Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 1

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

Every term will have a maximum power of 2 and all possible combinations will be considered. So, essentially we need to find the number of nonnegative integral solutions to a + b + c + ... + x + y + z = 2
For any equation, a + b + c + … + r terms = n
Number of nonnegative integral solutions will be (n + r – 1) C (n – 1)
Here it will be 27c25 which will be 351.

Q13) The cost price of four articles A, B, C and D are ‘a’, ‘b’, ‘c’ and ‘d’ respectively. A, B, C and D are sold at profits of 10%, 20%, 30% and 40% respectively. If the net profit on the sale of these four articles is 25%, ‘a’, ‘b’, ‘c’ and ‘d’ cannot be in the ratio
(a) 4 : 1 : 4 : 3
(b) 1 : 2 : 2 : 1
(c) 2 : 3 : 6 : 1
(d) 5 : 2 : 7 : 3

I did it from the options: 4 * 10 + 1 * 20 + 4 * 30 + 3 * 40 = 300 divided by 12 (ie. 4 + 1 + 4 + 3) is 25 and so on.
Not as tedious as it looks like and can be solved without using your rough sheet.

Q14) How many noncongruent triangles are possible with perimeter 19, whose two sides are even and one side is odd?

6 cases => (3,8,8)(5,6,8)(7,6,6)(7,8,4)(9,8,2)(9,6,4).
As the perimeter is less, it would be better to do it manually. Just have to remember that
(i) the largest side of a triangle is less than the semi perimeter of the triangle and is greater than or equal to one third of the perimeter of the triangle.
(ii) Sum of any two sides of a triangle is always greater than the third side.What if perimeter is a large value, say 41?
We can still do it manually and with an eye on the pattern of combinations. Longest side will be between 14 and 20.
14>(14,13)
15>(14,12)
16>(16,9)(15,10)(14,11)(13,12)
17>(16,8)(14,10)(12,12)
18>(18,5)(17,6)(16,7)(15,8)(14,9)(13,10)(12,11)
19>(18,4)(16,6)(14,8)(12,10)
20>(20,1)(19,2)...(11,10)Again, you need not write everything down and can simply do it by observing the trend. Generally there won't be many mindless questions in cat designed to make you while away your time. If some question comes which seems impossible, better to skip it.
PS: I am well aware of the (p+3)^2/48 formula but simply because there is an odd and even condition here, I would do it the normal way.

Q15) When working alone A, B and C can complete a piece of work in 8, 12 and 30 days respectively. At the most only two people can work on each day and nobody works for more than two consecutive days. What is the minimum number of days that they will take to finish the work?