Solved CAT Questions (Arithmetic) - Set 8
Q1. (CAT 1995)
In a race of 200 m run, A beats S by 20 m and N by 40 m. If S and N are running a race of 100 m with exactly same speed as before, then by how many metres will S beat N?
a. 11.11 m
b. 10 m
c. 12 m
d. 25 m
When A runs 200 m, S covers 200 - 20 = 180 and N covers 200 - 40 = 160
So when S covers 1 m, N covers 160/180 m
So when S covers 100 m, N covers 160/180 x 100 = 88.89 m
So S will beat N by 11.11 m (No need to calculate here as 160 x 100/180 is not an integer!)
Q2. (CAT 1995)
Two typists undertake to do a job. The second typist begins working one hour after the first. Three hours after the first typist has begun working, there is still 9/20 of the work to be done. When the assignment is completed, it turns out that each typist has done half the work. How many hours would it take each one to do the whole job individually ?
a. 12 hr and 8 hr
b. 8 hr and 5.6 hr
c. 10 hr and 8 hr
d. 5 hr and 4 hr
Let say, First typist takes F hours and Second one takes S hours to complete the whole job.
When the assignment was completed both did half of the job
To complete half the job, F/2 = S/2 + 1 (As Second one started an hour late)
Just use options and only C satisfies.
Q3. (CAT 1995)
I live X floors above the ground floor of a high-rise building. It takes me 30 s per floor to walk down the steps and 2 s per floor to ride the lift. What is X, if the time taken to walk down the steps to the ground floor is the same as to wait for the lift for 7 min and then ride down?
We gain 28 seconds per floor when we take life instead of walking.
Now we waited (or lost) for 7 minutes = 420 seconds which is equivalent to the amount of time we gained (which is 28 * X seconds)
28 * X = 420
X = 15
Q4. (CAT 1995)
A stockist wants to make some profit by selling sugar. He contemplates about various methods . Which of the following would maximize his profit?
I. Sell sugar at 10% profit.
II. Use 900 g of weight instead of 1 kg.
III. Mix 10% impurities in sugar and selling sugar at cost price.
IV. Increase the price by 5% and reduce weights by 5%.
a. I or III
c. II, III and IV
d. Profits are same
Let say CP for 1 Kg = 100 Rs
So Option 1 -> He sold 1000 g for 110 Rs (Profit = 10%)
Option 2 -> He sold 900 g for 100 Rs (Profit = 11.11%)
Option 3 -> He sold 1000 g of sugar for 110 Rs (Profit = 10%)
Option 4 -> He sold 950 g of sugar for 105 Rs (Profit = 10.52%)
So Option 2 gives the maximum profit.
Q5. (CAT 1995)
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 will he take to walk up the escalator, when it is not moving?
a. 30 s
b. 45 s
c. 60 s
d. 90 s
Length of the escalator = L
Speed of the escalator = E
Speed of Man = M
When the escalator and the man are moving up (same direction), relative speed = E + M
30 = L/(E + M) --- (1)
Similarly, When the escalator goes up and man walks down (opposite direction), relative speed = M - E
90 = L/(M - E) --- (2)
(2) / (1)
3 = (E + M)/(M - E)
3M - 3E = E + M
2M = 4E
E = M/2
As we need to find L/M, put E = M/2 in (1)
30 = L/(3M/2)
L/M = 45
The correct option is B.
Q6. (CAT 1995)
The rate of inflation was 1000%. Then what will be the cost of an article, which costs 6 units of currency now, 2 years from now?
1000% increase is 11 times the original price.
So 6 becomes 66 at the end of the first year and 66 becomes 726 at the end of the second year.
Q7. (CAT 1995)
Ram purchased a flat at Rs. 1 lakh and Prem purchased a plot of land worth Rs. 1.1 lakh. The respective annual rates at which the prices of the flat and the plot increased were 10% and 5%. After two years they exchanged their belongings and one paid the other the difference. Then
a. Ram paid Rs. 275 to Prem
b. Ram paid Rs. 475 to Prem
c. Ram paid Rs. 375 to Prem
d. Prem paid Rs. 475 to Ram
Price of the Flat after 2 years = 1 x 1.1 x 1.1 = 1.21 L
Price of the Plot after 2 years = 1.1 x 1.05 x 1.05 = 1.21275 L
Ram paid Prem 275 Rupees.
Q8. CAT 1994
One man can do as much work in one day as a woman can do in 2 days. A child does one third the work in a day as a woman. If an estate-owner hires 39 pairs of hands, men, women and children in the ratio 6 : 5 : 2 and pays them in all Rs. 1113 at the end of the days work. What must the daily wages of a child be, if the wages are proportional to the amount of work done?
Work done by a man in one day = x
work done by a woman in 1 day = x/2
work done by a child in 1 day = x/6
Number of men = 39 × 6 / 13 = 18
Number of women = 39 × 5 / 13 = 15
Number of children = 39 × 2 / 13 = 6
Work done by 18 men in 1 day = 18x
Work done by 15 women in 1 day = 15x/2
Work done by 6 children in 1 day = 6x/6 = x
Total amount given to 6 children = 1113 × x / ( 18x + 15x/2 + x) = 2226 / (36 + 15 + 2 ) = 42.
Daily wage of a child = 42 / 6 = Rs. 7.
Q9. (CAT 1994)
A water tank has three taps A, B, and C. A fills four buckets in 24 mins, B fills 8 buckets in 1 hour and C fills 2 buckets in 20 minutes. If all the taps are opened together a full tank is emptied in 2 hours. If a bucket can hold 5 litres of water, what is the capacity of the tank?
(a) 120 litres
(b) 240 litres
(c) 180 litres
(d) 60 litres
Tap A - 20 L of water in 24 minutes (5/6 L in 1 min)
Tap B - 40 L of water in 60 minutes (2/3 L in 1 min)
Tap C - 10 L of water in 20 minutes (1/2 L in 1 min)
Tap A + Tap B + Tap C - (5/6 + 2/3 + 1/2) in 1 minute = 2 L in 1 minute
So in 2 hours, 120 x 2 = 240 L - which is the capacity of the tank.
Q10. (CAT 1994)
Shyam went from Delhi to Shimla via Chandigarh by car. The distance from Delhi to Chandigarh is 3/4 times the distance from Chandigarh to Shimla. The average speed from Delhi to Chandigarh was half as much again as that from Chandigarh to Shimla. If the average speed for the entire journey was 49 kmph. What was the average speed from Chandigarh to Shimla?
(a) 39.2 kmph
(b) 63 kmph
(c) 42 kmph
(d) None of these
Distance - Delhi --------(3x/4 km) -------- Chandigarh ---------(x km)------- Shimla
Speed - Delhi --------(3y/2 kmph) -------- Chandigarh ---------(y kmph)------- Shimla
Time - Delhi --------(x/2y) -------- Chandigarh ---------(x/y km)------- Shimla
Average speed = Total distance/Total time
49 = (x + 3x/4)/(x/2y + x/y) = (7x/4)/(3x/2y) = 7y/6
y = 42 kmph