# Selection with Condition Questions for CAT LRDI

**Selection With**Â **Conditions** is one of the most important topics in the **CAT** **LRDI **section. If you’re weak in this topic, make sure you are aware of the **Concept of Selection with Conditions**. You can check out these CAT Selection with Condition questions from the **CAT Previous year papers**. Practice a good number of sets from CAT Selection with Condition in the **latest format of CAT LRDI Section**. This post will look into some important Selections with Condition questions for CAT. You can download these Important **Selection with Condition Questions for CAT** (with detailed answers) **PDF** along with the video solutions below, which are completely Free.

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**Instructions**

Directions for the following four questions: Answer the questions on the basis of the information given below.

Twenty one participants from four continents (Africa, Americas, Australasia, and Europe) attended a United Nations conference. Each participant was an expert in one of four fields, labour, health, population studies, and refugee relocation. The following five facts about the participants are given.

(a) The number of labour experts in the camp was exactly half the number of experts in each of the three other categories.

(b) Africa did not send any labour expert. Otherwise, every continent, including Africa, sent at least one expert for each category.

(c) None of the continents sent more than three experts in any category.

(d) If there had been one less Australasian expert, then the Americas would have had twice as many experts as each of the other continents.

(e) Mike and Alfanso are leading experts of population studies who attended the conference. They are from Australasia.

**Question 1:Â **Which of the following combinations is NOT possible?

a)Â 2 experts in population studies from the Americas and 2 health experts from Africa attended the conference.

b)Â 2 experts in population studies from the Americas and 1 health expert from Africa attended the conference.

c)Â 3 experts in refugee relocation from the Americas and 1 health expert from Africa attended the conference.

d)Â Africa and America each had 1 expert in population studies attending the conference.

**1)Â AnswerÂ (D)**

**Solution:**

According to given conditions, the possible solutions are as given in the tables below:

From the above tables we can see that there was no possible case in which Africa and America had 1 expert each in population studies. Thus, statement D is false.

**Question 2:Â **If Ramos is the lone American expert in population studies, which of the following is NOT true about the numbers of experts in the conference from the four continents?

a)Â There is one expert in health from Africa.

b)Â There is one expert in refugee relocation from Africa.

c)Â There are two experts in health from the Americas.

d)Â There are three experts in refugee relocation from the Americas.

**2)Â AnswerÂ (C)**

**Solution:**

From the first condition a, assume the number of experts in labour = x, then the number of experts in each of the other three categories =2x

Now, x+2x+2x+2x+2x = 21Â => x = 3 and 2x = 6

So the number of experts in labour = 3 and the number of experts in each of the three categories = 6

From condition d, if the number of experts of Astralaisa = y, then the number of experts of Americas = 2(y-1), the number of experts of Europe = y-1, the number of experts of Africa= y-1

Now, y+y-1+y-1+2(y-1)=21Â => 5y-4=21Â => y=5

Hence the table can be filled as follows:

Using condition b, the number of labour experts of Africa = 0 and the number of labour experts of the rest of the continents will be 1 each.

Consider the column of Europe, since Europe sent at least 1 in each category, each category will have 1 expert. (Since the total sum is 4)

From the condition e, the population studies of Australasia will contain 2 experts. So all the other categories will have to take at 1 expert because the total sum is 5 and no value can be 0. The table can be filled as follows:

Now, according to the condition c, the possible solutions are as given in the tables below:

From the tables we can see that when there is only 1 population expert from America, there are 3 American Health Experts. Hence option C is not true.

**Question 3:Â **Alex, an American expert in refugee relocation, was the first keynote speaker in the conference. What can be inferred about the number of American experts in refugee relocation in the conference, excluding Alex?

i. At least one

ii. At most two

a)Â Only i and not ii

b)Â Only ii and not I

c)Â Both i and ii

d)Â Neither i nor ii

**3)Â AnswerÂ (C)**

**Solution:**

According to given conditions, the possible solutions are as given in the tables below:

If an American expert in refugee relocation, was the first keynote speaker in the conference. Then apart from him there is atleast 1 and atmost 2 american expert in refugee relocation. Hence option C.

**Question 4:Â **Which of the following numbers cannot be determined from the information given?

a)Â Number of labour experts from the Americas.

b)Â Number of health experts from Europe.

c)Â Number of health experts from Australasia.

d)Â Number of experts in refugee relocation from Africa.

**4)Â AnswerÂ (D)**

**Solution:**

According to given conditions, the possible solutions are as given in the tables below:

We can see that there are 2 possibilities for number of experts in refugee relocation from Africa.Hence it cannot be determined.

