If a ∶ b = c ∶ d = e ∶ f = 5 ∶ 7, then what is the ratio (3a + 5c + 11e) ∶ (3b + 5d + 11f)?

Correct — D. Write each pair with its own multiplier, since only the ratios are fixed:
a = 5k, b = 7k
c = 5m, d = 7m
e = 5n, f = 7n

3a + 5c + 11e = 15k + 25m + 55n = 5(3k + 5m + 11n)

3b + 5d + 11f = 21k + 35m + 77n = 7(3k + 5m + 11n)

The bracket is identical in both, so it cancels and the ratio is 5 : 7 → option (d)

Why the others are wrong

• (a)
  11 : 7 — 11 : 7 keeps the largest coefficient as if it survived the cancellation. Every coefficient multiplies a term and its partner alike, so none of them reaches the answer.
• (b)
  3 : 7 — 3 : 7 keeps the first coefficient, 3, against the ratio's 7. But 3 multiplies both a and b, so it cannot tilt a ratio.
• (c)
  7 : 11 — 7 : 11 pairs the ratio's 7 with the coefficient 11, two numbers that never meet: 11 multiplies e and f, and those already stand in 5 : 7.

When a chain of ratios is equal, a weighted sum of the numerators over the same weighted sum of the denominators returns the ratio itself.

If a⁄b = c⁄d = e⁄f = k, then (pa + qc + re)⁄(pb + qd + rf) = k for any weights p, q and r that leave the denominator non-zero. This is the addendo rule.

The substitution a = 5k, b = 7k makes it visible: the weights end up inside a bracket that is common to the top and the bottom, so it divides out.

Each pair needs its own letter — k, m, n — because nothing says a, c and e are equal, only that each stands to its partner as 5 to 7.

Key facts

• If a⁄b = c⁄d = e⁄f = k, then (a + c + e)⁄(b + d + f) = k.
• The rule survives weighting: (pa + qc + re)⁄(pb + qd + rf) = k as well.
• Here k = 5⁄7, so the asked ratio is 5 : 7 whatever the weights are.

Common traps

• Folding the coefficients 3, 5 and 11 into the answer as if they weighted the ratio.
• Taking a, c and e to be equal to one another rather than each in 5 : 7 with its partner.
• Reversing the answer to 7 : 5 by reading the second term first.

No directly related past PYQ was found.