Quantitative Aptitude Shortcuts

January 1, 2018 | Author: Vishwajit Kale | Category: Speed, Train, Equations, Metre, Walking
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Quantitative Aptitude Shortcuts...


Quantitative Aptitude Shortcuts Published on Friday, September 18, 2015  Share on Facebook  Share on Twitter  Share on Google+

In a series of providing free resources for preparation of IBPS PO, Clerk, Specialist officers, SBI PO, SBI clerk and other government jobs exams. I have started Quantitative Aptitude preparation series. I will try to cover every topic of Quantitative section, please ask your in comments section, In case you are finding any kind of difficulty in any chapter just leave a reply. I will try to write a post on that topic as soon as possible.


Shortcut technique with concept examples {Video lecture} Practice questions (Easy) Questions from previous papers

Time and Distance   

Concept video with examples {Video} Basic concepts Practice questions

Number system  

Rules and techniques Formulae with examples

Simplifications  

Technique and Tricks {PDF} Technique to solve complex equations

Boats and Streams

Detailed Video lecture

Problems based on Trains 

Concept with examples

Simple and Compound Interest    

Tricks to solve {PDF} Basic concepts Practice questions from previous papers Practice questions set 2 (Moderate difficulty)

Profit and Loss    

Concepts and tricks {Video lecture} Important questions (Set 2) Practice questions Set 1, Set 2, Set 3 Using rule of Fraction to solve Profit and Loss questions

Average 

Tricks and practice questions

Ratio and Proportions  

Basic concepts and techniques Practice questions

Percentage problems   

7 Tricks to solve questions Technique to questions {PDF} Practice questions (using shortcut tricks)

Data Interpretation

I have made a separate Data Interpretation Preparation guide

Problems based on Ages   

Techniques to solve Questions from previous papers Practice questions solved using shortcut methods

Permutation and Combinations  

Concept and techniques Practice questions solved using shortcuts

Probability       

Concepts and tricks {Video lectures} Formulae and technique Venn diagrams Dice problems (2 dices rolled) Dice problems (3 dices rolled) Practice questions Practice questions set 2

Geometry   

Triangles Circles Lines and Angles


QT Cheatsheet

Quantitative Aptitude Made Easy ebook

Preparation DVD course

Time and Work - Shortcuts and Tricks

One simple technique is using days in denominator while solving questions. For example, A can do a job in 3 days and B can do the same job in 6 days. In how much time they can do the job together. Solution - 1/3 + 1/6 = 1/2, hence 2 days is the answer. Examiner can set the question in opposite way and can ask you how much time A or B alone will take to complete the job. It is quite easy to calculate said question by putting values in equation we arrived in above question. You need to understand one simple concept - If A can do a job in 10 day then in one day A can do 1/10th of job.

SHORTCUT Best trick that I use in exams myself is by finding the efficiency of workers in percent. If A can do a job in 2 days then he can do 50% in a day.

Number of days required to complete the work

Work that can be done per day

Efficiency in Percent





































Now let's solve questions with this trick Question - A take 5 days to complete a job and B takes 10 days to complete the same job. In how much time they will complete the job together ? Solution - A's efficiency = 20%, B's efficiency = 10%. If they work together they can do 30% of the job in a day. To complete the job they need 3.33 days.

Question - A is twice as efficient as B and can complete a job 30 days before B. In how much they can complete the job together ? Solution - Let efficiency percentage as x A's efficiency = 2x and B's efficiency = x A is twice efficient and can complete the job 30 days before B. So, A can complete the job in 30 days and B can complete the job in 60 days A's efficiency = 1/30 = 3.33% B's efficiency = 1/60 = 1.66% Both can do 5% ( 3.33% + 1.66% ) of the job in 1 day. So the can complete the whole job in 20 days (100/5)

Question - A tank can be filled in 20 minutes. There is a leakage which can empty it in 60 minutes. In how many minutes tank can be filled? Solution Method 1 ⇒ Efficiency of filling pipe = 20 minutes = 1/3 hour = 300% ⇒ Efficiency of leakage = 60 minutes = 100% We need to deduct efficiency of leakage so final efficiency is 200%. We are taking 100% = 1 Hour as base so answer is 30 minutes. Update - 09-09-2013 ( As Shobhna and Aswin are facing problem in solving this question, I am solving this question with second method which is also very easy, hope this will make the solution lot easier.) Method 2 ⇒ Efficiency of filling pipe = 100/20 = 5% ⇒ Efficiency of leakage pipe = 100/60 = 1.66% ⇒ Net filling efficiency = 3.33% So tank can be filled in = 100/3.33% = 30 minutes

You can change the base to minutes or even seconds. You can solve every time and work question with this trick. In above examples I wrote even simple calculations. While in exams you can do these calculations mentally and save lots of time. You can find more tricks like this in quantitative aptitude section. Comment below in case of any query, I promise to reply within 24 hours.

