a week ago Grade -10 (SEE) Math

Find the values of B1, B2, B3 and B4 from the box – Math Puzzle

Find the value of B1, B2, B3 and B4 from the figure below where B1, B2, B3 and B4 satisfies the given equations!Math Puzzle

2 Answers

a week ago Tribhuvan University > Institute of Science and Technology > B.Sc. (CSIT)

Let’s quickly write the equations

B1 + B2 = 14 

B1 + B3 = 15 

B2 + B4 = 16 

B3 – B4 = 10 

by adding all equations

B1 + B2 + B3 = (14 + 15 + 16 + 10 )/2 = 55/2 = 27.5

B3 = 27.5 – (B1 + B2) = 27.5 – 14 = 13.5 => B3 = 13.5

B2 = 27.5 – (B1 +B3) = 27.5 – 15 = 12.5 => B2 = 12.5

B1 = 14 – B2 = 14 – 12.5 = 1.5 => B1 = 1.5

B4 = B3 – 10 = 13.5 – 10 = 3.5 => B4 = 3.5

a week ago

For easier let us suppose B1, B2, B3 & B4 be a, b,c & d respectively.

here we have

a + b = 14 .......... (i )

a + C = 15..........(ii)

b + d= 16...........(iii)    

 

c - d = 10 ..........(iv) this equation is considering as a focused equation for now.

 

from equation i

a + b = 14     can be written as

a + b = 10  + 4    

a + b = c-d + 4   ( from equation iv) we gonna use this in equation ii & iii in similar way.

a + b - c + d = 4  .............. (v)

Similarly,

             Combining ii & iv.

we will get as,

   a + d = 5 ...........(vi)

 

Again,

             Combining iii & iv.

we will get as,

   b - c + 2d = 6 ...........(vii)

 

Now from equation v & vi, we get,

       b - c = -1

i.e.  c= 1 + b  ..... (viii)

Substituting value of b in equation vii, we get

i.e.  b - 1 -b + 2d = 6

       2d = 7

        d =  7/2

Substituting value of d in equation iii

a =16 - 7/2

   = (32-7)/2

  a = 25/2

 

Substituting value of a in equation i

b = 14 - 25/2

   = (28 - 25)/2

b = 3/2

Similary while substituting value of b in equation ii 

we will get

c = 27/2.

 

Therefore, value of B1, B2, B3 & B4 is 3/2, 25/2, 27/2 & 7/2 respectively.

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