Question

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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

581 is divided into x,y,z such that

4x=5y=7z

From (1) 4x=5y

4/5x=y ___(2)

From (1)4x=7z

4/7x=z ___(3)

581=x+y+z

581=x+4/5x+4/7x

581=35x+28x+30x/35

581=83x/35

83x=20335

x=245

From (2)

4/5x=y

4/5×245=y

196=y

From (3)

4/7x=z

4/7×245=z

140=z

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