The post Aptitude Simple Interest Online Test, Simple Interest Mock Test appeared first on Online Test.
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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 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 Work Online Test, Free Aptitude Quiz, Online Aptitude Time and Work Test. Aptitude Time and Work Question and Answers 2019. Aptitude Time and Work Quiz. Aptitude Time and Work Free Mock Test 2019. Aptitude Time and Work Question and Answers in PDF. The Aptitude Time and Work 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 Work Question and Answers in Hindi and English. Aptitude Time and Work Mock test for topic via Online Mode. Here we are providing Aptitude Time and Work Mock Test in Hindi. Now Test your self for “Aptitude Time and Work Online Test in Hindi” Exam by using below quiz…
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A can do a work in 15 days and B in 20 days. If they work on it together for 4 days, then the fraction of the work that is left is :
A’s 1 day’s work = 1/15 ;
B’s 1 day’s work = 1/20 ;
(A + B)’s 1 day’s work = (1/15 + 1/20) = 7/60
(A + B)’s 4 day’s work = (7/60 × 4) = 7/15
Therefore, Remaining work = (17/15) = 8/15
A’s 1 day’s work = 1/15 ;
B’s 1 day’s work = 1/20 ;
(A + B)’s 1 day’s work = (1/15 + 1/20) = 7/60
(A + B)’s 4 day’s work = (7/60 × 4) = 7/15
Therefore, Remaining work = (17/15) = 8/15
A, B and C can do a piece of work in 20, 30 and 60 days respectively. In how many days can A do the work if he is assisted by B and C on every third day?
A is thrice as good as workman as B and therefore is able to finish a job in 60 days less than B. Working together, they can do it in:
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 – 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes (3/2 × 60) = 90 days.
So, A takes 30 days to do the work.
A’s 1 day’s work = 1/30
B’s 1 day’s work = 1/90
(A + B)’s 1 day’s work = (1/30 + 1/90) = 4/90 = 2/45
A and B together can do the work in 45/2 = 22.5 days.
Ratio of times taken by A and B = 1 : 3.
The time difference is (3 – 1) 2 days while B take 3 days and A takes 1 day.
If difference of time is 2 days, B takes 3 days.
If difference of time is 60 days, B takes (3/2 × 60) = 90 days.
So, A takes 30 days to do the work.
A’s 1 day’s work = 1/30
B’s 1 day’s work = 1/90
(A + B)’s 1 day’s work = (1/30 + 1/90) = 4/90 = 2/45
A and B together can do the work in 45/2 = 22.5 days.
A alone can do a piece of work in 6 days and B alone in 8 days. A and B undertook to do it for Rs. 3200. With the help of C, they completed the work in 3 days. How much is to be paid to C?
If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be:
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2 .
Solving these two equations, we get : x = 1/100 and y = 1/200 .
(15 men + 20 boy)’s 1 day’s work = (15/100 + 20/200) = 1/4
∴ 15 men and 20 boys can do the work in 4 days.
Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y.
Then, 6x + 8y = 1/10 and 26x + 48y = 1/2 .
Solving these two equations, we get : x = 1/100 and y = 1/200 .
(15 men + 20 boy)’s 1 day’s work = (15/100 + 20/200) = 1/4
∴ 15 men and 20 boys can do the work in 4 days.
A can do a piece of work in 4 hours; B and C together can do it in 3 hours, while A and C together can do it in 2 hours. How long will B alone take to do it?
A’s 1 hour’s work = 1/4 ;
(B + C)’s 1 hour’s work = 1/3 ;
(A + C)’s 1 hour’s work = 1/2 .
(A + B + C)’s 1 hour’s work = (1/4 + 1/3) = 7/20
B’s 1 hour’s work = (7/12 – 1/2) = 1/12.
∴ B alone will take 12 hours to do the work.
A’s 1 hour’s work = 1/4 ;
(B + C)’s 1 hour’s work = 1/3 ;
(A + C)’s 1 hour’s work = 1/2 .
(A + B + C)’s 1 hour’s work = (1/4 + 1/3) = 7/20
B’s 1 hour’s work = (7/12 – 1/2) = 1/12.
∴ B alone will take 12 hours to do the work.
A can do a certain work in the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in:
(A + B)’s 1 day’s work = 1/10
C’s 1 day’s work = 1/50
(A + B + C)’s 1 day’s work = (1/10 + 1/50) = 6/50 = 3/25 ….(i)
A’s 1 day’s work = (B + C)’s 1 day’s work …. (ii)
From (i) and (ii), we get: 2 x (A’s 1 day’s work) = 3/25
A’s 1 day’s work = 3/50 .
