{"id":3360,"date":"2018-12-14T08:07:21","date_gmt":"2018-12-14T02:37:21","guid":{"rendered":"https:\/\/iquanta.in\/blog\/?p=3360"},"modified":"2021-12-28T23:25:04","modified_gmt":"2021-12-28T17:55:04","slug":"cat-2018-lrdi-solutions-by-iquanta-slot-1","status":"publish","type":"post","link":"https:\/\/www.iquanta.in\/blog\/cat-2018-lrdi-solutions-by-iquanta-slot-1\/","title":{"rendered":"CAT 2018 LRDI Solutions by iQuanta : Slot 1"},"content":{"rendered":"\r\n
\"\"<\/figure>\r\n\r\n\r\n\r\n

Set 1<\/p>\r\n\r\n\r\n\r\n

Fuel contamination levels at each of 20 petrol pumps P1, P2, \u2026, P20 were recorded as either high, medium, or low.
1. Contamination levels at three pumps among P1 \u2013 P5 were recorded as high.
2. P6 was the only pump among P1 \u2013 P10 where the contamination level was recorded as low.
3. P7 and P8were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 \u2013 P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.<\/p>\r\n\r\n\r\n\r\n

Solution<\/p>\r\n\r\n\r\n\r\n

\r\n
\"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

Explanation:<\/strong><\/p>\r\n\r\n\r\n\r\n

By point 3, we know that P7 and P8 are recorded at same level. Except these two, no other consecutive pumps have same level of contamination.<\/p>\r\n\r\n\r\n\r\n

By point 2, we know that only P6 is recorded as Low and all other are recorded as either Medium or High.<\/p>\r\n\r\n\r\n\r\n

By point 1, We know that P1 to P5 have 3 high and 2 medium pumps. As we know that the 3 high pumps cannot be consecutive then P1, P3 and P5 must be highly Contaminate and P2 and P4 must be medium.<\/p>\r\n\r\n\r\n\r\n

Next P6 is recorded as Low hence we have two possibilities for P7, P8, P9 and P10 as High, High, Medium, High or Medium, Medium, High, Medium.<\/p>\r\n\r\n\r\n\r\n

In case 1, we have 6 highly contaminated pumps and in case 2 we have 4 highly contaminated pumps.<\/strong><\/p>\r\n\r\n\r\n\r\n

Point 6 states that Highly contaminated pumps are twice than Lowly contaminated.<\/p>\r\n\r\n\r\n\r\n

The sum of these two must be a multiple of 3.<\/p>\r\n\r\n\r\n\r\n

So there can only be 8 medium rated pumps making 8 high rated and 4 low rated.<\/p>\r\n\r\n\r\n\r\n

In case we assume medium to be 5 and high and low as 10 and 5, the situation will violate the case then there must be every other pump rated as high while P16 to P20 has no high pump.<\/p>\r\n\r\n\r\n\r\n

Now, if we take the two cases stated above, in 2nd<\/sup> case we have 4 highly contaminated pumps. We know there are 8 highly contaminated pumps in total so now P11 to P15 will have to have 4 contaminated pumps which is not possible because no two consecutive pumps can have some level of contamination.<\/p>\r\n\r\n\r\n\r\n

After eliminating case 2, we have first 10 pumps as<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
P1<\/td>\r\nP2<\/td>\r\nP3<\/td>\r\nP4<\/td>\r\nP5<\/td>\r\nP6<\/td>\r\nP7<\/td>\r\nP8<\/td>\r\nP9<\/td>\r\nP10<\/td>\r\n<\/tr>\r\n
High<\/td>\r\nMed<\/td>\r\nHigh<\/td>\r\nMed<\/td>\r\nHigh<\/td>\r\nLow<\/td>\r\nHigh<\/td>\r\nHigh<\/td>\r\nMed<\/td>\r\nHigh<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n

P16 to P20 have no High contaminated Pumps so the two cases will be M, L, M, L, M or L, M, L, M, L.<\/p>\r\n\r\n\r\n\r\n

