# CAT 1998 QA

Paper year paper questions for CAT 1998 QA

**Direction: **Choose the appropriate alternative.

**1. CAT 1998 QA | Arithmetic - Time & Work**

A company has a job to prepare certain number cans and there are three machines A, B and C for this job. A can complete the job in 3 days, B can complete the job in 4 days, and C can complete the job in 6 days. How many days will the company take to complete the job if all the machines are used simultaneously?

- A.
4 days

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

- C.
3 days

- D.
12 days

Answer: Option B

**Explanation** :

In one day, A would do $\frac{1}{3}$ of the job, B would do $\frac{1}{4}$ of the job and C would do $\frac{1}{6}$ of the job.

Hence, if all three of them work simultaneously, in one day they would do $\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\right)=\frac{3}{4}$ of the job.

Hence, to complete the entire job together they would take $\frac{4}{3}$ days.

**Shortcut:**

A can complete the job in 3 days. So A, B and C combined will take less days than A alone to finish the job. So straightway option (b).

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**2. CAT 1998 QA | Algebra - Number Theory**

n^{3} is odd. Which of the following statement(s) is(are) true?

I. n is odd.

II. n^{2} is odd.

III. n^{2} is even.

- A.
I only

- B.
II only

- C.
I and II

- D.
I and III

Answer: Option C

**Explanation** :

If n^{3} is odd, then n should also be odd. Hence, n^{2} should also be odd, not even. So only I and II are true.

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

Production pattern for number of units (in cubic feet) per day.

For a truck that can carry 2,000 cubic ft, hiring cost per day is Rs. 1,000. Storing cost per cubic feet is Rs. 5 per day.

**3. CAT 1998 QA | Miscellaneous**

If all the units should be sent to the market, then on which days should the trucks be hired to minimize the cost?

- A.
2nd, 4th, 6th, 7th

- B.
7th

- C.
2nd, 4th, 5th, 7th

- D.
None of these

Answer: Option C

**Explanation** :

We can see that there are two types of cost: (i) Transportation cost (viz. hiring truck) and (ii) storing cost. It can be seen that the daily production is far less than the capacity of the truck. So a truck can be hired to carry multiple days production at one go. So as long as the storing cost is less than the cost of hiring the truck (i.e. Rs. 1,000), it makes sense to store the production. When the storing cost exceeds Rs. 1,000, it is best that the entire lot be sent to the market. The cost pattern is as given in the following table:

* In spite of the fact that storing is cheaper than hiring truck on the last day, we have to do with the latter option because everything that is manufactured has to be sent to the market. So according to this table, if the truck is hired on 2nd, 4th, 6th and 7th days, total cost = Rs. 6,150. But is this the most cost-effective scheme? It can be seen that we are hiring truck on two consecutive days (6th and 7th). Hence, since everything that is manufactured has to be sent to the market, we have yet another option of hiring the truck on the 5th day and sending the 6th and 7th days production together on the last day. In that case, the cost on 5th day would be Rs. 1,000 (i.e. Rs. 200 more than the present cost), the cost on the 6th day would be (120 × 5) = Rs. 600 (i.e.

Rs. 400 less than the present cost) and the cost on the 7th day would be Rs. 1,000 (the same as the present cost). Hence, we can say that the total cost would actually come down by (+200 – 400 = – 200) Rs. 200.

Hence, this becomes the most cost-effective scheme. So we should hire trucks on 2nd, 4th, 5th and 7th days.

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**4. CAT 1998 QA | Miscellaneous**

If the storage cost is reduced to Re 0.80 per cubic feet per day, then on which day(s), should the truck be hired?

- A.
4th

- B.
7th

- C.
4th and 7th

- D.
None of these

Answer: Option B

**Explanation** :

If the storage cost is reduced to Re 0.8 per cubic feet, then the cost pattern is as given in the following table.

** In spite of the fact that storing is cheaper than hiring truck on the last day, we have to do with the latter option because everything that is manufactured has to be sent to the market.

Hence, the most cost-effective scheme would be sending the entire production on the 7th day.

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**5. CAT 1998 QA | Arithmetic - Percentage**

One bacterium splits into eight bacteria of the next generation. But due to environmental condition only 50% survives and remaining 50% dies after producing next generation. If the seventh generation number is 4,096 million, what is the number in first generation?

- A.
1 million

- B.
2 million

- C.
4 million

- D.
8 million

Answer: Option A

**Explanation** :

Let there be x bacteria in the first generation i.e. n_{1} = x

∴ n₂ = 8x, but only 50% survives

⇒ n_{2, survived} = $\frac{8x}{2}=4x$

n_{3} = 8(4x), but only 50% survives

n_{3, survived} = 4^{2} x

n_{3, survived} = 4^{7-1} x = 4096 million

∴ x = 1 million

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**6. CAT 1998 QA | Geometry - Circles**

Three circles, each of radius 20, have centres at P, Q and R. Further, AB = 5, CD = 10 and EF = 12. What is the perimeter of ΔPQR?

