Quant Boosters  Hemant Yadav  Set 1

a) boxes and balls are all distinct and only 1 ball in one box
Its like permuting a 5 digit number, so 5! Waysb) boxes are same and balls are distict and only 1 ball per box
Since all the boxes are same and one ball per box, so just one wayc) balls are same and boxes are different and only 1 ball per box
Here all the balls are same and every box has one ball, so again one wayd) balls and boxes are same and only 1 ball per box
Here also only only waye) repeat all above questions with any number of balls in any box
i) 5^5 ways, as all the balls and all the boxes are different.
ii) Here boxes are same but balls are different.
(5, 0, 0, 0, 0) – 1 way
(4, 1, 0, 0, 0) – C(5, 4) = 5 ways
(3, 2, 0, 0, 0) – C(5, 3) = 10 ways
(3, 1, 1, 0, 0) – C(5, 3) = 10 ways, we don’t have to choose one of other two as they will automatically choose any of the two boxes (all the boxes are similar)
(2, 2, 1, 0, 0) – C(5, 2)*C(3, 2)/2 = 15, here we need to divide by 2 as as it doesn’t matter if I put the first pair in box 1 and second pair in box 2 or viceversa
(2, 1, 1, 1, 0) – C(5, 2) = 10 ways
(1, 1, 1, 1, 1) – 1 way
So, total = 1 + 5 + 10 + 10 + 15 + 10 + 1 = 52 waysiii) x1 + x2 + x3 + x4 + x5 = 5
So, C(9, 5) = 126 waysiv) Since all balls and boxes are same, we just need to find what different combinations are possible:
(5, 0, 0, 0, 0)
(4, 1, 0, 0, 0)
(3, 2, 0, 0, 0)
(3, 1, 1, 0, 0)
(2, 2, 1, 0, 0)
(2, 1, 1, 1, 0)
(1, 1, 1, 1, 1)
So, 7 ways

Q17) Find the largest possible value of k for which 3^11 is expressible as the sum of k consecutive positive integers. What will be the answer if 3^11 is expressed as sum of k consecutive integers.
[OA : 486, 2 * 3^11]

Let the first of the consecutive integers be a + 1, then the consecutive integers will be a + 1, a + 2, ..., a + k
So, (a + 1) + (a + 2) + .... + (a + k) = 3^11
k(k + 2a + 1) = 2 * 3^11Now for the first part all the integers are positive, so both (k + 2a + 1) > k
Hence maximum value of k will be 2 * 3^5 = 486For the second part there is no such constraint, so maximum possible value of k will be 2 * 3^11

Q18) Find the sum of 6 * 1! + 13 * 2! + 22 * 3! + 33 * 4! + 46 * 5! + ... n terms

It is (n +2)! + 2[(n + 1)!]  4 = (n + 4)[(n + 1)!]  4.
Key lies in simply finding the nth term of the sequence which is (n + 1)[(n + 1)!] + 2n[n!].

Q19) How many natural numbers N < 40000 such that cube of sum of digits of N gives back the same number N.
For eg, 8³ = 512, sum of digits is 8, same as cube root of 512.[OA : 6]

Since N < 40000, N < 35³
It is clear from the given condition that the sum of digits and cube of sum of digits should leave same remainder when divided by 9.
Hence sum of digits should be of form 9k + 1 or 9k  1 or 9k
1³ = 1 (sum of digits is 1, so possible)
8³ = 512 (sum of digits is 8, so possible)
9³ = 729 (sum of digits is 18, so not possible)
10³ = 1000 (not possible)
19³ = 6859 (sum of digit is not 19, so not possible)
18³ = 5832 (sum of digits is 18, so possible)
17³ = 4913 (sum of digit is 17, so possible)
26³ = 17576 (possible)
27³ = 19683 (possible)
28³ = 21952 (not possible)So, 1, 512, 4913, 5932, 17576 and 19683 are the only such numbers

Q20) How many lines in a three dimensional rectangular coordinate system pass through four distinct points of the form i, j, k where i, j and k are positive integers not exceeding four?
a) 60
b) 72
c) 64
d) 76[OA : 76]

We can see that there are 12 planes (4 paralle to xy plane, 4 parallel to yz plane and 4 parallel to zx plane) In every plane there would be 2 diagonals, so 24 lines. 4 diagonals of the cube and 16 lines parallel to every axis. So, total 4 + 24 + 48 = 76

