# Previous CAT Questions - Solved (Logical Reasoning) - Set 1

• (CAT 2000)
Persons X, Y, Z and Q live in red, green, yellow or blue colored houses placed in a sequence on a street. Z lives in a yellow house. The green house is adjacent to the blue house. X does not live adjacent to Z. The yellow house is in between the green and red houses. The color of the house X lives in is:
a. blue
b. green
c. red
d. not possible to determine

It is always better to 1st solve one pattern, then the next. (i.e. Here, 1st solve the clues related to color, then form an order out of color, and then match it with the persons. Don’t try to do simultaneously. It will lead to confusion, and you will lose time.)
We have X, Y, Z, Q (persons) and R, G, Y, B (colors). I will solve the colors first.
BG or GB (as they are together)
GYR or RYG (yellow between green and red)
So it is BGYR or RYGB (the order) (understand here that 1 order is the opposite of other)
Now match it with people. The clues here are Z lives in yellow, and X does not live near Z.

B – Can be X R
G Y - Z
Y – Z G
R B – Can be X.

Whatever be the case, X stays in Blue.

(CAT 2000)
My bag can carry no more than ten books; I must carry at least one book each of management, mathematics, physics and fiction. Also, for every management book I carry I must carry two or more fiction books, and for every mathematics book I carry I must carry two or more physics books. I earn 4, 3, 2 and 1 points for each management, mathematics, physics and fiction book, respectively, I carry in my bag. I want to maximize the points I can earn by carrying the most appropriate combination of books in my bag. The maximum points that I can earn are:
a. 20
b. 21
c. 22
d. 23

Link the books that are together. Here it will be Mng – Fic, Mth – Phy.
Since total 10.
Hence, we can have 2 books of Management or Math, 4 books of its pair (total 6), and 4 other books, which should include the other book pair.
So what we have with now is we have fixed either 2 management, 1 math or 1 management 2 math.
Lets take 2 management, 1 math. 2 management, means 4 fiction. 1 math means 2 physics. So totally, now we have 9 books worth 19 points. And we can add 1 fiction or 1 physics. Obviously, we add 1 physics, and totally we have 10 books with 21 points.
Lets take 2 math and 1 management. 2 math means 4 physics, 1 management means 2 fiction. So we have 9 books with 20 points. We can add 1 fiction or 1 physics. Adding 1 physics makes the total points 22.
So, the maximum point possible is 22.

(CAT 2000)
There are ten animals — two each of lions, panthers, bison, bears, and deer — in a zoo. The enclosures in the zoo are named X, Y, Z, P and Q and each enclosure is allotted to one of the following attendants: Jack, Mohan, Shalini, Suman and Rita. Two animals of different species are housed in each enclosure. A lion and a deer cannot be together. A panther cannot be with either a deer or a bison. Suman attends to animals from among bison, deer, bear and panther only. Mohan attends to a lion and a panther. Jack does not attend to deer, lion or bison. X, Y and Z are allotted to Mohan, Jack and Rita respectively. X and Q enclosures have one animal of the same species. Z and P have the same pair of animals. The animals attended by Shalini are:
a. bear & bison
b. bison & deer
c. bear & lion
d. bear & panther

Use tabular method, whenever you have 2+ use cases. Here you have animals, enclosures, and people assigned to enclosures. You can use tabular method, when there are 2 use cases also, but that might sometimes take away your time. (If the question is too easy). Read the question fully and thoroughly and strike off anything that is not important and concentrate on the “fixed” given details, before concentrating on the condition.

Given (marked in red)

1. Jack does not have deer, lion, bison. So he has bear and panther. He is in charge of Y as given.
2. Mohan has Lion and Panther and he has X (given)
3. Suman has 2 among bison, bear, panther or deer.
4. Rita has Z.

Conclusions (marked in blue)
5) Since X and Q has one animal of same species, and panther is already allocated, we have only Lion which can be present both in X and Q. And since Suman cannot handle lion, Shalini is in charge of Q, and she has Lion as one animal.
6) Now Z and P have same animals. And it should be the ones that are not in the table till now. The only options are deer and bison which are not allocated till now. So Z and P has Deer and Bison.
7) The only animal left is 1 bear, and it belongs to Shalini along with the Lion.

c is the right option.

If you see here, in the solution, we never had to use the use cases “lion and deer never together” and “panther cannot be with deer/bison”. This is the advantage of the method that was suggested in the 1st question, where we try to group and do, instead of taking all the clues in a go. Here, 1st we tried to assign fixed animals to the people, then the cage.

(CAT 2000)
Eighty kilograms (kg) of store material is to be transported to a location 10 km away. Any number of couriers can be used to transport the material. The material can be packed in any number of units of 10, 20 or 40kg. Courier charges are Rs. 10 per hour. Couriers travel at the speed of 10 km/ hr if they are not carrying any load, at 5 km/hr if carrying 10kg, at 2 km/hr if carrying 20kg and at 1 km/hr if carrying 40 kg. A courier cannot carry more than 40 kg of load. The minimum cost at which 80kg of store material can be transported to its distinction will be:
a. Rs.180
b. Rs.160
c. Rs.140
d. Rs.120

Fastest approach will be trial and error here, since calculation here is pretty simple and straightforward. From the problem, if we travel using 10kg pack, it takes 2hrs, and hence 20rs. Likewise for 20kg, it will be 50rs and for 40kg, it will be 100Rs.
Here, the following cases can be considered.
8 * 10kgs – 160 Rs
4 * 20kgs – 200 Rs
2 * 40kgs – 200 Rs.
10 * 2, 20 * 1, 40 * 1 = 190 Rs
10 * 4, 40 * 1 = 180 Rs
10 * 6, 20 *1 = 170 Rs
20 * 2, 40 *1 = 200 Rs
20 * 3, 10 *1 = 170 Rs
Hence the cheapest will be 160Rs.

Alternate approach:
Calculate the individual costs, i.e. the 1st 3 steps shown above. Using only 10kg or 20kg or 40kg. Using one 40kg will cost 100Rs, and we have to still transport 40kgs more. Since we have 40 kg left, we will incur more cost than 160Rs for sure, however we transport. Hence we can rule out transport using 40kg. What is remaining, will be, transport using 10/20 kg combo.

Again, here if we use two 20kgs, it is equivalent to using 1 40kg, as it costs the same 100Rs. If we use three 20kgs, it is 150Rs and 20kg more to transport. If we use one 20kg, then it will be 50rs, and we have 60 more kg to transport, again which will be more. Other than ruling out the 40kg option, this is more or less the same trial and error method. However, if you rule out the 40kg option, the number of trial and errors you need to perform will be much lesser.

(CAT 2000)
Five persons with names P, M, U, T and X live separately in any one of the following: a palace, a hut, a fort, a house or a hotel. Each one likes two different colors from among the following: blue, black, red, yellow and green. U likes red and blue. T likes black. The person living in a palace does not like black or blue. P likes blue and red. M likes yellow. X lives in a hotel. M lives in a:
a. hut
b. palace
c. fort
d. house

This is one kind of question which you can solve easily, because of 2 reasons.
No ambiguity in option. Option does not have “none of the above”, or “not possible to determine”.
No detail is given about the people living in hut, fort or house. And it says, X lives in a hotel. The only detail is about the person living in the palace, that he does not like black or blue. Hence the answer will be “palace” 99% of the time. (Just because, that is the only “place of stay” explained)
For the not-so convinced kind, let us solve it. Again form a table.

I formed the table with the values given in the question. P, U and T like blue/black.
Hence they cannot live in the palace. X lives in the hotel. So M lives in the palace. :-)

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