Aptitude

Number System       LCM and GCD       Problems on Ages       Averages       Ratios and Proportions       Mixtures and Alligations       Percentages       Profit and Loss       Simple and Compound Interest       Time and Work       Pipes and Cisterns       Time and Distance       Problems on Trains       Boats and Streams       Partnership       Permutations and Combinations       Probability       Races and Games       Logarithms       Mensuration or Areas and Volumes

Averages:

Definition:
Average means equal distribution among all the observations.
For ex: A’s salary is Rs. 30, B’s salary is Rs. 40, C’s salary is Rs. 50 then average of three persons salary is Rs. 40.
Example : If you are asked to tell about the marks you scored in your final exams, you would either say I scored 85% or scored 85 marks on an average.
Lets say your marks in 5 subjects were 82, 87, 83, 84, 89 Sum of all the marks = 82 + 87 + 83 + 84 + 89 = 425
On dividing the total sum (Sum of all the marks scored) with 5 (number of subjects), we arrive at the average = 425/5 = 85
Formula:
Average or Mean = (Sum of all the items in the group/Number of items in the group)
Note :
• The average of a group will always lie between the smallest and the largest value
• The average will always be more than the smallest value in the group (In this case 82), but will be less than the largest value in the group (In this case 89)
Points to Remember:
1. If the value of each item is increased by a same value ‘x’, then the average of the group will also increase by ‘x’
Example : In the above example, if each number is increased by 5, the numbers will be (82 + 5), (87 + 5), (83 + 5), (84 + 5), (89 + 5)
Average = (87 + 92 + 88 + 89 + 94)/5 = 450/5 = 90
Note : It can be seen that the average of the group increased by 5. Similar logic applies to the points below.
2. If the value of each item is decreased by a same value ‘x’, then the average of the group will also decrease by ‘x’
3. If the value of each item is multiplied by a same value ‘x’, then the average of the group will also be multiplied by ‘x’
4. If the value of each item is divided by a same value ‘x’, then the average of the group will also be divided by ‘x’
5. If the value of each item is increased by a same value ‘x’, then the average of the group will also increase by ‘x’
Short cut technique to find Average:
Problem : Find the average of 82, 87, 83, 84, 89?
Solution :
Consider any arbitrary number close to any of the numbers in the group lets say 90 (You can take anything which you want)
Subtract this arbitrary number from each of the numbers in the group and add their average to the arbitrary number as shown below
90 + [(82 – 90) + (87 – 90) + (83 – 90) + (84 – 90) + (89 – 90)]/5
= 90 + [(-8) + (-3) + (-7) + (-6) + (-1)]/5 = 90 + (-25/5) = 90 – 5 = 85
This method is very much useful when dealing with huge number of numbers in the group.
Weighted Average:
We have seen how to find average when a single group is involved. In case there are multiple groups whose average needs to be defined, then we use the concept of weighted average.
Example :
If a average age of ‘x’ number of students in a class is ‘a’,
average age of ‘y’ number of students in a class is ‘b’,
average age of ‘z’ number of students in a class is ‘c’ and so on,
then the weighted average of the entire group is given by the formula
Weighted Average = (xa + yb + zc +………)/(x + y + z +……)
Note : Same formula can be used in case of factors similar to Age like Height, Weight etc
Example:
The average height of 60 students in class A is 150 and the average height of 40 students in the class is 160, find the weighted average of the entire class?
Solution:
Using the formula, we get weighted average = [(60 × 150) + (40 × 160)]/(60 + 40)
= (6 × 15) + (4 × 16) = 90 + 64 = 154

 

Ratios and Proportions:

Definition:
If A and B are two quantities and we need to provide the relation between two quantities of the form A/B then it is called ratio. We can write it as A:B. The first term of the ratio 'a' is called antecedent and second term 'b' is called consequent.
for Let as assume there are 9 pens and 10 pencils then the ratio between pens to pencils is 9:10.
Types of Ratios:
1. a square(a^2) : b square(b^2) is a duplicate ratio of a : b
2. va : vb is a sub-duplicate ration of a : b
3. a cube(a^3) : b cube(b^3) is a triplicate ratio of a : b
4. (a) power 1/3 : (b) power 1/3 is a sub-triplicate ratio of a : b
5. The compounded ratio of the ratios (a : b), (c : d), (e : f) is (ace : bdf)
6. The inverse ratio of A:B is B:A

 

Mixtures and Alligations:

Definiton:
Alligation is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture at a given price.

Mean Price: The cost price of a quantity of the mixture is called the mean price.
Rule of Alligation: If two ingredients are mixed, then:

or Another way of representation is

Practice Examples:
1. In what ratio must two kinds of sugar at Rs. 1.15 and Rs. 1.24 per kg be mixed so that by selling at Rs. 1.50 per kg, 25% may be gained?
a. 4: 5 b. 5: 4 c. 1: 1 d. 2: 3 e. None of these
2. How much water be mixed in 36 litre of milk worth Rs. 5.00 per litre, so that value of mixture is Rs. 3.60 per litre?
a. 10 litres b. 12 litres c. 11 litres d. 14 litres e. 17 litres
3. How many kg of Sugar at 50 P per kg must a man mix with 25 kg of sugar at 34P per kg so that by selling the mixture at 44P per kg he gains 10% on the outlay?
a. 10 kg b. 15 kg c. 20 kg d. 16 kg e. 18 kg