Update 09 October 2013 - Question requested by Chitra Salin Question - 4 men and 6 women working together can complete the work within 10 days. 3 men and 7 women working together will complete the same work within 8 days. In how many days 10 women will complete this work ? Solution - Let number of men =x, number of women = y ⇒ Efficiency of 4 men and 6 women = 100/10 = 10% ⇒ so, 4x+6y = 10 Above equation means 4 men and 6 women can do 10% of a the job in one day. ⇒ Efficiency of 3 men and 7 women = 100/8 = 12.5% so, 3x+7y = 12.5 By solving both equations we get, x = -0.5 and y = 2 ⇒ Efficiency of 1 woman(y) = 2% per day ⇒ Efficiency of 10 women per day = 20% So 10 women can complete the job in 100/20 = 5 days Update 11-11-2013 - Question requested by Praisy Question - A and B together can complete a task in 20 days. B and C together can complete the same task in 30 days. A and C together can complete the same task in 30 days. What is the respective ratio of the number of days taken by A when completing the same task alone to the number of days taken by C when completing the same task alone? Solution ⇒ Efficiency of A and B = 1/20 per day = 5% per day ________________1 ⇒ Efficiency of B and C = 1/30 per day = 3.33% per day______________2 ⇒ Efficiency of C and A = 1/30 per day = 3.33% per day______________3

Taking equation 2 and 3 together ⇒ B + C = 3.33% and C + A = 3.33% ⇒ C and 3.33% will be removed. Hence A = B ⇒ Efficiency of A = B = 5%/2 = 2.5% = 1/40 ⇒ Efficiency of C = 3.33% - 2.5% = 0.833% = 1/120 ⇒ A can do the job in 40 days and C can do the job in 120 days he they work alone. ⇒ Ratio of number of days in which A and C can complete the job 1:3. Post your queries in the comments section.

#1 A and B together can do a job in 4 days.If A can do job in 12 days if he works alone ,then how many days B alone take to complete the job?

#2 A and B can do job in 8 days. B and C can do same job in 12 days.A,B and C together can do same job in 6 days.In how many days A and C together can complete the job?

#3 A is twice as efficient as B and can complete a job in 30 days before B.In how much they can complete the job together?


A tank can be filled in 20 minutes.There is a leakage which can empty in 60 minutes.In how many minutes tank can be filled?

#5 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?

#6 A work twice as fast as B. I f B can complete a work in 12 days independently,the number of days in which A and B can together finish the work in?

#7 20 Women can do a work in 16 days.Sixteen men can complete the same work in 15 days.What is the ration between the capacity of a man and a woman?

#8 A and B together can complete a piece of work in 4 days.If A alone can complete the same work in 12 days,in how many days can B alone complete that work?

#9 A,Band C can do a piece of work in 36,54 and 72 days respectively.they started the work but A left 8 days before the completion of the work while B left 12 days before the completion.The number of days for which C worked is?

#10 Ten 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?

#11 12 Men complete a work in 9 days.After they have worked for 6 days,6 more men joined them.How many days will they take to complete the remaining work?

#12 A,B and C together earn Rs.300 per day,while A and C together earn Rs.188 and B and C together earn Rs.152.the daily earning of C is?

Time and Distance - Concept and Questions {Video Lecture}

Time and Distance is an important chapter from examination point of view.

FORMULAE: 1. 2. 3. 4.

Speed = Distance/Time Time = Distance/Speed Distance = Speed × Time If the speed of a body is changed in the ratio a : b, then the ratio of the time taken changes in the ratio b : a. 5. m km/hr = [m × 5/18] m/sec. 6. m metres/sec = [m × 18/5] km/hr. I recommend you to watch the following concept video before solving the questions.

