A sum of 1835 was split among A- B- and C such that A- s share is equal to 11921 times of C- s share and B- s share is 1815 of A- s share- Which of the following is the share of A-

A=11921C A=4021C C=2140A Again, B=1815A Now, nbsp; A + B + C = 1835 A+1815A+2140A=1835 A+2315A+2140A=1835 A ( 120+23×8+21×3120 )=1835 [Distributive pro ...

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A sum of 1835...

Question

A sum of $$ 1835$ was split among $A, B,$ and $C$ such that $A^{'}$ s share is equal to $1 \frac{19}{21}$ times of $C^{'}$ s share and $B^{'}$ s share is $1 \frac{8}{15}$ of $A^{'}$ s share. Which of the following is the share of $A$ ?

$ 700

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$ 920

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$ 600

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$ 315

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Solution

The correct option is C $$ 600$
$A = 1 \frac{19}{21} C$

$A = \frac{40}{21} C$

$C = \frac{21}{40} A$

Again, $B = 1 \frac{8}{15} A$

Now,
$A + B + C = 1835$

$A + 1 \frac{8}{15} A + \frac{21}{40} A = 1835$

$A + \frac{23}{15} A + \frac{21}{40} A = 1835$

$A (\frac{120 + 23 \times 8 + 21 \times 3}{120}) = 1835 [Distributive property]$

$A (\frac{120 + 184 + 63}{120}) = 1835$

$A (\frac{367}{120}) = 1835$

$A = \frac{1835 \times 120}{367}$

$A = 5 \times 120$

$A = $ 600$

Thus, the share of A is $$ 600.$

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A sum of $1835 was split among A,B, and C such that A′ s share is equal to 11921 times of C′ s share and B′ s share is 1815 of A′ s share. Which of the following is the share of A?

The correct option is C $600A=11921C A=4021C C=2140A Again, B=1815A Now, A+B+C=1835 A+1815A+2140A=1835 A+2315A+2140A=1835 A(120+23×8+21×3120)=1835[Distributive property] A(120+184+63120)=1835 A(367120)=1835 A=1835×120367 A=5×120 A=$600 Thus, the share of A is $600.

A sum of $$ 1835$ was split among $A, B,$ and $C$ such that $A^{'}$ s share is equal to $1 \frac{19}{21}$ times of $C^{'}$ s share and $B^{'}$ s share is $1 \frac{8}{15}$ of $A^{'}$ s share. Which of the following is the share of $A$ ?