# CAT 1993 QA | Previous Year CAT Paper

Previous year paper questions for CAT 1993 QA

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**1. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

Given that X and Y are non-negative. What is the value of X?

I. 2X + 2Y ≤ 40

II. X − 2Y ≥ 20

Answer: 3

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**Explanation** :

From statement I: 2X + 2Y ≤ 40 or X + Y ≤ 20

This statement alone cannot give the value of X.

From statement II: X − 2Y ≥ 20

This statement also alone cannot give the value of X.

On combining statements I and II:

Multiplying the second statement by –1 and adding

both the statements, we get

3Y ≤ 0 i.e., Y ≤ 0, but it is given that Y is non negative.

∴Y = 0 and X = 20

Hence, the answer is (c).

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**2. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

What are the values of 3 integers a, b, and c?

I. ab = 8

II. bc = 9

Answer: 3

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**Explanation** :

From statement I:

(a, b) can be (1, 8), (2, 4), (4, 2) and (8, 1).

Therefore, statement I alone cannot give the value of

a, b and c.

From statement II:

(b, c) can be (1, 9), (3, 3) and (9, 1).

On combining statements I and II:

b = 1, a = 8 and c = 9

Hence, the answer is (c).

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**3. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

Is the average of the largest and the smallest of four given numbers greater than the average of the four numbers?

I. The difference between the largest and the second largest numbers is greater than the difference

between the second smallest and the smallest numbers.

II. The difference between the largest and the second largest numbers is less than the difference

between the second largest and the second smallest numbers.

Answer: 1

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**Explanation** :

If the numbers are a, b, c and d such that a < b < c < d, then from statement I, we get (d – c) > (b – a).

So we can say, (d + a) > (b + c) or (d + a) + (d + a) > (b + c) + (d + a). Dividing both the sides by 4, we get

$\frac{(d+a)}{2}>\frac{(a+b+c+d)}{4}$.

This shows that the average of the largest and the smallest of four numbers is indeed greater than the average of all the 4 numbers. Hence, we can answer the question using first statement only.

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**4. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

What are the ages of the three brothers?

I. The product of their ages is 21.

II. The sum of their ages is not divisible by 3.

Answer: 4

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**Explanation** :

From statement I, the ages could be either (1, 3, 7) or (1, 1, 21). Statement II doesn’t simplify this further as none of the above combinations when added is divisible

by 3.

Hence, the answer is (d).

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**5. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

Two types of widgets, namely type A and type B, are produced on a machine. The number of machine hours available per week is 80. How many widgets of type A must be produced?

I. One unit of type A widget requires 2 machine hours and one unit of type B widget requires 4 machine hours.

II. The widget dealer wants supply of at least 10 units of type A widget per week and he would not accept less than 15 units of type B widget.

Answer: 3

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**Explanation** :

**From statement I: **From this statement, exact number of widgets produced by machine A cannot be

determined.

**From statement II:** From this statement also exact number of widgets produced by machine A cannot be determined.

On combining statements I and II: Dealer produced minimum 10 units of widget A and 15 units of widget B and for that he requires 10 × 2 + 15 × 4 = 80 machine hours and number of machine hours available per week is also 80 hours.

Hence, he produced 10 units of widget A.

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**6. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

What is the area of a regular hexagon?

I. The length of the boundary line of the hexagon is 36 cm.

II. The area of the hexagon is 6 times the area of an equilateral triangle formed on one of the sides.

Answer: 1

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**Explanation** :

**From statement I: **We can say that the perimeter of the hexagon is 36 cm, or the length of each side is 6 cm. From this we can find its area. So this statement alone is sufficient to answer the question.

**From statement II:** It does not provide any other data, but merely states the property of a regular hexagon. So, this statement alone is not sufficient to answer the question.

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**7. CAT 1993 QA | Data Sufficiency**

**Each of these questions is followed by two statements. As the answer,**

Mark (a), If the question can be answered with the help of statement I alone,

Mark (b), If the question can be answered with the help of statement II, alone,

Mark (c), If both, statement I and statement II are needed to answer the question, and

Mark (d), If the question cannot be answered even with the help of both the statements.

What is the price of mangoes per kg?

I. Ten kg of mangoes and two dozens of oranges cost Rs.252.

II. Two kg of mangoes could be bought in exchange for one dozen oranges.

Answer: 3

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**Explanation** :

**From statement I: **Let price per kg of mangoes be

Rs.x and price per dozen of oranges be Rs.y.

∴10x + 2y = 252

From this statement, we cannot find x.

**From statement II: **2x = y

From this statement also, we cannot find the price per kg of mangoes.

