Elitmus
Exam
Numerical Ability
Number System
consider a two digit number "ab". square the number and divide the resultant by half of the numer. then add 18 in the resultant. and then divide by 2 the resultant. How many numbers are possible that it reversing the digit ex. "ba".
option:
A. 7
B. 8
C. 9
D. one more option I don't remember
Read Solution (Total 8)
-
- If the number is ab. Then it will be 10a+b.
Square it => (10a+b)^2
Divide by half the number => [(10a+b)^2] / [(10a+b)/2] => 2(10a+b)
Add 18 => 2(10a+b) + 18
Divide by 2 => (10a+b) + 9
This is equal to reverse of the number so => 10b+a
Equate them-
10a+b+9 = 10b+a
9a-9b = -9
b - a = 1
So we see the condition is :- (one's digit) - (tens digit) = 1
So there are 8 possibilities- 12, 23, 34, 45, 56, 67, 78, 89. - 7 years agoHelpfull: Yes(36) No(5)
- simple....
1. x=the number
2. x^2 and divide it by x/2
3. we get 2x
4. add 18 we get 2x+18
5. divide by 2 we get x+9
6. now use hit and trial
7. 12+9=21
23+9=32
34+9=43
.
.
.
89+9=98
8. so total possible number are=8 - 7 years agoHelpfull: Yes(3) No(6)
- B.....My answer is my explaination
- 7 years agoHelpfull: Yes(1) No(8)
- ans: 9
ab+9=ba
01+9=10
12+9=21
23+9=32
34+9=43
.
.
89+9=98 - 7 years agoHelpfull: Yes(1) No(4)
- D...................NO EXPLANATION
- 7 years agoHelpfull: Yes(0) No(6)
- Answer: 8
((ab)^2/ab + 18)/ 2 = ab + 9
now ab + 9 = ba
ab = 10a + b
ba = 10b + a
10a + b + 9 = 10b + a
9a - 9b + 9 = 0
b = a + 1
now assuming base 10 number system is used
so a can be [1, 8] not including 0 and 9 so to satisfy above conditions and a cant be 0 as to make it two digit number
so a can assume only 8 possible values between 0 to 8 - 7 years agoHelpfull: Yes(0) No(2)
- B.
possible numbers are
12,23,34,45,56,67,78,89 - 7 years agoHelpfull: Yes(0) No(1)
- anwer will be 8 as we cannot consider 01+9=10 coz den 01 will not be 2 digit number. read the 1st sentence its written ab is 2 digit no.
- 6 years agoHelpfull: Yes(0) No(0)
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