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

Question : 21

The probability that A can solve a problem is $2/3$ and B can solve it is $3/4$. If both attempt the problem, what is the probability that the problem gets solved ?

a) $7/{12}$

b) $5/{12}$

c) ${11}/{12}$

d) $9/{12}$

Answer: (c)

The probability that A cannot solve the problem

= 1 - $2/3 = 1/3$

The probability that B cannot solve the problem

= 1 - $3/4 = 1/4$

The probability that both A and B cannot solve the problem = $1/3 × 1/4 = 1/{12}$

∴ The probability that at least one of A and B can solve the problem = 1 - $1/{12} = {11}/{12}$

∴ The probability that the problem is solved = ${11}/{12}$

Question : 22

If three vertices of a regular hexagon are chosen at random, then the chance that they form an equilateral triangle is :

a) $1/5$

b) $1/{10}$

c) $1/3$

d) $1/2$

Answer: (b)

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

advance-math-aptitude-mcq

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

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

Question : 23

A and B play a game where each is asked to select a number from 1 to 5. If the two numbers match, both of them win a prize. The probability that they will not win a prize in a single trial is

a) 24/25

b) 2/25

c) 1/25

d) none of these

Answer: (d)

1,11,21,31,41,5
2,12,22,32,42,5
3,13,23,33,43,5
4,14,24,34,34,5
5,15,25,35,45,5

No. of total events = 25.

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

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

Question : 24

A box contains 10 balls out of which 3 are red and the rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour ?

a) 168

b) 189

c) 105

d) 120

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

Question : 25

ABCD is a convex quadrilateral. 3, 4, 5 and 6 points are marked on the sides AB, BC, CD and DA respectively . The number of triangles with vertices on different sides is

a) 320

b) 282

c) 270

d) 342

Answer: (d)

Triangles with vertices on AB, BC and CD are

3×4×5 = 60

Triangles with vertices on AB, BC and DA are

3×4×6 = 72

Triangles with vertices on AB, CD and DA are

3×5×6 = 90

Triangles with vertices on BC, CD and DA are

4×5×6 = 120

∴ Total no. of triangles = 60 + 72 + 90 + 120 = 342

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