⇒ B’s 1 day’s work = (1/10 – 3/50) = 2/50 = 1/25.
∴ So, B alone could do the work in 25 days.
(A + B)’s 1 day’s work = 1/10
C’s 1 day’s work = 1/50
(A + B + C)’s 1 day’s work = (1/10 + 1/50) = 6/50 = 3/25 ….(i)
A’s 1 day’s work = (B + C)’s 1 day’s work …. (ii)
From (i) and (ii), we get: 2 x (A’s 1 day’s work) = 3/25
A’s 1 day’s work = 3/50 .
⇒ B’s 1 day’s work = (1/10 – 3/50) = 2/50 = 1/25.
∴ So, B alone could do the work in 25 days.
A does 80% of a work in 20 days. He then calls in B and they together finish the remaining work in 3 days. How long B alone would take to do the whole work?
Whole work is done by A in ( 20 x 5/4 ) = 25 days.
Now, (1 – 4/5) i.e., 1/5 work is done by A and B in 3 days.
Whole work will be done by A and B in (3 x 5) = 15 days.
A’s 1 day’s work = 1/25 , (A + B)’s 1 day’s work = 1/15 .
Therefore B’s 1 day’s work = (1/15 – 1/25) = 4/150 = 2/75.
So, B alone would do the work in 75/2 = 37.5 days.
Whole work is done by A in ( 20 x 5/4 ) = 25 days.
Now, (1 – 4/5) i.e., 1/5 work is done by A and B in 3 days.
Whole work will be done by A and B in (3 x 5) = 15 days.
A’s 1 day’s work = 1/25 , (A + B)’s 1 day’s work = 1/15 .
Therefore B’s 1 day’s work = (1/15 – 1/25) = 4/150 = 2/75.
So, B alone would do the work in 75/2 = 37.5 days.
A machine P can print one lakh books in 8 hours, machine Q can print the same number of books in 10 hours while machine R can print them in 12 hours. All the machines are started at 9 A.M. while machine P is closed at 11 A.M. and the remaining two machines complete work. Approximately at what time will the work (to print one lakh books) be finished ?
A can finish a work in 18 days and B can do the same work in 15 days. B worked for 10 days and left the job. In how many days, A alone can finish the remaining work?
B’s 10 day’s work = (1/15 × 10) = 2/3
Remaining work = (12/3) = 1/3.
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in ( 18 x 1/3 ) = 6 days.
B’s 10 day’s work = (1/15 × 10) = 2/3
Remaining work = (12/3) = 1/3.
Now, 1/18 work is done by A in 1 day.
1/3 work is done by A in ( 18 x 1/3 ) = 6 days.
4 men and 6 women can complete a work in 8 days, while 3 men and 7 women can complete it in 10 days. In how many days will 10 women complete it?
Let 1 man’s 1 day’s work = x and 1 woman’s 1 day’s work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10 .
Solving the two equations, we get: x = 11/400 , y = 1/400
∴ 1 woman’s 1 day’s work = 1/400 .
⇒ 10 women’s 1 day’s work = (1/400 × 10) = 1/40.
Hence, 10 women will complete the work in 40 days.
Let 1 man’s 1 day’s work = x and 1 woman’s 1 day’s work = y.
Then, 4x + 6y = 1/8 and 3x + 7y = 1/10 .
Solving the two equations, we get: x = 11/400 , y = 1/400
∴ 1 woman’s 1 day’s work = 1/400 .
⇒ 10 women’s 1 day’s work = (1/400 × 10) = 1/40.
Hence, 10 women will complete the work in 40 days.
A and B can together finish a work 30 days. They worked together for 20 days and then B left. After another 20 days, A finished the remaining work. In how many days A alone can finish the work?
(A + B)’s 20 day’s work = ( 1/30 x 20 ) = 2/3 .
Remaining work = ( 1 – 2/3 ) = 1/3 .
Now, 1/3 work is done by A in 20 days.
Therefore, the whole work will be done by A in (20 x 3) = 60 days.
(A + B)’s 20 day’s work = ( 1/30 x 20 ) = 2/3 .
Remaining work = ( 1 – 2/3 ) = 1/3 .
Now, 1/3 work is done by A in 20 days.
