Quant Boosters  Rajesh Balasubramanian  Set 3

Number of Questions  30
Topic  Quant Mixed Bag
Solved ?  Yes
Source  2IIM

Q1) Triangle PQR has integer sides and a perimeter of 15. How many such triangles are possible ?

Let us say the three sides are p, q and r, and we write p < q < r (let's say, they are in an ascending order).
Then sum of any two sides must be larger than the third. It is enough to check if the sum of the two smaller sides is greater than the third. So let us try to find the largest possible value for r.
Now, p + q > r, so p + q must be greater than half of the perimeter. Or, we can say that p+q = 8, and r = 7, is the maximum possible value for r.
Let r = 7, (p+q = 8)
then p = 1, q = 7 OR
p =2, q = 6 OR
p = 3, q = 5 OR
p = 4, q = 4
Let r = 6, (p + q = 9)then p =1, q = 8 (don't count this. q must be less than or equal to r)
p = 2, q = 7 (nope)
p = 3, q = 6 (possible)
p = 4, q = 5 (possible)
Let r = 5, (p+q = 10)
only one value works. equilateral triangle, (5, 5, 5)
So 7 possible values  (1, 7, 7), (2, 6, 7), (3, 5, 7), (4, 4, 7), (3, 6, 6), (4, 5, 6) and (5, 5, 5)
Systematic thinking can come to the rescueBonus shortcut :
Number of Triangles with Integer sides for a given perimeter.
If the perimeter p is even then, total triangles is [p^2]/48.
If the perimeter p is odd then, total triangles is [(p+3)^2]/48Here p = 15, so required answer is [18^2 /48] = 7

Q2) How many triangles exist with integer sides, perimeter 14 that are not isosceles ?

Triangle is not isosceles  triangle is either equilateral or scalene.
But since the sides have to be integers, an equilateral triangle is only possible if the perimeter is a multiple of 3 (can't divide 14 equally into three integers)
So we are looking for any possible scalene triangle. I am a one trick pony. Same method (for those who were patient enough to read my solution to the first question) which starts by considering the largest possible value for r (let the sides be p, q, r)
Max possible value is 6 (r cannot be 7, because then p+q = 7, and we don't have a triangle)
when r = 6, p + q = 8
p = 1, q = 7 (not possible, remember we consider them in an ascending order to be sure we are counting all possible cases)
p = 2, q = 6 (this is an isosceles triangle)
p = 3, q = 5 (bingo! scalene triangle)
p = 4, q = 4 (isosceles again)
when r = 5, p + q = 9
only possible triangle is (4,5,5) clearly isosceles.
r cannot drop farther and still be the largest side. So we've exhausted all possible triangles.
Only one triangle exists  (3,5,6)Bonus shortcut :
Number of scalene Triangles with Integer sides for a given perimeter.
If the perimeter p is even then, total triangles is [(p  6)^2]/48.
If the perimeter p is odd then, total triangles is [(p  3)^2]/48Main point to remember is that an equilateral triangle is considered isosceles

Q3) How many triangles with perimeter 18 are isosceles ?

Pick the equal side first. We cannot haev 9, 9, 0.
So, we start from 8, 8, 2.
Then we move to 7, 7, 4..then 6, 6, 6,.. Then 5, 5, 8.. 4, 4, 10 is not possible. 4 possible triangles.Going with the previously mentioned formula : [18^2/48]  [(18  6)^2/48] = 7  3 = 4.

Q4) Triangle ABC has integer sides with product of two sides being 12. Further we know that the triangle is isosceles and acute. How many such triangles are possible? (Slightly tougher one)

Let the three sides be a,b,c, written in an ascending order (a < =b < =c)
Since the triangle is acute, a^2 + b^2 > c^2
Also, the triangle is isosceles, so we have to worry only about those triangles which have two sides equal.
Now, since the product of two sides is given to be 12, let us look at possible integer sides that can multiply to give 12. We can make this list:
12 x 1
6 x 2
4 x 3
So two of the sides must be among these values. For the first one, clearly we cannot have (1, 1, 12), since that will not be a triangle. So we can have (12, 12, 1) only. we can verify that 12^2 + 1^2 is indeed greater than 12^2.
For the second one, (2,2,6) is not possible. (6,6,2) is indeed an acute angled triangle.
For the last one, we have (4,4,3) or (3,3,4). Both are acute angled triangles.
So four triangles possible in total.

