Question

A person gains 25 calories by eating a cake and loses 80 calories by exercising for 30 mins. He decides to lose 500 calories per day. He exercises for some time in the morning and then eats some number of cakes. In the afternoon he does exercise for twice the time he did in the morning and eats three times the number of cakes he ate in the morning. In the evening he does exercise 1.5 times he did in the morning and eats four times the number of cakes he had in the morning. Had the man not ate any cakes he would have exceeded his target by 220 calories. But now he falls short by 180 calories. How many cakes did the man have in the evening?
[a] 12
[b] 8
[c] 16
[d] 20

Answer 1

Hint: Assume that the number of cakes he eats in the morning be x and the time for which he exercises in the morning be y. Determine the total calories lost by the person in case he did not eat cake and equate the expression to 720 and hence form an equation in x and y. Again find the total amount of calories lost by the person and equate the expression to 320 and hence form an equation in x and y. Solve the system by any of the known methods and hence find the number of cakes eaten by the man in the morning and hence find the number of cakes ate in the evening

Complete step-by-step solution -
Let the number of cakes the man eats in the morning be x and the time spent exercising in the morning be y minutes.
Hence the time spent exercising in the afternoon = 2y and the time spent exercising in evening = 1.5y
Also, the number of cakes the man eats in the afternoon =3x and the number of cakes he eats in evening =4x.
 Hence the total time spent exercising = $y+2y+1.5y=4.5y$
For every 30 mins of exercise the calories lost by the person $=80$
Hence the calories lost for every min of exercise $=\dfrac{80}{30}=\dfrac{8}{3}$
Hence the calories lost in 4.5y minutes $=\dfrac{8}{3}\times 4.5y=12y$
Given that if the man did not eat cake he will exceed his target by 220 calories, we have
The total amount of calories the man will lose given that he does not eat any cake = 500+220 = 720
Hence, we have
$12y=720$
Dividing both sides by 12, we get
$y=60$
Also the total number of cakes ate by the person $x+3x+4x=8x$
For every one cake the calories gained by man = 25
Hence the calories gained by eating 8x cakes = $25\times 8x=200x$
But since the man falls short of the target by 180 calories, we have
The total calories lost $=500-180=320$
Hence, we have
$12y-200x=320$
Substituting the value of y, we get
$\begin{align}
  & 720-200x=320 \\
 & \Rightarrow 200x=400 \\
\end{align}$
Dividing both sides by 200, we get
$x=2$
Hence the total number of cakes ate by the person in the morning = 2 and hence the number of cakes ate by the person in evening $=4\times 2=8$
Hence option [b] is correct.

Note: Alternative solution:
We know the man will have exceeded his target by 220 calories had he not eaten any cake.
Hence the total number of calories lost by the man through exercise = 500+220 = 720
Now the man gains some calories by eating cake and falls short of the target by 180 calories.
Hence after eating the cake the total calories lost = 500-180 =320
Hence the total calories gained because of eating cake = 720 -320 = 400
Hence the total number of cakes ate by man $=\dfrac{400}{25}=16$
Hence, we have
$\begin{align}
  & 8x=16 \\
 & \Rightarrow x=2 \\
\end{align}$
Hence the cakes eaten by the man in evening is 8 which is the same as obtained above.