Therefore, the whole work will be done by A in (20 x 3) = 60 days.
10 women can complete a work in 7 days and 10 children take 14 days to complete the work. How many days will 5 women and 10 children take to complete the work?
1 woman’s 1 day’s work = 1/70
1 child’s 1 day’s work = 1/140
(5 women + 10 children)’s day’s work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7
∴ 5 women and 10 children will complete the work in 7 days.
1 woman’s 1 day’s work = 1/70
1 child’s 1 day’s work = 1/140
(5 women + 10 children)’s day’s work = (5/70 + 10/140) = (1/14 + 1/14) = 1/7
∴ 5 women and 10 children will complete the work in 7 days.
X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last?
A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done in 23 days?
Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.
Then, 10 : 13 :: 23 : x ⇒ x = (23×13 / 10) ⇒ x = 299/10.
A’s 1 day’s work = 1/23 ;
B’s 1 day’s work = 10/299.
(A + B)’s 1 day’s work = (1/23 + 10/299) = 23/299 = 1/13.
Therefore, A and B together can complete the work in 13 days.
Ratio of times taken by A and B = 100 : 130 = 10 : 13.
Suppose B takes x days to do the work.
Then, 10 : 13 :: 23 : x ⇒ x = (23×13 / 10) ⇒ x = 299/10.
A’s 1 day’s work = 1/23 ;
B’s 1 day’s work = 10/299.
(A + B)’s 1 day’s work = (1/23 + 10/299) = 23/299 = 1/13.
Therefore, A and B together can complete the work in 13 days.
Ravi and Kumar are working on an assignment. Ravi takes 6 hours to type 32 pages on a computer, while Kumar takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
Number of pages typed by Ravi in 1 hour = 32/6 = 16/3 .
Number of pages typed by Kumar in 1 hour = 40/5 = 8.
Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3.
Time taken by both to type 110 pages = (110 × 3/40) hours
= 8 1/4 hours (or) 8 hours 15 minutes.
Number of pages typed by Ravi in 1 hour = 32/6 = 16/3 .
Number of pages typed by Kumar in 1 hour = 40/5 = 8.
Number of pages typed by both in 1 hour = (16/3 + 8) = 40/3.
Time taken by both to type 110 pages = (110 × 3/40) hours
= 8 1/4 hours (or) 8 hours 15 minutes.
Sakshi can do a piece of work in 20 days. Tanya is 25% more efficient than Sakshi. The number of days taken by Tanya to do the same piece of work is:
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x ⇒ x = (4 x 20/5)
⇒ x = 16 days.
Hence, Tanya takes 16 days to complete the work.
Ratio of times taken by Sakshi and Tanya = 125 : 100 = 5 : 4.
Suppose Tanya takes x days to do the work.
5 : 4 :: 20 : x ⇒ x = (4 x 20/5)
⇒ x = 16 days.
Hence, Tanya takes 16 days to complete the work.
A takes twice as much time as B or thrice as much time as C to finish a piece of work. Working together, they can finish the work in 2 days. B can do the work alone in:
Suppose A, B and C take x, x/2 and x/3 days respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
⇒ 6/x = 1/2
⇒ x = 12.
So, B takes (12/2) = 6 days to finish the work.
Suppose A, B and C take x, x/2 and x/3 days respectively to finish the work.
Then, (1/x + 2/x + 3/x) = 1/2
⇒ 6/x = 1/2
⇒ x = 12.
So, B takes (12/2) = 6 days to finish the work.
A and B can complete a work in 15 days and 10 days respectively. They started doing the work together but after 2 days B had to leave and A alone completed the remaining work. The whole work was completed in :
(A + B)’s 1 day’s work = (1/15 + 1/10) = 1/6
Work done by A and B in 2 days = (1/6 × 2) = 1/3.
Remaining work = (1 – 1/3) = 2/3
Now, 1/15 work is done by A in 1 day.
∴ 2/3 work will be done by a in (15 × 2/3) = 10 days.
Hence, the total time taken = (10 + 2) = 12 days.
(A + B)’s 1 day’s work = (1/15 + 1/10) = 1/6
Work done by A and B in 2 days = (1/6 × 2) = 1/3.
Remaining work = (1 – 1/3) = 2/3
Now, 1/15 work is done by A in 1 day.
∴ 2/3 work will be done by a in (15 × 2/3) = 10 days.
Hence, the total time taken = (10 + 2) = 12 days.
