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Arithmetic: Percentage Test -5
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Question 1 |
A person bought two tables for Rs. 2200. He sells one at 5% loss and the other at 6% profit and thus on the whole he neither gains nor loses. Find the cost price of each table.
Rs 1500, Rs 700 | |
Rs 2000, Rs 200 | |
Rs 1200, Rs 1000 | |
Rs 1100, Rs 1100 |
Question 1 Explanation:
Let the cost of table = t
Then the cost price of other will be = (2200 –t)
=> t x (95/100) + (2200 – t) x 106/100 =2200
=> 95t + 233200 – 106 t =220000
=> 11t = 13200
=> t = 1200 Rs and 2200 – t =1000 Rs
Then the cost price of other will be = (2200 –t)
=> t x (95/100) + (2200 – t) x 106/100 =2200
=> 95t + 233200 – 106 t =220000
=> 11t = 13200
=> t = 1200 Rs and 2200 – t =1000 Rs
Question 2 |
A tree was planted three years ago. The rate of its growth is 30% per annum. If at present, the height of the tree is 670 cm, what was it when the tree was planted?
305 cm | |
500 cm | |
405 cm | |
625 cm |
Question 2 Explanation:
Let the height of the tree be (h)
=> h x {1+(30/100)3} =670
=> h = 670 x {( 10 x 10 x10 )/(13 x 13 x 13)} = 305 approx.
=> h x {1+(30/100)3} =670
=> h = 670 x {( 10 x 10 x10 )/(13 x 13 x 13)} = 305 approx.
Question 3 |
When the price of sugar was increased by 32%, a family reduced its consumption in such a way that the expenditure on sugar was only 10% more than before. If 30 kg per month were consumed before, find the new monthly consumption.
42 kg | |
35 kg | |
25 kg | |
16 kg |
Question 3 Explanation:
Let the price of the sugar be Rs(p) / kg
=> Initial expenditure = 30p
=> new expenditure = 33p
=> new monthly consumption = 33p/1.32p = 25kg
=> Initial expenditure = 30p
=> new expenditure = 33p
=> new monthly consumption = 33p/1.32p = 25kg
Question 4 |
A man's income is increased by Rs. 1200 and at the same time, the rate of tax to be paid is reduced from 12% to 10%. He now pays the same amount of tax as before. What is his increased income, if 20% of his income is exempted from tax in both cases?
Rs 6300 | |
Rs 7200 | |
Rs 4500 | |
Rs 6500 |
Question 4 Explanation:
Let the increased income be P
=> (P – 1200) x 80/100 x 12/100 = P x 80/100 x10/100
=> 12P – 14400 =10p
=> P = Rs. 7200
=> (P – 1200) x 80/100 x 12/100 = P x 80/100 x10/100
=> 12P – 14400 =10p
=> P = Rs. 7200
Question 5 |
Mr. Bharat Toni buys a generator for Rs. 100000 and rents it. He puts 12.5% of each month's rent aside for upkeep and repairs, pays Rs. 325 per year as taxes and realizes 5.5% annually on his investment. Find the monthly rent.
Rs 634.76 | |
Rs 654.76 | |
Rs 554.76 | |
Rs 456.32 |
Question 5 Explanation:
Let Monthly rent be = (r)
=> 12r – 12r x (12.5/100) -325 = 100000 x (5.5/100)
=> 12r – 1.5r – 325 = 5500
=> r = (5500 +325)/ 10.5 = Rs 554.76
=> 12r – 12r x (12.5/100) -325 = 100000 x (5.5/100)
=> 12r – 1.5r – 325 = 5500
=> r = (5500 +325)/ 10.5 = Rs 554.76
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