4. How much sugar at Rs. 9.5 a kg should be added to 17 kg of tea at Rs. 20 a kg so that the mixture be worth Rs. 13 a kg.?
a. 11 kg b. 17 kg c. 21 kg d. 34 kg e. None of these
5. A hospital uses a mixture of salt and water at Rs. 7.62/litre. This mixture contains 5 % salt. Another mixture containing 75 % water costs Rs. 7.82/litre. How much does the patient pay if he buys 5 litres of mixture containing 18% salt?
a. Rs. 83.75 b. Rs. 73.85 c. Rs. 37.85 d. Rs. 38.75 e. None of these
6. How many kg of custard powder costing Rs. 42 per kg must be mixed with 16 kg of custard powder costing Rs. 60 per kg so that 20 % may be gained by selling the mixture at Rs. 60 per kg?
a. 11 kg b. 14 kg c. 12 kg d. 20 kg e. None of these
7. A vessel is full of mixture of spirit and water in which there is 20 per cent of spirit. 5 litres are drawn off and the vessel is filled up with water. If the spirit is now 12 %, find the total quantity in the vessel (in ltrs).
a. 722.1 b.5 c.60 d. 7.2 e. None of these
8. How much chicory at Rs. 5 a kg should be added to 20 kg of coffee at Rs. 12 a kg so that the mixture be worth Rs. 7.50 a kg.?
a. 21 kg b. 15 kg c. 36 kg d. 42 kg e. None of these
9. An alloy of copper and nickel contains 65 % copper. A second alloy contains copper and nickel in the ratio 17 : 3. In what ratio should the two alloys be mixed so that the new mixture contains 4 times as much copper as nickel?
a. 4 : 5 b. 5 : 4 c. 1 : 3 d. 2 : 3 e. None of these
10. Five litres of wine is removed from a cask full of wine and is replaced with water. Five litres of this mixture is then removed and replaced with water. If the ratio of wine to water in the cask is now 16 : 9, how much wine did the cask hold ?
a. 25 litres b. 50 litres c. 100 litres d. 150 litres e. 175 litres

 

Percentages:

Definition:
Percentage is defined as Per Cent means ”Out of 100”or ”For every 100”.
Example : x% = (x/100) this is read as ‘x’ percent is equal to x by 100
Commonly used Percentage formulas:
x% = (x/100)
(x × y)% = (x × y)/100
x% of y% = [(x × y)/100]% = (x/100) × (y/100) [Concept of Percentage]
x% of y = (x/100) × y = (x × y)/100
y% of x = (y/100) × x = (y × x)/100
Examples:
1) What is 45% of 480?
Solution:
Method 1 :
45% of 480 = (45/100) × 480 = (45 × 480)/100 = 21600/100 = 216
Method 2 :
We know that from the given question 100% = 480 ? 10% = 48
Therefore 45% = (40% + 5%) = (4 × 10%) + (10%/2) = (4 × 48) + (48/2) = 192 + 24 = 216
Note: The method 2 is very much useful to break down a difficult percentage into parts and arriving at the final answer.
2) What is 36% of 50% of 2000?
Solution:
36% of 50% of 2000 = [(36 × 50)/100]% × 2000 = 18% × 2000 = (18/100) × 2000 = 360
Memorize standard Percentages:
Its always better to remember standard value of few percentages in order to save time during problem solving. Below is the list:
(1/2) = 50%
(1/3) = 33 1/3 % or 33.33%
(1/4) = 25%
(1/5) = 20%
(1/6) = 16 2/3% or 16.66%
(1/7) = 14 2/7% or 14.287%
(1/8) = 12 1/2% or 12.5%
(1/9) = 11 1/9% or 11.11%
(1/10) = 10%
(1/11) = 9 1/11% or 9.09%
(1/12) = 8 1/3% or 8.33%
Note : The list can be extended further, but suggestion is to make sure you memorize the above percentage values without fail !!
Examples:
(3/8) = 3 × (1/8) = 3 × 12.5% = 37.5%
(9/11) = 9 × (1/11) = 9 × 9.09% = 81.81%
(4/9) = 4 × (1/9) = 4 × 11.11% = 44.44%
Percentage Change (Increase/Decrease):
This concept is considered to be the bread and butter for someone who wants to expertise in Percentages and has the major usage in real world.
Important Formulae:
1) Percentage Increase (Final Value > Initial Value) = [Actual Increase/(Initial or Original Value] × 100
where Actual Increase = Final Value – Initial or Original Value
2) Percentage Decrease (Actual Value > Final Value) = [Actual Decrease/(Initial or Original Value] × 100
where Actual Increase = Initial or Original Value – Final Value
3) If the percentage change (p%) and the initial value is given, then the formula for
If ‘p%’ increase, Final Value = Initial Value (1 + p%) = Initial Value [1 + (p/100)]
If ‘p%’ decrease, Final Value = Initial Value (1 – p%) = Initial Value [1 – (p/100)]
Note : Use the above formula when the problems deals with percentage change, Initial and Final Value.
Example Problems:
1) If the production of rice went up from 200 Tons to 250 Tons, then what is the percentage change?
Solution:
Actual Quantity = 200 Tons, Final Quantity = 250 Tons
Since Final Quantity > Actual Quantity, its a percentage increase
Using the formula, Percentage Increase = [(250 – 200)/200] × 100 = (50/200) × 100 = 25%
2) If the production of rice went down from 250 Tons to 200 Tons, then what is the percentage change?
Solution:
Actual Quantity = 200 Tons , Final Quantity = 250 Tons
Since Final Quantity < Actual Quantity, its a percentage decrease
Using the formula, Percentage Decrease = [(250 – 200)/250] × 100 = (50/250) × 100 = 20%
3) If the consumption of rice increases by 30% from the current consumption of 400 Tons, what would be the consumption after the increase?
Solution :
Final Consumption = Initial Consumption (1 + p/100)
400 [1 + (30/100] = 400 × (1 + 0.3) = 400 × 1.3 = 520 Tons
4) If the consumption of rice decreases by 30% from the current consumption of 400 Tons, what would be the consumption after the decrease?
Solution :
Final Consumption = Initial Consumption (1 – p/100)
400 [1 – (30/100] = 400 × (1 – 0.3) = 400 × 0.7 = 280 Tons
Some Short cut Techniques:
If the value of an item goes up by x% , the percentage reduction to now made to bring it back to the original value is 100x/(100 + x)%
If the value of an item goes down by x% , the percentage increment to now made to bring it back to the original value is 100x/(100 – x)%
If A is x% more than B, then B is 100x/(100 + x)% less than A
If A is x% less than B, then B is 100x/(100 – x)% more than A
Note : We will see the usage of these techniques when solving example problems.
Concept of Percentage Points:
It is the difference of two percentage figures.
Example :
If rice forms 30% of total food grains production in year 1 and 50% of total food grain production in year 2, then the percentage points = 50% – 30% = 20%