**On combining statements I and II:** 14x = 252 i.e., x = 18

Hence, the answer is (c).

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**8. CAT 1993 QA | Algebra - Simple Equations**

Two oranges, three bananas and four apples cost Rs.15. Three oranges, two bananas and one apple cost Rs 10. I bought 3 oranges, 3 bananas and 3 apples. How much did I pay?

- A.
Rs. 10

- B.
Rs. 8

- C.
Rs. 15

- D.
Cannot be determined

Answer: Option C

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**Explanation** :

The two equations are : 2o + 3b + 4a = 15 and 3o + 2b + a = 10.

Adding the two equations, we get

5o + 5b + 5a = 25

⇒ o + b + a = 5

∴ 3o + 3b + 3a = 15.

Hence, option (c).

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**9. CAT 1993 QA | Arithmetic - Percentage**

The rate of increase of the price of sugar is observed to be two percent more than the inflation rate expressed in percentage. The price of sugar, on January 1, 1994, is Rs. 20 per kg. The inflation rate for the years 1994 and 1995 are expected to be 8% each. The expected price of sugar on January 1, 1996 would be

- A.
Rs. 23.60

- B.
Rs. 24.00

- C.
Rs. 24.20

- D.
Rs. 24.60

Answer: Option C

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**Explanation** :

Since the inflation rate is 8% in both the years 1994 and 1995, therefore, the rate of increase of the price of sugar is 10%.

∴ Price of sugar on January 1, 1996 = Price of sugar on January 1, 1994 × ${\left(1+\frac{10}{100}\right)}^{2}$ = 20 × 1.21 = Rs. 24.20 per kg.

Hence, option (c).

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**10. CAT 1993 QA | Modern Math - Permutation & Combination**

An intelligence agency decides on a code of 2 digits selected from 0, 1, 2, …. , 9. But the slip on which the code is hand–written allows confusion between top and bottom, because these are indistinguishable. Thus, for example, the code 91 could be confused with 16. How many codes are there such that there is no possibility of any confusion?

- A.
25

- B.
75

- C.
80

- D.
None of these

Answer: Option C

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**Explanation** :

Total number of two digit codes that can be formed is 10 × 10 = 100

Out of them 0,1,6,8,9 can create confusion.

Using these five digits, total number of two digit numbers that can be made is 5 × 5 = 25.

But out of these 25 numbers 00,11,88,69 and 96 will not make any confusion.

Hence, the required answer is 100 – 25 + 5 = 80.

Hence, option (c).

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**11. CAT 1993 QA | Algebra - Number Theory**

Suppose one wishes to find distinct positive integers x, y such that (x + y)/xy is also a positive integer. Identify the correct alternative.

- A.
This is never possible.

- B.
This is possible and the pair (x,y) satisfying the stated condition is unique.

- C.
This is possible and there exist more than one but a finite number of ways of choosing the pair (x,y).

- D.
This is possible and the pair (x,y) can be chosen in infinite ways.

Answer: Option A

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**Explanation** :

It can be very easy to figure out that (x + y) will always be greater than xy, only if one of them is 1. For eg. If x = 1 and y =2, then (x + y) = 3 and xy = 2.

Hence, (x + y) > xy.

Other than this, for all other values of x & y, (x + y) will always be less than xy, and hence, the ratio of $\frac{(x+y)}{xy}<1,$ and hence, cannot be an integer. Also, even if one of the values is 1, $\frac{(x+y)}{xy}$ will never be an integer.

Hence, option (a).

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**12. CAT 1993 QA | Algebra - Number Theory**

Given odd positive integers x, y and z, which of the following is not necessarily true?

- A.
x

^{2 }y^{2}z^{2 }is odd - B.
3(x

^{2}+ y^{3})z^{2}is even. - C.
5x + y + z

^{4}is odd - D.
z

^{2}(x^{4}+ y^{4})/2 is even

Answer: Option D

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**Explanation** :

You can solve this by assuming values of x, y and z.

Let the three numbers be 1, 3 & 5.

Option (a) is 1^{2} 3^{2} 5^{2} = 225, which is odd.

Option (b) is 3(1^{2} + 3^{3})5^{2} = 2100, which is even.

Option (c) is 5 + 3 + 5^{4} = 633, which is odd.

Option (d) is ${5}^{2}\frac{(14+34)}{2}=$ 1025, which is not even and hence, the answer.

Hence, option (d).

