Let the price of a pen be 'x' and that of pencil be 'y' Then 87x + 29y = 783Multiplying eqn (i) by 30/29 we get (87x) x 30/29 + (29y) 30/29 = 783 x 30/29∴ 90x + 30y = 810
Since the difference between the divisors and the respective remainders is not constant, back substitution is the convenient method. None of the given numbers is satisfying the condition.
Let the digits be x and yTherefore, x + y = 12 .............(1)(10y + x) - (10x + y) Therefore, y - x = 4............. (2)Solving (1) and (2), y = 8 Therefore, x = 4There are two possible numbers 48 and 84. So the lowest no. is 48.
132 = 11 * 3 * 4Clearly, 968 is not divisible by 3None of 462 and 2178 is divisible by 4And, 5284 is not divisible by 11Each one of the remaining four numbers is divisible by each one of 4, 3 and 11.So, there are 4 such numbers.