Car A takes 1 hour more than car B, which travels at a speed of 60 km per hour, to cover some fixed distance. If car A had doubled its speed, it could cover the distance in 1 hour less time than car B travelling at 60 km per hour. What is the original speed of car A in km per hour ?

Correct — C, 45. Call the fixed distance D kilometres and car A's original speed a. Car B covers the distance in D/60 hours. The first sentence gives D/a = D/60 + 1, and the second gives D/(2a) = D/60 − 1. Subtracting the second from the first wipes out the D/60 term on the right and leaves 2 on that side, while the left side becomes D/a − D/(2a) = D/(2a). So D/(2a) = 2, that is D = 4a, which says car A takes exactly 4 hours at its original speed. Feeding that back into the first relation, 4 = D/60 + 1, so D/60 = 3 and D = 180 km. The original speed is therefore 180 ÷ 4 = 45 km per hour. Every part of the story checks out: A takes 4 hours, B takes 3, and A at a doubled 90 km per hour takes 2 hours, which is one hour less than B.

• (a)30 — Doubling 30 gives exactly 60, which is car B's own speed — so the second sentence would have car A arriving at the same time as car B, not an hour earlier. The first sentence alone would fix the distance at 60 km, and 60 km at 60 km per hour is one hour, against the required two.
• (b)40 — Fixing the distance from the first sentence gives D = 60 × 40 ÷ 20 = 120 km. At a doubled 80 km per hour that is 1.5 hours, while car B takes 2 hours, so car A would gain only half an hour rather than a full one.
• (d)50 — Here the first sentence forces D = 60 × 50 ÷ 10 = 300 km. At a doubled 100 km per hour that is 3 hours against car B's 5, a saving of two hours where the question allows exactly one.

Every question of this family rests on time equals distance divided by speed, and on the fact that the same distance travelled at two speeds gives two times whose difference is stated in the problem. Doubling a speed halves the time, which is what turns the second sentence into a second usable relation rather than a repetition of the first. With one unknown distance and one unknown speed, two relations are exactly enough.

Two shortcuts save time in the hall. The first is the subtraction used above: the two relations differ only in the sign of the 1, so subtracting them removes the D/60 term entirely and leaves a single clean statement, D = 4a. The second is back-substitution, which needs no algebra at all. Each offered speed fixes a distance through the first sentence, using D = 60a ÷ (60 − a), and that distance can then be tested against the second sentence. Running the four options through that test gives 60, 120, 180 and 300 km, and only 180 km survives. On a paper where the algebra is going badly, four quick divisions are a complete substitute for it.

• Time equals distance divided by speed, so doubling the speed halves the time for a fixed distance.
• The two statements give D/a = D/60 + 1 and D/(2a) = D/60 − 1; subtracting them leaves D/(2a) = 2.
• That reduces to D = 4a, which says car A's original journey takes exactly 4 hours.
• The distance works out to 180 km, so car A runs at 45 and car B at 60 km per hour.
• Back-substitution works as an independent check: for any candidate speed a, the first statement forces D = 60a ÷ (60 − a).

• Treating the two sentences as one relation and finding only one equation for two unknowns.
• Doubling the time instead of halving it when the speed is doubled.
• Forgetting that the two cars cover the same distance, which is the only thing tying the relations together.

Asked as a two-relation word problem where the second sentence supplies the extra equation that makes an apparently underdetermined problem solvable.

• practice — not a real PYQ

  A cyclist covers a fixed distance in 3 hours. If he raises his speed by 5 km per hour he would cover it in 2.5 hours. The distance is

  • (a)60 km
  • (b)75 km
  • (c)90 km
  • (d)120 km

  Answer(b) 75 km — if the original speed is v then 3v = 2.5(v + 5), so 0.5v = 12.5 and v = 25, giving a distance of 3 × 25 = 75 km.

• practice — not a real PYQ

  A car takes 2 hours less to cover a distance when its speed is doubled. The time taken at the original speed is

  • (a)2 hours
  • (b)3 hours
  • (c)4 hours
  • (d)6 hours

  Answer(c) 4 hours — doubling the speed halves the time, so the saving equals half the original time; half of it is 2 hours, making the original 4 hours.