Download LR Puzzles Questions for CAT<\/a><\/p>\n Instructions<\/b><\/p>\n Ten objects o1, o2, \u2026, o10 were distributed among Amar, Barat, Charles, Disha, and Elise. Each item went to exactly one person. Each person got exactly two of the items, and this pair of objects is called her\/his bundle.<\/p>\n The following table shows how each person values each object.<\/p>\n The value of any bundle by a person is the sum of that person\u2019s values of the objects in that bundle. A person X envies another person Y if X values Y\u2019s bundle more than X\u2019s own bundle.<\/p>\n For example, hypothetically suppose Amar\u2019s bundle consists of o1 and o2, and Barat\u2019s bundle consists of o3 and o4. Then Amar values his own bundle at 4 + 9 = 13 and Barat\u2019s bundle at 9 + 3 = 12. Hence Amar does not envy Barat. On the other hand, Barat values his own bundle at 7 + 5 = 12 and Amar\u2019s bundle at 5 + 9 = 14. Hence Barat envies Amar.<\/p>\n The following facts are known about the actual distribution of the objects among the five people. Question 1:\u00a0<\/b>What BEST can be said about the distribution of object o1?<\/p>\n a)\u00a0o1 was given to Disha<\/p>\n b)\u00a0o1 was given to Charles<\/p>\n c)\u00a0o1 was given to Charles, Disha, or Elise<\/p>\n d)\u00a0o1 was given to Charles or Disha<\/p>\n 1)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have the following table :<\/p>\n o10 is given to Elise and o9 is given to Bharat . Case 2 Elise receives o5 or o7 . Amar – o2,o6<\/p>\n Bharat -o9,o7<\/p>\n Charu -o3,o8 So o1 is given to Disha<\/p>\n Question 2:\u00a0<\/b>Object o5 was given to<\/p>\n a)\u00a0Disha<\/p>\n b)\u00a0Elise<\/p>\n c)\u00a0Amar<\/p>\n d)\u00a0Charles<\/p>\n 2)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have the following table :<\/p>\n o10 is given to Elise and o9 is given to Bharat . Case 2 Elise receives o5 or o7 . Amar – o2,o6<\/p>\n Bharat -o9,o7<\/p>\n Charu -o3,o8 o5 is given to Elise<\/p>\n Question 3:\u00a0<\/b>Object o4 was given to<\/p>\n a)\u00a0Elise<\/p>\n b)\u00a0Barat<\/p>\n c)\u00a0Charles<\/p>\n d)\u00a0Disha<\/p>\n 3)\u00a0Answer\u00a0(D)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have the following table :<\/p>\n o10 is given to Elise and o9 is given to Bharat . Case 2 Elise receives o5 or o7 . Amar – o2,o6<\/p>\n Bharat -o9,o7<\/p>\n Charu -o3,o8 o4 is given to Disha<\/p>\n Question 4:\u00a0<\/b>What is Amar\u2019s value for his own bundle?<\/p>\n 4)\u00a0Answer:\u00a012<\/b><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have the following table :<\/p>\n o10 is given to Elise and o9 is given to Bharat . Case 2 Elise receives o5 or o7 . Amar – o2,o6<\/p>\n Bharat -o9,o7<\/p>\n Charu -o3,o8 Amar’s own valuation = 9+3 =12<\/p>\n Question 5:\u00a0<\/b>Who among the following envies someone else?<\/p>\n a)\u00a0Barat<\/p>\n b)\u00a0Charles<\/p>\n c)\u00a0Amar<\/p>\n d)\u00a0Elise<\/p>\n 5)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have the following table :<\/p>\n o10 is given to Elise and o9 is given to Bharat . Case 2 Elise receives o5 or o7 . Amar – o2,o6<\/p>\n Bharat -o9,o7<\/p>\n Charu -o3,o8 Checkout: CAT Free Practice Questions and Videos<\/strong><\/a><\/em><\/p>\n Question 6:\u00a0<\/b>What BEST can be said about object o8?