In these Maths Picture Puzzles, there are maths equations. However, in each of the equations, the Maths variable has been replaced with pictures. Solve these picture puzzle equations to get the value of the last equation.
Answers to these math riddle equations are given at the end of this post along with a detailed explanation. Please do write your answers and feedback in the comments section of this post.
Answers to these math riddle equations are given at the end of this post along with a detailed explanation. Please do write your answers and feedback in the comments section of this post.
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Click to check Answers to Maths Picture Puzzles Riddles
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The hidden answer to these math picture puzzle riddles is given below. Do select the text between the numbers to read the corresponding answer.
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A1.
Let us use the following symbols:
G................grapes
M................Mango
O................Orange
Given:
G. G. G = 8.......(1)
G. G . M = 12......(2)
M. O = 15.....(3)
From Equation (1),
G = 2
Substituting Value of G in (2)
M = 12/4 = 3
Substituting the Value of M in equation (3)
O = 5.
G . M . O= 30.
A2.
Let us use the following symbols:
Donald ...........D
Micky.............M
Jerry.............J
Given:
D+2M-2J=5.........(1)
2M-2J = 4 ,.......(2)
D+4J= -7.............(3)
Substituting Value of 2M-2J from 2 in 1
D +4=5, D= 1
Substituting Value of D in eq 3.
4J= -7-1, J= -2
Substituting Value of J in 2
2M + 4= 4 , M= 0
D +M+ J= -1
A3. 3
Let us use the following symbols:
Owl..............O
Pigeon...........P
Cock.............C
Given:
2O +2P- 2C= 6...................(1)
2O-2C= 8........................(2)
P+ C= -2.........................(3)
Substituting Value of 2O-2C from 2 in 1
2P = -2
P = -1
Substituting Value of P in 3
C =-1
From equation 2, O = 3
therefore,
O+P-C = 3
A4.
10
Let us use the following symbols:
Dog ..........D
Bear..........B
Tiger.........T
3D+2B = -6...........(1)
12B- 3T = -12............(2)
-2D+15B = 4..............(3)
Multiply eq (1)x 2 and (2) x 3 and adding we get
49B= 0, B=0
Substituting the value of B in (1)
3D = -6 , D= -2
Substituting in (2) T= 4,
Hence , D+15B+3T = 10
A5.
6
Let us use the following symbols:
Pineapple............A
peach...............P
banana...........B
AxB+ PxB= 8...........(1)
A+P = 4...................(2)
4A -2P = -2.............(3)
Multiply (2)x 2 and add (2) and (3)
6A = 6
A = 1,
Substituting the value of A in (2)
P = 3
Substituting the value of A and P in (1)
B+3B = 8
4B= 8
B= 2
A+B+P = 6
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