- A.
120

- B.
66

- C.
93

- D.
87

Answer: Option C

**Explanation** :

PQ = PE + FQ – FE

= radius of circle 1 + radius of circle 2 – FE = 20 + 20 – 12 = 28

Similarly, QR = 20 + 20 – CD = 40 – 10 = 30 and PR = 20 + 20 – AB = 40 – 5 = 35

So perimeter of ΔPQR = 28 + 30 + 35 = 93

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**7. CAT 1998 QA | Algebra - Number Theory**

(BE)^{2} = MPB, where B, E, M and P are distinct integers. Then M =

- A.
2

- B.
3

- C.
9

- D.
None of these

Answer: Option B

**Explanation** :

Since MPB is a three-digit number, and also the square of a two-digit number, it can have a maximum value of 961 viz. 31^{2}. This means that the number BE should be less than or equal to 31. ⇒ B can be 0, 1, 2 or 3. Since the last digit of MPB is also B, it can only be 0 or 1 (as none of the squares end in 2 or 3).

The only squares that end in 0 are 100, 400 and 900. But for this to occur the last digit of BE also has to be 0. Since E and B are distinct integers, both of them cannot be 0. Hence, B has to be 1. BE can be a number between 11 and 19 (as we have also ruled out 10), with its square also ending in 1.

Hence, the number BE can only be 11 or 19. 11^{2} = 121. This is not possible as this will mean that M is also equal to 1. Hence, our actual numbers are 19^{2} = 361.

Hence, M = 3.

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**8. CAT 1998 QA | Algebra - Number Theory**

Five-digit numbers are formed using only 0, 1, 2, 3, 4 exactly once. What is the difference between the maximum and minimum number that can be formed?

- A.
19800

- B.
41976

- C.
32976

- D.
None of these

Answer: Option C

**Explanation** :

The maximum and the minimum five-digit numbers that can be formed using only 0, 1, 2, 3, 4 exactly once are 43210 and 10234 respectively. The difference between them is 43210 – 10234 = 32976.

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**9. CAT 1998 QA | Modern Math - Permutation & Combination**

How many numbers can be formed from 1, 2, 3, 4, 5, without repetition, when the digit at the unit’s place must be greater than that in the ten’s place?

- A.
54

- B.
60

- C.
17

- D.
2 × 4!

Answer: Option B

**Explanation** :

$\frac{}{}\frac{}{}\frac{}{}\frac{t}{Ten\text{'}splace}\frac{u}{Unit\text{'}splace}$

u > t

Case (i): t = 1

u can be 2, 3, 4, 5 ⇒ 4 possibilities

So total possible number that can be formed

= 4 × 3 × 2 × 1

Case (ii): t = 2

u can be 3, 4, 5 ⇒ 3 possibilities

So total possible number that can be formed

= 3 × 3 × 2 × 1

Case (iii): t = 3

u can be 4, 5 ⇒ 2 possibilities

So total possible number that can be formed

= 2 × 3 × 2 × 1

Case (iv): t = 4

u can be 5 ⇒ 1 possibility

So total possible number that can be formed

= 1 × 3 × 2 × 1

Hence, total possible number = (4 + 3 + 2 + 1) $\overline{)\overline{)3}}$ = 60

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**10. CAT 1998 QA | Arithmetic - Time, Speed & Distance**

Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed of 4 km/hr. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When will

they meet each other?

- A.
7 hr

- B.
10 hr

- C.
35 km from A

- D.
Mid-way between A and B

Answer: Option D

**Explanation** :

Since A and B are moving in opposite directions, we will add their speeds to calculate the effective speeds. In other words, in the first hour they would effectively cover a distance of (4 + 2) = 6 km towards each other. In the second hours, they would effectively cover a distance of (4 + 2.5) = 6.5 km towards each other. In the third hour, (4 + 3) = 7 km.

In the fourth hour, (4 + 3.5) = 7.5 km and so on. We can see that the distances that they cover in each hour are in AP, viz. 6, 6.5, 7, 7.5 ... with a = 6 and d = 0.5. Since they have to effectively cover a distance of 72 km, the time taken to cover this much distance would be the time taken to meet each other. So the sum of the first n terms of our AP has to be 72.

If we are to express this in our equation of sum of first n terms of the AP, we will get S_{n} = $\frac{n}{2}\times $ [2a + (n – 1)d].

Substituting our values, we will get

72 = $\frac{n}{2}$ × [12 + 0.5(n – 1)]

Solving this, we get n = 9 hr. In 9 hr A would have covered (9 × 4) = 36 km.