<\/p>\n a)\u00a0o8 was given to Amar, Charles, or Disha<\/p>\n b)\u00a0o8 was given to Disha<\/p>\n c)\u00a0o8 was given to Charles<\/p>\n d)\u00a0o8 was given to Charles or Disha<\/p>\n 6)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n We have the following table :<\/p>\n o10 is given to Elise and o9 is given to Bharat . Case 2 Elise receives o5 or o7 . Amar – o2,o6<\/p>\n Bharat -o9,o7<\/p>\n Charu -o3,o8 So o8 is given to Charu.<\/p>\n Instructions<\/b><\/p>\n Sixteen patients in a hospital must undergo a blood test for a disease. It is known that exactly\u00a0one of them has the disease. The hospital has only eight testing kits and has decided to pool\u00a0blood samples of patients into eight vials for the tests. The patients are numbered 1 through\u00a016, and the vials are labelled A, B, C, D, E, F, G, and H. The following table shows the vials\u00a0into which each patient\u2019s blood sample is distributed.<\/p>\n If a patient has the disease, then each vial containing his\/her blood sample will test positive. If\u00a0a vial tests positive, one of the patients whose blood samples were mixed in the vial has the\u00a0disease. If a vial tests negative, then none of the patients whose blood samples were mixed in\u00a0the vial has the disease.<\/p>\n Question 7:\u00a0<\/b>Suppose one of the lab assistants accidentally mixed two patients’ blood\u00a0samples before they were distributed to the vials. Which of the following\u00a0correctly represents the set of all possible numbers of positive test\u00a0results out of the eight vials?<\/p>\n a)\u00a0{5,6,7,8}<\/p>\n b)\u00a0{4,5,6,7}<\/p>\n c)\u00a0{4,5,6,7,8}<\/p>\n d)\u00a0{4,5}<\/p>\n 7)\u00a0Answer\u00a0(C)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n Let one of the patients, patient 1 or patient 16 has the disease and his blood is mixed with other them all 8 vials will tests positive. \u21d2<\/p>\n 8 has to be one of the answers.<\/p>\n If patient 2 and patients 16\u2019s blood is mixed of one of them has the disease then 7 of the 8 vials will test positive. So 7 has to be there If 1 has the disease and 1, 7 are mixed then 6 out the 8 vials tests positive.<\/p>\n IF 1 has the disease and 1,9 are mixed then 5 of the 8 vials tests positive,<\/p>\n Now, let us assume that patient 1 has the disease if his blood is not mixed, Hence 3 is the correct answer.<\/p>\n Question 8:\u00a0<\/b>Which of the following combinations of test results is NOT possible?<\/p>\n a)\u00a0Vials A and E positive, vials C and D negative<\/p>\n b)\u00a0Vial B positive, vials C, F and H negative<\/p>\n c)\u00a0Vials A and G positive, vials D and E negative<\/p>\n d)\u00a0Vials B and D positive, vials F and H negative<\/p>\n 8)\u00a0Answer\u00a0(A)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The patients in<\/p>\n Vial A: 9, 10, 11, 12, 13, 14, 15, 16<\/p>\n Vial B: 1, 2, 3, 4, 5, 6, 7, 8.<\/p>\n Vial C: 5,6,7,8,13,14,15,16<\/p>\n Vial D:1,2,3,4,9,10,11,12<\/p>\n Vial E:3,4,7,8,11,12,15,16<\/p>\n Vial F:1,2,5,6,9,10,13,14<\/p>\n Vial G:2,4,6,8,10,12,14,16<\/p>\n Vial H:1,3,5,7,9,11,13,15<\/p>\n If vials C and D negative then no patient could test negative. Hence A is correct answer.<\/p>\n Question 9:\u00a0<\/b>Suppose vial A tests positive and vials D and G test negative. Which of the\u00a0following vials should we test next to identify the patient with the disease?