A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
A works twice as fast as B. If B can complete a work in 12 days independently, the number of days in which A and B can together finish the work in :
Ratio of rates of working of A and B = 2 : 1.
So, ratio of times taken = 1 : 2.
B’s 1 day’s work = 1/12 .
∴ A’s 1 day’s work = 1/6 ; (2 times of B’s work)
(A + B)’s 1 day’s work = (1/6 + 1/12) = 3/12 = 1/4.
So, A and B together can finish the work in 4 days.
Ratio of rates of working of A and B = 2 : 1.
So, ratio of times taken = 1 : 2.
B’s 1 day’s work = 1/12 .
∴ A’s 1 day’s work = 1/6 ; (2 times of B’s work)
(A + B)’s 1 day’s work = (1/6 + 1/12) = 3/12 = 1/4.
So, A and B together can finish the work in 4 days.
Twenty women can do a work in sixteen days. Sixteen men can complete the same work in fifteen days. What is the ratio between the capacity of a man and a woman?
(20 x 16) women can complete the work in 1 day.
1 woman’s 1 day’s work = 1/320 .
(16 x 15) men can complete the work in 1 day.
1 man’s 1 day’s work = 1/240
So, required ratio = 1/240 : 1/320
= 1/3 : 1/4
= 4 : 3 (cross multiplied)
(20 x 16) women can complete the work in 1 day.
1 woman’s 1 day’s work = 1/320 .
(16 x 15) men can complete the work in 1 day.
1 man’s 1 day’s work = 1/240
So, required ratio = 1/240 : 1/320
= 1/3 : 1/4
= 4 : 3 (cross multiplied)
A and B can do a work in 8 days, B and C can do the same work in 12 days. A, B and C together can finish it in 6 days. A and C together will do it in :
(A + B + C)’s 1 day’s work = 1/6 ;
(A + B)’s 1 day’s work = 1/8 ;
(B + C)’s 1 day’s work = 1/12 .
So, A and C together will do the work in 8 days.
(A + B + C)’s 1 day’s work = 1/6 ;
(A + B)’s 1 day’s work = 1/8 ;
(B + C)’s 1 day’s work = 1/12 .
So, A and C together will do the work in 8 days.
A can finish a work in 24 days, B in 9 days and C in 12 days. B and C start the work but are forced to leave after 3 days. The remaining work was done by A in:
(B + C)’s 1 day’s work = (1/9 + 1/12) = 7/36.
Work done by B and C in 3 days = (7/36 × 3) = 7/12.
Remaining work = (17/12) = 5/12.
Now, 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in (24 × 5/12) = 10 days.
(B + C)’s 1 day’s work = (1/9 + 1/12) = 7/36.
Work done by B and C in 3 days = (7/36 × 3) = 7/12.
Remaining work = (17/12) = 5/12.
Now, 1/24 work is done by A in 1 day.
So, 5/12 work is done by A in (24 × 5/12) = 10 days.
A and B together can do a piece of work in 30 days. A having worked for 16 days, B finishes the remaining work alone in 44 days. In how many days shall B finish the whole work alone?
Let A’s 1 day’s work = x and B’s 1 day’s work = y.
Then, x + y = 1/30 and 16x + 44y = 1.
Solving these two equations, we get: x = 1/60 and y = 1/60
B’s 1 day’s work = 1/60 .
Hence, B alone shall finish the whole work in 60 days.
Let A’s 1 day’s work = x and B’s 1 day’s work = y.
Then, x + y = 1/30 and 16x + 44y = 1.
Solving these two equations, we get: x = 1/60 and y = 1/60
B’s 1 day’s work = 1/60 .
Hence, B alone shall finish the whole work in 60 days.
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Father is aged three times more than his son Ronit. After 8 years, he would be two and a half times of Ronit’s age. After further 8 years, how many times would he be of Ronit’s age?
Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.
∴ (4x + 8) = 5/2 (x + 8)
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
⇒ x = 8.
Hence, required ratio = (4x+16) / (x+16) = 48/24 = 2.
Let Ronit’s present age be x years. Then, father’s present age =(x + 3x) years = 4x years.
∴ (4x + 8) = 5/2 (x + 8)
⇒ 8x + 16 = 5x + 40
⇒ 3x = 24
⇒ x = 8.
Hence, required ratio = (4x+16) / (x+16) = 48/24 = 2.
The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
⇒ 5x = 20
⇒ x = 4.
