Quant Boosters  Rajesh Balasubramanian  Set 1

The equation has distinct real roots
> b^2 – 4ac > 0
> b = 8.
> 64 – 4ac > 0
> 4ac < 64
> ac < 16
If a = 1, c can be 1, 2, 3, 4, 5, 6, 7, 8, 9, or
10 – 10 possibilities
When a = 2, c can be 1, 2, 3, 4, 5, 6, or 7 > 7 possibilities
When a = 3, c can be 1, 2, 3, 4, 5 > 5 possibilities
When a = 4, c can be 1, 2, 3 > 3 possibilities
When a = 5, c can be 1, 2, 3 > 3 possibilities
When a = 6, c can be 1, 2 > 2 possibilities
When a = 7, c can be 1, 2 > 2 possibilities
When a = 8, c can be 1 > 1 possibility
When a = 9, c can be 1 > 1 possibility
When a = 10, c can be 1 > 1 possibility
Totally 35 pairs of values

Q21) 2a + 5b = 103. How many pairs of positive integer values can a, b take such that a > b?

Let us find the one pair of values for a, b. a = 4, b = 19 satisfies this equation. 2×4 + 5×19 = 103. Now, if we increase ‘a’ by 5 and decrease ‘b’ by 2 we should get the next set of numbers. We can keep repeating this to get all values.
Let us think about why we increase ‘a’ by 5 and decrease b by 2. a = 4, b = 19 works. Let us say, we increase ‘a’ by n, then the increase would be 2n. This has to be offset by a corresponding decrease in b. Let us say we decrease b by ‘m’. This would result in a net drop of 5m. In order for the total to be same, 2n should be equal to 5m. The smallest value of m, n for this to work would be 2, 5.
a = 4, b = 19
a = 9, b = 17
a = 14, b = 15
..
And so on till
a = 49, b = 1
We are also told that ‘a’ should be greater than ‘b’, then we have all combinations from (19, 13) … (49, 1).
7 pairs totally.

Q22) Using the vertices of a regular hexagon as vertices, triangles of how many different areas can be formed?

Let us number the vertices from 1 to 6. There are three different types of triangles that can be formed.
Type I: (1, 2, 3): This type of triangle is also seen with (2, 3, 4), (3, 4, 5) etc.
Type II: (1, 2, 4): (1, 2, 5) has the same shape and area. So, do (2, 3, 5), (3, 4, 6) etc.
Type III: (1, 3, 5): (2, 4, 6) has the same shape and area.
So, there will be 3 triangles of different areas that can be formed from a regular hexagon.
There are 6 triangles of Type I, 12 of type II and 2 of type III, adding up to 20 triangles totally.
The total number of triangles possible = 6C3 = 20.

Q23) Diagonals of a square ABCD of side 35 cms intersect at O. A rhombus of perimeter 52 exists such that P lies on AO, Q on BO, R on CO and S on DO. What is the maximum number of circles with integer radii that can be drawn with center O such that the rhombus is inside the circle and the circle is inside the square?

We are trying to draw circles that are inside the square. So, radius should be less than
35/2 = 17.5. Or, maximum integer radius should be 17.
We are trying to draw circles such that the rhombus should be inside the circle. So, the diameter of the circle should be greater than the longest diagonal of the rhombus. Or, the longer diagonal of the rhombus should be as short as possible. For a rhombus of the given perimeter, this is possible only if it is a square. A square of perimeter 52cm will have diagonals of length 13√2.
Diameter of circle > 13√2
Or radius > 13/√2
Or, radius > 10
Possible values of radius = 10, 11, 12, 13, 14, 15, 16, 17 = 8 different values.

Q24) The diagonals of Hexagon intersect at n distinct points inside the hexagon. What is the maximum value n can take?

In any hexagon, there are n (n  3)/2 = 9 diagonals.
First let us draw the diagonals and try to visualise this diagramThere are six ‘short’ diagonals AC, AE, CE, BD, BF, and DF. These intersect with other diagonals at 3 points each.
There are 3 long diagonals – AD, BE and CF. These intersect with other diagonals at 4 points each.
Note that the ‘short’ diagonals need not be shorter than the ‘long’ diagonal.
So, the total number of points of intersection should be 6 × 3 + 3 × 4 = 30.
But in this case, we would count every point of intersection twice. So, number of points of intersection would be exactly half of this = 30/2 = 15 points.

Q25) Consider a circle of radius 6 cms. What is the maximum number of chords of length 6 cms that can be drawn in the circle such that no two chords intersect or have points of contact?