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**13. CAT 1993 QA | Miscellaneous**

139 persons have signed up for an elimination tournament. All players are to be paired up for the first round, but because 139 is an odd number one player gets a bye, which promotes him to the second round, without actually playing in the first round. The pairing continues on the next round, with a bye to any player left over. If the schedule is planned so that a minimum number of matches is required to determine the champion, the number of matches which must be played is

- A.
136

- B.
137

- C.
138

- D.
139

Answer: Option C

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**Explanation** :

This can be logically done in the following manner.

There are 139 players in all. We want to determine 1 champion among them. So all except the Champion should lose. A player can lose only once and since any match produces only one loser, to produce 138 losers, there should be 138 matches that should be played.

Hence, option (c).

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**14. CAT 1993 QA | Miscellaneous**

There are ten 50 paise coins placed on a table. Six of these show tails, four show heads. A coin is chosen at random and flipped over (not tossed). This operation is performed seven times. One of the coins is then covered. Of the remaining nine coins, five show tails and four show heads. The covered coin shows

- A.
a head

- B.
a tail

- C.
more likely a head

- D.
more likely a tail

Answer: Option A

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**Explanation** :

The initial reading for 10 coins is : 6 Tails and 4 Heads After repeating the process of flipping one coin at random for 7 times, the final reading for 9 coins is: 5 Tails and 4 Heads.

Therefore, possible final reading for 10 coins is: 6 Tails and 4 Heads or 5 Tails and 5 Heads. If the final reading is 6T and 4H, it is same as the initial one. However, this is not possible as the process of flipping a coin has taken place an odd number of times, so there has to be atleast one change in the final reading.

Therefore, the final reading is 5T and 5H.

So the covered coin will certainly be a Head.

Hence, option (a).

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**15. CAT 1993 QA | Arithmetic - Ratio, Proportion & Variation**

From each of the two given numbers, half the smaller number is subtracted. Of the resulting numbers the larger one is three times as large as the smaller. What is the ratio of the two numbers?

- A.
2 : 1

- B.
3 : 1

- C.
3 : 2

- D.
None

Answer: Option A

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**Explanation** :

Let the two given numbers be x and y such that x > y.

According to the question,

x - $\frac{y}{2}=3\left(y-\frac{y}{2}\right)$

⇒ 2x - y = 6y - 3y

⇒ 2x = 4y

$\Rightarrow \frac{x}{y}=\frac{2}{1}.$

Hence, option (a).

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**16. CAT 1993 QA | Geometry - Circles**

Three identical cones with base radius r are placed on their bases so that each is touching the other two. The radius of the circle drawn through their vertices is

- A.
smaller than r

- B.
equal to r

- C.
larger than r

- D.
depends on the height of the cones.

Answer: Option C

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**Explanation** :

It can be seen that, if we place the 3 cones in such a way that they touch each other, it will be similar to placing 3 circles touching, with vertices of the cone corresponding to the centers of the circles. The centers of the circle form an equilateral triangle with each side being 2r. A circle that passes through the centers will be the circumcircle to such a triangle. The radius of the circumcircle of an equilateral triangle is $\left(\frac{1}{\sqrt{3}}\right)$ times its side.

Hence, in our case it would be be $\left(\frac{2r}{\sqrt{3}}\right)$ and $\left(\frac{2r}{\sqrt{3}}\right)$ > r, since $\sqrt{3}$ = 1.73 (approx).

Hence, option (c).

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**17. CAT 1993 QA | Geometry - Circles**

The line AB is 6 metres in length and is tangent to the inner one of the two concentric circles at point C. It is known that the radii of the two circles are integers. The radius of the outer circle is

- A.
5 metres

- B.
4 metres

- C.
6 metres

- D.
3 metres

Answer: Option A

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**Explanation** :

Let x meters and y meters be the radius of the outer and the inner circles respectively and O be their center.

In right angled Δ OCB,

CB^{2} = OB^{2} - OC^{2}

⇒ 9 = x^{2} – y^{2}

⇒ (x + y) ( x – y) = 9 × 1

As x and y are integers, therefore, x + y = 9 and x – y = 1.

Thus, x = 5.

Hence, radius of the outer circle is 5 meters.

Hence, option (a).

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**18. CAT 1993 QA | Modern Math - Permutation & Combination**

Four cities are connected by a road network as shown in the figure. In how many ways can you start from any city and come back to it without travelling on the same road more than once?

- A.
8

- B.
12

- C.
16

- D.
20

Answer: Option B

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**Explanation** :

It can be seen that every city is connected to all the other 3 cities.

If we start from city A, there are 3 ways in which we can proceed, viz. AB, AD or AC.

