data Sufficiency**

**Direction: Each of these items has a question followed by two statements, I and II. Mark the answer**

2. if the question can be answered with the help of any one statement independently.

3. if the question can be answered with the help of both statements together.

4. if the question cannot be answered even with the help of both statements together.

**After what time will the two persons Tez and Gati meet while moving around the circular track? Both of them start at the same point and at the same time.**

I. Tez moves at a constant speed of 5 m/s, while Gati starts at a speed of 2 m/s and increases his speed by 0.5 m/s at the end of every second thereafter.

II. Gati can complete one entire lap in exactly 10 s.

Answer: 4

**Explanation** :

None of the statements specifies the direction in which Tez and Gati are moving, which is very significant.

Workspace:

## Feedback

**Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing
us your valuable feedback about Apti4All and how it can be improved.**