Assume first child (the youngest) get = Rs. x

According to the question ;
each son having Rs. 30 more than the younger one

Second child will get = Rs. x + 30
Third child will get = Rs. x + 30 + 30 = x + 60
Forth child will get = Rs. x + 30 + 30 + 30 = x + 90
Fifth child will get = Rs. x + 30 + 30 + 30 + 30 = x + 120

Total amount they got = Rs. 2000

x + (x+30) + (x+60) + (x+90) + (x+120) = 2000
5x + 300 = 2000
5x = 1700
x = Rs. 340
So the youngest child will get Rs. 340.

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are k, l, m, n & o respectively, then
k + l + m + n = 58 x 4 = 232 ---- (i) &
l + m + n + o = 60 x 4 = 240 ---- (ii)
Subtracting (i) from (ii), we get
o-k = 8 ---(iii)
Given, k/o = 8/9 ----(iv)
So from (iii) & (iv), we get
a = 64, e = 72
Therefore, number of visitors on 5th day is 72.

Average speed = 2aba+b, here a=25 b=4
= 2 x 25 x 4/(25+4) = 200/29 km/hr.

Distance covered in 5 hours 48 minutes
= Speed x time = (200/29)x (29/5) Distance covered in 5 hours 48 minutes = 40kms.

Distance of the post office from the village = (40/2) = 20 km.

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8.

Seats for Mathematics, Physics and Biology in a school are 5x, 7x and 8x.

There is a proposal to increase these seats by 40%, 50% and 75% respectively.

Then Seats for Mathematics, Physics and Biology in a school will be

1.4 X 5x, 1.5 X 7x and 1.75 X 8x and

ratio will be 7 : 10.5 : 14  => 2:3:4.

One day work of 6 boys and 8 girls is given as 6b + 8g = 1/10 -------->(I)

One day work of 26 boys and 48 women is given as 26b + 48w = 1/2 -------->(II)

Divide both sides by 2 in (I) and then multiply both sides by 5

Now we get, 15b + 20g = 1/4.

Therefore, 15 boys and 20 girls can do the same work in 4 days.

We know 60 min = 1 hr
Total northward Laxmi's distance = 20kmph x 1hr = 20 km
Total southward Prasanna's distance = 30kmph x 1hr = 30 km
Total distance between Prasanna and Laxmi is = 20 + 30 = 50 km.

Let initially 'X' ticket has been sold.
So now in 2nd week 20% increases, so
=> X x 120/100
In 3rd week 16% increases, so
=> X x (120/100) x (116/100)
In 4th week 20% decrease, so
=> X x (120/100) x (116/100) x (80/100) = 1392
X = 1250.

Let 'd' be the distance between A and B
K -time = d/10 + d/9 = 19d/90 hours
L -time = 2d/12 = d/6 hours
We know that, 10 min = 1/6 hours
Thus, time difference between K and L is given as 10 minutes.
=> 19d/90 - d/6 = 1/6
=> (19d-15d)/90 = 1/6
=> 4d/90 = 1/6
Thus,
d= 15/4 km = 3.75 km.

hence the distance between A and B is 3.75 km.