Once we are at any of these cities, each one of them is connected to the other 3 cities. But since we cannot go back to city A, there are only 2 ways in which we can proceed from here.

If we are at B, we can take either paths BD or BC.

From this point, we have a choice of going directly to A (thus skipping 4th city) or go to 4th city and come back to A. Eg. If we are at D, we can either take DA or DCA. So there are 2 more ways to go from here.

Hence, required number of ways = 3 × 2 × 2 = 12.

Hence, option (b).

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**Use the following information:**

Eighty five children went to an amusement park where they could ride on the merry – go round, roller coaster, and Ferris wheel. It was known that 20 of them took all three rides, and 55 of them took at least two of the three rides. Each ride cost Re.1, and the total receipt of the amusement park was Rs.145.

**19. CAT 1993 QA | Algebra - Simple Equations**

How many children did not try any of the rides?

- A.
5

- B.
10

- C.
15

- D.
20

Answer: Option C

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**Explanation** :

Let x, y and z be the number of children who took 1 rides, 2 rides and 3 rides respectively.

Since z = 20 and y + z = 55, y = 35.

Then, total number of rides = x + 2y + 3z = 145

⇒ x + 2 × 35 + 3 × 20 = 145

⇒ x = 15

Number of children, who did not try any of the rides

= 85 – (x + y + z)

= 85 – (15 + 35 + 20) = 15

Hence, option (c).

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**20. CAT 1993 QA | Algebra - Simple Equations**

How many children took exactly one ride?

- A.
5

- B.
10

- C.
15

- D.
20

Answer: Option C

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**Explanation** :

Consider the solution to previous question of this set.

Number of children taking exactly one ride = 15.

Hence, option (c).

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**21. CAT 1993 QA | Algebra - Simple Equations**

John bought five mangoes and ten oranges together for forty rupees. Subsequently, he returned one mango and got two oranges in exchange. The price of an orange would be

- A.
Rs. 1

- B.
Rs. 2

- C.
Rs. 3

- D.
Rs. 4

Answer: Option B

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**Explanation** :

The price of 1 mango is equal to the price of 2 oranges.

Hence, 5 mangoes will be equivalent to 10 oranges.

So, 20 oranges cost Rs.40, therefore one orange will cost Rs.2.

Hence, option (b).

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**22. CAT 1993 QA | Algebra - Number Theory**

The number of positive integers not greater than 100, which are not divisible by 2, 3 or 5 is

- A.
26

- B.
18

- C.
31

- D.
None

Answer: Option A

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**Explanation** :

The following Venn diagram shows the distribution of numbers between 1 and 100 that are divisible by 2, 3, 5 or a combination of two or more of them.

Number of numbers divisible by 2 are Quotient of 100/2 = 50

Number of numbers divisible by 3 are Quotient of 100/3 = 33

Number of numbers divisible by 5 are Quotient of 100/5 = 20

Number of numbers divisible by 2 & 5 are Quotient of 100/10 = 10

Number of numbers divisible by 5 & 3 are Quotient of 100/15 = 6

Number of numbers divisible by 3 & 2 are Quotient of 100/6 = 16

Number of numbers divisible by 2, 3 & 5 are Quotient of 100/30 = 3

Total number of numbers that are divisible by one or more among 2, 3 and 5 = 27 + 14 + 7 + 13 + 3 + 7 + 3 = 74.

Hence, the required number = 100 – 74 = 26.

Hence, option (a).

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**23. CAT 1993 QA | Algebra - Progressions**

Let U_{n+1} = 2U_{n} + 1 (n = 0, 1, 2, ...) and u_{0} = 0. Then u_{10} is nearest to

- A.
1023

- B.
2047

- C.
4095

- D.
8195

Answer: Option A

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**Explanation** :

U_{0 }= 0 = 2^{0} – 1 = 0

U_{1} = 2 × 0 + 1 = 1 = 2^{1} – 1

U_{2} = 2 × 1 + 1 = 3 = 2^{2} – 1

U_{3} = 2 × 3 + 1 = 7 = 2^{3} – 1 and so on.

⇒ U_{n} = 2^{n} - 1

∴ U_{10} = 2^{10} – 1 = 1023.

Hence, option (a).

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**Answer the next 2 questions based on the information given below:**

A function f(x) is said to be even if f(–x) = f(x), and odd if f(–x) = –f(x). Thus, for example, the function given by f(x) = x^{2} is even, while the function given by f(x) = x^{3} is odd. Using this definition, answer the following questions.

