TIME DISTANCE Example 1. A man travels a distance of 61 km in 9 hours partly on foot at the rate of 4 km/h and partly on bicycle at 9 km/h. How m uch distance does he cover on foot?
Let the distance covered on foot be x. Then the distance covered by bicycle is 61 -x. The total distance is covered in 9 hours. Time for which the man travels by foot = Distance/ velocity = x/4 hrs Time for which the man travels by bicycle = (61 -x) / 9 hrs x/4 + (61 - x)/9 = 9 9x + 244 -4x = 9 36 5x = 324 -244 x = 16 km Example 2. Brian covers a distance of 200 200 km traveling with a uniform speed of 'S' km/h. He could have covered the same distance in 2 hours less had the speed been (s + 5) km/h. Find the value of s.
Let 't' be the time that Brian takes to cover 200 km at 'S' km/hr. km/hr. Then t = 200/s At a speed of (s + 5) km/hr, the distance of 200 km gets covered in (t -2) hours. Then , (t -2) = 200 / (s + 5) Substituting the value of t = 200/s in the above equation, we obtain, (200/s -2) = 200/(s + 5) 200 -2s = s (s + 5) (200 -2s) (s + 5) = 200s 200s -2s + 1000 -10s = 200s 200s 2s + 10s -1000 = 0 (Refer to the chapter on Quadratic Equations) S + 5s -500 -500 = 0 S + 25s -20s -500 = 0 s (s + 25) -20 (s + 25) = 0 (s -20) (s + 25) = 0 s = 20 or s = -25 Since speed cannot be negative, we discard s = -25 Hence our solution, s = 20. Example 3. How long will a train 75m long moving at 60 km/h take to pass a platform 200m long?
To completely completely pass a platform, the train will have to cover a distance equal to the sum of its own length and the length of the platform. The total length to be covered by the train is 75 + 200 = 275 mtrs. The velocity of the train is 60 km/hr, which needs to be converted into mtrs per sec. 60 km/hr can be written as 60,000 mtrs per 60 * 60 secs. Or 100/6 mtrs/sec. Using the formula velocity = Distance/Time, we get Time = Distance/Velocity T ime to cross the platform = 275 = 275 * 6 = 33/2 = 16.5 secs. 100/6 100
Example 4. A 75 m long train moving at 60 km/h can pass another another train 100 m long, moving at 65 km/h in the opposite direction in:
Such problems can be solved using the formula velocity = distance/time. It's necessary to make sure that similar units are used in the formula. To completely pass each other, other, the trains have to cover a distance equal to the sum of the Iengths of the two trains, 75 + 100 = 175 mtrs. When travelling in opposite directions, the velocity with which this distance gets covered is the sum of the 2 velocities. 60 + 65 = 125 Kmph or 125 * 1000 mtrs per 60 * 60 Secs or 625/18 mtrs/sec. mtrs/sec. Therefore time to cross each other = 175 = 5.04 secs. 625/18 Example 5. Train A is 120 m Iong traveling traveling at 90 km/h. It overtakes train B which is 130 mtrs Iong traveling traveling in the same direction at 72 km/h. How long does it take for train A to overtake train B.
To completely pass each other, other, the trains have to cover a distance equal to the sum of the Iengths of the two trains, 120 + 130 = 250 mtrs. When travelling in the same direction, the velocity with which this distance gets covered is the difference between the 2 velocities. velocities. 90 - 72 = 18 Kmph or 18 * 1000 mtrs per per 60 * 60 Secs or 5 mtrs/sec. Therefore time to cross each other = = 50 secs. 5 Example 6. Find the speed of the current if a boy rows 13 km upstream upstream and 28 km downstream taking 5 hours each time.
Let the speed of the current be x km/hr and that of the boy when he rows in still water be y km/hr. Then the relative speed when the boy rows upstream = y -x km/hr The relative speed when the boy rows downstream = y + x km/hr The time taken for the 13 km long upstream journey is 5 hours. Therefore we can write this as Speed = Distance/Time y -x = 13/5 13/5 ------ I The time taken for the 28 km long downstream journey is also 5 hours, therefore y + x = 28/5 ----- IIII Subtracting equations ll from equation I, we obtain, Subtracting equations ll from equation I, we obtain, 2x = 28/5 -13/5 2x = (28 -13) / 5 2x = 15/5 x = 3/2 = 1.5 km/hr, km/hr, the speed of the the current Example 7. A boat moves with a speed speed of 8 km/h in still water. Find the rate rate of the stream if the boat can travel 20 km downstream in the same time as it can traveI 12 km upstream.
