Here from 4th December 1999 to 3rd January 2000,
there is a difference of 30 days.
Number of odd days = 30/7 = 2 odd days

Then no. of odd days = 2 => tuesday and wednesday(from monday).

Hence on 3rd january it is wednesday.

Let number of 20 ps coins = x and

number of 25 ps coins = y

Given total coins in the bag = 50

x + y = 50.......(1)

But the total money in the bag = Rs. 10.25

0.20x + 0.25y = 10.25

20x + 25y = 1025.........(2)

Now multiplying (1) by 25 we get

25x+25y=1250.............(3)

By solving (2) and (3)

20x + 25y = 1025;

=> x = 45;

Then, the no. of 20 ps coins are 45.

Let the the age of the elder boy = E

Let the the age of the younger boy = Y

Given that Y = cube root of EY

=> Y3 = EY => E = Y2 .....(1)
By the condition of number replacement the age of the father is YE

The Mother's age = EY/2

But she is 3 years less than father => EY/2 + 3 = YE
2YE = EY + 6 ......(2)

Then now from the given options we can identify which satisfies the all the conditions.

Here Y =2 and E = 4 satisfies all the conditions.

Let there are L men

job requires LK man hours.

job completed in first 4 hrs = L x 4 = 4L

job completed in next 4 hrs = 4 x L/2 = 2L

job completed in next 4 hrs = 4 x L/4 = L

job completed in last 4 hrs = 4 x L/8 = L/2

4L + 2L + L + L/2 = KL

K = 7+1/2 = 7.5 hours.

The market was full of people wearing attractive clothes of different styles.

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