# CAT 1995 QA

Paper year paper questions for CAT 1995 QA

**1. CAT 1995 QA | Geometry - Triangles**

ABCD is a square of area 4, which is divided into four non-over lapping triangles as shown in figure. Then the sum of the perimeters of the triangles is

- A.
8(2 + $\sqrt{2}$)

- B.
8(1 + $\sqrt{2}$)

- C.
4(1 + $\sqrt{2}$)

- D.
4(2 + $\sqrt{2}$)

Answer: Option B

**Explanation** :

Since base of each triangle will be counted once, Sum of perimeters of the triangles = Perimeter of the

square + 2 × (Sum of its diagonals).

But each of the other two sides of the triangles is common to two triangles, so it will be counted twice.

Area of the square = 4, therefore length of its side = 2 and perimeter = 8.

Also its diagonal = $2\sqrt{2}$

So the required perimeter = (8 + 2 × 4$\sqrt{2}$) = 8(1 + $\sqrt{2}$).

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**2. CAT 1995 QA | Algebra - Number Theory**

5^{6} - 1 is divisible by

- A.
13

- B.
31

- C.
5

- D.
None of these

Answer: Option B

**Explanation** :

5^{6} – 1 = (5^{3})^{2} – 1 = (125)^{2} – (1)^{2} = (125 + 1) (125 – 1) = 124 × 126 = 31 × 4 × 126.

So among the given answer choices, it is divisible by 31.

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**3. CAT 1995 QA | Arithmetic - Percentage**

Ram purchased a flat at Rs.1 lakh and Prem purchased a plot of land worth Rs.1.1 lakh. The respective annual rates at which the prices of the flat and the plot increased were 10% and 5%. After two years they exchanged their belongings and one paid the other the difference. Then

- A.
Ram paid Rs.275 to Prem

- B.
Ram paid Rs.475 to Prem

- C.
Ram paid Rs.375 to Prem

- D.
Prem paid Rs.475 to Ram

Answer: Option A

**Explanation** :

After 2 years, the price of the flat will be (1)(1.10)^{2} = Rs.1.21 lakh.

Correspondingly the price of the land will be (1.1)(1.05)^{2} = Rs.1.21275 lakh.

Hence, the price of the plot = Rs.(1.21275 – 1.21) lakh = Rs.275 more than that of the flat.

Hence, if they exchange, Ram will have to pay this amount to Prem .

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

Four sisters — Suvarna, Tara, Uma and Vibha are playing a game such that the loser doubles the money of each of the other players from her share. They played four games and each sister lost one game in alphabetical order. At the end of fourth game, each sister had Rs.32.

**4. CAT 1995 QA | Algebra - Simple Equations**

How many rupees did Suvarna start with?

- A.
Rs. 60

- B.
Rs. 34

- C.
Rs. 66

- D.
Rs. 28

Answer: Option C

**Explanation** :

Please note that the best way to solve this question is by working backwards.

E.g. after the 4th round, each one of them had Rs.32. Since it is Vibha who lost in this round, all the remaining three must have doubled their share.

In other words, they would have had Rs.16 each after the 3rd round.

Since the increase is of Rs.16 in each one’s share, i.e., Rs.48 overall which comes from Vibha's share, her share before the 4th round was (32 + 48) = Rs.80, after the 3rd round.

Working backwards in this manner, we can get the following table.

Suvarna started with Rs.66.

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**5. CAT 1995 QA | Algebra - Simple Equations**

Who started with the lowest amount?

- A.
Suvarna

- B.
Tara

- C.
Uma

- D.
Vibha

Answer: Option D

**Explanation** :

Please note that the best way to solve this question is by working backwards.

E.g. after the 4th round, each one of them had Rs.32. Since it is Vibha who lost in this round, all the remaining three must have doubled their share.

In other words, they would have had Rs.16 each after the 3rd round.

Since the increase is of Rs.16 in each one’s share, i.e., Rs.48 overall which comes from Vibha's share, her share before the 4th round was (32 + 48) = Rs.80, after the 3rd round.

Working backwards in this manner, we can get the following table.

It was Vibha who started with the lowest amount, viz. Rs.10.

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**6. CAT 1995 QA | Algebra - Simple Equations**

Who started with the highest amount?

- A.
Suvarna

- B.
Tara

- C.
Uma

- D.
Vibha

Answer: Option A

**Explanation** :

Please note that the best way to solve this question is by working backwards.

E.g. after the 4th round, each one of them had Rs.32. Since it is Vibha who lost in this round, all the remaining three must have doubled their share.