So B would also have covered (72 – 36) = 36 km.

Hence, they would meet mid-way between A and B.

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

A cow is tethered at point A by a rope. Neither the rope nor the cow is allowed to enter ΔABC.

∠BAC = 30°

I(AB) = I(AC) = 10 m

**11. CAT 1998 QA | Geometry - Mensuration**

What is the area that can be grazed by the cow if the length of the rope is 8 m?

- A.
134π $\frac{1}{3}$ sq.m

- B.
121π sq.m

- C.
132π sq.m

- D.
$\frac{176\pi}{3}$ sq.m

Answer: Option D

**Explanation** :

It can be seen that if the length of the rope is 8 m, then the cow will be able to graze an area equal to (the area of the circle with radius 8m) – (Area of the sector of the same circle with angle 30°) = π(8)² - $\frac{30}{360}$π(8)^{2}

= 64π - $\frac{1}{12}$ (64π) = 64π $\left(\frac{11}{12}\right)$ = $\frac{176\pi}{3}$ sq.m

**Shortcut:**

Area grazed without restriction is 64π m^{2} it should be less than 64π sq.m. with restriction. So choice (d).

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**12. CAT 1998 QA | Geometry - Mensuration**

What is the area that can be grazed by the cow if the length of the rope is 12 m?

- A.
133π $\frac{1}{3}$ sq.m

- B.
121π sq.m

- C.
132π sq.m

- D.
$\frac{176\pi}{3}$ sq.m

Answer: Option A

**Explanation** :

If the length of the rope is 12 m, then the total area that can be grazed by the cow is as depicted in the diagram.

Area 1 is (the area of the circle with radius 12) – (Area of the sector of the same circle with angle 30º)

So area 1 = π(12)² - $\frac{30}{360}$π(12)² = 132π

Since the length of the rope is higher than the sides of the triangle (viz. AB and AC), if the cow reaches point B or C, there would still be a part of the rope (12 – 10) = 2 m in length. With this extra length available the cow can further graze an area equivalent to some part of the circle with radius = 2 m from both points, i.e. B and C. This is depicted as area 2 and area 3 in the diagram. Hence, the actual area grazed will be slightly more than 132π . The only answer choice that supports this is (a).

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**13. CAT 1998 QA | Algebra - Simple Equations**

A yearly payment to the servant is Rs. 90 plus one turban. The servant leaves the job after 9 months and receives Rs. 65 and a turban. Then find the price of the turban.

- A.
Rs. 10

- B.
Rs. 15

- C.
Rs. 7.50

- D.
Cannot be determined

Answer: Option A

**Explanation** :

Let the cost of the turban be T.

Total payment for one year = Rs. 90 + T. So the payment for 9 months should be $\frac{3}{4}$ (90 + T). But this is equal to (65 + T). Equating the two, we get T = Rs. 10.

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**14. CAT 1998 QA | Geometry - Quadrilaterals & Polygons**

Four identical coins are placed in a square. For each coin the ratio of area to circumference is same as the ratio of circumference to area. Then find the area of the square that is not covered by the coins.

- A.
16(π - 1)

- B.
16(8 - π)

- C.
16(4 - π)

- D.
$16\left(4-\frac{\pi}{2}\right)$

Answer: Option C

**Explanation** :

Let R be the radius of each circle. Then 2 $\frac{\pi {R}^{2}}{2\pi R}=\frac{2\pi R}{\pi {R}^{2}}$ which implies that $\frac{R}{2}=\frac{2}{R},$ i.e., R^{2} = 4, i.e. R = 2.

Then the length of the square is 8. Thus, the area of the square is 64, while the area covered by each coin is π2^{2} = 4π. Since there are four coins, the area covered by coins is 4(4π ) = 16π.

Hence, the area not covered by the coins is 64 – 16π = 16(4 – π ).

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**15. CAT 1998 QA | Algebra - Number Theory**

Three wheels can complete 60, 36 and 24 revolutions per minute. There is a red spot on each wheel that touches the ground at time zero. After how much time, all these spots will simultaneously touch the ground again?

- A.
$\frac{5}{2}s$

- B.
$\frac{5}{3}s$

- C.
6 s

- D.
7.5 s

Answer: Option C

**Explanation** :

The time taken by the white spots on all three wheels to simultaneously touch the ground again will be equal to the LCM of the times taken by the three wheels to complete one revolution.

The first wheel complete 60 revolutions per minute.

This means that to complete one revolution, it takes $\left(\frac{60}{60}\right)$ = 1 s.

Similarly, the second wheel takes $\left(\frac{36}{60}\right)=\frac{3}{5}$ s to complete one revolution.

Similarly, the third wheel takes $\frac{24}{60}=\frac{2}{5}$ s to complete one revolution.