Let the speed of the stream be x km/hr And the time that it takes to traveI 20 km downstream or 12 km upstream be t hrs. Then Time t when moving upstream is given as Time = Distance/Velocity Distance/Velocity t = 12/(8 -x) -------- I The time t when moving downstream is given as t = 20/(8 + x) ----------- II
Combining I and II, 12/(8 -x) = 20/(8 + x) 12(8 + x) = 20(8 -x)
Since the work is the same, the same no. of man-hours are required. Since now, 48 men work for 7 days, Iet's assum e they work for Y hours a d ay. Then, 48 * 7 * Y = 20 * 8 * 21 Y = 20 * 8 * 21 = 10 hours. 48 * 7 Example 3. Abu and Bill together can do a piece of work in 10 days, but Abu alone can do it in 15 days. In how many days would Bill alone do the same job? Let 'b' be the no. of days that Bill takes to do the job alone.
12(8 + x) = 20(8 -x) 96 + 12x = 160 -20x 32x = 64 x = 2 km/hr, the speed of the stream
WORK
If one person can do a work in 6 days, then two people having the same efficiency can do the same workin 3 days, or 3 persons can do that work in 2 days. This follows that a job requires a certain number of man-days to complete. Man-days is the product of the number of persons working on the job and the number of days they take to complete the job. For a particular job the man-days required to complete it remain constant. In the above example the man-days required are 1 * 6 = 6 man-days, or 2 * 3 = 6 man-days, or 3 * 2 = 6 man-days. For a particular job, if the number of people on the job increases, then the number of days that they take to complete the job reduces proportionately, such that the number of man-days remains constant. Also note that if a person can complete a certain job in 10 days, then the amount of work that he does in one day is 1/10 of the total. Problems on Pipes
If a pipe can fill a tank in 10 minutes, then it fills 1/10th of the tank in 1 minute. If two different pipes can fill the same tank in 10 minutes and 15 minutes separately, then the amount that they will fill in one minute, running together, is 1/10 + 1/15 = 1/6 . And the time that they will take, while running together, to fill the tank is 1/(1/6) minutes or 6 minutes. These are the basics that you need to understand to tackle all problems related to Time and Work. Solved Examples Example 1. If sixteen men can do a piece of work in 12 days, working 14 hours a day, how long wilI 28 men, working 12 hours a day, take to do the same work?
In the first scenario, the number of man-hours required to complete the work = 16 * 12 * 14 Since it is the same job, the man-hours required to complete the job remains the same. The man-hours in the second scenario = 28 * 12 * Y, where Y is the number of days required. Therefore, 16 * 12 * 14 = 28 * 12 * Y Y = 8 days Example 2. Twenty men working 8 hours per day can complete a piece of work in 21 days. How many hours per day must 48 men work to complete the same piece of work in 7 days?
20 men working 8 hours a day can complete a piece of work in 21 days. This implies that 20 * 8 * 21 hours is required to complete the work. Or 20 * 8 * 21 man-hours of work.
Then the amount of work that Bill can do in one day = 1/b The amount of work that Abu can complete in one day = 1/15 Since the two working together can complete the work in 10 days, the amount of work that the two can complete in one day = 1/10 Thus, 1/b + 1/15 = 1/10 Or 1/b = 1/10 -1/15 1/b = 1/30 or b = 30 Example 4. It takes 21 days for 36 men to build a walI 140 m long. How many men would require 18 days to build a similar walI 50 m in length? It requires 21 * 36 man-days to build a 140 mtr long walI. To build a 50 mtr long walI, 21 * 36 * 50 man-days are required. 