Question

A man is riding a bike with front and back wheel circumference of 40 inches and 70 inches respectively. If the man rides the bike on a straight road without slippage, how many inches will the man have travelled when the front wheel has made 15 revolutions more than the back wheel?

A) 1100 B) 1300 C) 1400 D) 1200

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Let us Assume That, Total Revolutions by back wheel are x .

Than,

→ Total revolutions by front wheel = (x + 15) .

Now, we know That,

☛ Distance covered = circumference of wheel * Total No. of Revolutions.

Now, In case of Distance travelled by bike :-

☛ Distance covered by back wheel = Distance covered by front wheel.

⟿ x * 70 = (x + 15) * 40

⟿ 70x = 40x + 600

⟿ 70x - 40x = 600

⟿ 30x = 600

⟿ x = 20 Revolutions.

Hence,

☛ Total Distance covered = 20*70 = 35*40 = 1400m.(C) (Ans.)

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• Circumference of front wheel = 40inches
• Circumference of back wheel = 70 inches
• the front wheel has made 15 revolutions more than back wheel.

${\bf{\blue{\underline{To\:Find:}}}}$

• Distance travelled =?

${\bf{\blue{\underline{Now:}}}}$

• Let the number of revolution made by front wheel be m.
• and the number of revolution made by back wheel be m-15

$\star \: \underline {\underline{\frak{according \: to \: the \: ques : }}}$

$: \implies{\sf{70(m - 15) = 40m}}$

$: \implies{\sf{70m - 1050= 40m}}$

$: \implies{\sf{70m - 40m=1050}}$

$: \implies{\sf{70m - 40m=1050}}$

$: \implies{\sf{30m=1050}}$

$: \implies{\sf{m= \frac{1050}{30} }}$

$: \implies{\sf{m= \frac{ \cancel{1050}}{ \cancel{30}} }}$

$: \implies\boxed{\sf{m=35}}$

___________________________________

$: \implies{\sf{ distance \: travelled = 40(35) = 70(35) - 1050}}$

$\dagger \: \: \boxed{\sf{ \purple{distance \: travelled = 1400 \: inches}}}$