<\/p>\n a)\u00a0Vial B<\/p>\n b)\u00a0Vial E<\/p>\n c)\u00a0Vial C<\/p>\n d)\u00a0Vial H<\/p>\n 9)\u00a0Answer\u00a0(B)<\/strong><\/p>\n View Video Solution<\/a><\/p>\n Solution:<\/b><\/p>\n The patients in<\/p>\n Vial A: 9, 10, 11, 12, 13, 14, 15, 16<\/p>\n Vial B: 1, 2, 3, 4, 5, 6, 7, 8.<\/p>\n Vial C: 5,6,7,8,13,14,15,16<\/p>\n Vial D:1,2,3,4,9,10,11,12<\/p>\n Vial E:3,4,7,8,11,12,15,16<\/p>\n Vial F:1,2,5,6,9,10,13,14<\/p>\n Vial G:2,4,6,8,10,12,14,16<\/p>\n Vial H:1,3,5,7,9,11,13,15<\/p>\n Suppose vial A tests positive and vials D and G test negative then the patient who tested positive must be one of patient 13 or 15.<\/p>\n Patient 13 or 15 are not present in vial B. So, A is not the answer.<\/p>\n Both patients present in vial C. Even if tested positive or negative we can’t know who has got the disease.\u00a0So, C is not the answer.<\/p>\n Both patients present in vial H. Even if tested positive or negative we can’t know who has got the disease.\u00a0So, H is not the answer.<\/p>\n only patient 15 is present in vial E, if tested positive then patient 15 has the disease else patient 13 as disease.<\/p>\n Hence Option 2 is correct.<\/p>\n Question 10:\u00a0<\/b>Suppose vial C tests positive and vials A, E and H test negative. Which patient\u00a0has the disease?<\/p>\n a)\u00a0Patient 14<\/p>\n b)\u00a0Patient 8<\/p>\n c)\u00a0Patient 6<\/p>\n d)\u00a0Patient 2<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_yQ9zfzQ.png\")
<\/p>\n
\n1. If someone\u2019s value for an object is 10, then she\/he received that object.
\n2. Objects o1, o2, and o3 were given to three different people.
\n3. Objects o1 and o8 were given to different people.
\n4. Three people value their own bundles at 16. No one values her\/his own bundle at a number higher than 16.
\n5. Disha values her own bundle at an odd number. All others value their own bundles at an even number.
\n6. Some people who value their own bundles less than 16 envy some other people who value their own bundle at 16. No one else envies others.<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_SOFJEXK.png\")
<\/p>\n
\nNow as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7.
\nCase 1 :
\no1 is given to Elise
\nNow the total valuation of Elise = 12
\nValuation of Disha is an odd number
\nSo we can say Amar , Bharat and Charles values their bundles at 16 .
\nSo for Bharat the valuation to be 16, o7 will be given to him
\nso we get
\nBharat – o9 and o7 and Elise -o10 and o1
\nFor charles to have valuation 16
\nthe only way = 8+8
\nso we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together )
\nNow for Amart to have a valuation of 16
\nthe only way possible = 9+7
\nNow so we can say
\nAmar will receive either o2 or o3 and o5 .
\nNow we are left with 04 and o6
\nSo if Disha receives o4 and o6
\nThe valuation of Disha will be 5+3 =8 which is not an odd number
\nso this case is discarded.<\/p>\n
\nNow Valuation of Elise = 16 .
\nAnd Elise receives o10 and o5\/o7.
\nBharat
\nreceived o9 and we know the evaluation of Bharat is an even number and
\nthe minimum even number possible for valuation of Bharat is 16 and no
\none can have evaluation more than 16 so Bharat received o7 .
\nSo Elise received o5 .
\nSo we have
\nBharat – o9 ,o7
\nElise -o10,o5.
\nNow as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha .
\nNow As per Amar
\nhe values Bharat at 17 so he envy him
\nSo Amar will value his bundle less than 16
\nSo the only possibility for Amar to value his bundle less than 16 = 12 =9+3.