140 To build the 50 mtr long walI, Iet Y be the no. of men required for 18 days. Or 18 * mandays. Thus 18 Y = 21 * 36 * 50 140 or Y = 15 Example 5. P, Q and R can do a certain piece of work in 12, 15 and 20 days respectively. They began to work but R Ieaves after 2 days. In how many days would P and Q be able to finish the work together? The amount of work P, Q and R working together can complete in one day 1/12 + 1/15 + 1/20 = 5 + 4 + 3 = Ñ 60 60 5 The amount of work, P and Q working together can complete in 1 day is 1/12 + 1/15 = 9/60 = 3/20 Before R Ieaves, in 2 days before his departure the three of them working together complete 2/5 of the work (1/5 * 2 = 2/5) The amount of work to be completed by P and Q working together after the departure of R = 1 -2/5 = 3/5 Let Y be the number of days that P and Q work to complete the job. Then 3/20 x Y = 3/5 Or Y = 4 Example 6. Anan, Bardle and Cicely can finish a piece of work in 18 days; Anan and Cicely together work twice as much as Bardle, Anan and Bardle together work thrice as much as Cicely. In what time can each do it individually? Let a, b and c be the no. of days that Anan, Bardle and Cicely take to do the job individually. Then 1/a, 1/b and 1/c are the amount of work each can finish in one day. As per the question, 1/a + 1/c = 2 * 1/b 1/a + 1/b = 3 * 1/c
and we also know that all of them working together can complete the work in 18 days. Therefore, 1/a + 1/b + 1/c = 1/18 Thus we have 3 equations and 3 variables, which can be solved to obtain the values of each variable. 1/c -1/b = 2/b -3/c 3/c + 1/c = 1/b + 2/b 4/c = 3/b 3c = 4b Put c = 4/3 b in 1/a + 1/c = 2/b 1/a + 3/4b = 2/b 1/a = 2/b -3/4b 1/a = 5/4b Substitute the values of 'a' and 'c' in the third equation to obtain, 5/4b + 1/b + 3/4b = 1/18 on solving, we obtain 1/b = 1/54, b = 54 This implies that Bardle can complete the entire work alone in 54 days. Now substituting the value of 'b' in 1/a = 5/4b, We obtain 1/a = 5/(4 * 54) 1/a = 5/216 or a = 43 1/5 Similarly solving for 'c' by using the expression 3c = 4b c = 4 * 54/3 = 72 Example 7. Pipe P fills a tank in 24 minutes. Pipe Q fills the same tank in 30 minutes. Both the pipes are turned on together, and after 8 minutes, pipe P is turned off. In how many minutes would pipe Q fill the remaining tank? P can fill the tank in 24 minutes or it fills 1/24 of the tank in 1 minute. Q can fill the tank in 30 minutes or it fills 1/30 of the tank in 1 minute. Running together the two pipes will filI 1/24 + 1/30 of the tank in 1 minute. In 8 minutes the two pipes together will filI 8 (1/24 + 1/30) of the tank, The amount left to be filled in the tank after 8 minutes = 1 - 8 (1/24 + 1/30) The time that it would take Pipe Q to fill the remaining capacity of the tank: = 1 8 1/24 + 1/30 1/30 1-8 5+4 -
=
120
1/30
=
1 -8 9/120 1/30
= 30 120 -72 120
= 30 * 48 = 48 / 4 = 12 minutes 120 Example 8. It takes 16 and 32 minutes for pipes A and B respectively to fill a cistern. Another pipe C at the bottom of the cistern can empty the full cistern in 64 minutes. If all the pipes are opened together, how Iong will it take for the cistern to filI? The part of the cistern that A alone can fill in one minute is 1/16 The part of the cistern that B alone can fill in one minute is 1/32 The part of the cistern that C can empty in one minute is 1/64 When all the pipes are opened, in one minute, the volume of the tank that gets filled is 1/16 + 1/32 1/64 = (4 + 2 1) / 64 = 5/64 If 5/64 of the tank gets filled in one minute, then the time that it takes to fill the tank completely is = 64/5 = 12.8 minutes. 5/64 -
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MIXTURE: Example 1: A 50 Iitre mixture of wine and water contains 20% wine. How much more wine should be added to the
mixture to increase the concentration of wine to 50%? Solution: The quantity of wine in the mixture = (20/100) * 50 = 10 Iitres. Let the quantity of wine added to the mixture be x litres. Then the total quantity of wine in the mixture is (10 + x) Iitres and the total quantity of mixture is 50 + x litres. The proportion of wine in the new mixture is (10 + x) / (50 + x) We require this proportion to be equal to 50%. Therefore, (10 + x) / (50 + x) = 50/100 = . 