Skip to main contentSkip to solution

(47)(\frac{4}{7}) of a pole is in the mud. When (13)(\frac{1}{3}) of the pole is pulled out, 250 cm is still in the mud. The length of the pole is

Solution

✅ Correct Option: 1

Let total length of pole be xx cm

Initially, 47x\frac{4}{7}x is in mud.

After pulling out 13\frac{1}{3} of the pole, 250250 cm remains in mud.

Therefore:

47x−13x=250\frac{4}{7}x - \frac{1}{3}x = 250

1221x−721x=250\frac{12}{21}x - \frac{7}{21}x = 250

521x=250\frac{5}{21}x = 250

x=250×215x = 250 × \frac{21}{5}

x=1050x = 1050 cm

Therefore, the total length of the pole is 1050 cm.

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question