\nNow we can say Charu will have 16 as his own valuation so he will get 8+8 .
\nNow
\no8 will be given to Charu, and he cannot have o1 , also he cannot have
\no2 because if he has o2 he will value Bharat\u2019s bundle as 17 and will
\nenvy him which is not possible so Charu will have o3,o8
\nNow Amar will have o2 and Disha will have o1.
\nNow
\nAmar will not have o4 because in that case Charles will envy Amar and
\nis not possible so we can say Amar will have o6 and Disha will have o4.
\nSo we have the following :<\/p>\n
\nDisha o1,o4
\nElise -o10,o5<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_PPYhuuh.png\")
<\/p>\n
\nNow as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7.
\nCase 1 :
\no1 is given to Elise
\nNow the total valuation of Elise = 12
\nValuation of Disha is an odd number
\nSo we can say Amar , Bharat and Charles values their bundles at 16 .
\nSo for Bharat the valuation to be 16, o7 will be given to him
\nso we get
\nBharat – o9 and o7 and Elise -o10 and o1
\nFor charles to have valuation 16
\nthe only way = 8+8
\nso we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together )
\nNow for Amart to have a valuation of 16
\nthe only way possible = 9+7
\nNow so we can say
\nAmar will receive either o2 or o3 and o5 .
\nNow we are left with 04 and o6
\nSo if Disha receives o4 and o6
\nThe valuation of Disha will be 5+3 =8 which is not an odd number
\nso this case is discarded.<\/p>\n
\nNow Valuation of Elise = 16 .
\nAnd Elise receives o10 and o5\/o7.
\nBharat
\nreceived o9 and we know the evaluation of Bharat is an even number and
\nthe minimum even number possible for valuation of Bharat is 16 and no
\none can have evaluation more than 16 so Bharat received o7 .
\nSo Elise received o5 .
\nSo we have
\nBharat – o9 ,o7
\nElise -o10,o5.
\nNow as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha .
\nNow As per Amar
\nhe values Bharat at 17 so he envy him
\nSo Amar will value his bundle less than 16
\nSo the only possibility for Amar to value his bundle less than 16 = 12 =9+3.
\nNow we can say Charu will have 16 as his own valuation so he will get 8+8 .
\nNow
\no8 will be given to Charu, and he cannot have o1 , also he cannot have
\no2 because if he has o2 he will value Bharat\u2019s bundle as 17 and will
\nenvy him which is not possible so Charu will have o3,o8
\nNow Amar will have o2 and Disha will have o1.
\nNow
\nAmar will not have o4 because in that case Charles will envy Amar and
\nis not possible so we can say Amar will have o6 and Disha will have o4.
\nSo we have the following :<\/p>\n
\nDisha o1,o4
\nElise -o10,o5<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_LcMGmbd.png\")
<\/p>\n
\nNow as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7.
\nCase 1 :
\no1 is given to Elise
\nNow the total valuation of Elise = 12
\nValuation of Disha is an odd number
\nSo we can say Amar , Bharat and Charles values their bundles at 16 .
\nSo for Bharat the valuation to be 16, o7 will be given to him
\nso we get
\nBharat – o9 and o7 and Elise -o10 and o1
\nFor charles to have valuation 16
\nthe only way = 8+8
\nso we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together )
\nNow for Amart to have a valuation of 16
\nthe only way possible = 9+7
\nNow so we can say
\nAmar will receive either o2 or o3 and o5 .
\nNow we are left with 04 and o6
\nSo if Disha receives o4 and o6
\nThe valuation of Disha will be 5+3 =8 which is not an odd number
\nso this case is discarded.<\/p>\n
\nNow Valuation of Elise = 16 .
\nAnd Elise receives o10 and o5\/o7.
\nBharat
\nreceived o9 and we know the evaluation of Bharat is an even number and
\nthe minimum even number possible for valuation of Bharat is 16 and no
\none can have evaluation more than 16 so Bharat received o7 .
