Find all pairs of consecutive odd positive integers both of which are smaller than 10 such that their sum is more than 11,

Solution: Let the two consecutive odd positive integers be x and x + 2. x < 10, x + 2 < 10, x + (x + 2) > 11 x + 2 <...

Solution: Let the two consecutive even positive integers be x and x + 2. x > 5, x + 2 > 5, x + (x + 2) < 23 x > 5, x...

Correct answer is option (C) 0 ≤ r < b

Let the three consecutive integers bex, x+1and x+2. According to the question, x+x+1+x+2 = 51 => 3x+3=51 => 3x+3-3=51-3 [Subtracting 3 from both sides] => 3x=48 => 3x/3 = 48/3 [Dividing both sides by 3] X=16 Hence, first...

Let, a = 2p+1, p∈N and b = 2q+1 , q∈N\(\cup\){0} \(\because\) a > b ⇒ 2p+1 > 2q+1 ⇒ p > q \(\frac{a+b}{2}\) = \(\frac{2p+1+2q+1}{2}\) = p+q+1 \(\frac{a-b}{2}\) = \(\frac{2p+1-(2q+1)}{2}\) = p-q Case-I: \(\frac{a+b}{2}\) is odd which implies p+q+1 is odd \(\Rightarrow\) p+q is even (\(\because\) odd - 1 = even) \(\Rightarrow\) p+q-2q...

Let x be the smaller of the two consecutive even positive integers. Then, the other integer is x + 2.  Since both the integers are larger than 5,  x > 5 ..........................

(b) A + A + 2 + A + 4 + A + 6 = 180 4A + 12 = 180 A = 42. Next four consecutive even numbers are 50 + 52 +...

(a) Out of the given alternatives, 137 × 139 = 19043 Required smaller number = 137

 (a) According to the question,  x + x + 2 + x + 4 + x + 6 + x + 8 = 140  or, 5x + 20 = 140  or, 5x =...

(d) Lowest number of set A = 280/5 – 4 = 52  Lowest number of other set = 52 × 2 – 71 = 33   Required sum = 33 + 34...