 

Profit and Loss:

Cost Price: The price (amount) paid to purchase a product or the cost incurred in manufacturing a product is known as the cost price (CP) of that product.
Types of cost
1. Fixed cost: It is a type of cost which is fixed under all conditions and does not vary according to the number of units produced.
2. Variable cost: Variable cost is a type of cost which varies according to the number of units. This is quite easy to understand.
3. Semi-variable cost: As the name suggests, these costs are the ones that are fixed in part and variable in part. Effectively, this is the case that we see most often.Imagine the scenario in a factory. There is a capital cost, which remains the same under all conditions (fixed cost) and a variable cost of the product, which in turn depends upon various factors.
Selling Price: The price at which a product is sold is called the selling price (SP) of the product.
Marked Price: Do shopkeepers put up price on the label that they wish to sell on or they put up an inflated price? If you think closely, majority of the shopkeepers mark-up their products, in anticipation of the discounts they would have to offer. This is a clever way of operating. Mark-up the price in advance, offer a discount and make the customer feel happy, and then sell the product. Pretty effective, eh?
List Price: List price or the tag price is the price that is printed on the tag of the article. For all practical purposes, we assume it to be same as the marked-price.
Profit: If a man sold an article at a price which is more than the actual price is called Profit or gain.
Loss: If a man sold an article which is less than actual price is called Loss
Margin: The profit percentage on selling price is known as MARGIN. Practice Examples:
Example : Pranav sell an article at a discount of 80% and get a profit of 60% on that article calculate the mark up over the cost price?
Solution:
Let us put the above definitions in use.
Let us assume that Cost Price = Rs 100.
So, Selling Price = Rs 160.
Now, after giving a discount of 80% over MP, Rs 160 is the SP.
Let the Marked Price: MP
SP = 20% of MP
160 = 20% of MP
MP = 800
% Mark Up = (700/100) × 100 = 700 %.
Formulae
a. Profit = Selling Price – Cost Price
b. Profit % = [{(Selling Price -Cost Price)/Cost Price} x 100]
c. Loss = Cost Price – Selling Price
d. Loss % = [{(Cost Price-Selling Price)/Cost Price} x 100]
e. If there is a profit of P %, Cost Price = CP
Then, SP = {(100+P)/100} x CP
If there is a loss of L %, Cost Price = CP
Then SP = {(100-L)/100} x CP
f. If there is a profit of P %, Cost Price = CP Sale price= SP
Then,
CP = [{100/(100+P)} x SP]
If there is a loss of L %,
Then
CP =[{100/(100+L)} x SP]
Example: Mehak and Pranav sells some article for Rs 8000 each. Mehak calculates her profit per cent on his CP and Pranav calculates his profit per cent wrongly on SP. What is the difference in their actual profit if both claim to have a profit of 60%?
Solution:
For Mehak
SP = Rs 8,000
Profit = 60% of CP
CP = Rs 5000
Profit = Rs 3000
For Pranav
SP = Rs 8,000
Profit = 60% of SP = 60% of 8000 = 4800
CP = Rs 3,200
Profit = Rs 4800
So, the difference in profit = Rs 1800
Example: Megha and Richa sold two articles at Rs 12,000 each. One is sold at a profit of 20% and another one at a loss of 20%. What is the net loss?
Solution:
For article-1
SP 1 = Rs 12,000
CP1 = Rs 12,000/1.2 = Rs 10,000
For article-2
SP2 = Rs 12,000
CP2 = Rs 12,000/0.8 = Rs 15,000
So, total CP = Rs 25,000 and total SP = Rs 24,000
So, loss = Rs 1,000.
Examples:
1. An article is sold at a loss of 10%. Had it been sold for Rs. 9 more, there would have been a gain of 25/2% on it. The cost price of the article is :
(a) Rs. 40 (b) Rs. 45 (c) Rs. 50 (d) Rs. 35
2. A man sold an article at a loss of 20%. If he has sold that article for Rs. 12 more he would have gained 10%. Find the cost price of that article :
(a) Rs. 60 (b) Rs. 40 (c) Rs. 30 (d) Rs. 22
3. If an article is sold for Rs. 178 at a loss of 11%, what should be its selling price in order to earn a profit of 11%?
(a) Rs. 222.50 (b) Rs. 267 (c) Rs. 222 (d) Rs. 220
4. A businessman sells a commodity at 10% profit. If he had bought it at 10% less and sold it for Rs. 2 less, then he would have gained 50/3%. The cost price of the commodity is
(a) Rs. 32 (b) Rs. 36 (c) Rs. 40 (d) Rs. 48
5. A car worth Rs. 1,50,000 was sold by X to Y at 5% profit. Y sold the car back to X at 2% loss. In the entire transaction.
(a) X gained Rs. 4,350 (b) Y lost Rs. 4,350 (c) X gained Rs. 3,150 (d) X lost Rs. 3,150
6. If an article is sold at 5% gain instead of 5% loss, the man gains Rs. 5 more. Find the cost price of that article
(1) Rs. 100 (2) Rs. 105 (3) Rs. 50 (4) Rs. 110
7. In terms of percentage profit, which is the best transaction?
C.P. (in Rs.) Profit (In Rs.)
Transaction (I) 36 17
Transaction(II) 50 24
Transaction(III) 40 19
Transaction(IV) 60 29
(1) I (2) II (3) III (4) IV
8. A merchant buys an article for Rs. 27 and sells it at a profit of 10% of the selling price. The sell¬ing price of the article is:
(1) Rs. 29.70 (2) Rs. 30 (3) Rs. 37 (4) Rs. 32
9. Krishnan bought a camera and paid 20% less than its original price. He sold it at 40% profit on the price he had paid. The per¬centage of profit earned by Krish¬nan on the original price was
(1) 22% (2) 32% (3) 12% (4) 15%
10. By selling a basket for Rs. 19.50, a shopkeeper gains 30%. For how much should he sell it to gain 40%?
(1) Rs. 21 (2) Rs. 21.50 (3) Rs. 24 (4) Rs. 23