In other words, they would have had Rs.16 each after the 3rd round.

Since the increase is of Rs.16 in each one’s share, i.e., Rs.48 overall which comes from Vibha's share, her share before the 4th round was (32 + 48) = Rs.80, after the 3rd round.

Working backwards in this manner, we can get the following table.

It was Suvarna who started with the highest amount, viz. Rs.66.

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**7. CAT 1995 QA | Algebra - Simple Equations**

What was the amount with Uma at the end of the second round?

- A.
36

- B.
72

- C.
16

- D.
None of these

Answer: Option B

**Explanation** :

E.g. after the 4th round, each one of them had Rs.32. Since it is Vibha who lost in this round, all the remaining three must have doubled their share.

In other words, they would have had Rs.16 each after the 3rd round.

Since the increase is of Rs.16 in each one’s share, i.e., Rs.48 overall which comes from Vibha's share, her share before the 4th round was (32 + 48) = Rs.80, after the 3rd round.

Working backwards in this manner, we can get the following table.

At the end of the second round, Uma had Rs.72.

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**Direction: **Answer the questions independently.

**8. CAT 1995 QA | Algebra - Number Theory**

72 hens cost Rs.__ 96.7__. Then what does each hen cost, where two digits in place of ‘__’ are not visible or are written in illegible hand?

- A.
Rs. 3.23

- B.
Rs. 5.11

- C.
Rs. 5.51

- D.
Rs. 7.22

Answer: Option C

**Explanation** :

**Hint: **The best way to solve this question is to multiply the alternatives by 72 and find which one gives you the middle three digits as 96.7.

To save time, you can multiply 72 by integer values only.

E.g. 72 × 3 = 216, 72 × 5 = 360 and 72 × 7 = 504.

It is to be noted that when the decimal part of the answer will be multiplied by 72, the actual answer will increase.

Let us now roughly multiply the decimal values of the options also by 72. E.g. 72 x 0.2 = 14.4, 72 × 0.1 = 7.2 and 72 × 0.5 = 36.

So option (a) will yield (216 + 14) = 230

(approximately), (b) will yield (360 + 7) = 367

(approximately), (c) will yield (360 + 36) = 396

(approximately) and (d) will yield (504 + 14) = 528

(approximately).

Of these, only option (c) satisfies our requirement of 2nd and 3rd digits being 96.

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**9. CAT 1995 QA | Arithmetic - Percentage**

A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance he purchases oranges. Number of oranges he can purchase is

- A.
36

- B.
40

- C.
15

- D.
20

Answer: Option D

**Explanation** :

Let us assume that the person has Rs.100.

With this, he can buy 50 oranges or 40 mangoes.

In other words, the price of an orange is Rs.2 and that of a mango is Rs.2.50.

If he decides to keep 10% of his money for taxi fare, he would be left with Rs.90.

Now if he buys 20 mangoes, he would spend Rs.50 and will be left with Rs.40.

Thus, he can buy 20 oranges.

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**10. CAT 1995 QA | Algebra - Number Theory**

The value of $\frac{{55}^{3}+{45}^{3}}{{55}^{2}-55\times 45+{45}^{2}}$ is

- A.
100

- B.
105

- C.
125

- D.
75

Answer: Option A

**Explanation** :

The given expression is of the form $\frac{[{x}^{3}+{y}^{3}]}{[{x}^{2}-xy+{y}^{2}]}$.

We know, x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2}).

Thus, required value = (x + y) = (55 + 45) = 100.

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**11. CAT 1995 QA | Algebra - Number Theory**

For the product n(n + 1)(2n + 1), n ∈ N, which one of the following is not necessarily true?

- A.
It is even

- B.
Divisible by 3

- C.
Divisible by the sum of the square of first n natural numbers

- D.
Never divisible by 237

Answer: Option D

**Explanation** :

Since n(n + 1) are two consecutive integers, one of them will be even and thus their the product will always be even.

Also, sum of the squares of first ‘n’ natural numbers is given by $\frac{n(n+1)(2n+1)}{6}.$

Hence, our product will always be divisible by this.

Also you will find that the product is always divisible by 3 (you can use any value of n to verify this).

However, we can find that option (d) is not necessarily true. E.g. If n = 118, (2n + 1) = 237 or if n = 236, then (n + 1) = 237 or if n itself is 237, etc.

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**12. CAT 1995 QA | Algebra - Number Theory**

The remainder obtained when a prime number greater than 6 is divided by 6 is

- A.
1 or 3

- B.
1 or 5

- C.
3 or 5

- D.
4 or 5

Answer: Option B

**Explanation** :

The best way to solve this question is by the method of simulation. Choose any prime number greater than 6 and verify the result.