Question 1. Express a speed of 18 km/hr in metres per second.

Solution: 18 km/hr = [18 × 5/18] m/sec. = 5 metres/sec. Question 2. Express 10 m/sec. in km/hr.

Solution: 10 metres/sec = [10 × 18/5] km/hr. = 36 km/hr.

THEOREM If a certain distance is covered at m km/hr and the same distance is covered at n km/hr then the average speed during the whole journey is 2mn/(m+n) km/hr.

Let the distance be A km.

Time taken to travel the distance at a speed of m km/hr = A/m hrs. Time taken to travel the distance at a speed of n km/hr = A/n hrs. we see that the total distance of 2A km is travelled in (A)/m+A/n hrs.

Question 3. Rakesh covers a certain distance by car driving at 70 km/hr and he returns to the starting point riding on a scooter at 55 km/hr. Find his average speed for the whole journey.

Question 4. Raju covers distance between his house and office on scooter. Having an average speed of 30 km/hr, he is late by 10 min. However, with a speed of 40 km/hr, he reaches his office 5 min earlier. Find the distance between his house and office.

Question 5. A man walking with a speed of 5 km/hr reaches his target 5 minutes late. If he walks at a speed at a speed of 6 km/hr, he reaches on time. Find the distance of his target from his house.

Question 6. A boy goes to school at a speed of 3 km/hr and returns to the village at a of 2 km/hr. if he takes 5 hrs in all, what is the distance between the village and the school?

CONCEPTS 1) There is a relationship between speed, distance and time: Speed = Distance / Time OR Distance = Speed* Time 2) Average Speed = 2xy / x+y where x km/hr is a speed for certain distance and y km/hr is a speed at for same distance covered. **** Remember that average speed is not just an average of two speeds i.e. x+y/2. It is equal to 2xy / x+y

3) Always remember that during solving questions units must be same. Units can be km/hr, m/sec etc. **** Conversion of km/ hr to m/ sec and m/ sec to km/ hr x km/ hr = (x* 5/18) m/sec i.e. u just need to multiply 5/18 Similarly, x m/sec = (x*18/5) km/sec 4) As we know, Speed = Distance/ Time. Now, if in questions Distance is constant then speed will be inversely proportional to time i.e. if speed increases ,time taken will decrease and vice versa.

TIME AND DISTANCE PROBLEMS Problem 1: A man covers a distance of 600m in 2min 30sec. What will be the speed in km/hr? Solution: Speed =Distance / Time ⇒ Distance covered = 600m, Time taken = 2min 30sec = 150sec Therefore, Speed= 600 / 150 = 4 m/sec ⇒ 4m/sec = (4*18/5) km/hr = 14.4 km/ hr.

Problem 2: A boy travelling from his home to school at 25 km/hr and came back at 4 km/hr. If whole journey took 5 hours 48 min. Find the distance of home and school.

Solution: In this question, distance for both speed is constant. ⇒ Average speed = (2xy/ x+y) km/hr, where x and y are speeds ⇒ Average speed = (2*25*4)/ 25+4 =200/29 km/hr Time = 5hours 48min= 29/5 hours Now, Distance travelled = Average speed * Time ⇒ Distance Travelled = (200/29)*(29/5) = 40 km Therefore distance of school from home = 40/2 = 20km.

Problem 3: Two men start from opposite ends A and B of a linear track respectively and meet at point 60m from A. If AB= 100m. What will be the ratio of speed of both men? Solution: According to this question, time is constant. Therefore, speed is directly proportional to distance. Speed∝Distance

⇒ Ratio of distance covered by both men = 60:40 = 3:2 ⇒ Therefore, Ratio of speeds of both men = 3:2

Problem 4: A car travels along four sides of a square at speeds of 200, 400, 600 and 800 km/hr. Find average speed. Solution: Let x km be the side of square and y km/hr be average speed Using basic formula, Time = Total Distance / Average Speed x/200 + x/400 + x/600 + x/800 = 4x/y ⇒ 25x/ 2400 = 4x/ y⇒ y= 384 ⇒ Average speed = 384 km/hr

1. How many minutes Raman will take to cover a distance of 400 meters if he runs at a speed of 20 km/hr ? A.

2 mins


1⅕ mins


1.5 mins


2.5 mins


None of these Answer & Explanation

Answer : [C] Explanation : ⇒ Raman's speed = 20 km/hr = 20 × 5/18 = 50/9 m/sec ⇒ 400 × 9/50 = 1⅕ mins

2. John travelled from his town to city. John went to city by bicycle at the speed of 25 km/h and came back at the speed of 4 km/h. If John took 5 hours and 48 min to complete his journey, what is the distance between town and city ? A.