∴ Age of the youngest child = x = 4 years.
Let the ages of children be x, (x + 3), (x + 6), (x + 9) and (x + 12) years.
Then, x + (x + 3) + (x + 6) + (x + 9) + (x + 12) = 50
⇒ 5x = 20
⇒ x = 4.
∴ Age of the youngest child = x = 4 years.
A father said to his son, “I was as old as you are at the present at the time of your birth”. If the father’s age is 38 years now, the son’s age five years back was:
Let the son’s present age be x years. Then, (38 – x) = x
⇒ 2x = 38.
⇒ x = 19.
∴ Son’s age 5 years back (19 – 5) = 14 years.
Let the son’s present age be x years. Then, (38 – x) = x
⇒ 2x = 38.
⇒ x = 19.
∴ Son’s age 5 years back (19 – 5) = 14 years.
A is two years older than B who is twice as old as C. If the total of the ages of A, B and C be 27, the how old is B?
Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.
∴ (2x + 2) + 2x + x = 27
⇒ 5x = 25
⇒ x = 5.
Hence, B’s age = 2x = 10 years.
Let C’s age be x years. Then, B’s age = 2x years. A’s age = (2x + 2) years.
∴ (2x + 2) + 2x + x = 27
⇒ 5x = 25
⇒ x = 5.
Hence, B’s age = 2x = 10 years.
Present ages of Sameer and Anand are in the ratio of 5 : 4 respectively. Three years hence, the ratio of their ages will become 11 : 9 respectively. What is Anand’s present age in years?
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, 5x+3 / 4x+3 = 11/9
⇒ 9(5x + 3) = 11(4x + 3)
⇒ 45x + 27 = 44x + 33
⇒ 45x – 44x = 33 – 27
⇒ x = 6.
∴ Anand’s present age = 4x = 24 years.
Let the present ages of Sameer and Anand be 5x years and 4x years respectively.
Then, 5x+3 / 4x+3 = 11/9
⇒ 9(5x + 3) = 11(4x + 3)
⇒ 45x + 27 = 44x + 33
⇒ 45x – 44x = 33 – 27
⇒ x = 6.
∴ Anand’s present age = 4x = 24 years.
A man is 24 years older than his son. In two years, his age will be twice the age of his son. The present age of his son is:
Let the son’s present age be x years. Then, man’s present age = (x + 24) years.
∴ (x + 24) + 2 = 2(x + 2)
⇒ x + 26 = 2x + 4
⇒ x = 22.
Let the son’s present age be x years. Then, man’s present age = (x + 24) years.
∴ (x + 24) + 2 = 2(x + 2)
⇒ x + 26 = 2x + 4
⇒ x = 22.
Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11 : 10. What is Sagar’s age at present?
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then,(6x+6)+4 / (5x+6)+4 = 11/10
⇒ 10(6x + 10) = 11(5x + 10)
⇒ 5x = 10
⇒ x = 2.
∴ Sagar’s present age = (5x + 6) = 16 years.
Let the ages of Kunal and Sagar 6 years ago be 6x and 5x years respectively.
Then,(6x+6)+4 / (5x+6)+4 = 11/10
⇒ 10(6x + 10) = 11(5x + 10)
⇒ 5x = 10
⇒ x = 2.
∴ Sagar’s present age = (5x + 6) = 16 years.
The sum of the present ages of a father and his son is 60 years. Six years ago, father’s age was five times the age of the son. After 6 years, son’s age will be:
Let the present ages of son and father be x and (60 x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
⇒ 54 – x = 5x – 30
⇒ 6x = 84
⇒ x = 14.
∴ Son’s age after 6 years = (x+ 6) = 20 years..
Let the present ages of son and father be x and (60 x) years respectively.
Then, (60 – x) – 6 = 5(x – 6)
⇒ 54 – x = 5x – 30
⇒ 6x = 84
⇒ x = 14.
∴ Son’s age after 6 years = (x+ 6) = 20 years..
At present, the ratio between the ages of Arun and Deepak is 4 : 3. After 6 years, Arun’s age will be 26 years. What is the age of Deepak at present ?
Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,
4x + 6 = 26 ⇔ 4x = 20
x = 5.
∴ Deepak’s age = 3x = 15 years.
Let the present ages of Arun and Deepak be 4x years and 3x years respectively. Then,
4x + 6 = 26 ⇔ 4x = 20
x = 5.