**Instructions**

Two traders, Chetan and Michael, were involved in the buying and selling Of MCS shares over five trading days. At the beginning of the first day, the MCS share was priced at Rs 100, while at the end of the fifth day it was priced at Rs 110. At the end of each day, the MCS share price either went up by Rs 10, or else, it came down by Rs 10. Both Chetan and Michael took buying and selling decisions at the end of each trading day. The beginning price of MCS share on a given day was the same as the ending price of the previous day. Chetan and Michael started with the same number of shares and amount of cash, and had enough of both. Below are some additional facts about how Chetan and Michael traded over the five trading days.

â€¢ Each day if the price went up, Chetan sold 10 shares of MCS at the closing price. On the other hand, each day if the price went down, he bought 10 shares at the closing price.

â€¢ If on any day, the closing price was above Rs 110, then Michael sold 10 shares of MCS, while if it was below Rs 90, he bought 10 shares, all at the closing price.

**Question 5:Â **If Chetan sold 10 shares of MCS on three consecutive days, while Michael sold 10 shares only once during the five days, what was the price of MCS at the end of day 3?

a)Â Rs 90

b)Â Rs 100

c)Â Rs 110

d)Â Rs 120

e)Â Rs 130

**5)Â AnswerÂ (C)**

**Solution:**

The above table includes the values of the share price at the end of each day. Chetan and Michael column shows the number of shares at the end of 5th day with Chetan and Michael respectively. (-10 means he has sold 10 shares, +10 means he has bought 10 shares)

there are 10 different possible cases according to the initial and final share price.

The question asks aboutÂ the case where Chetan has sold 3 times and Michael sells only once.

Starting with Michael, for exactly one sell, the price should touch 120 only once as Michael sells the share only at price greater than 110.

if the price touches 120 twice or more, Michael will sell the share more than once which is not a desirable case.

Also, chetan has to sell thrice consecutively which is only possible if the share price is 90 at one instance and rises to 120 in straight 3 days.

This is only possible in case 2. Hence the price on 3rd day’s endÂ in case 2 is 110.

**Question 6:Â **If Michael ended up with Rs 100 less cash than Chetan at the end of day 5, what was the difference in the number of shares possessed by Michael and Chetan (at the end of day 5)?

a)Â Michael had 10 less shares than Chetan.

b)Â Michael had 10 more shares than Chetan.

c)Â Chetan had 10 more shares than Michael.

d)Â Chetan had 20 more shares than Michael.

e)Â Both had the same number of shares.

**6)Â AnswerÂ (E)**

**Solution:**

The above table includes the values of the share price at the end of each day. Chetan and Michael column shows the number of shares at the end of 5th day with Chetan and Michael respectively. (-10 means he has sold 10 shares, +10 means he has bought 10 shares)

there are 10 different possible cases according to the initial and final share price.

Please note that Chetan will always be having -10 shares(10 shares sold) andÂ 1300 as cash.

This is because Chetan buys for every fall in price and sells for every rise in price. But the fluctuation in share price is constant as it starts from 100 and closes at 110 on day 5. So, in total chetan will always sell 10 shares in all 5 days combined.

As chetan has sold 10 shares , he’ll get 110*10 =1100 cash because of it in every case. Also chetan earns Rs. 200 in every case because of buying at low and selling at high. So total cash chetan will always have after 5 days = 1100+200 =1300.

The question asks aboutÂ the case where Michael ended up with Rs 100 less cash than Chetan at the end of day 5.

So we have to look for the case where Michael has 1200 cash which is only possible when Michael has -10 number of shares in the end of day 5 and also has made profit of Rs. 100.

This is possible in case 7 as Michael has sold the shares at rs 120 but did not sell those shares as price never went below 90.

In the end Michael will have -10 shares at a price 110 which is Rs. 10 less than the price he sold at.

So he makes profit of 100 rs through it and will have the cash of 1100 through the sold shares.

Their difference in 7th case is 1300-1200=100

Also, in this case they have same number of shares sold at the end of the 5th day. So E is the answer.

Checkout: **CAT Free Practice Questions and Videos**

**Question 7:Â **If Chetan ended up with Rs 1300 more cash than Michael at the end of day 5, what was the price of MCS share at the end of day 4?

a)Â Rs 90

b)Â Rs 100

c)Â Rs 110

d)Â Rs 120

e)Â Not uniquely determinable

**7)Â AnswerÂ (B)**

**Solution:**

The above table includes the values of the share price at the end of each day. Chetan and Michael column shows the number of shares at the end of 5th day with Chetan and Michael respectively. (-10 means he has sold 10 shares, +10 means he has bought 10 shares)

there are 10 different possible cases according to the initial and final share price.

