Problem 33
If the cost price of 11 articles is equal to selling price of 10 articles, what is the profit percentage? (asked by Pretty)
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Problem 32
Zamera purchased 21 articles for $18 and sold 20 of them at the rate of 6 articles for $7. The remaining 1 article is damaged. What is the profit/loss percentage? (asked by Pretty)
Solution:
Modify the question by multiplying it with 3.
Zamera purchased 63 articles for $54 and sold 60 of them at 6 articles for $7. The remaining 3 articles are damaged.
Zamera’s cost price is $54 and selling price is $70.
Therefore, profit percentage is (16/54)*100 = 30% approx.
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Problem 31
A man purchases 9 pens for $10 and sells them at 6 pens for $7. What is the profit/loss per cent? (asked by Pretty)
Solution:
You can make this problem look simple by changing the numbers.
Find the LCM of 9 and 6 (18). Now, read the problem after changing the numbers.
A man purchases 18 pens for $20 and sells them at 18 for $21. What is the profit/loss per cent?
It looks simple, right?
For an investment of $20, the profit is $1. Therefore, for an investment of $100, the profit will be $5. Hence, the profit per cent is 5.
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Problem 30
A man sells two horses, one at 20% profit and the other at 20% loss. If the selling prices of the two horses are the same, what is the profit/loss percentage on the whole transaction? (asked by Pretty)
Solution:
You can use the formula I have given in the previous problem. But that is bit lengthy. If the profit and loss percentages are same, then you can simply use the other formula: x+y+x*y/100
20-20-(20*20/100) = -4 (negative sign indicates loss)
The answer is 4% loss.
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Problem 29
The selling prices of two tables are equal. There is a profit of 20% on one table and a loss of 5% on the other table. Approximately, what is the overall profit percentage? (asked by Pretty)
Solution:
The conventional method is bit lengthy. I provide you here a formula that may reduce your time taken to solve such problems. We assume selling price is $100 each.
To find the total cost price: 100(200+x+y)/(100+x+y+x*y/100)
In the above problem x=+20 and y=-5
100(215)/114 = $188.6 approximately.
Since total selling price is more than total cost price, there is a overall profit. Now, find the overall profit percentage.
11.4*100/188.6 = 6% approximately.
To find the overall profit/loss percentage:
100(x+y+2*x*y/100)/(200+x+y)
You can directly substitute the values of x and y and find the percentage.
100(20-5-2)/(200+20-5)
1300/215 = 6% approximately.
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Problem 28
A trader loses 25%, if an article is sold at $243. Find the selling price of that article in order to get a profit of 25% (asked by Pretty).
Solution:
The selling price of $243 is the cost price less the loss of 25% on the cost price.
That is the ratio of cost price to selling price is 100:75 or 4:3.
=> cost price is $324
Therefore, the required selling price is 324*1.25 = $405
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Problem 27
Peter gains 20 per cent by selling an article for $840. What is his cost price? (asked by Pretty)
Solution:
$840 includes cost price and 20% profit on cost price.
That is, 120% is 840. Therefore, what is 100%.
=> cost price is $700.
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Problem 26
If 40 per cent of x is 40 more than 40 per cent of 40, then how much is 140 per cent of x? (asked by Pretty)
Solution:
This is a standard question. The answer is 14*14 = 196
If it is 30 and 130, then the answer is 13*13 = 169
However, I provide you the conventional method:
.4*x = 40+.4*40 = 56 => x = 140; 1.4*140 = 196.
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Problem 25
In an exam 60 per cent of the boys passed whereas 30 per cent of the girls failed. If the number of girls who passed is 42, which is one-fourth the number of boys who failed. What is the total number of students who appeared for the exam? (asked by Pretty)
Solution:
From the first statement, we know that 70% of girls passed, which corresponds to an absolute value of 42. That means the total number of girls is 60. Also, 40% of boys failed, which corresponds to an absolute value of 4*42=168. That means the total number of boys is 420 (168*5/2). Therefore, total number of students appeared for the examination is 480.
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Problem 24
The population of a town increases annually at 25%. If the present population is 10 million, what is the difference between the population 3 years ago and 2 years ago? (asked by Pretty)
Solution:
If the present population is 125, then 1 year ago it was 100, 2 years ago it was 80, and 3 years ago it was 64. In this case, the difference between 3 years ago and 2 years ago is 16.
125 corresponds to 16
10 million corresponds to (16*10,000,000)/125 = 1,280,000.
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Problem 23
Stephen spends 70 per cent of his income. His income increases by 20 per cent and hence he increases his expenditure by 10 per cent. If Stephen’s initial expenditure was $17,500, then what is his present savings? (asked by Pretty)
Solution:
Let the income be $100; Stephen spends $70 and saves $30.
Now, his income is increased to $120; he spends $77 and saves $43.
$70 corresponds to $17,500
$43 corresponds to $10,750
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Problem 22
In a farm, there are some cows and hens. The number of cows is equal to 60 per cent of the number of hens. The total number of cows and hens is 48. Find the number of cows (asked by Pretty).