\nSo Elise received o5 .
\nSo we have
\nBharat – o9 ,o7
\nElise -o10,o5.
\nNow as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha .
\nNow As per Amar
\nhe values Bharat at 17 so he envy him
\nSo Amar will value his bundle less than 16
\nSo the only possibility for Amar to value his bundle less than 16 = 12 =9+3.
\nNow we can say Charu will have 16 as his own valuation so he will get 8+8 .
\nNow
\no8 will be given to Charu, and he cannot have o1 , also he cannot have
\no2 because if he has o2 he will value Bharat\u2019s bundle as 17 and will
\nenvy him which is not possible so Charu will have o3,o8
\nNow Amar will have o2 and Disha will have o1.
\nNow
\nAmar will not have o4 because in that case Charles will envy Amar and
\nis not possible so we can say Amar will have o6 and Disha will have o4.
\nSo we have the following :<\/p>\n
\nDisha o1,o4
\nElise -o10,o5<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_tpoEbtA.png\")
<\/p>\n
\nNow as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7.
\nCase 1 :
\no1 is given to Elise
\nNow the total valuation of Elise = 12
\nValuation of Disha is an odd number
\nSo we can say Amar , Bharat and Charles values their bundles at 16 .
\nSo for Bharat the valuation to be 16, o7 will be given to him
\nso we get
\nBharat – o9 and o7 and Elise -o10 and o1
\nFor charles to have valuation 16
\nthe only way = 8+8
\nso we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together )
\nNow for Amart to have a valuation of 16
\nthe only way possible = 9+7
\nNow so we can say
\nAmar will receive either o2 or o3 and o5 .
\nNow we are left with 04 and o6
\nSo if Disha receives o4 and o6
\nThe valuation of Disha will be 5+3 =8 which is not an odd number
\nso this case is discarded.<\/p>\n
\nNow Valuation of Elise = 16 .
\nAnd Elise receives o10 and o5\/o7.
\nBharat
\nreceived o9 and we know the evaluation of Bharat is an even number and
\nthe minimum even number possible for valuation of Bharat is 16 and no
\none can have evaluation more than 16 so Bharat received o7 .
\nSo Elise received o5 .
\nSo we have
\nBharat – o9 ,o7
\nElise -o10,o5.
\nNow as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha .
\nNow As per Amar
\nhe values Bharat at 17 so he envy him
\nSo Amar will value his bundle less than 16
\nSo the only possibility for Amar to value his bundle less than 16 = 12 =9+3.
\nNow we can say Charu will have 16 as his own valuation so he will get 8+8 .
\nNow
\no8 will be given to Charu, and he cannot have o1 , also he cannot have
\no2 because if he has o2 he will value Bharat\u2019s bundle as 17 and will
\nenvy him which is not possible so Charu will have o3,o8
\nNow Amar will have o2 and Disha will have o1.
\nNow
\nAmar will not have o4 because in that case Charles will envy Amar and
\nis not possible so we can say Amar will have o6 and Disha will have o4.
\nSo we have the following :<\/p>\n
\nDisha o1,o4
\nElise -o10,o5<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_5g3gbkg.png\")
<\/p>\n
\nNow as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7.
\nCase 1 :
\no1 is given to Elise
\nNow the total valuation of Elise = 12
\nValuation of Disha is an odd number
\nSo we can say Amar , Bharat and Charles values their bundles at 16 .
\nSo for Bharat the valuation to be 16, o7 will be given to him
\nso we get
\nBharat – o9 and o7 and Elise -o10 and o1
\nFor charles to have valuation 16
\nthe only way = 8+8
\nso we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together )
\nNow for Amart to have a valuation of 16
\nthe only way possible = 9+7
\nNow so we can say
\nAmar will receive either o2 or o3 and o5 .
\nNow we are left with 04 and o6
\nSo if Disha receives o4 and o6
\nThe valuation of Disha will be 5+3 =8 which is not an odd number
\nso this case is discarded.<\/p>\n
\nNow Valuation of Elise = 16 .
