Quant Boosters  Hemant Yadav  Set 1

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]

8/15 < n/(n + k) < 7/13
=> 13/7 < (n + k)/n < 15/8
=> 6/7 < k/n < 7/8Now, LCM of 7 and 8 is 56
=> 48/56 < k/n < 49/56
=> 96/112 < k/n < 98/112Here we can see that k can be only 97, now of we increase n then k will have 2 values.
So, 112 is maximum possible value of n for which k will have a unique value.

Q27) Find the number of nonnegative integral solutions for the equation 3a + 4b + 12c = 432
[OA : 703]

Method 1 :
Clearly a is a multiple of 4 and b is a multiple of 3. So, let’s say, a = 4k and b = 3n
Hence, 12k + 12b + 12c = 432
k + b + c = 36
So, C(38, 36) = 703 solutionsMethod 2 :
3a + 4b + 12c = 432 (1)
RHS is multiple of 12 so must be LHS, that means
3a + 4b = 12k (2)
So, 12k + 12c = 432 = 12*36
or, k + c = 36 (3)(3) has 37 solutions where k varies from 0 to 36. So we are to find the sum of number of solutions of (2) for this range of k.
Now (2) has k + 1 number of solutions as a is a multiple of 4 varies from 0 to 4k.
So basically we need to find the sum of an AP of 37 terms whose kth term is k + 1 and k varies from 0 to 36
i.e. 1 + 2 + 3 + .... + 37 = 37 * 19 = 703

Q28) Find the number of ordered triplets (a,b,c) of positive integers such that
LCM (a,b) = 1000
LCM (b,c) = 2000 and
LCM (c,a) = 2000[OA : 70]

LCM(a, b) = 1000 = 2^3 * 5^3
LCM(b, c) = 2000 = 2^4 * 5^3
LCM(a, c) = 2000 = 2^4 * 5^3a = 2^x1 * 5^y1
b = 2^x2 * 5^x2
c = 2^x3 * 5^x3So, x3 = 4
=> At least one of x1, x2 has to be 3, else LCM(a, b) will not be 1000.
SO, 4^2  3^2 = 7 waysNow, at least two of y1, y2, y3 should be 3, else one of the LCM's will not be as desired.
So, 3 * 4  2 = 10 waysSo, answer will be 70 ways.

Q29) H.C.F. of three positive integers – A, B and C – is 6. If the sum of the three integers is 48, how many triplets (A, B and C) are possible?
[OA : 18]

a + b + c = 48
6k + 6n + 6m = 48
k + n + m = 8, where gcd(k, n, m) = 1
for k + n + m = 8, gcd(k, n, m) can take only two values 1 or 2total solutions of the equation k + n + m = 8 is 7C2
when gcd is 2, p + q + r = 4, so 3C2 = 3 waysso, 7C2  3 = 18 triplets

Q30) a + 1/b = 7/3
b + 1/c = 4
c + 1/a = 1
Find abc[OA : 1]

Multiply them to get
(a + 1/b)(b + 1/c)(c + 1/a) = 28/3
=> a + b + c + 1/a + 1/b + 1/c + 2abc = 28/3
=> 22/3 + 2abc = 28/3
=> abc = 1

If h1, h2, h3 are the heights corresponding to the sides a, b, c, then
a * h1 = b * h2 = c * h3 = 2A (where A is area of the triangle)
=> a = 2A/h1, b = 2A/h2 and c = 2A/h3
Now, a + b > c
=> 2A/h1 + 2A/h2 > 2A/h3
=> 1/h1 + 1/h2 > 1/h3
=> 1/2 + 1/3 > 1/h3
=> h3 > 6/5Also, b + c > a
=> 2A/h2 + 2A/h3 > 2A/h1
=> 1/h2 + 1/h3 > 1/h1
=> 1/3 + 1/h3 > 1/2
=> h3 < 6So, 6/5 < h3 < 6
=> Integral values that h3 can take are 2, 3, 4, 5

minimum of x+x17...+x440
median of 0,17,40,85,173,440 : (40+85)/2 = 62.5
put x=62.5..so 641
[swetabh kumar]

for 1st meet: ad/(a+b)= 400
distance by A til 2nd meet: a/(a+b) * 3d = d + 300
so 1200 = d + 300, d = 900
[swetabh kumar]

ab  a  b = 35
=> (a  1)(b  1) = 36
(a, b) = (2, 36), (3, 19), (4, 13), (5, 10), (7, 7)
In last two cases a and b are not coprime

case 1: when there is no power of 2
so 3 * 3 * 3=27 cases
case 2: when there is only 1(2) in it
one of the dice will have 2 or 6... and other two dices will have odd numbers.. so
2 * 3 * 3 * 3c1=54
case 3:
when there are exactly two (2's)
two dices will have (2,6) and one dice will have (1,3,5)
so 2 * 2 * 3 * 3c1=36
part2:
exactly one dice will have 4 in it.. other two will have (1,3,5)
so 3 * 3 * 3c1=27adding all 144 cases which will not have a multiple of 8.. remaining 72..
so 72/216=1/3[jasneet dua]