Solution:
If the number of hens is 100, then number of cows is 60; the ratio is 5:3.
Therefore, the number of cows is 48*(3/8) = 18
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Problem 21
A reduction of 37.5 per cent in the price of apples per dozen enables a customer to buy 6 dozens more for $20. Find the original price of apples per dozen (asked by Pretty).
Solution
Method 1
Let the original price be p per dozen and reduced price be .625p; let the original quantity be n dozens.
By supposition, n*p = (n+6)*.625p = $20.
Solving, we get p = $2
Method 2
We can use the formula, x+y+xy/100
-37.5+y-0.375y = 0, zero indicates no change in expenditure.
=>y = +60%, that is , 60 per cent increase in consumption
60 per cent increase in consumption corresponds to 6 dozen increase in absolute value.
Therefore, 100 per cent corresponds to 10 dozen. Hence the original consumption is 10 dozen.
Method 3
We can use another formula, 100*r/(100-r), where r is percentage decrease in price. The formula gives percentage increase in consumption while keeping the expenditure constant.
The formula gives 60% increase in consumption. For subsequent steps, refer method 2.
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Problem 20
A salesman receives salary of $500 per month and a commission of 5% on annual turnover. What should his annual sales be such that he earns an average income of $1250 per month? (asked by Pretty)
Solution
His annual income is $1250*12 = $15000, which includes his salary and commission.
His salary, 12*500 = $6000 and commission = $9000.
Therefore, his annual sales turnover is 9000(100/5) = $180,000.
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Problem 19
A solution of acid and water, contains 70 per cent acid and 30 per cent water. If 10 per cent of the solution is taken out and is replaced with water of same quantity, then find the percentage of water in the mixture. (asked by Pretty)
Solution
Let the solution is 100 liters; 70 liters acid and 30 liters water. If 10% of water is removed then only 27 liters of water remains in the solution. Again, if 10 liters of water is added then there will be 37 liters of water in the solution. It makes up 37 % of water in the resultant solution.
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Problem 18
In a certain examination there are 160 objective questions. Each correct option will get 3 marks, an incorrect option attracts 1 negative mark and a question not attempted attracts 1/2 negative mark. A student attempts 62.5% of the total questions and 20% of them are incorrect. Find the percentage of marks scored by the student (asked by Pretty)
Solution
62.5% of 160 = 100 (questions attended)
20% of 100 = 20 (incorrect); therefore, 80 are correct.
80 correct questions fetch her 240 marks; 20 incorrect attract 20 negative marks; 60 questions not attended attract 30 negative marks. The net score is 190.
Percentage of marks scored is (190/480)*100 = 39.58%
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Problem 17
In a class of 300 students, 75% of them passed in an examination. By what per cent the number of students who failed less than those who passed? (asked by Pretty)
Solution
75% of the students passed and 25% failed in the examination.
We have to find: (50/75)*100 = 66.67%
Note: You don’t need to find the actual number of students.
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Problem 16
A, B, C, and D are four quantities. A is 20% more than B, B is 100/3 per cent more than D. If D is 50/3 per cent less than C, then C is what per cent of A? (asked by Pretty)
Solution
If D = 300 B = 400; A = 480
If C is 300 then D is 250; ratio of C to D is 6 to 5. Therefore, if D is 300 then C is 360.
Now we got all values. We have to find C = x% of A.
360 = (x/100)*480 => x = (360*100)/480 = 75%
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Problem 15
In an election, 80 per cent of 1500 persons were ‘For’ for a resolution. If the number of voters increased by 100 per cent and the resolution was to be passed by a 2/3rd majority, what percentage of new voters should vote ‘For’ for the resolution? (asked by Pretty)
Solution
80% of 1500 = 1200
Present number of voters, after 100% increase = 3000
2/3 rd majority = 2000
Required new voters to pass the resolution = 2000-1200 = 800
x% of new voters = 800; (x/100)*1500 = 800
x = (800*100)/1500 =53.33%
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Problem 14
A worker spends 20% of his income on house rent, 30 % of the remaining on food, 40% of the remaining on education, 50% of the remaining on miscellaneous items and saves the remaining amount every month. What per cent of the total income does he save every month? (asked by Priya)
Solution
Let the worker’s total income be 1000
house rent = 200; remaining = 800
food = 240; remaining = 560
education = 224; remaining = 336
misc = 168; saving = 168
Therefore, the worker saves 16.8%
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Problem 13
Two candidates contested in an election. The candidate who got 40 per cent of votes was defeated by 3210 votes. Find the total number of votes polled if no votes were invalid (asked by Priya).
Solution
The percentage difference between the two candidates is 20% and this represents 3210 votes. We have to find 100%, that is, the total votes. The total votes are 5 times of 3210 = 16050.
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Problem 12
In a class 30 per cent of the students are boys. If in an examination, 40 per cent of the boys and 70 per cent of the total students have passed, then a) what is the per cent of the girls who passed in the examination?, b) Of the girls, what is the pass percentage?, c) what is the per cent of boys who passed in the examination?
