# Previous CAT Questions - Solved (Logical Reasoning) - Set 2

• (CAT 2001)
Four friends Ashok, Bashir, Chirag and Deepak are out shopping. Ashok has less money than three times the amount that Bashir has. Chirag has more money than Bashir. Deepak has an amount equal to the difference of amounts with Bashir and Chirag. Ashok has three times the money with Deepak. They each have to buy at least one shirt, or one shawl, or one sweater, or one jacket, that are priced Rs 200, Rs 400, Rs 600 and Rs 1000 apiece, respectively. Chirag borrows Rs 300 from Ashok and buys a jacket. Bashir buys a sweater after borrowing Rs 100 from Ashok and is left with no money. Ashok buys three shirts. What is the costliest item that Deepak could buy with his own money?
(a) A shirt
(b) A shawl
(c) A sweater
(d) A jacket

From the clues, we have
A – 3y / B – x / C – x+y / D – y
Amount with A < 3x.
We know, x + 100 = 600. (Bashir buys sweater with 100rs from Ashok and no money left)
x = 500.
Amount with A < 1500. (Since amount with ashok is less than 3 times amount with Bashir)
Also, x + y + 300 >= 1000. (Chirag buys sweater after borrowing 300rs from Ashok)
500 + y + 300 > = 1000.
y > = 200.
Amount with Ashok = 300 (lent to C) + 100 (lent to B ) + 3y >= 1000
1000 < = Ashok Amount < = 1500.
333.3 < = Deepak < = 500. So he can buy a shawl.

Note: There may be a confusion that D = C - B = 700 - 500 = 200. But we dont know whether C has 700 rs or not.. it is not said anything that whthr C is left with any money after buying the jacket.. but for B we know it is 500 as it is said that B is left with no money after buying a sweater.

(CAT 2001)
In a family gathering there are two males who are grandfathers and four males who are fathers. In the same gathering there are two females who are grandmothers and four females who are mothers. There is at least one grandson or a grand-daughter present in this gathering. There are two husband-wife pairs in this group. These can either be a grandfather and a grandmother, or a father and a mother. The single grandfather (whose wife is not present) has two grandsons and a son present. The single grandmother (whose husband is not present) has two grand-daughters and a daughter present. A grandfather or a grandmother present with their spouses does not have any grandson or grand-daughter present. What is the minimum number of people present in this gathering?
(a) 10
(b) 12
(c) 14
(d) 16

Always approach these kind of questions with the “tree – method”, and use different symbols/alphabets at least for male and female, and if possible for the different sub trees. 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 – 1 single GF has 1 Son and 2 Grandsons present.

Given - 1 single GM has 1 daughter and 2 granddaughters present.

Now the “condition” – 4 fathers and 4 mothers. From the 1st two trees, we have 2 fathers and 2 mothers (including the GF and GM, who are also fathers/mothers). Now we need 2 more fathers and 2 more mothers. Let us try “spouse-ing” the father and mother above.

The above diagram makes for only 3 fathers and 3 mothers, and we still have to draw 1 GF and 1 GM (as there are 2 GF and 2 GM in the question). We can draw 1 GF, 1 GM, and they can have 1 Son and 1 daughter as well. However, this will result in a total of 16 people. Now, since we need the minimum number of people, let us think if some other way is possible.

Now, what about assigning the 2nd GF and GM with a son and having a spouse for him, instead of attaching spouses to the already “existing” father and mother. The diagram will look like this.

This satisfies all the conditions. We have 2 GF, 2GM, 2 married couple, 4 fathers and 4 mothers, and only 12 people!

