Question:

A car driver increases the average speed of his car by 3 km/hr every hour. The total distance travelled in 7 hours if the distance covered in first hour was 30 km, is

Show Hint

When speed (or distance per hour) increases by a fixed amount each hour, use the sum of an AP: $S_n=\frac{n}{2}[2a+(n-1)d]$.
Updated On: Jul 15, 2026
  • 266 km
  • 273 km
  • 280 km
  • 287 km
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Approach Solution - 1

Distance covered each hour forms an arithmetic progression:
First hour $=30$ km; common difference $=3$ km.
Seven hourly distances: $30,33,36,39,42,45,48$.
Sum of AP: $S_n=\dfrac{n}{2}\,[2a+(n-1)d]$.
$S_7=\dfrac{7}{2}\,[2\cdot30+(7-1)\cdot3]=\dfrac{7}{2}\,(60+18)=\dfrac{7}{2}\times78=273$ km.
Was this answer helpful?
0
0
Show Solution
collegedunia
Verified By Collegedunia

Approach Solution -2

A car's hourly distance increases by 3 km/hr each hour, starting at 30 km, and we need the total distance over 7 hours. The hourly distances form an arithmetic sequence, so we can check which option matches the sum of that sequence.

  1. 266 km: This needs an average hourly distance of \( \frac{266}{7}=38 \) km, but the actual sequence 30, 33, 36, 39, 42, 45, 48 averages \( \frac{30+48}{2}=39 \) km, so this is too low.
  2. 273 km: This needs an average of \( \frac{273}{7}=39 \) km, which matches the actual average of the first and last hourly distances, \( \frac{30+48}{2}=39 \) km, exactly.
  3. 280 km: This needs an average of \( \frac{280}{7}=40 \) km, higher than the actual average of 39 km, so this is too high.
  4. 287 km: This needs an average of \( \frac{287}{7}=41 \) km, even further from the actual average of 39 km, so this is too high.

The hourly distances run from 30 km in the first hour to 48 km in the seventh hour, averaging 39 km per hour over 7 hours, which totals 273 km.

Therefore, the correct answer is 273 km.

Was this answer helpful?
0
0
Show Solution
collegedunia
Verified By Collegedunia

Approach Solution -3

The distance covered in the first six hours, before any option comes into play, is fixed: \( 30+33+36+39+42+45=225 \) km. Whatever the seventh hour's distance turns out to be, following the pattern of increasing by 3 km each hour, it must be \( 45+3=48 \) km. We can check each option by seeing what seventh-hour distance it would require, given the fixed 225 km from the first six hours.

  1. 266 km: This would require a seventh-hour distance of \( 266-225=41 \) km, but the actual pattern demands 48 km, so this does not fit.
  2. 273 km: This requires a seventh-hour distance of \( 273-225=48 \) km, matching the pattern exactly.
  3. 280 km: This would require a seventh-hour distance of \( 280-225=55 \) km, more than the pattern allows.
  4. 287 km: This would require a seventh-hour distance of \( 287-225=62 \) km, far more than the pattern allows.

Since the first six hours fix a running total of 225 km and the seventh hour must add exactly 48 km by the given pattern, the total comes to 273 km.

Therefore, the correct answer is 273 km.

Was this answer helpful?
0
0

Top CLAT Quantitative Aptitude Questions

View More Questions

Top CLAT Questions

View More Questions