In case 1, we have 6 medium rated and 3 Low rated pumps. In case 2, we have 5 medium and 4 (All) low rated pumps.<\/p>\r\n\r\n\r\n\r\n

Nothing much can be said about the rest so we move to the questions.<\/p>\r\n\r\n\r\n\r\n

Question 1<\/p>\r\n\r\n\r\n\r\n

Q.1 Which of the following MUST be true?<\/p>\r\n\r\n\r\n\r\n

1. The contamination level at P13 was recorded as low.<\/p>\r\n\r\n\r\n\r\n

2. The contamination level at P10 was recorded as high.<\/p>\r\n\r\n\r\n\r\n

3. The contamination level at P20 was recorded as medium.<\/p>\r\n\r\n\r\n\r\n

4. The contamination level at P12 was recorded as high.<\/p>\r\n\r\n\r\n\r\n

Answer: we know that P10 is recorded as High. Others may or may not be true. So the answer is option 2.<\/p>\r\n\r\n\r\n\r\n

Question 2<\/p>\r\n\r\n\r\n\r\n

What best can be said about the number of pumps at which the contamination levels were recorded\u00a0 as medium?\u00a0\u00a0<\/p>\r\n\r\n\r\n\r\n

1. At most 9<\/p>\r\n\r\n\r\n\r\n

2. More than 4<\/p>\r\n\r\n\r\n\r\n

3. At least 8<\/p>\r\n\r\n\r\n\r\n

4. Exactly 8<\/p>\r\n\r\n\r\n\r\n

Answer: There are 8 medium level contaminated pumps.<\/p>\r\n\r\n\r\n\r\n

Question 3<\/p>\r\n\r\n\r\n\r\n

If the contamination level at P11 was recorded as low, then which of the following MUST be true?<\/p>\r\n\r\n\r\n\r\n

1. The contamination level at P18 was recorded as low.<\/p>\r\n\r\n\r\n\r\n

2. The contamination level at P12 was recorded as high.<\/p>\r\n\r\n\r\n\r\n

3. The contamination level at P15 was recorded as medium.<\/p>\r\n\r\n\r\n\r\n

4. The contamination level at P14 was recorded as medium.<\/p>\r\n\r\n\r\n\r\n

Answer: In this we know that P11 was recorded as low. This is possible in case 1 where P16 to P20 had only 2 Low level contaminated pumps. So they must have been recorded as Med, Low, Med, Low and Med respectively. This eliminates option 1 as P18 is rated as Medium.<\/p>\r\n\r\n\r\n\r\n

Option 3 is eliminated because P16 is recorded as medium and so P15 must be High (Cant be low as all the low recorded pumps have found already).<\/p>\r\n\r\n\r\n\r\n

But here we see that if P15 is recorded as Highly contaminated pump then P14 can only be Medium contaminated. So the answer is option 4.<\/p>\r\n\r\n\r\n\r\n

Question 4<\/p>\r\n\r\n\r\n\r\n

If contamination level at P15 was recorded as medium, then which of the following MUST be FALSE?<\/p>\r\n\r\n\r\n\r\n

1. Contamination levels at P10 and P14 were recorded as the same.<\/p>\r\n\r\n\r\n\r\n

2. Contamination levels at P11 and P16 were recorded as the same.<\/p>\r\n\r\n\r\n\r\n

3. Contamination level at P14 was recorded to be higher than that at P15.<\/p>\r\n\r\n\r\n\r\n

4. Contamination levels at P13 and P17 were recorded as the same.<\/p>\r\n\r\n\r\n\r\n

Answer: This gives us that P15 is rated as Medium and so P16 will be rated as low and our case 2 will be applicable here that is no Low recorded contamination in P11 to P15.<\/p>\r\n\r\n\r\n\r\n

If P15 is medium so P14 must be high. P10, as we already know, is high. Option 1 is true hence not our answer.<\/p>\r\n\r\n\r\n\r\n

For the same reason as above, Option 3 is true hence not our answer.<\/p>\r\n\r\n\r\n\r\n

P17 is Medium contaminated pump and P13 can also be Medium contaminated so not necessarily false.<\/p>\r\n\r\n\r\n\r\n

