CAT 1999 | Question: 57

Question

Let a, b, c be distinct digits. Consider a two-digit number ‘ab’ and a three-digit number ‘ccb’, both defined under the usual decimal number system, if $(ab)^2= ccb > 300$ then the value of b is

• A. 1
• B. 0
• C. 5
• D. 6

Answer 1

here 2 conditions are given

1) $(ab)^{2}$= ccb

2) ccb>300

so clearly we "ab " will be greater than 17 bcoz (17)^2 = 289

and here last digit in ab , which b and last digit in ccb which is b are same  ..

(18)^2= 324  here 8 not equal to 2

only unit digit 0, 1 , 5 , 6  give same unit digit when we square the number .

so $(20)^{2}$ = 400 but here c =4 , c=0 , b= 0 so c not equal to c .

$(21)^{2}$ = 441 , here b = 1 , and c= 4 which follow all conditions  means 1 should be the value of b ..

so ans should be A)

Answer 2

first of all :unit digit of two digit  number and its square(which is of 3 digit) should be same...so  unit digit must be either 1 or 0 or 5 or 6...
now square of number should be greater than 300 ...and less than or equal to 999...
so number will be between 17 and 32...
now we will check all number and their square between 17 and 32...having unit digit 0 or 1 or 5 or 6....
20^2=400
21^2=441
25^2=625
26^2=676
30^2=900
31^2=961
in our question given that hundred and tenth place of three digit no are same so...here we have only one option 441
and required number is 21...so b should be 1...