15 km


20 km


 22 km


 25 km 

Answer & Explanation

Answer : [C] Explanation : ⇒ Average speed of John = 2xy/x+y = 2 × 25 × 4 / 25 + 4= 200/29 km/h ⇒ Distance traveled = Speed × Time = 200/29 × 29/5 = 40 Km ⇒ Distance between city and town = 40/2 = 20 km

3. Speed of a train is 20 meters per second. It can cross a pole in 10 seconds. What is the length of train ? 

A. 150 m


200 m


 250 m


 300 m 

Answer & Explanation

Answer : [C] Explanation : ⇒ Lenght of train = 20 × 10 = 200 meters

4. Ram walks at a speed of 12 km/h. Today the day was very hot so walked at ⅚ of his average speed. He arrived his school 10 minutes late. Find the usual time he takes to cover distance between his school and home ? A.

40 mins


50 mins


45 mins


60 mins 

Answer & Explanation

Answer : [C] Explanation : ⇒ ⇒ ⇒ ⇒

If Ram is walking at ⅚ of his usual speed that means he is taking 6/5 of using time. 6/5 of usual time - usual time = 10 mins 1/5 of usual time = 10 mins Usual time = 50 mins

5. A car running at 65 km/h takes one hour to cover a distance. If the speed is reduced by 15 km/hour then in how much time it will cover the distance ? A.

72 mins


76 mins


78 mins


None of these

Answer & Explanation

Answer : [B] Explanation : ⇒Reduced speed = 65-15 = 50 km/h ⇒ Now car will take 65/50 × 60 mins = 78 mins

6. In a 100 m race A runs at a speed of 1.66 m/s. If A gives a start of 4m to B and still beats him by 12 seconds. What is the speed of B ? A.

1 m/s


1.25 m/s


1.33 m/s


Rs 1.5 m/s 

Answer & Explanation

Answer : [B] Explanation : ⇒Time taken by A to cover 100 meters = 60 seconds ⇒ Since A gives a start of 4 seconds then time takes by B = 72 seconds ⇒ B takes 72 seconds to cover 96 meters ⇒ Speed of B = 96/72 = 1.33 m/s

7. In a kilometer race, A beats B by 100 meters. B beats C by 100 meters. By how much meters does A beat C in the same race ? A.

200 meters


190 meters


180 meters


210 meters 

Answer & Explanation

Answer : [C] Explanation : ⇒ While A covers 1000 meters, B can cover 900 meters ⇒ While B covers 1000 meters, C can cover 900 meters

⇒ Lets assume that all three of them are running same race. So when B runs 900 meters, C can run 900 × 9/10 =810 ⇒ So A can beat C by 190 meters.

Problems on Trains - Types of Problems with Examples

Problems on Trains chapter is almost same as TIME AND DISTANCE. The only difference is that the length of the moving object is also considered.

Some important things to be noticed: (i) When two trains are moving in opposite directions, their speeds should be added to find the relative speed. (ii) When they are moving in the same direction, the relative speed is the difference of their speeds. (iii) When a train passes a platform, it should travel the length equal to the sum of the lengths of trains & platform both.

Trains passing a telegraph post or a stationary man 1.How many seconds will a train 100 metres long running at the rate of 36 km an hour take to pass a certain telegraph post? Solution: In passing the post the train must travel its own length. Now,

36 km/hr


36 ×5/18


10 m/sc.

∴ Required time = 100/10 = 10 seconds.

Trains crossing a bridge or passing a railway station 2.How long does a train 110 metres long running at the rate of 36 km/hr take to cross a bridge 132 metres in length?

Trains running in opposite direction 3.Two trains 121 metres and 99 metres in length respectively are running in opposite directions, one at the rate of 40 km and the other at the rate of 32 km an hour. In what time will they be completely clear of each other from the moment they meet?

