Three consecutive positive integers k, l and m are such that l\(^2\) = 3(k+m). Find the value of m.
4
5
6
7
Correct answer is D
l\(^2\) = 3 (k + m)
Since they are consecutive positive numbers, we have
l = k+1, m = k+2.
\(\to\) (k+1)\(^2\) = 3(k + k + 2)
k\(^2\) + 2k + 1 = 3(2k + 2)
k\(^2\) + 2k + 1 = 6k + 6
k\(^2\) + 2k - 6k + 1 - 6 = 0
k\(^2\) - 4k - 5 = 0
k\(^2\) - 5k + k - 5 = 0
k(k - 5) + 1(k - 5) = 0
k = -1 or 5
Since k, l and m are positive, then k = 5.
m = k + 2 = 5 + 2
= 7.
4≤X≤5
5≤X≤8
5≤X≤10
8≤X≤10
Correct answer is C
Number of oranges = X; Costing N5X.
Number of mangoes = 2X; Costing N8X.
\(\therefore 65 \leq 5X + 8X \leq 130\)
\(5 \leq X \leq 10\)
Factorize completely \(x^{2} + 2xy + y^{2} + 3x + 3y - 18\).
(x+y+6)(x+y-3)
(x-y-6)(x-y+3)
(x-y+6)(x-y-3)
(x+y-6)(x+y+3)
Correct answer is A
Expand the options and collect like terms to check which gives the presented expression.
Expansion of Options A:
(x+y+6)(x+y-3) → \(x^2+xy-3x + xy+y^2-3y +6x+6y-18\)
Collect like terms: \(x^2 + xy + xy + y^2 +6x - 3x +6y - 3y - 18\)
= \(x^2 + 2xy + y^2 + 3x + 3y - 18\)
2/3
1/2
-1/2
-2/3
Correct answer is D
\(a*a^{-1} = aa^{-1} + a + a^{-1} = e\)
if e = 0
\(2.2^{-1} + 2 + 2^{-1} = 0\)
collecting like terms, we have:
\(3.2^{-1} + 2 = 0\)
= \(2^{-1}\) = -\(\frac{2}{3}\)
p/2
3p/2
5p/2
3p
Correct answer is B
Let the numbers be x and y
x+y = 2p.....(i)
x-y = p......(ii)
2x = 3p
x = 3p/2