## Monday, December 9, 2013

### ARITHMETIC SHORT CUT METHODS WITH EXAMPLES

This material compiled by Akula. Praveen Kumar, SPM, Papannapet Sub Office-502 303, Medak Division, Andhra Pradesh (9849636361, 8019549939)

Disclaimer:- Author of blog does not accepts any responsibility in relation to the accuracy, completeness, usefulness or otherwise, of the contents.

Time and Work: Short Cuts
Short Cut 1:
If A and B can do a piece of work in “x” days, B and C in y days and C and A in z days. Then A, B and C can together do the work in:
Example 1:
If A and B can do a piece of work in 12 days, B and C in 15 days and C and A in 20 days. In how many days will, all of them working together would take complete the work?
Solution:
Here x= 12, y= 15 and z= 20,

Putting values in formula:

Short Cut 2:

If A can do a work in x days, B can do the same work in y days. Together A and B can do the same work in:

Example:2
If Ram can do a piece of work in 20 days, Shayam can do the same work in 30 days, in how many days Ram and Shayam will together do the work?

Solution:  Here x= 20 and y= 30, putting values in formula we get:
Short Cut 3:

If A, B and C respectively can finish a work in x, y and z days, then all of them together would finish the work in following days:

Example 3:

Amar, Akbar and Anthony can finish a work in 10, 15 and 20 days respectively. If they all work together, then in how many days, they will be able to finish the work?
Solution:     Here x= 10, y=15 and z= 20       Putting the values in the formula:
We get:

If a worker takes T units of time to complete a work W, then the rate of work R is given by R = W/T and so T = W/R.
W and R are directly proportional, when T is constant
W and T are directly proportional when R is const
Formulas:

tA = time taken by A to complete the work
tB = time taken by B to complete the work
t(A+B)  = time taken by  both A and B to complete the work
• t(A+B) = (tA * tB)/(tA+tB)
• tB = (tA * tA+B)/tA – (tA+B)
• tA+B+C =L/[(L/tA) + (L/tB) + (L/tC) ];L =>L.C.M of tA, tB, tC.
• If A+B, B+C, A+C are given then A+B+C=?
§  tA+B+C = 2L/[ (L/tA+B) + (L/tB+C) + (L/tC+A) ]
§  tA = [(tA+B+C)*(tB+C)]/[ (tB+C)- (tA+B+C) ]
§  tB = [(tA+B+C)*(tA+C)]/[ (tA+C)- (tA+B+C) ]
§  tC = [(tA+B+C)*(tA+B)]/[ (tA+B)- (tA+B+C) ]
• (M1*T1) /W1 = (M2*T2)/W2
§  M be the number of men’s
§  T be the number of days/ hours
§  W be the work done by the men
§
• w(A+B  = wA*wB /(wA+wB)
Example 1:
If a worker takes 10 hours to complete a work, his rate of work or the work done per hour is 1/10 of the work.
Similarly, if the rate of work is 1/15, it would take 15 hours to complete the work.
Example 2:
Worker A takes 8 hours to do a job. Worker B takes 10 hours to do the same job. How long should it take both A and B, working together but independently, to do the same job?
By formula 1,
t(A+B) = (tA * tB)/(tA+tB) = 8*10/(8+10) = 40/9 hours.
Example 3:
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?
By formula 2,
tB = (12*4 )/(12-4) = 6 days.

Example 4:
A and B can do a piece of work in 18 days: B and C can do it in 24 days; A and C can do it in 36 days. In how many days will A,B and C finish it, working together  and separately?
By formula 4,
Working together,
• tA+B+C = 2L/[ (L/tA+B) + (L/tB+C) + (L/tC+A) ]
t(A+B+C) = 2*48/[(48/18)+(48/24)+(48/36)] =16 days
Working separately,
• tA = [(tA+B+C)*(tB+C)]/[ (tB+C)- (tA+B+C) ]
tA = [16*24]/[24-16] = 48 days,
tB = [16*36]/[36-16] = 144/5  days,
tC = [16*18]/[18-16] = 144 days.
Example 5:
The IT giant Tirnop has recently crossed a head count of 150000 and earnings of \$7 billion. As one of the forerunners in the technology front, Tirnop continues to lead the way in products and services in India. At Tirnop, all programmers are equal in every respect. They receive identical salaries and also write code at the same rate. Suppose 12 such programmers take 12 minutes to write 12 lines of code in total. How long will it take 72 programmers to write 72 lines of code in total?
By formula : (M1 * T1)/ W1 = (M2 * T2)/ W2
12*12/12 = 72*X/72  ::  X = 12
Example 6:
A and B working separately can do a piece of work in 9 and 12 days. If they work for a day alternatively, A beginning, in how many days, the work will be completed?
Explanation:
t(A+B) = (tA * tB)/(tA+tB)
Work done by A in one day, tA= 1/9 day
Work done by B in one day, tB = 1/12 day
They are working alternatively,
Ist day, A’s work is 1/9 and 2nd day, B’s work is 1/12.
In two days work done by A and B , t (A+B) = 7/36.
Every two days work done is increasing evenly, so we can calculate in pair of days.
Work done in 5 pair of days = 5* (7/36) =35/36.
Remaining work = 1-35/36 = 1/36.
On 11th day, A’s turn:
1/ 9 work is done by A in one day.
1/36 work is done by A in (9*1/36)=1/4 day.
Therefore time taken to complete the work =10 ¼ days.

Short Cut: Clocks
Short Cut 1
Between x and (x+1) o’clock, the two hands will be together at :
Minutes past x.
Example:

At what time between 11 and 12 o’ clock, the minute hand and hour hand would be together?
Solution:

Here x= 4 so putting the formula we get:

The both hands will be together at 60 minutes past 11 o’clock.
Short Cut 2:
Between x and (x+1) O’ clock, the two hands will be ‘t’ minutes apart at

Past x.