 

Simple and Compound Interest:

A Interest is the amount paid by someone who borrows a certain amount of money. This term is commonly used in banks, loans, installments and investments. It is associated with percent, rate and the length of time, for which the amount of money is borrowed.
There are many types of interest that can be applied. Simple interest is the simplest and most common type of interest. This type of interest is applicable for a short-term duration, usually in days, weeks, months or even a few years with not so large amounts of money.
Factors of Simple Interest
There are only 3 common factors to be considered with regards to simple interest.
1. Principal
This is the amount of money being borrowed.This could be loaned from a bank or any loaning establishment or borrowed from a person. This will be the basis of how much will be paid with the additional compensation for borrowing.
2. Rate of Interest
This is the percent to be used to calculate the additional amount to be paid along with the principal. Common rates of interest ranges from 1 to 10% but it can also be higher depending on the agreement between the parties.
3. Time
This is the period from the beginning when the money was borrowed to the period that when the money should be returned with the additional amount (interest). This can also be called a term or deadline. This should properly and strictly be observed especially in huge amount of loans.
Compound Interest
Compound interest is a type of interest where in the interest added to the principal amount also earns interest starting from the time when it is added to the principal amount.Payments or investments using compound interest are done in a periodical manner. The interest earned in the period will be added to the principal amount and both will earn interest in the next period. This manner is called compounding wherein accumulated interest from previous periods will also earn interest as it goes along the succeeding payments.
Terms related with Compound Interest
Similar to simple interest, the same common factors affect compound interest but there are some additional things that should be taken into account.
Principal : This is the amount which is being invested or borrowed. This amount increases periodically as interest accumulated in each period is added to the principal as the new principal amount.
Compounding : Unlike simple interest, the rate of interest (nominal) should be divided by the number of periods that the interest would be compounded in a year.
For example:
Rate of Interest : This is the percent rate to be used to calculate the additional amount to be earned or paid for using the money. The rate of interest to be used for each period depends on the manner that the interest is being compounded. The formula for calculating the rate of compound interest is:
Rate per Period =Nominal Rate/Periods per Year
Note: The nominal rate of interestis the annual interest rate.
Number of Periods The period in compound interest is not the entire duration of the transaction but it is the time interval in which the compound interest will be applied. The formula for getting the number of periods is:
Number of Periods = Periods per Year x Number of Years
For example:
10% compounded monthly for 2 years Compound interest is applied every month. Therefore, the number of periods would be 12 months/year x 2 years = 24.
Formulae