P11 must beMedium because P10 is High and P16, as per our case, is Low. This must befalse.<\/p>\r\n\r\n\r\n\r\n

\"CAT<\/a><\/p>\r\n

Set 2<\/p>\r\n\r\n\r\n\r\n

An ATM dispenses exactly Rs. 5000 per withdrawal using 100, 200 and 500 rupee notes. The ATM requires every customer to give her\u00a0 preference for one of the three denominations of notes. It then dispenses notes such that the number of notes of the customer\u2019s preferred denomination exceeds the total number of notes of other denominations dispensed to her.<\/p>\r\n\r\n\r\n\r\n

\r\n
\"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

Explanation:<\/strong><\/p>\r\n\r\n\r\n\r\n

Question 1<\/p>\r\n\r\n\r\n\r\n

In how many different ways can the ATM serve a customer who gives 500 rupee notes as her preference?<\/p>\r\n\r\n\r\n\r\n

Answer: we can take 100x + 200y + 500z = 5000 where x, y and z are the number of notes.<\/p>\r\n\r\n\r\n\r\n

Or say, x + 2y + 5z = 50.<\/p>\r\n\r\n\r\n\r\n

Now the condition we have here is x + y < z<\/p>\r\n\r\n\r\n\r\n

If z=10, only 1 case as 5*10 = 50.<\/p>\r\n\r\n\r\n\r\n

If z=9, we get x + 2y + 5*9 = 50 => x + 2y = 5. This can be done in 3 ways at x=1, 3 and 5. All these values will be less than z.<\/p>\r\n\r\n\r\n\r\n

If z=8, we get x + 2y + 5*8 = 50 => x + 2y = 10. This can be done at x = 0, 2, 4, 6, 8 and 10.<\/p>\r\n\r\n\r\n\r\n

But in this case, x + y < 8 is valid only at x = 0, 2 and 4 so 3 cases.<\/p>\r\n\r\n\r\n\r\n

Total possible cases are 7.<\/p>\r\n\r\n\r\n\r\n

Question 2<\/p>\r\n\r\n\r\n\r\n

If the ATM could serve only 10 customers with a stock of fifty 500 rupee notes and a sufficient number of notes of other denominations, what is the maximum number of customers among these 10 who could have given 500 rupee notes as their preferences?<\/p>\r\n\r\n\r\n\r\n

Answer: To maximize the number of people getting 500 rupee notes, we need to minimize the 500 rupee notes in every transaction. As seen in the previous question, a person giving 500 rupee note as his preference will receive at least 8 500 rupee notes. So if everyone receives 8 500 rupee notes then maximum number of people who can get these 50 notes is 50 \/ 8 = 6.25.<\/p>\r\n\r\n\r\n\r\n

Maximum number of people who give 500 as their preference are 6.<\/p>\r\n\r\n\r\n\r\n

Question 3<\/p>\r\n\r\n\r\n\r\n

What is the maximum number of customers that the ATM can serve with a stock of fifty 500rupee notes and a sufficient number of notes of other denominations, if all the customers are to be served with at most 20 notes per withdrawal?<\/p>\r\n\r\n\r\n\r\n

    \r\n
  1. 10<\/li>\r\n
  2. 13<\/li>\r\n
  3. 12<\/li>\r\n
  4. 16<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

    Answer: we only have a limit on 500 rupee notes so we will be using a maximum of 200 rupee notes.<\/p>\r\n\r\n\r\n\r\n

    By using 200 rupee notes only, we need 25 notes to make 5000, 20 notes to make 4000 and 15 notes to make 3000.<\/p>\r\n\r\n\r\n\r\n

    We can make 5,000 rupees with 15 200 rupee notes and 4 500 rupee notes, that is, with a total of 19 notes.<\/p>\r\n\r\n\r\n\r\n

    Now we are using 4 500 rupee notes in every transaction and we have 50 in total so the maximum number of customers that can be served like this are 50 \/ 4 = 12.5 that is 12 customers.<\/p>\r\n\r\n\r\n\r\n

