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

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

_______________________________________________________________________

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

_______________________________________________________________________

REFERENCES : https://youtu.be/PIMaHCt2wL0

For further details visit the page given below :

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

_______________________________________________________________________