The post Aptitude Profit and Loss Online Test, Aptitude Profit and Loss Mock Test appeared first on Online Test.
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Aptitude Profit and Loss Online Test, Free Aptitude Quiz, Online Aptitude Profit and Loss Test. Aptitude Profit and Loss Question and Answers 2019. Aptitude Profit and Loss Quiz. Aptitude Profit and Loss Free Mock Test 2019. Aptitude Profit and Loss Question and Answers in PDF. The Aptitude Profit and Loss online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Profit and Loss Question and Answers in Hindi and English. Aptitude Profit and Loss Mock test for topic via Online Mode. Here we are providing Aptitude Profit and Loss Mock Test in Hindi. Now Test your self for “Aptitude Profit and Loss Online Test in Hindi” Exam by using below quiz…
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Alfred buys an old scooter for Rs. 4700 and spends Rs. 800 on its repairs. If he sells the scooter for Rs. 5800, his gain percent is:
Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.
Selling Price (S.P.) = Rs. 5800.
Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.
Gain % = (300/5500 x 100)% =
Cost Price (C.P.) = Rs. (4700 + 800) = Rs. 5500.
Selling Price (S.P.) = Rs. 5800.
Gain = (S.P.) – (C.P.) = Rs.(5800 – 5500) = Rs. 300.
Gain % = (300/5500 x 100)% =
The cost price of 20 articles is the same as the selling price of x articles. If the profit is 25%, then the value of x is:
Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.
S.P. of x articles = Rs. 20.
Profit = Rs. (20 – x).
∴ 20x/x × 100 = 25
⇒ 2000 – 100x = 25x
125x = 2000
⇒ x = 16.
Let C.P. of each article be Re. 1 C.P. of x articles = Rs. x.
S.P. of x articles = Rs. 20.
Profit = Rs. (20 – x).
∴ 20x/x × 100 = 25
⇒ 2000 – 100x = 25x
125x = 2000
⇒ x = 16.
If selling price is doubled, the profit triples. Find the profit percent.
Let C.P. be Rs. x and S.P. be Rs. y.
Then, 3(y – x) = (2y – x) y = 2x.
Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.
∴ Profit % = (x/x × 100) % = 100%
Let C.P. be Rs. x and S.P. be Rs. y.
Then, 3(y – x) = (2y – x) y = 2x.
Profit = Rs. (y – x) = Rs. (2x – x) = Rs. x.
∴ Profit % = (x/x × 100) % = 100%
In a certain store, the profit is 320% of the cost. If the cost increases by 25% but the selling price remains constant, approximately what percentage of the selling price is the profit?
Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 – 125) = Rs. 295.
∴ Required percentage = (295/420 × 100) % = 1475/21 % = 70% (approximately).
Let C.P.= Rs. 100. Then, Profit = Rs. 320, S.P. = Rs. 420.
New C.P. = 125% of Rs. 100 = Rs. 125
New S.P. = Rs. 420.
Profit = Rs. (420 – 125) = Rs. 295.
∴ Required percentage = (295/420 × 100) % = 1475/21 % = 70% (approximately).
A vendor bought toffees at 6 for a rupee. How many for a rupee must he sell to gain 20%?
C.P. of 6 toffees = Re. 1
S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5
For Rs. 6/5 , toffees sold = 6.
For Re. 1, toffees sold = (6 × 5/6) = 5.
C.P. of 6 toffees = Re. 1
S.P. of 6 toffees = 120% of Re. 1 = Rs. 6/5
For Rs. 6/5 , toffees sold = 6.
For Re. 1, toffees sold = (6 × 5/6) = 5.
The percentage profit earned by selling an article for Rs. 1920 is equal to the percentage loss incurred by selling the same article for Rs. 1280. At what price should the article be sold to make 25% profit?
Let C.P. be Rs. x.
Then, 1920x/x × 100 = x1280/x × 100
⇒ 1920 – x = x – 1280
⇒ 2x = 3200
⇒ x = 1600
∴ Required S.P. = 125% of Rs. 1600 = Rs. (125/100 × 1600) = Rs 2000.
Let C.P. be Rs. x.
Then, 1920x/x × 100 = x1280/x × 100
⇒ 1920 – x = x – 1280
⇒ 2x = 3200
⇒ x = 1600
∴ Required S.P. = 125% of Rs. 1600 = Rs. (125/100 × 1600) = Rs 2000.
A shopkeeper expects a gain of 22.5% on his cost price. If in a week, his sale was of Rs. 392, what was his profit?
C.P. = Rs. (100/122.5 × 392) = Rs. (1000/1225 × 392) = Rs. 320
Profit = Rs. (392 – 320) = Rs. 72.
C.P. = Rs. (100/122.5 × 392) = Rs. (1000/1225 × 392) = Rs. 320
Profit = Rs. (392 – 320) = Rs. 72.
A man buys a cycle for Rs. 1400 and sells it at a loss of 15%. What is the selling price of the cycle?
S.P. = 85% of Rs. 1400 = Rs. (85/100 x 1400) = Rs. 1190
S.P. = 85% of Rs. 1400 = Rs. (85/100 x 1400) = Rs. 1190
Sam purchased 20 dozens of toys at the rate of Rs. 375 per dozen. He sold each one of them at the rate of Rs. 33. What was his percentage profit?
Cost Price of 1 toy = Rs. (375/12) = Rs. 31.25
Selling Price of 1 toy = Rs. 33
So, Gain = Rs. (33 – 31.25) = Rs. 1.75
Profit % = (1.75/31.25 × 100) % = 28/5 % = 5.6%
Cost Price of 1 toy = Rs. (375/12) = Rs. 31.25
Selling Price of 1 toy = Rs. 33
So, Gain = Rs. (33 – 31.25) = Rs. 1.75
Profit % = (1.75/31.25 × 100) % = 28/5 % = 5.6%
Some articles were bought at 6 articles for Rs. 5 and sold at 5 articles for Rs. 6. Gain percent is:
Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.
C.P. of 30 articles = Rs. (5/6 x 30) = Rs. 25.
S.P. of 30 articles = Rs. (6/5 x 30) = Rs. 36.
∴ Gain % = (11/25 x 100) % = 44%.
Suppose, number of articles bought = L.C.M. of 6 and 5 = 30.
C.P. of 30 articles = Rs. (5/6 x 30) = Rs. 25.
S.P. of 30 articles = Rs. (6/5 x 30) = Rs. 36.
∴ Gain % = (11/25 x 100) % = 44%.
On selling 17 balls at Rs. 720, there is a loss equal to the cost price of 5 balls. The cost price of a ball is:
(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)
⇒ C.P. of 12 balls = S.P. of 17 balls = Rs.720.
⇒ C.P. of 1 ball = Rs. (720/12) = Rs. 60.
(C.P. of 17 balls) – (S.P. of 17 balls) = (C.P. of 5 balls)
⇒ C.P. of 12 balls = S.P. of 17 balls = Rs.720.
⇒ C.P. of 1 ball = Rs. (720/12) = Rs. 60.
When a plot is sold for Rs. 18,700, the owner loses 15%. At what price must that plot be sold in order to gain 15%?
85 : 18700 = 115 : x
x = (18700×115 / 85) = 25300.
Hence, S.P. = Rs. 25,300.
85 : 18700 = 115 : x
x = (18700×115 / 85) = 25300.
Hence, S.P. = Rs. 25,300.
100 oranges are bought at the rate of Rs. 350 and sold at the rate of Rs. 48 per dozen. The percentage of profit or loss is:
A shopkeeper sells one transistor for Rs. 840 at a gain of 20% and another for Rs. 960 at a loss of 4%. His total gain or loss percent is:
A trader mixes 26 kg of rice at Rs. 20 per kg with 30 kg of rice of other variety at Rs. 36 per kg and sells the mixture at Rs. 30 per kg. His profit percent is:
C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.
