Palindromic Distance and Average Speed

CAT 2023 Slot 1 · QA · Medium · Arithmetic

All CAT 2023 Slot 1 questionsMore Arithmetic questionsAll CAT QA

Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of xx km till then, where xx is a whole number and is palindromic. At the end of the trip, the car had travelled a total of 26862 km till then, this number again being palindromic. If Brishti never drove at more than 110 km/h, then the greatest possible average speed at which she drove during the trip, in km/h, was

  1. A.

    90

  2. B.

    100

  3. C.

    80

  4. D.

    110

Answer

B

Explanation

Trip duration =8= 8 hours. Maximum speed ≤110\le 110 km/h   ⟹  \implies Maximum distance travelled ≤8×110=880\le 8 \times 110 = 880 km. So, x≥26862−880=25982x \ge 26862 - 880 = 25982.

To maximize the average speed, we must maximize the distance travelled, which corresponds to finding the smallest palindromic integer x≥25982x \ge 25982.

Palindromes of 5 digits starting with 26000:

  • The smallest palindrome ≥25982\ge 25982 is 2606226062.

Distance travelled =26862−26062=800= 26862 - 26062 = 800 km. Average speed =8008=100= \frac{800}{8} = 100 km/h, which is ≤110\le 110 km/h.

Practise this under exam conditions

Sign in to solve it with a live timer, the on-screen CAT calculator, and streak and accuracy tracking across every question you attempt.

Solve in the workspace