Hence, LCM of 1, $\frac{3}{5},\frac{2}{5}=\frac{L.C.M.(1,3,2)}{H.C.F.(1,5,5)}=\frac{6}{1}=6$ s

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**16. CAT 1998 QA | Algebra - Number Theory**

A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29.

- A.
5

- B.
4

- C.
1

- D.
Cannot be determined

Answer: Option A

**Explanation** :

Since 899 is divisible by 29, so you can directly divide the remainder of 63 by 29, so $\frac{63}{29}$ will give 5 as a remainder, option (a).

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**17. CAT 1998 QA | Algebra - Number Theory**

A is the set of positive integers such that when divided by 2, 3, 4, 5, 6 leaves the remainders 1, 2, 3, 4, 5 respectively. How many integers between 0 and 100 belong to set A?

- A.
0

- B.
1

- C.
2

- D.
None of these

Answer: Option B

**Explanation** :

Note that the difference between the divisors and the remainders is constant.

2 – 1 = 3 – 2 = 4 – 3 = 5 – 4 = 6 – 5 = 1

In such a case, the required number will always be [a multiple of LCM of (2, 3, 4, 5, 6) – (The constant difference)].

LCM of (2, 3, 4, 5, 6) = 60

Hence, the required number will be 60n – 1.

Thus, we can see that the smallest such number is (60 × 1) – 1 = 59

The second smallest is (60 × 2) – 1 = 119

So between 1 and 100, there is only one such number, viz. 59

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

A, B, C and D collected one-rupee coins following the given pattern.

Together they collected 100 coins.

Each one of them collected even number of coins.

Each one of them collected at least 10 coins.

No two of them collected the same number of coins.

**18. CAT 1998 QA | Algebra - Number Theory**

The maximum number of coins collected by any one of them cannot exceed

- A.
64

- B.
36

- C.
54

- D.
`None of these

Answer: Option A

**Explanation** :

For, if any one of them collects the maximum number of coins, the remaining three should collect the minimum number of coins. To have distinct, even, atleast 10 coins; they will have to collect 10, 12, 14 coins. So if the three of them collect (10 + 12 + 14) = 36 coins, the fourth one has to collect (100 – 36) = 64 coins which has to be the maximum by any one person

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**19. CAT 1998 QA | Algebra - Number Theory**

If A collected 54 coins, then the difference in the number of coins between the one who collected maximum number of coins and the one who collected the second highest number of coins must be at least

- A.
12

- B.
24

- C.
30

- D.
None of these

Answer: Option C

**Explanation** :

Since A has collected 54 coins out of 100, he should obviously be the person who collected the maximum number of coins. For the difference between him and the second highest person to be minimum, the second highest person should collect the maximum number of coins possible under the given conditions. And for this to happen, the remaining two should collect the minimum number of coins. So if the two of them collect 10 and 12 coins, i.e. 22 coins between themselves, the third person would have to collect (100 – 54 – 22) = 24 coins. Hence, the difference between him and the highest person should at least be (54 – 24) = 30.

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**20. CAT 1998 QA | Algebra - Number Theory**

If A collected 54 coins and B collected two more coins than twice the number of coins collected by C, then the number of coins collected by B could be

- A.
28

- B.
20

- C.
26

- D.
22

Answer: Option D

**Explanation** :

If A has collected 54 coins, the remaining three of them should collect (100 – 54) = 46 coins between themselves.

Let us assume that C has collected 10 coins. So B will collect (2 × 10) + 2 = 22. So A will collect (46 – 10 – 22) = 14 coins, which is a possible combination.

Let us now assume that C picks up 12 coins. So B should pick up (2 × 12) + 2 = 26. So A will have to collect (46 – 12 – 26) = 8 coins.

This combination is not possible. It can be concluded that C cannot pick up more than 10 coins and hence B has to pick up 22 coins to satisfy the given condition.

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**21. CAT 1998 QA | Algebra - Number Theory**

Number of students who have opted for subjects A, B and C are 60, 84 and 108 respectively. The examination is to be conducted for these students such that only the students of the same subject are allowed in one room. Also the number of students in each room must be same. What is the minimum number of rooms that should be arranged to meet all these conditions?

- A.
28

- B.
60

- C.
12

- D.
21

Answer: Option D

**Explanation** :

HCF of 60, 84 and 108 is 12. Hence, 12 students should be seated in each room. So for subject A we would require $\left(\frac{60}{12}\right)$ = 5 rooms, for subject B we would require $\left(\frac{84}{12}\right)$ = 7 rooms and for subject C we would require $\left(\frac{108}{12}\right)$ = 9 rooms. Hence, minimum

number of rooms required to satisfy our condition = (5 + 7 + 9) = 21 rooms.