Q5) In how many ways, can we rearrange the word MONSOON such that no two adjacent positions are taken by the same letter? (Tougher than what we will see in CAT)

First up, lets get the facts out of the way – The three O’s need to be kept apart, then the 2 N’s.
Let us focus on the three O’s.We can place the three O’s in some blanks around the other letters. Or, three O’s can be placed in 3 slots out of the 5 in M__N__S__N . This can be done in 5C3, or 10 ways.
Or, the O’s can be in slots {1, 3, 5} or {1, 3, 6} or {1, 3, 7} or {1, 4, 6} or {1, 4, 7} or {1, 5, 7} or {2, 4, 6} or {2, 4, 7} or {2, 5, 7} or {3, 5, 7} – Whew.
Now, for each of these arrangements, there are 4!/2! = 12 arrangements for the other 4 letters. But the one thing we need to keep in mind now is the fact that 2 N’s could be adjacent in these arrangements. We will need to eliminate these.O’s in slots {1, 3, 5} or O__O__O__ __ – Ns could be in the last two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10
O’s in slots {1, 3, 6} or O__O__ O – Ns could be two slots 4 and 5. There are 2! = 2
words like this. So, number of words that we need to count = 12 – 2 = 10O’s in slots {1, 3, 7} or O__O__ __ __O – Ns could be in the slots {4, 5} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 – 4 = 8
O’s in slots {1, 4, 6} or O__ O__O – Ns could be in the slots {2, 3}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10
O’s in slots {1, 4, 7} or O__ O __O – Ns could be in slots {2, 3} or {5, 6}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 – 4 = 8
O’s in slots {1, 5, 7} or O__ __ O O – Ns could be in the slots {2, 3} or {3, 4}. There are totally 4 words that we need to subtract. So, number of words that we need to count = 12 – 4 = 8
O’s in slots {2, 4, 6} or O__O__O – Tehre are no possible slots for N. So, we count all 12 words on this list.
O’s in slots {2, 4, 7} or O__O __ O – Ns could be in slots {5, 6}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10
O’s in slots {2, 5, 7} or O O O – Ns could be in slots {3, 4}. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10.
O’s in slots {3, 5, 7} or __ __ O__O__O – Ns could be in the first two slots. There are 2! = 2 words like this. So, number of words that we need to count = 12 – 2 = 10
Total number of words = 10 + 10 + 8 + 10 + 8 + 8 + 12 + 10 + 10 + 10 = 96.
Phew!.There is a far more elegant method for accounting for the words where the 2 N’s appear together. This one came from Mukund Sukumar.
We need to account for the number of possibilities of N,N being together. So to subtract that part, consider ‘NN’ being together as one letter and place O’s. In 3 out of the 4 slots in _ M _ N N _ S _
The O’s can be selected in 4C3 ways. The MNNS can be rearranged in 3! Ways if the N’s have to appear together.Or, we get 4C3 * 3! = 24 ways. So, we have 120  24 = 96 ways totally.

Q6) How many 3digit numbers 'abc' exist such that
 a > b > c
 a < b < c

 Select 3 digits from 0 to 9.
and then think about arranging them.
These can be arranged in only one way, answer is 10C3.  Bear in mind that this number cannot have a zero in it, so the answer is 9C3.
Dont listen to anyone who says P and C is not fun!!
 Select 3 digits from 0 to 9.

Q7) From a class of 10 students, in how many ways can we select
 3 people  one to play football, one to play cricket and one to play tennis?
 3 people to play cricket?
 3 people  2 to play cricket and 1 to play football?

The difference between 1st and 2nd one is the idea of order. Qn 1, answer is 10 * 9 * 8, while the second one is 10C3. The third part can be done as 10C3 and then sending 2 players for one sport. So, 10C3 * 3, which is nothing but 10C2 * 8

Q8) How many 6digit numbers can we have that have only digits 2 and 4 and are divisible by 6?

We can have all 2's, all 4's or 3 twos and 3 fours.
So, we can have 1 + 1 + 6!/3!/3! = 1 + 1 + 20 = 22 numbers

Q9) How many 7digit numbers exist comprising only digits 5 and 6 that are multiples of 15?

Multiples of 15  Or we are looking for multiples of 3 and 5. Last digit should be 5. Sum of the digits should be a multiple of 3. Or, the number of 5's should be a multiple of 3 ( Think about this, this is significant). We can have six 5's, three 5's or zero 5's. zero 5's is not possible. So, we can have six 5's or three 5's. 6 + 15 = 21 numbers

Q10) Product of the distinct digits of a natural number is 60. How many such numbers are possible?