20 + 2x = 50 + x x = 30 Hence, 30 Iitres of wine when added to 50 Iitres of the mixture, results in 80 Iitres of the mixture with 50% concentration of wine. That is, 10 Iitres of wine was already in the original mixture and another 30 Iitres when added makes the total quantity of wine as 40 Iitres in 80 Iitres of the mixture, which is 50% of the total volume of the mixture. Example 2: 10 kg of grade 'A' of Tea costing $15 per kg is mixed with 30 kg of grade 'B' of Tea costing $10 per kg. What is the cost per kg of the mixture? Solution: Cost of 10 kg of grade 'A = 10 * 15 = $150 Cost of 30 kg of grade 'B = 30 * 10 = $300 TotaI Cost of 30 kgs of the mixture = $150 + $300 = $450. Cost of the mixture = $450/40 = $11.25 per kg. Example 3: In what proportion should tea costing $1.00 per kg be mixed with tea costing $2.00 per kg to get a mixture that costs $1.35 per kg? Solution: Let the quantity of tea costing $1.00 per kg be x and that of tea costing $2.00 per kg be y. Then, the quantity of the mixture costing $1.35 per kg would be x + y. Therefore, 1 * x + 2 * y = 1.35 * (x + y) x + 2y = 1.35x + 1.35y 0.65y = 0.35x x/y = 0.65/0.35 = 13/7 which means, 13 parts of tea costing $1.00 per kg and 7 parts of tea costing $2.00 per kg are needed to make a mixture costing $1.35 per kg. Shortcut Problems such as these can be solved without involving variables like x, y etc. GRAPHIC HERE PIace the two cost prices, 1 and 2 in the top corners as shown. PIace the cost price of the mixture in between the 2 prices, then take the difference of the cost price and the price of the mixture and place the results as shown by the arrow. Now, simplify the two differences to obtain the required proportion. The above shortcut can be used whenever quantities before the mixing and after the mixing are both given
in the problem, and it is the proportion of their mixing that is required to be found. The quantity could be price, profit or percentage content of an ingredient. We have taken an example where the cost price was provided. The next two examples are based on profits and % age content of an ingredient. Example 4: A store owner has a stock of 100 shirts. He sold some shirts for a 10 day period at a profit of 20%. Thereafter, he put the shirts on a sale, and sold the shirts at a Iower profit of 10%. If he made an overall profit of 13%, how many shirts did he sell in the sale? Example 4: A store owner has a stock of 100 shirts. He sold some shirts for a 10 day period at a profit of 20%. Thereafter, he put the shirts on a sale, and sold the shirts at a Iower profit of 10%. If he made an overall profit of 13%, how many shirts did he sell in the sale? GRPAH Example 5: A bartender removed from a bottle a certain quantity of a drink, which had 25% alcohol content and replaced the same quantity with another drink that had an alcohol content of 10%. The alcohol content in the drink after replacement is 20%. What was the quantity of the drink that he replaced, if there was 1 Iitre of the drink in the bottle? graphic here Using the shortcut, 1/3 * 1 Iitre = 1/3 Iitres Example 6: A barrel contains 100 Iitres of an acid. 10 Iitres of the acid is taken out and replaced by water. This process is repeated 2 more times. What is the amount of acid in the barrel now? Solution: The quantity of acid, after the first replacement is, 90/100 of the totaI. After the second operation, it is, (90/100) * (90/100) = (90/100) Similarly, after the third operation, (90/100) = 729/1000 Since there was 100 Iitres of acid to start with, the acid in the barrel after 3 replacements is: (729/1000) * 100 = 72.9 Iitres. Shortcut Acid left in the barrel after n replacements / starting quantity of acid = [(p _q) / p]n Where, p = starting quantity of acid q = the replaced quantity n = number of replacements In the above example: P = 100. q = 10. n = 3. In the above example: P = 100; q = 10; n = 3. Hence [(p _q)/p]n = [(100 _10) / 100] = (90/100) = 729/1000 PERCENT
Percent means per cent or per 100 or for every 100. A fraction whose denominator is 100 is called a percentage and the numerator of the fraction is called the rate percent. It is denoted by the symbol "%". Therefore, 12% denotes 12/100. A percent can be expressed as a fraction and fraction as a percentage. The fraction 3/25 can be expressed as a percentage by multiplying and dividing the fraction by 100 and retaining the 100 in the denominator. Thus, [(3/25) / 100] * 100 = (300/25) / 100 = 12/100 or 12%. And 12% can be expressed as a fraction by dividing the percent by 100 and simplifying; 12% = 12/100 = 3/25 Example: What percentage is the marks of a student who scored 12/20 in a test? Solution: Multiply and divide 12/20 by 100, and retain 100 in the denominator as shown: {(12/20) * 100} / 100 = 60/100 = 60% Percentages are used; A. to determine an increase or decrease in a quantity B. to define a part of a quantity C. to compare two different quantities Let us take an example of each: A. An increase or decrease in a quantity. If a number, say 30 is increased by 25%, then 25% of 30 is (25/100) * 30 = 15/2 = 7.5. This implies that increasing 30 by 25% is the same as increasing 30 by 7.5. The number after the increase becomes; 30 + 7.5 = 37.5, or this can be found by using the formula: 30 [(100 + r)/100], where r is the percentage increase. 