Solution
Let the number of students be 1oo; => Boys = 30; Girls = 70
Boys who passed = 12; Girls who passed = 58
a) percentage of girls who passed = 58%
b) Of the girls, the percentage of girls who passed = (58/70)*100 = 82.85%
c) percentage of boys who passed = 12%
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Solution for problem 11
The percentage difference between the two students is 10%
The actual difference in marks between the two is 40
Therefore, the maximum marks (100%) is 400
Hence, the pass marks are 30% of 400+30 or 40% of 400-10, that is , 150.
Problem 11
In an examination, a student secured 30 per cent of the maximum marks and failed by 30 marks. Another student secured 40 per cent of the maximum marks which is 10 marks more than the pass mark. Find the pass mark (asked by Preeti).
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Solution for problem 10
Let the total votes be 100; therefore valid votes = 65
The ratio of votes secured by X and Y is 160:100 or 8:5
X secured 40% and Y secured 25% of the total votes.
Problem 10
In an election, there were only two contestants X and Y. It is found that 35 per cent of total votes are invalid. If X secured 60% more valid votes than Y, then find the percentage of the total votes secured by X (asked by Preeti).
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Solution for problem 09
Use the formula, x+y+xy/100.
25-y-.25y = -10
=>y = 28%
Problem 09
The price of an article is increased by 25%. By what percentage must its consumption be reduced so that the expenditure on it is reduced by 10 percent? (asked by Preeti)
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Solution for problem 08
The fractions are 1/6, 3/5, and 8/9. Take the LCM of the denominators: 6, 5, and 9. The LCM is 90. Start with 90 as a convenient number. It gives 15 mathematics, 45 fictions and 80/3 history books, which is a fraction. Therefore, 90 cannot be the answer.
Let’s take the next multiple, that is, 180. It gives 30 maths, 90 fictions and 160/3 history books, which is again a fraction.
Let’s take the next multiple, that is, 270. It gives 45 maths books, 135 fictions, 80 history books, and 10 science books. Hence, 270 is the least number of books in the library.
Problem 08
One sixth of the number of books in a library consist of Mathematics, 3/5 of the remainder of Fiction, 8/9 of what still remains of History, and the remaining books are on science. What should be the least number of books in the library to satisfy these conditions? (asked by Preeti)
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Solution for problem 07
The first pendulum takes 58/57 seconds for the next tick
The second pendulum takes 609/608 seconds for the next tick
LCM of 58/57 and 609/608 is 1218/19
They together tick after 1218/19 seconds
Therefore, in 1 hour or 3600 seconds, they tick together 3600*19/1218=56
Including the first tick 57 times.
Problem 07
One pendulum ticks 57 times in 58 seconds, another 608 times in 609 seconds. If they are started together, how often will they have ticked together in the first hour including the first stroke at start? (asked by Preeti)
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Solution for Problem 06
HCF of any two numbers = HCF of (sum of the two numbers, LCM)
Therefore, HCF of 126 and 180 = 18
Let the two numbers be 18x and 18y
18x*18y = 18*180 =>x*y = 10
The possible pairs of (x,y) = (1,10), (2,5)
Out of these pairs, only (2,5) fits the solution
The numbers are 36 and 90
Problem 06
The sum of two numbers is 126 and their LCM is 180. Find the numbers (asked by Preeti)
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Solution for Problem 05
Product of two numbers= HCF*LCM (of the two numbers)
Therefore, the product of the two numbers = 1260/15=84
Now, you have to find the set of coprime numbers whose product is 84
(1,84), (3,28), (4,21), (7, 12)
Problem 05
The least common multiple of two numbers is 1260 and their highest common factor is 15. How many pairs of such numbers can be found? (asked by Preeti)
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Solution for problem 04
The rule: You have to find the highest common factor of (175-151), (235-175), and (235-151)
HCF of 24, 60, and 84 = 12
Problem 04
Find the greatest number that will divide 151, 175, and 235 leaving the same remainder in each case (asked by Preeti).
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Solution for problem 03
The rule: subtract 3 from each number and then find highest common factor (HCF).
HCF of 210, 238, and 294 is 14.
Problem 03
Find the greatest number that will divide 213, 241, and 297 leaving remainder 3 in each case (asked by Preeti).
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Solution for Problem 02
S1 = 5; S2 = 1; Speed upstream = 4; speed down stream = 6
Let distance = d
By supposition, d/4 + d/6 = 1 => d= 2.4 mph
Problem 02
A man can row at 5 mph in still water. If the speed the river is 1 mph and it takes him 1 hour to row to a place and come back, how far is the place?
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Solution for problem 01
Let ‘S1’ and ‘S2’ are the speed of boat in still water and the speed of stream respectively.
S1 = 10 mph; S2 is not given.
Upstream speed = S1-S2; Downstream speed = S1+S2
By supposition, 36/(10-S2) = 36/(10+S2) + 3/2
Solving we get S2 = 2 mph
Today’s problem:01
A boat takes 90 minutes less to travel 36 miles downstream than to travel the same distance upstream. If the speed of the boat in still water is 10 miles per hour, the speed of the stream is:
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