Logical Reasoning capsules by Vikas Saini - Set 6

  • Set 1

    Five friends Pawan, Qureshi, Rajan, Sultan and Tango are Musician, Architect, Doctor, Engineer and Artist by profession and like White, Blue, Red, Yellow and Green colour but not necessarily in that order. Their hobbies are Net Surfing, Gardening, Reading, Painting and Dancing but not necessarily in the same order.

    The person whose hobby is dancing preferred lemonade to cola while others preferred cola to lemonade in beverages. The four friends who took cola were Pawan, the one who is an Engineer, the person whose favourite colour is Green and the one whose hobby is net surfing. Sultan did not take lemonade and his favourite colour is White. Qureshi’s favourite colour is Blue. He did not like lemonade. Tango’s hobby is not painting, reading or gardening. Sultan clicks a picture of his friend who is an Engineer. The person whose favourite colour is Red likes painting and the person who is artist likes gardening. Sultan is not a doctor. The person who is a doctor takes cola. The person who is an Engineer likes Blue colour. The musician’s favourite colour is not Yellow. Rajan’s favourite colour is Green.

    Q1. Who among the following is a Doctor?
    a) Rajan
    b) Pawan
    c) Sultan
    d) Can’t say
    e) None of these

    Q2. Qureshi’s hobby is
    a) Reading
    b) Painting
    c) Gardening
    d) Can’t say
    e) None of these

    Q3. The person who likes Blue colour is a/an
    a) Architect
    b) Musician
    c) Engineer
    d) Can’t say
    e) None of these

    Q4. Whose favourite colour is Yellow?
    a) Tango
    b) Rajan
    c) The one who is an artist
    d) Can’t say
    e) None of these

    Q5. Which of the following combinations is not correctly matched?
    a) Tango-Architect-Yellow-Dancing-Cola
    b) Rajan-Artist-Green-Gardening-Cola
    c) Qureshi-Engineer-Blue-Reading-Cola
    d) Pawan-Doctor-Red-Painting-Cola

    Solution :-

    Q1) B
    Q2) A
    Q3) C
    Q4) A
    Q5) A

    Set 2

    There are 200 students in a class. 20% students are there who don’t like any subject. 50% students like physics, 60% students like Chemistry and 70% students like Mathematics.

    1. How many maximum students are there who like all three subjects.
    2. How many minimum students are there who like all three subjects.

    Solution :-

    1. 20% studenst don’t like any subjects.
      I + II + III = 80..................(1)
      I+2II+3III= 50+60+70 = 180..........(2)
      Equ (2) – Equ (1)
      II + 2III = 100.
      To make III maximum, we need to put II = 0.
      2 III = 100.
      III(max) = 50.
      50% of 200 = 100.
      There are maximum 100 people can be there who like all three subjects.

    2. From equation 1
      II + III = 80 – I...........(3)
      From equation 2
      2II + 3III = 180 – I.......(4)
      Equ (4) – 2 x equ (3)
      III = I + 20.
      Put I = 0.
      III(min) = 20.
      20% of 200 =40.
      There are minimum 40 people who like all three subjects.

    Set 3

    In a shop, five articles – P,Q,R,S,T are to be sold. The cost price & the selling price of each of the 5 articles among 650,700,750,800 & 900. The cost price of each of the articles is different and also the selling price each of the articles is different. For any article, the selling price is not equal to its cost price.

    1. The cost price of article R is equal to the selling R as well as T the shopkeeper incurred a loss.
    2. The cost price of Q is more than that of S and the shopkeeper obtained a profit by selling Q.
    3. The profit made by selling any article is more than rs 50. The profit made on any two articles not the same. The loss incurred on any two articles is not the same.
    4. On only two articles,the shopkeeper made a profit. The profit/loss made on any article is not rs 150.

    Q1) What is the selling price of P ?
    Q2) What is the difference between the cost price and the selling price of article ?
    Q3) Which article’s selling price is 700 ?

    Solution :-

    From the given data

    1. CP of R = SP of T (both incurred a loss)
    2. CP of Q > CP of S, CP of Q < SP of Q.
    3. Profit = 50+, profit and loss for all articles are not same.
    4. On two articles shopkeeper made profit. No profit/loss is 150 rs.

