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Cryparithmetic Division Problem 6

Solve this Problem, the first person to solve this problem and give the best detailed Solution to find all values will get 500Rs from PrepInsta. To submit answers text us on Facebook our Facebook page here – www.facebook.com/prepinsta

       A C Q S
_____________
Q B A |B Z Q A R C
X P C
-------
S Y B A
S Z Y P
-------
S A C R
S S X Y
----------
Y C C
Q B A
-------
A Z A

Step 1

The problem can be broken into –

Eq 1

  Q B A
x A
-------
X P C

and

Eq 2

  Q B A
x C
-------
S Z Y P

Eq 3

Q B A
x Q
-------
S S X Y

Eq 4

  Q B A
x S
-------
Q B A
       A C Q 1
_____________
Q B A |B Z Q A R C
X P C
-------
1 Y B A
1 Z Y P
-------
1 A C R
1 1 X Y
----------
Y C C
Q B A
-------
A Z A

Values:
S = 1, A = 2

Step 2

Eq 4

  Q B A
x     S
-------
  Q B A

If we look at equation 4 it is clear that S = 1.

Also,

 1 A C R
 1 1 X Y
--------
     Y C

A – 1 = 0

Now, this is only possible if A = 2

  • If step C – X is borrowing 1 from A.
  • A was 2 and 1 borrowed so A becomes 1
       2 C Q 1
       _____________
Q B 2 |B Z Q 2 R C
       X P C
       -------
       1 Y B 2
       1 Z Y P
         -------
         1 2 C R
         1 1 X Y
         ----------
             Y C C
             Q B 2
              -------
             2 Z 2

Values:
S = 1, A = 2, C = 4

Step 3

Eq 4

  Q B A
x     A
-------
  X P C
  • A x A = C
  • Thus C = 2 x 2 = 4

Also, from following subtraction –

  1 Y B 2 
- 1 Z Y P
-------
1 2 C

After replacing C = 4

  1 Y B 2 
- 1 Z Y P
-------
1 2 4

The only way 4 can be obtained is –

  • If, P = 8
  • 2 takes borrow from B and 12 – 8 = 4
       2 4 Q 1
       _____________
Q B 2 |B Z Q 2 R 4
       X 8 4
       -------
       1 Y B 2
       1 Z Y 8
         -------
         1 2 4 R
         1 1 X Y
         ----------
             Y 4 4
             Q B 2
              -------
             2 Z 2

Values:
S = 1, A = 2, C = 4, P = 8

Step 4

Eq 4

  Q B 2
x     2
-------
  X 8 4
  • Now, B x 2 = _8
    • This, is only possible when
    • B = 2 (Not possible A = 2)
    • B = 9
    • 9 x 2 = 8 (1 carry to next step)

Also, Q x 2 + 1 (carry from previous step)= X

  • Thus, resultant of above is single digit
  • Possible values for Q to make this happen

Q = {1, 2, 3, 4}

  • Values, 1, 2 and 4 are already taken by S, A and C respectively
  • Thus, Q = 3

Also, X = (Q x 2) + 1 carry = 7

X = 7

       2 4 3 1
       _____________
3 9 2 |9 Z 3 2 R 4
       7 8 4
       -------
       1 Y 9 2
       1 Z Y 8
         -------
         1 2 4 R
         1 1 7 Y
         ----------
             Y 4 4
             3 9 2
              -------
             2 Z 2

Values:
S = 1, A = 2, Q = 3, C = 4, X = 7, P = 8, B = 9

Step 5

Eq 4

  3 9 2
x     4
-------
1 Z Y 8
  • Clearly, Y = 9 x 4 = 6 (3 carry to next step
  • Y = 6
  • Also, Z = 3 x 4 + 3(carry) = 12 + 3 = 15
  • Z = 5

Finally, step 

 1 2 4 R 
1 1 7 6
----------
6 4

Here,

  • R – 6 = 4
  • Only possible when R takes 1 borrow from 4
  • (1R) – 6 = 4
  • Thus, R = 0
       2 4 3 1
_____________
3 9 2 |9 5 3 2 0 4
7 8 4
-------
1 6 9 2
1 5 6 8
-------
1 2 4 0
1 1 7 6
----------
6 4 4
3 9 2
------
2 5 2

Final Step 

These are the final values for equation.