S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
∴ Gain = (80/1600 × 100) % = 5%.
C.P. of 56 kg rice = Rs. (26 x 20 + 30 x 36) = Rs. (520 + 1080) = Rs. 1600.
S.P. of 56 kg rice = Rs. (56 x 30) = Rs. 1680.
∴ Gain = (80/1600 × 100) % = 5%.
A shopkeeper sells some articles at the profit of 25% on the original price. What is the exact amount of profit? To find the answer, which of the following information given in Statements I and II is/are necessary?
I. Sale price of the article
II. Number of articles sold
Gain = 25% of C.P.
In order to find gain, we must know the sale price of each article and the number of articles sold.
∴ Correct answer is (D).
Gain = 25% of C.P.
In order to find gain, we must know the sale price of each article and the number of articles sold.
∴ Correct answer is (D).
A shopkeeper sells some toys at Rs. 250 each. What percent profit does he make? To find the answer, which of the following information given in Statements I and II is/are necessary?
I. Number of toys sold.
II. Cost price of each toy.
S.P. = Rs. 250 each.
To find gain percent, we must know the C.P. of each.
∴ Correct answer is (B).
S.P. = Rs. 250 each.
To find gain percent, we must know the C.P. of each.
∴ Correct answer is (B).
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
By selling an article what is the profit percent gained?
I. 5% discount is given on list price.
II. If discount is not given, 20% profit is gained.
III. The cost price of the articles is Rs. 5000.
I. Let the list price be Rs. x.
Then, S.P. = 95% of Rs. x = Rs. (x × 95/100) = Rs. (19x/20)
II. When S.P. = Rs. x and gain = 20%.
Then, C.P. = Rs. (100/120 × x) = Rs. 5x/6
∴ Gain = (19x/20 – 5x/6) = (57x50x/60) = 7x/60
∴ Gain % = (7x/60 × 6/5x × 100) % = 14%.
Thus, I and II only give the answer.
∴ Correct answer is (A).
I. Let the list price be Rs. x.
Then, S.P. = 95% of Rs. x = Rs. (x × 95/100) = Rs. (19x/20)
II. When S.P. = Rs. x and gain = 20%.
Then, C.P. = Rs. (100/120 × x) = Rs. 5x/6
∴ Gain = (19x/20 – 5x/6) = (57x50x/60) = 7x/60
∴ Gain % = (7x/60 × 6/5x × 100) % = 14%.
Thus, I and II only give the answer.
∴ Correct answer is (A).
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What was the percentage of discount given?
I. 23.5% profit was earned by selling an almirah for Rs. 12,350.
II. If there were no discount, the earned profit would have been 30%.
III. The cost price of the almirah was Rs. 10,000.
I. S.P. = Rs. 12350, Gain = 23.5%
∴ C.P. = Rs. (100/123.5 x 12350) = Rs. 10,000.
II. M.P. = 130% of C.P. = 130% of Rs. 10,000 = Rs. 13,000.
From I and II, discount = Rs. (13000 – 12350) = Rs. 650.
Discount % = (650/13000 x 100) % = 5%.
Thus, I and II give the answer.
II and III can not give the answer. Because we require profit percentage with discount and profit percentage without discount. So II and III are not sufficient.
Since III gives C.P. = Rs. 10,000, I and III give the answer.
Therefore, I and II [or] I and III give the answer.
∴ Correct answer is (E).
I. S.P. = Rs. 12350, Gain = 23.5%
∴ C.P. = Rs. (100/123.5 x 12350) = Rs. 10,000.
II. M.P. = 130% of C.P. = 130% of Rs. 10,000 = Rs. 13,000.
From I and II, discount = Rs. (13000 – 12350) = Rs. 650.
Discount % = (650/13000 x 100) % = 5%.
Thus, I and II give the answer.
II and III can not give the answer. Because we require profit percentage with discount and profit percentage without discount. So II and III are not sufficient.
Since III gives C.P. = Rs. 10,000, I and III give the answer.
Therefore, I and II [or] I and III give the answer.
∴ Correct answer is (E).
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
What is the percent profit earned by the shopkeeper on selling the articles in his shop?
I. Labeled price of the articles sold was 130% of the cost price.
II. Cost price of each article was Rs. 550.
III. A discount of 10% on labeled price was offered.
I. Let C.P. be Rs. x.
Then, M.P. = 130% of x = Rs. (13x/10) .
III. S.P. = 90% of M.P.
Thus, I and III give, S.P. = Rs. (90/100 x 13x/10) = Rs. (117x/100)
Gain = Rs. (117x/100 – x) = Rs. (17x/100)
Thus, from I and III, gain % can be obtained.
Clearly, II is redundant.
I. Let C.P. be Rs. x.
Then, M.P. = 130% of x = Rs. (13x/10) .
III. S.P. = 90% of M.P.
Thus, I and III give, S.P. = Rs. (90/100 x 13x/10) = Rs. (117x/100)
Gain = Rs. (117x/100 – x) = Rs. (17x/100)
Thus, from I and III, gain % can be obtained.
Clearly, II is redundant.
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Aptitude Simple Interest Online Test, Free Aptitude Quiz, Online Aptitude Simple Interest Test. Aptitude Simple Interest Question and Answers 2019. Aptitude Simple Interest Quiz. Aptitude Simple Interest Free Mock Test 2019. Aptitude Simple Interest Question and Answers in PDF. The Aptitude Simple Interest online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Simple Interest Question and Answers in Hindi and English. Aptitude Simple Interest Mock test for topic via Online Mode. Here we are providing Aptitude Simple Interest Mock Test in Hindi. Now Test your self for “Aptitude Simple Interest Online Test in Hindi” Exam by using below quiz…
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A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is:
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.
S.I. for 1 year = Rs. (854 – 815) = Rs. 39.
S.I. for 3 years = Rs.(39 x 3) = Rs. 117.
Principal = Rs. (815 – 117) = Rs. 698.
Mr. Thomas invested an amount of Rs. 13,900 divided in two different schemes A and B at the simple interest rate of 14% p.a. and 11% p.a. respectively. If the total amount of simple interest earned in 2 years be Rs. 3508, what was the amount invested in Scheme B?
A sum fetched a total simple interest of Rs. 4016.25 at the rate of 9 p.c.p.a. in 5 years. What is the sum?
Principal = Rs. (100 × 4016.25)
= Rs. (401625/45)
= Rs. 8925.
Principal = Rs. (100 × 4016.25)
= Rs. (401625/45)
= Rs. 8925.
How much time will it take for an amount of Rs. 450 to yield Rs. 81 as interest at 4.5% per annum of simple interest?
Time = (100×81 / 450×4.5) years = 4 years.
Time = (100×81 / 450×4.5) years = 4 years.
Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?
Let rate = R% and time = R years.
Then, (1200×R×R / 100) = 432
⇒ 12R^{2} = 432
⇒ R^{2} = 36
⇒ R = 6.
Let rate = R% and time = R years.
Then, (1200×R×R / 100) = 432
⇒ 12R^{2} = 432
⇒ R^{2} = 36
⇒ R = 6.
A sum of Rs. 12,500 amounts to Rs. 15,500 in 4 years at the rate of simple interest. What is the rate of interest?
S.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = (100 × 3000/12500 × 4)% = 6%
S.I. = Rs. (15500 – 12500) = Rs. 3000.