    Let’s make a table as per given information above.


    Cost price of S is less than Q hence it must be also in profit.
    Now P will incurred loss.
    CP of S is 650 rs. Selling price of it shall be 750/850/900.
    Case 1.If CP of Q is 700 then SP of it should be 900 only.
    Case2. If CP of Q is 750 then SP of it 850 only.
    But only possible way of profit is +250 and 100.
    Only possible way of loss is 50,100,200.
    Hence S must get profit of 250 rs and Q must get profit of 100 rs.
    CPs left now 700,850,900 and respective SPs are 650,750,700.
    CP of R is 700 and SP of T is 700.
    Now this can be done by making a table.

    Profit/loss (+/-)-100100-50250-200


    Q1) 750
    Q2) 250
    Q3) T

    Set 4

    A team is to be selected from among ten persons – A,B,C,D,E,F,G,H,I and J subject to the following conditions

    (i) Exactly two among E,J,C and I must be selected.
    (ii) If F is selected then J cannot be selected.
    (iii) Exactly one among A & C must be selected.
    (iv) Unless a is selected, E cannot be selected.
    (v) If and only if G is selected, d must not be selected.
    (vi) If D is not selected, then H must be selected.

    The size of a team is defined as the number of teams.

    Q1) Who among the following cannot be a team of size 4 ?
    a) E
    b) H
    c) F
    d) None of these

    Q2) What should be the size of a team that includes both F & H ?
    a) 8
    b) 3
    c) 4
    d) 5

    Q3) In how many ways can the team of size 6 be selected if it includes E ?
    a) 5
    b) 6
    c) 7
    d) 8

    Q4) What is the largest possible size of the team
    a) 5
    b) 6
    c) 7
    d) 8

    Solution :-

    (i) E,J,I,C -> EJ/EI/EC/JI/JC/IC
    (ii) F,J -> F/J/C
    (iii) A,C -> A/C
    (iv) E -> A
    (v) G,D -> G/D
    (vi) D,H -> D/H/HD

    From (ii), among F and J only J can be selected or none of them can be selected from (iv), among A & E, either both can be selected or none of them can be selected.

    Q1) The following teams includes E or H or F but violate none of them given conditions

    1. E,A,I,G
    2. J,H,C,D
    3. F,I,C,D

    Hence option 4

    Q2) As both F & H are selected. Exactly two among E,I and C must be selected and exactly on among G and D must be selected.
    The team which includes both F & H, can not be the size of 3 or 4.
    The maximum possible size of the team is 7.
    It can not be 8.
    But the size of team could be 5, as follows F,H,I,C,G.
    Hence option 4.

    Q3) If E is selected, then A must also be selected. C can not be selected.

    In the following ways, the team can be selected

    1. A,E,J,G/D,H,B --------- 2 ways.
    2. A,E,I,F,G/D.H ----------2 ways.
    3. A,E,I,G/D,H,B----------2 ways.
    4. A,E,F,D,B ---------------1 way.

    There are total 7 ways.
    Henc e option 3.

    Q4) To maximize the size of the team. We should follow the steps given below
    (i) Among F & J, let us select F.
    (ii) Among A & C, let us select A.
    (iii) As A is selected, let us selected E also.
    (iv) Among G,D and H, let us select D and H or G & H.
    (v) Among E,I,J and C as E is already selected and J and C cannot be selected.I should be selected.
    (vi) Let us include b also in the team.The possible team is F,A,e,D,H,I,B
    (vii) The maximum possible size of team is 7.
    Hence option 3

  • Sir, didn't get the part 2 of set 2 'how many minimum numbers of students who like all three subjects."

  • @surya_prakash There are 200 students in a class. 20% students are there who don’t like any subject. 50% students like physics, 60% students like Chemistry and 70% students like Mathematics.
    I + II + III = 100 - 20 = 80......(1)
    I + 2II + 3III = 50+60+70 = 180.....(2)
    Now if we put I = 0 then
    II + III = 80....(3)
    2II + 3III = 180....(4)
    II should be maximum so that III cold be minimum.
    Now equation (4) - 2 x equation (3)
    III = 20.
    This 20 is percent actually.
    20% of 200 = 40.
    I hope it's clear now.

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