Question 1049002
This is a disguised rate problem. Even though
you don't see miles/hr or km/hr, there are still
rates involved. 
Put both of their rates in terms of frog jumps
Sherman's rate: {{{ 2*2 / 1  = 4/1 }}} frog jumps / unknown time unit
Cyrus's rate: {{{ 3 / 1 }}} frog jumps / unknown time unit
I could call time unit anything, so I'll just say it is 1 minute
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Cyrus's headstart is a distance of {{{ 10 }}} frog jumps
Start a stopwatch when Sherman leaves
Stop the stopwatch when Sherman catches Cyrus
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Let {{{ f }}} = distance in frog jumps
Let {{{ t }}} = time in minutes
Sherman's equation:
(1) {{{ f = ( 4/1)*t }}}
Cyrus's equation:
(2) {{{ f - 10 = (3/1)*t }}}
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(1) {{{ t = (1/4)*f }}}
Substitute this into (2)
(2) {{{ f - 10 = (3/4)*f }}}
(2) {{{ (1/4)*f = 10 }}}
(2) {{{ f = 40 }}}
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This tells me that Sherman makes {{{ 40 }}} frog jumps
from start to stop of stopwatch.
In that time, Cyrus has made:
{{{ f - 10 = 40 - 10 }}}
{{{ f - 10 = 30 }}} frog jumps which is the answer
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check the answer:
(1) {{{ f = ( 4/1)*t }}}
(1) {{{ 40 = 4t }}}
(1) {{{ t = 10 }}} min
and
(2) {{{ f - 10 = (3/1)*t }}}
(2) {{{ 40 - 10 = 3t }}}
(2) {{{ 3t = 30 }}}
(2) {{{ t = 10 }}} min
So, 10 min shows on the stopwatch- not the best 
time unit to choose, but you can pick any.
OK