CAT LRDI Questions PDF [Most Important with solutions]

The Logical Reasoning and Data Interpretation (LRDI) section in the CAT Exam is often a deciding factor for many students. Neglecting this section could prove costly and, thus, requires thorough preparation. If you’re new to this section, you can check out these CAT LRDI Questions from the CAT previous year papers. This article provides essential LRDI questions in PDF format, complete with solutions and crucial tips for solving them. These CAT LRDI questions are also available for download, including detailed solutions and tricks to aid in solving these questions effectively.

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Instructions

DIRECTIONS for the following questions: These questions are based on the situation given below: A young girl Roopa leaves home with x flowers, goes to the bank of a nearby river. On the bank of the river, there are four places of worship, standing in a row. She dips all the x flowers into the river. The number of flowers doubles. Then she enters the first place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the second place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the third place of worship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number of flowers doubles. She goes to the fourth place of worship, offers y flowers to the deity. Now she is left with no flowers in hand.

Question 1: If Roopa leaves home with 30 flowers, the number of flowers she offers to each deity is:

a) 30

b) 31

c) 32

d) 33

1) Answer (C)

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Solution:

Number of flowers after the first dipping = 60

Number of flowers after the second dipping = 2(60-y) = 120-2y

Number of flowers after the third dipping = 2(120-2y-y) = 240-6y

Number of flowers after the fourth dipping = 2(240-6y-y) = 480-14y

y = 480 – 14y

15y = 480

y = 32

Question 2: The minimum number of flowers that could be offered to each deity is:

a) 0

b) 15

c) 16

d) Cannot be determined

2) Answer (C)

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Solution:

Number of flowers after the first dipping = 2x

Number of flowers after the second dipping = 2(2x-y) = 4x-2y

Number of flowers after the third dipping = 2(4x-2y-y) = 8x-6y

Number of flowers after the fourth dipping = 2(8x-6y-y) = 16x-14y

16x-14y = y

y = 16x/15

Minimum value of y = 16 when x = 15

Question 3: The minimum number of flowers with which Roopa leaves home is:

a) 16

b) 15

c) 0

d) Cannot be determined

3) Answer (B)

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Solution:

Number of flowers after the first dipping = 2x

Number of flowers after the second dipping = 2(2x-y) = 4x-2y

Number of flowers after the third dipping = 2(4x-2y-y) = 8x-6y

Number of flowers after the fourth dipping = 2(8x-6y-y) = 16x-14y

16x-14y = y

y = 16x/15

Minimum value of y = 16 when x = 15

Instructions

DIRECTIONS for the following questions: These questions are based on the situation given below:

There are m blue vessels with known volumes $V1, V2 , …., V_m$, arranged in ascending order of volume, where $v_1 > 0.5$ litre, and $V_m < 1$ litre. Each of these is full of water initially. The water from each of these is emptied into a minimum number of empty white vessels, each having volume 1 litre. The water from a blue vessel is not emptied into a white vessel unless the white vessel has enough empty volume to hold all the water of the blue vessel. The number of white vessels required to empty all the blue vessels according to the above rules was n.

Question 4: Among the four values given below, which is the least upper bound on e, where e is the total empty volume in the m white vessels at the end of the above process?

a) $mv_m$

b) $m(1 – v_m)$

c) $mv_1$

d) $m(1 – v_1)$

4) Answer (B)

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Solution:

$v_1 > 0.5$

$v_m < 1$

As we cannot empty a blue vessel into a white vessel unless there is enough space to hold all the volume in blue vessel in the white vessel, we have to use m white vessels. This is because we cannot empty more than 1 blue vessel into 1 white vessel.

The minimum volume that can be left in each white vessel is $1-v_m$.

=> The minimum volume that can be left in m white vessels is $m(1-v_m)$, which is the least upper bound.

Question 5: Let the number of white vessels needed be n1 for the emptying process described above, if the volume of each white vessel is 2 liters. Among the following values, which is the least upper bound on n1?

a) m/4

b) smallest integer greater than or equal to (m/2)

c) n

d) greatest integer less than or equal to (m/2)

5) Answer (B)

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Solution:

To find the limiting value, let us consider that all the blue vessels are of 1 litre capacity.

If all the blue vessels have 1 litre capacity, then the number of white vessels required is smallest integer greater than or equal to (m/2).

But, blue vessels are all not of the same capacity => the above value of n1 is the upper bound.

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Instructions

DIRECTIONS for the following four questions:

A low-cost airline company connects ten India cities, A to J. The table below gives the distance between a pair of airports and the corresponding price charged by the company. Travel is permitted only from a departure airport to an arrival airport. The customers do not travel by a route where they have to stop at more than two intermediate airports.

<img “=”” alt=”” class=”img-responsive” src=”https://cracku.in/media/questionGroup/DI_6_3.png”/>

Question 6: What is the lowest possible fare, in rupees, from A to J?

a) 2275

b) 2850

c) 2890

d) 2930

e) 3340

6) Answer (A)

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Solution:

From the table we can see that, the lowest price would be from A to H and H to J.

The cost of travel from A to H = Rs 1850

The cost of travel from H to J = Rs 425

Total cost = 1850 + 425 = Rs 2275.

Question 7: The company plans to introduce a direct flight between A and J. The market research results indicate that all its existing passengers travelling between A and J will use this direct flight if it is priced 5% below the minimum price that they pay at present. What should the company charge approximately, in rupees, for this direct flight?

a) 1991

b) 2161

c) 2707

d) 2745

e) 2783

7) Answer (B)

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Solution:

From the table we can see that, the lowest price would be from A to H and H to J.

The cost of travel from A to H = Rs 1850

The cost of travel from H to J = Rs 425

Total cost = 1850 + 425 = Rs 2275

Lowest price = Rs 2275

95% of 2275 = Rs 2161

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Question 8: If the airports C, D and H are closed down owing to security reasons, what would be the minimum price, in rupees, to be paid by a passenger travelling from A to J?

a) 2275

b) 2615

c) 2850

d) 2945

e) 3190

8) Answer (C)

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Solution:

If the airports C, D and H are closed down  the minimum price to be paid by a passenger travelling from A to J would be by first travelling to F and then from F to J.

The cost of travel from A to F = Rs 1700

The cost of travel from F to J = Rs 1150

Total cost = 1700 + 1150 = Rs 2850

Question 9: If the prices include a margin of 10% over the total cost that the company incurs, what is the minimum cost per kilometer that the company incurs in flying from A to J?

a) 0.77

b) 0.88

c) 0.99

d) 1.06

e) 1.08

9) Answer (B)

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Solution:

The minimum cost from A to J we know is 2275.

Let the CP to company be C

Since 10% over actual CP is the total price  i.e.  $\text{CP}\times1.1 = 2275  \rightarrow  CP =  \frac{2275}{1.1}$

The total distance is 1950+1400=2350 Km.

Cost per Km = $\dfrac{\frac{2275}{1.1}}{2350}$ = Rs 0.88/Km

Question 10: If the prices include a margin of 15% over the total cost that the company incurs, which among the following is the distance to be covered in flying from A to J that minimizes the total cost of travel for the company?

a) 2170

b) 2180

c) 2315

d) 2350

e) 2390

10) Answer (D)

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Solution:

Even if If the prices include a margin of 15% over the total cost that the company incurs, the total cost company incurs would be minimum for route AHJ i.e 2350 km. Hence option D.

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