**24. CAT 1993 QA | Algebra - Functions & Graphs**

The function given by f(x) = |x|^{3} is

- A.
even

- B.
odd

- C.
neither

- D.
both

Answer: Option A

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**Explanation** :

f(x) = |x|^{3}

∴ f(–x) = |–x|^{3 }= |x|^{3 }= f(x).

Since, f(-x) = f(x), the given function is even.

Hence, option (a).

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**25. CAT 1993 QA | Algebra - Functions & Graphs**

The sum of two odd functions

- A.
is always an even functions

- B.
is always an odd function

- C.
is sometimes odd and sometimes even

- D.
may be neither odd nor even

Answer: Option B

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**Explanation** :

Let f(x) = g(x) + h(x), where g and h are odd functions.

∴ f(–x) = g(–x) + h(–x) = –g(x) – h(x) = –f(x).

Hence, f(x) is a odd function.

Hence, option (b).

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**26. CAT 1993 QA | Modern Math - Permutation & Combination**

A five digit number is formed using digits 1, 3, 5, 7 and 9 without repeating any one of them. What is the sum of all such possible numbers?

- A.
6666600

- B.
6666660

- C.
6666666

- D.
None

Answer: Option A

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**Explanation** :

If we assume that any digit is in a fixed position, then the remaining four digits can be arranged in 4! = 24 ways. So each of the 5 digits will appear in each of the five places 24 times. So the sum of the digits in each position is 24(1 + 3 + 5 + 7 + 9) = 600.

Hence, the sum of all such numbers will be 600(1 + 10 + 100 + 1000 + 10000) = 6666600.

Hence, option (a).

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**27. CAT 1993 QA | Modern Math - Probability**

A box contains 6 red balls, 7 green balls and 5 blue balls. Each ball is of a different size. The probability that the red ball selected is the smallest red ball, is

- A.
1/18

- B.
1/3

- C.
1/6

- D.
2/3

Answer: Option C

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**Explanation** :

Since there are 6 red balls and all six of them are of different sizes, probability of choosing the smallest among them is $\frac{1}{6}$.

Hence, option (c).

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**Answer the following questions based on the information given below:**

ABC forms an equilateral triangle in which B is 2 km from A. A person starts walking from B in a direction parallel to AC and stops when he reaches a point D directly east of C. He, then, reverses direction and walks till he reaches a point E directly south of C.

**28. CAT 1993 QA | Geometry - Triangles**

Then D is

- A.
3 km east and 1 km north of A

- B.
3 km east and $\sqrt{3}$ km north of A

- C.
$\sqrt{3}$ km east and 1 km south of A

- D.
$\sqrt{3}$ km west and 3 km north of A

Answer: Option B

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**Explanation** :

Since Δ ABC is an equilateral triangle with length of the side 2 km, so its altitude will be $\sqrt{3}$ km.

The horizontal distance of D from A = half of AB + CD = 1 + 2 = 3 kms

The vertical distance of D from A = height of the triangle ABC = √3 kms

∴ D is 3 km east and $\sqrt{3}$ km north of A.

Hence, option (b).

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**29. CAT 1993 QA | Geometry - Triangles**

The total distance walked by the person is

- A.
3 km

- B.
4 km

- C.
2$\sqrt{3}$ km

- D.
6 km

Answer: Option D

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**Explanation** :

Consider the solution to previous question of this set.

Hence, the total distance walked by the person = BD + DB + BE = 2 + 2 + 2 = 6 km.

Hence, option (d).

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**30. CAT 1993 QA | Geometry - Mensuration**

A slab of ice 8 inches in length, 11 inches in breadth, and 2 inches thick was melted and resolidified into the form of a rod of 8 inches diameter. The length of such a rod, in inches, is nearest to

- A.
3

- B.
3.5

- C.
4

- D.
4.5

Answer: Option B

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**Explanation** :

Let l be the length of the rod, then

Volume of slab = Volume of rod

⇒ 8 × 11 × 2 = π × ${\left(\frac{8}{2}\right)}^{2}$ × l

⇒ l = 11/π = 3.5 inches.

Hence, option (b).

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**31. CAT 1993 QA | Algebra - Progressions**

Let x < 0.50, 0 < y < 1, z > 1. Given a set of numbers, the middle number, when they are arranged in ascending order, is called the median. So the median of the numbers x, y, and z would be

- A.
less than one

- B.
between 0 and 1

- C.
greater than 1

- D.
cannot say

Answer: Option B

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**Explanation** :

Since there are two numbers which are less than 1 (viz. x and y), it is obvious that the median will be less than 1.

Hence, option (c) cannot be the answer.

Since x < 0.5 and 0 < y < 1, the median will not be less than 0.

