Advance Math Model Questions Set 1 Practice Questions Answers Test with Solutions & More Shortcuts
Advance Math PRACTICE TEST [2 - EXERCISES]
Advance Math Model Questions Set 1
Advance Math Model Questions Set 2
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 »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 »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 = $2/{^6C_3} = 2/{20} = 1/{10}$
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 »Answer: (d)
1,1 | 1,2 | 1,3 | 1,4 | 1,5 |
2,1 | 2,2 | 2,3 | 2,4 | 2,5 |
3,1 | 3,2 | 3,3 | 3,4 | 3,5 |
4,1 | 4,2 | 4,3 | 4,3 | 4,5 |
5,1 | 5,2 | 5,3 | 5,4 | 5,5 |
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 »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 »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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