Merchant Taylor Maths Test Question 31

July 31, 2015 Madeleine K

How to do this 11+ simultaneous equations problem and similar questions.


From Merchant Taylor's 11+ Maths Sample Paper

Available on their website


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This particular type of maths problem comes up on most 11+ exams. It is, essentially, simultaneous equations. If your student has a strong grasp on algebra, then teaching them how to do simultaneous equations will help with this problem type. However, it’s not necessary. Here’s how to solve the last question on the Merchant Taylor’s sample paper, which can be found on their website.


First, it’s absolutely key to recognise that you can work out the price of 1 drink.

Help your student to see this by having them look at the difference in stuff you buy (2 crisps and 2 drinks versus 2 crisps and 1 drink), and the difference in cost. Now point out that these difference are equal (in other words, the 1 extra drink costs the difference in price, 40p).


Now that you know the price of a drink, you can work out the price of crisps.

2 crisps + 1 Drink = 1.10. Subtract 1 drink (40p).

2 crisps = 70p. Divide by 2 to find 1 crisp

1 crisp = 35p.


Now that you know the price of each item, you can solve the problem in 2 ways.


Method 1 (Preferred)


Multiply the price of the single items by how many you need (price of crisps x3, price of drink x4).

35px3= 1.05

40px4= 1.60


Now add the total price of 3 crisps to the total price of 4 drinks.

1.05+1.60= 2.65


Method 2


It’s also possible to solve this with less multiplying and more adding.

You know the price of 2 packets of crisps (70p), so once you work out the price of 1 (35p), add this to the price of 2.

.70+.35 = 1.05


The drink price requires a few more steps. You can take the total of 2 crisps and 2 drinks, and subtract the price of 2 crisps (1.50-.70) to get the price of 2 drinks (80p).

Now double that price for 4 drinks

.80 +.80=1.60




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