First let us think about the angle subtended at the center by this chord. In a circle of radius 6cms, a chord of length 6 cms subtends an angle of 60° at the center. This chord along with the two radii forms an equilateral triangle.
So, we can place six such equilateral triangles at the center to account for 360°. This would form a regular hexagon. But in this scenario, the chords would have points of contact. So, the maximum number of chords that can be drawn such that there are no points of contact is 5.
More generally, if a chord makes an angle q at the center, then we can draw 360/q such chords around the circle. The maximum number of chords that can be drawn such that they do not touch each other = (360/q) – 1.

Q26) x (x – 3) (x +2 ) < 200, and x is an integer such that x < 20, how many different values can n take?

Let us start with a trial and error. The expression is zero for x = 0, x = 3 and x = –2
x = 3, the above value = 0
x = 4, the above value would be 4 × 1 × 6 = 24
x = 5, the above value would be 5 × 2 × 7 = 70
x = 6, the above value would be 6 × 3 × 8 = 144
x = 7, the above value would be 7 × 4 × 9 > 200
So, the equation holds good for x = 3, 4, 5, 6.
For x = 2 and 1, the above value is negative.
So, the above inequality holds good for x = 6, 5, 4, 3, 2, 1, 0.
For, x = –1, the value would be –1 × –4 × 1 = 4.
For, x = –2, the value would be 0.
So, this works for x = 6, 5, 4, 3, 2, 1, 0, –1, –2.
For, x = –3, the expression is negative, so holds good. For all negative values < –3, this holds good.
The smallest value x can take is –19.
So, the above inequality it holds good for –19, –18, –17…..–1, 0, 1 ……6, a total of 26 values.

Q27) In how many ways can we pick three cards from a card pack such that they form a sequence of consecutive cards, not all cards belong to the same suit, and nor do all cards belong to distinct suits? Consider Ace to be the card following King in each suit. So, Ace can be taken to precede ‘2’ and succeed ‘King’. So, QKA would be a sequence, so would be A23. However, KA2 is not a sequence.

First let us see how many sequences of 3 we can form. We can have A23, 234…..JQK, QKA – a total of 12 sets of 3.
If cards should not be of the same suit, and nor should all three be of different suits, then we should have two cards from one suit and one from another.
So, cards should be from two suits. The two suits can be selected in 4C2 ways. Now, from these two suits, one suit should have two cards. The suit that has two cards can be selected in 2C1 ways. Now, out of the three cards, the two cards that have to be from the suit that repeats can be selected in 3C2 ways.
So, total number of possibilities = 12 × 4C2 × 2C1 × 3C2 = 12 × 6 × 2 × 3 = 432.

Q28) Given that k < 15, how many integer values can k take if the equation x^2 – 6x + k = 0 has exactly 2 real roots?

The equation can be rewritten as x^2 – 6x + k = 0. This is a quadratic in x. This can have 2 real roots, 1 real root or 0 real roots.
If we have x = positive value, we have two possible values for x.
If we have x = negative value, we have no possible values for x.
If we have x = 0, we would have 1 possible value for x.
So, for the equation to have 2 values of x, we should have 1 positive root for x.
Scenario I: x^2 – 6x + k has exactly one real root (and that root is positive). b^2 – 4ac = 0 => k = 9. If k = 9, x = 3, x can be 3 or –3
Scenario II: x^2 – 6x + k has two real roots and exactly one of them is positive. This tells us that the product of the roots is negative. => k has to be negative. K has to be less than 15.
= > k can take values –14, –13, –12, ….–1 : 14 different values
Total possibilities = –14, –13, –12, ….–1 and k = 9; 15 different values

Q29) In how many ways can 6 boys be accommodated in 4 rooms such that no room is empty and all boys are accommodated?

No room is empty, so the boys can be seated as 1113 in some order or 1122 in some order.
Scenario I: 1113. We can do this as a two–step process.Step I: Select the three boys – 6C3.
Step II: Put the 4 groups in 4 rooms – 4! ways
Total number of ways = 20 × 4!
Scenario I: 1122. This is slightly tricky. So let us approach this slightly differently.
Step I: Let us select the 4 people who are going to be broken as 2 + 2; this can be done in 6C4 ways. Now, these 6C4 groups of 4 can be broken into 2 groups of two each in 4C2 / 2 ways. So, the total number of ways of getting 2 groups of 2 is 6C4 × 4C2/2 = 15 × 6/2 = 45 ways
Step II: Now, we need to place 2, 2, 1, 1 in four different groups. This can be done in 4! ways.
The total number of ways = 20 × 4! + 45 × 4! = 4! (20 + 45) = 24 × 65 = 1560 ways.

Q30) In how many ways can 4 boys and 4 girls be made to sit around a circular table if no two boys sit adjacent to each other?