Trains running in the same direction 4. In example above. If the trains were running in the same direction, in what time will they be clear each other?

Trains passing a man who is walking 5. A train 110 metres in length travels at 60 km/hr. In what time will it pass a man who is walking at 6 km an hour (i) against it; (ii) in the same direction?

Solution: This question is to be solved like the above examples 3 and 4, the only difference being that the length of the man is zero.

6. Two trains are moving in the same direction at 50 km/hr and 30 km/hr. The faster train crosses a man in the slower train in 18 seconds. Find the length of the faster train.

7. A train running at 25 km/hr takes 18 seconds to pass a platform. Next, it takes 12 seconds to pass a man walking at 5 km/hr in the opposite direction. Find the length of the train and that of the platform.

Problems Based on Ages - Techniques to Solve Quickly

Only if you let it be a problem! Solve it and no problem anymore! ‘Problems based on Ages’ – is a very popular question in Clerical Exams; it can come in either Reasoning or in Quantitative Aptitude. But it will come. I know some people leave this one out because its way too confusing to ‘em – well, let us try to see how to not get confused.   

 

The important thing in any kind of Age Problem, is to decide which age – present or past or future – to be taken as ‘x’! Let us make a simple rule for ourselves – the ‘x’ should be the present age always. In most cases, taking the present age as ‘x’, i.e., the base year works just fine. Past will become, say (x-5) years, and future can be denoted as (x+5). But sometimes, ‘present age’ is not directly given in words. Then, take ‘x’ to be the age you are supposed to find. You can also try putting yourself in someone’s place and try to calculate the age!

Also, sometimes – when nothing works and you’re stuck on an age question in the last 4 minutes of the exam – just look at the options and solve it through back calculations! Works just fine! Also, please know – keep it simple – the age problems are easily solved. 

Quantitative Aptitude Techniques Book Some practice questions for your benefit: Question 1: The age of Rekha is twelve times that of her daughter Avani. If the age of Avani is 3 years, what is the age of Rekha? Solution. Who’s age do we need? Rekha’s. Present age required to be found? Yes! Okay, so, Rekha’s present age = x Rekha’a age is 12 times her daughter’s age. Daughter’s age = 3. Therefore, 12 times of 3 = x 12 x 3 = x = 36 years = Rekha’s age. Question 2: The father’s age four years ago was 8 times the age of his son. At present the father’s age is 4 times that of his son. Find the son’s present age. Solution. Okay, who’s present age do we need to find? Son’s. Therefore, Son’s present age = x We’ll first make the present age’s equation with dad and son – only ‘cause it makes things easier to proceed from the present! ‘At present the father’s age is 4 times that of his son’, i.e., Father’s present age = 4x. ‘Father’s age four years ago was 8 times the age of his son’ – ‘four years ago’ will simple mean, subtracting 4 from the present ages of BOTH father and son;

and the whole equation will stand as : (4x – 4) = 8 x (x – 4) Solving it – x = 7 years, The son’s present age = 7 years. Easy does it, doesn’t it?

Question 3: At present, the ratio between the ages of Amar and Norman is 4:3. After 6 years, Amar’s age will be 26 years. What is the age of Norman at present? Solution. Who’s ‘present’ age do we need to find? Norman’s. But they have also have given the ration of present ages, 4:3. So we can use ‘x’ to denote both their present ages to be 4x and 3x, i.e., Amar’s and Norman’s respectively. Next, ‘Amar’s age 6 years later’, = (4x + 6) = 26. x = 5 years. Norman’s present age = 3x = 3 x 5 = 15 years. Question 4: The ratio of the father’s age to the son’s age is 4:1 the product of their ages is 196. What will the ratio of their ages after 5 years? Ans.: 11:4

Question 5: Sam’s age is one-fourth of his father’s age and two-third of his sister Rita’s. What is the ratio of the ages of Sam, Rita and their father respectively? Ans.: 2:3:8

Question 6: The average age of boys in a class of 30 is 15 years. If 10 more boys join the class, the average of the whole class gets reduced by a year. What is the average age of newcomers? [This mixes age and average – a real IBPS

worthy question!] Ans.: 11years

Question 7: Present age of Vinny is 8 lesser than Amu’s present age. If 3 years ago Amu’s age was X, what will be Vinny’s present age? Ans.: (X-5) years

