Math, asked by chsndreshwar99, 1 year ago

A,B,C and D are positive integers less than 10.A is 3 times of B.The difference between B and D is 2. C is 1 less than D.Which is the smallest number?


Anonymous: Actually there are no such integers which satisfy above conditions
Anonymous: If 0 was included then integers would be A=9 B=3 C=0 D=1
geornegoblins: yeah ur right

Answers

Answered by TPS
0
A is 3 times B, thus B is smaller than A
B = A/3 
So B can be 1,2 or 3 as A is less than 10.

Difference between B and D is 2
D can be either B+2 or B-2.

If B=1 or 2, D will be B+2 (as B-2 is -1 and 0, but we know that D is a positive integer). C will be D-1 or B+1. So B will be smallest.

If B=3, then D will be either 1(B-2) or 5(B+2).
C is one less than D, thus C is smaller than D.
C is 0 or 4. But all are positive. So C is 4.
Hence B(=3) is the smallest.

Thus B is smallest.

kvnmurty: can u delete the other answer pls... i want to write the answer in a different way
Answered by kvnmurty
0
   1 <= A, B, C, D <= 9

         A = 3 B    =>        1 <= B < 3   and  3 <= A <= 9
 
         C = D - 1      =>     1 <= C <= 8    and     2 <= D <= 9
 
         | B - D |  =  2   
  
       B is maximum 3, D is minimum 2  =>      B - D <= 1, if   B > D.
  
        Hence   D > B  and   D - B = 2       =>  B = D -2 =  C - 1

     B = A/3,  C - 1 and   D - 2 .     Hence  B is the smallest.


TPS: nice solution sir!!
kvnmurty: thanx
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