An algebra problem by Javed Siddiqui

Algebra Level 2

Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:

(a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much

e) a third as much d) half as much, (a) three times as much, c) the same, (b) twice as much

Javed Siddiqui by Javed Siddiqui , 23, Pakistan

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1 solution

Hung Woei Neoh
Aug 19, 2016

Let Jones' old speed be a a . His old distance is 50 50 miles

We know that

Speed = Distance Time ⟹ Time = Distance Speed \text{Speed}=\dfrac{\text{Distance}}{\text{Time}} \implies \text{Time} = \dfrac{\text{Distance}}{\text{Speed}}

Old time = 50 a =\dfrac{50}{a}

His new speed is 3 a 3a , and his new distance is 300 300 miles

New time = 300 3 a = 100 a = 2 ( 50 a ) = 2 × =\dfrac{300}{3a} = \dfrac{100}{a} = 2\left(\dfrac{50}{a}\right) = 2 \times Old time

Therefore, his new time is b) twice as much \boxed{\text{b) twice as much}} as his old time

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