Question Bank - 100 Arithmetic Questions From Previous CAT Papers (Solved)
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
Q46. (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%.
d. None of these
for 4000, 2000x/100 + 2000y/100 = 700
for 5000, 2000x/100 + 1000y/100 = 900
10y = 200, y = 20
x = 15
Q47. (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?
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!)
Q48. (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.
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
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
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
Part 1 : 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.
Part 2 : Total time taken for return journey = Total time for onward journey - 1 = 25 - 1 = 24 hours
Part 3 : Not sufficient data here.
Q49. (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.
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
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?
Part 1 : 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
Part 2 : 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
Q50. (CAT 1996)
Instead of a metre scale, a cloth merchant uses a 120 cm scale while buying, but uses an 80 cm scale while selling the same cloth. If he offers a discount of 20% on cash payment, what is his overall profit percentage?
Let price of 1 meter is 100
Merchant bought 1200 cm for 100 and he sells 800 cm for 80
Selling price of 1200 cm is 1200 x 80/800 = 120
Profit percentage is 20%
Q51. (CAT 1996)
The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs. 70,000 less. Find the original price of the diamond.
a. Rs. 1.4 lakh
b. Rs. 2 lakh
c. Rs. 1 lakh
d. Rs. 2.1 lakh
If we assume weight of diamond 10x. Well I came up with 10x because the ratio given 1 : 2 : 3 : 4 adds upto 10.
So individual weights of diamonds will be x, 2x, 3x and 4x
original price = k(10x)^2
Price for pieces = k(x^2 + 4x^2 + 9x^2 + 16x^2) = k30x^2
Thus change in price we see is 100kx^2 - 30kx^2 = 70kx^2 which is given as 70,000
Thus the original price 100kx^2 will be 1,00,000
will be 1,00,000
Q52. (CAT 1996)
In a mile race, Akshay can be given a start of 128 m by Bhairav. If Bhairav can give Chinmay a start of 4 m in a 100 m dash, then who out of Akshay and chinmay will win a race of one and half miles and what will be the final lead given by the winner to the loser? (One mile is 1600 m)
(a) Akshay, 1/2 mile
(b) Chinmay, 1/32 mile
(c) Akshay, 1/24 mile
(d) Chinmay, 1/16 mile
B will cover 1600m when A covers 1600 – 128 = 1472m
Let the time taken be t
Speed of B= 1600/t
Speed of A= 1472/t
Speed of B/ Speed of A= 1600 / 1472 = 25/23 --------- (1)
Speed of B / Speed of C = 100/96 = 25/24 --------- (2)
From (1) and (2)
Speed of A / Speed of C = 23/24
Let speed of A = 23x, then Speed of C = 24x
Time taken by C to complete one and half miles race = (1600 x 1.5)/24x = 100/x
In same time distance covered by A = 100/x * 23x = 2300m
Therefore lead = 2400 – 2300
= 100 m
= 1/16 mile
Therefore winner is C by 1/16 mile
Q53. (CAT 1996)
Two liquids A and B are in the ratio 5 : 1 in container 1 and 1 : 3 in container 2. In what ratio should the contents of the two containers be mixed so as to obtain a mixture of A and B in the ratio 1 : 1?
a. 2 : 3
b. 4 : 3
c. 3 : 2
d. 3 : 4
Take 1 portion from container 1 and N portions from container 2.
Then we collect total 5A/6 + B/6 + N(3B/4 + A/4)
Ratio of A and B in the new mix is (5/6 + N/4)A : (3N/4 + 1/6)B
for A : B = 1 : 1, the coefficients of A and B must be equal
(5/6 + N/4) = (3N/4 + 1/6)
N = 4/3
So the required ratio is 1 : 4/3 = 3 : 4
Q54. (CAT 1996)
A man travels three-fifths of a distance AB at a speed 3a, and the remaining at a speed 2b. If he goes from B to A and return at a speed 5c in the same time, then
a) 1/a + 1/b = 1/c
b) a + b = c
c) 1/a + 1/b = 2/c
d) None of these
Let distance between A and B = d
Time taken to cover 3d/5 = (3d/5)/3a = d/5a
Time taken to cover remaining distance = (2d/5)/2b = d/5b
Time taken to go from B to A and then return to B = 2d/5c
Now, d/5a + d/5b = 2d/5c
1/a + 1/b = 2/c
Q55. (CAT 1996)
A man travels from 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. 6 hr
c. 2 hr
d. 4 hr