# Solved CAT Questions (Arithmetic) - Set 5

• Q1 - Q2 (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%.

Q1. What is the maximum current gain possible?
a. 10%
b. 12.5%
c. 0%
d. 7.5%

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.

Q2. 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%

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%. )

Q3. (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.

Q4. (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.

Q5. (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 - hopefully it's correct!)

Q6. (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!)

Q7. (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

Let the fixed amount be F and the partly varying amount be p.
F + 25p = 700 * 25
F + 50p = 600 * 50
On solving these equations we get p = 500 and F = 5000
So, for 100 people, we get 5000 + 100(500) = 55000
Hence, average expense = 55000/100 = 550.

Q8 - Q9 (CAT 1997)
A thief, after committing the burglary, started fleeing at 12 noon, at a speed of 60 km/hr. He was then chased by a policeman X. X started the chase, 15 min after the thief had started, at a speed of 65 km/hr.

Q8. At what time did X catch the thief?
a. 3.30 p.m.
b. 3 p.m.
c. 3.15 p.m.
d. None of these

In the 15 minutes, thief gets a head start of 15 km.
Every hour Police man gains 65 - 60 = 5 km
So to cover 15 km, Police man needs 15/5 = 3 hours.
So Thief will be caught at 12.15 + 3 = 3.15 PM

Q9. If another policeman had started the same chase along with X, but at a speed of 60 km/hr, then how far behind was he when X caught the thief?
a. 18.75 km
b. 15 km
c. 21 km
d. 37.5 km

Thief's speed is same as of second police man and Thief already has a 15 km head start. So this distance won't change between them.

Q10. (CAT 1997)
After allowing a discount of 11.11%, a trader still makes a gain of 14.28%. At how many percentage above the cost price does he mark on his goods?
a. 28.56%
b. 35%
c. 22.22%
d. None of these

Let the cost price be 100
SP = 114.28
114.28 = MP * (100 - 11.11)/100
MP = 114.28 x 100 / (100 - 11.11) = 128.56
which is 28.56% more than the cost price.

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