# Solved CAT Questions (Arithmetic) - Set 6

• Q1. (CAT 1997)
A dealer buys dry fruits at Rs. 100, Rs. 80 and Rs. 60 per kilogram. He mixes them in the ratio 3 : 4 : 5 by weight, and sells at a profit of 50%. At what price per kilogram does he sell the dry fruit?
a. Rs. 80
b. Rs. 100
c. Rs. 95
d. None of these

cost price = 300 + 320 + 300 = 920 for 12 kg
profit = 460
SP = 920 + 460 = 1380
price per kg = 1380/12 = 115 ( None of the above)

Q2. (CAT 1997)
An express train travelling at 80 km/hr overtakes a goods train, twice as long and going at 40 km/hr on a parallel track, in 54 s. How long will the express train take to cross a platform of 400 m long?
a. 36 s
b. 45 s
c. 27 s
d. None of these

Let's take train length be d.
distance covered = d + 2d
relative velocity = 80 - 40 = 40 kmph = 40 x 5/18 ( 1 km/h = 5/18 m/s)
3d = 40 x 5/18 x 54 = 600
d = 600/3 = 200
To cross 400 m platform - total length covered = platform length + train length = 400 + 200 = 600
600 = t x 80 x 5/18
t = 600 x 18 / ( 5 x 80 ) = 27 seconds.

Q3. (CAT 1997)
The average marks of a student in 10 papers are 80. If the highest and the lowest scores are not considered, the average is 81. If his highest score is 92, find the lowest.
a. 55
b. 60
c. 62
d. Cannot be determined

Total marks = 10 x 80 = 800
Excluding highest and lowest, total marks = 648
sum of highest and lowest = 800 - 648 = 152
lowest = 152 - 92 = 60

Q4. (CAT 1997)
A man earns x% on the first Rs. 2,000 and y% on the rest of his income. If he earns Rs. 700 from income of Rs. 4,000 and Rs. 900 from if his income is Rs. 5,000, find x%.
a. 20%
b. 15%
c. 25%
d. None of these

for 4000, 2000x/100 + 2000y/100 = 700
for 5000, 2000x/100 + 1000y/100 = 900
10y = 200, y = 20
x = 15

Q5. (CAT 2002)
Two boys are playing on a ground. Both the boys are less than 10 years old. Age of the younger boy is equal to the cube root of the product of the age of the two boys. If we place the digit representing the age of the younger boy to the left of the digit representing the age of the elder boy, we get the age of father of the younger boy. Similarly, if we place the digit representing the age of the elder boy to the left of the digit representing the age of the younger boy and divide the figure by 2, we get the age of mother of the younger boy. The mother of the younger boy is younger to his father by 3 years. Then, what is the age of the younger boy?
a. 3
b. 4
c. 2
d. None of these

Let the ages of the two boys are p and q ( p > q )
Given that q = (pq)^1/3
Age of the father = 10q + p
Age of the mother = (10p + q)/2
10q + p = (10p + q)/2 + 3
19q - 8p - 6 = 0

Both p and q are single digit numbers (age < 10)
and from the equation, p = q^2
So if we go with the options,
if q = 3, then p = 9, 19q - 8p - 6 # 0 (not ok. should have been 0)
if q = 4 then p = 16 (not ok. p should be single digit)
if q = 2 then p = 4, 19q - 8p - 6 = 0 (perfect!)

Q6 - Q8 (CAT 1997)
Boston is 4 hr ahead of Frankfurt and 2 hr behind India. X leaves Frankfurt at 6 p.m. on Friday and reaches Boston the next day. After waiting there for 2 hr, he leaves exactly at noon and reaches India at 1 a.m. On his return journey, he takes the same route as before, but halts at Boston for 1hr less than his previous halt there. He then proceeds to Frankfurt.

Q6. If his journey, including stoppage, is covered at an average speed of 180 mph, what is the distance between Frankfurt and India?
a. 3,600 miles
b. 4,500 miles
c. 5,580 miles
d. Data insufficient

X reaches Boston at 10 AM. At the same moment, time in Frankfurt is 6 AM. Hence, the journey time was 12 hours.
X reaches India at 1 AM. At the same moment, time in Boston is 11 PM. Hence, the journey time was 11 hours.
Waiting time = 2 hours.
So, total time = 12 + 2 + 11 = 25 hours.
Avg speed = 180 mph
Distance = 180 x 25 = 4,500 miles.

Q7 . If X had started the return journey from India at 2.55 a.m. on the same day that he reached there, after how much time would he reach Frankfurt?
a. 24 hr
b. 25 hr
c. 26 hr
d. Data insufficient

Total time taken for return journey = Total time for onward journey - 1 = 25 - 1 = 24 hours

Q8. What is X's average speed for the entire journey (to and fro)?
a. 176 mph
b. 180 mph
c. 165 mph
d. Data insufficient

Not sufficient data here.

Q9 - Q10 (CAT 1996)
A watch dealer incurs an expense of Rs. 150 for producing every watch. He also incurs an additional expenditure of Rs. 30,000, which is independent of the number of watches produced. If he is able to sell a watch during the season, he sells it for Rs. 250. If he fails to do so, he has to sell each watch for Rs. 100.

Q9. If he is able to sell only 1,200 out of 1,500 watches he has made in the season, then he has made a profit of
a. Rs. 90,000
b. Rs. 75,000
c. Rs. 45,000
d. Rs. 60,000

For the 1200 watches he managed to sell in the season will give him 100 rs profit a piece. (250 - 150)
Remaining 300 watches will bring him a loss of 50 rupees a piece (as they can be sold at 100 only)
So total earning = 120000 - 15000 = 105000
out of which 30000 is additional expense (fixed)
So total profit = 105000 - 30000 = 75000

Q10. If he produces 1,500 watches, what is the number of watches that he must sell during the season in order to break-even, given that he is able to sell all the watches produced?
a. 500
b. 700
c. 800
d. 1,000

Cost Price = 150 x 1500 + 30000 = 255000
Selling price (if he sold x items during the season) = x * 250 + (1500 - x) * 100 = 150x + 150000
For break even, 255000 = 150x + 150000
150x = 105000
x = 700

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