# Word Problem Wednesday – Pears and Oranges

Summer’s here, but you’re missing your math? Don’t despair – we’ve got you covered. Check the site each week for one whopper of a word problem that’s sure to challenge!

This week’s problem comes from Challenging Problems in Primary Schools – Intermediate by Dr Y H Leong, published in 2004 by SNP Panpac Pte Ltd.( Intermediate is for students in Primary 4 and 5.)

##### 3 pears and 4 oranges cost \$3.80. If 1 pear and 1 orange together cost \$1.10, find the cost of 1 pear.

Submit your solutions and we’ll post all interesting strategies next week.

Last week’s problem and solution:
Gavin has 356 cards. He has 286 more cards than Howie. How many cards must Gavin give to Howie so that they both have the same number of cards?

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Passionate about Singapore Math + Teacher Trainer and Coach + Treasure Hunter + Learner. Answer to the ultimate question? 42.

1. I don’t know how to draw bars on the computer…

PO=\$1.10
PPPOOOO= \$3.80
PPPOOO=\$3.30 (3 x 1.10)
therefore one O = .50 (\$3.80-\$3.30)

Back to original…
PO =\$1.10
if O=..50, P=.60 (1.10-.50)

check: 3x.60=1.80 (pears)
4x.50= \$2.00 (oranges)
\$2.00 + \$1.80= \$3.80

One pear costs 60 cents.

And whoever “Americanized” this should have used mangoes because O’s look like 0’s in the model.
🙂

• Cassy says

Hi Cindy! I took it word for word from the book published in Singapore. Perhaps mangosteens and jackfruits next time?

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