    Question 4<\/p>\r\n\r\n\r\n\r\n

    What is the number of 500 rupee notes required to serve 50 customers with 500 rupee notes as their preferences and another 50 customers with 100 rupee notes as their preferences, if the total number of notes to be dispensed is the smallest possible?<\/p>\r\n\r\n\r\n\r\n

      \r\n
    1. 800<\/li>\r\n
    2. 750<\/li>\r\n
    3. 900<\/li>\r\n
    4. 1400<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

      Answer: We have to minimize the number of notes required so we maximize the usage of 500 rupee notes.<\/p>\r\n\r\n\r\n\r\n

      For customers having 500 rupee note as their preference, we can give them 10 500 rupee notes each making total requirement of 500 rupee notes here as 50 * 10 = 500.<\/p>\r\n\r\n\r\n\r\n

      For customers having 100 rupee note as their preference, they can be given 10 100 rupee notes and 8 500 rupee notes for our purpose. The requirement of 500 rupee notes, in this case becomes 50 * 8 = 400.<\/p>\r\n\r\n\r\n\r\n

      Total requirement = 500 + 400 = 900.<\/p>\r\n\r\n\r\n\r\n

      \"CAT<\/a><\/p>\r\n

      Set 3<\/p>\r\n\r\n\r\n\r\n

      The multi-layered pie-chart below shows the sales of LED television sets for a big retail electronics outlet during 2016 and 2017. The outer layer shows the monthly sales during this period, with each label showing the month followed by sales figure of that month. For some months, the sales figures are not given in the chart. The middle-layer shows quarter- wise aggregate sales figures (in some cases, aggregate quarter-wise sales numbers are not given next to the quarter). The innermost layer shows annual sales. It is known that the sales figures during the three months of the second quarter (April, May, June) of2016 form an arithmetic progression, as do the three monthly sales figures in the fourth quarter (October, November, December) of that year.<\/p>\r\n\r\n\r\n\r\n

      \r\n
      \"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

      Solution<\/p>\r\n\r\n\r\n\r\n

      \r\n
      \"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

      Explanation:<\/strong><\/p>\r\n\r\n\r\n\r\n

      In this, we have been given sales of each months (except some), sum of 3 months of each quarter (except some).<\/p>\r\n\r\n\r\n\r\n

      Thing to do here is that we have to do the sum of the months of each quarter and find the missing values.<\/p>\r\n\r\n\r\n\r\n

      In quarter 1 of 2016, sum of the quarter will be 80 + 60 + 100 = 240<\/p>\r\n\r\n\r\n\r\n

      In quarter 2 of 2016, sum of the quarter is 180. In the set, it is also given that sale of April, May and June is in AP. So the revenue of April, May and June will be 40, 60 and 80 respectively.<\/p>\r\n\r\n\r\n\r\n

      In quarter 3 of 2016, sale of July, August and September is 75, 120 and 55 respectively. Then the sum of the quarter will be 250<\/p>\r\n\r\n\r\n\r\n

      Similarly, we keep on finding the values of each month and quarter.<\/p>\r\n\r\n\r\n\r\n