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**22. CAT 1998 QA | Modern Math - Permutation & Combination**

How many five-digit numbers can be formed using the digits 2, 3, 8, 7, 5 exactly once such that the number is divisible by 125?

- A.
0

- B.
1

- C.
4

- D.
3

Answer: Option C

**Explanation** :

Let us find some of the smaller multiples of 125. They are 125, 250, 375, 500, 625, 750, 875, 1000 ... A five-digit number is divisible by 125, if the last three digits are divisible by 125.

So the possibilities are 375 and 875, 5 should come in unit’s place, and 7 should come in ten’s place. Thousand’s place should contain 3 or 8. We can do it in 2! ways. Remaining first two digits, we can arrange in 2! ways. So we can have 2! × 2! = 4 such numbers.

There are: 23875, 32875, 28375, 82375.

Hence, option 3.

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**23. CAT 1998 QA | Algebra - Number Theory**

You can collect as many rubies and emeralds as you can. Each ruby is worth Rs. 4 crore and each emerald is worth Rs. 5 crore. Each ruby weighs 0.3 kg. And each emerald weighs 0.4 kg. Your bag can carry at the most 12 kg. What should you collect to get the maximum wealth?

- A.
20 rubies and 15 emeralds

- B.
40 rubies

- C.
28 rubies and 9 emeralds

- D.
None of these

Answer: Option B

**Explanation** :

To maximise the value of the wealth, we must carry more of the one whose value per kilogram is more. Value per kilogram of ruby = $\left(\frac{4}{0.3}\right)$ = Rs. 13.33 crore, and value per rupee of each emerald = $\left(\frac{5}{0.4}\right)$ = Rs. 12.5 crore.

It is obvious that we should carry entire 12 kg of ruby. This would amount to $\left(\frac{12}{0.3}\right)$ = 40 rubies.

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**24. CAT 1998 QA | Arithmetic - Ratio, Proportion & Variation**

I have one-rupee coins, 50-paisa coins and 25-paisa coins. The number of coins are in the ratio 2.5 : 3 : 4. If the total amount with me is Rs. 210, find the number of one-rupee coins.

- A.
90

- B.
85

- C.
100

- D.
105

Answer: Option D

**Explanation** :

Since the number of coins are in the ratio 2.5 : 3 : 4, the values of the coins will be in the ratio

(1 × 2.5) : (0.5 × 3) : (0.25 × 4) = 2.5 : 1.5 : 1 or 5 : 3 : 2

Since they totally amount to Rs. 210, if the value of each type of coins are assumed to be 5x, 3x and 2x, the average value per coin will be $\frac{210}{10x}$.

So the total value of one-rupee coins will be $5\times \left(\frac{210}{10x}\right)$ = Rs. 105

So the total number of one-rupee coins will be 105.

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**25. CAT 1998 QA | Algebra - Simple Equations**

My son adores chocolates. He likes biscuits. But he hates apples. I told him that he can buy as many chocolates he wishes. But then he must have biscuits twice the number of chocolates and should have apples more than biscuits and chocolates together. Each chocolate cost Re 1. The cost of apple is twice the chocolate and four biscuits are worth one apple. Then which of the following can be the amount that I spent on that evening on my son if number of chocolates, biscuits and apples brought were all integers?

- A.
Rs. 34

- B.
Rs. 33

- C.
Rs. 8

- D.
None of these

Answer: Option A

**Explanation** :

Let the number of chocolates be c.

Number of apples has to be more than 3c, lets say 3c + k, k is a positive integer.

Total spend = 8c + 2k

for c = 4, k = 2

Total spend = 34

Hence (a) is the answer.

The cost of each chocolate is Re 1. So the cost of apple should be Rs. 2 and that of one biscuit should be Re 0.5. Thus, if he eats x chocolates, he has to eat 2x biscuits. Hence, the total value of chocolates will be Rs. x and that of biscuits will be (0.5)(2x) = Rs. x. Hence, we see that the value of chocolates is to the value of biscuits will always be 1 : 1. As per our assumption he will have to eat more than (x + 2x) = 3x apples and hence the total value of the apples will be more than (2)(3x) = 6x. In other words, the ratio of value chocolates to apples or biscuits to apples will be more than 1 : 6. In other words, if the value of chocolates and biscuits is Re 1 each, then the value of apples has to be more than Rs. 6, or the combined value will be more than Rs. 8. This means that the

value of apples will always constitute more than $\frac{6}{8}$ or $\frac{3}{4}$ of the entire bill. It can further be observed that the total value of chocolates and biscuits together will always be an even integer and so will be the value of apples. This means that the combined value of all three of them has to be even and not odd. So Rs. 33 cannot be the answer. Also Rs. 8 cannot be the answer as, if we take the value of chocolates and biscuits to be minimum, i.e. Re 1 each, then the value of apples can be a minimum of Rs. 8. Hence, the total value will always be Rs. 10 or higher. The only option possible is Rs. 34. To verify this let us find two even numbers (one of them higher than $\frac{3}{4}$ of 34) which adds 34.

