# Logical Reasoning capsules by Vikas Saini - Set 11

• Set 1

If A & B are playing a coin picking game. It is decided between them that who will pick the last coin will lose the game. Each can pick minimum 2 coins and maximum 6 coins.

Q1) If the game starts with the 52 coins and it’s turn to play of A, then how many coins should he pick to win ?
a) 1
b) 4
c) 2 or 3
d) 2

Q2) If the number of coins to be picked up by B is 5 in the term of win, then which of the following can be number of coins on table ?
a) 22
b) 42
c) 72
d) None of these

Q3) Which of the number of coins, before A’s turn to play, if it is known that B will win the game, no matter how many coins A picks ?
a) 72
b) 62
c) 89
d) None of these

Solution :-

Suppose no of coins on the table N =1,2...... so on.
If A has to start the game then no of coins
N = 1, then A will lose.
N = 2, then A will lose.
N=3, then A will win. A will pick only 2 coins.
N =4, then A will win, will pick only 2 coins.
N = 5, then A will win, will pick 3 coins.
N = 6 ,7, 8 then also A will win, will pick any number of coins as his intelligence.
Then for N = 9,10. A will lose again.
Then again N = 11,12,13,14,15,16. A will win again.
This makes a pattern.
For 8k+1 and 8k+2 A will lose.
Here 8 is nothing but sum of minimum coin pick and maximum coin picking.

Q1)There is 52 coins over there. Then to win this game A will have to pick the coins in a manner so that B gets coins as 8k+1 and 8k+2.
Nearest 8k+1 and 8k+2 of 52 is 49 and 50 respectively.
52 – 49 = 3.
52 – 50 = 2.
A will have to pick the coins either 2 or 3. (Option C)

Q2) B has picked number of coins 5. In order to wining of B, no of coins should remain 8k+1 and 8k+2 form.
Let’s check option by option.
22 – 5 = 17 (8k+1)
42 – 5 = 37 (8k + 5 form)
72 – 5 = 67 (8k + 3 form)
option A.

Q3) 72 is in 8k form.
62 is in 8k-2 form.
89 is in 8k+1 form.
Hence option C.

Set 2

The table gives the breakup of percentage of Vitamin A , B, C and D contained in four vitamin drinks - 1 , 2 , 3 and 4 . The cost of the drinks 1 , 2 , 3 and 4 are respectively 100, 250 , 200 and 150 per litre . Two or more of the drinks can be mixed in Any ratio to produce a drink containing the vitamins in the required ratio

VitaminABCDRate per litre
Drink 120301040100
Drink 210403020250
Drink 315253525200
Drink 440153510150

Q1) if a drink which contains at least 20% of each of vitamins A , B , C and D is to be prepared , then what would be least cost / litre incurred ??
a) 100
b) 120
c) 125
d) 140

Q2) if a drink which contains 25% of vitamin C is to be prepared at the least cost per litre then the ratio in which any two drinks can be mixed is ??
a) 1:2
b) 3:4
c) 2: 3
d) 4:1

Q3) if not more than 2 litres of each drink is available , then what can be the maximum percentage of Vitamin B in a drink obtained by mixing 2 or more drinks, if the cost / litre of resulting drink should not exceed 175
a) 22.5
b) 25
c) 27.5
d) none of these

Solution

Q1) At least 20% of each vitamins should be there at least rate. We will use then drink 1 and drink 4 due to least rate. Vitamin A is there 20 & 40 already more than 20%. Vitamin B is 30 and 15 there.
30................15
.........20..........
5...................10
1 : 2
Drink 1 : Drink 4.
Vitamin C is 10 and 35 there.
10...................35
..........20...........
15....................10
3 : 2
Drink 1 : Drink 4.
Vitamin D is 40 and 10 there.
40.....................10
.............20...........
10......................20
1 : 2.

For vitamin B and D ratio 1 : 2.
We need to go for vitamin C where ratio is 3 : 2.
3 x 100 + 2 x 150 / (3 + 2) = 120.
Option B.

Q2) Vitamin C should be at least 25% at least rate.
We need to take drink 1 and drink 4 due to least rate.
Drink 1..............Drink 4
10.........................35
.............25..............
10..........................15
2 : 3.
Ratio must be 2 : 3. Option C.

Q3)
Cost per litre < = 175
Percentage of vitamin B should be maximum.
Only drink 1 and drink 2 are best option.
30 x 2 + 40 x 2 / 2 x 2 = 35.
Rate = 100 x 2 + 250 x 2 / 4 = 175.
Option D.

Set 3

In a Business school of 120 students each student has to opt for either one or two or three areas of specializations out of Marketing, Finance and Strategy. The number of students taking exactly two out of three areas of specializations is 92. The number of students taking exactly one area of specialization is six times the number of students taking all specializations. Out of students taking exactly two areas of specializations, those taking Finance and strategy is two times those taking Marketing and Finance. Of those students who opted for only one area of specialization those taking strategy are two more than those taking Marketing and two less than those taking Finance. Also, students opting for strategy specialization is 22 more than those taking marketing specialization. Following table lists average salary for students opting exactly one, exactly two, exactly three and all students achieved at the placement rounds of the Business School. I. The number of students opting for Finance as area of specialization is
II. The number of students who opted for Finance and Strategy(both) as area of specialization is
III. Average Salary of students who opted for all three areas of specialization is
IV. If for students opting for only one area of specialization the ratio of average salaries of students opting for Marketing, Finance and Strategy are in the ration 1:2:3 then the average salary of students with Finance only as specialization is.

Solution As per given data
d + e + f = 92.
a + b + c = 6g.
a + b +c + g = 120 – 92 = 28.
6g + g = 28.
g = 4.
a + b + c = 24.
2f = e.
c = a + 2.
=> a = c -2.
c = b – 2.
=> b = c + 2.
Now a + b + c = 24.
c – 2 + c + 2 + c = 24.
c = 8.
a = 6.
b = 10.
a + f + g + d + 22 = d + g + e + c.
a + f + 22 = e + c.
6 + f + 22 = 2f + 8.
f = 20.
e = 40.
d = 32. I. The number of students opting for Finance as area of specialization is 74.
II. The number of students who opted for Finance and Strategy(both) as area of specialization is 44.
III. Average Salary of students who opted for all three areas of specialization is ?
10 x 24 + 15 x 92 + 4 p = 14.1 x 120
=> 4p = 72
=> P = 18 lac.
IV. If for students opting for only one area of specialization the ratio of average salaries of students opting for Marketing, Finance and Strategy are in the ration 1:2:3 then the average salary of students with Finance only as specialization is.
6 x k + 10 x 2k + 8 x 3k = 10 x 24.
50 k =240.
k = 4.8
2k = 9.6 lac.

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