Problem 1. A sum of Rs. 25000 becomes Rs. 27250 at the end of 3 years when calculated at simple interest. Find the rate of interest.
Solution:
Simple interest = 27250 – 25000 = 2250
Time = 3 years.
SI = PTR / 100 ? R = SI * 100 / PT
R = 2250 * 100 / 25000 * 3 ? R = 3%.
Problem 2. Find the present worth of Rs. 78000 due in 4 years at 5% interest per year.
Solution:
Amount with interest after 4 years = Rs. 78000
Therefore, simple interest = 78000 – Principal.
Let the principal amount be p.
78000 – p = p*4*5/100 ? p=13000
Principal = 78000 – 13000 = Rs. 65000
Problem 3. A certain principal amounts to Rs. 15000 in 2.5 years and to Rs. 16500 in 4 years at the same rate of interest. Find the rate of interest.
Solution:
Amount becomes 15000 in 2.5 years and 16500 in 4 years.
Simple interest for (4-2.5) years = 16500 – 15000
Therefore, SI for 1.5 years = Rs. 1500.
SI for 2.5 years = 1500/1.5 * 2.5 = 2500
Principal amount = 15000 – 2500 = Rs. 12500.
Rate of Interest = 2500 * 100 / 12500 * 2.5 ? R = 8%.
Problem 4. Find the compound interest on Rs. 3000 at 5% for 2 years, compounded annually.
Solution:
Amount with CI = 3000 (1+ 5/100)2 = Rs. 3307.5
Therefore, CI = 3307.5 – 3000 = Rs. 307.5
Problem 5. Find the compound interest on Rs. 10000 at 12% rate of interest for 1 year, compounded half-yearly.
Solution:
Amount with CI = 10000 [1+ (12/2 * 100)]2 = Rs. 11236
Therefore, CI = 11236 – 10000 = Rs. 1236
Problem 6. The difference between SI and CI compounded annually on a certain sum of money for 2 years at 8% per annum is Rs. 12.80. Find the principal.
Solution:
Let the principal amount be x.
SI = x * 2 * 8 / 100 = 4x/25
CI = x[1+ 8/100]2 – x ? 104x/625
Therefore, 104x/625 – 4x/25 = 12.80
Solving which gives x, Principal = Rs. 2000.
Problem 7. Find the simple interest on Rs. 5000 at a certain rate if the compound interest on the same amount for 2 years is Rs. 253.125.
Solution:
Let the rate of interest be r.
5000[1+ r/100]2 = 5000+253.125
? [1+r/100]2 = 5253.125/5000
Solving which gives
[1+ r/100]2 = 1681/1600
? 1+r/100 = 41/40
? r = 2.5
Therefore, SI = 5000* 2 * 2.5/ 100 = Rs. 250.
Problem 8. A certain amount becomes Rs. 5760 in 2 years and Rs. 6912 in 3 years. What is the principal amount and the rate of interest?
Solution:
SI on Rs. 5760 for 1 year = 6912 – 5760 = Rs. 1152
Therefore, Rate of interest for 1 year = 100*1152/5760*1 = 20%
Let the principal be p.
Then, Principal = p[1+ 20/100]2 = 5760
Solving which gives Principal = Rs. 4000
Problem 9. How long will it take a certain amount to increase by 30% at the rate of 15% simple interest?
Solution:
Let the principal be Rs. x
Simple interest = x*30/100 = 3x/10
T = 100*SI/PR = 100*3x/10 / x*15 = 2%
Alternatively, this can be solved by considering principal amount to be Rs. 100. Then simple interest becomes Rs. 30.
Then, T = 100*30/100*15 = 2%

 

Time and Work:

Introduction: Time And Work
Work is defined as something which has an effect or outcome; often the one desired or expected. The basic concept of Time and Work is similar to that across all Arithmetic topics, i.e. the concept of Proportionality.
Efficiency is inversely proportional to the Time taken when the amount of work done is constant.
Illustrations:
Here are some basic questions to illustrate time and work shortcut tricks.
Example 1)
Rahim can finish a work in 10 days and Ram can finish the same work in 40 days. If Ram and Rahim both work together then what is the total number of days taken?
Solution:
The problem can be solved in two different approaches.
Approach 1: Using Fractions Ram can finish the work in 10 days i.e. in one day he will do 1/10th of the work.
Rahim can finish the work in 40 days i.e. in one day he will do 1/40th of the work.
So, in one day, both working together can finish= (1/10) + (1/40) = 5/40 = 1/8th of the work. So, to complete the work they will take 8 days.
Approach 2: Using Percentage (Shortcut- Recommended) Rahim can finish 100 % of work in 10 days i.e. in one day he finishes 10% of the work.
Ram can finish 100% of the work in 40 days i.e. in one day he finishes 2.5 % of the work.
So, working together, in a single day they can finish 12.5% of the work. So, to complete 100% of the work, they will take 100/12.5 = 8 days.
Example 2)
Ravi can do a job in 10 days. Raman can do the same job in 20 days. They together start doing the job but after 4 days Raman leaves. How many more days will be required by Ravi to complete this job alone?
Solution:
Ravi can finish a job in 10 days i.e in one day he can finish 10 % of the job.
Raman can finish the same job in 20 days i.e. in one day he can finish 5 % of the job.
So, working together, in a day they can do 10 + 5 = 15 % of the job.
In the 4 days, if they worked together, they would have finished 4 × 15% = 60% of the job.
So, job left = 100 – 60 = 40%. This work has to be done by Ravi who does 10 % of the job in a day. So, to finish the remaining 40%, he will take 40/ 10 = 4 more days.
Example: 1) Pipe A can fill a tank in ‘an’ hours. On account of a leak at the bottom of the tank, it takes thrice as long to fill the tank. How long will the leak at the bottom of the tank take to empty a full tank, when pipe A is kept closed? a) (3/2)hours b) (2/3)hours c) (4/3)hours d) (3/4)hours Solution: Method 1: Using variables The pipe can fill (1/a)^th of the tank in an hour. Because of the leak, it can only fill 1/3a of the tank per hour. Let X be the Time in which the leak can completely empty the tank, hence 1/x=1/a-1/3a = x=3a/2hrs. Option (a) Answer: Option (a) Method 2: Using Numbers (Shortcut) Assume a value for “a”- say 10 hours. Because of the leak, it will take 30 hours. Now, this means that in these 30 hours, the filling will occur at a rate of 3.33% and the leaking will slow down the process by 6.66% every hour Thus, the Time taken to empty a full tank= 100/6.66= 15 hours. Answer: Only option (a) satisfies this value. Example:2) A Cistern has three pipes A, B, and C. Pipe A can fill a Cistern in 10 hrs, Pipe B can fill a Cistern in 5 hrs while Pipe C can empty the Cistern in 20 hrs. If they are switched on at the same Time; in how many hours will the Cistern be filled? Solution: In one hour Pipe A can fill 100/10= 10% of the Cistern. In one hour Pipe B can fill 100/5= 20 % of the Cistern. In one hour Pipe C can empty 100/20= 5 % of the Cistern. If all three are working together, (10 + 20 – 5) =25% of the Cistern will get filled in one hour, so it will take 4 hrs for the Cistern to fill. Example:3) A tank can be filled by tap A in 6 hrs and by tap B in 3 hrs. But when they are open simultaneously to fill an empty tank they take 3 hrs more than their normal Time. A hole is later discovered as the reason for the delay. Find the Time taken by the hole to empty the tank if it is completely filled. Solutions: Tap A takes 6 hrs to fill, so in one hour it will fill 16.67 % of the tank. Tap B takes 3 hrs to fill, so in one hour it will fill 33.33% of the tank. Together they will fill 16.67% + 33.33% = 50 % of the tank. They should take 100/50= 2 hrs to fill the tank. But they take 3 hrs more, because of the hole; they totally take 5 hrs to fill i.e. they fill 20% in an hour. This is possible if the hole empties 50-20 = 30 % of the tank in an hour. So to completely empty the full tank; the hole will take 100/30 hrs = 3.33 hrs = 3 hrs and 20 mins. Inverse Proportionality of Efficiency and Time Taken If the product of the two variables is always constant, the two are said to be inversely proportional. Efficiency and Time taken are inversely proportional implies that If A is twice as good as B then A will take half the Time that B will take. If the efficiencies are in the ratio m: n then Time taken will be in the ratio, n: m. i.e. If A is thrice is as good as B then A will take (1/3)^rd of the Time.
Illustration:
Example 1:
C takes 30 days to finish a work and A is thrice as good as C and twice as good as B. If A, B, and C work together, then how many days they will take to finish the work?
Solution:
Approach 1:
C takes 30 days to finish the work. Since A is thrice as good as C, A will take 10 days to finish the work Since A is twice as good as B, B will take 20 days to finish the work. Together A, B, and C will take 1/(1/10+1/20+1/30) = 5.4545 days.
Approach 2: Using Percentages (Shortcut)
1 day = 3.33% work by C
1 day= 10% work by A
1 day= 5% work by B
Together, in one day = 18.33% of work. Thus they will take 100/18.33 days= 5.4545
Example 2:
A is twice as efficient as B. if they take 6 days to finish a certain job together, how much Time will they individually take to complete the work?
Solution:
Using the percentage approach It is given that A is twice as efficient as B. If they are finishing the work in 6 days,
then in one day they will do 100/6 % of work= 16.66% as A is twice as efficient, work done by A in one day will be twice the work done by B. So work done by A in one day = 2/3 * 100/6 % = 100/9 %= 11.11% Work done by B in one day = 100/18 %= 5.55%. So, A will take 9 days to finish the job individually while B will take 18 days.
Practice Examples:
1) Ram starts working on a job and works on it for 12 days and completes 40% of the work. To help him complete the work, he employs Ravi and together they work for another 12 days and the work gets completed. How much more efficient is Ram than Ravi?
