Mary had some 20cent and some 50 cent coins. The number of 20cent coins is 3/7 of the number of 50 cent coins she has. She saved another 20 coins of each type of coins. Now the number of 20cent coins is 7/11 of number of 50cent coins. How many 20cent coins does she has at first?
Note: my bro doesnt understand algebra methods. Can someone solve it using heuretics or model approach? Thanks a lot
Originally posted by ultimatenolifer:Mary had some 20cent and some 50 cent coins. The number of 20cent coins is 3/7 of the number of 50 cent coins she has. She saved another 20 coins of each type of coins. Now the number of 20cent coins is 7/11 of number of 50cent coins. How many 20cent coins does she has at first?
Note: my bro doesnt understand algebra methods. Can someone solve it using heuretics or model approach? Thanks a lot
20 cent l__l__l__l__l__l__l__l
50 cent l__l__l__l__l__l__l__l__l__l__l__l
Black = Original
Purple = The 20 coins
Comparing the two, you will realise that 20 coins = 4 units.
Thus, 1 unit = 5 coins.
Amount of 20 cent coins at first = 3 units = 3 X 5 = 15 coins