Rate = (100 × 3000/12500 × 4)% = 6%
An automobile financier claims to be lending money at simple interest, but he includes the interest every six months for calculating the principal. If he is charging an interest of 10%, the effective rate of interest becomes:
Let the sum be Rs. 100. Then,
S.I. for first 6 months = Rs. (100 × 10 × 1 / 100 × 2) = Rs. 5
S.I. for last 6 months = Rs. (105 × 10 × 1 / 100 × 2) = Rs. 5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
∴ Effective rate = (110.25 – 100) = 10.25%
Let the sum be Rs. 100. Then,
S.I. for first 6 months = Rs. (100 × 10 × 1 / 100 × 2) = Rs. 5
S.I. for last 6 months = Rs. (105 × 10 × 1 / 100 × 2) = Rs. 5.25
So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25
∴ Effective rate = (110.25 – 100) = 10.25%
A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in all from both of them as interest. The rate of interest per annum is:
A sum of Rs. 725 is lent in the beginning of a year at a certain rate of interest. After 8 months, a sum of Rs. 362.50 more is lent but at the rate twice the former. At the end of the year, Rs. 33.50 is earned as interest from both the loans. What was the original rate of interest?
Let the original rate be R%. Then, new rate = (2R)%.
Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s).
Let the original rate be R%. Then, new rate = (2R)%.
Note:
Here, original rate is for 1 year(s); the new rate is for only 4 months i.e. 1/3 year(s).
A man took loan from a bank at the rate of 12% p.a. simple interest. After 3 years he had to pay Rs. 5400 interest only for the period. The principal amount borrowed by him was:
Principal = Rs. (100 × 5400 / 12 × 3) = Rs. 15000.
Principal = Rs. (100 × 5400 / 12 × 3) = Rs. 15000.
A sum of money amounts to Rs. 9800 after 5 years and Rs. 12005 after 8 years at the same rate of simple interest. The rate of interest per annum is:
S.I. for 3 years = Rs. (12005 – 9800) = Rs. 2205.
S.I. for 5 years = Rs. (2205/3 × 5) = Rs. 3675
Principal = Rs. (9800 – 3675) = Rs. 6125.
Hence, rate = (100 × 3675 / 6125 × 5)% = 12%
S.I. for 3 years = Rs. (12005 – 9800) = Rs. 2205.
S.I. for 5 years = Rs. (2205/3 × 5) = Rs. 3675
Principal = Rs. (9800 – 3675) = Rs. 6125.
Hence, rate = (100 × 3675 / 6125 × 5)% = 12%
What will be the ratio of simple interest earned by certain amount at the same rate of interest for 6 years and that for 9 years?
Let the principal be P and rate of interest be R%.
Let the principal be P and rate of interest be R%.
A certain amount earns simple interest of Rs. 1750 after 7 years. Had the interest been 2% more, how much more interest would it have earned?
We need to know the S.I., principal and time to find the rate.
Since the principal is not given, so data is inadequate.
We need to know the S.I., principal and time to find the rate.
Since the principal is not given, so data is inadequate.
A person borrows Rs. 5000 for 2 years at 4% p.a. simple interest. He immediately lends it to another person at 6 1/4 p.a for 2 years. Find his gain in the transaction per year.
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Aptitude Compound Interest Online Test, Free Aptitude Quiz, Online Aptitude Compound Interest Test. Aptitude Compound Interest Question and Answers 2019. Aptitude Compound Interest Quiz. Aptitude Compound Interest Free Mock Test 2019. Aptitude Compound Interest Question and Answers in PDF. The Aptitude Compound Interest online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Compound Interest Question and Answers in Hindi and English. Aptitude Compound Interest Mock test for topic via Online Mode. Here we are providing Aptitude Compound Interest Mock Test in Hindi. Now Test your self for “Aptitude Compound Interest Online Test in Hindi” Exam by using below quiz…
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A bank offers 5% compound interest calculated on halfyearly basis. A customer deposits Rs. 1600 each on 1^{st}January and 1^{st} July of a year. At the end of the year, the amount he would have gained by way of interest is:
The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 4% per annum is Re. 1. The sum (in Rs.) is:
There is 60% increase in an amount in 6 years at simple interest. What will be the compound interest of Rs. 12,000 after 3 years at the same rate?
What is the difference between the compound interests on Rs. 5000 for 1.5 years at 4% per annum compounded yearly and halfyearly?
The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:
What will be the compound interest on a sum of Rs. 25,000 after 3 years at the rate of 12 p.c.p.a.?
At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?
The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:
Albert invested an amount of Rs. 8000 in a fixed deposit scheme for 2 years at compound interest rate 5 p.c.p.a. How much amount will Albert get on maturity of the fixed deposit?
The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable halfyearly is:
Simple interest on a certain sum of money for 3 years at 8% per annum is half the compound interest on Rs. 4000 for 2 years at 10% per annum. The sum placed on simple interest is:
If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same at the same rate and for the same time?
The difference between simple interest and compound on Rs. 1200 for one year at 10% per annum reckoned halfyearly is:
The difference between compound interest and simple interest on an amount of Rs. 15,000 for 2 years is Rs. 96. What is the rate of interest per annum?
The compound interest on a certain sum for 2 years at 10% per annum is Rs. 525. The simple interest on the same sum for double the time at half the rate percent per annum is:
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What is the rate of interest p.c.p.a.?
I. An amount doubles itself in 5 years on simple interest.
II. Difference between the compound interest and the simple interest earned on a certain amount in 2 years is Rs. 400.
III. Simple interest earned per annum is Rs. 2000.
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What will be the compound interest earned on an amount of Rs. 5000 in 2 years?
I. The simple interest on the same amount at the same rate of interest in 5 years is Rs. 2000.
II. The compound interest and the simple interest earned in one year is the same.
III. The amount becomed more than double on compound interest in 10 years.
P = Rs. 5000 & T = 2 years.
I. S.I. on Rs. 5000 in 5 years is Rs. 2000.
5000 × R × 5/100 = 2000 ⇒ R = 8.
Thus I only gives the answer.
∴ Correct answer is (A).
P = Rs. 5000 & T = 2 years.
I. S.I. on Rs. 5000 in 5 years is Rs. 2000.
5000 × R × 5/100 = 2000 ⇒ R = 8.
Thus I only gives the answer.
∴ Correct answer is (A).
In each of the following questions, a question is asked and is followed by three statements. While answering the question, you may or may not require the data provided in all the statements. You have to read the question and the three statements and then decide whether the question can be answered with any one or two of the statements or all the three statements are required to answer the question. The answer number bearing the statements, which can be dispensed with, if any, while answering the question is your answer.
Mr. Gupta borrowed a sum of money on compound interest. What will be the amount to be repaid if he is repaying the entire amount at the end of 2 years?
I. The rate of interest is 5 p.c.p.a.
II. Simple interest fetched on the same amount in one year is Rs. 600.
III. The amount borrowed is 10 times the simple interest in 2 years.
I. gives, Rate = 5% p.a.
II. gives, S.I. for 1 year = Rs. 600.
III. gives, sum = 10 x (S.I. for 2 years).
Now I, and II give the sum.
For this sum, C.I. and hence amount can be obtained.
Thus, III is redundant.
Again, II gives S.I. for 2 years = Rs. (600 × 2) = Rs. 1200.
Now, from III, Sum = Rs. (10 × 1200) = Rs . 12000.
Thus, Rate = 100 × 1200/2 × 12000 = 5% p.a.
Thus, C.I. for 2 years and therefore, amount can be obtained.
Thus, I is redundant.
Hence, I or III redundant.
I. gives, Rate = 5% p.a.
II. gives, S.I. for 1 year = Rs. 600.
III. gives, sum = 10 x (S.I. for 2 years).
Now I, and II give the sum.
For this sum, C.I. and hence amount can be obtained.
Thus, III is redundant.
Again, II gives S.I. for 2 years = Rs. (600 × 2) = Rs. 1200.
Now, from III, Sum = Rs. (10 × 1200) = Rs . 12000.
Thus, Rate = 100 × 1200/2 × 12000 = 5% p.a.
Thus, C.I. for 2 years and therefore, amount can be obtained.
Thus, I is redundant.
Hence, I or III redundant.