When 7 is divided by 6, it gives a remainder 1. So our answer could be (a) or (b). When 11 is divided by 6, it gives a remainder 5. Hence, our answer is (b).

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**13. CAT 1995 QA | Modern Math - Permutation & Combination**

Boxes numbered 1, 2, 3, 4 and 5 are kept in a row, and they are to be filled with either a red or a blue ball, such that no two adjacent boxes can be filled with blue balls. Then how many different arrangements are possible, given that all balls of a given colour are exactly identical in all respects?

- A.
8

- B.
10

- C.
15

- D.
22

- E.
13

Answer: Option E

**Explanation** :

There cannot be four or more blue balls.

Case 1:

If there are three blue balls, then they can be only in box 1, 3 and 5.

Case 2:

If there are two blue balls, then total number of cases = ^{5}C_{2} = 10

But in 4 cases the blue ball will be in adjacent boxes.

These cases are when blue balls in boxes 1 and 2 or 2 and 3 or 3 and 4 or 4 and 5.

Therefore, total number of cases when there are two blue balls = 10 – 4 = 6

Case 3:

If there are one blue ball, then total number of cases = ^{5}C_{1 }= 5

Case 4:

If there are no blue ball, then total number of cases = ^{5}C_{5} = 1

Hence, total number of cases = 1 + 6 + 5 + 1 = 13.

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**14. CAT 1995 QA | Geometry - Triangles**

AB ⊥ BC, BD ⊥ AC and CE bisects ∠C, ∠A = 30°. Then what is ∠CED?

- A.
30°

- B.
60°

- C.
45°

- D.
65°

Answer: Option B

**Explanation** :

In ∆ABC, ∆ACB = 180 – 90 – 30 = 60°.

∴ ∠DCE = 30°, since ∠CDE = 90°.

In ∆CED, ∠CED = 180 – 90 – 30 = 60°.

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**15. CAT 1995 QA | Arithmetic - Simple & Compound Interest**

A man invests Rs.3,000 at the rate of 5% per annum. How much more should he invest at the rate of 8%, so that he can earn a total of 6% per annum?

- A.
Rs. 1,200

- B.
Rs. 1,300

- C.
Rs. 1,500

- D.
Rs. 2,000

Answer: Option C

**Explanation** :

Using alligation, the ratio of the amounts invested at both the rates = 2 : 1.

Since he has invested Rs.3,000 at 5%, he should further invest Rs.1,500 at 8% to earn a total interest of 6% per annum.

**Alternative method:**

Let the amount invested at 8% be Rs. x.

Then, $3000\times \frac{105}{100}+x\times \frac{108}{100}=(3000+x)\frac{106}{100}$

⇒ 0.02x = 30 ⇒ x = 1,500

∴ He should further invest Rs.1,500 at 5% to earn a total interest of 6% per annum.

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**16. CAT 1995 QA | Arithmetic - Percentage**

$\frac{2}{5}$ of the voters promise to vote for P and the rest promised to vote for Q. Of these, on the last day 15% of the voters went back of their promise to vote for P and 25% of voters went back of their promise to vote for Q, and P lost by 2 votes. Then the total number of voters is

- A.
100

- B.
110

- C.
90

- D.
95

Answer: Option A

**Explanation** :

Let there be 100 voters in all.

Initially, 40 of these promised to vote for P, while 60 of them promised to vote for Q.

On the last day, (15% of 40) = 6 voters went back of their promise and voted for Q.

Also, 25% of 60 = 15 voters shifted their interest from Q to P.

So finally, P end up getting (40 – 6 + 15) = 49 votes and

Q end up getting (60 – 15 + 6) = 51 votes.

Hence, margin of victory for Q = (51 – 49) = 2, which is true. Hence, there were 100 voters in all.

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**17. CAT 1995 QA | Geometry - Quadrilaterals & Polygons**

PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR = OS. Then the ratio of area of the circle to the area of the square is

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

- B.
$\frac{11}{7}$

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

- D.
$\frac{7}{11}$

Answer: Option A

**Explanation** :

In the given figure, the area of the circle = pr^{2}.

To find out the area of the circle, we need to find out the length of the side of the square.

We know, OR = OT + TR = OT + OS = 2r.

In right-angled triangle ORS, OR = 2r and OS = r.

So SR^{2} = OR^{2} – OS^{2}.

But SR^{2} = Area of the square = 4r^{2} – r^{2} = 3r^{2}.