      The final table for this will be<\/p>\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n\r\n
      Year<\/td>\r\nMonth<\/td>\r\nMonthly Sale<\/td>\r\nQuarterly Sale<\/td>\r\nYearly Sale<\/td>\r\n<\/tr>\r\n
      2016<\/td>\r\nJanuary<\/td>\r\n80<\/td>\r\n240<\/td>\r\n1030<\/td>\r\n<\/tr>\r\n
      February<\/td>\r\n60<\/td>\r\n<\/tr>\r\n
      March<\/td>\r\n100<\/td>\r\n<\/tr>\r\n
      April<\/td>\r\n40<\/td>\r\n180<\/td>\r\n<\/tr>\r\n
      May<\/td>\r\n60<\/td>\r\n<\/tr>\r\n
      June<\/td>\r\n80<\/td>\r\n<\/tr>\r\n
      July<\/td>\r\n75<\/td>\r\n250<\/td>\r\n<\/tr>\r\n
      August<\/td>\r\n120<\/td>\r\n<\/tr>\r\n
      September<\/td>\r\n55<\/td>\r\n<\/tr>\r\n
      October<\/td>\r\n100<\/td>\r\n360<\/td>\r\n<\/tr>\r\n
      November<\/td>\r\n120<\/td>\r\n<\/tr>\r\n
      December<\/td>\r\n140<\/td>\r\n<\/tr>\r\n
      2017<\/td>\r\nJanuary<\/td>\r\n120<\/td>\r\n380<\/td>\r\n1300<\/td>\r\n<\/tr>\r\n
      February<\/td>\r\n100<\/td>\r\n<\/tr>\r\n
      March<\/td>\r\n160<\/td>\r\n<\/tr>\r\n
      April<\/td>\r\n60<\/td>\r\n200<\/td>\r\n<\/tr>\r\n
      May<\/td>\r\n75<\/td>\r\n<\/tr>\r\n
      June<\/td>\r\n65<\/td>\r\n<\/tr>\r\n
      July<\/td>\r\n60<\/td>\r\n220<\/td>\r\n<\/tr>\r\n
      August<\/td>\r\n90<\/td>\r\n<\/tr>\r\n
      September<\/td>\r\n70<\/td>\r\n<\/tr>\r\n
      October<\/td>\r\n150<\/td>\r\n500<\/td>\r\n<\/tr>\r\n
      November<\/td>\r\n170<\/td>\r\n<\/tr>\r\n
      December<\/td>\r\n180<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n\r\n\r\n\r\n

      Question 1<\/p>\r\n\r\n\r\n\r\n

      What is the percentage increase in sales in December 2017 as compared to the sales inDecember 2016?<\/p>\r\n\r\n\r\n\r\n

        \r\n
      1. 22.22<\/li>\r\n
      2. 28.57<\/li>\r\n
      3. 50.00<\/li>\r\n
      4. 38.46<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

        Answer: Sales in December 2017 increased from December 2016 that is became 180 from 140.<\/p>\r\n\r\n\r\n\r\n

        This increase is 28.57%.<\/p>\r\n\r\n\r\n\r\n

        Question 2<\/p>\r\n\r\n\r\n\r\n

        In which quarter of 2017 was the percentage increase in sales from the same quarter of 2016 the highest?<\/p>\r\n\r\n\r\n\r\n

          \r\n
        1. Q3<\/li>\r\n
        2. Q1<\/li>\r\n
        3. Q4<\/li>\r\n
        4. Q2<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

          Answer: By the table, percentage increase in quarterly sales was 58.33% for the first quarter which was highest. The answer is option 2, quarter 1.<\/p>\r\n\r\n\r\n\r\n

          Question 3<\/p>\r\n\r\n\r\n\r\n

          During which quarter was the percentage decrease in sales from the previous quarter\u2019s sales the highest?<\/p>\r\n\r\n\r\n\r\n

          1. Q2 of 2016<\/p>\r\n\r\n\r\n\r\n

          2. Q2 of 2017<\/p>\r\n\r\n\r\n\r\n

          3. Q4 of 2017<\/p>\r\n\r\n\r\n\r\n

          4. Q1 of 2017<\/p>\r\n\r\n\r\n\r\n

          Answer: Sale decrease percent in Quarter 2 in 2017 was highest in Quarter 2 of 2017 which is47.37%.<\/p>\r\n

          \"CAT<\/a><\/p>\r\n\r\n\r\n\r\n\r\n\r\n

          Question 4<\/p>\r\n\r\n\r\n\r\n

          During which month was the percentage increase in sales from the previous month\u2019s sales the highest?<\/p>\r\n\r\n\r\n\r\n

          1. October of 2016<\/p>\r\n\r\n\r\n\r\n

          2. October of 2017<\/p>\r\n\r\n\r\n\r\n

          3. March of 2016<\/p>\r\n\r\n\r\n\r\n

          4. March of 2017<\/p>\r\n\r\n\r\n\r\n

          Answer:Percentage increase in sales compared to previous month is Highest in October2017 that is 114.29%.<\/p>\r\n\r\n\r\n\r\n