In each of the following questions, a question is asked and is followed by three statements. While answering the question, you may or may not require the data provided in all the statements. You have to read the question and the three statements and then decide whether the question can be answered with any one or two of the statements or all the three statements are required to answer the question. The answer number bearing the statements, which can be dispensed with, if any, while answering the question is your answer.
What is the compound interest earned at the end of 3 years?
I. Simple interest earned on that amount at the same rate and for the same period is Rs. 4500.
II. The rate of interest is 10 p.c.p.a.
III. Compound interest for 3 years is more than the simple interest for that period by Rs. 465.
I. gives, S.I for 3 years = Rs. 4500.
II. gives, Rate = 10% p.a.
III. gives, (C.I.) – (S.I.) = Rs. 465.
Clearly, using I and III we get C.I. = Rs. (465 + 4500).
Thus, II is redundant.
Also, from I and II, we get sum = (100 × 4500/10 × 3) = 15000.
Now C.I. on Rs. 15000 at 10% p.a. for 3 years may be obtained.
Thus, III is redundant.
∴ Either II or III is redundant.
I. gives, S.I for 3 years = Rs. 4500.
II. gives, Rate = 10% p.a.
III. gives, (C.I.) – (S.I.) = Rs. 465.
Clearly, using I and III we get C.I. = Rs. (465 + 4500).
Thus, II is redundant.
Also, from I and II, we get sum = (100 × 4500/10 × 3) = 15000.
Now C.I. on Rs. 15000 at 10% p.a. for 3 years may be obtained.
Thus, III is redundant.
∴ Either II or III is redundant.
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
What is the rate of compound interest?
I. The principal was invested for 4 years.
II. The earned interest was Rs. 1491.
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Aptitude Percentage Online Test, Free Aptitude Quiz, Online Aptitude Percentage Test. Aptitude Percentage Question and Answers 2019. Aptitude Percentage Quiz. Aptitude Percentage Free Mock Test 2019. Aptitude Percentage Question and Answers in PDF. The Aptitude Percentage online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Percentage Question and Answers in Hindi and English. Aptitude Percentage Mock test for topic via Online Mode. Here we are providing Aptitude Percentage Mock Test in Hindi. Now Test your self for “Aptitude Percentage Online Test in Hindi” Exam by using below quiz…
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A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = (50/110 × 100)% =
Number of runs made by running = 110 – (3 x 4 + 8 x 6)
= 110 – (60)
= 50.
Required percentage = (50/110 × 100)% =
Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100 (x + 9 + x)
⇒ 25(x + 9) = 14(2x + 9)
⇒ 3x = 99
⇒ x = 33
So, their marks are 42 and 33.
Let their marks be (x + 9) and x.
Then, x + 9 = 56/100 (x + 9 + x)
⇒ 25(x + 9) = 14(2x + 9)
⇒ 3x = 99
⇒ x = 33
So, their marks are 42 and 33.
A fruit seller had some apples. He sells 40% apples and still has 420 apples. Originally, he had:
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
⇒ 60/100 x x = 420
⇒ x = 420×100 / 60) = 700.
Suppose originally he had x apples.
Then, (100 – 40)% of x = 420.
⇒ 60/100 x x = 420
⇒ x = 420×100 / 60) = 700.
What percentage of numbers from 1 to 70 have 1 or 9 in the unit’s digit?
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number = 14
Required percentage = (14/70 × 100)% = 20%.
Clearly, the numbers which have 1 or 9 in the unit’s digit, have squares that end in the digit 1. Such numbers from 1 to 70 are 1, 9, 11, 19, 21, 29, 31, 39, 41, 49, 51, 59, 61, 69.
Number of such number = 14
Required percentage = (14/70 × 100)% = 20%.
If A = x% of y and B = y% of x, then which of the following is true?
x% of y = (x/100 × y) = (y/100 × x) = y% of x
∴ A = B.
x% of y = (x/100 × y) = (y/100 × x) = y% of x
∴ A = B.
If 20% of a = b, then b% of 20 is the same as:
20% of a = b ⇒ 20/100 a = b.
b% of 20 = (b/100 × 20) = (20/100 a × 1/100 × 20) = 4/100 a = 4% of a.
20% of a = b ⇒ 20/100 a = b.
b% of 20 = (b/100 × 20) = (20/100 a × 1/100 × 20) = 4/100 a = 4% of a.
In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
Let the number of students be x. Then,
Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
∴ 80% of x = 48 + 2/3 of 48
⇒ 80/100 x = 80
⇒ x = 100.
Let the number of students be x. Then,
Number of students above 8 years of age = (100 – 20)% of x = 80% of x.
∴ 80% of x = 48 + 2/3 of 48
⇒ 80/100 x = 80
⇒ x = 100.
Two numbers A and B are such that the sum of 5% of A and 4% of B is twothird of the sum of 6% of A and 8% of B. Find the ratio of A : B.
A student multiplied a number by 3/5 instead of 5/3 .
What is the percentage error in the calculation?
Let the number be x.
Then, error = 5/3 x – 3/5 x = 16/15 x
Error% = 16x/15 × 3/5x × 100)% = 64%
Let the number be x.
Then, error = 5/3 x – 3/5 x = 16/15 x
Error% = 16x/15 × 3/5x × 100)% = 64%
In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes was 7500, the number of valid votes that the other candidate got, was:
Number of valid votes = 80% of 7500 = 6000.
∴ Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.
Number of valid votes = 80% of 7500 = 6000.
∴ Valid votes polled by other candidate = 45% of 6000
= (45/100 x 6000) = 2700.
Three candidates contested an election and received 1136, 7636 and 11628 votes respectively. What percentage of the total votes did the winning candidate get?
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 × 100)% = 57%
Total number of votes polled = (1136 + 7636 + 11628) = 20400.
Required percentage = (11628/20400 × 100)% = 57%
Two tailors X and Y are paid a total of Rs. 550 per week by their employer. If X is paid 120 percent of the sum paid to Y, how much is Y paid per week?
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
⇒ z + 120/100 z = 550
⇒ 11/5 z = 550
⇒ z = (550 × 5/11) = 250.
Let the sum paid to Y per week be Rs. z.
Then, z + 120% of z = 550.
⇒ z + 120/100 z = 550
⇒ 11/5 z = 550
⇒ z = (550 × 5/11) = 250.
Gauri went to the stationers and bought things worth Rs. 25, out of which 30 paise went on sales tax on taxable purchases. If the tax rate was 6%, then what was the cost of the tax free items?
Let the amount taxable purchases be Rs. x.
Then, 6% of x = 30/100
x = (30/100 × 100/6) = 5.
∴ Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70
Let the amount taxable purchases be Rs. x.
Then, 6% of x = 30/100
x = (30/100 × 100/6) = 5.
∴ Cost of tax free items = Rs. [25 – (5 + 0.30)] = Rs. 19.70
Rajeev buys good worth Rs. 6650. He gets a rebate of 6% on it. After getting the rebate, he pays sales tax @ 10%. Find the amount he will have to pay for the goods.
Rebate = 6% of Rs. 6650 = Rs. (6/100 x 6650) = Rs. 399.
Sales tax = 10% of Rs. (6650 – 399) = Rs. (10/100 x 6251) = Rs. 625.10
∴ Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
Rebate = 6% of Rs. 6650 = Rs. (6/100 x 6650) = Rs. 399.
Sales tax = 10% of Rs. (6650 – 399) = Rs. (10/100 x 6251) = Rs. 625.10
∴ Final amount = Rs. (6251 + 625.10) = Rs. 6876.10
The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:
Increase in 10 years = (262500 – 175000) = 87500.
Increase% = (87500/175000) x 100 % = 50%.
Required average = (50/10) % = 5%.
Increase in 10 years = (262500 – 175000) = 87500.