Hence, the required ratio = $\frac{\pi}{3}.$

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**18. CAT 1995 QA | Arithmetic - Time, Speed & Distance**

In a race of 200 m run, A beats S by 20 m and N by 40 m. If S and N are running a race of 100 m with exactly same speed as before, then by how many metres will S beat N?

- A.
11.11 m

- B.
10 m

- C.
12 m

- D.
25 m

Answer: Option A

**Explanation** :

In the same time as A runs 200 m in the race, S runs 180 m and N runs 160 m.

In other words, in the same time as S runs 180 m, N runs 160 m.

So in the same time as S runs 100 m, N will run $\left(100\times \frac{160}{180}\right)$ = 88.89 m.

Hence, in a 100 m race, S will beat N by (100 – 88.89) = 11.11 m.

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**19. CAT 1995 QA | Algebra - Simple Equations**

Three consecutive positive even numbers are such that thrice the first number exceeds double the third by 2, then the third number is

- A.
10

- B.
14

- C.
16

- D.
12

Answer: Option B

**Explanation** :

If the numbers are (x – 2), x and (x + 2), then 3(x – 2) – 2 = 2(x + 2).

∴ x + 2 = 14.

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**20. CAT 1995 QA | Arithmetic - Time & Work**

A group of men decided to do a job in 8 days. But since 10 men dropped out every day, the job got completed at the end of the 12th day. How many men were there at the beginning?

- A.
165

- B.
175

- C.
80

- D.
90

Answer: Option A

**Explanation** :

If x men were there on day one, there would be (x – 110) men on the 12th day.

Hence, on an average, there were (x – 55) men.

The job takes $\frac{3}{2}$ times the normal time.

Hence, the average number of people = $\frac{2}{3}.$

$\Rightarrow x-55=\frac{2}{3}x$

Hence, x = 165

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**21. CAT 1995 QA | Modern Math - Probability**

If a 4 digit number is formed with digits 1, 2, 3 and 5. What is the probability that the number is divisible by 25, if repetition of digits is not allowed?

- A.
$\frac{1}{12}$

- B.
$\frac{1}{4}$

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

- D.
None of these

Answer: Option A

**Explanation** :

Total number of four-digit numbers that can be formed = 4!.

If the number is divisible by 25, then the last two digits are 25.

So the first two digits can be arranged in 2! ways.

Hence, required probability = $\frac{2!}{4!}=\frac{1}{12}.$

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**22. CAT 1995 QA | Arithmetic - Time & Work**

Two typists undertake to do a job. The second typist begins working one hour after the first. Three hours after the first typist has begun working, there is still $\frac{9}{20}$ of the work to be done. When the assignment is completed, it turns out that each typist has done half the work. How many hours would it take each one to do the whole job individually ?

- A.
12 hr and 8 hr

- B.
8 hr and 5.6 hr

- C.
10 hr and 8 hr

- D.
5 hr and 4 hr

Answer: Option C

**Explanation** :

Let the first typist takes X hours and the second takes Y hours to do the whole job.

When the first was busy typing for 3 hr, the second was busy only for 2 hr.

Both of them did $\frac{11}{20}$ of the whole work.

$\therefore \frac{3}{X}+\frac{2}{Y}=\frac{11}{20}.$

When the assignment was completed, it was revealed that each typist had done half the work.

∴ The first one spent $\frac{X}{2}$ hr, and the second $\frac{Y}{2}$ hr.

And since the first had begun one hour before the second, we have $\frac{X}{2}-\frac{Y}{2}=1$

⇒ X = 10 hr, Y = 8 hr.

Hence, option (c).

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**23. CAT 1995 QA | Arithmetic - Time, Speed & Distance**

I live X floors above the ground floor of a high-rise building. It takes me 30 s per floor to walk down the steps and 2 s per floor to ride the lift. What is X, if the time taken to walk down the steps to the ground floor is the same as to wait for the lift for 7 min and then ride down?

- A.
4

- B.
7

- C.
14

- D.
15

Answer: Option D

**Explanation** :

Since I live X floors above the ground floor and it takes me 30 s per floor to walk down and 2 s per floor to ride the lift, it takes 30X s to walk down and 2X s to ride the lift after waiting 420 s.

⇒ 30X = 2X + 420 ⇒ X = 15.

**Alternative method:**

X > 14 as time taken to walk has to be greater than 7 min.

