- This is an assessment test.
- To draw maximum benefit, study the concepts for the topic concerned.
- Kindly take the tests in this series with a pre-defined schedule.
Arithmetic : Level 3 Test -2
Congratulations - you have completed Arithmetic : Level 3 Test -2.You scored %%SCORE%% out of %%TOTAL%%.You correct answer percentage: %%PERCENTAGE%% .Your performance has been rated as %%RATING%%
Your answers are highlighted below.
Question 1 |
A person who has a certain amount with him goes to market. He can buy 50 oranges or 40 mangoes. He retains 10% of the amount for taxi fares and buys 20 mangoes and of the balance he purchases oranges. Number of oranges he can purchase is
36 | |
40. | |
15 | |
20 |
Question 1 Explanation:
Let us assume that he has Rs. 100.
Fare = 10% = Rs. 10
Therefore money left is Rs. 90
In Rs. 100 he can buy 50 oranges or 40 mangoes.
The price of an orange is Rs. 2 and that of a mango is Rs. 2.50.
Now if he buys 20 mangoes, he would spend Rs. 50
and will be left with Rs. 40, in which he can buy 20 oranges.
Fare = 10% = Rs. 10
Therefore money left is Rs. 90
In Rs. 100 he can buy 50 oranges or 40 mangoes.
The price of an orange is Rs. 2 and that of a mango is Rs. 2.50.
Now if he buys 20 mangoes, he would spend Rs. 50
and will be left with Rs. 40, in which he can buy 20 oranges.
Question 2 |
The price of a Maruti car rises by 30% while the sales of the car come down by 20%. What is the percentage change in the total revenue?
–4% | |
–2% | |
+4% | |
+2% |
Question 2 Explanation:
Lets suppose that the price was Rs 100
And the sale was also 100
Total revenue = 10000
Price raised by 30% i.e new price = Rs. 130
New sale = 80
This can simply be solved by multiplying the two
multiplication factors to get the effective New revenue = 10400
Which is 400 more than the old one so there is increase in the revenue by 4%
And the sale was also 100
Total revenue = 10000
Price raised by 30% i.e new price = Rs. 130
New sale = 80
This can simply be solved by multiplying the two
multiplication factors to get the effective New revenue = 10400
Which is 400 more than the old one so there is increase in the revenue by 4%
Question 3 |
A man earns r% on the first Rs. 2,000 and s% on the rest of his income. If he earns Rs. 700 from income of Rs. 4,000 and Rs. 900 from Rs. 5,000 of Income , find r%.
20% | |
15% | |
25% | |
None of these |
Question 3 Explanation:
The two equations can be written
2000 (r/100) + 2000(s/100) = 700
and 2000(r/1000) + 3000(s/100) = 900
The equations can be simplified to
r + s = 35 and 2r + 3s = 90. Solving these two equations
Simultaneously, we get r = 15%.
2000 (r/100) + 2000(s/100) = 700
and 2000(r/1000) + 3000(s/100) = 900
The equations can be simplified to
r + s = 35 and 2r + 3s = 90. Solving these two equations
Simultaneously, we get r = 15%.
Question 4 |
A man buys spirit at Rs. 60 per liter, adds water to it and then sells it at Rs. 75 per liter. What is the ratio of spirit to water if his profit in the deal is 37.5%?
9 : 1 | |
10 : 1 | |
11 : 1 | |
None of these |
Question 4 Explanation:
Since his SP of (spirit + water) = Rs.75/l
Profit of 37.5%,
CP of (spirit + water) = 75/1.375 = Rs.54.54.
Assume that the cost price of water = 0
So if we alligate,
We come to know that
Ratio will be 10 : 1
Spirit (60) - 54.54 = 5.545
Water(0) – 54.54 = 54.54
So the ratio of spirit to water is 10 : 1
Profit of 37.5%,
CP of (spirit + water) = 75/1.375 = Rs.54.54.
Assume that the cost price of water = 0
So if we alligate,
We come to know that
Ratio will be 10 : 1
Spirit (60) - 54.54 = 5.545
Water(0) – 54.54 = 54.54
So the ratio of spirit to water is 10 : 1
Question 5 |
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 > B | |
A < B | |
A = B | |
Cannot be determined |
Question 5 Explanation:
Let the capacity of each cup be 100 ml.
The ratio of mixture remained in the vessel
after taking out mixture from first and by pouring it to the second is 5 : 3.
Hence, proportion of alcohol in B = 3/8
Now after the second process of taking out 300 ml of mixture from the second container
The quantity of water will be (300 x 5/18 ) = 187.5 ml of water
The quantity of alcohol will be (300 x 3/18 ) = 112.5 ml of alcohol.
Now by pouring this mixture in the second vessel the quantity would be
(200 + 112.5) = 312.5 ml of alcohol and 187.5 ml of water.
Hence, ratio of alcohol to water in this container = 312.5: 187.5 = 5 : 3
Hence, proportion of water = A = 3/8 Hence, we find that A = B
The ratio of mixture remained in the vessel
after taking out mixture from first and by pouring it to the second is 5 : 3.
Hence, proportion of alcohol in B = 3/8
Now after the second process of taking out 300 ml of mixture from the second container
The quantity of water will be (300 x 5/18 ) = 187.5 ml of water
The quantity of alcohol will be (300 x 3/18 ) = 112.5 ml of alcohol.
Now by pouring this mixture in the second vessel the quantity would be
(200 + 112.5) = 312.5 ml of alcohol and 187.5 ml of water.
Hence, ratio of alcohol to water in this container = 312.5: 187.5 = 5 : 3
Hence, proportion of water = A = 3/8 Hence, we find that A = B
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
There are 5 questions to complete.
List |