We can find many such numbers e.g. 32 + 2, 30 + 4, 28 + 6 and 26 + 8. All of these could be a possible combination.

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

A company purchases components A and B from Germany and USA respectively. A and B form 30% and 50% of the total production cost. Current gain is 20%. Due to change in the international scenario, cost of the German mark increased by 30% and that of USA dollar increased by 22%. Due to market conditions, the selling price cannot be increased beyond 10%.

**26. CAT 1998 QA | Arithmetic - Profit & Loss**

What is the maximum current gain possible?

- A.
10%

- B.
12.5%

- C.
0%

- D.
7.5%

Answer: Option A

**Explanation** :

Let CP = 100

Current gain = 20

⇒ SP = 120

CP = Cost of A + Cost of B + other

= 30% + 50% + 20%

Since cost of German mark (A) increase by 30%

New cost of A = 30 + 30% of 30 = 39

Similarly

New cost of B = 50 + 22% of 50 = 61

New CP = 39 + 61 + 20 = 120

New SP = 120 + 10% of 120 = 132

Maximum profit = $\frac{12}{20}=10\%$

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**27. CAT 1998 QA | Arithmetic - Profit & Loss**

If the USA dollar becomes cheap by 12% over its original cost and the cost of German mark increased by 20%, what will be the gain? (The selling price is not altered.)

- A.
10%

- B.
20%

- C.
15%

- D.
7.5%

Answer: Option B

**Explanation** :

Let CP = 100

Current gain = 20

⇒ SP = 120

CP = Cost of A + Cost of B + other

= 30% + 50% + 20%

New cost of A = 30 + 20% of 30 = 36

New cost of B = 50 – 12% of 50 = 44

New CP = 36 = 44 + 20 = 100

Gain = 20%

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**28. CAT 1998 QA | Arithmetic - Time, Speed & Distance**

I started climbing up the hill at 6 a.m. and reached the top of the temple at 6 p.m. Next day I started coming down at 6 a.m. and reached the foothill at 6 p.m. I walked on the same road. The road is so short that only one person can walk on it. Although I varied my pace on my way, I never stopped on my way. Then which of the following must be true?

- A.
My average speed downhill was greater than that of uphill

- B.
At noon, I was at the same spot on both the days.

- C.
There must be a point where I reached at the same time on both the days.

- D.
There cannot be a spot where I reached at the same time on both the days.

Answer: Option C

**Explanation** :

Since the distance travelled was the same both ways and also it was covered in the same time, the average speed will be the same uphill and downhill. Hence, statement (a) is false. Statement (b) need not be true. It would be true and had the speeds (and not average speed) been the same both ways. But it is clearly indicated that he varied his pace throughout. Now it has to be noted that the journey uphill and the journey downhill started at the same time, i.e. 6 a.m. and also ended at the same time 6 p.m. (though on different days). So if we were to assume a hypothetical case in which one person starts downhill at 6 a.m. and

other one starts uphill at 6 a.m., along the same path, then there would be a point on the path where they would meet (i.e. they would reach at the same time), irrespective of their speeds. Our case is similar to that, except for the fact that here, we have only one person moving both ways. So there has to be a point on the path, where he reached at the same time on both days.

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**29. CAT 1998 QA | Algebra - Number Theory**

What is the digit in the unit’s place of 2^{51}?

- A.
2

- B.
8

- C.
1

- D.
4

Answer: Option B

**Explanation** :

Since 2 has a cyclicity of 4,

i.e. 2^{1} = 2, 2^{2} = 4, 2^{3} = 8, 2^{4} = 16, 2^{5} = 32, 2^{6} = 64 ..., the last digits (2, 4, 8, 6) are in four cycles.

∴ On dividing $\frac{51}{4}$, we get the remainder as 3.

∴ The last digits has tobe 2^{3} = 8

**Shortcut:**

Since cyclicity of the power of 2 is 4, so 2^{51} can be written in 2^{4(12) + 3} or unit digit will be 2^{3} = 8.

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**30. CAT 1998 QA | Arithmetic - Mixture, Alligation, Removal & Replacement**

There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the

second container. Then

- A.
A > B

- B.
A < B

- C.
A = B

- D.
Cannot be determined

Answer: Option C

**Explanation** :

Let the capacity of each cup be 100 ml. So 300 ml of alcohol is taken out from the first container and poured into the second one. So the first vessel will have 200 ml of alcohol and the second one will have 500 ml of water and 300 ml of alcohol. So the ratio of water to alcohol in the second vessel is 5 : 3.

