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

Answer: (a)

Let us take books A and B as one i.e., they are always continuous.

Now, number of books = 4 – 2 + 1 = 3

These three books can be arranged in 3! ways and also A and B can be arranged in 2 ways among themselves.

So, number of ways when books A and B are always continuous = 2 × 3!

Total number of ways of arrangement of A, B, C and D = 4!

Hence, number of ways when A and B are never continuous = Total number of ways – number of ways when A and B always continuous

= 4! – 2 × 3! = 12

Answer: (a)

The probability that the person hits the target = 0.3

∴ The probability that he does not hit the target in a trial = 1 – 0.3 = 0.7

∴ The probability that he does not hit the target in any of the ten trials = $(0.7)^{10}$

∴ Probability that he hits the target

= Probability that at least one of the trials succeceds

= 1 – $(0.7)^{10}$.

Answer: (b)

We have in all 12 points. Since, 3 points are used to form a triangle, therefore the total number of triangles including the triangles formed by collinear points on AB, BC and CA is $^{12}C_3$ = 220. But this includes the following :

The number of triangles formed by 3 points on AB = $^3C_3$ = 1

The number of triangles formed by 4 points on BC = $^4C_3$ = 4.

The number of triangles formed by 5 points on CA = $^5C_3$ = 10.

Hence, required number of triangles = 220 – (10 + 4 + 1) = 205.

Answer: (c)

No. of way to answer.

= $^6C_2 ^6C_5 + ^6C_3 ^6C_4 + ^6C_4 ^6C_3 + ^6C_5 ^6C_2$

= $2({^6}C_2 ^6C_5 + ^6C_3^6C_4) = 2({^6}C_2 ^6C_1 + ^6C_3 ^6C_2)$

= 2$({6 × 5}/2 × 6 + {6 × 4 × 5}/{3 × 2} × {6 × 5}/2)$

= 2 (90 + 300) = 780

Answer: (b)

At least one black ball can be drawn in the following ways:

(i) one black and two other colour balls

(ii) two black and one other colour balls, and

(iii) all the three black balls

Therefore the required number of ways is

$^3C_1 × ^6C_2 + ^3C_2 × ^6C_1 + ^3C_3$ = 64.