Quant Boosters  Soumya Chakraborty  Set 3

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
I hope everyone understand continuous proportion.
In this case, it simply means that
a:b = b:c = c:d, taking a,b,c,d as the four numbers
Further, we know that these are each equal to 3:5
After this, it is a straightforward, simple combining of ratios problem.
a:b = b:c = 3:5
Combining, we have a : b : c = 9 : 15 : 25
Also, we have c:d = 3:5
Combining these two, we have a : b : c : d = 27 : 45 : 75 : 125
As, this is the reduced ratio and a,b,c,d are NATURAL NUMBERS, in order to minimize the sum, we need to minimize the values of each one of them, which cannot be reduce further from the above ratio
So, the minimum sum = 27+45+75+125 = 272

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Q26) Two natural numbers are taken, such that their mean proportional is 12 and their third proportional is 324. Find the numbers.

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Please understand that if few numbers are in continuous proportion, they MUST be in GP series as well.
The important part of the question is to understand that there are two continuous proportions
A, 12, B & A, B, 324
We can then treat these two as GP series as well
Let us say the common ratio of the first series is 'r' ...
implying B = A * r^2
And let us say the common ratio of the second series be 'R' ...
implying B = A * R
equating the B,
we have R = r^2
Now, obviously we cannot have the same value of r and R, as they are definitely different series ... otherwise B has to be both 12 and 324 simultaneously (which is funny)
Let us take few more values that can satisfy: If r = 2, R= 4
Then B, according to the first series = 12 * 2 = 24, and in the 2nd series B = 324/4 which is not 24
Taking, r = 3, R = 9
first series, B = 12 * 3 = 36
second series, B =324/9 = 36 ... BANG, we have our solution
So, A = 12/r = 12/3 = 4
Thus, the two numbers are 4 and 36

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Q27) The sides of a right angled triangle are a, a + 17x and a + 18x. If 'a' and 'x' are both positive, find a/x.

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Everyone is aware of Pythagoras theorem. The question is: How many Pythagorean triplets are you aware of?
Please remember the following five:
3,4,5; 5,12,13; 7,24,25; 8,15,17; and 9,40,41
If we consider the triplet in our scenario:
a, a+17x, a+18x: we find that the difference between the first two is 17x and difference between the last two is x, where the differences are in the ratio 17:1, which i satisfied by the triplet 7,24,25
So, if we take our 'x' to be 1, our 'a' must be 7, implying a/x = 7

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Q28) A , B and C , three friends are enjoying a bonfire. A is contributing 5 wood logs , B is contributing 4 wood logs for the bonfire. C is not contributing any wood logs. Hence he is giving Rs 27 to A and B . What amount of money should be taken by A and B respectively? ( all the wood logs are identical)

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
First thing that we need to understand is what would C pay for ?
C is not going to pay for the woods A and B provided to the group. Actually C is going to pay for the woods A and B have provided to C
So, let's figure that out A gives 5 wood, B gives 4. So total contribution is 9. But, that is equally shared among the three.
So, each one of them consumes 3 in return
So, A gives 5, but consumes 3 himself, providing 53 = 2 to C
And, B gives 4, but consumes 3 himself, providing 43 = 1 to C
So, A and B gives wood to C in the ratio 2:1, their payment should also be in the ratio 2:1
the Rs. 27 distributed in the ratio 2:1, would be 18 and 9 ... These should be the amounts

Soumya Chakraborty last edited by Soumya Chakraborty
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Q29) Find ordered & unordered solutions possible for a + b + c = 120 if HCF (a, b, c) = 6

Soumya Chakraborty last edited by Soumya Chakraborty
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Ordered 
Unordered 

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
Q30) Which is the largest perfect square that can be expressed in the form of a^2 + ab + b^2 where a and b are positive prime numbers

Soumya Chakraborty last edited by
CAT 2017  QA 100 Percentile  IIM Lucknow  Managing Partner  MathOratory
a^2 + ab + b^2 = n^2
(a+b)^2 = n^2 + ab
(a+b+n)(a+bn) = abas 'a' and 'b' are prime numbers
ab has 4 factors: 1, a, b, ab
obviously, a + b + n > a + b  nCase 1:
a + b + n = ab
a + b  n = 1Case 2:
a + b + n = a
a + b  n = b ... considering a>bBut, case 2 is not possible.
b = n
a = n,
a = b ... which is impossibleCase 1:
adding the two equations
2a + 2b = ab + 1
2a  ab + 2b = 1
a(2b)  2(2b) = 3
(a2)(b2) = 3only possibility for 'a' and 'b' being prime and satisfying the above condition is: 5 and 3
giving us he only possibility 49