10, 25, 56, 70, 85, 95, 125
56
avg of 10nos.=23==>23*10=230
if each no increased by 4 ==> 4*10=40
then new avg is giveen by : 230+40=270
270/10=27
hence the new avg =27
Each PAIR of stations means a PAIR of tickets (A to B and B to A)
2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46
(N * X) + (X * (X – 1)) = 23
factoring ___ X (N + X – 1) = 23
23 is a prime number with only two factors, 1 and 23
so N = 23 and X = 1
A is travelling at 50kmph, B is travelling at 40
kmph……..according to the formula
time taken to meet = distance between them
———————-
relative speed of two vehicles
so, time taken to meet= 15/(50-40)=15/10=3/2hrs
3days
Yes answer is 18 days
GIVEN: 2A(B+C)+AC-2C(A-B)
THEREFORE 2AB+2AC+AC-2AC+2BC
2AB+AC+2BC
2(AB+BC)+AC
LET b1=AB b2=BC b3=AC
STEP1: b1 = b1+b2
so b1 = AB+BC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AC
STEP 2: b3 = b1+b3
so b3 = AB+BC+AC
THEREFORE
NOW: b1 = AB+BC b2 = BC b3 = AB+BC+AC
STEP3:
NOW: b1 = b1+b3
so b1 = AB+BC+AB+BC+AC
=2(AB+BC)+ AC
AB BC AC
STEP1 AB+BC BC AC
STEP2 AC BC AB+BC+AC
STEP3 AB+BC+AB+BC+AC BC AB+BC+AC
i.e 2(AB+BC)+AC BC AB+BC+AC
1600 years contain 0 odd day.
300 years contain 1 odd day.
94 years = (23 leap years + 71 ordinary years)
= (46 + 71) odd days
= 117 odd days, i.e., 5 odd days
Days from 1st January 1995 to 28th February 1995
= (31 + 28) days = 59 days
= (8 weeks + 3 days) = 3 odd days
∴ Total number of odd days
= (0 + 1 + 5 + 3) = 9 odd days i.e., 2 odd days.
So, the required day is Tuesday.
$57.30
s1(1+x/100)+s2(1+x/100)+s3(1+x/100)
31
All books can be arranged in 10! ways. A single pair of books can be taken as a unit and arranged among the 8 others in 9! ways. The pair of books can also be interchanged and therefore rearranged in 2! ways. Thus the probability of the pair always being together is (9!*2!)/10!
Given y/x = 1/3, x+2y =10
3y=x
Then substitute x=3y in x+2y=10
3y+2y=10
5y=10
y=2
Then substitute y=2 in x=3y
x=3*2
x=6
: x=6, y=2