Hence, option (b).

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**32. CAT 1993 QA | Algebra - Functions & Graphs**

The maximum possible value of y = min (1/2 – 3x^{2}/4, 5x^{2}/4) for the range 0 < x < 1 is

- A.
1/3

- B.
1/2

- C.
5/27

- D.
5/16

Answer: Option D

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**Explanation** :

So maximum possible value will be at the point of intersection of the two graphs.

$\therefore \frac{1}{2}-\frac{3{x}^{2}}{4}=\frac{5{x}^{2}}{4}\phantom{\rule{0ex}{0ex}}\Rightarrow {x}^{2}=\frac{1}{4}$

Hence, required maximum value = $\frac{5{x}^{2}}{4}=\frac{5}{4}\times \frac{1}{4}=\frac{5}{16}.$

Hence, option (d).

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**33. CAT 1993 QA | Arithmetic - Time & Work**

A group of workers was put on a job. From the second day onwards, one worker was withdrawn each day. The job was finished when the last worker was withdrawn. Had no worker been withdrawn at any stage, the group would have finished the job in two-third the time. How many workers were there in the group?

- A.
2

- B.
3

- C.
5

- D.
11

Answer: Option B

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**Explanation** :

Let the work done by a worker be x units, w be the total work and n be the number of workers in the group.

Then, w = Work done on the nth day i.e. last day + Work done on the second last day + … + Work done on the first day

⇒ w = x + 2x + ... + nx = $\frac{n(n+1)x}{2}$ ...(i)

When none of the workers is removed, then

w = nx × $\frac{2n}{3}=\frac{2{n}^{2}x}{3}$ ...(ii)

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

$\frac{n(n+1)x}{2}=\frac{2{n}^{2}x}{3}$

⇒ n = 3.

Hence, option (b).

Workspace:

**34. CAT 1993 QA | Geometry - Quadrilaterals & Polygons**

Consider the five points comprising of the vertices of a square and the intersection point of its diagonals. How many triangles can be formed using these points?

- A.
4

- B.
6

- C.
8

- D.
10

Answer: Option C

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**Explanation** :

We can form a triangle with any 3 points which are not collinear.

3 points out of 5 can be chosen in ^{5}C_{3} = 10 ways.

But of these, the three points lying on the two diagonals will be collinear.

So 10 – 2 = 8 triangles can be formed.

Hence, option (c).

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**35. CAT 1993 QA | Venn Diagram**

Out of 100 families in the neighbourhood, 45 own radios, 75 have TVs, 25 have VCRs. Only 10 families have all three and each VCR owner also has a TV. If 25 families have radio only, how many have only TV?

- A.
30

- B.
35

- C.
40

- D.
45

Answer: Option C

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**Explanation** :

Since each VCR owner also has a TV, therefore, 15 families own both TV and VCR but not Radio.

Since 25 families have radio only, therefore, 10 families own both TV and Radio but not VCR.

Hence, number of families having only TV = 75 – 10 – 10 – 15 = 40.

Hence, option (c).

Workspace:

**The following functions have been defined for three numbers A, B and C:**

@ (A, B) = average of A and B

*(A, B) = product of A and B

/(A, B) = A divided by B

Answer these questions with the above data.

**36. CAT 1993 QA | Miscellaneous**

If A = 2 and B = 4, the value of @(/(*(A, B), B), A) would be

- A.
2

- B.
4

- C.
6

- D.
16

Answer: Option A

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**Explanation** :

@(/(*(2, 4), 4), 2) = @(/(8, 4), 2) = @(2, 2) = 2.

Hence, option (a).

Workspace:

**37. CAT 1993 QA | Miscellaneous**

The sum of A and B is given by

- A.
*(@(A, B), 2)

- B.
/(@(A, B), 2)

- C.
@(*(A, B), 2)

- D.
@(/(A, B), 2)

Answer: Option A

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**Explanation** :

A + B = 2((A + B)/2) = 2(@(A, B)) = *(@(A, B), 2).

Hence, option (a).

Workspace:

**38. CAT 1993 QA | Miscellaneous**

The sum of A, B, and C is given by

- A.
*(@(*(@(B, A), 2), C), 3)

- B.
/(@(*(@(B, A), 3), C), 2)

- C.
/(*(@(*(B, A), 2), C), 3)

- D.
None of these

Answer: Option D

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**Explanation** :

(HINT : the best way is to simplify the answer choices and work backwards.)

Hence, option (d).