30 [(100 + 25) / 100] = 30 * (5/4) = 150/4 = 37.5. B. Define a part of a quantity. Out of 100,000 people in a village, 52% are women. Find the number of men in the village. Number of women in the village = (52/100) * 100,000 = 52,000 Number of men in the village = 100,000 _52,000 = 48,000. C. Compare two quantities: If Ben earns $1000 per week, and Keith earns 15% more than Ben, then what is Keith's income? Keith's income = 1000 * (100 + 15) / 100 = $1150. *** Time goes. Kleptomania is expiring. Order now at http://www.structurise.com/kleptomania/order.htm *** Example 1. Max receives an annual salary of $8800 in the year 1965. This is 10% more than his salary in 1964. What was his salary in 1964? If Max's salary in 1964 was 100, then his salary in 1965 is 100 + 10% of 100 That is, 100 + * 100 = 110 100 Thus when his salary in 1965 is 110, his salary in 1964 was 100. When his salary in 1965 is 1, then his salary in 1964 = 110 Now, when his salary in 1965 is $8800, then his salary in 1964 = * 8800 = $8000 110 Example 2. To pass a certain examination, a student needs to obtain 35% of the total marks. He gives three papers. In the first paper, he gets 62 out of 120, and in the second paper he obtains just 35 out of 150. In order to just qualify for pass, how much must he obtain out of 180 in the third paper? The maximum marks out of the three papers = 120 + 150 + 180 = 450 The pass percentage is 35% Passing marks = 35 * 450 = 157.5 100
Marks already scored in the two papers = 62 + 35 = 97 Marks to be scored in the third paper = 157.5 _97 = 60.50
*** Copy file lists and folder trees from Explorer *** Example 3. 11/18th of the population of a village are males, and the rest are females. If the total population of the village is 4500 and 40% of the males are married, find the number of married males. The number of males in the village = Ñ11 * 4500 = 2750 18 The number of married males = * 2750 = 1100 100 Example 4. The total produce of a farm in the year 1965 was 150 quintals. It increased by 10% in 1966 and then by 8% in 1967. Calculate the total produce in 1967. The produce in the year 1966 = 150 (100 + 10) = 165 quintals 100 The produce in the year 1967 = 165 (100 + 8) = 178.2 quintals 100 Example 5. A and B contested an election. The winning candidate secured 57% of the total valid votes polled and won by a majority of 42000 votes. Calculate the number of total votes polled. If the winning candidate secured 57% of valid votes, then the votes polled by the Iosing candidate = 100 _57 = 43% Difference between the votes polled by the two candidates = 57 _ 43 = 14%, which is the majority percentage. In other words when 14 is the majority, total votes polled is 100. When 1 is the majority, total votes polled is 100/14. When 42000 is the majority, total votes polled is 100/14 * 42000 = 3,00,000 votes. Example 6. A person Iost 12% of his money in gambling and 5% of the remainder was picked from his pocket. If the total money he now has is $8360, what was the original sum of money he had before gambling? Let the original sum of money he had be x. After losing 12% of his money on gambling, he is left with (100 _12)% or 88% of x. He then loses 5% of the balance, which is 5% of 88% or 4.4% of the original amount x. The balance that he has left with him is 83.6% of x or 0.836 x. This is given to be equal to $8360. Therefore, 0.836 x = 8360 Or x = 8360/0.836 = $10,000 PROBABILITY Probability defines the chances of an event occurring. An event is any desired outcome, example, rain on a particular day, heads on tossing of a coin, etc. The probability of an event occurring Iies between the numbers 0 and 1. A 0 probability makes the event impossible to occur and a probability of 1 defines an event that is certain to occur. The higher the probability, the higher its chances of occurring.
