CAT 2025 Lesson : Proportion & Variation - Past Questions

Observation/Strategy

$1$) The two sets of equations have the $b$ term in common. A combined ratio can be written by equating the equations to a constant $K$.

Let $a = 6b = 12c = K$

⇒ $a = K, b = \dfrac{K}{6}, c = \dfrac{K}{12}$

∴ $a : b : c = K : \dfrac{K}{6} : \dfrac{K}{12} = 12 : 2 : 1 \longrightarrow (1)$

Let $2b = 9d = 12e = M$
⇒ $b = \dfrac{M}{2}, d = \dfrac{M}{9}, e = \dfrac{M}{12}$

∴$b : d : e = \dfrac{M}{2} : \dfrac{M}{9} : \dfrac{M}{12} = 18 : 4 : 3 \longrightarrow (2)$

$(1)$ can be written as $a : b : c = 108 : 18 : 9 \longrightarrow (3)$

Combining $(2)$ and $(3)$,
$a : b : c : d : e = 108 : 18 : 9 : 4 : 3$

$\dfrac{c}{d} = \dfrac{9}{4} = 2.25$ is not an integer.

Answer: ($4$) $[ \dfrac{a}{6} , \dfrac{c}{d} ]$

Observation/Strategy

1) The reduction in speed (and not the speed itself) directly varies with the square root of the number of wagons.
2) To find the greatest number of wagons with which the train can move, we shall find the number of wagons when it just comes to a halt and then reduce by $1$.

Let the $f$ be the fall in speed and $n$ be the number of wagons attached.

$f = K \sqrt{n}$
With $9$ wagons, the fall in speed is $15$ km/hr.

$15 = K \sqrt{9} ⇒ K = 5$

The train will come to a halt when $\bm{f = 45}$.
⇒ $45 = 5 \times \sqrt{n}$
⇒ $\bm{n = 81}$

With $81$ wagons, the train comes to a complete halt. Therefore, $81 - 1 =$ 80 is the greatest number of wagons with which the train will just be able to move.

Answer: (3) $80$

Observation/Strategy

1) The answer options are a constant. So, we can just substitute values for the variables.
2) The variables substituted can be directly taken from the ratios given.

Let $m, n, r$ and $t$ be $4, 3, 9$ and $14$ respectively.

$\dfrac{3 \space mr - nt}{4 \space nt - 7 \space mr} = \dfrac{(3 \times 4 \times 9) - (3 \times 14)}{(4 \times 3 \times 14) - (7 \times 4 \times 9)}$

$= \dfrac{108 - 42}{168 - 252}$

$= - \dfrac{66}{84} = - \dfrac{11}{14}$

Answer: (3) $- \dfrac{11}{14}$