Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 1

This one looks difficult but you simply need to visualise and get past the lack of confidence barrier. For those who are clueless about this, start small and then generalise. You simply need to know that a nonagon has 9 sides. You can start with a basic pentagon or hexagon and make a star (either the five pointed one or the six pointed one by overlapping two equilateral triangles) and figure out the angle of the points. In case of the 5 pointed star, we get internal angle =108, so external angle = 72. Two external angles make 144 and so, the angle of the point will be 36.
Similarly, for a 9 sided figure, each internal angle will be equal to (9  2) * 180/9=140 degrees.
So, external angle = 40 degrees and so, angle at each point of the start will be 180  2 * 40 = 100.

Q3) 1 * 1! + 2 * 2! + 3 * 3! + 4 * 4! + ... + 7 * 7! = ?

We can rewrite the series as
(21) * 1! + (31) * 2! + (41) * 3! + ... + (81) * 7!
2!  1! + 3!  2! + 4!  3!... 8!  7!
8! 1 = 40319 is the answer.

Q4) If x and y are positive real numbers such that 12xy + 21x + 20y = 46, find the minimum value of 3x + 4y.
a. 6
b. 7
c. 8
d. 9

12xy + 21x + 20y + 35 = 81 (we add this 35 to make sure that we can take some common factor out)
3x(4y + 7) + 5(4y + 7) = 81
(3x + 5)(4y + 7) = 81
Using the AM > = GM property,
(3x + 5 + 4y + 7)/2 > = sqrt((3x + 5)(4y + 7))
3x + 4y + 12 > = 18
3x + 4y > = 6
So minimum value is 6.

Q5) N is an eight digit number having sum of digits as 66. How many such numbers are there?
a. 1216
b. 2716
c. 216
d. 1716

The largest sum one can get from an 8 digit number is 9 * 8=72
We are falling short by 6 in this case and this 6 has to be taken from one or many of the 8 digits in question.
a + b + c + d + e + f + g + h = 6
13c7 solutions = 13 * 12 * 11 * 10 * 9 * 8/720 = 1716 cases.
Also, you have to understand that none of the digits can be 0 as then the sum of digits won't exceed 63 and so, we are saved of additional steps.

Q6) Three friends Ram, Shyam and Mohan have decided to complete a work together. The time taken by Ram alone to the complete the work is 8(1/3)% more than the time taken by Shyam and Mohan together to complete the work. The time taken by Shyam alone is 25% more than the time taken by Ram and Mohan together to complete the work. If Mohan alone takes 75 days to complete the work, find the time taken by all three of them together to complete the work.
a. 5(2/3) days
b. 5(1/3) days
c. 7(1/3) days
d. None of these

R takes 8(1/3)% more than M+S. On simplifying the fraction, we can see that the ratios of time taken by R alone and M+S together is 13/12. It means that whatever part R finishes alone in 13 days, M+S can finish off in 12 days working together. So, if all three are working together, R's contribution to the total work will be 12/25th. Similarly, using the second statement, we can say that S's contribution to the total work will be 4/9th. So, contribution of M will be 112/254/9 = 17/225. But M needs 75 days to complete the work by himself. So, to complete 17/225 of the work, he will require 17/225*75 or 5(2/3) days.
The concept to understand here is that R is taking 13 days to finish off a piece of work that M+S are doing in 12 days. So, considering that total work is 156 units, R will do 12 units of work per day which is his efficiency and M+S will do 13 units of work per day which is their combined efficiency. So, if one day's work is considered a total of 25 units will be done out of which R has contributed 12 and so, his contribution will be 12/25.

Q7) Three boys A, B and C start running at constant speeds from the same point P along the circumference of a circular track. The speeds of A, B and C are in the ratio 5:1:1. A and B run clockwise while C runs in the anticlockwise direction. Each time A meets B or C on the track he gives them a card. What is the difference in the number of cards received by B and C if A distributes 33 cards in all?
(a) 3
(b) 7
(c) 5
(d) 11

Let the circumference be 120 units and speed of A, B, C be 5 units, 1 unit and 1 unit respectively. So, A and B will meet once every 20 seconds and A and C will meet once every 30 seconds. So, in 1 minute, A gives B 3 flags and gives C 2 flags. So, we can safely say that in the first 30 distributions, B gets 18 flags and C gets 12 flags. Post that, after 20 seconds, B gets a flag, at the 30th second C gets a flag and at the 40th second, B gets another flag. So, in total, B gets 20 flags and C gets 13 flags. So, difference is 7.

Q8) abc denotes a 3 digit number. If a and b are interchanged, the value of the original number decreases by 90. How many possible pairs (a, b) exist?

Original number is 100a+10b+c. New number is 100b+10a+c
100a+10b+c(100b+10a+c)=90
90a90b=90
ab=1
a can take values from 1 to 9 and b can take corresponding values from 0 to 8. So total of 9 cases.

Q9) In how many ways can 60 identical balls be distributed among A B and C such that A gets more than B and B gets more than C?
a. 1800
b. 1891
c. 300
d. 330

a + b + c = 60
We have to eliminate the cases which have repetitions and then check the cases that have all unequal integral solutions.
Total solutions are 62c2 = 61 * 31 = 1891
When 2 people get similar number of balls  2a + c = 60, a will vary from 0 to 30 i.e. 31 cases one among which will be 20 + 20 + 20. Ignoring that and ordering the remaining among a, b, c we get 30 * 3 = 90 solutions. Adding the one 20 + 20 + 20 case we get 91 cases in total with at least two terms being similar to each other.
1800 cases will be equally distributed among 6 cases which are a > b > c, a > c > b, b > a > c, b > c > a, c > a > b and c > b > a. So, 1/6th of 1800 satisfies our required condition of a > b > c and so the answer is 300.

Q10) If a^2 + b^2 + c^2 = x^2 + y^2 + z^2 = 1, Find the maximum value of ax + by + cz

Consider a^2 and x^2. Using AM > = GM, we get
(a^2 + x^2)/2 > = sqrt(a^2 * x^2)
(a^2 + x^2) > = 2ax.
Similarly,
(b^2 + y^2) > = 2by
(c^2 + z^2) > = 2cz
Adding all three,
2 > = 2(ax+by+cz)
So, max value of ax + by + cz = 1.Alternatively, you can check for a = b = c = x = y = z = 1/sqrt(3) and solve.

Q11) A toy consists of a base that is the section of a sphere and a conical top. The volume of the conical top is 30 Pi Cubic units and its height is 10 units. The total height of the toy is 19 units. The volume of the sphere (in cubic units) from which the base has been extracted is

Equating the volume of the cone with 30pi, we get the radius of the cone to be 3 units. If we look at the combination properly, we can see that we have two intersecting chords of 3+3 on one side and 9+x on the other. So, 3 * 3 = 9 * x and so, x=1. So, diameter of the sphere is 9+1=10. So, radius is 5 units. So volume is 4pi(5^3)/3 = 500pi/3.

Q12) Find the no of terms (a + b + c + ... + x + y + z)^2