\nAnd Elise receives o10 and o5\/o7.
\nBharat
\nreceived o9 and we know the evaluation of Bharat is an even number and
\nthe minimum even number possible for valuation of Bharat is 16 and no
\none can have evaluation more than 16 so Bharat received o7 .
\nSo Elise received o5 .
\nSo we have
\nBharat – o9 ,o7
\nElise -o10,o5.
\nNow as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha .
\nNow As per Amar
\nhe values Bharat at 17 so he envy him
\nSo Amar will value his bundle less than 16
\nSo the only possibility for Amar to value his bundle less than 16 = 12 =9+3.
\nNow we can say Charu will have 16 as his own valuation so he will get 8+8 .
\nNow
\no8 will be given to Charu, and he cannot have o1 , also he cannot have
\no2 because if he has o2 he will value Bharat\u2019s bundle as 17 and will
\nenvy him which is not possible so Charu will have o3,o8
\nNow Amar will have o2 and Disha will have o1.
\nNow
\nAmar will not have o4 because in that case Charles will envy Amar and
\nis not possible so we can say Amar will have o6 and Disha will have o4.
\nSo we have the following :<\/p>\n
\nDisha o1,o4
\nElise -o10,o5
\nSo Amar envies someone else<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/Screenshot_489_DJqpFlV.png\")
<\/p>\n
\nNow as Elise values his own bundle at an even number so the only two objects which can be given to Elise is o1 or o5 or o7.
\nCase 1 :
\no1 is given to Elise
\nNow the total valuation of Elise = 12
\nValuation of Disha is an odd number
\nSo we can say Amar , Bharat and Charles values their bundles at 16 .
\nSo for Bharat the valuation to be 16, o7 will be given to him
\nso we get
\nBharat – o9 and o7 and Elise -o10 and o1
\nFor charles to have valuation 16
\nthe only way = 8+8
\nso we can say o8 is given to charles along with either o2 or o3 .(o1 and o8 cannot be together )
\nNow for Amart to have a valuation of 16
\nthe only way possible = 9+7
\nNow so we can say
\nAmar will receive either o2 or o3 and o5 .
\nNow we are left with 04 and o6
\nSo if Disha receives o4 and o6
\nThe valuation of Disha will be 5+3 =8 which is not an odd number
\nso this case is discarded.<\/p>\n
\nNow Valuation of Elise = 16 .
\nAnd Elise receives o10 and o5\/o7.
\nBharat received o9 and we know the evaluation of Bharat is an even number and the minimum even number possible for valuation of Bharat is 16 and no one can have evaluation more than 16 so Bharat received o7 .
\nSo Elise received o5 .
\nSo we have
\nBharat – o9 ,o7
\nElise -o10,o5.
\nNow as we know o1 ,o2 and o3 are given to three different persons so they are Amar, Charles and Disha .
\nNow As per Amar
\nhe values Bharat at 17 so he envy him
\nSo Amar will value his bundle less than 16
\nSo the only possibility for Amar to value his bundle less than 16 = 12 =9+3.
\nNow we can say Charu will have 16 as his own valuation so he will get 8+8 .
\nNow o8 will be given to Charu, and he cannot have o1 , also he cannot have o2 because if he has o2 he will value Bharat\u2019s bundle as 17 and will envy him which is not possible so Charu will have o3,o8
\nNow Amar will have o2 and Disha will have o1.
\nNow Amar will not have o4 because in that case Charles will envy Amar and is not possible so we can say Amar will have o6 and Disha will have o4.
\nSo we have the following :<\/p>\n
\nDisha o1,o4
\nElise -o10,o5<\/p>\n![](\"https:\/\/cracku.in\/media\/uploads\/348394.png\")
<\/figure>\n
\nin the option.<\/p>\n
\nthen 4 vials will definitely show positive.<\/p>\n