Example:  At what time between 4 and 5 o’clock the hands are 2 minutes space apart?
Solution:     Here x= 4
So, Case 1:    = 24 minutes past 11
Case 2:    = 19.63 minutes past 11
Short Cut 3
Between x and x+1 o’clock the two hands are at right angle at :
Minutes Past x
Ratio & Proportion
Ratios :
• Two quantities are in the ratio a : b => if the first quantity is ax, then the second quantity is bx.
• The ratio a : b is the same as a/b.
• If two quantities are in the ratio of a : b, then the first quantity is a/(a + b) times the sum of the two quantities and the second quantity is b/(a + b) times the sum of the two quantities.
Steps in comparison of Ratios:
• Convert the ratios into fractions.
• Compare the fractions by equalizing the denominator.
Proportions:
• Product of means = Product of Extremes.
A: B :: C:D <==> B*C = A*D
• Mean proportional: Mean proportional between a and b is sqrt (ab).
• Third proportional: If a:b = b:c, then c is called the Third Proportional to a,b.
• Fourth Proportional:If a:b = c:d, then d is called the Fourth Proportional to a,b,c.
Componendo and dividendo:
• if (a/b) = (c/d), then [(a+b)/(a-b)] = [(c+d)/(c-d)]
Some other tricks :
(a+b)/b = (c+d)/d
• (a-b)/b = (c-d)/d
•  a/c = b/d
Example 1:
The statement, ‘Father’s salary and the son’s salary are in the ratio 5 : 3.’ implies the following:
If father’s salary is Rs.5000, the son’s salary is Rs.3000.
1. If father’s salary is Rs.7500, the son’s salary is Rs.4500.
2. If father’s salary is 5x, the son’s salary is 3x.
Example 2:
The weights of two rods are in the ratio 9:7. Their total weight is 256g, find the weight of each rod?
Given total weight: 256g
Ratio is 9:7
From the Formula,
Weight of first rod = [9/(9+7)]*(256) = 144
Weight of Second rod = [7/(9+7)]*(256) = 112
So the weight of each rod is 144g and 112g.
Example 3:
Two alloys copper and tin are in the ratio of 5:6 and 7:11.Which of them contain more copper ?
Ratio of copper and tin in 1st alloy 5:6 = 5/6
Ratio of copper and tin in 2nd alloy 7:11 = 7/11
The denominator values of the two ratios are 6,11 respectively. Now we have to wquate the denominator value in both the ratios.
(5/6) * (11/11) = 55/66
(7/11)*(6/6) = 42/66
So 5/6 >7/11.
Hence , I alloy contains more copper than II alloy.
Example 3:
If x:y = 2:3, y:z = 4:3, then find x:y:z ?
Here. Y is common to both the ratios. So we have to make equal value of y in both the ratios. In the first and second ratio the values of y are 3 and 4 respectively. Multiply the first ratio by 4 and second ratio by 2. i.e.,
x:y = 2*4 : 3:4 = 8:12
y:z = 4*3 : 3*3 = 12:9
hence x:y:z = 8:12:9
Example 4:
If two numbers are in the ratio 3:5,when 6 is added to each term of the ratio , it becomes 2:3, then the numbers are ?
Let the numbers be a,b respectively. Given that a:b = 3:5
a=3k,b=5k , where k is a common factor of both a,b.
(3k + 6)/(5k+6) = 2/3
9k+18 = 10k + 12==> k= 6
Hence the numbers are (3*6) = 18 and (5*6) = 30 respectively.
Short Cut 1
If the sum of two numbers is A and their difference is a, then the ratio of the numbers is given by:
A+a:A-a
Example 1
The sum of two numbers is 36 and their difference is 6. What is the ratio of the numbers?
Solution 1
Here, A= 36 and a = 6,

Putting the value in equation:
A+a:A-a
Short Cut 2:
A number which when added to the terms of the ratio a:b makes it equal to the ratio c:d is:
Example 2:
Find the number which when added to the terms of the ratio 13:28 makes it equal to the ratio 1:2.
Solution 2
Here a: b= 13:28 and c:d= 1:2
Putting the values in the equation:
Short Cut: 1

The number which when multiplied by x is increased by “y” is given by:
Example: 1

Find the number which when multiplied by 10 is increased by 261?

Solution   Here, x= 10 and y= 261. Putting the values in equation:
Short Cut 2:
In case of partnership, if investment of three partners are in ratio a:b:c and the timings are in ratio x:y:z then the profit are in the ratio:
ax: by: cz
Example 2:
A,B and C invested capitals in the ratio 5:6:8. The investment timing is in ratio 2:3:4. Find the ratio of the profits?
Solution:
Here, a=5, b=6, c=8 and x=2, y=3, z=4. Putting the values in equation:
ax: by: cz

Averages
Short Cut 1
The average of marks obtained by ‘n’ candidates in a certain examination is ‘T’. If the average marks of passed candidates is “P” and that of failed candidates is “F”. Then the number of candidates who passed the exam is given by:

Example 1:
The average marks obtained by 90 candidates in a certain examination is 38, if the average marks of passed candidates is 40 and that of the failed candidates is 30, what is the number of candidates who passed the exam?
Solution:
Here: n= 90, T=38, F= 30 and P= 40, putting the values in the equation:

Short Cut 2
The average of marks obtained by ‘n’ candidates in a certain examination is ‘T’. If the average marks of passed candidates is “P” and that of failed candidates is “F”. Then the number of candidates who passed the exam is given by:

Example 2:
The average marks obtained by 90 candidates in a certain examination is 38, if the average marks of passed candidates is 40 and that of the failed candidates is 30, what is the number of candidates who failed  the exam?
Solution:
Here: n= 90, T=38, F= 30 and P= 40, putting the values in the equation:

Short Cut 1:
If the average age/weight/height etc of “n” objects is “x” and the average of “m” objects is “y” then the average of all of them put together in a single group is:

Example 1:
The average age of 20 girls in a college is 21 years  while the average age of 25 boys of the same same college is 22 years. Find the combine average of the group.
Solution:
Applying the formula:

n= 20, m= 25, x= 21 and y = 22
Short Cut 2:
In a group if the average age/weight/height etc of “n” objects is “x”, and the average age/weight/height etc of “m” objects out them in the group is “y”, then the average age/weight/height etc of remaining “n-m” objects is::

Example 2:
A group of 20 girls has average age of 12 years. Average age of first 12 from the same group is 13 years. What is the average age of remaining 8 girls in the group.
Solution:
Applying the formula:

where n= 20, m= 12, x= 12, y = 13.
Short Cut 3:
If the average age/weight/height etc of “n” objects in a group is “x” and out of them 1 object is added, as a result of this addition the average of the group becomes “y”. Then the age/weight/height etc of the new entrant is:

Example 3:
The average age of 30 boys in a class is equal to 13 years. When the age of the class teacher is included the average become 15 years. Find the age of the teacher.
Solution:
The age of the entrant is given by:
Here, n=30; y= 15; x= 13
Putting the formula we get;

(30(15-13) + 15) = 75
Example 4:
In an old age home the average age of 10 inmates is 65 years. On Sunday the grandson of an inmate visited the old age home. The average age with the arrival of grandson becomes 60 years. What was the age of the grandson?
Solution:
The age of the entrant is given by:
Here, n=10; y= 60; x= 65
Putting the formula we get;