Workspace:

**39. CAT 1993 QA | Arithmetic - Percentage**

A report consists of 20 sheets each of 55 lines and each such line consist of 65 characters. This report is retyped into sheets each of 65 lines such that each line consists of 70 characters. The percentage reduction in number of sheets is closest to

- A.
20

- B.
5

- C.
30

- D.
35

Answer: Option A

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**Explanation** :

The total number of the characters = (20 × 55 × 65).

Let the number of pages in the new format be n.

Thus, the total number of the characters = (65 × 70 × n).

Since the total number of the characters remains same,

⇒ 20 × 55 × 65 = 65 × 70 × n

⇒ n ≈ 16.

Hence, the required percentage $=\frac{20-16}{20}\times $ 100 = 20%.

Hence, option (a).

Workspace:

**40. CAT 1993 QA | Algebra - Number Theory**

Let x < 0, 0 < y < 1, z > 1. Which of the following may be false?

- A.
(x

^{2}– z^{2}) has to be positive. - B.
yz can be less than one.

- C.
xy can never be zero.

- D.
(y

^{2}– z^{2}) is always negative.

Answer: Option A

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**Explanation** :

Let us evaluate each option.

Option (a): As x < 0 and z > 1, let x = – 1 and z = 3, then (x^{2} –z^{2}) = – 8. Hence, this option is not true.

Option (b): As 0 < y < 1 and z > 1, let y = $\frac{1}{4}$ and z = 2, then yz = $\frac{1}{4}\times 2=\frac{1}{2}.$

Therefore, yz can be less than 1.

Option (c): Since, none of the x and y is equal to zero, therefore xy can never be zero.

Option (d): 0 < y < 1 and z > 1, therefore (y^{2} – z^{2}) is always negative.

Hence, option (a).

Workspace:

**41. CAT 1993 QA | Modern Math - Permutation & Combination**

A young girl counted in the following way on the fingers of her left hand. She started calling the thumb 1, the index finger 2, middle finger 3, ring finger 4, little finger 5, then reversed direction, calling the ring finger 6, middle finger 7, index finger 8, thumb 9, then back to the index finger for 10, middle finger for 11, and so on. She counted up to 1994. She ended on her.

- A.
thumb

- B.
index finger

- C.
middle finger

- D.
ring finger

Answer: Option B

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**Explanation** :

The counting is done in the following manner:

thumb - index finger - middle finger - ring finger - little finger

1 2 3 4 5

9 8 7 6

10 11 12 13

17 16 15 14

∴ The numbers, counted on thumb are 1, 9, 17 and so on. This is an arithmetic progression whose first term is 1 and common difference is 8.

⇒ A number counted on thumb can be represented as = 1 + (n - 1) × 8 = 8n - 7

Now, let us figure out highest number less than or equal to 1994 that will be counted on thumb.

⇒ 8n - 7 ≤ 1994

⇒ n ≤ 2001/8 = 125.125

Highest number counted on thumb is when n is 125 = 8 × 125 - 7 = 1993

∴ Next number 1994 will be counted on index finger.

Hence, option (b).

Workspace:

**Directions for next 3 questions:**

Q started to move from point B towards point A exactly an hour after P started from A in the opposite direction. Q’s speed was twice that of P. When P had covered one-sixth of the distance between the points A and B, Q had also covered the same distance.

**42. CAT 1993 QA | Arithmetic - Time, Speed & Distance**

The point where P and Q would meet is

- A.
Closer to A

- B.
Exactly between A and B

- C.
Closer to B

- D.
P and Q will not meet at all

Answer: Option A

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**Explanation** :

It is clear that after a particular amount of time P and Q are equidistant from A and B respectively and speed of Q is twice the speed of P, therefore, in the remaining time distance moved by Q will be twice than P.

Hence, they would meet closer to A.

Hence, option (a).

Workspace:

**43. CAT 1993 QA | Arithmetic - Time, Speed & Distance**

How many hours would P take to reach B?

- A.
2

- B.
5

- C.
6

- D.
12

Answer: Option D

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**Explanation** :

Let the speed of P be x and the distance between A and B be d, so the speed of Q will be 2x.

According to the question, (1 + t)v = 2vt = $\frac{d}{6}$ (Let t be the travel time of Q)

⇒ t = 1, and d = 12v

Hence, the time taken by P to reach to B = $\frac{d}{v}$ = 12 hours.

Hence, option (d).

Workspace:

**44. CAT 1993 QA | Arithmetic - Time, Speed & Distance**

How many more hours would P (compared to Q) take to complete his journey?

- A.
4

- B.
5

- C.
6

- D.
7

Answer: Option C

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**Explanation** :

As P takes 12 hours to complete his journey, so Q moving with twice the speed of P will take $\frac{12}{2}=6$ hours to complete his journey.