Increase% = (87500/175000) x 100 % = 50%.
Required average = (50/10) % = 5%.
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Aptitude Partnership Online Test, Free Aptitude Quiz, Online Aptitude Partnership Test. Aptitude Partnership Question and Answers 2019. Aptitude Partnership Quiz. Aptitude Partnership Free Mock Test 2019. Aptitude Partnership Question and Answers in PDF. The Aptitude Partnership online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Partnership Question and Answers in Hindi and English. Aptitude Partnership Mock test for topic via Online Mode. Here we are providing Aptitude Partnership Mock Test in Hindi. Now Test your self for “Aptitude Partnership Online Test in Hindi” Exam by using below quiz…
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A and B invest in a business in the ratio 3 : 2. If 5% of the total profit goes to charity and A’s share is Rs. 855, the total profit is:
Let the total profit be Rs. 100.
After paying to charity, A’s share = Rs. (95 × 3/5) = Rs. 57.
If A’s share is Rs. 57, total profit = Rs. 100.
If A’s share Rs. 855, total profit = (100/57 × 855) = 1500.
Let the total profit be Rs. 100.
After paying to charity, A’s share = Rs. (95 × 3/5) = Rs. 57.
If A’s share is Rs. 57, total profit = Rs. 100.
If A’s share Rs. 855, total profit = (100/57 × 855) = 1500.
A, B and C jointly thought of engaging themselves in a business venture. It was agreed that A would invest Rs. 6500 for 6 months, B, Rs. 8400 for 5 months and C, Rs. 10,000 for 3 months. A wants to be the working member for which, he was to receive 5% of the profits. The profit earned was Rs. 7400. Calculate the share of B in the profit.
For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 – 370) = Rs. 7030.
Ratio of their investments = (6500 × 6) : (8400 × 5) : (10000 × 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
∴ B’s share = Rs. (7030 × 14/37) = Rs. 2660.
For managing, A received = 5% of Rs. 7400 = Rs. 370.
Balance = Rs. (7400 – 370) = Rs. 7030.
Ratio of their investments = (6500 × 6) : (8400 × 5) : (10000 × 3)
= 39000 : 42000 : 30000
= 13 : 14 : 10
∴ B’s share = Rs. (7030 × 14/37) = Rs. 2660.
A, B and C enter into a partnership in the ratio 7/2 : 4/3 : 6/5. After 4 months, A increases his share 50%. If the total profit at the end of one year be Rs. 21,600, then B’s share in the profit is:
A, B, C subscribe Rs. 50,000 for a business. A subscribes Rs. 4000 more than B and B Rs. 5000 more than C. Out of a total profit of Rs. 35,000, A receives:
Let C = x.
Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.
So, x + x + 5000 + x + 9000 = 50000
⇒ 3x = 36000
⇒ x = 12000
A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.
∴ A’s share = Rs. (35000 x 21 /50) = Rs. 14,700.
Let C = x.
Then, B = x + 5000 and A = x + 5000 + 4000 = x + 9000.
So, x + x + 5000 + x + 9000 = 50000
⇒ 3x = 36000
⇒ x = 12000
A : B : C = 21000 : 17000 : 12000 = 21 : 17 : 12.
∴ A’s share = Rs. (35000 x 21 /50) = Rs. 14,700.
Three partners shared the profit in a business in the ratio 5 : 7 : 8. They had partnered for 14 months, 8 months and 7 months respectively. What was the ratio of their investments?
Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.
Let their investments be Rs. x for 14 months, Rs. y for 8 months and Rs. z for 7 months respectively.
A starts business with Rs. 3500 and after 5 months, B joins with A as his partner. After a year, the profit is divided in the ratio 2 : 3. What is B’s contribution in the capital?
Let B’s capital be Rs. x.
Then, (3500 × 12/7x = 2/3)
⇒ 14x = 126000
⇒ x = 9000.
Let B’s capital be Rs. x.
Then, (3500 × 12/7x = 2/3)
⇒ 14x = 126000
⇒ x = 9000.
A and B entered into partnership with capitals in the ratio 4 : 5. After 3 months, A withdrew 1/4 of his capital and B withdrew 1/5 of his capital. The gain at the end of 10 months was Rs. 760. A’s share in this profit is:
A and B started a partnership business investing some amount in the ratio of 3 : 5. C joined then after six months with an amount equal to that of B. In what proportion should the profit at the end of one year be distributed among A, B and C?
Let the initial investments of A and B be 3x and 5x.
A : B : C = (3x × 12) : (5x × 12) : (5x × 6) = 36 : 60 : 30 = 6 : 10 : 5.
Let the initial investments of A and B be 3x and 5x.
A : B : C = (3x × 12) : (5x × 12) : (5x × 6) = 36 : 60 : 30 = 6 : 10 : 5.
A, B, C rent a pasture. A puts 10 oxen for 7 months, B puts 12 oxen for 5 months and C puts 15 oxen for 3 months for grazing. If the rent of the pasture is Rs. 175, how much must C pay as his share of rent?
A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.
∴ C’s rent = Rs. (175 × 9/35) = Rs. 45.
A : B : C = (10 x 7) : (12 x 5) : (15 x 3) = 70 : 60 : 45 = 14 : 12 : 9.
∴ C’s rent = Rs. (175 × 9/35) = Rs. 45.
A and B started a business in partnership investing Rs. 20,000 and Rs. 15,000 respectively. After six months, C joined them with Rs. 20,000. What will be B’s share in total profit of Rs. 25,000 earned at the end of 2 years from the starting of the business?
A : B : C = (20,000 × 24) : (15,000 × 24) : (20,000 × 18) = 4 : 3 : 3.
∴ B’s share = Rs. (25000 × 3/10) = Rs. 7,500.
A : B : C = (20,000 × 24) : (15,000 × 24) : (20,000 × 18) = 4 : 3 : 3.
∴ B’s share = Rs. (25000 × 3/10) = Rs. 7,500.
A began a business with Rs. 85,000. He was joined afterwards by B with Rs. 42,500. For how much period does B join, if the profits at the end of the year are divided in the ratio of 3 : 1?
Suppose B joined for x months. Then,
Then, (85000 × 12/42500 × x = 3/1)
⇒ x = 85000 × 12/42500 × 3 = 8.
So, B joined for 8 months.
Suppose B joined for x months. Then,
Then, (85000 × 12/42500 × x = 3/1)
⇒ x = 85000 × 12/42500 × 3 = 8.
So, B joined for 8 months.
Aman started a business investing Rs. 70,000. Rakhi joined him after six months with an amount of Rs.. 1,05,000 and Sagar joined them with Rs. 1.4 lakhs after another six months. The amount of profit earned should be distributed in what ratio among Aman, Rakhi and Sagar respectively, 3 years after Aman started the business?
Aman : Rakhi : Sagar = (70,000 x 36) : (1,05,000 x 30) : (1,40,000 x 24) = 12 : 15 : 16.
Aman : Rakhi : Sagar = (70,000 x 36) : (1,05,000 x 30) : (1,40,000 x 24) = 12 : 15 : 16.
Arun, Kamal and Vinay invested Rs. 8000, Rs. 4000 and Rs. 8000 respectively in a business. Arun left after six months. If after eight months, there was a gain of Rs. 4005, then what will be the share of Kamal?
Arun : Kamal : Vinay = (8,000 x 6) : (4,000 x 8) : (8,000 x 8)
= 48 : 32 : 64
= 3 : 2 : 4.
∴ Kamal’s share = Rs. (4005 x 2/9) = Rs. 890.
Arun : Kamal : Vinay = (8,000 x 6) : (4,000 x 8) : (8,000 x 8)
= 48 : 32 : 64
= 3 : 2 : 4.
∴ Kamal’s share = Rs. (4005 x 2/9) = Rs. 890.
