Advertisements
Advertisements
Question
Three consecutive integers, when taken in increasing order and multiplied by 2, 3 and 4 respectively, total up to 74. Find the three numbers.
Advertisements
Solution
Let the 3 consecutive integers be ‘x’, ‘x + 1’ and ‘x + 2’
Given that when multiplied by 2, 3 and 4 respectively and added up, we get 74
i.e. 2 × x + 3 × (x + 1) + 4 × (x + 2) = 74
Simplifying the equation, we get
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 63
9x = 63 ⇒ x = `63/9` = 7
First number = 7
Second numbers = x + 1 ⇒ 7 + 1 = 8
Third numbers = x + 2 ⇒ 7 + 2 = 9
∴ The numbers are 7, 8 and 9
APPEARS IN
RELATED QUESTIONS
Find the product of the following pair of monomial.
4p, 0
Obtain the product of xy, yz, zx.
Obtain the product of a, 2b, 3c, 6abc.
Express each of the following product as a monomials and verify the result for x = 1, y = 2:
\[\left( \frac{4}{9}ab c^3 \right) \times \left( - \frac{27}{5} a^3 b^2 \right) \times \left( - 8 b^3 c \right)\]
Multiply: abx, −3a2x and 7b2x3
Solve: (-12x) × 3y2
Multiply (4x2 + 9) and (3x – 2)
Product of the following monomials 4p, –7q3, –7pq is ______.
Multiply the following:
–7pq2r3, –13p3q2r
Multiply the following:
7pqr, (p – q + r)
