Number of Questions - 30

Topic - Games & Tournaments

Answer key available? - Yes

Source - Curated Content

Number of Questions - 30

Topic - Games & Tournaments

Answer key available? - Yes

Source - Curated Content

(i) There were 49 coins and the person picking the last coin won the game

(ii) There were 36 coins and the person picking the last coin won the game

(iii) There were 42 coins and the person picking the last coin lost the game

(iv) There were 25 coins and the person picking the last coin lost the game

a Only (i) and (iv)

b Only (iv)

c Only (iii) and (i)

d Only (ii) and (iii)

[OA: C]

]]>a. 26, 23

b. 27, 24

c. 28, 25

d. 29, 26

[OA: 26, 23]

]]>a) 12

b) 48

c) 36

d) 42

[OA: 42]

]]>[OA: Anthony]

]]>[OA: 11]

]]>- Who won the first round? ‘
- Who won the second round?
- How many points did B have at the start of the game?

Assume that the player who picks the last coin win the game.

[OA: 2]

]]>A) 280 runs with 7 wickets lost

B) 290 runs with 6 wickets lost

C) 271 runs with 8 wickets lost

D) 259 runs with 9 wickets lost

[OA: C]

]]>For each wicket lost, this score decreases by 11 (6 runs not scored plus 5 runs lost due to the wicket). For example, if 3 wickets are lost, the maximum possible score is 327 [360 − (3 × 11)], which could be achieved by hitting sixes on all the remaining 57 balls.

For 7, 6, 9 and 8 wickets lost (as given in the options), the maximum possible scores are 283, 294, 261 and 272 respectively.

The scores in options 1 and 2 can be achieved if a player hits 3 or 2 runs instead of 6 on one ball. However, option 3 is not possible, as for that, a player would have to hit 5 runs instead of 6 on one ball, which is not possible in this game.

Hence, option C.

[Credits : Lokesh Agarwal] ]]>

(a) 256

(b) 128

(c) 255

(d) 127

[OA: 256]

]]>a) 1

b) 2

c) 3

d) 4

e) 5

[OA: 1]

]]>2 males can be selected in 4c2 = 6 ways.

We cannot select the wives of already selected males and should go for others.

so 2c2=1 way.

So in total 2 males and 2 females can be selected in 4c2 * 2c2 = 6 ways..

now the teams can be interchanged as well.

so 6 * 2 = 12 games possible.

case 2: when exactly one couple is selected.

so u can select that couple in 4c1=4 ways.

Now after this u need to select 1 male and 1 female more but they should not be a couple.

so selecting 1 male from remaining 3 males in 3c1 way and now for a female u have 2 options left..

so 4c1 * 3c1 * 2 = 24 ways

case 3: when the 2 couples are selected. so 4c2 = 6 ways

Total = 42

Note : Assume A1 B1 and A2 B2 are the two couples selected. Now they cannot be in same team but can play in the same game. A1B2 in one team and A2B1 in other team

[Credits: Jasneet Dua]

]]>Given that Ashwin starts the game what no should he choose first ?

(a) 4

(b) 5

(c) 3

(d) he can never winAshwin starts first and chooses 10; what should Vijay choose next in order to win?

(a) 14

(b) 11

(c) 12

(e) None of the above

Aim = One attempt of shooting at a balloon.

Shot = One instance of shooting down a balloon.

Miss = One wasted aim.

The rules of the game are as follows:

(i) Each person will be given a maximum of four rounds of aims, the first round comprising three aims. A person gets the second round of aims only if he scores at least one shot in the first (i.e. the previous) round and so on.

(ii) ( If in any round, the number of shots by a player is 50% or more but less than 100% of the number of aims he had in that round, then he gets one extra aim in each of the remaining rounds. If the number of shots by the player is 100% of the number of aims he had in that round, he gets two extra aims in each of the remaining rounds.)

(iii) For each shot, a player is awarded five points and for each miss, he earns two negative points.

If the number of shots in P’s first round = that in Q’s third round = that in S’s second round = that in R’s fourth round, and P scored eight points in the second round, then what is the maximum possible score by Q in the fourth round?

]]>a.14

b. 15

c. 16

d. 17

[OA: 17]

]]>Total points without draw = 42 * 3 = 126

with draw point reduces by 1

because win-lose - point distribution = 3

draw -draw= point distribution=2

Let the total points of A,C,E be a and that of B and G be b

3a + 2b + 36 = Total points [ also a < b ]

Total points can be 126 or 125 or 124 and so on

But our objective is to maximize a we have to take value of total as high as possible.

3a + 2b = 126 then max = 16

but if we take 3a + 2b = 125 then max(a) = 17

so 17

[Credits : @hemant_malhotra]

]]>(a) 3

(b) 67

(c) 195

(d) 323

(e) 451

[OA: 67]

]]>a) Rs. 23.25

b) Rs. 27.00

c) Rs. 18.75

d) Rs. 16.50

[OA: 18.75]

]]>What is the probability that a person wins on the 2nd throw?

a) 100/1296

b) 50/1296

c) 100/216

d) 50/216

e) None of these

What is the probability that a person loses on the 2nd throw ?

a) 100/1296

b) 120/1296

c) 144/1296

d) 72/1296

e) None of these