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2 Bumps

# Any math experts?

Wondering if anyone on here could solve this?

Janet invested \$26,000 part at 6% and part at 3%. If the total interest at the end of the year is \$1080, how much did she invest at 6%?

Asked by Anonymous at 10:31 PM on Mar. 6, 2011 in Just for Fun

• ouch my head hurts, Janet should have gone on a spending spree instead

Answer by myheartx4 at 10:34 PM on Mar. 6, 2011

• LOL do your own math homework. This is not just for fun.

Answer by parajumper3 at 10:36 PM on Mar. 6, 2011

• 0.06 x m + 0.03 x (26000 - m) = 10 800. Solve for m. (m is the about invested at 6%)

Answer by judimary at 10:43 PM on Mar. 6, 2011

• Assuming they are straight interest, no componding. \$16000 at 3% and \$10,000 at 6%. The formula you are solving for is (26000-X) * 6% + X*3% = 1080. You solve for X.

Answer by Olivia4116 at 10:47 PM on Mar. 6, 2011

• HUH ???

Answer by myheartx4 at 10:54 PM on Mar. 6, 2011

• In other words, she invested \$1 080 at 6% pa. (assuming simple interest).

Answer by judimary at 11:18 PM on Mar. 6, 2011

• You're wrong, Olivia4116, that is the equation for finding how much was invested at 3%.

Answer by judimary at 11:20 PM on Mar. 6, 2011

• Sorry, my final answer above is wrong. You solve .06 x m + 0.03(26000 - m) = 1080 for m, and you get m is \$10 000.

Answer by judimary at 11:26 PM on Mar. 6, 2011

• Hahaha.....not just for fun. Could solve this, but it makes me want to do something else....like cry. LOL. Good luck.

Answer by Kword at 12:21 AM on Mar. 7, 2011

• What? Did you mention something abt shopping? Unable to read your query...... Okay jokes apart my Maths is only okay; hoped it to be otherwise... I certainly hope one of us to help you.... :-)

Answer by AnuMeha at 1:32 AM on Mar. 7, 2011

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