Aptitude - Problems on Numbers
- Problems on Numbers - Formulas
- Problems on Numbers - General Questions
- Problems on Numbers - Data Sufficiency 1
- Problems on Numbers - Data Sufficiency 2
Let the ten's digit be x and unit's digit be y.
Then, x + y = 15 and x - y = 3 or y - x = 3.
Solving x + y = 15 and x - y = 3, we get: x = 9, y = 6.
Solving x + y = 15 and y - x = 3, we get: x = 6, y = 9.
So, the number is either 96 or 69.
Hence, the number cannot be determined.
Let the numbers be a, b and c.
Then, a2 + b2 + c2 = 138 and (ab + bc + ca) = 131.
(a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) = 138 + 2 x 131 = 400.
(a + b + c) = 400 = 20.
Video Explanation: https://youtu.be/qmJ-0X8j_xQ
Let the ten's digit be x and unit's digit be y.
Then, number = 10x + y.
Number obtained by interchanging the digits = 10y + x.
(10x + y) + (10y + x) = 11(x + y), which is divisible by 11.
Video Explanation: https://youtu.be/lytJE8GqRvM
Let the ten's digit be x.
Then, unit's digit = x + 2.
Number = 10x + (x + 2) = 11x + 2.
Sum of digits = x + (x + 2) = 2x + 2.
(11x + 2)(2x + 2) = 144
22x2 + 26x - 140 = 0
11x2 + 13x - 70 = 0
(x - 2)(11x + 35) = 0
x = 2.
Hence, required number = 11x + 2 = 24.
Let the number be x.
Then, x + 17 = | 60 |
x |
x2 + 17x - 60 = 0
(x + 20)(x - 3) = 0
x = 3.