Advance Math Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (c)

Let there be n participants in the beginning. Then the number of games played by (n – 2) players = $^{n - 2}C_2$

∴ $^{n - 2} C_2 + 6 = 84$

(Two players played three games each)

⇒$^{n - 2}C_2$ = 78⇒(n - 2)(n - 3) = 156

⇒$n^2$ - 5n - 150 = 0

⇒$n^2$ - 15n + 10n = 150 = 0

⇒n(n - 15) + 10(n - 15) = 0

(n - 15)(n + 10) = 0

n = 15, –10

n cannot be –ve

Therefore n = 15.

Answer: (b)

Let us start with Red colour

Where, R = Red, B = Black, W = White

There are eight such arrangements, if we start with Red ball. Similarly, there are 8 arrangements, if we start with black or white ball.

Hence, No. of arrangements = 8 + 8 + 8 = 24

Answer: (d)

Any 3 numbers out of 9 can be selected in $^9C_3$ ways.

Now, these three numbers can be arranged among themselves in ascending order in only 1 way.

Hence, total no. of ways = $^9C_3$ × 1 = 84

Answer: (b)

Starting with the letter A, and arranging the other four letters, there are 4! = 24 words. These are the first 24 words. Then starting with G, and arranging A, A, I, and

N in different ways, there are ${4!}/{2!1!1!} = {24}/2$ = 12 words.

Hence, total 36 words.

Next, the 37th word starts with I. There are 12 words starting with I. This accounts up to the 48th word. The 49th word is NAAGI.

The 50th word is NAAIG.

Answer: (d)

Let the total no. of participants be 'n' at the beginning.

Players remaining after sometime = n – 3

Now, $^{n–3}C_2$ + (3 × 3) = 75

${(n - 3)!}/{2!(n - 5)!}$ + 9 = 75

$n^2$ – 7n – 120 = 0

(n + 8) (n – 15) = 0

neglecting n = – 8, n = 15