(CAT 2001)
Eight people carrying food baskets are going for a picnic on motorcycles. Their names are A, B, C, D, E, F, G and H. They have four motorcycles M1, M2, M3 and M4 among them. They also have four food baskets O, P, Q and R of different size and shapes and which can be carried only on motorcycles M1, M2, M3 or M4 respectively. No more than two persons can travel on a motorcycle and no more than one basket can be carried on a motorcycle. There are two husband-wife pairs in this group of eight people and each pair will ride on a motorcycle together. C cannot travel with A or B. E cannot travel with B or F. G cannot travel with F, or H, or D. The husband-wife pairs must carry baskets O and P. Q is with A and P is with D. F travels on M1 and E travels on M2 motorcycles. G is with Q and B cannot go with R. Who is travelling with H?
(a) A
(b) B
(c) C
(d) D

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. Here, we see that the husband-wife part is unnecessary, as the question/arrangement is not based on it. Let us represent the given details, in a tabular form.

O, P, Q, R is with M1, M2, M3, M4 respectively.
Q is with A, and G is with Q. That means M3 which has basket Q has A & G.
P is with D, and E is in M2. So M2 which has basket P has D & E.
F is in M1, and B cannot go with R. That means, the only other vacant place B can go is with F in M1.
So, M4 which has basket R will have C & H.

The question is solved, just by considering the given details, and with a single condition, that the bike can have only 2 people. Now, if you cross verify the conditions, you can see that all the conditions are satisfied. The time saved will be humongous!

(CAT 2001)
I have a total of Rs 1000. Item A costs Rs 110, item B costs Rs 90, item C costs Rs 70, item D costs Rs 40 and item E costs Rs 45. For every item D that I purchase, I must also buy two of item B. For every item A, I must buy one of item C. For every item E, I must also buy two of item D and one of item B. For every time purchased I earn 1000 points and for every rupee not spent I earn a penalty of 150 points. My objective is to maximize the points I earn. What is the number of items that I must purchase to maximize my points?
(a) 13
(b) 14
(c) 15
(d) 16

Group whichever items are together. i.e.
AC will be together if I buy A.
B & C can be brought individually.
D will be with 2B.
E will be with 2D and 1B. But D is equivalent to 2B. So E will have to be brought with 2D and 4B (you need not have an extra 1B here, because already you have 4B with E). So E’s combo will be E-2D-4B, total of 7 items.
I decide to buy A, I have to buy C. AC – 180Rs. 2 items – 180rs.
Average for 1 item = 90rs.
If I buy B, 1 item = 90rs.
If I buy C, 1 item =70Rs.
If I buy D, D 2B, 3 items = 220, average for 1 item = 73.3Rs
If I buy E, E 2D 4B, 7 items = 485Rs, average for 1 item = 69.28Rs.

We see that the least average price is for buying E. We can buy 2 “sets” of E for 970Rs, but 30rs will be pending, and it means 30*150 will be deducted from 14000 (gained by buying 14 items)

Now, we can go to C, which is the 2nd in the list. We can buy 14 items of C, with 20 Rs pending. Else, we can buy 13 items of C with 90Rs pending, and with 90rs, we can but an item of B. Hence, totally 14items can be brought, and 0rs pending, a total of 14000 can be earned.

Any other combo can be tried, but will not give this much points.
Hence the answer will be 14 items, 13 of C and 1 of B.

(CAT 2001)
On her walk through the park, Hamsa collected 50 colored leaves, all either maple or oak. She sorted them by category when she got home, and found the following:

• The number of red oak leaves with spots is even and positive.
• The number of red oak leaves without any spot equals the number of red maple leaves without spots. All non-red oak leaves have spots, and there are five times as many of them as there are red spotted oak leaves.
• There are no spotted maple leaves that are not red.
• There are exactly 6 red spotted maple leaves.
• There are exactly 22 maple leaves that are neither spotted nor red.**

How many oak leaves did she collect?
(a) 22
(b) 17
(c) 25
(d) 18

Red = R, Not Red = NR,
Oak = O, Maple = M,
Spot = S, No Spot = NS
R.O.NS = x
R.O.S. = y.
NR.O.S = 5y
R.M.NS = x
R.M.S = 6
NR.M.NS = 22
Total = 50.
Adding up, 2x + 6y = 22, x + 3y = 11. y is even and positive. So y can only be 2. x is 5.
total Oak leaves - 17.

1

1

1

1

1

2

7

1