a) 50%
b) 200%
c) 60%
d) 100%
2) If A and B work together, they will complete a job in 7.5 days. However, if A works alone and completes half the job and then B takes over and completes the remaining half alone, they will be able to complete the job in 20 days. How long will B alone take to do the job if A is more efficient than B?
a) 20 days
b) 40 days
c) 30 days
d) 24 days
3) A tank is fitted with 8 pipes, some of them that fill the tank and others that are waste pipe meant to empty the tank. Each of the pipes that fill the tank can fill it in 8 hours, while each of those that empty the tank can empty it in 6 hours. If all the pipes are kept open when the tank is full, it will take exactly 6 hours for the tank to empty. How many of these are filling pipes?
a) 2
b) 4
c) 6
d) 4. 5
4) A can finish a work in 12 days while B can finish the same work in 15 days. If both work together, then calculate the Time taken to complete the work.
a) 6 2/3 days
b) 6 days
c) 8 days
d) 4 days
5) Arun can finish a work in 12 days; Ajit can finish the same work in 15 days while Amit can finish the same work in 20 days. Find the total Time taken when all three work together to complete the work.
a) 6 days
b) 5 days
c) 4 days
d) 3 days
6) As a worker, Ajit is thrice as good as Dev. If both working together can finish the work in 5 days, then determine the Time taken by Ajit to complete the work alone.
a) 10/3 days
b) 5 days
c) 4 days
d) 20/3 days
7) A and B can do a work in 12 days, B and C can do the same work in 15 days and A and C can complete the work in 20 days. In how many days A, B and C working together can complete the whole work?
a) 10 days
b) 12 days
c) 8 days
d) 14 days
8) A can finish the work in 16 days while B can finish the work in 8 days. After A started the work alone, B joined him after 4 days. Find out the total Time taken to finish the work.
a) 4 days
b) 8 days
c) 6 days
d) 14 days
9) Amit takes 20 days to complete a certain work. Amit started the work and Suraj joined him 4 days before the work was completed. Find out the number of days for which Amit worked alone if Suraj’s efficiency is 25% more than that of Amit’s.
a) 10 days
b) 8 days
c) 12 days
d) 11 days
10) The ratio of number of days taken by B is to C is 2: 3. The ratio of efficiency of A is to C is 5: 3. A takes 4 days less than C, when A and C complete the work individually. A, B and C started the work and B & C left after 2 days. The number of days taken by A to complete the remaining work is:
a) 1 day
b) 2 days
c) 3 days
d) 5 days
11) A takes 4 days less than B and 2 days more than C to do a job. A and B together can do the job in the same Time as C. Determine the ratio of number of days taken by A and B to complete the job individually.
a) 2: 3
b) (1 + v3): (3 + v3)
c) 1: v2
d) (1 + v3): (2 + v6)
12) Two companies GIL and NCC are working together to build a flyover in Delhi. GIL working on its own would have finished the project in 5 months but working with NCC it is able to finish the project in 4 months. NCC has two teams, one headed by Ramesh and other headed by Sanjeev working on this project. They are doing the same kind of work but efficiency of Sanjeev’s team is 75% of the efficiency of Ramesh’s team. If the total value of the contract for building this flyover is Rs. 35, 00, 000 then determine the money obtained by Sanjeev’s team.
a) Rs. 4, 00, 000
b) Rs. 3, 50, 000
c) Rs 2,50, 000
d) Rs. 3, 00, 000
13) 30 people can do a piece of work in 24 days. Find out the number of days taken by 40 people to complete the same amount of work.
a) 20 days
b) 25 days
c) 18 days
d) 22 days
14)24 men can complete a work in 20 days. How many more men are required to complete the work in 12 days?
a) 40
b) 12
c) 16
d) 14
15)20 men can complete a work in 20 days. After how many days should the number of men be increased by 50% so that the work is completed in 75% of the actual Time?
a) 6
b) 4
c) 3
d) 5
16) 15 workers working 4 hours a day for 25 days can build a platform of width 120 meters, length 10 meter and height 14 meters. How many days will 12 workers working 5 hours a day will take if they have to build a platform of width 600 meters, length 14 metres and height 12 metres?
a) 150
b) 130
c) 125
d) 120
17) 5 men or 8 women can reap a field in 12 days. Find the number of days (approximately) taken by 3 men and 4 women to reap the same field.
a) 10
b) 11
c)12
d) 13
18) 6 children and 2 men complete a job in 6 days. Each child takes twice as much Time a man takes to complete the same amount of job. In how many days will 4 men complete the same job.
a) 7
b) 8
c) 6
d) 7.5
19) Three women and four men can complete a work in 4 days. Two men and five women can complete the same work in 5 days. Find out the money received by a man for his work if a woman is paid Rs. 60 for her work.
a) Rs. 120
b) Rs. 150
c) Rs. 130
d) Rs. 75
20) (x – 2) men can do a job in x days and (x+ 7) men can do a job in ¾ of the same work in (x – 10) days. Calculate the number of days in which (x + 4) men can finish the job.
a) 12
b) 14
c) 8
d) 15
21) One Indian, one Chinese and one Japanese worked for a company for the same period. The Indian is twice as efficient as Chinese and the Chinese is thrice as efficient as a Japanese. If Rs. 10,00,000 were given to all the three together, then calculate the amount received by the Chinese and the Japanese together.
a) Rs. 8, 00, 000
b) Rs. 4, 00, 000
c) Rs. 9, 00, 000
d) Rs. 6, 50, 000
22) 4 men and 2 boys can do a job in 6 2/3 days. 3 women and 4 boys can finish the same job in 5 days. Also 2 men and 3 women can finish the same job in 4 days. In how many days can 1 man, 1 woman and 1 boy finish the work at their double efficiency?
a) 5
b) 6
c) 4
d) 7
23) Pipe A can fill a tank in 4 hours while Pipe B can empty the same tank completely in 5 hours. If both the pipes are working simultaneously, then determine the Time in which an empty tank will be completely filled.
a) 6
b) 9
c) 20
d) 13
24) A tap can fill a tank in 12 hours, but because of a hole in the bottom of the tank, it fills the tank in 15 hours. Determine the Time it will take to empty the tank if it is completely filled once and tap is closed.
a) 30 hrs
b) 60 hrs
c) 20 hrs
d) 25 hrs
25) If one pipe A can fill the tank in 10 hours then 4 pipes each of efficiency 20% as that of Pipe A can fill the same tank in how many hours?
a) 15 hrs
b) 12 hrs
c) 10 hrs
d) 12.5 hrs