Q21) Circles A, B and C are externally tangent to each other, and internally tangent to a bigger circle D. Circles B and C are congruent. Circle A has radius 1 and passes through the center of D. What is the radius of circle B?
[OA : 8/9]

O is the centre of the bigger circle. and A,B,C are the centres of the remaining circles. Circles with centres B and C have the same radius. Radius of circle with centre A = 1.
Obviously, the radius of the bigger circle is 2.
Let OD = x (D is the midpoint of BC)
Then we have: r^2 + x^2 = (2r)^2
and (x+1)^2 + r^2 = (1+r)^2
Taking the diff of the above equations gives x = 3r  2
using which we get r = 8/9

Q22) For any positive integer n , f(n) is the highest power of 2 that divides n!
Find f(1) + f(2) + f(3) + ..... + f(1023)[OA : 518656]

It is similar to finding the highest power of 2 in 1! * 2! * 3! * ...... * 1023!
First we will count that how many numbers are divisible by 2 in every factorial, then by 4, then by 8, ...., and lastly by 512
So, S = (0 + 1 + 1 + 2 + 2 + ..... + 511 + 511) + (0 + 0 + 0 + 1 + 1 + 1 + 1 + .... + 255) + ..... + (0 + 0 + .... + 1 + 1 + 1 (512 times))
= 511 * 512 + 512 * 255 + 512 * 127 + 512 * 63 + 512 * 31 + 512 * 15 + 512 * 7 + 512 * 3 + 512
= 512(511 + 255 + 127 + 63 + 31 + 15 + 7 + 3 + 1) = 512 * 1013 = 518656=> f(1) + f(2) + f(3) + ..... + f(1023) = 518656

Q23) The sides of a triangle form an arithmetic progression. The altitudes also form an arithmetic progression. Then what can be said about the triangle
a) It's a rt angle triangle
b) It's an equilateral triangle
c) It's an obtuse angle triangle
d) Nothing can be said about it

Method 1 :
Say the sides are A B C and the heights are a,b,c respectively one those sides.
So Aa = Bb = Cc
Now IF B/A should be a/b and also their difference should be same B  A = a  b
So it's only possible when A = B = C.Method 2:
Since, Aa = Bb = Cc = k
So, a = k/A, b = k/B and c = k/C
From this we can say that A, B, C are in AP as well as HP. That means A = B = C. Hence an equilateral triangle.Method 3 :
Let the sides be (a – x), a, (a + x) and corresponding altitudes be (h – y), h, (h + y)
Since, (a – x)(h – y) = ah
ay + hy = xy ..........(1)Also, (a + x)(h + y) = ah
ay + hy = xy .........(2)From (1) and (2), we can say that xy= 0
Hence atleast one of x and y is zero.
Say, x = 0, then from eq (1) we will get y = 0Hence, x = y = 0 and the triangle is equilateral triangle

Q24) A pyramidshaped box has internal volume of 256 cubic cms and a square base of 8 cm * 8 cm. How many small solid pyramids, each with a volume of 4 cubic cms and a 2cm * 2cm square base ,can we pack wholly inside the box ?
[OA : 44]

On the top, we can place 1 pyramid
then in the next layer 2 * 2 + 1 Pyramids
in the 3rd layer we can place 3 * 3 + 4 Pyramids
and in the last layer we can place 4 * 4 + 9 Pyramids
So Total = 1 + 5 + 13 + 25 = 44 Pyramids

Q25) ∆PQR has median lengths as 10, 24 and 26. What will be the area of ∆PQR?
[OA : 160]

If medians are PX, QY and RZ and centroid be O.
Extend OX to W such that OX = XWIn triangle OWR, we can see that OW = (2/3)PX, OR = (2/3)RZ and RW = (2/3)QY
So, triangle OWR will have 4/9th area of a triangle having sides equal to PX, QY and RZ.So, triangle OXR will have 2/9th area of triangle having sides equal to PX, QY and RZ.
But triangle OXR has 1/6th area of triangle PQRHence, triangle PQR has 4/3rd area of triangle sides equal to PX, QY and RZ.
As, PX = 10, QY = 24 and RZ = 26
Hence, area of triangle PQR = (4/3) * 10 * 24/2 = 160

Q26) What is the largest positive integer 'n' for which there is a unique integer 'k' such that 8/15 < n/(n + k) < 7/13
a) 49
b) 56
c) 98
d) 112
e) 168[OA : 112]