Question

Divide 3740 in three parts in such a way that half of the first part,one-third of the second part and one-sixth of the third part are equal

Answer 1

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Answer 2

Divide 3740 in three parts in such a way that half of the first part,one-third of the second part and one-sixth of the third part are equal is 680, 1020, 2040

Solution:

Given that we have to divide 3740 in three parts in such a way that half of the first part, one-third of the second part and one-sixth of the third part are equal

Let the first part be "a"

Let the second part be "b"

Then the third part is 3740 - (a + b)

By given information,

Half of first part = one-third of the second part = one-sixth of the third part

$\frac{a}{2} = \frac{b}{3} = \frac{3740 - (a + b)}{6}$

Equating first and third,

$\frac{a}{2} = \frac{3740 - (a + b)}{6}\\\\3a = 3740 - a - b\\\\4a + b = 3740$

4a + b = 3740 ---- eqn 1

Equating second and third,

$\frac{b}{3} = \frac{3740 - (a + b)}{6}\\\\2b = 3740 - (a + b)\\\\2b = 3740 - a - b\\\\a + 3b = 3740$

a + 3b = 3740 --------- eqn 2

Multiply eqn 2 by 4

4a + 12b = 14960 ---- eqn 3

Subtract eqn 1 from eqn 3

4a + 12b = 14960

4a + b = 3740

(-) ------------------

11b = 11220

b = 1020

Substitute b = 1020 in eqn 1

4a + 1020 = 3740

4a = 2720

a = 680

First part = a = 680

Second part = b = 1020

Third part = 3740 - (a + b) = 3740 - (680 + 1020) = 2040

Thus 3740 is divided into 680, 1020, 2040

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