Find out the correct number

Find out the correct number to replace the question mark.

Source: Beijing Aidi International School summer test (2018-08-21). Author: 坦途国际教育

• The hats on the third row and the hat on the fourth row are different. Is that intentional? Oh hang on: its two hats isn't it. – JimM Sep 20 at 7:44

You are correct. The answer is 9.

Explanation

The first line has 3 bears with hats totaling 21. So a bear with a hat is 21/3 = 7.

Then

The next line is 2 sets of a bear in a bus plus a bear in a hat totaling 19. So we know a bear in a bus is (19-7)/2 = 6.

Then

Next is a hat, a bear in a bus, and a bear in a hat totaling 15. So we know a hat is 15-6-7=2.

Then

Now that we know a hat is 2, a bear must be 5 (7-2). If a bear is 5, than a bus is 1 (6-5)

Thus

A hat is 2, a bear is 5, and a bus is 1. That makes the last line 5 + 4*1 which does equal 9.

• Apparently, this is the earliest correct answer. It deserves more votes. – xhienne Sep 19 at 22:14
• Yeah I think the asker saw I had left a comment before his was posted saying I’d found the error and was fixing it and was just doing formatting so that’s why it was posted before and hadn’t decided who to give it to. Or maybe I’m just reading into things too much. I am fine either way fair is fair. – gabbo1092 Sep 20 at 1:47

Adding neater formatting but here is solution:

The key to this puzzle is attention to detail about what each image truly is

For the first line:

We have 3 bears and 3 hats. This adds up to 21 meaning a bear + a hat = 7.

Next we have to look at the third line:

In this line we have a bear with a hat (7), a hat and a bus with a bear. In total then we have a bus of unknown value and 2 bears with hats valued at 7 each. To get the total sum of 15 an empty bus must equal 1 (15-(2x7)).

Back to the second line now:

There are 2 buses with a hat-less bear and one bear with a hat (7). Since each bus is worth 1 and the bear with hat is worth 7, subtracting 9 from 19 leaves 10 for the value of two bears and 5 for the value of each individual bear.

Value of a hat:

Since a hat and a bear is 7 and a bear is 5, a hat is worth 2.

Final Line:

To recap: A bear without hat is 5, two hats will be 4 and a bus is 1. 1x4 (BEDMAS/PEDMAS) is still 4. All that is left is 4 + 5 which equals 9.

• Nope you didn't notice the last equation properly. Bear is hatless – Unza Sep 19 at 20:52
• I found a different error fixing in updated solution – gabbo1092 Sep 19 at 20:57
• If the individual bear is worth 4 then according to the second line the sum should have been 17 not 19. (4×3)+2+3=17 – Unza Sep 19 at 21:03
• Yeah as I mentioned in my more recent comment I caught and was fixing that. That was fixed in my most recent update. – gabbo1092 Sep 19 at 21:05
• Let me object: the total of the two hats on last line is 22 (youtube.com/watch?v=Zh3Yz3PiXZw) – xhienne Sep 19 at 22:08

Let t be the number of teddy bears, h hats and b buses. Then the equations are:

$3t+3h=21 \tag1$
$3t+h+2b=19 \tag2$
$2t+2h+b=15 \tag3$

and we want to know:

$t+(2h\times b)=t+2hb$

Gaussian elimination tells us the answer is:

$t=5, h=2, b=1$, therefore $t+2hb=9$.

but there is a quick way:

From (1) $t+h=7$. From (3) $b=1$. From (2) and (3), $t-h=3 \implies t=5, h=2$.

Pig (with a hat) = 7
Bus (with a pig) = 6
Hat = 2
Pig = 5
Bus = 1
Hence pig (without hat) + (2 * hat) * bus (without a pig) = 5 + 4 * 1 = 9