Instructions

DIRECTIONS for the following questions: These questions are based on the situation given below: There are fifty integers $$a_1, a_2,...,a_{50}$$, not all of them necessarily different. Let the greatest integer of these fifty integers be referred to as $$G$$, and the smallest integer be referred to as $$L$$. The integers $$a_1$$ through $$a_{24}$$ form sequence $$S1$$, and the rest form sequence $$S2$$. Each member of $$S1$$ is less than or equal to each member of $$S2$$.

CAT 1999 - Question 41

Elements of $$S1$$ are in ascending order, and those of $$S2$$ are in descending order. $$a_{24}$$ and $$a_{25}$$ are interchanged. Then, which of the following statements is true?

Video Solution

Solution

We know that $$a_{24}$$ is less than $$a_{25}$$.

So, even if $$a_{25}$$ replaces $$a_{24}$$, the ascending order still exists in S1.

But, $$a_{25}$$ is less than $$a_{26}$$. Hence, the descending order does not exist in S2 anymore.

CAT 1999 - Question 42

Every element of S1 is made greater than or equal to every element of S2 by adding to each element of S1 an integer x. Then x cannot be less than:

Video Solution

Solution

For the least element L in $$S1$$ to be greater than the greatest element or equal to G in $$S2$$, the number that is added to L cannot be less than G - L.

Instructions

DIRECTIONS for the following questions: These questions are based on the situation given below: Let x and y be real numbers

f(x, y) = | x + y |

F(f(x, y)) = -f(x, y)

G(f(x, y)) = -F(f(x, y))

CAT 1999 - Question 43

Which of the following statements is true?

Video Solution

Solution

f(x,y) = |x+y|

F(f(x,y)) = -f(x,y) = -|x+y|

G(f(x,y)) = -F(f(x,y)) = |x+y|

Option A: F(f(x, y)) . G(f(x, y)) = -F(f(x, y)) . G(f(x, y)) =>LHS = -|x+y|$$^2$$, RHS = |x+y|$$^2$$ Hence false.

Option B: F(f(x, y)) . G(f(x, y)) > -F(f(x, y)) . G(f(x, y)), Since, LHS is smaller than RHS. False

Option C: F(f(x, y)) . G(f(x, y)) $$\neq $$ G(f(x, y)) . F(f(.x, y)), Here LHS=RHS. Hence false.

Option D:

=> G(f(x,y)) + F(f(x,y)) = 0

f(x,y) = f(-x,-y)

=> G(f(x,y)) + F(f(x,y)) + f(x,y) = f(-x.-y)

CAT 1999 - Question 44

What is the value of f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2))))?

Video Solution

Solution

F(f(x,y)) = -f(x,y)

G(f(x,y)) = -F(f(x,y)) = f(x,y)

G(f(1,0)) = 1

F(f(1,2)) = -3

G(f(1,2)) = 3

f(F(f(1,2)),G(f(1,2))) = 0

f(G(f(1, 0)), f(F(f(1, 2)), G(f(1, 2)))) = 1 + 0 = 1

CAT 1999 - Question 45

Which of the following expressions yields $$x^2$$ as its result?

Video Solution

Solution

F(f(x,-x)) = 0 and G(f(x,-x)) = 0

F(f(x,x)) = -2x and G(f(x,x)) = 2x

F(f(x,x)).G(f(x,x)) = -$$4x^2$$

$$\log_216$$ = 4

=> $$\frac{-F(f(x,x)).G(f(x,x))}{\log_216}$$ = $$x^2$$

Instructions

DIRECTIONS for the following questions:

These questions are based on the situation given below: A robot moves on a graph sheet with x and y-axes. The robot is moved by feeding it with a sequence of instructions. The different instructions that can be used in moving it, and their meanings are: Instruction Meaning GOTO(x,y) move to point with coordinates (x, y) no matter where you are currently WALKX(P) Move parallel to the x-axis through a distance of p, in the positive direction if p is positive, and in the negative direction if p is negative WALKY(P) Move parallel to the y-axis through a distance of p, in the positive direction if p is positive, and in the negative direction if p is negative.

CAT 1999 - Question 46

The robot reaches point (6, 6) when a sequence of three instructions is executed, the first of which is a GOTO(x, y) instruction, the second is WALKX(2) and the third is WALKY(4). What are the values of x and y?

Video Solution

Solution

Before, the third instruction, the point on which the robot is present is (6,2).

Before, the second instruction, the point on which the robot is present is (4,2).

Hence, the values of x and y are 4 and 2 respectively.

CAT 1999 - Question 47

The robot is initially at (x, y), x > 0 and y < 0. The minimum number of instructions needed to be executed to bring it to the origin (0,0) if you are prohibited from using the GOTO instruction is:

Video Solution

Solution

WALKX(-x) and WALKY(y) are the commands to be used for the robot to reach origin in any order.

Hence, the answer is 2.

Instructions

DIRECTIONS for the following three questions

These questions are based on the situation given below:

A road network (shown in the figure below) connects cities A, B, C and D. All road segments are straight lines. D is the midpoint on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long. Ms. X and Mr. Y leave A at 8:00 am, take different routes to city C and reach at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 km per hour. Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.

CAT 1999 - Question 48

What is the average speed of Y in km per hour?

Video Solution

Solution

Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.

AD=DC =a(D is the midpoint on the road connecting A and C.)

The time taken by Y to reach D= a/45

Further, the time taken by Y to reach C= a/55

Y has travelled equal distances are constant speed.

=> Average speed = $$\frac{2a}{\frac{a}{45}+\frac{a}{55}}$$=$$\frac{2*45*55}{100}$$ = 49.5 kmph

CAT 1999 - Question 49

The total distance traveled by Y during the journey is approximately

Video Solution

Solution

$$BC^2$$ + 10000 = $$AC^2$$.

$$\frac{BC+100}{61.875}$$ = $$\frac{AC}{49.5}$$ => $$\frac{BC+100}{AC}$$ = 1.25 = > BC = 1.25AC - 100

On solving these equations, we get AC as 105 and BC a 31.

=> Y traveled 105 km.

CAT 1999 - Question 50

What is the length of the road segment BD?

Video Solution

Solution

$$BC^2$$ + 10000 = $$AC^2$$.

$$\frac{BC+100}{61.875}$$ = $$\frac{AC}{49.5}$$ => $$\frac{BC+100}{AC}$$ = 1.25 = > BC = 1.25AC - 100

On solving these equations, we get AC as 105 and BC a 31.

Let the point B be (0,0) => A = (0, -100) and C = (31,0)

D = Mid point of AC = (15.5, -50)

BD = $$\sqrt{15.5^2+50^2}$$ = 52.5 (approximately).

CAT Content

CAT 1999 Question Paper

CAT 1999 - Question 41

CAT 1999 - Question 42

CAT 1999 - Question 43

CAT 1999 - Question 44

CAT 1999 - Question 45

CAT 1999 - Question 46

CAT 1999 - Question 47

CAT 1999 - Question 48

CAT 1999 - Question 49

CAT 1999 - Question 50

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CAT Content

CAT 1999 Question Paper

CAT 1999 - Question 41

CAT 1999 - Question 42

CAT 1999 - Question 43

CAT 1999 - Question 44

CAT 1999 - Question 45

CAT 1999 - Question 46

CAT 1999 - Question 47

CAT 1999 - Question 48

CAT 1999 - Question 49

CAT 1999 - Question 50

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