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?

How did you do?

About Cassy

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

Comments

  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.
    🙂

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