If an event A can happen in m different ways and cannot happen in n ways, and when the occurrence of all events is equally Iikely to happen and the happening of any one does not in any way rule out the possibility of another event happening, then, Probability of event A happening, P(A) = m / (m + n) The number of ways in which the event A can happen P(A) = Number of ways in which event A can happen+Number of ways in which event A cannot happen Since an event can either happen or it cannot happen. m + n = r, where r is the total number of possible outcomes. Therefore, P(A) = m/r Solved Examples : 1. What is the probability of heads turning up on a single toss of a coin? The total number of possible outcomes on a single toss of a coin is 2 _a heads or a tails. The number of ways in which the required event is likely to happen is 1 _a heads showing up. Probability of heads turning up = P (Heads) = 2. What is the probability of 2 showing up on a single throw of a dice with 6 faces? On a single throw of a dice, the total number of possible outcomes is 6 - (1, 2, 3, 4, 5 or 6 showing up)
The number of ways in which the required event is likely to happen is 1 _(2 showing up) Probability of 2 showing up = P (2) = 1/6 3. What is the probability of a number greater than 3 showing up on a single throw of a dice? On a single throw of a dice, the total number of possible outcomes is 6 - (1, 2, 3, 4, 5 or 6 showing up)
The number of ways in which a number greater than 3 will show up is 3 _(4, 5, 6 showing up) Probability of a number greater than 3 showing up = P (number > 3) = 3/6 = 1/2 4.
What is the probability of drawing a King from a pack of playing cards?
The total number of cards in a pack of playing cards is 52. Since all 52 cards are different the total number of possible outcomes is 52. There are 4 kings in a pack of playing cards, therefore number of ways that a King can be drawn is 4. Probability of drawing a king = P (King) = 4/52 = 1/13 5.
If 2 die are thrown, what is the probability of getting a 10?
The total number of different combinations possible with a throw of 2 dice is 6 * 6 = 36 The number of ways in which 10 can show up is : Die 1 Die 2 Total 6 4 10 5 5 10 4 6 10 Therefore in 3 different ways Probability of getting a 10 = P (10) = 3/36 = 1/12 Note : If the probability of an event occurring is p then the probability of it,s not occurring is 1p Therefore in the above problem if Probability of getting a 10 is 1/12, then the probability of not getting a 10 is 1 - 1/12 = 11/12
6. Jenny, Bill and Mark are assigned a task, the solution to which has to be worked out by them working together. If the probability of their finding the solution is 1/3, 1/4 and 1/5 respectively, find the probability of their finding a solution. The required probability is the product of their individual probabilities. P (Solution) = 1/3 * 1/4 * 1/5 = 1/60 Note : The probability of occurrence of two or more independent events which occur at the same time is the product of their individual probabilities. In the above examp e Jenny, Bill and Mark are working on the task independently, hence the probability of their completing the task is the product of their individual probabilities. 7. In the above question if Jenny, Bill and Mark are working on the task independently and the task is considered complete when any one of them can complete the task, then what is the probability of completion of the task? The probability of any one of them completing the task is computed as shown: Probability of Jenny not completing the task = 1 - 1/3 = 2/3 Probability of Bill not completing the task = 1 _1/4 = 3/4 Probability of Mark not completing the task = 1 _1/5 = 4/5 The probability of the task not being completed = 2/3 * 3/4 * 4/5 = 2/5 And the probability of the task being completed = 1 _ Probability of task not being completed = 1 _2/5 = 3/5 Note : It's easier to solve the problem by following the method above. The alternate way to solve the problem is P (task being completed) = P (any one of them can complete the task) + P (any two of them can complete the task) + P (AII three can complete the task). Where P (any one of them can c omplete the task) is given as = P (Jenny completes the task and the other 2 don't) + P (Bill can complete the task and the other 2 don't) + P (Mark can complete the task and the other 2 don't) *** Request more tips by using Kleptomania *** And P (Jenny completes the task and the other 2 don't) = 1/3 * 3/4 * 4/5 = 1/5 Similarly you can compute the other probabilities to achieve the same result that we computed above, i.e., P (task being completed) = 3/5 8. There are 2 boxes, the first contains 5 blue and 3 red balls and the second contains 4 blue and 6 red balls. What is the probability of drawing a blue ball if one ball is drawn from a bag that is selected at random? The probability of selecting a particular bag is , since there are 2 bags. The probability of selecting a blue ball from the first bag: No. of blue balls in the first bag = 5 Total number of balls in the first bag = 5 + 3 = 8 Probability of drawing a blue balI = 5/8 Probability of drawing a blue ball from the first bag = (Probability of selecting the first bag) * (Probability of selecting a blue balI) = 1/2 * 5/8 = 5/16 The probability of selecting a blue ball from the second bag: No. of blue balls in the Second bag = 4 Total number of balls in the second bag = 4 + 6 = 10