           <\/p>\r\n

          Set 4<\/p>\r\n\r\n\r\n\r\n

          1600 satellites were sent up by a country for several purposes. The purposes are classified as broadcasting (B),communication (C), surveillance (S), and others (O). A satellite can serve multiple purposes; however a satellite serving either B, or C, or S does not serve O. The following facts are known about the satellites:
          1. The numbers of satellites serving B, C,and S (though may be not exclusively) are in the ratio 2:1:1.
          2. The number of satellites serving all three of B, C, and S is 100.
          3. The number of satellites exclusively serving C is the same as the number of satellites exclusively serving S. This number is 30% of the number of satellites exclusively serving B.
          4. The number of satellites serving O is the same as the number of satellites serving both C and S but not B.<\/p>\r\n\r\n\r\n\r\n

          Solution<\/p>\r\n\r\n\r\n\r\n

          \r\n
          \"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

          Explanation:<\/strong><\/p>\r\n\r\n\r\n\r\n

          In this, we have a satellite that can serve 4 purposes, one or more than one at a time. So this set looks like a 4 set venn diagram question but as we read, no satellite serving B, C or S, can serve purpose O. So we can make a 3 set venn with external part representing O.<\/p>\r\n\r\n\r\n\r\n

          By point 3, we have satellites serving exclusively C and satellites serving exclusively S are equal and each of these numbers is 30% of exclusively B.<\/p>\r\n\r\n\r\n\r\n

          So we take Exclusively B as 10a and exclusively C and exclusively S as 3a each which is 30% of 10a.<\/p>\r\n\r\n\r\n\r\n

          By point 2, we have all three as 100.<\/p>\r\n\r\n\r\n\r\n

          By point 4, we have satellites serving Oare equal to satellite serving both C and S but not B. So we take both of thesevalues as \u2018b\u2019.<\/p>\r\n\r\n\r\n\r\n

          \r\n
          \"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

          By point 1, satellites serving B are twice the number satellites serving C as well as satellites serving S. And satellites serving C and S are equal.<\/p>\r\n\r\n\r\n\r\n

          As total of C and S are equal and the three values that we know of them are also equal (3a, b, 100) then the 4th<\/sup> value that is satellites serving B and C but not S and satellites serving B and S but not C must also be equal. We write them as \u2018c\u2019.<\/p>\r\n\r\n\r\n\r\n

          Now, we also know that total number of satellites serving B is twice than total number of satellites serving C (or S).<\/p>\r\n\r\n\r\n\r\n

          If we equate them, we get<\/p>\r\n\r\n\r\n\r\n

          10a + 2c + 100 = 6a + 2b + 2c + 200<\/p>\r\n\r\n\r\n\r\n

          Solving, we get<\/p>\r\n\r\n\r\n\r\n

          4a \u2013 100 = 2b or<\/p>\r\n\r\n\r\n\r\n

          2a \u2013 50 = b.<\/p>\r\n\r\n\r\n\r\n

          Now we put the value of b in the venn diagram we have.<\/p>\r\n\r\n\r\n\r\n

          \r\n
          \"\"<\/figure>\r\n<\/div>\r\n\r\n\r\n\r\n

          We have all the values in terms of a and c so we can do here is<\/p>\r\n\r\n\r\n\r\n

          Sum of all of them is equal to 1600 as total number of satellites in 1600. We get<\/p>\r\n\r\n\r\n\r\n

          10a + c + c + 100 + 3a + 2a \u2013 50 + 3a + 2a \u2013 50 = 1600<\/p>\r\n\r\n\r\n\r\n

          That is 20a + 2c = 1600 or,<\/p>\r\n\r\n\r\n\r\n

          10a + c = 800.<\/p>\r\n\r\n\r\n\r\n

          Using this and using b = 2a \u2013 50, we get the minimum and maximum values of a as 25 and 80 respectively.<\/p>\r\n\r\n\r\n\r\n