∴ Deepak’s age = 3x = 15 years.
Sachin is younger than Rahul by 7 years. If their ages are in the respective ratio of 7 : 9, how old is Sachin?
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
∴ x7/x = 7/9
⇒ 9x – 63 = 7x
⇒ 2x = 63
⇒ x = 31.5
Hence, Sachin’s age =(x – 7) = 24.5 years.
Let Rahul’s age be x years.
Then, Sachin’s age = (x – 7) years.
∴ x7/x = 7/9
⇒ 9x – 63 = 7x
⇒ 2x = 63
⇒ x = 31.5
Hence, Sachin’s age =(x – 7) = 24.5 years.
The present ages of three persons in proportions 4 : 7 : 9. Eight years ago, the sum of their ages was 56. Find their present ages (in years).
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 56
⇒ 20x = 80
⇒ x = 4.
∴ Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
Let their present ages be 4x, 7x and 9x years respectively.
Then, (4x – 8) + (7x – 8) + (9x – 8) = 56
⇒ 20x = 80
⇒ x = 4.
∴ Their present ages are 4x = 16 years, 7x = 28 years and 9x = 36 years respectively.
Ayesha’s father was 38 years of age when she was born while her mother was 36 years old when her brother four years younger to her was born. What is the difference between the ages of her parents?
Mother’s age when Ayesha’s brother was born = 36 years.
Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.
∴ Required difference = (42 – 36) years = 6 years.
Mother’s age when Ayesha’s brother was born = 36 years.
Father’s age when Ayesha’s brother was born = (38 + 4) years = 42 years.
∴ Required difference = (42 – 36) years = 6 years.
A person’s present age is twofifth of the age of his mother. After 8 years, he will be onehalf of the age of his mother. How old is the mother at present?
Let the mother’s present age be x years.
Then, the person’s present age = (2/5x)years.
∴ (2/5x+8) = 1/2(x+8)
⇒ 2(2x + 40) = 5(x + 8)
⇒ x = 40.
Let the mother’s present age be x years.
Then, the person’s present age = (2/5x)years.
∴ (2/5x+8) = 1/2(x+8)
⇒ 2(2x + 40) = 5(x + 8)
⇒ x = 40.
Q is as much younger than R as he is older than T. If the sum of the ages of R and T is 50 years, what is definitely the difference between R and Q’s age?
Given that:
1. The difference of age b/w R and Q = The difference of age b/w Q and T.
2. Sum of age of R and T is 50 i.e. (R + T) = 50.
Question: R – Q = ?.
Explanation:
R – Q = Q – T
(R + T) = 2Q
Now given that, (R + T) = 50
So, 50 = 2Q and therefore Q = 25.
Question is (R – Q) = ?
Here we know the value(age) of Q (25), but we don’t know the age of R.
Therefore, (RQ) cannot be determined.
Given that:
1. The difference of age b/w R and Q = The difference of age b/w Q and T.
2. Sum of age of R and T is 50 i.e. (R + T) = 50.
Question: R – Q = ?.
Explanation:
R – Q = Q – T
(R + T) = 2Q
Now given that, (R + T) = 50
So, 50 = 2Q and therefore Q = 25.
Question is (R – Q) = ?
Here we know the value(age) of Q (25), but we don’t know the age of R.
Therefore, (RQ) cannot be determined.
The age of father 10 years ago was thrice the age of his son. Ten years hence, father’s age will be twice that of his son. The ratio of their present ages is:
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10]
⇒ 3x + 20 = 2x + 40
⇒ x = 20.
∴ Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
Let the ages of father and son 10 years ago be 3x and x years respectively.
Then, (3x + 10) + 10 = 2[(x + 10) + 10]
⇒ 3x + 20 = 2x + 40
⇒ x = 20.
∴ Required ratio = (3x + 10) : (x + 10) = 70 : 30 = 7 : 3.
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 Sonia’s present age?
I. Sonia’s present age is five times Deepak’s present age.
II. Five years ago her age was twentyfive times Deepak’s age at that time.
I. S = 5D D = S/5 ….(i)
II. S – 5 = 25 (D – 5) ⇔ S = 25D – 120 ….(ii)
Using (i) in (ii), we get S = (25 × S/5) – 120
⇒ 4S = 120.
⇒ S = 30.
Thus, I and II both together give the answer. So, correct answer is (E).