Probability of drawing a blue ball from the second bag = 4/10 = 2/5 Probability of drawing a blue ball from the second bag = (Probability of selecting the second bag) * (Probability of selecting a blue balI) = 1/2 * 2/5 = 1/5 The probability of drawing a blue balI = P (BIue ball from first bag) + P (BIue ball from second bag) P (BIue BalI) = 5/16 + 1/5 = (25 + 16) / 80 = 41/80 Note : The probability of occurrence of ù Ê of the two or more events is the sum of their individual probabilities. In the above example a blue ball can be drawn from the first or the second bag and the required probability is the sum of the probabilities of drawing a blue ball from the first bag and from the second bag. You may note that the individual probabilities have to be added if one of the many events have to occur. And they have to be multiplied if all the events can occur simultaneously. SIMULATANEOUS EQUATIONS To solve for two variables, 2 equations are required to be worked upon simultaneously. Example: Solve for x and y; 2x + 3y = 12 3x + 5y = 19 To solve the above problem, multiply each equation with a number such that either x or y term becomes equal in both the equations. In the given example, multiply the first equation by 3 and the second by 2. Thus, 6x + 9y = 36 6x + 10y = 38 Subtracting the second equation from the first one; + 9y = 36 + 10 = 38 -y = -2 or y = 2 Now, substitute the value of y in any one of the equations to obtain the value of x. Thus, 2x + 3 * 2 = 12 2x = 12 -6 x = 6/2 = 3 AIternate method for solving simultaneous equations (The Substitution method) Let's take the same example as above. Solve for x and y; 2x + 3y = 12 3x + 5y = 19 Take the first equation and find the value of x in terms of y. 2x = 12 -3y x = (12 -3y)/2 Now substitute the value of x computed above in the second equation. 3x + 5y = 19 3 (12 -3y)/2 + 5y = 19 18 -9y/2 + 5y = 19 5y -9y/2 = 1 10y -9y = 2
y=2 Now substitute y = 2 in any of the equations to get the value of x. 2x + 3.2 = 12 2x = 12 -6 x = 6/2 = 3 Therefore x = 3 and y = 2 Solved Examples Example 1. Two numbers are such that the sum of twice the first and thrice the second is 18, while the sum of thrice the first and twice the second is 17. Find the numbers. Let the two numbers be x and y. Then as per the question 2x + 3y = 18 3x + 2y = 17 Solve for x and y, Multiply both sides of the first equation by 3 and of the second equation by 2. 6x + 9y = 54 6x + 4y = 34 Subtracting the second equation from the first, we obtain, 6x + 9y = 54 6x + 4y = 34 5y = 20 or y = 4 __ = 18 -3. 4 = 6/2 = 3 2 2 Thus the two numbers are 3 and 4. Example 2. If 2 is subtracted from both the numerator and the denominator of a fraction, it becomes 1/3, and if 1 is added to both the numerator and the denominator, it becomes 2/3. Find the fraction. Let the fraction be x/y, then = 1, y _2 3 3x -6 = y -2, or 3x -y = 4 x + 1 = 2, y+1 3 3x + 3 = 2y + 2, or 3x -2y = -1 Solve for x and y using the 2 equations, 3x _y = 4 3x _2y = -1 Subtracting the second equation from the first, y=5 x = (4 + y)/3 = (4 + 5)/3 = 9/3 = 3 Therefore the required fraction is, x/y = 3/5 PROFIT AND LOSS When a good is bought for $100 and sold for $110, there is a profit of $10. Profit = Selling Price (SP) _Cost Price (CP). When SP is less than CP, there is a loss. Loss = CP _SP Profit and loss are generally denoted in terms of percent, of cost price. % age Profit = [(SP _CP) / CP] * 100 % age Loss = [(CP _SP) / CP] * 100
A trader offers a discount. The discount is offered on a price that is displayed to customers, called the marked price (MP). The selling price on a discounted good is always lesser than the marked price. For a discounted good, with D as the percentage discount, SP = MP (100 _D) / 100 and MP = SP * [100 / (100 _D)] Solved Examples Example 1. Suleman bought 240 roses at $9 per dozen and sold all of them at $1 each. Find the profit percentage of Suleman? Cost Price per rose = $9/12 Sale price per rose = $1 Profit percentage = SP _CP * 100 = (1 _9/12 100 = 3/12 100 CP 9/12 9/12 = 3 * 100 = = 33 1/3 % 9 3 Example 2. A bookseller allows a discount of 10% on the advertised price of a particular book. What price must be marked on the book which costs him $600 to make a profit of 20%? Cost price