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**24. CAT 1995 QA | Geometry - Triangles**

The sides of a triangle are 5, 12 and 13 units. A rectangle is constructed, which is equal in area to the triangle, and has a width of 10 units. Then the perimeter of the rectangle is

- A.
30 units

- B.
36 units

- C.
13 units

- D.
None of these

Answer: Option D

**Explanation** :

Since 5-12-13 forms a Pythagorean triplet, the triangle under consideration is a right-angled triangle with height 12 and base 5.

So area of the triangle = $\left(\frac{1}{2}\right)$(12)(5) = 30 sq. units.

If area of the rectangle with width 10 units is 30 sq. units, its length = 3 units.

Hence, its perimeter = 2(10 + 3) = 26 units.

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**25. CAT 1995 QA | Geometry - Quadrilaterals & Polygons**

In the adjoining figure, AC+ AB = 5AD and AC – AD = 8. Then the area of the rectangle ABCD is

- A.
36

- B.
50

- C.
60

- D.
Cannot be answered

Answer: Option C

**Explanation** :

Since AD = BC (Opposite sides of a rectangle are equal.)

AB + AC = 5BC and AC – BC = 8 or AC = BC + 8

∴ AB = 4(BC – 2)

By Pythagoras’ Theorem, AB^{2} + BC^{2} = AC^{2}

Expressing AB and AC in terms of BC we get, BC = 5.

∴ AB = 12 and AC = 13

So area of the rectangle = 5 × 12 = 60.

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**26. CAT 1995 QA | Algebra - Quadratic Equations**

One root of x^{2} + kx – 8 = 0 is square of the other. Then the value of k is

- A.
2

- B.
8

- C.
-8

- D.
-2

Answer: Option D

**Explanation** :

If the roots are a and a^{2}, the product of roots = a^{3} = –8.

∴ a = –2.

Hence, sum of the roots = -k = (a + a^{2}) = (–2 + 4) = 2.

∴ k = -2.

Hence, option (d).

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**27. CAT 1995 QA | Geometry - Circles**

In the given figure, AB is diameter of the circle and points C and D are on the circumference such that ∠CAD = 30° and ∠CBA = 70°. What is the measure of ∠ACD?

- A.
40°

- B.
50°

- C.
30°

- D.
90°

Answer: Option A

**Explanation** :

If we draw the imaginary lines AC and BD, we find that

∠ CAD and ∠CBD are subtended by the same chord DC.

∴ ∠CAD = ∠CBD = 30°.

Thus, ∠DBA = (70 – 30) = 40°.

Also, ∠DBA and ∠ACD are subtended by the same chord DA.

Hence, ∠ACD = DBA = ∠40°.

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**28. CAT 1995 QA | Geometry - Triangles**

The length of a ladder is exactly equal to the height of the wall it is learning against. If lower end of the ladder is kept on a stool of height 3 m and the stool is kept 9 m away from the wall, the upper end of the ladder coincides with the top of the wall. Then the height of the wall is

- A.
12 m

- B.
15 m

- C.
18 m

- D.
11 m

Answer: Option B

**Explanation** :

The figure can be drawn as shown above.

Height of the wall = AD = AC = (AB + 3) or AB = (AC – 3).

In right-angled triangle ABC, AB^{2} + BC^{2} = AC^{2}.

Thus, (AC – 3)^{2} + 81 = AC^{2}.

∴ AC = 15 m.

Hence, height of the wall = 15 m.

Hence, option (b).

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**29. CAT 1995 QA | Arithmetic - Profit & Loss**

A stockist wants to make some profit by selling sugar. He contemplates about various methods . Which of the following would maximise his profit?

I. Sell sugar at 10% profit.

II. Use 900 g of weight instead of 1 kg.

III. Mix 10% impurities in sugar and selling sugar at cost price.

IV. Increase the price by 5% and reduce weights by 5%.

- A.
I or II

- B.
II

- C.
II, III and IV

- D.
Profits are same

Answer: Option B

**Explanation** :

Profit percentage in each case is

(i) 10%

(ii) $\frac{(100\times 100)}{900}=\frac{100}{9}\%$

(iii) $\frac{100\times 1.1.1-100}{100}\times 100=10\%$

(iv) $\frac{(10\times 100)}{95}=\frac{200}{19}\%$

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**30. CAT 1995 QA | Arithmetic - Time, Speed & Distance**

A man can walk up a moving ‘up’ escalator in 30 s. The same man can walk down this moving ‘up’ escalator in 90 s. Assume that his walking speed is same upwards and downwards. How much time will he take to walk up the escalator, when it is not moving?

- A.
30 s

- B.
45 s

- C.
60 s

- D.
90 s

Answer: Option B

**Explanation** :

Let the length of the escalator be 90 ft.

