Let the first number be K and second number be L
Then, 2K + 3L = 100.....(1)
3K + 2L = 120.......(2)
solving (1)&(2), we get
K = 32 and L = 12

Hence the largest number is 32.

From March 6th to November 11th,
there are 250 days and
Thus 250 x 24 hours
So from Mar 6th 9AM to Nov 11th 8PM, there will be 250 x 24 + 11 hours = 6011 hrs.

Let the average bill paid by twenty members = 'x'

But 19 men paid each = Rs. 70

20th man paid Rs. 90.25 more than the avg bill of 20 = x + 90.25

20x = 19(70) + x + 90.25
19x = 1330 + 90.25
19x = 1420.25
x = 1420.25/19 = Rs. 74.75

But the total bill = 20 x 74.75 = Rs. 1495.

If we consider the third term to be ‘x”
The 15th term will be (x + 12d)
6th term will be (x + 3d)
11th term will be (x + 8d) and
13th term will be (x + 10d).
Thus, as per the given condition, 2x + 12d = 3x + 21d.Or x + 9d = 0.
x + 9d will be the 12th term.

Thus, 12th term of the A.P will be zero.

As these three strategies P, Q and R are independent so these will cut cost one after the other.
If initial cost is Rs 100, then
20% cost is cut after initializing strategy P, then cost will remain 80% = 80
further 30% cost is cut after strategy Q, then cost will remain 70% of 80 = 56
further 10% cost is cut after strategy R, then cost will remain 90% of 56 = 50.4

Thus final cost remains 50.4 % of the original cost. Hence net saving is 100 - 50.4 = 49.6 %.

Speed = 108 km/hr
enter a tunnel,, means that the train covers a distance equal to its own length
so (108 x 5/18) = x/4 where x= length of train
hence x = 120 mtrs
to pass completely train has to travel a distance = (x + tunnel length) that is 1120 mtrs.
so (1120/t) = (108x5/18)
t = 37.33 sec

Given monday population = 3000
Per day 20% increase in population.

Tuesday population = 3000 + 3000x20/100 = 3000 + 600 = 3600

Wednesday population =3600 + 3600x20/100 = 3600 + 720 = 4320

Thursday population = 4320 + 4320x20/100 = 4320 + 864 = 5184

So that day is Thursday.

Draw the two chords as shown in the figure. Let O be the center of the circle. Draw OC
perpendicular to both chords. That divides the two chords in half.
So CD = 10 and AB = 14. Draw radii OA and OD, both equal to radius r.
We are given that BC = 5, the distance between the two chords. Let
OB = x.

We use the Pythagorean theorem on right triangle ABO

AO² = AB² + OB²
r² = 14² + x²

We use the Pythagorean theorem on right triangle DCO

DO² = CD² + OC²

We see that OC = OB+BC = x+5, so

r² = 10² + (x+5)²

So we have a system of two equations:

r² = 14² + x²
r² = 10² + (x+5)²

Since both left sides equal r², set the right sides
equal to each other.

14² + x² = 10² + (x+5)²
196 + x² = 100 + x² + 10x + 25
196 = 125 + 10x
71 = 10x
7.1 = x

r² = 14² + x²
r² = 196 + (7.1)²
r² = 196 + 50.41
r² = 246.41
r = √246.41
r = 15.69745202 cm