(10(60-65) + 60) = 10
Average = (Sum of observations / Number of observations)
Some important formulas to remember :
• Average of first ‘n’ natural numbers = (n+1)/2
• Average of the squares of the first ‘n’ natural numbers = [(n+1)(2n+1)]/6
• Average of the cubes of the first ‘n’ natural numbers = [n2(n+1)2]/4
• Average of the first ‘n’ even natural numbers = n
• Average of the first ‘n’ odd natural numbers = (n+1)
• A man travels a certain distance at ‘x’ kmph and an equal distance at ‘y’ kmph. Then the average speed during whole journey = (2xy/(x+y))kmph.
• Average of ‘n’ average is not equal to the { (sum of ‘n’ averages )/ ‘n’ }.
Properties :
• If all values are increased or decreased by a certain quantity, the average also increases or decreases by the same quantity.
• If all values are multiplied or divided by a certain quantity, the average also gets multiplied or divided by the same quantity.
• When a new value X is added to a set of B values with average A, if the average increases by Y, then the added value X = A + (B + 1)Y.
• When a new value X is added to a set of B values with average A, if the average decreases by Y, then the added value X = A – (B + 1)Y.
• When a value X is removed from a set of B values with average A, if the average increases by Y, then the the deleted value X = A – (B – 1)Y.
• When a value X is removed from a set of B values with average A, if the average decreases by Y, then the deleted value X = A + (B – 1)Y.
Example 1:Find the average of first 20 natural numbers ?
By formula 1, Average of first 20 natural numbers = 21/2 = 10.5
Example 2:A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th innings?
Hints: These type of problems , we need to find the sum values and then we have to do the calculation.
Initially batsman average be X i.e., average of 16 innings. After he scores 87 runs in 17th innings , the average increases to X+3.
Score obtained by batsman in 16 innings = 16X ………1
Score obtained by batsman in 17 innings = 17(X+3) ………2
or Score obtained by batsman in 17 innings = 16X + 87 ………3
Equating 2 and 3 , we get : 17(X+3) = 16X + 87
X = 36; Average after 17th innings (X+3) = 39
Shortcut : By property 4:    X = A + (B + 1)Y
X be the added value = 87
A be the average before adding value = C
B be the number of innings before his average gets increased = 16
Y be the value which the average increases = 3
we get A = 36 ,then the final value (C+3) is 39.
Profit & Loss
Short Cut 1:
If a seller by selling N articles, gains or losses the cost price of n articles, then the gain or loss percentage is calculated as:

Example 1:
By selling 240 mangoes a fruit seller gains the cost price of 40 mangoes. What is the profit percentage?
Solution:
Here, N= 240 and n= 40, putting the values in equation:

Short Cut 2:
If a seller sells an article at Rs “A” after giving x% discount on the mark price. Had he not given the discount, he would have earned a profit of y% on the cost price. The cost price is given by:

Example 2:
A shopkeeper sold certain articles at Rs 425 each after giving a discount of 15% on the marked price. Had he not given the discount, he would have earned a profit of 25% on the cost price, find the Cost Price of each article?
Solution:
Here A= 425, x= 15, y= 25
Putting values in equation:

Short Cut 1:
If the cost price of x articles is equal to the selling price of y articles, then the profit percentage is given by:

Example 1:
If Sudha sells 18 articles at the cost price of 24 articles. Find her profit percentage?
Solution:
Here, x= 24 and y= 18
Putting the values in the formula

Short Cut 2:
A man purchases a certain number of articles at x a rupee and the same number at y a rupee. He mixes them together and sells them at z per rupee. Then his gain or loss percentage is given by:

Example 2:
A man purchases a certain number of oranges at 7 per rupee and the same number at 15 per rupee. He mixes them together and sells them at 12 per rupee. Determine his profit or loss percentage.
Solution:
x= 7, y= 15 and z= 12, putting the values in the equation:

Short Cut 3:
By selling an article for Rs A, a dealer makes a profit of x%. If he wants to make profit of y%, then he must increase his selling price by Rs:

The new selling price is given by:

Short Cut 1
If a shopkeeper wants to earn x% profit on an article after offering y% discount to the customer, to arrive at label price, the marketed price should be increased by following percentage:

Example
If a dealer wants to earn 20% profit on an article after offering 50% discount to the customer, by what percentage should he increase his marked price to arrive at a label price?
Solution:
Here, x=20%, y=50%
Putting the formula:

Short Cut-2
If a seller uses “X” gm in place of 1000 gm to sell his goods and gains a profit of y% on cost price, then his actual gain or less percentage is:

According as the sign is negative or positive.
Example:
A seller uses 840 gm in place of 1 kg to sell his goods. Find his actual % profit or loss, when he sells his article on 4% gain on cost price.
Solution:
X = 840 gm, y=4
Putting values:

We get= 18.6%
Short Cut 3
A dishonest seller sells the goods at x% loss on the cost price but uses y% less weight to weigh the goods, than his percentage profit or loss is:

Example:
A dishonest dealer sells the goods at 10% loss on cost price but uses 20% less weight. What is the percentage profit or loss?
Solution:
Here x= 10%, y= 20%
Putting values:

We get= 12.5%
Formulas such as ,
1. Profit = S.P – C.P :: S.P >C.P
2. Loss = C.P – S.P :: C.P > S.P
3. Profit % = (Profit / C.P )*100 or
4. Profit % = {(S.P – C.P)/C.P }*100
5. Loss % = (Loss / C.P )*100 or
6. Loss % = {(C.P – S.P)/C.P }*100
7. S.P = {(100+Gain % )/100}*C.P or
8. C.P = {100 / (100+gain %)}*S.P
9. S.P = {(100-Loss % )/100}*C.P or
10. C.P = {100 / (100-Loss %)}*S.P
11. When a Person sells two similar items, one at a gain of x% and the other at a loss of x%, then the seller always incurs a loss of is :
Loss % = {(Common Loss and Gain %)/ 10 }2 = (x/10)2
1. When a Person buys two similar items, sells one at a gain of x% and the other at a loss of x%, then the seller incurs no gain no loss
2. If a trader profess to sell his goods at cost Price, But uses false weights, then
Gain % = {(error)/ ((True Value)-(Error))}*100 %
Important Points to remember that :
• Profit % and Loss % is fully based on Cost Price alone .
Profit =10% i.e. 10% of Cost Price is Profit .This is the meaning of 10% Profit.
• Discount is fully based on the Market Price/Retail Price/List Price.
Discount = 10% i.e. 10% on Market Price is Discount .This is the meaning of 10% Discount.
• Doubling the price and then reducing it by 50% does not yield 50% profit – the net effect is no-profit-no-loss.
• Successive discounts of 10%, 20% and 30% does not yield an overall 60% discount –the actual total is only 49.6%.
• Successive discounts of 25%, 10% and 5% is not the same as successive discounts of 20%, 15% and 5% although both add up to 40%. The actual total discounts are 35.875% and 35.4% respectively.
• A is 200% of B => A = 2B. But A is 200% more of B = >A = 3B. Similarly, P is twice as old as Q => P’s age = 2 × Q’s age, but A is twice older than B = >A’s age = 3 × B’s age.
Example:
1.A man buys an article for Rs.27.50 and sells it for Rs.28.60.Find his gain percent ?
Buys denotes the cost price , c.p = 27.50
Sells denotes the selling price, s.p = 28.60
For finding the profit % by two methods.
1. find the profit and apply on (3) profit % formula
2. Directly apply in 7 th formula.
Decide which is easy for you .
First method
Profit = 28.60 – 27.50 = 1.10
Profit % = (1.10 / 27.50)*100 = 4%
Second Method
60.  = ((100 + gain% )/100 }27.50
(28.60*100 )/27.50 = 100 + gain %
Gain % = 104-100 = 4%
2.John bought a satellite radio for rs 4,000 and sold it at a loss of 5% due to unavoidable circumstances.Find his Selling Price ?
Cost Price = 4000
Loss = 5 %
From the formula 7 ,we can get it directly ,
Selling Price = rs3800.
3.If the cost Price of two articles is 1000 each, one of them is sold at 10% profit and the other at 10% loss. Find the percentage of profit or loss on the whole transaction ?
From formula 12 ,we can say directly No Loss No Gain .Lets Check out,
Total Cost Price = Rs.2000.
Need to find the total Selling Price ,
By the Formula 7,
Selling Price of one article (10% profit) = 1100
By the Formula 8,
Selling Price of another article (10% loss) = 900
So the total selling price = 1100+900 = Rs.2000.
Selling Price = Cost Price .So No Loss No Gain in whole transaction.
4.If the Selling Price of two articles is 1000 each, sold one at 10% profit and the other at 10% loss. Find the percentage of profit or loss on the whole transaction ?
From formula 11, we can say that loss % = (10/10)2= 1%
Lets check out in another way ,
Total Selling Price = Rs.2000
By the Formula 7,
Cost Price of one article (sold at 10% profit) = 909.09
By the Formula 8,
Cost Price of another article (sold at 10% loss) = 1111.11
Total Cost Price = 909.09+1111.11 = Rs.2020.20
here C.P > S.P . So loss incurs in whole transaction.
Loss % = {(2020.20 – 2000 )/ 2020.20 }*100 = 1.00%
We have verified the answer in both the way. So we can use formula directly
5.The selling price of 15 chairs is equals to the cost price of 20 chairs. Find the Profit or Loss % ?
Given , S.P of 15 chairs = C.P of 20 chairs.
From the above eqn we can conclude that we can get profit. Because Selling 15 chair itself we will get the total cost of the 20 chairs. So we get the profit of selling 5 chairs.
First method,
Using Profit% formula we can find the answer,
Profit% = (Profit/ Cost Price )*100
Profit % = {(S.P of 5 chairs)/(S.P of 15 chairs or C.P of 20 chairs) }*100
Profit % = (5/15)* 100 = 33.33%
Second method,
Let we take cost price be Rs 1,selling price be Rs x.
from the eqn, we can write 15x = 20 ; x = Rs. 1.3333
S.P > C.P , So we will get profit only. Profit = 1.3333 – 1.00 = Rs. 0.3333
Profit % = (0.3333/1 )*100 =33.33%
Third method,
Let we take cost price be ‘rs x’ and selling price be ‘rs y’. We can do this way also. Try this method
Simple Interest
Short Cut 1:
If a sum of money become “x” times in “t” years, at Simple Interest, then the rate of interest is given by:

Example 1:
A scheme in Muthoot Finance claim to double any amount in 5 years at simple interest. What is the rate of effective rate of interest?
Solution 1:
Here, x=2 and t= 5 and putting the values in equation:

Short Cut 2:
A certain sum of money amounts to Rs A1 in t years at r% per annum, then the time in which it will amount to Rs A2 at the same rate of interest is given by:

Example 2:
A certain sum of money amounts to Rs 4800 in 5 years at 4% per annum. In how many years it will amount to Rs 5120 at the same rate.
Solution 2:
Here, A2= 5120, A1= 4800, t= 5 and r= 4
Putting the values in the equation

Short Cut 3:
If a sum of money becomes “n” times in “t” years at Simple Interest, then the time in which it will amount to “m” times is given by:

Example 3:
If a sum of money doubles itself in 4 years at simple interest. How long will it take for the amount to be 10 times?
Solution 3:
Here, n= 2, t= 4 and m= 10
Putting values in equation:

In simple Interest,
Let
P be the Principal/Sum,
R be the Interest,(denotes in %)
N Number of Years,
SI Simple Interest.( Interest amount of certain sum for certain period)
SI = [P*N*R]/ 100
Note:
1.If a sum of money become “X” times in “T” years, at Simple Interest, then the rate of interest “R%” is given by:
R=100(X-1)/T %
2.Diff. between CI and SI be Rs.X  for 2 years for a rate of R%.
To find the Principal Amount,
Y = X (100/R)2
For 3 years , Principal Amount,
Y =X*1003/ [R2 (300+R)]
Ex:
1.Calculate the simple interest amount for principal of rs 10000 on rate of 6% for 5 years.
P = 10000 N = 5 R = 6%
SI = [10000*5*6]/100
= 3000
Interest amount is Rs 3000.
Compound Interest
Let
P be the Principal/Sum,
R be the Interest,(denotes in %)
N be the Number of Years,
A be the Amount,
CI be the Compound Interest.
Compound Interest = Amount – Principal
1.When the interest is calculated Annually,
Amount = P{ 1 + R/100}N
Example:
1.Find the compound Interest on Rs.7500 at 4% per annum for 2 years compounded Annually?
A = 7500 { 1+ 4/100 }2
A= Rs.8112
CI = A – P
= 8112 – 7500
= Rs.612
2.When the interest is calculated Half-Yearly,
Amount = P{ 1 + (R/2)/100}2N
Example:
1.Find the compound Interest on Rs.7500 at 4% per annum for 2 years compounded Half-Yearly?
A = 7500 { 1+ (4/2)/100 }2*2
= 7500 { (102/100)4 }
= Rs.8118.24
CI = A – P
= 8118.24 – 7500
= Rs.618.24
3.When the interest is calculated Quarterly,
Amount = P{ 1 + (R/4)/100}4N
Example:
1.Find the compound Interest on Rs.7500 at 4% per annum for 2 years compounded Quarterly?
A = 7500 { 1+ (4/4)/100 }2*4
= 7500 {(101/100)8}
A= Rs.8121.43
CI = A – P
= 8121.43 – 7500
= Rs.621.43
4.When the Rates R1,R2,R3… are different for different years N1,N2,N3,……
Amount = P{ 1 + R1/100} { 1 + R2/100} { 1 + R3/100}……….
Example:
1.Find the compound Interest on Rs.7500 at 4%,5%,6% per annum for 1,2,3 year respectively compounded Annually?
A = 7500 { 1+ 4/100 }{ 1+ 5/100 }{ 1+ 6/100 }
A= Rs.8681.4
CI = A – P
= 8681.4 – 7500
= Rs.1181.4
Points to Remember:
If the interest of R% and a sum of Rs P and no of years N , the interest for SI is same for every year .But differ for every year in CI. Bcoz In CI, (Suppose we compounded annually) The amount obtain after first year becomes principal for second year,the amount obtain after second year becomes principal for third year and so on…
Simple Shortcuts:
If Sn and Cn represent the simple and the compound interest respectively for the nth year, then
1. Sn is the same for all values of n. i.e. simple interest is the same for all years.
2. C1 = S1
3. Cn = Cn – 1(1 + i). i.e. compound interest for any year is (1+ i) times the previous year’s compound interest.
4. Cn = Cn – 1(1 + i) = Cn – 2(1 + i)2 = Cn – 3(1 + i)3………..
5. Cn – Cn – 1 = Cn – 1 × i. i.e. extra compound interest for any year is i times the previous year’s compound interest.
i represents the interest (not in percentage)
Example:
A sum invested for two years yields an interest of Rs.1200 under simple interest
and Rs.1260 under compound interest. What is the rate of interest?
From 1: SI for 2 years is Rs.1200;
For 1 year 1200/2 = Rs.600.
From 2: CI for 2 years is Rs.1260;
For 1 year Rs.600 Bcoz C1 = SI1=600
C2 = 1260-600 = 660.
Use 5 : Cn – Cn – 1 = Cn – 1 × i.
Here n = 2,
C2 – C2-1= C2-1 * i
C2 – C1 =C1 *i
660 – 600= 600 * i
60 = 600*i
i = 60/600
i = 0.1
So the interest is 0.1*100 = 10%
Streams
Short Cut 1:
If the speed of boat in still water is x km/hr and the speed of stream is y km/hr. If the boat travels for “T” hours, then distance covered is given by:
Distance travelled upstream in time “T” is given by:
(x-y)*T
Distance travelled downstream in Time “T” is given by:
(x+y)*T
Example 1:
If the speed of boat in still water is 10 km/hr and the speed of stream is 5 km/hr. If the boat travels for 3 hours upstream and 2 hours downstream, what is the total distance covered?
Solution 1:
Here, x= 10, y=5 and T= 2
Distance travelled upstream is:
(x-y)*T
Distance travelled upstream is: 10 km
Distance travelled downstream is:
(x+y)*T
Distance travelled upstream is: 30 km
Answer: Total distance travelled is: 40 km
Short Cut 2:
A man can row x km/hr in still water. If in a stream which is flowing at y km/hr, it takes him T hrs to row to a place and back. The distance between two places is given by:

Example 2:
A man can row 5km/hr in still water. If the river is running at 1.5 km/hr, it takes him 1 hour to row to a place and back. How far is the place?
Solution 2:
Here x= 5, y= 1.5, T= 1
Putting the value:

Short Cut 3:
A man rows a certain distance downstream x hours and returns the same distance in y hours. If the stream flows at the rate of z km/hr then the speed of man in still water is given by:

Example 3:
Ram can row a certain distance downstream in 6 hours and return the same distance in 9 hours. If the stream flows at the rate of 3 km/hr find the speed of Ram in still water.
Solution 3:
x=6, y=9 and z=3

Downstream (along the current) speed (D) = Boat speed (B) + current (stream) speed (C).         D=B+C
Upstream (against the current) speed (U) = Boat speed – current (stream) speed. U=B–C
Speed of the boat = average of downstream and upstream speeds B = (D + U)/2
Speed of the current = half the difference of downstream and upstream speeds    C = (D – U)/2
Example:
1.A boat takes 5 hours to go from A to B and 8 hours to return to A. If AB distance is 40 km, find the speed of (a) the boat and (b) the current.
Since B to A takes more time, it is upstream and hence AB is downstream. Downstream speed = 40/5 = 8 kmph.
Upstream speed= 40/8 = 5 kmph.
Boat speed = (8 + 5)/2 = 6.5 kmph.
Current speed = (8 – 5)/2 = 1.5 kmph.
2.A man cn row a boat at 20 kmph in still water.If the speed of the stream is 6 kmph, what is the time taken to row a distance of 60 km downstream ?
Speed of downstream = boat speed + stream speed = 20 + 6 = 26 kmph
Time required to cover 60 km downstream = d/s = 60/26 = (30/13) hours.
3.The time taken by a man to row his boat upstream is twice the time taken by him to row the same distance downstream. If the speed of the boat in still water is 42 kmph, find the speed of the stream ?
The time taken to row his boat upstream is twice the time taken by him to row the same distance downstream. Therefore, the ratio of the time taken is (2:1). So, the ratio of the speed of the boat in still water to the speed of the stream = (2+1)/(2-1) = 3:1 .Thus, Speed of the stream = (42)/3 = 14 kmph.
Boats and Streams problems are very easy. If you having any difficult questions, we will discuss on that. Plz post here:)
Escalator Problem
The escalator problem is identical to the boat stream problem, escalator replacing the stream and person replacing the boat. The speed and distance are expressed in terms of the number of steps of the escalator.
Example:
1.A couple and their son get into an escalator going up. The father reaches the top in 15 seconds taking 4 steps, but the son reaches 5 seconds ahead of his father. The mother who did not take any step of her own reaches the top 5 seconds after her husband reached.
Determine (a) the number of steps taken by the son, (b) the distance between the two points and (c) the speed of the escalator.
Let e be the speed of the escalator in steps per second, s the number of steps climbed by the son and n the number of steps between the points. Father takes 15 seconds to reach the top and during this time he climbs 4 steps  and the escalator moves by 15e steps. So, the total number of steps is (4 + 15e) which by the above assumption is n.
Thus, n = 4 + 15e………. (1)
By the same logic for the son,
n = s + 10e……….. (2)
Since the mother does not take any steps of her own, n = 20e ………….. (3)
From (2) and (3), s = 10e…………… (4)
From (1) and (2), s = 5e + 4………… (5)
From (4) and (5), 5e = 4 or e = 0.8……………. (6)
(6) in (4): s = 10 × 0.8 = 8 and
(6) in (3): n = 20 × 0.8 = 16.
Thus, the number of steps taken by the son = 8
the distance between the points = 16
the speed of the escalator = 0.8 step per second.
Percentage
Short Cut 1:
If the value of a number is first increased by x% and later decreased by x% then the resultant value is always less than the original value. The percent decrease is given by:

Example 1:
The salary of a worker is first increased by 10% and thereafter it was reduced by 10%. What was the net change in the salary?
Solution:
The net effect on the salary is decrease in following percentage:
Where x=10

Short Cut 2:
If first value is r% more then the second value, then the second value is lesser by first by following percentage:

Example 2:
If the salary of Manish is 25% greater than Abhishek, than by how much percentage in the salary of Abhishek less then Manish?
Solution 2:
Here r= 25
Thus, putting the value we get:

Short Cut 3:
If first value is r% less then the second value, then the second value is lesser by first by following percentage:

Example 3:
If the salary of Manish is 20% lesser than Summit, than by how much percentage in the salary of Abhishek less then Manish?
Solution: Putting r= 20

Sign Conventions:
For increase in percentage on a number or an amount put +ve Sign.
For increase in percentage on a number or an amount put -ve Sign
Apply the following formula , the net increase or decrease in percentage or a number is given by:

Where:
x% and y% are the successive increase or decrease in the percentage on a number or account, follow the sign conventions for the increase or decreaseto get the result.
Case 1:
If a number is increase successively by a% and b% the net effect on the original number is:

Case 2:
If a number is decrease successively by a% and b% the net effect on the original number is:

The negative sign indicates net decrease in the orignal amount or number
Case 3:
If a number is first decrease by a% and then increase by b% the net effect on the original number is:

The sign of result in such case would indicate that if the original number was increased or decreased.
Example 1:
The tax on a commodity was reduced by 20% as a result its consumption increase by 15%. Find the net effect on the revenue.
Solution:
It is the case of decrease in percentage followed by an increase.
Applying the formula here x= -20, y= 15
Thus net change in the orignal revenue is:

Example 2:
Peter England a well know brand of Clothes offer 30% and 70% discounts successively on Christmas on Men Shirt. What is the actual discount offered by Peter England?
Solution:
The formula for successive discount comes under the category of succesive decrease. The following in the formula
Where a= 30, b= 70;

Thus the net discount in this case is 79%.
• To convert the given fraction into percentage we have to multiply it by 100.To express a/b as a percent : a/b = {(a/b)*100 }%
• Example 1:  1/4 or 0.25 = (1/4)*100 or 0.25*100 =25%
• To convert the given percentage into fraction we have to divide it by 100. To express x% as a fraction : a% = a/100.
• Example 2:  20% = 20/100 = 1/5
• If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is                                      {(R/(100+R))*100}%
• If the price of a commodity decreases by R%, then the increase in consumption so as not to decrease the expenditure is                            {(R/(100-R))*100}%
• If A is R% more than B, then B is less than A by {R/(100+R))*100}%
• If A is R% less than B, then B is more than A by { (R/(100-R))*100}%
• If the value or price of an item goes up/down by p%, then the percentage reduction/increment to be made to bring it back to the original level is given by, [(100p)/(100+p)]% or [(100p)/(100-p)]%
• If the value or price of an item goes up/down by p%, then the quantity consumed should be reduced/increased by [(100p)/(100+p)]% or [(100p)/(100-p)]% so that the total expenditure remains the same.
Example 3: In a class of 75 students 32% are girls.Find the number of boys in that class ?
The total number of students in the class is 75. In which, 32% are girls.
Therefore the remaining 68% of students are boys.
Hence, the number of boys in the class = (68/100)*75 = 51
Population:
Let the population of a town be P now and suppose it increases at the rate of R% per annum, then:
Population after n years = P(1+R/100)n
Population n years ago = P/(1+R/100)n
Depreciation :
Let the present value of a machine be P.Suppose it depreciates at the rate of R% per annum. Then:
Value of the machine after n years = P(1-R/100)n
Value of the machine n years ago = P/(1-R/100)n
Percentage Increase:
If the final value is more than the initial value, then there is an increase. i.e., Increased value = Final value – Initial Value
Percentage increase = {Increased Value / Initial Value }*100
Percentage Decrease:
If the final value is more than the initial value, then there is an decrease. i.e., Decreased value = Final value – Initial Value
Percentage decrease = {Decreased Value / Initial Value }*100
Example 4:If in an examination Prabhu got 80% of marks and Sathish got 90% of marks , then find the percentage by which Sathish got more than Prabhu ?
Given that Sathish = 90%, Prabhu got = 80%
In Question, the initial value (referred value) is Prabhu’s score and the final value is Sathish’s score. By using the formula we get,
% increase = (Increase value/ Initial Value)*100
={(90-80)/80}*100
=12.5%
Therefore , the Sathish got 12.5% of marks more than Prabhu.
Successive Increase:
If there are two successive increases of p%,q% then the effective percentage increase is [{(100+p)/100}{(100+q)/100}-1]*100
Successive Decrease:
If there are two successive decreases of p%,q% then the effective percentage increase is
[1-{(100+p)/100}{(100+q)/100}]*100
Increase followed by Decrease :
If there is an increase of P% and followed by a decrease q%, then the effective percentage increase/decrease is
[{(100+p)/100}{(100-q)/100}-1]*100
Decrease followed by Increase :
If there is an Decrease of P% and followed by a Increase q%, then the effective percentage increase/decrease is
[1-{(100+p)/100}{(100-q)/100}]*100
Example 5:
Raman’s salary was decreased by 50% and subsequently increased by 50% .How much Percentage does he lose ?
By formula , p= 50, q=50,
Percentage, he lose = [1-{(150/100)}{(50/100)}]*100
= (1-75/100)*100  =25%
Pipes and Cisterns
Short Cut: 1
If a pipe fills the tank in x hours, and another fills the same tank in y hours and the third fills the tank in z hour. Then the time required, if all the three pipes are open together to fill the tank is:

Example:
Three pipes can fill the tank in 20, 30 and 40 hours respectively. Find the time required to fill the tank if all the pipes are filled simultaneously.
Solution
Here; x= 20, y= 30 and z= 40 hours
Putting the values in

Short Cut: 2
If pipe A can fill a tank in x minutes, pipe B can fill the same in y minutes, there is also an outlet C in the Tank. All these are opened simultaneously and the tank takes “T” minutes to get filled. The time in which C can empty the tank in minutes is given by:

Example:
Two pipes can fill a cistern in 60 minutes and 75 minutes respectively. There is also an outlet C, if all the three pipes are opened together, the tanks get filled in 50 minutes. Find the time taken by C to empty the full tank?
Here, x= 60, y= 75 and T= 50 minutes, putting the values in equation:

Short Cut 3
A tap M can empty a tank in x minutes, while another tap N can empty it in y minutes. If both the emptying taps are opened together, then the time taken to empty the full tank is:

Example
A tap can empty the tank in 10 minutes; another tap can do the same in 5 minutes. Find the time required by both the taps to empty the tank simultaneously?
Solution
Here x= 10 and y= 5, putting the values we get:

Inlet :
A pipe connected with a tank or a cistern or a reservoir, that fills it, is known as an inlet.
Outlet:
A pipe connected with a tank or a cistern or a reservoir, emptying it,is known as an outlet.
Properties:
• If a pipe can fill a tank in x hours, then:
part filled in 1 hour = 1/x
• If a pipe can empty a tank in y hours, then:
part filled in 1 hour = 1/y
• If a pipe can fill a tank in x hours and another pipe can empty the tank the full tank in y hours ( where y>x), there on opening both the pipes, the net part filled in 1 hour = { 1/x – 1/y }
• If a pipe can fill a tank in x hours and another pipe can empty the tank the full tank in y hours ( where y
E represents Total time taken to empty the tank
F represents Total time taken to fill the tank
L represents the L.C.M
e represents time taken to empty the tank
f represents time taken to fill the tank
• Time for emptying , (emptying pipe is bigger in size.)
E = (f * e)/(f – e)
• Time for filling , (Filling pipe is bigger in size.)
F = (e * f)/(e – f)
• Pipes ‘A’ & ‘B’ can fill a tank in f1 hrs & f2 hrs respectively. Another pipe ‘C’ can empty the full tank in ‘e’ hrs. If the three pipes are opened simultaneously then the tank is filled in ,
F = L/[(L/f1) + (L/f2) - (L/e)]
• Two taps ‘A’ & ‘B’ can fill a tank in ‘t1′ & ‘t2′ hrs respectively. Another pipe ‘C’ can empty the full tank in ‘e’ hrs. If the tank is full & all the three pipes are opened simultaneously . Then the tank will be emptied in,
E = L/[(L/e) - (L/f1) - (L/f2)]
• Capacity of the tank is , F = (f * e)/(e – f)
• A filling tap can fill a tank in ‘f’ hrs. But it takes ‘e’ hrs longer due to a leak at the bottom. The leak will empty the full tank in ,
E = [ t(f + e) * tf ] / [ t(f + e) – tf ]
TCS pattern:
Example :
If a pipe can fill the tank in 6 hrs but unfortunately there was a leak in the tank due to which it took 30 more minutes .Now if the tank was full how much time will it take to get emptied through the leak?
(a) 39 hrs (b) 78 hrs (c) 72 hrs (d) 70 hrs
By last property,
t(f+e) = 6+0.5 =6.5hrs
tf = 6 hrs
E = 6.5*6 / (6.5 – 6)
= 78 hrs .
Number System
Short Cut 1:
To find the greatest number that will divide given numbers say x1, x2, x3…….xn so as to leave the same remainder in each case, we find the HCF of the positive difference of numbers i.e. HCF of:
Ix1-x2I, Ix2-x3I, Ix3-x4I and……
Example 1:
Find the greatest number which when divide 55, 127 and 175 leaves the same remainder in each case?
Solution 1:
Here x1= 55, x2= 127 and x3= 175
The values of: Ix1-x2I, Ix2-x3I, Ix3-x1I are 72, 48, 120
HCF of 72, 48 and 120 is 24
Short Cut 2:
To find the greatest number which will divide x, y and z leaving remainders a, b and c respectively is given by:
H.C.F of (x-a), (y-b) and (z-c).
Example 2:
What is the greatest number that will divide 29, 60 and 103 and will leave remainders 5, 12 and 7 respectively?
Solution 2:
Here, x= 29, y= 60 and z=103
Also, a= 5, b=12 and c=7
Thus, (x-a), (y-b) and (z-c) are 24, 48 and 96. HCF of these numbers is 24.
Short Cut 3:
To find the least number which, when divided by “x”, “y” and “z” leaves the same remainder “r” in each case is given by:
L.C.M of (x,y,z) + r
Example 3:
Find the least number which when divided by 2, 3, 5 and 7 leaves the remainder 1.
Solution 3:
The least number is given by L.C.M of (2,3,5,7) + 1
Short Cut 1:
If the sum of a number and its square is x, then the number is given by:

Example:
If the sum of a number and its square is 156, find the number
Solution:
Here x= 156
Putting the value in the formula:

Short Cut 2:
If the sum of two numbers is “s” and their difference is “d”. Then the product of the number is given by:

Example:
The sum of two numbers is 38 and the difference is 4, find the product of the two numbers:
Solution:
Here s= 38 and d= 4, putting the values in the formula:

Short Cut 3
If the ratio of the sum and the difference of two number is a:b, then the ratio of these two numbers is given by:

Example 3:
Ratio of the sum and the differences of the two numbers is 13:3, find the ratio of these two numbers
Solution:
Here, a=13 and b=3
Putting the value:

For Calculating Squares
To  Square a Number Ending in 1
Example  :  41  x  41
Solution by steps:
(a)  Square the next lower number (always ends with zero)  40  x   40  =  1,600
(b)  Obtain the sum of the next lower number and the number being squared:  40  +  41  =  81
(c )  Add (a) and (b):  1,600  +  81  =  1,681,  answer
Short way:  40  x  40  =  1,600;  40  +  41  =  81;  answer  1,681
To Square a Number Ending in 9
Example :    39  x  39
Solution by steps  :
(a)  Square the next higher  number 40  x   40  =  1,600
(b)  Obtain the sum of the number being squared and the next higher number:  39  +  40  =  79
(c )  Subtract  (a) and (b):  1,600  - 79  =  1,521,  Answer
Short way:  40  x  40  =  1,600;  -(39  +  40)  =  1,521
To Square a number Ending in 5
Example : 35  x  35
Solution by steps :
(a) Multiply the first digit by the digit plus 1 : 3  x  4 = 12
Short way :   3  x  4 = 12;answer  1,225
Population
Short Cut 1
If the present population of a place is P, and it annually increases at the rate r%. The population at the end of “n” years is given by:

Short Cut 2
If the population of a place in the current year is P, it increases at a rate of x% for the first year, increases at the rate of y% for the second year and z% for the third year. The population after third is given by:

Short Cut 3
If the population of a place in the current year is P, it decreases at a rate of x% for the first year, increases at the rate of y% for the second year and z% for the third year. The population after third is given by:

Short Cut 4
If the population of a place in the current year is P, it increases at a rate of x% for the first year, decreases at the rate of y% for the second year and again increases at z% for the third year. The population after third is given by:

Distance and Speed
Short Cut 1
If a person travels half a distance in a journey at x km/hr and remaining half at the speed of y km/hr, the average speed of the whole journey is given by:

Example 1
A person travels half of journey at the speed of 30 km/hr and the next half at a speed of 15 km/hr. What is the average speed of the person during the whole journey?
Solution:
Here; x= 30 and y= 15
Putting the values we get:

Short Cut 2
If a person travels three equal distances in a journey at a speed of x km/hr, y km/hr and z km/hr. The average speed is given by:

Example 2
Ravi starts from Delhi to Agra a distance of about 300 kms. He divided his journey into a distance of thee equal parts in terms of distance and covered them with the speed of 30 km/hr, 60 km/hr and 90 km/hr, calculate his average speed during the journey:
Solution:
Here: x= 30, y= 60, z= 90.

Putting the values we get:
Short Cut 3
If a person travels Mth part of a distance at x km/hr, Nth part at y km/hr and the remaining Pth part at z km/hr, then his average speed in km/hr is given by:

If instead of proportion, the parts of the distance are given in percentage i.e. M%, N% and P% respectively, then the formula becomes:

Example 3
Hari travels 20% of a distance at an average speed of 20 km/hr, next 30% distance at an average speed of 30 km/hr and the remaining 50% of distance at the average speed of 50 km/hr. Find his average speed during the whole journey.
Solution:
Here M=20, N=30 and P= 50 also x= 20, y= 30 and z= 50
Putting the values we get:
Train problems broadly center around the following three types.
1. Time taken by a train to cross a stationary point
2. Time taken by a train to cross a stationary length
3. Time taken by a train to cross a moving length
• Time taken by a train to cross a stationary point = train length/train speed
• Time taken by a train to cross a stationary length = (train length + stationary length)/train speed
• Time taken by a train to cross another train moving in the same direction = sum of length of the two trains/difference in their speeds
• Time taken by a train to cross another train moving towards it = sum of length of the two trains/sum of their speeds
Example:
1.If a train going at 90 kmph takes 28 seconds to go past a lamp post, but 80seconds to cross a platform, what is the length of (a) the train and (b) the platform?
90 kmph= (90 × 5/18) = 25 m/s.
Solution for (a):
Given the time to cross the lamp post as 28 seconds,
Formula:
(train length/25 )= 28 or train length = 700 m.
Solution for (b):
Given the time to cross the platform as 80 seconds,
Formula:
{(train length + platform length)/25 } = 80
or (train length + platform length) = 2000 m.
Since train length is 700 m, platform length = 1300 m or 1.3 km.
2.How long does it take for a train of length 800 m moving at 80 kmph to cross a train of length 1200 m coming at a speed of 100 kmph from the opposite direction?
The relative speed = 80 + 100 (Moving opposite direction)
= 180 kmph = 50 m/s.
The distance to be covered = 800 + 1200 ( Sum of the train length)
= 2000 m.
Formula:
Time taken for crossing = 2000/50 (Time = Sum of the train length / Relative Speed)         = 40 seconds.
Above two example problems are basic type. Here i will give three more examples with little bit different.
Example 3:
Two trains of length 100m and 200m are 100m apart.They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph.After how much time will the trains meet ?
In this problem, we need to find the time taken by trains to meet each other. So no need to consider the  train length.
Distance between the trains = 100m
Relative speed = 54+72 kmph (Trains are in opposite direction)
= 126 kmph = 35mps
Time taken = Distance/ Relative speed
= 100/35 = 20/7 seconds
Example 4:
Two trains of length 100m and 200m are 100m apart.They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph.After how much time will the trains cross each other ?
In this problem, we need to find the time taken by trains to cross each other. So we need to consider the train length and also distance between the trains.
Distance to be covered = 100+200 +100 (Train length 1 + Train length 2 +distance between them)
Relative Speed = 35 mps
Time taken = Distance/ Relative speed
= 400/35 = 80/7 seconds
Example 5:
Two trains of length 100m and 200m are 100m apart.They start moving towards each other on parallel tracks, at speed 54 kmph and 72 kmph.After how much time will the trains be 100m apart again ?
In this problem, Trains have to cross each other and then be at 100m apart.
For distance , we need to consider initial distance , train length and their final distance between them.
Relative Speed = 35 mps
Distance to be covered = 100+100+200+100 = 500m
Time taken = 500/35 = 100/7 seconds.
Consumption and Cost
In a number of questions the increase or decrease in Consumption or cost is given in percentage and its effect on decrease or increase in Cost or Consumption is asked in percentage. The two cases are as:
Case 1 : Increase in Cost leading to decrease in Consumption
If the price of a commodity is increased by x%, then reduction in consumption so as not to increase the expenditure is:
Example:
The price of cooking oil has increased by 25%. The percentage change of reduction that a family should effect in the use of cooking oil so as not to increase the expenditure is:
Solution:

Applying the formula, here x% = 25%
So,
Case 2: Decrease in Cost leading to increase in Consumption
If the price of a commodity is decreased by x%, then increase in consumption so as not to increase the expenditure is:

Example:
The price of cooking oil has decreased by 25%. The percentage change of increase in consumption so that a family should effect in the use of cooking oil so as not to decrease the expenditure is:
Solution:
Applying the formula, here x% = 25%
Ages :
The basic principle involved in age problems is the age difference between any two persons remains the same at all points of time.
Example 1:
My present age is 25 and my brothers age is 30 . The age difference between us is 5. Ten years before, my age is 15 and my brothers age is 20. At that time also, Age difference between us is 5 only.
Example 2: The ages of two persons differ by 16 years.If 6 years ago, the elder one be 3 times as old as the younger one, Find their present ages?
Let the age of younger be x years.Then the age of elder be x+16 years.
(x+16)-6 = 3(x-6) :: solving this equation we get x = 14 .
Then their present ages are 14 years and 30 years.
Example 3: Sachin is younger than Rahul by 4 years. If their ages are in the respective ratio of 7:9, how old is Sachin ?
Given: R= S+4; R:S = 7:9
To find Sachin age, R/S = (S+4)/S = 7/9
solving above equation we get , S =18years.
Example 4:The sum of the ages of father and his son is 60 years. Six years ago, fathers age was five times the age of the son. After 6 years, son’s age will be :
Given:
f+s = 60; (f-6)= 5(s-6) <==> f-5s = -24;
solving above eqn we get , f=46 , s=14 ;
After 6 years, son’s age will be : 20years.