Simran started a software business by investing Rs. 50,000. After six months, Nanda joined her with a capital of Rs. 80,000. After 3 years, they earned a profit of Rs. 24,500. What was Simran’s share in the profit?
Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.
∴ Simran’s share = Rs. (24500 x 3/7) = Rs. 10,500.
Simran : Nanda = (50000 x 36) : (80000 x 30) = 3 : 4.
∴ Simran’s share = Rs. (24500 x 3/7) = Rs. 10,500.
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
How much did Rohit get as profit at the yearend in the business done by Nitin, Rohit and Kunal?
I. Kunal invested Rs. 8000 for nine months, his profit was 3/2 times that of Rohit’s and his investment was four times that of Nitin.
II. Nitin and Rohit invested for one year in the proportion 1 : 2 respectively.
III. The three together got Rs. 1000 as profit at the year end.
I and II give:
K = Rs. (8000 × 9) for 1 month = Rs. 72000 for 1 month.
N = Rs. (1/4 × 8000 × 12) for 1 month = Rs. 24000 for 1 month.
R = Rs. 48000 for 1 month.
∴ K : N : R = 72000 : 24000 : 48000 = 3 : 1 : 2.
III gives, total profit = Rs. 1000.
∴ Rohit’s share = Rs. (1000 × 2/6) = Rs. 333.33 1
∴ Correct answer is (D).
I and II give:
K = Rs. (8000 × 9) for 1 month = Rs. 72000 for 1 month.
N = Rs. (1/4 × 8000 × 12) for 1 month = Rs. 24000 for 1 month.
R = Rs. 48000 for 1 month.
∴ K : N : R = 72000 : 24000 : 48000 = 3 : 1 : 2.
III gives, total profit = Rs. 1000.
∴ Rohit’s share = Rs. (1000 × 2/6) = Rs. 333.33 1
∴ Correct answer is (D).
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What is R’s share of profit in a joit venture?
I. Q started business investing Rs. 80,000.
II. R joined him after 3 months.
III. P joined after 4 months with a capital of Rs. 1,20,000 and got Rs. 6000 as his share profit.
From I, II and III, we get P : Q : R = (120000 x 8) : (80000 x 12) : (x x 9).
Since R’s investment is not given, the above ratio cannot be give.
∴ Given data is inadequate.
From I, II and III, we get P : Q : R = (120000 x 8) : (80000 x 12) : (x x 9).
Since R’s investment is not given, the above ratio cannot be give.
∴ Given data is inadequate.
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
Three friends, P, Q and R started a partnership business investing money in the ratio of 5 : 4 : 2 respectively for a period of 3 years. What is the amount received by P as his share profit?
I. Total amount invested in the business in Rs. 22,000.
II. Profit earned at the end of 3 years is 3/8 of the total investment.
III. The average amount of profit earned per year is Rs. 2750.
I and II give, profit after 3 years = Rs. (3/8 x 22000) = Rs. 8250.
From III also, profit after 3 years = Rs. (2750 x 3) = Rs. 8250.
∴ P’s share = Rs. (8250 x 5/11) = Rs. 3750.
Thus, (either III is redundant [or] I and II are redundant).
∴ Correct answer is (B).
I and II give, profit after 3 years = Rs. (3/8 x 22000) = Rs. 8250.
From III also, profit after 3 years = Rs. (2750 x 3) = Rs. 8250.
∴ P’s share = Rs. (8250 x 5/11) = Rs. 3750.
Thus, (either III is redundant [or] I and II are redundant).
∴ Correct answer is (B).
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
Ravi, Gagan and Nitin are running a business firm in partnership. What is Gagan’s share in the profit earned by them?
I. Ravi, Gagan and Nitin invested the amounts in the ratio of 2 : 4 : 7.
II. Nitin’s share in the profit is Rs. 8750.
Let us name Ravi, Gagan and Nitin by R, G and N respectively.
I. R : G : N = 2 : 4 : 7.
II. N = 8750..
From I and II, we get:
When N = 7, then G = 4.
When N = 8750, then G = (4/7 x 8750) = 5000.
Thus, both I and II are needed to get the answer.
∴ Correct answer is (E).
Let us name Ravi, Gagan and Nitin by R, G and N respectively.
I. R : G : N = 2 : 4 : 7.
II. N = 8750..
From I and II, we get:
When N = 7, then G = 4.
When N = 8750, then G = (4/7 x 8750) = 5000.
Thus, both I and II are needed to get the answer.
∴ Correct answer is (E).
Each of the questions given below consists of a statement and / or a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statement(s) is / are sufficient to answer the given question. Read the both statements and
Rahul, Anurag and Vivek started a business together. In what proportion would the annual profit be distributed among them?
I. Rahul got onefourth of the profit.
II. Rahul and Vivek contributed 75% of the total investment.
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Aptitude Area Online Test, Free Aptitude Quiz, Online Aptitude Area Test. Aptitude Area Question and Answers 2019. Aptitude Area Quiz. Aptitude Area Free Mock Test 2019. Aptitude Area Question and Answers in PDF. The Aptitude Area online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Area Question and Answers in Hindi and English. Aptitude Area Mock test for topic via Online Mode. Here we are providing Aptitude Area Mock Test in Hindi. Now Test your self for “Aptitude Area Online Test in Hindi” Exam by using below quiz…
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The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then the area of the park (in sq. m) is:
Perimeter = Distance covered in 8 min. = (12000/60 × 8) m = 1600 m.
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 × 320) m^{2} = 153600 m^{2}.
Perimeter = Distance covered in 8 min. = (12000/60 × 8) m = 1600 m.
Let length = 3x metres and breadth = 2x metres.
Then, 2(3x + 2x) = 1600 or x = 160.
∴ Length = 480 m and Breadth = 320 m.
∴ Area = (480 × 320) m^{2} = 153600 m^{2}.
An error 2% in excess is made while measuring the side of a square. The percentage of error in the calculated area of the square is:
100 cm is read as 102 cm.
∴ A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2} – A_{1}) = [(102)^{2} – (100)^{2}]
= (102 + 100) x (102 – 100)
= 404 cm^{2}.
∴ Percentage error = (404/100×100 × 100)x % = 4.04%
100 cm is read as 102 cm.
∴ A_{1} = (100 x 100) cm^{2} and A_{2} (102 x 102) cm^{2}.
(A_{2} – A_{1}) = [(102)^{2} – (100)^{2}]
= (102 + 100) x (102 – 100)
= 404 cm^{2}.
∴ Percentage error = (404/100×100 × 100)x % = 4.04%
The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle?
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq. m, then what is the width of the road?
Area of the park = (60 x 40) m^{2} = 2400 m^{2}.
Area of the lawn = 2109 m^{2}.
∴ Area of the crossroads = (2400 – 2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x – x^{2} = 291
⇒ x^{2} – 100x + 291 = 0
⇒ (x – 97)(x – 3) = 0
⇒ x = 3.
Area of the park = (60 x 40) m^{2} = 2400 m^{2}.
Area of the lawn = 2109 m^{2}.
∴ Area of the crossroads = (2400 – 2109) m^{2} = 291 m^{2}.
Let the width of the road be x metres. Then,
60x + 40x – x^{2} = 291
⇒ x^{2} – 100x + 291 = 0
⇒ (x – 97)(x – 3) = 0
⇒ x = 3.
The diagonal of the floor of a rectangular closet is 7.5 feet. The shorter side of the closet is 4.5 feet. What is the area of the closet in square feet?
A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is:
A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?
The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:
What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 × 41) cm^{2}.
Required number of tiles = (1517 × 902/41 × 41) = 814.
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile = (41 × 41) cm^{2}.
Required number of tiles = (1517 × 902/41 × 41) = 814.
The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is:
We have: (l – b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m^{2} = 2520 m^{2}.
We have: (l – b) = 23 and 2(l + b) = 206 or (l + b) = 103.