Question 8: The ratio between the present ages of P and R is 5:3. The ratio between P’s age four years ago and R’s age four years hence is 1:1. What is the ratio between P’s age 4 years hence and B’s age four years ago? [‘Hence’ means future. And this is one gem of a question!] Ans.: 3:1

Question 9: The ratio between the present ages of Doom and Room is 2:3. 4 years ago the ratio between their ages was 5:8. What will be Doom’s age after 7 years? Ans.: 31 years

Question 10: The age of Rani is 5 times the age of her daughter. After 12 years the age of Rani will be thrice the age of her daughter. Find the present age of Rani’s daughter. Ans.: 12 years.

Age Problems - Concepts and Tricks

Age problems are one of the most common topics in IBPS, CAT, GMAT and otherbanks exams. Students waste lots of time in this question as it look very

simple but when they start solving with triditional methods, it takes a lot of time. Today I am solving few solving with shortcut trick. In case of any problem, please comment below.

Best way to solve Age questions is to assume fixed period with which further conditions will be compared. For example taking 2000 as fixed year.

APPLICATION OF THIS RULE Example 1 Raman's age after 15 years will be 5 times his age 5 years back. What is his present age ? Solution - Let's assume right now it is year 2000

Age of Raman in 1995 = x Age of Raman in 2015 = 5x Present age of Raman (in 2000) = x+5 or 5x-15

we will solve these two equation to find x.

X= 5. Then Raman's present age becomes = x +5 = 10

Example 2 Rahul was 4 times old as his son 8 years back and he will be 2 times old as his son after 8 years. Calculate Rahul and his son's age.

Assume that currently it is year 2000. In 1992 Rahul's age = 4x, Age of Rahul's son = x In 2008 Rahul's age = 2y and Age of Rahul's son = y

Now we get two equations 2y - 4x = 16 and y - x = 16 By solving this equation x = 8, so Rahul' son's current age = 16 years and Rahul's age = 40 years.

Problems based on Ages - Tricks

To solve the problems based on ages, you need to have knowledge of linear equations. This method needs some basic concepts as well as some more time than it deserves. Sometimes it is easier to solve the problems by taking the given choices in account. But this hit-and-trial method proves costly sometimes, when we reach our solution much later. We have tried to evaluate some easier as well as quicker methods to solve this type of questions. Although we are not able to cover each type of questions in this section, our attempt is to minimize your difficulties. I have already shared a trick related to this chapter here.

Have a look at the following questions:

Ques 1. The age of the father 3 years ago was 7 times the age of his son. At present the father’s age is five times that of his son. What are the present ages of the father and son?

Ques 2.

At present the age of the father is five times that age of his son.Three years hence, the father’s age would be four times that of his son. Find the present ages of the father and son.

Ques 3. Three years earlier the father was 7 times as old as his son. Three years hence the father’s age would be four times that of his son. What are the present ages of the father and son?

Solving using Conventional Method: Solution: 1 Let the present age of son = x yrs. Then, the present age of father = 5x yrs.

3 years ago, 7(x – 3) = 5x – 3 Or, 7x – 21 = 5x – 3 Or, 2x = 18 Therefore, son’s age = 9 yrs. Fathers’ age = 45 years.

Solution: 2. Let the present age of son = x yrs.

Then, the present age of father = 5x yrs.

3 years hence, 4(x + 3) = 5x + 3 Or, 4x + 12 = 5x + 3 ∴ X = 9 yrs.

Therefore, son’s age = 9 yrs. Fathers’ age = 45 years.

Solution: 3. Let the present age of son = x yrs. And the present age of father = y yrs. 3 yrs earlier

7(x – 3) = y – 3 or, 7x – y = 18……….(i)

3 yrs hence

4(x + 3) = y + 3 or, 4x – y = -9………..(ii)

Solving (i) & (ii) we get, x = 9 yrs & y = 45 yrs.

Quicker Method:

Ques 4. 10 yrs ago, Sita’s mother was 4 times older than her daughter. After 10 yrs, the mother will be twice older than the daughter. What is the present age of Sita?

Ques 5. One years ago the ratio between Samir’s age and Ashok’s age was 4 : 3. One year hence the ratio of their ages will be 5 : 4. What is the sum of their present ages in yrs.

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