TRICKY MATHS PUZZLES – WITH ANSWERS – PART – II
Here in this blog we are discussing about TRICKY MATHS PUZZLES – WITH ANSWERS – PART – II . Some of them are illustrated below :
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
Three red balloons = Rs. 21
3 = 21
Then the cost of one red balloon = Rs. 7
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
Thus the cost of one pair of blue balloon = Rs. 12
We already knew that the cost of
=> = 7
+ + = 31
one pair of blue balloons is Rs. 12
Here the cost of one pair blue and two orange balloons = Rs. 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
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 :
( TRICKY MATHS PUZZLES – PART – I )
https://payilagam.com/blogs/tricky-maths-puzzles-part-i/
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COURTESY : https://www.youtube.com/results?search_query=maths+tricky+puzzles
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