Hence, proportion of alcohol in B = 3 : 8

Now if 300 ml of mixture is removed from the second container, it will have $\left(300\times \frac{5}{8}\right)$ = 187.5 ml of water and $\left(300\times \frac{3}{8}\right)$ = 112.5 ml of alcohol. Now if this mixture is poured in the second vessel, that vessel would have (200 + 112.5) = 312.5 ml of alcohol and 187.5 ml of water. Hence, ratio of alcohol to water in this container = 312.5 : 187.5 = 5 : 3

Hence, proportion of water = A = 3 : 8

Hence, we find that A = B

**Note: **This result will be independent of the capacity of the cup.

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**31. CAT 1998 QA | Algebra - Number Theory**

A number is formed by writing first 54 natural numbers in front of each other as 12345678910111213 ... Find the remainder when this number is divided by 8.

- A.
7

- B.
1

- C.
2

- D.
0

Answer: Option C

**Explanation** :

The number formed by the last 3 digits of the main number is 354. The remainder is 2 if we divide 354 by 8. So the remainder of the main number is also 2 if we divide it by 8.

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**32. CAT 1998 QA | Geometry - Circles | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

Find the length of AB if ∠YBC = ∠CAX = ∠YOX = 90°.

I. Radius of the arc is given.

II. OA = 5

Answer: 4

**Explanation** :

Do not make the mistake of assuming O to be the centre of the circle. Since the centre is not known, knowing radius is not of great help. It can be observed that ∠BCA is also 90° , as in the quadrilateral OBCA, the remaining three angles are 90° . So the quadrilateral can either be a square or a rectangle. As we do not know even this, we cannot make use of the second statement as well. Hence, both the statements are not sufficient to answer the question.

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**33. CAT 1998 QA | Algebra - Number Theory | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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 n odd?

I. n is divisible by 3, 5, 7 and 9.

II. 0 < n < 400

Answer: 3

**Explanation** :

LCM of 3, 5, 7, 9 = 315. Hence, all the multiples of 315 will be divisible by 3, 5, 7 and 9. These may be even or odd. Hence, the statement I in itself is not sufficient to answer the question.

The statement II however suggests that the number is 315 itself (as it is the only multiple that lies between 0 and 400). Hence, n is indeed odd. We require both the statements together to answer this.

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**34. CAT 1998 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

Find 2⊗3, where 2 ⊗ 3 need not be equal to 3 ⊗ 2 .

I. 1⊗ 2 = 3

II. a ⊗ b = $\frac{(a+b)}{a},$ where a and b are positive.

Answer: 1

**Explanation** :

It is clear that statement II alone is enough to answer the question. This statement gives the value of the function a ⊗b, so we can find the value of 2⊗3 .

So 2⊗3 = $\frac{(2+3)}{2}=2.5$

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**35. CAT 1998 QA | Arithmetic - Percentage | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

Radha and Rani appeared in an examination. What was the total number of questions?

I. Radha and Rani together solved 20% of the paper.

II. Radha alone solved $\frac{3}{5}$ of the paper solved by Rani.

Answer: 4

**Explanation** :

Even by using both the statements, we can only find out the proportion of the paper solved by Radha and Rani. In the light of the fact that the number of questions solved by either or both of them is not given, we cannot answer the question asked.

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**36. CAT 1998 QA | Algebra - Simple Equations | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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 price of tea?

I. Price of coffee is Rs. 5 more than that of tea.

II. Price of coffee is Rs. 5 less than the price of a cold drink which cost three times the price of tea.

Answer: 3

**Explanation** :

From the statement II, we can frame the equation that:

(Cold drink) = 3(Tea) and (Coffee) = (Cold drink) – 5 = 3(Tea) – 5. So we have one equation in terms of prices of tea and coffee. Although, this alone may not be sufficient to answer the question, in the light of the equation provided by the first statement, viz. (Coffee) = (Tea) + 5, we can solve the two equations simultaneously and get the price of tea.

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**37. CAT 1998 QA | Arithmetic - Ratio, Proportion & Variation | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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’?

I. Ratio of a and b is 3 : 5, where b is positive.

II. Ratio of 2a and b is $\frac{12}{10},$ where a is positive.

Answer: 4

**Explanation** :

Note that both the statements give the same piece of information that a : b = 3 : 5 and that a and b are both positive. But none of the statements either in itself or together can give the value of a.

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**38. CAT 1998 QA | Arithmetic - Mixture, Alligation, Removal & Replacement | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

In a group of 150 students, find the number of girls.

