# Question Bank - 100 Arithmetic Questions From Previous CAT Papers (Solved)

• I know this inspector and before solving the question will tell you the something that happened in the interview of this traffic controller.
Our guy meets the inspector for interview.
Inspector puts this question to him: "What would you do if you realized that two trains were heading for each other on the same track?"
He says, "I would switch the points for one of the trains."
"What if the lever broke?" asked the inspector.
"Then I'd dash down out of the signal box," he said, "and I'd use the manual lever over there."
"What if that had been struck by lightning?" - Inspector didn't let him go that easily!
"Then," he continues, "I'd run back into the signal box and phone the next signal box."
"What if the phone was engaged?" - asked the inspector.
"Well in that case, I'd rush down out of the box and use the public emergency phone at the level crossing up there."
"What if that was vandalized?" - Inspector strikes again.
"Oh well, then I'd ask my uncle to come to station as soon as possible."
This puzzles the inspector, so he asks, "Why would you do that?"
"Because he's never seen a train crash." ;)

Now back to the question.
At 7:30, train leaves from A to B would have covered 50 km while train from B to A would have covered 20 km since it travels for only half an hour.
Distance remaining between the trains = 100 - (50+20) i.e. 30km and the relative speed = 90kmph.
Therefore, time = 30/90 hours = 20 mins.

• Q31. (CAT 1999)
A road network (shown in the figure below) connects cities A, B, C and D. All road segments are straight lines. D is the mid-point on 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 100 km long. Ms X and Mr Y leave A at 8.00 a.m., take different routes to city C and reach at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 kmph. Y takes the direct route AC and travels at 45 kmph on segment AD. Y’s speed on segment DC is 55 kmph

What is the average speed of Y?
a. 47.5 kmph
b. 49.5 kmph
c. 50 kmph
d. 52 kmph

The total distance travelled by Y during the journey is approximately
a. 105 km
b. 150 km
c. 130 km
d. Cannot be determined

What is the length of the road segment BD?
a. 50 km
b. 52.5 km
c. 55 km
d. Cannot be determined

• Part 1 : Route of X = A - B - C ; Average speed = 61.785 kmph
Route of Y = A - C ; Travels 45 kmph on segment AD and at 55 kmph on DC
Average speed (distance in both cases are same) = 2 x 45 x 55 / (45 + 55) = 49.50

Note: When one travels at speed a for half the distance and speed b for other half of the distance, Average speed = 2ab/(a + b)

Part 2 : Total distance travelled by Y = AC
Apply Pythagoras theorem : 100^2 + BC^2 = AC^2 -- (1)
Now another condition given is that both X and Y reach at the same time ( T = D/S)
So (100 + BC)/61.785 = AC/49.50 ( 49.50 is the avg speed for Y we calculated before)
(100 + BC)/AC = 61.785 / 49.50 ~ 1.25
BC = 1.25AC - 100 -- (2)
Solve (1) and (2), AC = 105 and BC = 31
Y travelled 105 KM.

Part 3 : The midpoint of the hypotenuse (D) is equidistant from all the three vertices.
So BD = AD = CD = AC/2 = 105/2 = 52.5 KM

• Q32. (CAT 1999)
Rajiv reaches city B from city A in 4 hours, driving at speed of 35 kmph for the first two hour and at 45 kmph for the next two hours. Aditi follows the same route, but drives at three different speeds: 30, 40 and 50 kmph, covering an equal distance in each speed segment. The two cars are similar with petrol consumption characteristics (km per litre) shown in the figure below.

Zoheb would like to drive Aditi’s car over the same route from A to B and minimize the petrol consumption for the trip. What is the quantity of petrol required by him?
a. 6.67 L
b. 7 L
c. 6.33 L
d. 6.0 L

The quantity of petrol consumed by Aditi for the journey is
a. 8.3 L
b. 8.6 L
c. 8.9 L
d. 9.2 L

The quantity of petrol consumed by Aditi for the journey is
a. 8.3 L
b. 8.6 L
c. 8.9 L
d. 9.2 L

• Part 1 : To minimize the petrol consumption, Zoheb should maximize the mileage. He needs to drive at 40 kmph which gives him 24 km per litre.
Total distance covered will be 35 x 2 + 45 x 2 = 160 km.
Petrol consumption = 160/24 = 6.67 Litres.

Part 2 : Aditi follows the same route so the distance is same as calculated before - 160 KM.
Petrol consumption = (160/3)/16 + (160/3)/24 + (160/3)/50 = 8.9 litres

• Q33. (CAT 1998)
A company has a job to prepare certain number cans and there are three machines A, B and C for this job. A can complete the job in 3 days, B can complete the job in 4 days, and C can complete the job in 6 days. How many days will the company take to complete the job if all the machines are used simultaneously?
a. 4 days
b. 4/3 days
c. 3 days
d. 12 days

• It is easy to assume the work package consist of 12 cans ( LCM of 3, 4 and 6)
A will make 4 cans, B will make 3 cans and C will make 2 cans per day.
A, B and C together will make 9 cans a day
So for 12 cans, 12/9 = 4/3 days.

