Free Arithmetic 01 Practice Test - CAT 

The ship is 18 miles from the shore, as the speed is in the ratio of 1:10 the distance will also be in the ratio of 1:10.

When the ship travels 1 mile the plane will travel 10 miles

At this point the ship is 19 miles from the shore and the plane is 10 miles away from the shore.

Next when the ship moves by 1 mile covering 20 miles totally the plane will also cover the same distance of 20 miles hence will meet the ship

Approach 1: Convetional :
(b) Time taken by A in covering (27-x) km is same as time taken by Bin covering (27+x) km.

$\frac{27-x}{5}=\frac{27+x}{7}$
$x=\frac{27}{6}$
$\therefore XZ=27-\frac{27}{6}=27\times \frac{5}{6}=22.5km.$

Approach 2 : Alternate Approach (Using Ratios):
The total distance they will have to travel will be 2d or 54 kms. As their speeds are in the ratio of 5:7 the distances will also be in the same ratio. Hence,54 kms will have to be split in the ratio of 5:7. (i.e.,)22.5 and 41.5,hence they will meet at 22.5km.

Approach 3 : Going from answer options :
The total distance they will have to travel will be 2d or 54kms.

Now,going by the answer options :
Let us assume 25 to be the right option. Hence,A will take 5 hrs in the same time B would have covered a distance of 35 kms.hence the total distance covered will be 60, which is not what is given in the question (we need to get 54.)
Now assume 22.5 to be the right answer,hence A will take 4.5 hrs in the same time B would have covered a distance of 31.5kms,the tally of which is 54 hence that is the right option.
 

Approach 1: Conventional Approach:
(b) Let the distance of race be x metres which is covered by A in t seconds. Then, in the same time B covers (x-12)metres and C covers (x-18)metres.

$\therefore SpeedofA=\frac{x}{t}m/sec,$
$SpeedofB=\frac{x-12}{t}m/sec,$
$SpeedofC=\frac{x-18}{t}m/sec.$

Time taken by B to finish the race $=\frac{x}{\frac{x-12}{t}}=\frac{xt}{x-12}$ sec.

Now distance travelled by C in this time,
$=x\times t\times \frac{x-18}{(x-12)t}=x-8$
$\to \frac{x(x-18)}{x-12}=x-8\to x=48m.$ 

Approach 2: Reverse Gear Approach

Going from answer options:

Note the ratio of distance covered should be the same:

Assume option (c) to be correct: then we get the distances travelled to be in the ratio of 60:48:40 and for B and C we get the ratio to be 60:52, here we observe for B and C that the ratio is not the same.

For option B we get A:B:C = 48:36:30 or 8:6:5 for B and C we get 48:40 or 6:5 we see that for both the options we have the same ratio of distance travelled hence this the correct answer option!

Approach 1: Conventional Approach :
Let the speed of the train be T, and let it be at a distance of X from the tunnel AB.
Let the speed of the cat be C, and let the distance of the tunnel be Y, hence the cat is at a distance of $\frac{3}{8}Y$  from A.

Now 2 conditions are given:
 When the train meets the cat at B it would have travelled a distance of X+Y while the cat will travel  $\frac{5}{8}Y$ at its respective speeds,
$\frac{X+Y}{T}=\left(\frac{5}{8}\right)\frac{Y}{C}.......\left(1\right)$
 When the train meets the cat at A it would have travelled a distance of X while the cat will travel  $\frac{3}{8}Y$ at its respective speeds.
Also,  $\frac{X}{Y}=\frac{\left(\frac{5}{8}\right)Y}{C}.....\left(2\right)$
From (1)
$\frac{X}{T}+\frac{Y}{T}=\frac{5Y}{5C}$
Sub form (2)
$\frac{\left(\frac{5}{8}\right)Y}{C}+\frac{Y}{T}=\frac{5Y}{8C}$
$=\frac{Y}{T}=\frac{2Y}{8C}$
or $T:C=4:1$

Approach 2: Shortcut- Assumption

Assume the lengths of the tunnel to be 8 km and the speed of the cat be 8 km/hr

Time taken for the cat to reach A is 3 hr.