(There is no loss of generality in making this assumption.)

Let the speed of the escalator be y ft per second and the man’s walking speed be x ft per second.

According to the question, we get

$\frac{90}{30}=$ x + y

$\frac{90}{90}=$ x - y

Adding the above equations, we get 2x = 4, i.e., x = 2.

∴ Time taken by the man to walk up the escalator

when it is not moving = $\frac{90}{2}$ or 45 s.

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**31. CAT 1995 QA | Algebra - Quadratic Equations**

Two positive integers differ by 4 and sum of their reciprocals is $\frac{10}{21}$. Then one of the numbers is

- A.
3

- B.
1

- C.
5

- D.
21

Answer: Option A

**Explanation** :

Let one of the numbers be x. So the other number would be (x + 4).

According to the question, we have $\frac{1}{x}+\frac{1}{(x+4)}=$21 or x = 3.

**Hint:**

****Please note that the sum of reciprocals is basically $=\frac{(Sumoftheintegers)}{(Productoftheintegers)}$.

So we have to find two integers whose sum is 10 and whose product is 21.

So x + (x + 4) = 10 or x = 3.

Hence, option (a).

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**32. CAT 1995 QA | Algebra - Number Theory**

Three bells chime at an interval of 18 min, 24 min and 32 min. At a certain time they begin to chime together. What length of time will elapse before they chime together again?

- A.
2 hr and 24 min

- B.
4 hr and 48 min

- C.
1 hr and 36 min

- D.
5 hr

Answer: Option B

**Explanation** :

The bells will chime together after a time that is equal to the LCM of 18, 24 and 32 = 288 min = 4 hr and 48 min.

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**33. CAT 1995 QA | Algebra - Inequalities & Modulus**

What is the value of m which satisfies 3m^{2} – 21m + 30 < 0?

- A.
m < 2 or m > 5

- B.
m > 2

- C.
2 < m < 5

- D.
Both a and c

Answer: Option D

**Explanation** :

3m^{2} – 21m + 30 < 0

∴ m^{2} – 7m + 10 < 0 ⇒ (m – 5)(m – 2) < 0.

So either (m – 5) < 0 and (m – 2) > 0 or (m – 2) < 0 and (m – 5) > 0.

Hence, either m < 5 and m > 2, i.e., 2 < m < 5 or m < 2 and m > 5.

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**34. CAT 1995 QA | Arithmetic - Percentage**

The rate of inflation was 1000%. Then what will be the cost of an article, which costs 6 units of currency now, 2 years from now?

- A.
666

- B.
660

- C.
720

- D.
726

Answer: Option D

**Explanation** :

Required cost = $6{\left[1+\frac{1000}{100}\right]}^{2}$= 6(11)^{2} = 121 × 6 = 726.

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**35. CAT 1995 QA | Algebra - Functions & Graphs**

Largest value of min(2 + x^{2}, 6 – 3x), when x > 0, is

- A.
1

- B.
2

- C.
3

- D.
4

Answer: Option C

**Explanation** :

If x = 1, we have min(3, 3) = 3.

If x = 2, we have min(6, 0) = 0.

If x = 3, we have min(11, –3) = –3.

If x = 0.5, we have min(2.25, 4.5) = 2.25.

If x = 0.3, we have min(2.09, 5.1) = 2.09.

Thus, we find that as x increases above 1 and when it decreases below 1, the value of the function decreases.

It is maximum at x = 1 and the corresponding value = 3.

**Hint: **Please note that the highest value of the given fraction will be at a point where (2 + x^{2}) = (6 – 3x), as

even if one of the values increases beyond this, the other value will be the minimum value.

If we equate the two, we get x^{2} + 3x – 4 = 0. Solving this, we get x = 1 or x = –4.

Since x > 0, it has to be 1 and hence the result.

Workspace:

**36. CAT 1995 QA | Miscellaneous**

A, B, C and D are four towns, any three of which are non-collinear. Then the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is

- A.
7

- B.
8

- C.
9

- D.
24

Answer: Option D

**Explanation** :

Let us choose a town, say A.

If I were to consider this as the base town and construct two roads such that I connect any pair of towns, I get

the following combinations:

1. AB – BC, 2. AB – BD, 3. AC – CB, 4. AC – CD,

5. AD – DB and 6. AD – DC.

From any of these combinations, if I were to construct a road such that it again comes back to A, then it would form a triangle.

To avoid a triangle, the third road that I construct should not be connected to A but to the third town.

