# Quant Boosters - Swetabh Kumar - Set 1

• Q27) Find the number of zeroes in : 100^1 * 99^2 * 98^3 * 97^4 * ... * 1^100

• (96+91+86...6+1)+(76+51+26+1)
= (5 * 19 + 1 + 5 * 18 + 1 ... 5 * 0 + 1) + 154
= 20 + 5 * 190 + 154
= 1124

• Q28) In an arithmetic sequence, a1 = 4. If a17, a37 and a77 form a geometric sequence in that order, find a500.

• 503 or 4
(4+36d)^2 = (4+16d)(4+76d)
16+1296d^2 + 288d = 16+368d + 1216d^2
80d = 80
d=1 or d=0
a+499d= 4+499=503 or 4+0 = 4

• Q29) The numbers 6, m, n form an arithmetic progression and the numbers m, n, 16 form a geometric
progression. Find the maximum value of m + n.

• 2m=6+n or 16m=48+8n
and n^2 = 16m
n^2-8n-48 = 0
n=12 and m = 9 so 21

• Q30) The points (3,7), (6,2) and (2,k) are the vertices of a triangle. For how many values of k is the triangle a right triangle ?

• 4 values
A=(3,7), B(6,2), C(2,k)
AB^2 = 9+25 = 34
AC^2 = 1+(7-k)^2 = k^2-14k+50
BC^2 = 16+(2-k)^2 = k^2-4k+20
if AB is hyp. 2k^2-18k+70 = 34
k^2--9k+18= 0
2 real values here.
if AC is hyp: k^2-4k+54 = k^2-14k+50
10k=-4 so k=-0.4 so one value
if BC is hyp, k^2-14k+84=k^2-4k+20 so k=6.4 so one value.
so (2+1+1)= 4 values

• @swetabh_kumar

so cases here: 63
hence 66 * 4C2

shouldn't it be 63 *(4!/2!) ?????

• @swetabh_kumar
x = 3/2 minimum = 1

61

62

63

61

64

61

61

61