I. Each girl was given 50 paise, while each boy was given 25 paise to purchase goods totalling Rs. 49.

II. Girls and boys were given 30 paise each to buy goods totalling Rs. 45.

Answer: 1

**Explanation** :

Using the first statement alone, we can alligate and find the ratio of boys to girls and hence the number of girls, i.e. as shown in the adjacent diagram, 150 students when divided in the ratio 115 : 260, give 46 girls and 104 boys. The second statement, however, does not throw any further light on the data given in the question as it simply suggests 0.3B + 0.3G = 45 or B + G = 150, which is already known. Hence, only statement I is required to answer the question.

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**39. CAT 1998 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

There are four envelopes — E_{1}, E_{2}, E_{3} and E_{4}— in which one was supposed to put letters L_{1}, L_{2}, L_{3} and L_{4 }meant for persons C_{1}, C_{2}, C_{3} and C_{4} respectively, but by mistake the letters got jumbled up and went in wrong envelopes. Now if C_{2} is allowed to open an envelope at random, then how will he identify the envelope containing the letter for him?

I. L_{2} has been put in E_{1}.

II. The letter belonging to C_{3} has gone in the correct envelope.

Answer: 1

**Explanation** :

The issue at hand is to make C_{2} identify in which envelope is the letter L_{2}. The first statement actually tells him this. Hence, it alone is sufficient to answer the question. The second statement only implies that his letter would be in either E_{1}, E_{2} or E_{4} and hence is not sufficient to answer the question.

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**40. CAT 1998 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

There are four racks numbered 1, 2, 3, 4 and four books numbered 1, 2, 3, 4. If an even rack has to contain an odd-numbered book and an odd rack contains an even-numbered book, then what is the position of book 4?

I. Second book has been put in third rack.

II. Third book has been put in second rack.

Answer: 1

**Explanation** :

From the question itself, we can figure out that book 4 can either be in rack 1 or rack 3. The first statement says that book 2 has been kept in rack 3. Hence, book 4 has to be kept in rack 1. So this statement is sufficient to answer the question. The second statement, however, does not add any additional information to what we already know. As books 3 in rack 2 would still imply book 4 can be in rack 1 or 3.

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**41. CAT 1998 QA | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

Find the value of X in terms of ‘a’.

I. Arithmetic mean of X and Y is ’a’ while the geometric mean is also ‘a’.

II. $\frac{X}{Y}$ = R; X - Y = D.

Answer: 1

**Explanation** :

Statement II is not required at all as no way can we express X in terms of 'a'.

Statement I implies that X + Y = 2a and XY = a2. Solving these two, we can say that X = a.

Hence, statement I alone is sufficient.

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**42. CAT 1998 QA | Geometry - Circles | Data Sufficiency**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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.

There are two concentric circles C_{1} and C_{2} with radii r_{1} and r_{2}. The circles are such that C_{1} fully encloses C_{2}. Then what is the radius of C_{1}?

I. The difference of their circumference is k cm.

II. The difference of their areas is m sq. cm.

Answer: 3

**Explanation** :

The information given in the question implies that r_{1} > r_{2}.

Statement I suggests that (r_{1} – r_{2}) = $\frac{k}{\left(2\pi \right)}$. Hence, this statement alone does not give the value of r_{1}.

Statement II implies that (r_{1}^{2} – r_{2}^{2}) = $\frac{m}{\pi}.$ .

Hence, again this statement alone is not sufficient to answer the question. But in the second equation, we simplify (r_{1}^{2} – r_{2}^{2}) as (r_{1} + r_{2})(r_{1} – r_{2}) and then substitute the value of (r_{1} – r_{2}) from the first equation, we will get the value of (r_{1} + r_{2}). Now we have two equations in r_{1 }and r_{2}, which can be solved simultaneously to get the value of r_{1}. Hence, both the statements when together taken can answer the question.

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**43. CAT 1998 QA | Geometry - Circles**

**Direction: **Each question is followed by two statements, I and II. Answer the questions based on the statements and mark the answer as

1. if the question can be answered with the help of any one statement alone but not by the other statement.

2. if the question can be answered with the help of either of the statements taken individually.

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 circle circumscribes a square. What is the area of the square?

I. Radius of the circle is given.

II. Length of the tangent from a point 5 cm away from the centre of the circle is given.

Answer: 2

**Explanation** :

Statement I itself is sufficient to answer the question.

As, if we know the radius of the circle we can find out the length of the diagonal of the square (which will be the diameter) and if we know the diagonal of a square we can find the length of its sides and hence the area. Again the second statement in itself can answer the question. As, from the data that is given we can find

the radius of the circle and hence the area of the square (as given before). This can be explained from the diagram given. Since the tangent makes a right angle with the radius at the circumference, the triangle is a right-angled triangle. Hence, A^{2} = 5^{2} + r^{2}. Hence, knowing the value of A, we can find out r. Hence, both statements in itself can answer the question.

Therefore, the answer is (b).

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