Hence, the combination would be:

1. AB – BC – CD, 2. AB – BD – DC, 3. AC – CB – BD,

4. AC – CD – DB, 5. AD – DB – BC and 6. AD – DC – CB.

Thus, from each town, we can construct 6 such combinations.

Hence, total number of combinations that we can have from four towns = (6 × 4) = 24.

Workspace:

**Directions for next 2 questions: Answer the questions based on the following information.**

le(x, y) = Least of (x, y)

mo(x) = |x|

me(x, y) = Maximum of (x, y)

**37. CAT 1995 QA | Algebra - Functions & Graphs**

Find the value of me(a + mo(le(a, b)); mo(a + me(mo(a), mo(b))), at a = –2 and b = –3.

- A.
1

- B.
0

- C.
5

- D.
3

Answer: Option A

**Explanation** :

If a = –2 and b = –3, then our expression will be me(–2 + mo(le(–2, –3)), mo(–2 + me(mo(–2), mo(–3)))

= me(–2 + mo(–3), mo(–2 + me(2, 3)))

= me(–2 + 3, mo(–2 + 3))

= me(1, mo(1)) = me(1, 1) = 1.

Workspace:

**38. CAT 1995 QA | Algebra - Functions & Graphs**

Which of the following must always be correct for a, b > 0?

- A.
mo(le(a, b)) ≥ (me(mo(a), mo(b))

- B.
mo(le(a, b)) > (me(mo(a), mo(b))

- C.
mo(le(a, b)) < (le(mo(a), mo(b))

- D.
mo(le(a,b)) = le(mo(a), mo(b))

Answer: Option D

**Explanation** :

Please note that the fastest way to solve these sums is the method of simulation, i.e., select any arbitrary values in the range given and verify whether the option holds good. E.g. a = 2, b = 3.

In this case, option (a) LHS = mo(le(2, 3)) = mo(2) = 2.

RHS = (me(mo(2), mo(3)) = (me(2, 3)) = 3. Hence, LHS < RHS.

(b) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (me(mo(2), mo(3)) = (me(2, 3)) = 3. Hence, LHS < RHS.

(c) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (le(mo(2), mo(3)) = le(2, 3) = 2. Hence, LHS = RHS.

(d) LHS = mo(le(2, 3)) = mo(2) = 2. RHS = (le(mo(2), mo(3)) = le(2, 3) = 2. Hence, LHS = RHS.

Workspace:

**39. CAT 1995 QA | Algebra - Functions & Graphs**

For what values of ‘a’ is me(a^{2} – 3a, a – 3) < 0?

- A.
a > 3

- B.
0 < a < 3

- C.
a < 0

- D.
a = 3

Answer: Option B

**Explanation** :

Let us verify by taking arbitrary values of a in the range specified.

(a) a > 3. Let a = 4.

So me(a^{2} – 3a, a – 3) = me(4, 1) = 4 > 0.

(b) 0 < a < 3. Let a = 2.

So me(a^{2} – 3a, a – 3) = me(–2, –1) = –1 < 0.

(c) a < 0. Let a = –1.

So me(a^{2} – 3a, a – 3) = me(4, –4) = 4 > 0.

(d) a = 3, me(a^{2} – 3a, a – 3) = me(0, 0) = 0.

Workspace:

**40. CAT 1995 QA | Algebra - Functions & Graphs**

For what values of ‘a’ is le(a^{2} – 3a, a – 3) < 0?

- A.
a > 3

- B.
0 < a < 3

- C.
a < 0

- D.
Both b and c

Answer: Option D

**Explanation** :

We can work this on the above lines. E.g.

(a) a > 3. Let a = 4. So le(a^{2} – 3a, a – 3) = le(4, 1) = 1 > 0.

(b) 0 < a < 3. Let a = 2. So le(a^{2} – 3a, a – 3) = le(–2, –1) = –2 < 0.

(c) a < 0. Let a = –1. So le(a^{2} – 3a, a – 3) = le(4, –4) = –4 < 0.

Workspace:

**41. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

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

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**If x, y and z are real numbers, is z – x even or odd?**

I. xyz is odd.

II. xy + yz + zx is even.

Answer: 1

**Explanation** :

Statement I suggests that xyz is odd. This is only possible if all three of them are odd.

Hence, z – x is even.

So only statement I is required to answer the question.

Workspace:

**42. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

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

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the value of x, if x and y are consecutive positive even integers?**

I. (x – y)^{2} = 4

II. (x + y)^{2} < 100

Answer: 4

**Explanation** :

Statement I is useless as it only tells that if x and y are consecutive positive even integers, then (x – y)^{2} has to be equal to 4.