Solving the two equations, we get: l = 63 and b = 40.
∴ Area = (l x b) = (63 x 40) m^{2} = 2520 m^{2}.
The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?
Let original length = x and original breadth = y.
Original area = xy.
New length = x/2 .
New breadth = 3y.
New area = (x/2 × 3y) = 3/2 xy.
∴ Increase % = (1/2 xy × 1/xy × 100) % = 50%.
Let original length = x and original breadth = y.
Original area = xy.
New length = x/2 .
New breadth = 3y.
New area = (x/2 × 3y) = 3/2 xy.
∴ Increase % = (1/2 xy × 1/xy × 100) % = 50%.
The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ 26.50 per metre is Rs. 5300, what is the length of the plot in metres?
Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = (5300/26.50) m = 200 m.
∴ 2[(x + 20) + x] = 200
⇒ 2x + 20 = 100
⇒ 2x = 80
⇒ x = 40.
Hence, length = x + 20 = 60 m.
Let breadth = x metres.
Then, length = (x + 20) metres.
Perimeter = (5300/26.50) m = 200 m.
∴ 2[(x + 20) + x] = 200
⇒ 2x + 20 = 100
⇒ 2x = 80
⇒ x = 40.
Hence, length = x + 20 = 60 m.
A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required?
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
We have: l = 20 ft and lb = 680 sq. ft.
So, b = 34 ft.
∴ Length of fencing = (l + 2b) = (20 + 68) ft = 88 ft.
A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:
Area to be plastered = [2(l + b) × h] + (l × b)
= {[2(25 + 12) × 6] + (25 × 12)} m^{2}
= (444 + 300) m^{2}
= 744 m^{2}.
∴ Cost of plastering = Rs. (744 × 75/100) = Rs. 558.
Area to be plastered = [2(l + b) × h] + (l × b)
= {[2(25 + 12) × 6] + (25 × 12)} m^{2}
= (444 + 300) m^{2}
= 744 m^{2}.
∴ Cost of plastering = Rs. (744 × 75/100) = Rs. 558.
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What is the area of the hall?
I. Material cost of flooring per square metre is Rs. 2.50
II. Labour cost of flooring the hall is Rs. 3500
III. Total cost of flooring the hall is Rs. 14,500.
I. Material cost = Rs. 2.50 per m^{2}
II. Labour cost = Rs. 3500.
III. Total cost = Rs. 14,500.
Let the area be A sq. metres.
∴ Material cost = Rs. (14500 – 3500) = Rs. 11,000.
∴ 5A/2 = 11000 ⇔ A = 11000 × 2/5 = 4400 m^{2}.
Thus, all I, II and III are needed to get the answer.
∴ Correct answer is (C).
I. Material cost = Rs. 2.50 per m^{2}
II. Labour cost = Rs. 3500.
III. Total cost = Rs. 14,500.
Let the area be A sq. metres.
∴ Material cost = Rs. (14500 – 3500) = Rs. 11,000.
∴ 5A/2 = 11000 ⇔ A = 11000 × 2/5 = 4400 m^{2}.
Thus, all I, II and III are needed to get the answer.
∴ Correct answer is (C).
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What is the area of a rightangled triangle?
I. The perimeter of the triangle is 30 cm.
II. The ratio between the base and the height of the triangle is 5 : 12.
III. The area of the triangle is equal to the area of a rectangle of length 10 cm.
From II, base : height = 5 : 12.
Let base = 5x and height = 12x.
Then, hypotenuse = √(5x)^{2 }+ (12x)^{2 } = 13x.
From I, perimeter of the triangle = 30 cm.
∴ 5x + 12x + 13x = 30 ⇔ x = 1.
So, base = 5x = 5 cm, height = 12x = 12 cm.
∴ Area = (1/2 x 5 x 12)cm^{2} = 30 cm^{2}.
Thus, I and II together give the answer.
Clearly III is redundant, since the breadth of the rectangle is not given.
∴ Correct answer is (A).
From II, base : height = 5 : 12.
Let base = 5x and height = 12x.
Then, hypotenuse = √(5x)^{2 }+ (12x)^{2 } = 13x.
From I, perimeter of the triangle = 30 cm.
∴ 5x + 12x + 13x = 30 ⇔ x = 1.
So, base = 5x = 5 cm, height = 12x = 12 cm.
∴ Area = (1/2 x 5 x 12)cm^{2} = 30 cm^{2}.
Thus, I and II together give the answer.
Clearly III is redundant, since the breadth of the rectangle is not given.
∴ Correct answer is (A).
Each of the questions given below consists of a question followed by three statements. You have to study the question and the statements and decide which of the statement(s) is/are necessary to answer the question.
What is the area of rectangular field?
I. The perimeter of the field is 110 metres.
II. The length is 5 metres more than the width.
III. The ratio between length and width is 6 : 5 respectively.
I. 2(l + b) = 110 ⇒ l + b = 55.
II. l = (b + 5) ⇒ l – b = 5.
III. l/b = 6/5 ⇒ 5l – 6b = 0.
These are three equations in l and b. We may solve them pairwise.
∴ Any two of the three will give the answer.
∴ Correct answer is (B).
I. 2(l + b) = 110 ⇒ l + b = 55.
II. l = (b + 5) ⇒ l – b = 5.
III. l/b = 6/5 ⇒ 5l – 6b = 0.
These are three equations in l and b. We may solve them pairwise.
∴ Any two of the three will give the answer.
∴ Correct answer is (B).
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
What is the area of the given rectangle?
I. Perimeter of the rectangle is 60 cm.
II. Breadth of the rectangle is 12 cm.
III. Sum of two adjacent sides is 30 cm.
From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.
So, III is redundant.
Also, from II and III, we can find the length and breadth and therefore the area can be obtained.
So, I is redundant.
∴ Correct answer is “II and either I or III”.
From I and II, we can find the length and breadth of the rectangle and therefore the area can be obtained.
So, III is redundant.
Also, from II and III, we can find the length and breadth and therefore the area can be obtained.
So, I is redundant.
∴ Correct answer is “II and either I or III”.
Each of these questions is followed by three statements. You have to study the question and all the three statements given to decide whether any information provided in the statement(s) is redundant and can be dispensed with while answering the given question.
What is the cost painting the two adjacent walls of a hall at Rs. 5 per m2 which has no windows or doors?
I. The area of the hall is 24 sq. m.
II. The breadth, length and height of the hall are in the ratio of 4 : 6 : 5 respectively.
III. Area of one wall is 30 sq. m.
From II, let l = 4x, b = 6x and h = 5x.
Then, area of the hall = (24x^{2}) m^{2}.
From I. Area of the hall = 24 m^{2}.
From II and I, we get 24x^{2} = 24 ⇔ x = 1.
∴ l = 4 m, b = 6 and h = 5 m.
Thus, area of two adjacent walls = [(l x h) + (b x h)] m^{2} can be found out and so the cost of painting two adjacent walls may be found out.
Thus, III is redundant.
∴ Correct answer is (C).
From II, let l = 4x, b = 6x and h = 5x.
Then, area of the hall = (24x^{2}) m^{2}.
From I. Area of the hall = 24 m^{2}.
From II and I, we get 24x^{2} = 24 ⇔ x = 1.
∴ l = 4 m, b = 6 and h = 5 m.
Thus, area of two adjacent walls = [(l x h) + (b x h)] m^{2} can be found out and so the cost of painting two adjacent walls may be found out.
Thus, III is redundant.
∴ Correct answer is (C).