of the book = $600 Required profit percentage = 20% Required profit = 20/100 * 600 = $120 Required Selling Price = CP + profit = 600 + 120 = $720 Required selling price is the price at which the book is sold after giving a discount of 10%. Therefore the price of the book should be marked such that a discount of 10% on the marked price results in the book being sold at $720. If the marked price is Y, then Y * 100 _discount ercent = Required Selling Price 100 => Y * 100 _10 = 720 100 => Y = 720 * 100 90 => Y = $800 Or simply using the formula MP = SP * [100 / (100 _D)] MP = 720 * [100 / (100 _10)] MP = 720 (100/90) = $800 Example 3. Abraham sells a camera, which costs him $400 to Ben at a profit of 20%. Ben then sells it to Chang, making a profit of 10% on the price he paid to Abraham. What does Chang pay to Ben? Abraham's Selling price = Ben's cost price = 400 * 100 + 20 = $480 100 Ben's selling price = Chang's cost price = 480 * 100 + 10 = $528 100 Example 4. A shopkeeper allows two successive discounts of 20% and 10% on an article. If he gets $108 from the article, find its marked price. For successive discounts use the formula Marked Price = Selling Price * 100 * 100 100 _First Discount 100 _second discount Marked Price = 108 * 100 * 100 = 108 * 100 * 100 = 108 * 100 = 100 - 20 100 _10 80 * 90 72 Example 5. Kumar bought two cameras for $400. He sells one of the cameras at a profit of 20% and the other at a loss of 20%. find the total loss or profit percent.
Selling price of the first camera = 400 * 100 + 20 = 4 * 120 = $480 100 Profit on the first camera = 480 _400 = $80 Selling Price of the second camera = 400 x 100 _20 = 4 x 80 = $320 100 Loss on the second camera = 400 _320 = $80 Net profit or loss = 80 _80 = 0 You may note that when the cost price of the two articles is the same and the profit and loss percentages on their sale is also the same, then the net profit is always zero. Example 6. The cost price of each of the 1000 articles is $0.85. It is then found that only 700 articles may be sold. If the selling price is to be fixed such that a profit of 40% on the total cost price is obtained, find the selling price of one article. The total cost of 1000 articles is 0.85 * 1000 = $850 A profit of 40% on the total cost price = 850 * = $340 100 The required selling price of 700 articles = 850 + 340 = $1190 Selling price of each article = 1190 / 700 = $1.70 Example 7. A trader buys 180 articles at $40 each from a wholesaler. He fixes the selling price per article such as to yield a profit of 45% of the cost price, and sells 2/5th of the articles at this price. He then lowers the selling price per article so as to gain a profit of 20% of the cost price. Find his total profit if all the articles are sold. The total cost price for the trader = 180 * 40 = $7200 The first lot, 2/5th of the articles, 2/5 * 180 = 72 articles, are sold at a profit of 45%. Profit per article in the first lot = 40 * = $18 100 Profit in the first lot = 18 * 72 = $1296 The second lot of remaining articles, 180 _72 = 108, are sold at a profit of 20%. Profit per article in the second lot = 40 * = $8 100 Profit in the second lot = 8 * 108 = $864 The total profit on sale of 180 articles = 1296 + 864 = $2160 TENSES Nothing to get Tense about...... As noted earlier, the verb in a sentence denotes an action, as in the following sentences : 1 . He rarely goes to bed before midnight. 2. I Ieave for work at 9:00 A.M. 3. Make hay while the sun shines. In some cases, the verb may do an auxiliary function of completing the sentence. 4. Honesty is the best policy. 5. I am terrified at the thought. In all the examples above, the action is such that it happens all the time or repeatedly (examples [1] and [2]), or the sentence talks of general things which are truisms (examples [3] and [4]). In order to induce the element of time in a sentence, we make use of what are called tenses. Tenses -present, past, and future -indicate the chronological sequence of events that the sentence speaks of. Let's quickly run through the various forms of tenses with these examples : 6. She Ieaves for work at 9:00 A.M. (Simple Present) 7. She is leavin for work right now. (Present Continuous) 8. for work. (Present Perfect) 9. for work two hours ago. (Simple Past)
10. She was leavin for work when l reached. (Past Continuous) 11 . She had left for work when l called up. (Past Perfect) 12. She will leave for work in twenty minutes. (Simple Future) 13. She will be leavin for work soon. (Future Continuous) 14. She will have left for work by the time l reach. (Future Perfect) Notice how the verb changes its form with reference to the time when the action takes place. Using the appropriate tense is fundamental to getting many Sentence Correction questions right -the GMAT setters can make them trickier than you think!
