Paige’s investment in his savings account matured to $4,517.27 at the end of 135 days.
If the account was earning simple inte

1 thought on “Paige’s investment in his savings account matured to $4,517.27 at the end of 135 days.<br />If the account was earning simple inte”

1. [tex] {\pmb{\underline{\sf{ Required \ Solution … }}}} \\ [/tex]

   As We know that Paige’s invest as:

   • Amount = $4,517.27
   • Time (t) = 135 days (⅓ years)
   • Rate (r) = 3.50% (7/2)

   If the account was earning simple interest then we’ve to apply the Simple Interest Formula to obtain such results as:-

   [tex] \circ {\underline{\boxed{\sf{ Interest_{(Simple)} = \dfrac{PRT}{100} }}}} \\ [/tex]

   Let the Principal be x

   so, As We know that we have Amount of investment, then:-

   [tex] \colon\implies{\sf{ Interest = Amount – Principal }} \\ \\ \colon\implies{\sf{ 4,517.27-x _{(Interest)} }} \\ [/tex]

   [tex] \circ {\pmb{\underline{\sf{ According \ to \ Question: }}}} \\ \\ \\ \colon\implies{\sf{ 4,517.27-x = \dfrac{x \times 1 \times 7}{100 \times 3 \times 2} }} \\ \\ \\ \colon\implies{\sf{ 4,517.27-x = \dfrac{7x}{600} }} \\ \\ \\ \colon\implies{\sf{600( 4,517.27-x) = 7x }} \\ \\ \\ \colon\implies{\sf{ 2710362 – 600x = 7x }} \\ \\ \\ \colon\implies{\sf{ 2710362 = 600x+7x }} \\ \\ \\ \colon\implies{\sf{ 2710362 = 607x }} \\ \\ \\ \colon\implies{\sf{ \cancel{ \dfrac{2710362}{607} } = x }} \\ \\ \\ \colon\implies{\sf{ x = \$ \ 4465.17 }} [/tex]

   Hence,

   Paige’s initial investment was $ 4465.17.

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