## TRICKY MATHS PUZZLES –  WITH ANSWERS – PART – II

### We have already discussed about some problems in the previous blog TRICKY MATHS PUZZLES – PART – I .

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1.   SOLVE THE GIVEN TRICKY QUESTION .

GIVEN : The cost of three red balloons = Rs. 21

The  cost of one red and two pairs of blue balloons = Rs. 31

The cost of one pair of blue balloons and two orange balloons = Rs. 20

+              +                      =   21

+              +                         =    31

+      +              =   20

+                  x        =   ?

+             +           =  21

Three red balloons = Rs. 21

3                =   21

Then the cost of one red balloon = Rs. 7

=>           =  7

+                 +               =  31

Again the cost of one red and two pairs of blue balloons = Rs. 31

Also the cost of one red balloon = Rs. 7

So the cost of two pairs of blue balloons = 31 – 7

7      +                  +              =  31

2            = 31 – 7

= 24                              = 12

Thus the cost of one pair of blue balloon = Rs. 12

We already knew that the cost of one pair of blue balloons  is Rs. 12

Here the cost of one pair blue and two orange balloons = Rs. 20

+                +               =   20

12    +               +              =   20

Hence the cost of two orange balloons =  Rs . 20 – Rs . 12

=  Rs. 8

2               =  20 – 12   =   8

=  4

So the cost of orange balloon = Rs.4

+                    x                  =   ?

Coming to the given question , what is the cost of above ?

Cost of red balloon                = Rs. 7

Cost of pair of blue  balloon = Rs. 12

Cost of orange balloon         = Rs. 4

7      +        12      x        4         =   ?

7      +      (  12    x   4   )          =   ?

7    +    48      =  55

Thus the cost of the given balloons = Rs. 55

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REFERENCES : https://youtu.be/PIMaHCt2wL0

For further details visit the page given below :

#### https://payilagam.com/blogs/tricky-maths-puzzles-part-i/

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