Please note that Chetan will always be having -10 shares(10 shares sold) and 1300 as cash.

This is because Chetan buys for every fall in price and sells for every rise in price. But the fluctuation in share price is constant as it starts from 100 and closes at 110 on day 5. So, in total Chetan will always sell 10 shares in all 5 days combined.

As Chetan has sold 10 shares, he’ll get 110*10 =1100 cash because of it in every case. Also, Chetan earns Rs. 200 in every case because of buying at low and selling at high. So total cash Chetan will always have after 5 days = 1100+200 =1300.

The question asks about the case where Michael ended up with Rs 1300 cash less than Chetan.

As Chetan will have exactly 1300 cash in every case; we have to look for the cases where Michael does not make any profit and also does not have any sold shares at the end of 5 days.

This is possible in 4 cases ie. 3,4,5 and 9.

In each of these cases, the price fluctuates in the range 90-110 which does not allow Michael to buy or sell any shares.

Also, in all these cases the price of MCS is 100 at the end of day 4.

So the answer is 100.

**Question 8:Â **What could have been the maximum possible increase in combined cash balance of Chetan and Michael at the end of the fifth day?

a)Â Rs 3700

b)Â Rs 4000

c)Â Rs 4700

d)Â Rs 5000

e)Â Rs 6000

**8)Â AnswerÂ (D)**

**Solution:**

there are 10 different possible cases according to the initial and final share price.

Please note that Chetan will always be having -10 shares(10 shares sold) and 1300 as cash.

This is because Chetan buys for every fall in price and sells for every rise in price. But the fluctuation in share price is constant as it starts from 100 and closes at 110 on day 5. So, in total Chetan will always sell 10 shares in all 5 days combined.

As Chetan has sold 10 shares, he’ll get 110*10 =1100 cash because of it in every case. Also, Chetan earns Rs. 200 in every case because of buying at low and selling at high. So total cash Chetan will always have after 5 days = 1100+200 =1300.

The question asks about the case whereÂ there hasÂ been the maximum possible increase in the combined cash balance of Chetan and Michael at the end of the fifth day.

As we know chetan has 1300 cash in all the cases , so we have to maximize the case where Michael has the most cash.

Also, if we see clearly, the profit made by selling and buying is in hundreds while the cash received by selling the shares is much far in terms of cash.

As 10 shares sold give = 10*110 = 1100 cash. So we have to look at the case where Michael has sold most shares which is case 8.

In case 8Â the price rises from 100 to 110 ->120 ->130 ->120 ->110

in this case 110(0 new shares)->120(Michael sold 10 shares)->130(Michael sold 10 shares)120(Michael sold 10 shares)->110(Michael does nothing) = 30 shares sold in total

Cash =Â 20 shares at 120 and 10 shares at 130 =Â 120*20+130*10 = 2400+1300 =3700

As Chetan has 1300 cash and Michael has 3700 cash;

total cash they have is : 3700+1300 =5000

**Question 9:Â **If Michael ended up with 20 more shares than Chetan at the end of day 5, what was the price of the share at the end of day 3?

a)Â Rs 90

b)Â Rs 100

c)Â Rs 110

d)Â Rs 120

e)Â Rs 130

**9)Â AnswerÂ (A)**

**Solution:**

there are 10 different possible cases according to the initial and final share price.

Please note that Chetan will always be having -10 shares(10 shares sold) and 1300 as cash.

To have 20 more shares than Chetan, Michael has to buy 10 shares which is case 1.

In case 1, the share price at the end of day 3 is 90.

**Instructions**

Directions for the following three questions: Answer the questions based on the passage below.

A group of three or four has to be selected from seven persons. Among the seven are two women: Fiza and Kavita, and five men: Ram, Shyam, David, Peter and Rahim. Ram would not like to be in the group If Shyam is also selected. Shyam and Rahim want to be selected together in the group. Kavita would like to be in the group only if David is also there. David, if selected, would not like Peter in the group. Ram would like to be in the group only if Peter is also there. David insists that Fiza be selected in case he is there in the group.

**Question 10:Â **Which of the following is a feasible group of three?

a)Â David, Ram and Rahim

b)Â Peter, Shyam and Rahim

c)Â Kavita, David and Shyam

d)Â Fiza, David and Ram

**10)Â AnswerÂ (B)**

**Solution:**

Shyam and Rahim have to be selected together. This rules out options a) and c). Ram will be in the group only if Peter is also there. This rules out option d). Option b) is a feasible group