Time taken for the cat to reach B is 5 hr.

The difference in time of A and B gives the time taken by the train, which is 2 hours.

Hence the cat takes 8 hours while the train takes 2 hours,

Hence T:C = 4:1

Using answer options :
 Option (a) total time taken from A to D.
$\frac{12}{x}+x+\frac{12}{2x}+2x+\frac{12}{4x}=16$
$\to \frac{21}{x}+3x=16$
$\to 3x^2-16x+21=0$
$\to x=3,\frac{3}{7}$

Alternatively:
Going by the answer options:
Let us assume the answer to be 2 hrs.
 
$\begin{array}{ccccc}\text{FROM-TO}& \text{Speed}& \text{Time taken to cover 12 kms}& \text{Rest time}& \text{Cumulative time}\\ A-B& 2& 6& 2& 8\\ B-C& 4& 3& 4& 7\\ C-D& 8& 1.5& 0& 1.5\\ Totaltimetaken& & & & 16.5\end{array}$

Let us assume the answer to be 3 hrs.
$\begin{array}{ccccc}\text{FROM-TO}& \text{Speed}& \text{Time taken to cover 12 kms}& \text{Rest time}& \text{Cumulative time}\\ A-B& 3& 4& 3& 7\\ B-C& 6& 2& 6& 8\\ C-D& 12& 1& 0& 1\\ Totaltimetaken& & & & 16\end{array}$

Using  $\frac{1}{x}\&\frac{1}{x+1}$
We know that the speed is increasing by  $\frac{1}{2}$  hence the time has to decrease by  $\frac{1}{3}$  (using the  $\frac{1}{x}\&\frac{1}{x+1})$
shortcut (refer demo tutorial) but the time is actually decreasing by 2 hrs.
Hence the actual time of travel is 6 hrs hence he leaves at 7:00 AM.
The total distance travelled is  $10\times 6=60kms$
To reach at noon he will have to cover 60 km in 5 hours.  i.e., he has to travel at 12 km/hr.

Approach 2: Conventional Approach:
Let the person take t hrs to cover the distance at 10 km/hr.
If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m, which means he takes (t-2)hr.
Let the distance covered be D.
So we have  $15\times (t-2)=10\times \left(t\right)$
T=6 , or he started 6 hours before 1,  i.e., at 7.
Hence the total distance now is  $10\times 6=60$ km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.

Approach 1: Conventional Approach :
option (c) time taken to cover AC $=\frac{3x}{5}\times 3a=\frac{x}{5a}$ hr.
Time taken to cover CB $=\frac{x}{5b}$ hr.
Time taken to cover BA and back AB $=\frac{2x}{5c}$
Given  $\frac{x}{5a}+\frac{x}{5b}=\frac{2x}{5c}\to \frac{1}{a}+\frac{1}{b}=\frac{2}{c}$

Approach 2: Shortcut using Assumption !
As the questions in terms of variables we can assume values for all the given variables .
Let the distance AB be 5 kms:
The man travels three-fifth of a distance AB at a speed 3a. If we take  a=1  we see that he covers a  distance of 3 km. The remaining distance of 2 kms is covered in 2b time we get b=1. Also, he goes from B to A and returns at a speed 5c in the same time, we see that c=0.5. Sub these values into the options we find that only (c) fits in. (all the other options get eliminated) 

BD is half of AC. Solve for AC. You will get AC = 105.
Hence BD will be 52.5 or option (b)

Fuel consumption is given in litre per hour. It is, therefore, clear from the graph that in travelling 60 km fuel consumption is 4 litres. Hence in travelling 200 km fuel consumption will be = 13.33 litres

(b) At a speed of 40 km/hr,60 km/hr, and 80 km/hr distance travelled in 1 litre of petrol.

$=\frac{40}{2.5}=16km,\frac{60}{4}=15km,\frac{80}{7.9}=10.1km.$ respectively.
Hence at lower speed fuel consumption is less.Hence in order to minimize the fuel consumption, the speed should be decreased.