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# Mensa Math/Logic Problems...How to Solve?

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So many, don't have the energy to go through them all...

The "find the missing number" type of questions don't usually have a strategy, you just stare at them and try to draw connections, look for patterns like for example are the numbers rising or falling, usually the answer should be very simple like taking a number and summing it's two digits and stuff like that.

The sum of every 4 circles in a line is 21

Take the numbers in the left and right squares and reverse them then sum them, so for first row 31+12 becomes 14+21 which is 35

http://en.wikipedia.org/wiki/Binomial_coefficient

Also the chosen answer is wrong, it should be 120

We simply skip prime numbers.

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Multiply the numbers in the left and right circles then reverse the digits.

Replace the shapes with numbers, then write the numbers in English: six, three, five, four, four.

The next number is the number of letters in the previous number

Each circle is the XOR of the two circles below it.

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Thanks for all the answers, esp. 11/20! I forgot that equation from high school math.

I have an alternate solution for 9/20, the shapes problem.

6 3 5 4 4

\ \ / \ / /

\ \ / \ / /

\ \ / \ / /

\ \/ \ / /

\ /\ \ / /

\ / \ / \ /

\ / \ / \ /

-1 \ +1 -1 ? ----> +1 (so 4+1 = 5)

So are there multiple solutions to these problems?

Edited by marcussoori
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For the second 9/20, answer is 62? Multiply 2 numbers outside of circle, subtract the multiplication of the two remaining numbers, and you have the number in the middle of the circle.

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For the first image, 4/20, answer is 75. For each row, add the two numbers, and then reverse the digits. If the addition results in a number over 100, add the 1 to the last digit, then reverse the resulting two digit number.

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For the third image, 8/20, isn't there an error on the last row of the problem? I am using the XOR method you described (add images and remove overlapping areas). Anyway, I believe the solution is the fifth option.

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I still can't figure out 15/20...anyone?

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I also can't solve the 2nd 4/20

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For the 2nd 8/20, is the answer 7?

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Any solution for the 3rd 8/20?

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And what is the method to solve problems like 3/20?

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72; Add one to the next prime number

Edited by DeGe
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25; Sum of numbers in one area is 3x the sum of numbers in the area opposite to it

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*

First 4/20: 75
Add first & last numbers in each row, reverse the digits of the sum if below 100, and resulting number occupies the center place in the row. If the sum of the first and the last number in a row is more than 100, then remove the first digit and add it to the remaining two digit number and then reverse the digits of the number so obtained.
*First 5/20: 3
Sum of all the numbers, on each side of the triangle formed by the circles, is equal to 21.
*First 8/20: Answer is the last pentagon in the options given.
When two pentagons are overlapped, those circles within pentagons disappear which overlap each other. The other circles occupy their respective places, and form the figure on top of these two circles.
Note: Second pentagon in the last row does not seem to follow this logic. This pentagon may occupy only three circles then the above logic follows perfectly.
* First 10/20: 67
Add last digits of first and last numbers in each row - say the result is X, then add first digits of the first and last numbers of the row – say the result is Y, then the middle number in the row is XY.
* First 11/20: 120
[This can be solved with the help of simple mathematics. I adopted rather longer procedure. If someone asks for it I will post it.]
* First 12/20: 30
The prime numbers are not included in the series.
* Second 5/20: 8
Product of the first and the third number in each row forms the second number but with the digits of product are reversed.
* First 15/20: 72
Next number is the number obtained by adding ‘1’ to the next prime number.
* Second 4/20: 25
Sum of all the numbers in one sector of the circle is three times the product of all the numbers in another sector just opposite to it.
* Second 8/20: 7
Sum of the two larger numbers is divided by 9, to get the third smaller number.
* First 9/20: Again Rectangle.
Writing the number of lines (contained by each figure) in words, next figure is seen to contain the number of lines equal to the number of letters in the previous word.
* Third 8/20: 20
1st Fig: 14 – 12 = 2 and 11 – 9 = 2, then 17 + 2 + 2 = 21 (the number in centre).
2nd Fig: 6 – 5 = 1 and 6 – 5 = 1, then 18 + 1 + 1 = 20 (the number in centre)
Similarly, 3rd Fig: 6 – 8 = (-) 2 and 5 – 2 = 3, then 19 - 2 + 3 = 20.
* Second 9/20: 62
Multiply the numbers in front of each other, in the outer circles. And deduct smaller product from the larger product to get the number in the center of the circle.
* First 14/20: First figure of the options given.
When two figures overlap each other, then if any circle (except the outermost), line or white dot overlaps another, it disappears, while the other ones occupy their respective places. The figure so formed is the figure on top of these two figures.
* Last 13/20: 31
Add all the numbers on the corners of the triangle, reverse the digits of the sum, number obtained occupies the center space of the triangle.

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*Last but one (3/20): 83

Multiply the 2nd and 3rd number of a row.

Reverse the digits of first number in this row. Subtract the number so obtained from the product obtained above. Result is the fourth number of this row.

Edited by bhramarraj
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Spoiler 8/20

I think it's 7.(23+1)/6=4 (31-1)/6=5

(27-2)/5=5 (18+2)/5=4

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