# Free Arithmetic 01 Practice Test - CAT

A ship leaves on a long voyage. When it is 18 miles from the shore, a sea plane, whose speed is ten times that of ship, is sent to deliver mail. How far from the shore does the sea plane catch up with the ship? (1994)

A. 24 miles
B. 25 miles
C. 22 miles
D. 20   miles

#### SOLUTION

Solution : D

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

The winning relay team in a high school sports competition clocked 48 min for a distance of 13.2 km. Its runners A, B, C and D maintained speeds of 15 km/hr, 16 km/hr, 17 km/hr and 18 km/hr respectively. What is the ratio of the time taken by B to the time taken by D?

A. 5 : 16
B. 5 : 17
C. 9 : 8
D. 8 : 9

#### SOLUTION

Solution : C

(c) Time is inversely proportional to the speed.
Hence ratio of time taken by B and D = 18 : 16 = 9 : 8

A and B walk from X to Y, a distance of 27 km at 5 km/hr and 7 km/hr respectively. B reaches Y and immediately turns back meeting A at Z. What is the distance from X to Z? (1994)

A. 25km
B. 22.5km
C. 24km
D. 20km

#### SOLUTION

Solution : B

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

27x5=27+x7
x=276
XZ=27276=27×56=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.

Three runners A, B and C run a race, with runner A finishing 12 meters ahead of runner   B and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runner travels the entire distance at a constant speed. What was the length of the race? (2001)

A. 36 meters
B. 48 meters
C. 60 meters
D. 72 meters

#### SOLUTION

Solution : B

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.

Speed of A=xtm/sec,
Speed of B=x12tm/sec,
Speed of C=x18tm/sec.

Time taken by B to finish the race =xx12t=xtx12 sec.

Now distance travelled by C in this time,
=x×t×x18(x12)t=x8
x(x18)x12=x8x=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!

A train X departs from station A at 11.00 a. m. for station B, which is 180 km away. Another train Y departs from station B at 11 : 00 a. m. for station A. Train X travels at an average speed of 70 km/hr and not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 minutes at station C, which is 60 kms away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to the point where the train crosses each other ? (2001)

A. 112
B. 118
C. 120
D. None of these

#### SOLUTION

Solution : A

Approach 1:
Option (a) Time taken by B to cover 60 km  =6050=65 hours.
Time taken by Y at station C  =1560=14 hours.
Now distance travelled bt train X in  (65+14)=2920  hours =70×2920=101.5 km.
Distance between X and Y when Y starts from station C =180 -(101.5 +60)=18.5 km.
Relative speed =(70+50)=120 km/hr.
Hence, time taken by them in crossing one another  =18.5120=0.15 hours. Now, distance travelled by X in 0.15 hour =70×0.15=10.5 km.
Therefore, distance of X from station A, when they meet =(101.5+10.5)=112 km.

Approach 2:
By using a table to check the distance travelled by each of X and Y:

TextX distanceY distanceTotal12:00705012012:12141014412:2717.50161.5Tally101.5

The last 18.5 kms will have to be divided in the ratio of 70:50. Hence we see that the distance travelled by X is around 112 kms.

A train approaches a tunnels AB. Inside the tunnel a cat located at a point that is  38  of the distance measured from the entrance A. when the train whistles, the cat runs. If the cat moves to the entrance of the tunnel, A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. The speed of the train is greater than the speed of the cat by what order?

A. 3 : 1
B. 4 : 1
C. 5 : 1
D. None of these

#### SOLUTION

Solution : B

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 38Y  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  58Y at its respective speeds,
X+YT=(58)YC.......(1)
When the train meets the cat at A it would have travelled a distance of X while the cat will travel  38Y at its respective speeds.
Also,  XY=(58)YC.....(2)
From (1)
XT+YT=5Y5C
Sub form (2)
(58)YC+YT=5Y8C
=YT=2Y8C
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

A man can walk up a moving 'up' escalator in 30 s. The same man can walk down this moving 'up' escalator in 90 s. Assume that his walking speed is same upwards and downwards. How much time did he take to walk up the escalator, when it is not moving? (1995)

A. 30s
B. 45s
C. 60s
D. 90s

#### SOLUTION

Solution : B

Let the speed of escalator be y ft per second speed of man's be x ft per second .Let us assume that length of the escalator be 90 ft.
Then,  x+y=9030x+y=3............(i)

And  xy=9090xy=1..................(ii)

x=25 ft/sec.And time taken by the man to walk up bt the escalator when it is not moving  =902=45 second.

Also note : Options a and d can directly be eliminated.

