Advance Math Model Questions Set 1 Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 2 EXERCISES
The following question based on Advance Math topic of quantitative aptitude
(a) 320
(b) 282
(c) 270
(d) 342
The correct answers to the above question in:
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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Read more model questions set 1 Based Quantitative Aptitude Questions and Answers
Question : 1
In how many different ways can four books A, B, C and D be arranged one above another in a vertical order such that the books A and B are never in continuous position ?
a) 12
b) 14
c) 9
d) 18
Answer »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
Question : 2
The probability that a person will hit a target in shooting practice is 0.3. If he shoots 10 times, the probability that he hits the target is
a) 1 – $(0.7)^{10}$
b) $(0.7)^{10}$
c) 1
d) $(0.3)^{10}$
Answer »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 – $(0.7)^{10}$.
Question : 3
The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The total number of triangles that can be constructed by using these points as vertices is
a) 204
b) 205
c) 220
d) 195
Answer »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.
Question : 4
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 : 5
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 : 6
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 = $2/{^6C_3} = 2/{20} = 1/{10}$
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