Question

Divide 581 into three parts such that 4 times the first may equal to 5 times the second and 7 times the third?

Answer 1

QUESTION :-

Divide 581 into three parts such that 4 times the first may equal to 5 times the second and 7 times the third then find the three parts

GIVEN :-

• 581 is divided into three parts and each part is related as 4times the first part is equal to 5 times the second part and 7 times the third part

TO FIND :-

• The three parts

SOLUTION :-

Let the first part be 'a'

Let the second part be 'b'

Let the third part be 'c'

Then ,

• Four times the first part = 4a

• Five times the second part = 5b

• Seven times the third part = 7c ,

We are given ,

• 4a = 5b = 7c

$\sf {\underbrace{ {4a = 7c}}} = 5b \\ \\ \implies \sf \: 4a = 7c \\ \\ \implies {\underline{ \boxed {\pink{ \sf {\: c= \frac{4a}{7} }}}}} \longrightarrow \: eq(1)$

$\implies \sf \: 7c= \underbrace{5b= 4a} \\ \\ \implies \sf \: 5b = 4a \\ \\ \implies {\underline {\boxed {\pink{ \sf {\: b = \frac{4a}{5} }}}}} \longrightarrow eq(2)$

Here The sum of these three parts is equal to 581

$\implies \sf \: a+ b + c = 581 \\ \\ \implies \sf \: \frac{4a}{5} + \frac{4a}{7} + a = 581 \\ \\ \implies \sf \: \frac{5(4a) + (4a)7}{35} + a = 581 \\ \\ \implies \sf \: \frac{20a + 28a}{35} + a = 581 \\ \\ \implies \sf \: \frac{48a}{35} + a= 581 \\ \\ \implies \sf \: \frac{48a + 35a}{35} = 581 \\ \\ \implies \sf \: \frac{83a}{35} = 581$

$\implies \sf \: 83a = 581 \times 35 \\ \\ \implies \sf \: a= \frac{581 \times 35}{83} \\ \\ \implies {\underline {\boxed {\blue{\sf {\: a= 245}}}}}$

From eq(1) ,

$\large \sf \: c = \frac{4a}{7} \\ \\ \implies \sf \: c = \frac{4(245)}{7} \\ \\ \implies {\underline {\boxed {\blue {\sf {\: c= 140}}}}}$

From eq(2) ,

$\sf \: b = \frac{4a}{5} \\ \\ \implies \sf \: b = \frac{4(245)}{5} \\ \\ \implies {\underline {\boxed {\blue {\sf {\:b = 196}}}}}$

∴ The three parts are 245 , 196 and 140

VERIFICATION :-

(1) We have got ,

• a = 245
• b = 196
• c = 140

Four times first part = 245(4) = 980

Five times the second part = 196(5) = 980

Seven times the third part = 140(7) = 980

4a = 5b = 7c ✓, First condition is satisfied

(2) Sum of these three numbers must be 581

$\implies \sf \: 245 + 140 + 196 = 581$

Second conditions is also satisfied

$\dag\huge {\blue {\boxed {\green {\mathbf{HENCE\:VERIFIED}}}}}$

Answer 2

Answer:

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