          Consequently, the minimum and maximum value of c becomes 0 and 550.<\/p>\r\n\r\n\r\n\r\n

          Question 1<\/p>\r\n\r\n\r\n\r\n

          What best can be said about the number of satellites serving C?<\/p>\r\n\r\n\r\n\r\n

          1. Must be between 450 and 725<\/p>\r\n\r\n\r\n\r\n

          2. Cannot be more than 800<\/p>\r\n\r\n\r\n\r\n

          3. Must be at least 100<\/p>\r\n\r\n\r\n\r\n

          4. Must be between 400 and 800<\/p>\r\n\r\n\r\n\r\n

          Answer: satellites serving C is equal to 3a + 2a \u2013 50 + 100 + c that is<\/p>\r\n\r\n\r\n\r\n

          5a + c + 50<\/p>\r\n\r\n\r\n\r\n

          Putting the minimum and maximum value of a into it, we get the minimum and maximum number of satellites serving C as 450 and 725.<\/p>\r\n\r\n\r\n\r\n

          The answer would be option 1.<\/p>\r\n\r\n\r\n\r\n

          Question 2<\/p>\r\n\r\n\r\n\r\n

          What is the minimum possible number of satellites serving B exclusively?<\/p>\r\n\r\n\r\n\r\n

            \r\n
          1. 100<\/li>\r\n
          2. 250<\/li>\r\n
          3. 500<\/li>\r\n
          4. 200<\/li>\r\n<\/ol>\r\n\r\n\r\n\r\n

            Answer: From the venn, number of satellites serving exclusively B is 10a.<\/p>\r\n\r\n\r\n\r\n

            As we know that minimum possible value of a is 25, the answer would be 10a = 10 * 25 = 250.<\/p>\r\n\r\n\r\n\r\n

            Question 3<\/p>\r\n\r\n\r\n\r\n

            If at least 100 of the 1600 satellites we reserving O, what can be said about the number of satellites serving S?<\/p>\r\n\r\n\r\n\r\n

            1. At least 475<\/p>\r\n\r\n\r\n\r\n

            2. No conclusion is possible based on the given information<\/p>\r\n\r\n\r\n\r\n

            3. At most 475<\/p>\r\n\r\n\r\n\r\n

            4. Exactly 475<\/p>\r\n\r\n\r\n\r\n

            Answer: Satellites serving O are 2a \u2013 50, and as given in the question, this is at least 100.<\/p>\r\n\r\n\r\n\r\n

            So, 2a \u2013 50 >= 100 that means a >= 75 i.e. minimum value \u2018a\u2019 is revised as 75. As we know 10a + c = 800, Value of c at this point will be 50.<\/p>\r\n\r\n\r\n\r\n

            Number of satellites serving S are 3a + 2a + c \u2013 50 + 100, where minimum value of \u2018a\u2019 is 75 and maximum is 80.<\/p>\r\n\r\n\r\n\r\n

            Minimum and maximum number of satellites serving S would be 450 and 475.<\/p>\r\n\r\n\r\n\r\n

            The answer to the question is option 3, at most 475.<\/p>\r\n\r\n\r\n\r\n

            Question 4<\/p>\r\n\r\n\r\n\r\n

            If the number of satellites serving at least two among B, C, and S is 1200, which of the following MUST be FALSE?<\/p>\r\n\r\n\r\n\r\n

            1. All 1600 satellites serve B or C or S<\/p>\r\n\r\n\r\n\r\n

            2. The number of satellites serving B exclusively is exactly 250<\/p>\r\n\r\n\r\n\r\n

            3. The number of satellites serving C cannot be uniquely determined<\/p>\r\n\r\n\r\n\r\n

            4. The number of satellites serving B is more than 1000<\/p>\r\n\r\n\r\n\r\n

            Answer: Satellites serving at least two among B, C and S are 1200. From the venn, this number is 2a \u2013 50 + 100 + c + c that is 2a + 2c + 50.<\/p>\r\n\r\n\r\n\r\n

            Equating these we get, 2a + 2c + 50 = 1200<\/p>\r\n\r\n\r\n\r\n