Question

A company of 7300 soldiers formed of 4 battalions, so that 1/2 of the first = 2/3 of second = 3/4 of 3rd = 4/5 of the fourth are all composed of the same number. The difference between number of soldiers between the battalions with maximum and minimum number is?

Answer 1

1/2a=2/3b=3/4c=4/5d

a+b+c+d=7300

Answer 2

Answer:

The difference between no. of soldiers between the battalions with maximum and minimum no. is 900.

Step-by-step explanation:

Let the no. of soldiers in first, second, third and fourth battalions be p, q, r and s respectively.

Given :

p + q + r + s = 7300

$\frac{1}{2}p = \frac{2}{3} q = \frac{3}{4} r = \frac{4}{5}s$

Rewriting equations in terms of p

q = (3/4) p

r = (2/3) p

s = (5/8) p

Substituting the above values in the first equation, we have

$p + \frac{3}{4}p + \frac{2}{3}p + \frac{5}{8}p = 7300$

$p(1 + \frac{3}{4} + \frac{2}{3} + \frac{5}{8}) = 7300$

$p(\frac{24+18+16+15}{24} ) = 7300$

$p(\frac{73}{24} ) = 7300$

$p = 7300(\frac{24}{73} ) = 2400$

∴ p = 2400

Substituting the value of p in the equations, we have

q = (3/4)p = (3/4)(2400) = 1800

r = (2/3) p = (2/3) (2400) = 1600

s = (5/8) p = (5/8) (2400) = 1500

Of the four battalions, first battalion has the maximum no. of 2400 soldiers and the fourth battalion has the minimum no. of 1500 soldiers.

The difference between no. of soldiers between the battalions with maximum and minimum no. = (2400-1500) = 900

∴The difference between no. of soldiers between the battalions with maximum and minimum no. is 900.