Question
Find the smallest natural number which when divided by
‘4’, ‘6’ and ‘15’, leaves a remainder ‘3’ in each case.Solution
LCM of (4, 6 and 15) = 60 So, required number = 60 + 3 = 63
More No. System Questions
Equation 1: x² - 180x + 8100 = 0
Equation 2: y² - 170y + 7225 = 0
l). p² - 26p + 153 = 0
ll). q² - 17q + 72 = 0
How many values of x and y satisfy the equation 2x + 4y = 8 & 3x + 6y = 10.
Equation 1: x² + 16x + 63 = 0
Equation 2: y² + 10y + 21 = 0
I. (4x-5)3 + 1/(4x-5)3 = 2
II. 2[(y+1/y)2- 2]- 9(y+1/y)= -14
I. 5x² = 19x – 12
II. 5y² + 11y = 12
I. x2 – 10x + 21 = 0
II. y2 + 11y + 28 = 0
(i) 2x² – x – 3 = 0
(ii) 2y² – 6y + 4 = 0
I. 27x² + 120x + 77 = 0
II. 56y² + 117y + 36 = 0
I. 6 y² + 11 y – 7= 0
II. 21 x² + 5 x – 6 = 0
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