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Arithmetic : Level 3 Test -6
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Question 1 |
A stockiest 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%.
I or III | |
II | |
II, III and IV | |
Profits are same |
Question 1 Explanation:
Profit percentage in each case is
(i) 10%
(ii) {Error x 100}/wrong weight = (100 x 100)/ 900 =100/9 %
(iii) [{100 – (100/1.1)}/ 100/1.1] x 100 = 10 %
(iv) (105)/95 = 21/19%
Hence, b is the right option
(i) 10%
(ii) {Error x 100}/wrong weight = (100 x 100)/ 900 =100/9 %
(iii) [{100 – (100/1.1)}/ 100/1.1] x 100 = 10 %
(iv) (105)/95 = 21/19%
Hence, b is the right option
Question 2 |
A dealer offers a cash discount of 20% and still makes a profit of 20%, when he further allows 16 articles to a dozen to a particularly sticky bargainer. How much percent above the cost price were his wares listed?
100% | |
80% | |
75% | |
66 2/3% |
Question 2 Explanation:
Profit percentage = 20%.
Selling Price = 1.2 x CP (Cost Price).
This profit is made after the loss due to selling 16 articles at the price of 12.
The loss would be (16 – 12)/16 = 25%.
Therefore, SP x 0.75 = 1.2CP
SP = 1.6CP.
His actual Selling Price, SP = 0.8MP (Marked Price)
So, 0.8MP = 1.6CP or MP = 2CP.
Hence, Marked Price was 100% above his Cost Price.
Selling Price = 1.2 x CP (Cost Price).
This profit is made after the loss due to selling 16 articles at the price of 12.
The loss would be (16 – 12)/16 = 25%.
Therefore, SP x 0.75 = 1.2CP
SP = 1.6CP.
His actual Selling Price, SP = 0.8MP (Marked Price)
So, 0.8MP = 1.6CP or MP = 2CP.
Hence, Marked Price was 100% above his Cost Price.
Question 3 |
A ship leaves on a long voyage. When it is 18 miles from the shore, a seaplane, whose speed is ten times that of the ship, is sent to deliver mail. How far from the shore does the seaplane catch up with the ship?
24 miles | |
25 miles | |
22 miles | |
20 miles |
Question 3 Explanation:
Distance between the ship and the seaplane is 18 miles
Given that they are travelling in the same direction,
So, their relative speed would be 9 times the speed of the ship
Therefore, the time taken by seaplane = 18/9s = 2/s hours.
Now, the ship covers s miles in an hour. In 2/s hours it would cover 2 miles.
When the seaplane catches up with the ship, it would be 18+2 = 20 miles from the shore.
Given that they are travelling in the same direction,
So, their relative speed would be 9 times the speed of the ship
Therefore, the time taken by seaplane = 18/9s = 2/s hours.
Now, the ship covers s miles in an hour. In 2/s hours it would cover 2 miles.
When the seaplane catches up with the ship, it would be 18+2 = 20 miles from the shore.
Question 4 |
There is a leak in the bottom of the tank. This leak can empty a full tank in 8 hours. When the tank is full, a tap is opened into the tank which admits 6 liters per hour and the tank is now emptied in 12 hours. What is the capacity of the tank?
28.8 liters | |
36 liters | |
144 liters | |
Cannot be determined |
Question 4 Explanation:
The leak can empty the tank in 8 hours,
Work done by leak in one hour = 1/8.
And it’s given that the leak along with the tap can empty it in 12 hours.
So, 1/n – 1/8 = 1/12 (where n is the time taken by the tap to fill the tank).
On solving,
1/n = 1/24 or n = 24.
Therefore, we can say that tap can fill the tank in 24 hours.
Since the tap fill 6 liters per hour,
It will fill exactly fill (6 x 24) = 144 litres in 24 hours
The capacity of the tank is 144 Litres.
Work done by leak in one hour = 1/8.
And it’s given that the leak along with the tap can empty it in 12 hours.
So, 1/n – 1/8 = 1/12 (where n is the time taken by the tap to fill the tank).
On solving,
1/n = 1/24 or n = 24.
Therefore, we can say that tap can fill the tank in 24 hours.
Since the tap fill 6 liters per hour,
It will fill exactly fill (6 x 24) = 144 litres in 24 hours
The capacity of the tank is 144 Litres.
Question 5 |
The winning relay team in a high school sports competition clocked 48 minutes for a distance of 13.2 km. Its runners A, B, C and D maintained speeds of 15 kmph, 16 kmph, 17 kmph, and 18 kmph respectively. What is the ratio of the time taken by B to than taken by D?
5 : 16 | |
5 : 17 | |
9 : 8 | |
8 : 9 |
Question 5 Explanation:
As, time is inversely proportional to the speed and it is a relay race,
all the runners ran the same distance. Â For a same distance,
(Ratio of Times Taken) = 1/(Ratio of Speeds).
Hence, Ratio of times taken by B and D = 18 : 16 = 9 : 8.
all the runners ran the same distance. Â For a same distance,
(Ratio of Times Taken) = 1/(Ratio of Speeds).
Hence, Ratio of times taken by B and D = 18 : 16 = 9 : 8.
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