Hence P will take 6 hours more than Q to complete the journey.

Hence, option (c).

Workspace:

**45. CAT 1993 QA | Algebra - Number Theory**

The smallest number which when divided by 4, 6 or 7 leaves a remainder of 2, is

- A.
44

- B.
62

- C.
80

- D.
86

Answer: Option D

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**Explanation** :

Required number = LCM (4, 6, 7) + 2 = 86.

Hence, option (d).

**Concept**:

Workspace:

**46. CAT 1993 QA | Geometry - Mensuration**

The diameter of a hollow cone is equal to the diameter of a spherical ball. If the ball is placed at the base of the cone, what portion of the ball will be outside the cone?

- A.
50%

- B.
less than 50%

- C.
more than 50%

- D.
100%

Answer: Option C

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**Explanation** :

It can be seen that if a spherical ball is placed inside a hollow cone of same diameter, the ball won’t go up to the diameter. In other words, because of the slanting edges of the cone, only less than 50% of the ball would enter the cone. i.e, more than 50% of the ball would be outside the cone.

Hence, option (c).

Workspace:

**47. CAT 1993 QA | Arithmetic - Time, Speed & Distance**

A ship leave on a long voyage. When it is 18 miles from the shore, a seaplane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the seaplane catch up with the ship?

- A.
24 miles

- B.
25 miles

- C.
22 miles

- D.
20 miles

Answer: Option D

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**Explanation** :

The separation between the ship and the seaplane is 18 miles.

Since the two are travelling in the same direction, the relative speed would be 9 times the speed of the ship (If speed of ship is x miles/hour, speed of the seaplane would be 10x and 10x – x = 9x).

Hence, to catch up with the ship, the seaplane would take $\frac{18}{9x}=\frac{2}{x}$ hours.

Now, the ship covers x miles in an hour, so in $\frac{2}{x}$ hours it would cover 2 miles.

So when the seaplane catches up with the ship, it would be 18 + 2 = 20 miles from the shore.

Hence, option (d).

Workspace:

**48. CAT 1993 QA | Algebra - Number Theory**

The product of all integers from 1 to 100 will have the following numbers of zeros at the end.

- A.
20

- B.
24

- C.
19

- D.
22

Answer: Option B

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**Explanation** :

The number of zeros at the end will be same as the highest power of 10 in 100!.

Number of 2 in the product of all integers from 1 to 100 $=\frac{100}{2}+\frac{100}{4}+\frac{100}{8}+\frac{100}{16}+\frac{100}{32}+\frac{100}{64}$

= 50 + 25 +12 + 6 + 3 + 1= 97 and number of 5 in the product of all integers from 1 to 100 = $\frac{100}{5}+\frac{100}{25}$ = 20 + 4 = 24.

∴ 100! = 2^{97} × 5^{24} × ...

⇒ 100! = 2^{73} × (2 × 5^{24}) × ...

⇒ 100! = 2^{73} × (10^{24}) × ...

∴ Highest power of 10 in 100! is 24.

Hence, option (b).

**Concept**:

Workspace:

**49. CAT 1993 QA | Algebra - Number Theory**

Let x, y and z be distinct positive integers satisfying x < y < z and x + y + z = k. What is the smallest value of K that does not determine x, y, z uniquely?

- A.
9

- B.
6

- C.
7

- D.
8

Answer: Option D

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**Explanation** :

In this case since x, y and z are distinct positive integers, our aim is figure out which of the answer choices cannot be expressed as the sum of 3 integers uniquely.

For eg. 6 can only be expressed as (1 + 2 + 3).

7 can only be expressed as (1 + 2 + 4).

But 8 can be expressed as either (1, 2, 5) or (1, 3, 4).

Hence, option (d).

Workspace:

**50. CAT 1993 QA | Venn Diagram**

Amar, Akbar, and Anthony came from the same public school in the Himalayas. Every boy in that school either fishes for trout or plays frisbee. All fishermen like snow while no frisbee player likes rain. Amar dislikes whatever Akbar likes and likes whatever Akbar dislikes. Akbar likes rain and snow. Anthony likes whatever the other two like. Who is a fisherman but not a frisbee player?

- A.
Amar

- B.
Akbar

- C.
Anthony

- D.
None

Answer: Option B

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**Explanation** :

Since Akbar likes rain, he cannot be a frisbee player (as no frisbee player likes rain). And since every boy in the school does one of the two, Akbar has to be a fisherman.

Hence, option (b).

Workspace:

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