I. S = 5D D = S/5 ….(i)
II. S – 5 = 25 (D – 5) ⇔ S = 25D – 120 ….(ii)
Using (i) in (ii), we get S = (25 × S/5) – 120
⇒ 4S = 120.
⇒ S = 30.
Thus, I and II both together give the answer. So, 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
Average age of employees working in a department is 30 years. In the next year, ten workers will retire. What will be the average age in the next year?
I. Retirement age is 60 years.
II. There are 50 employees in the department.
I. Retirement age is 60 years.
II. There are 50 employees in the department.
Average age of 50 employees = 30 years.
Total age of 50 employees = (50 x 30) years = 1500 years.
Number of employees next year = 40.
Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.
Average age next year = 940/40 years = 23.5 years.
Thus, I and II together give the answer. So, correct answer is (E).
I. Retirement age is 60 years.
II. There are 50 employees in the department.
Average age of 50 employees = 30 years.
Total age of 50 employees = (50 x 30) years = 1500 years.
Number of employees next year = 40.
Total age of 40 employees next year (1500 + 40 – 60 x 10) = 940.
Average age next year = 940/40 years = 23.5 years.
Thus, I and II together give the answer. So, 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
Divya is twice as old as Shruti. What is the difference in their ages?
I. Five years hence, the ratio of their ages would be 9 : 5.
II. Ten years back, the ratio of their ages was 3 : 1.
Let Divya’s present age be D years and Shruti’s present age b S years
Then, D = 2 x S ⇔ D – 2S = 0 ….(i)
I. D + 5 / S + 5 = 9/5 ….(ii)
II. D – 10 / S – 10 = 3/1 ….(iii)
From (ii), we get : 5D + 25 = 9S + 45 ⇔ 5D – 9S = 20 ….(iv)
From (iii), we get : D – 10 = 3S – 30 ⇔ D – 3S = 20 ….(v)
Thus, from (i) and (ii), we get the answer.
Also, from (i) and (iii), we get the answer.
∴ I alone as well as II alone give the answer. Hence, the correct answer is (C).
Let Divya’s present age be D years and Shruti’s present age b S years
Then, D = 2 x S ⇔ D – 2S = 0 ….(i)
I. D + 5 / S + 5 = 9/5 ….(ii)
II. D – 10 / S – 10 = 3/1 ….(iii)
From (ii), we get : 5D + 25 = 9S + 45 ⇔ 5D – 9S = 20 ….(iv)
From (iii), we get : D – 10 = 3S – 30 ⇔ D – 3S = 20 ….(v)
Thus, from (i) and (ii), we get the answer.
Also, from (i) and (iii), we get the answer.
∴ I alone as well as II alone give the answer. Hence, the 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 Arun’s present age?
I. Five years ago, Arun’s age was double that of his son’s age at that time.
II. Present ages of Arun and his son are in the ratio of 11 : 6 respectively.
III. Five years hence, the respective ratio of Arun’s age and his son’s age will become 12 : 7.
I. 5 years ago, Arun’s age = 2 x His son’s age.
II. Let the present ages of Arun and his son be 11x and 6x years respectively.
III. 5 years hence, Arun’s Age / Son’s age = 12/7
Clearly, any two of the above will give Arun’s present age.
∴ Correct answer is (D).
I. 5 years ago, Arun’s age = 2 x His son’s age.
II. Let the present ages of Arun and his son be 11x and 6x years respectively.
III. 5 years hence, Arun’s Age / Son’s age = 12/7
Clearly, any two of the above will give Arun’s present age.
∴ 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 Ravi’s present age?
I. The present age of Ravi is half of that of his father.
II. After 5 years, the ratio of Ravi’s age to that of his father’s age will be 6 : 11.
III. Ravi is 5 years younger than his brother.
I. Let Ravi’s present age be x years. Then, his father’s present age = 2x years.
II. After 5 years, Ravi’s age / Father’s age = 6/11
III. Ravi is younger than his brother.
From I and II, we get x + 5 / 2x + 5 = 6/11 . This gives x, the answer.
Thus, I and II together give the answer. Clearly, III is redundant.
Correct answer is (A).
I. Let Ravi’s present age be x years. Then, his father’s present age = 2x years.
II. After 5 years, Ravi’s age / Father’s age = 6/11
III. Ravi is younger than his brother.
From I and II, we get x + 5 / 2x + 5 = 6/11 . This gives x, the answer.
Thus, I and II together give the answer. Clearly, III is redundant.
Correct answer is (A).
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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: