Advance Math Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

Three vertices can be selected in $^6C_3$ ways.

The only equilateral triangles possible are $A_1 A_3 A_5 \text"and" A_2 A_4 A_6$

p = $2/{^6C_3} = 2/{20} = 1/{10}$

Answer: (d)

No. of total events = 25.

Chance of winning in one trial = $5/25 = 1/5$

Hence, chance of not winning = $1 - 1/5 = 4/5$

Answer: (a)

Six balls can be selected in the following ways: one red balls and 5 blue balls or

Two red balls and 4 blue balls

Total number of ways

= $^3C_1 × ^7C_5 + ^3C_2 × ^7C_4$

= 3 × ${7 × 6}/{2 × 1} + 3 × {7 × 6 × 5}/{3 × 2 × 1}$

= 63 + 105 = 168

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

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)

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