• Q34. (CAT 1998)
Distance between A and B is 72 km. Two men started walking from A and B at the same time towards each other. The person who started from A travelled uniformly with average speed of 4 km/hr. While the other man travelled with varying speed as follows: in the first hour his speed was 2 km/hr, in the second hour it was 2.5 km/hr, in the third hour it was 3 km/hr, and so on. When will they meet each other?
a. 7 hr
b. 10 hr
c. 35 km from A
d. Mid-way between A and B

• M1 - Uniform speed of 4 km/hr
So distance covered in nth hour = 4n

M2 - distance covered in 1st, 2nd, 3rd hour.. is 2, 2.5, 3, 3.5 etc.. respectively
This is an AP with a = 2 and d = 0.5
Total distance covered in nth hour = Sum of n terms of the above AP = n/2 [ 4 + (n - 1)/2]

As A and B are travelling in opposite direction, we can say when they meet, they would cover 72 km together
n/2 [ 4 + (n - 1)/2] + 4n = 72
4n/2 + n(n - 1)/4 + 4n = 72
Solve and n = 9 satisfies. (So they will meet in the 9th hour)

Distance covered by M1 (or M2) when they meet = 4 x 9 = 36 (half way between A and B )

• Q35. (CAT 1998)
A company purchases components A and B from Germany and USA respectively. A and B form 30% and 50% of the total production cost. Current gain is 20%. Due to change in the international scenario, cost of the German mark increased by 30% and that of USA dollar increased by 22%. Due to market conditions, the selling price cannot be increased beyond 10%.

What is the maximum current gain possible?
a. 10%
b. 12.5%
c. 0%
d. 7.5%

If the USA dollar becomes cheap by 12% over its original cost and the cost of German mark increased by 20%, what will be the gain? (The selling price is not altered.)
a. 10%
b. 20%
c. 15%
d. 7.5%

• Part 1 : Let say total production cost = 100
A (German) is responsible for 30 and B (USA) is responsible for 50 and 20 is from other expenses.
Selling price = 100 + 20 = 120
Now, Mark increased by 30% => previous 30 is now 39
and Dollar increased by 22% => previous 50 is now 61
So current production cost = 39 + 61 + 20 = 120
Max selling price can be 120 + 12 = 132
Max gain possible = 132 - 120 = 12, which is 10% of the production cost.

Part 2 : 12% decrease in 50 and 20% increase in 30 will cancel each other and the cost price remains same. As selling price is also not altered the gain will be same as before ( 20%)

• Q36. (CAT 1997)
There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D. If all the students of the class have the same weight, then which of the following is false?
a. The average weight of all the four groups is the same
b. The total weight of A and C is twice the total weight of B
c. The average weight of D is greater than the average weight of A
d. The average weight of all the groups remains the same even if a number of students are shifted from one group to another

• All students has same weight so irrespective of the number of students in each group, average would be same.
So option c is wrong.

• Q37. (CAT 1998)
I started climbing up the hill at 6 a.m. and reached the top of the temple at 6 p.m. Next day I started coming down at 6 a.m. and reached the foothill at 6 p.m. I walked on the same road. The road is so short that only one person can walk on it. Although I varied my pace on my way, I never stopped on my way. Then which of the following must be true?
a. My average speed downhill was greater than that of uphill
b. At noon, I was at the same spot on both the days.
c. There must be a point where I reached at the same time on both the days.
d. There cannot be a spot where I reached at the same time on both the days.

• Both trip took the same time.
Same route so same distance.
a. False, as average speed is same.
b. We cannot say that as we don't know if the speed is uniform or not.
c and d are conflicting statement so only one can be true.
starting point and end point are at the same time (6 am and 6 pm respectively) so c can be correct.

• Q38. (CAT 1998)
There are two containers: the first contains 500 ml of alcohol, while the second contains 500 ml of water. Three cups of alcohol from the first container is taken out and is mixed well in the second container. Then three cups of this mixture is taken out and is mixed in the first container. Let A denote the proportion of water in the first container and B denote the proportion of alcohol in the second container. Then
a. A > B
b. A < B
c. A = B
d. Cannot be determined

• Let's assume that volume of cup is 10 ml.

Container 1Container 2
WaterAlcoholWaterAlcohol
Start05005000
Step 1047050030
Step 228.3471.7471.728.3

So A = B ( did the calculations in hurry - hope it's correct!)

• Q39. (CAT 1999)
The speed of a railway engine is 42 kmph when no compartment is attached, and the reduction in speed is directly proportional to the square root of the number of compartments attached. If the speed of the train carried by this engine is 24 kmph when 9 compartments are attached, the maximum number of compartments that can be carried by the engine is
a. 49
b. 48
c. 46
d. 47

• Speed when no compartments attached = 42
Speed with 9 compartments attached = 24
Reduced speed = 18
18 = k * sqrt(9)
k = 6
Train doesn't move when there are more compartments. so assuming that the speed is reduced by 42
42 = 6 * sqrt(n)
n = 49
For 49 compartments, Engine won't move. So the maximum possible number of compartments = 49 - 1 = 48
(And no surprise we have 49 also in the option!)

• Q40. (CAT 1999)
Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
a. 550
b. 580
c. 540
d. 570

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