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Aptitude Time and Distance Online Test, Free Aptitude Quiz, Online Aptitude Time and Distance Test. Aptitude Time and Distance Question and Answers 2019. Aptitude Time and Distance Quiz. Aptitude Time and Distance Free Mock Test 2019. Aptitude Time and Distance Question and Answers in PDF. The Aptitude Time and Distance online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Time and Distance Question and Answers in Hindi and English. Aptitude Time and Distance Mock test for topic via Online Mode. Here we are providing Aptitude Time and Distance Mock Test in Hindi. Now Test your self for “Aptitude Time and Distance Online Test in Hindi” Exam by using below quiz…
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A person crosses a 600 m long street in 5 minutes. What is his speed in km per hour?
An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours. To cover the same distance in 1 2/3, it must travel at a speed of:
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 2/3 hours as 5/3 hours]
Distance = (240 x 5) = 1200 km.
Speed = Distance/Time
Speed = 1200/(5/3) km/hr. [We can write 1 2/3 hours as 5/3 hours]
If a person walks at 14 km/hr instead of 10 km/hr, he would have walked 20 km more. The actual distance travelled by him is:
Let the actual distance travelled be x km.
Then, X/10 = X+20/14
⇒ 14x = 10x + 200
⇒ 4x = 200
⇒ x = 50 km.
Let the actual distance travelled be x km.
Then, X/10 = X+20/14
⇒ 14x = 10x + 200
⇒ 4x = 200
⇒ x = 50 km.
A train can travel 50% faster than a car. Both start from point A at the same time and reach point B 75 kms away from A at the same time. On the way, however, the train lost about 12.5 minutes while stopping at the stations. The speed of the car is:
Excluding stoppages, the speed of a bus is 54 kmph and including stoppages, it is 45 kmph. For how many minutes does the bus stop per hour?
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = (9/54 × 60) min = 10 min.
Due to stoppages, it covers 9 km less.
Time taken to cover 9 km = (9/54 × 60) min = 10 min.
In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. The duration of the flight is:
A man complete a journey in 10 hours. He travels first half of the journey at the rate of 21 km/hr and second half at the rate of 24 km/hr. Find the total journey in km.
The ratio between the speeds of two trains is 7 : 8. If the second train runs 400 km in 4 hours, then the speed of the first train is:
A man on tour travels first 160 km at 64 km/hr and the next 160 km at 80 km/hr. The average speed for the first 320 km of the tour is:
A car travelling with 5/7 of its actual speed covers 42 km in 1 hr 40 min 48 sec. Find the actual speed of the car.
In covering a distance of 30 km, Abhay takes 2 hours more than Sameer. If Abhay doubles his speed, then he would take 1 hour less than Sameer. Abhay’s speed is:
Let Abhay’s speed be x km/hr.
Then, 30/x – 30/2x = 3
⇒ 6x = 30
⇒ x = 5 km/hr.
Let Abhay’s speed be x km/hr.
Then, 30/x – 30/2x = 3
⇒ 6x = 30
⇒ x = 5 km/hr.
Robert is travelling on his cycle and has calculated to reach point A at 2 P.M. if he travels at 10 kmph, he will reach there at 12 noon if he travels at 15 kmph. At what speed must he travel to reach A at 1 P.M.?
Let the distance travelled by x km.
Then, x/10 – x/15 = 2
⇒ 3x – 2x = 60
⇒ x = 60 km.
Time taken to travel 60 km at 10 km/hr = (60/10) hrs = 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = (60/5) kmph. = 12 kmph.
Let the distance travelled by x km.
Then, x/10 – x/15 = 2
⇒ 3x – 2x = 60
⇒ x = 60 km.
Time taken to travel 60 km at 10 km/hr = (60/10) hrs = 6 hrs.
So, Robert started 6 hours before 2 P.M. i.e., at 8 A.M.
Required speed = (60/5) kmph. = 12 kmph.
It takes eight hours for a 600 km journey, if 120 km is done by train and the rest by car. It takes 20 minutes more, if 200 km is done by train and the rest by car. The ratio of the speed of the train to that of the cars is:
Let the speed of the train be x km/hr and that of the car be y km/hr.
Solving (i) and (ii), we get: x = 60 and y = 80.
∴ Ratio of speeds = 60 : 80 = 3 : 4.
Let the speed of the train be x km/hr and that of the car be y km/hr.
Solving (i) and (ii), we get: x = 60 and y = 80.
∴ Ratio of speeds = 60 : 80 = 3 : 4.
A farmer travelled a distance of 61 km in 9 hours. He travelled partly on foot @ 4 km/hr and partly on bicycle @ 9 km/hr. The distance travelled on foot is:
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 x) km.
So, x/4 + (61x)/9 = 9
⇒ 9x + 4(61 x) = 9 x 36
⇒ 5x = 80
⇒ x = 16 km.
Let the distance travelled on foot be x km.
Then, distance travelled on bicycle = (61 x) km.
So, x/4 + (61x)/9 = 9
⇒ 9x + 4(61 x) = 9 x 36
⇒ 5x = 80
⇒ x = 16 km.
A man covered a certain distance at some speed. Had he moved 3 kmph faster, he would have taken 40 minutes less. If he had moved 2 kmph slower, he would have taken 40 minutes more. The distance (in km) is:
Let distance = x km and usual rate = y kmph.
Then, x/y – x/y+3 = 30/60 ⇒ 2y(y+3) = 9x…..(i)
And, x/y2 – x/y = 40/60 ⇒ y(y2) = 3x…..(ii)
On dividing (i) by (ii), we get: x = 40.
Let distance = x km and usual rate = y kmph.
Then, x/y – x/y+3 = 30/60 ⇒ 2y(y+3) = 9x…..(i)
And, x/y2 – x/y = 40/60 ⇒ y(y2) = 3x…..(ii)
On dividing (i) by (ii), we get: x = 40.
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Aptitude Problems on Trains Online Test, Free Aptitude Quiz, Online Aptitude Problems on Trains Test. Aptitude Problems on Trains Question and Answers 2019. Aptitude Problems on Trains Quiz. Aptitude Problems on Trains Free Mock Test 2019. Aptitude Problems on Trains Question and Answers in PDF. The Aptitude Problems on Trains online mock test paper is free for all students.The below Aptitude questions and answers can improve your skills in order to face the Interview, Competitive examination, Govt Exams and various entrance test with full confidence. Aptitude online test is very useful for exam preparation and getting for Rank. Aptitude Problems on Trains Question and Answers in Hindi and English. Aptitude Problems on Trains Mock test for topic via Online Mode. Here we are providing Aptitude Problems on Trains Mock Test in Hindi. Now Test your self for “Aptitude Problems on Trains Online Test in Hindi” Exam by using below quiz…
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A train running at the speed of 60 km/hr crosses a pole in 9 seconds. What is the length of the train?
A train 125 m long passes a man, running at 5 km/hr in the same direction in which the train is going, in 10 seconds. The speed of the train is:
The length of the bridge, which a train 130 metres long and travelling at 45 km/hr can cross in 30 seconds, is:
Two trains running in opposite directions cross a man standing on the platform in 27 seconds and 17 seconds respectively and they cross each other in 23 seconds. The ratio of their speeds is:
A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
A train 240 m long passes a pole in 24 seconds. How long will it take to pass a platform 650 m long?
Two trains of equal length are running on parallel lines in the same direction at 46 km/hr and 36 km/hr. The faster train passes the slower train in 36 seconds. The length of each train is:
A train 360 m long is running at a speed of 45 km/hr. In what time will it pass a bridge 140 m long?
Two trains are moving in opposite directions @ 60 km/hr and 90 km/hr. Their lengths are 1.10 km and 0.9 km respectively. The time taken by the slower train to cross the faster train in seconds is:
A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
A 270 metres long train running at the speed of 120 kmph crosses another train running in opposite direction at the speed of 80 kmph in 9 seconds. What is the length of the other train?
A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
Two trains, each 100 m long, moving in opposite directions, cross each other in 8 seconds. If one is moving twice as fast the other, then the speed of the faster train is: