Who Can Solve This Tricky Mathematics? - Education - Nairaland
Nairaland Forum › Nairaland General › Education › Who Can Solve This Tricky Mathematics? (568 Views)
| Who Can Solve This Tricky Mathematics? by Stanweezy(op): 1:46pm On Sep 25, 2023 |
Who can solve this tricky illustration, gurus in the house let's see how good you are in mathematics.
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| Re: Who Can Solve This Tricky Mathematics? by meum: 1:52pm On Sep 25, 2023 |
27kg. ✅ This one no be guru dem mathematics |
| Re: Who Can Solve This Tricky Mathematics? by Stanweezy(op): 3:18pm On Sep 25, 2023 |
meum:Show all working |
| Re: Who Can Solve This Tricky Mathematics? by meum: 4:23pm On Sep 25, 2023 |
Stanweezy:For what you should have added options A,B,C,D & E E for us to choose from? Let me give it a shot. Assuming weight of Cat & Mouse are 7kg & 3kg respectively, Then dog will be 17kg since Dog & mouse equals 20kg. From the above values, weight of dog + weight of cat equals 24 (I.e. 17kg + 7kg) Therefore the weight of all three animals(mouse, cat & dog) equals 27kg (I.e. 3kg + 7kg + 17kg) QED😁 |
| Re: Who Can Solve This Tricky Mathematics? by Stanweezy(op): 4:51pm On Sep 25, 2023 |
meum:Hahahaha In mathematics we don't assume, you have to show how you arrived at the individual weight of those animals |
| Re: Who Can Solve This Tricky Mathematics? by akwesenana: 4:54pm On Sep 25, 2023 |
C= Cat D= Dog R = Rat. C + R = 10. D + R = 20. C + D = 24. C = 10 - R D = 20- R C + D = (10-R)+(20-R) I know say I dey do rubbish. |
| Re: Who Can Solve This Tricky Mathematics? by meum: 4:59pm On Sep 25, 2023 |
Stanweezy:Do you know about “if statements” & “let X be…” in computer programming & mathematics? These are assumptions. So assumptions are valid as long as you can prove your final answer based on it |
| Re: Who Can Solve This Tricky Mathematics? by Stanweezy(op): 5:01pm On Sep 25, 2023 |
meum:If you are using X, Y and Z or any other letter for equation, you have to show how you arrived at your answer In mathematics theory question, the instruction states "show all working" |
| Re: Who Can Solve This Tricky Mathematics? by Chuks0485: 7:49pm On Sep 25, 2023 |
Let x be mouse Let y be cat and z be dog. From the diagram x+y=10 ----------(1) x+z=20 -----------(2) y+z=24 -----------(3) (1) + (2) ===>. 2x + y + z= 30 -------(4) Put (3) in (4) ===>. 2x + 24= 30 ===> 2x=30 -24=6 .•. x= 3 Put x =3 in (1) Gives y= 10 - 3=7 Put y=7 in (3) Gives z=24 - 7 = 17 So x+y+z = 3 + 7 + 17= 27 |
| Re: Who Can Solve This Tricky Mathematics? by immortalmortal: 6:28am On Sep 26, 2023 |
D |
| Re: Who Can Solve This Tricky Mathematics? by Calabar1stSon: 6:47am On Sep 26, 2023 |
Las las na olodo be una president now and forever more.
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| Re: Who Can Solve This Tricky Mathematics? by Auraking: 10:24am On Sep 26, 2023 |
27kg. Cheap Data on my signature |
| Re: Who Can Solve This Tricky Mathematics? by willzy0909: 7:34pm On Sep 26, 2023*. Modified: 8:15pm On Sep 26, 2023 |
Aba The sum of the first three pictures is 2times the animals in the questions picture. If we sum the 3 picture and their weights, we get 2dogs, two cats and two rats which their sum is 57. This simply means the 1 dog, 1 cat and 1 rat will be half of 54.... Which is 27. It's a no brainer... |
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