sahilkak wrote:
Sumanta needs to type up her 1950-word paper by its 5 pm deadline. If she starts at least two hours in advance, her typing speed will be a constant 20 words per minute, but for every two minutes beyond 3 pm that she waits before starting, her constant typing speed will increase by one word per minute. What is the latest time at which Sumanta can begin typing in order to finish her paper by the deadline?
a)2:50 pm
b)3:23 pm
c)3:30 pm
d)4:30 pm
e) 4:42 pm
any experts please explain the solution-how to solve this using backsolving?
We will start by assuming the last time:
(E) 4:42 pm
If she starts typing 18 mins before the deadline, she has 18 mins to type 1950 words. Her speed at this time will be 20 + (the mins passed after 3)/2 words i.e. her speed at 4:42 will be
20 + (120 - 18)/2 = 71 words/min
She can type a total of 18 * 71 = 1278 words.
Not enough since she has to type 1950 words. She must start earlier.
(D) 4:30 pm
If she starts 30 mins before deadline, she has 30 mins to type 1950 words. Her speed at this time will be 20 + (120 - 30)/2 = 65 words/min
She can type a total of 30*65 = 1950 words
This is enough.
So she can start at 4:30
Answer (D)
Let's look at the algebraic solution too though backsolving is the best way way here.
Let's say she starts her paper T mins before 5 pm.
Speed = 20 + (120 - T)/2
Time to type = T min
Words typed = 80 - T/2
We need this expression to be not less than 1950
(80 - T/2)*T = 1950
160T - T^2 = 2*1950
T^2 - 160T + 2*1950 = 0
(T - 30)(T - 130) = 0
T can be 30 or 130. The only possible option when two hrs are left is 30 mins.
(Note that if we wouldn't have got integer values for T, it would have become complicated.)
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Karishma
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