Statement II suggests the possibility that the numbers could be 2 and 4. But it does not suggest which is x and

which is y.

Hence, we cannot find the value of x using either statements.

Workspace:

**43. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

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

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the profit percentage?**

I. The cost price is 80% of the selling price.

II. The profit is Rs.50.

Answer: 1

**Explanation** :

It is clear that only statement I is required to answer the question, if CP = 0.8 SP, then SP = $\left(\frac{1}{0.8}\right)$CP.

∴ SP = 1.25 CP

Thus, profit percentage is 25.

Workspace:

**44. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the area of the triangle?**

I. Two sides are 41 cm each.

II. The altitude to the third side is 9 cm long.

Answer: 3

**Explanation** :

Both statements are required to answer the question.

Statement I tells us that the triangle is an isosceles triangle.

In an isoceles triangle, the altitude is also the median and bisects the third side.

Hence, if we know the altitude length and the length of the congruent sides in an isoceles triangle, we can find its base. And if we know the base and the height of a triangle, we can find its area.

Workspace:

**45. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the price of bananas?**

I. With Rs.84, I can buy 14 bananas and 35 oranges.

II. If price of bananas is reduced by 50%, then we can buy 48 bananas in Rs.12.

Answer: 2

**Explanation** :

Statement II in itself suggests the price of a banana.

Since we can buy 48 bananas in Rs.12, price of a banana = Re.0.25. And since this price is after 50% reduction, the actual price of a banana = Re.0.5.

Workspace:

**46. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the first term of an arithmetic progression of positive integers?**

I. Sum of the squares of the first and the second term is 116.

II. The fifth term is divisible by 7.

Answer: 1

**Explanation** :

Since 116 is less than 11^{2}, it can be figured out that both the first two terms of the AP should be less than 10.

There is only one pair of positive integers whose squares add up to 116 and they are 10 and 4.

Thus, these two should be the first two terms of the AP.

Hence, the first term is 4, and can be obtained only from statement I.

Statement II merely suggests that the fifth term is of the form 7k. Nothing correct can be concluded from this.

Workspace:

**47. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the length of rectangle ABCD?**

I. Area of the rectangle is 48 square units.

II. Length of the diagonal is 10 units.

Answer: 3

**Explanation** :

This can be solved using both the statements together.

From statement I, we know that L × B = 48 or B = $\frac{L}{48}.$

From statement II, we know L^{2} + B^{2} = 100

Combining statements I and II,

${L}^{2}+{\left(\frac{L}{48}\right)}^{2}=100.$

L is the only unknown in this equation and can be found out.

Workspace:

**48. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**What is the number x?**

I. The LCM of x and 18 is 36.

II. The HCF of x and 18 is 2.

Answer: 3

**Explanation** :

We know that product of two numbers = LCM × HCF = 36 × 2 = 72.

So x = $\frac{72}{18}$ = 4.

Hence, both the statements are required to answer the question.

Workspace:

**49. CAT 1995 QA | Data Sufficiency**

**Direction: Each of these questions is followed by two statements, I and II. Mark the answer as**

2. if the question can be answered with the help of statement II alone.

3. if both statement I and statement II are needed to answer the question.

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

**Is x + y – z + t even?**

I. x + y + t is even.

II. t and z are odd.

Answer: 3

**Explanation** :

This can be answered using both the statements.

Statement II suggests that both t and z are odd.

Statement I suggests that (x + y + t) is even.

Since the difference between an even and an odd number is always odd, (x + y + t) – z will be odd.

Workspace:

**50. CAT 1995 QA | Geometry - Triangles**

Which one of the following cannot be the ratio of angles in a right-angled triangle?

- A.
1 : 2 : 3

- B.
1 : 1 : 2

- C.
1 : 3 : 6

- D.
None of these

Answer: Option C

**Explanation** :

The largest angle in a right-angled triangle is 90º, which corresponds to the highest part of the ratio.

Let us evaluate each option.

Option (a): Angles will be x, 2x and 3x

∴ x + 2x + 3x = 180°

⇒ x = 30° and 3x = 90°

Hence, option (a) is correct.

Option (b): Angles will be x, x and 2x

∴ x + x + 2x = 180°

⇒ x = 45° and 2x = 45°

Hence, option (b) is correct.

Option (c): Angles will be x, 3x and 6x

∴ x + 3x + 6x = 180°

⇒ x = 18° and neither 3x nor 6x = 90°

Hence, option (c) is not a right triangle.

Hence, option (c).

Workspace:

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