 

Pipes and Cisterns:

Introduction to Pipes and Cisterns:
Problems on pipes and cisterns are completely based on the concept of time and work. In pipes and cisterns, we mainly come across problems that are concerned with filling or emptying a tank.
Points to Remember:
• Inlet pipe : A pipe that fills a tank or cistern is the inlet pipe.
• Outlet pipe : A pipe that empties a tank or cistern is the outlet pipe.
• If a pipe takes x hours to fill a tank, then the portion of the tank filled in 1 hour is 1/x
• If a pipe takes x hours to empty a tank, then the portion of the tank emptied in 1 hour is 1/x
• If a pipe takes x hours to fill a tank and another pipe takes y hours (y>x) to empty a full tank, then the portion of the tank filled in 1 hour when both the pipes are open is 1/x –1/y
• If a pipe takes x hours to fill a tank and another pipe takes y (x>y) hours to empty a full tank, then the portion of the tank emptied in 1 hour when both the pipes are open is 1/y–1/x
Practice Examples:
1. Two pipes can fill a tank in 4 hours and 6 hours respectively. Find the time taken by both the pipes to fill the tank.
a. 2.4 hrs b. 1.2 hrs c. 1.8 hrs d. 3 hrs
2. Pipe A can fill a cistern in 3 hours while pipe B can empty the cistern in 5 hours. If both the pipes are open together find the total time taken to fill the tank.
a. 5 hours 20 minutes b. 6 hours 10 minutes c. 5 hours 20 minutes d. 5 hours 20 minutes
3. A pipe normally fills a tank in 30 minutes but due to a leak in the tank it takes 40 minutes to fill the tank. In how much time does the full tank empty due to the leak.
a. 1 hr b. 1.5 hrs c. 2 hrs d. 2.5 hrs
4. Pipes A and B can fill a tank in 9 hours and 6 hours respectively. If both the pipes are opened together and after 2 hours pipe B is turned off. Find the total time in which the tank will be full.
a. 5 hrs b. 6 hrs c. 3 hrs d. 4 hrs
5. Two pipes can fill a tank in 6 hours and 8 hours respectively. If they are turned on alternatively for 1 hour each, find the time in which the tank is full. (Assume pipe A is opened first)
a. 3 hours 30 minutes b. 4 hours 45 minutes c. 5 hours 30 minutes
d. 6 hours 45 minutes
6. Two pipes together can fill a tank in 4 hours. If one pipe takes double the time of the other pipe to fill the tank separately, find the time in which the faster pipe fills the tank.
a. 2 hrs b. 4 hrs c. 6 hrs d. 8 hrs
7. Two pipes P and Q can fill a tank in 30 and 45 minutes respectively. If both the pipes were open for few minutes after which Q was closed and the tank was full in 25 minutes, find the time for which pipe Q was open.
a. 5 minutes b. 7.5 minutes c. 9 minutes d. 11 minutes
8. Three pipes open 16 hours a day can empty a tank in 4 days. How many hours a day must 4 pipes be open to empty the tank in 2 days.
a. 8 hrs b. 6 hrs c. 10 hrs d. 12 hrs
9. Three pipes A, B and C attached to a tank can fill it completely in 6 hours. After opening together for 2 hours, C is closed. A and B fill the remaining part in 7 hours. The number of hours required by C to fill the tank is?
a. 14 hrs b. 12 hrs c. 10 hrs d. 16 hrs
10. A cistern is filled by 3 pipes P, Q and R with uniform flow. The second pipe Q takes 3232 times the time taken by P to fill the tank, while R takes twice the time taken by Q to fill the tank. If all the three pipes can fill the tank in 5 hours, find the time required by pipe A alone to fill the tank.
a. 6 hrs b. 8 hrs c. 10 hrs d. 12 hrs
11. Three pipes can fill a tank in 30, 40 and 50 minutes respectively. Find the time taken by all the three pipes to fill the tank together.
a. 11.5hrs b. 12.7 hrs c. 13.3 hrs d. 10.2 hrs
12. Two pipes P and Q can fill a tank in 3 and 4 hours respectively while a pipe R can empty the tank in 6 hours. If all the 3 pipes are opened together, find the time in which the tank is full.
a. 2 hours 24 minutes b. 1 hour 30 minutes c. 4 hours 24 minutes
d. 3 hours 30 minutes
13. Two pipes take 3 hours and 4 hours respectively to fill a tank. But due to a leak they take 5 hours to fill the tank. Find the time taken by the leak to empty a full tank.
a. 1 11/20 hrs b. 2 14/23 hrs c. 2 7/9 hrs d. 4 1/2 hrs
14. Pipes P, Q and R can fill a tank in 3, 4 and 5 hours respectively. If all the pipes are opened together and after 30 minutes pipes Q and R are turned off, find the total time in which the tank is full.
a. 2 13/40 hrs b. 1 2/3 hrs c. 2 10/11 hrs d. 1 13/40 hrs
15. Pipes A, B and C can fill a cistern in 20, 30 and 40 hours respectively. If pipes B and C are turned on alternatively for 1 hour (assume B is opened first) and A is turned on the whole time. Find the total time in which the tank is full.
a. 10 hours 10 minutes b. 12 hours 12 minutes c. 10 hours 36 minutes d. 12 hours 36 minutes

 

Time and Distance

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