A man travels A to B at a speed x km/hr. He then rests at B for x hours. He then travels from B to C at a speed 2x km/hr and rests for 2x hours. He moves further to D at a speed twice as that between B and C. He thus reaches D in 16 hr. If distances A-B, B-C and C-D are all equal to 12 km, the time for which he rested at B could be:

A. 3 hr
B. 6hr
C. 2hr
D. 4hr

#### SOLUTION

Solution : A

Using answer options :
Option (a) total time taken from A to D.
12x+x+122x+2x+124x=16
21x+3x=16
3x216x+21=0
x=3 , 37

Alternatively:
Going by the answer options:
Let us assume the answer to be 2 hrs.

FROM-TOSpeedTime taken to cover 12 kmsRest timeCumulative timeAB2628BC4347CD81.501.5Total time taken16.5

Let us assume the answer to be 3 hrs.
FROM-TOSpeedTime taken to cover 12 kmsRest timeCumulative timeAB3437BC6268CD12101Total time taken16

If a man cycles at 10 km/hr, then he arrives at a certain place at 1 p.m. If he cycles at 15 km/hr, he will arrive at the same place at 11 a.m. At what speed must he cycle to get there at noon?

A.

11 km/hr

B.

12 km/hr

C.

13 km/hr

D.

14 km/hr

#### SOLUTION

Solution : B

Using  1x & 1x+1
We know that the speed is increasing by  12  hence the time has to decrease by  13  (using the  1x & 1x+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  10×6=60 kms
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×(t2)=10×(t)
T=6 , or he started 6 hours before 1,  i.e., at 7.
Hence the total distance now is  10×6=60 km.
To reach at noon he will have to cover 60 km in 5 hours. He was to travel at 12 km/hr.

A man travels three-fifth of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and returns at a speed 5c in the same time, then :

A. 1a+1b=1c
B. a + b = c
C. 1a+1b=2c
D. None of these

#### SOLUTION

Solution : C

Approach 1: Conventional Approach :
option (c) time taken to cover AC =3x5×3a=x5a hr.
Time taken to cover CB =x5b hr.
Time taken to cover BA and back AB =2x5c
Given  x5a+x5b=2x5c1a+1b=2c

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)

Directions: Answer the questions based on the following information.

A road network (shown figure)connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long. Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.

What is the average speed of Y?

A. 47.5 km/hr
B. 49.5 km/hr
C. 50 km/hr
D. 52 km/hr

#### SOLUTION

Solution : B

(b) Time taken by Y to cover AC =x45+x55=x×10055×45

Average speed of Y between AC.
=2x×55×45x×100=49.5 km/hr.

or just take the harmonic mean of 45 and 55 that will give 49.5 which is the answer.

Directions: Answer the questions based on the following information.

A road network (shown figure)connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long. Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.

The total distance travelled by Y during the journey is approximately:

A. 105 km
B. 150 km
C. 130 km
D. Cannot be determined

#### SOLUTION

Solution : A

(a) Since, X and Y reach C at the same time ,

100+BC61.875=AC49.5

Also,  BC2=AC21002
Now we can assume from answer options, the value of AC.

(105)2(100)2=1025        BC=32

100+BC61.875=10549.5
Even in this calculation we see that BC is 32.Hence this is the right option.

Directions: Answer the questions based on the following information.

A road network (shown figure) connects cities A, B, C and D all road segments are the straight line. D is the mid-point of the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB, the segment AB is 100km long. Ms.X and Mr Y leave A at 8.00 a.m .., take different routes to city C and reach at the time. X takes the highway from A to B to C and travels at an average speed of 61.875 km/hr. Y takes the direct route AC and travels at 45 km/hr on segment AD. Y's speed on segment DC is 55 km/hr.

What is the length of road segment BD?

A. 50 km
B. 52.5 km
C. 55 km
D. Cannot be determined

#### SOLUTION

Solution : B

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

The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the graph below. Manasa makes a 200 km trip from Mumbai to Pune at a steady speed of 60 km per hour. What is the amount of petrol consumed for the journey?

A. 12.5 litres
B. 13.33 litres
C. 16 litres
D. 19.75 litres

#### SOLUTION

Solution : 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

The petrol consumption rate of a new model car 'Palto' depends on its speed and may be described by the graph below. Manasa would like to minimize the fuel consumption for the trip by driving at the appropriate speed. How should she change the speed?

A. Increase the speed
B. Decrease the speed
C. Maintain the speed at 60 km/hr
D. Cannot be determined

#### SOLUTION

Solution : B

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

=402.5=16 km , 604=15 km , 807.9=10.1 km. respectively.
Hence at lower speed fuel consumption is less.Hence in order to minimize the fuel consumption, the speed should be decreased.