Quant Boosters  Shashank Prabhu, CAT 100 Percentiler  Set 1

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?

Let total work be 120 units. A does 15 units per day, B does 10 units per day, C does 4 units per day. In a set of 3 days, they finish off 58 units of work. So, 120/58 = 2 sets of 3 days plus an additional 4 units to go which can be done by A and B together in the quickest time of 4/25 days. So total of 6 4/25.

Q16) Fourteen fruits and twenty two flowers are to be distributed among 10 people in such a way that each person gets something. Anyone who gets more than two flowers cannot get more than one fruit and anyone who gets more than one fruit cannot get more than three flowers. What is the maximum number of flowers that one can get?

One person can get 22 flowers and the rest of the fruits can easily be distributed. A question specifically designed for those who think that CAT does not ask questions that have obvious answers.

Q17) If N is a natural number less than 100, then for how many values of N are the numbers 6N + 1 and 15N + 2 relatively prime?
(a) 16
(b) 10
(c) 33
(d) None of these

Let there be two numbers ax and bx where x is the hcf. So, axbx=(ab)x will be divisible by x as ab will be an integer. Also, (ab)xbx = (a2b)x which will also be divisible by x as a2b is an integer.
So, if two numbers share a common factor, the same factor divides the difference (d) between the two numbers and also the difference between d and the smaller number. Say the two numbers are 111 and 153. The difference is 42 which is divisible by the hcf of 111 and 153 ie. 3. Also, 11142=69 is also divisible by 3. Remember that we are taking absolute values only.
Applying the same concept, If 6N + 1 and 15N + 2 are divisible by x, then their difference i.e. 9N + 1 will always be divisible by x. Similarly (9N + 1) – (6N + 1) i.e. 3N will also be divisible by x. If x divides 3N then it can also divide 6N. So, x divides 6N and 6N + 1 both i.e. two consecutive numbers. Hence x cannot be anything but 1. So for all the values of N, the given two numbers will be coprime.

Q18) A tank of capacity 165 liters, has N taps, numbered from 1 to N fitted to it. At the start of the nth minutes, where 1 < = n < = N, the tap numbered n, which fills the tank at the rate of n litres per minute, is opened. If it takes exactly N minutes to fill the tank in this manner, find N.
a. 8
b. 9
c. 10
d. 11