Showing posts with label Aptitude. Show all posts
Showing posts with label Aptitude. Show all posts
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Quantitative Aptitude: Variation-Chain rule concept

Variation-Chain rule
If 10 hens lay 10 eggs in 10 days, how many eggs will 1 hen lay in 1 day?

When one instance of a problem is given & a question is asked on another instance, it involves a rule called chain rule.

A very simple example 

4 carpenters make 20 chairs in 5 days. How many chairs will 8 carpenters make in 10 days?

4 carpenters make 20 chairs in 5 days. So 8 carpenters will make double the number of chairs in 5 days (same time as earlier) i.e. 20 × (8/4) = 40 chairs

So, 8 carpenters will make in 10 days= 40 × (10/5) = 80 chairs ==> Final Answer

Easy, isn't it. But don't underestimate these type of questions. When a long chain rule is applied, it becomes complicated to solve the question as it will involve many steps by this approach.

A General Approach to be followed 

To ease out the step by step process, there is a general approach. It is easier than above method & applicable for any number of steps involved, called Direct Proportionality rule.

Direct proportionality rule 

E.g: Eight men working 10 hrs everyday can completely build a wall of length 180 m, breadth 4 m and height 20 m in 20 days. In how many days can 12 men working 8 hrs a day build a wall of length 300 m, breadth 6 m and height 12 m?

Since we have to find the number of days, so we will use the proportion relative to the number of days.

In General, 
==> for less days to be involved, we use proportion making less than 1.
==> for more days to be involved, use proportion making greater than 1.

  • 8-->12 men means less days will be taken to do the work. So we use 8/12.
  • 10-->8 hrs per day means more days will be taken. So we use 10/8 this time.
  • 180-->300 m means more length-->more days. Thus we use 300/180
  • 4-->6 m means more breadth-->more days. Thus we use 6/4.
  • 20-->12 means less height-->less days. Thus we use 12/20.
In case of other problems, we will define proportions in terms of the quantity asked in the problem.

I hope this concept is clear to everyone & in general could be used to any problem involving chain variation.

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Quantitative Aptitude: Proportions- Concepts

This article explains the concept of proportions asked in various bank exams & other competitions.
Proportions

When two ratio are equal,  ac  , then a, b, c, d are said to be in proportion
                                             b   d 

If a, b, c and d are in proportion, then a × d = b × c

Operations on Proportions

If  a = c  , then following is also true:
     b   d 

 b = d    (Invertendo)
  a   c

 a = b    (Alterendo)
 c   d  

 a+b = c+d  (Componendo, by adding 1 to both sides of original equality)
   b        d 

 a-b = c-d   (Dividendo, by subtracting 1 to both sides of original equality)
   b       d 

 a+b = c+d   (Componendo and Dividendo, dividing above two equalities)
 a-b     c-d 


Law of Equal Ratios

If a/b= c/d= e/f = k then, by the law of equal ratios, a+c+e = k
                                                                                     b+d+e


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Quantitative Aptitude: Ratios- Concepts & Problems

This article explains all concepts related to Ratios & questions asked in Bank & other exams.

Ratios

When we talk of ratios, we are essentially commenting on the ‘relative’ amounts. 

E.g. If the ratio of boys and girls in a class is 4 : 3 and the total number of students in the class is 84, find the number of boys and girls.
Assuming the number of boys and girls to be 4k and 3k, we have 4k + 3k = 84 i.e. k = 12
Thus the number of boys and girls is 48 and 36 respectively.

Simple haan!

Dividing a given sum in a ratio

E.g. Divide Rs. 72 in the ratio 2 : 3 : 4.

Let the three parts be 2k, 3k and 4k respectively. The total of the three parts will be the total amount i.e. Rs. 72
Thus, 2k + 3k + 4k = 72 i.e. 9k = 72 i.e. k = 8
Thus the three parts will be 2 × 8 = 16; 3 × 8 = 24; and 4 × 8 = 32

When Ratios are in Fractions

A, B and C’s shares are in the ratio 1:  1:  1
                                                          2   3   4
We multiply with the LCM of denominators to get rid of the fractions. Thus, A, B and C’s share are in ratio 6 : 4 : 3

Given a : b and b : c, finding a : c

E.g. If a : b is 3 : 4 and b : c is 5 : 6, find the ratio a : b : c.

Since b is common to the two ratios, we should make the numeric value of b the same in both the ratios.
REMEMBER: When all terms of a ratio are multiplied with the same constant, 
the ratio does not change.

Thus in the first ratio b could be changed to any multiple of 4 and in the second ratio b could be 
changed to any multiple of 5. So, we should make b a multiple of 4 and 5 i.e. 20
  • a : b is 3 × 5 : 4 × 5 i.e. 15 : 20
  • b : c is 5 × 4 : 6 × 4 i.e. 20 : 24                
Thus, when b is 20, a is 15 and c is 24 and required ratio of a : b : c is 15 : 20 : 24.

A typical example

E.g. A bag has Re. 1, 50 paisa and 25 paisa coins. If the ratio of the number of coins of the respective denominations is 3 : 2 : 8 and the bag has a total amount of Rs. 156, find the number of 50 paise coins. 

Since the ratio is given of the number of coins, let the number of coins be 3k, 2k and 8k.
Remember these are the number of coins and equating 3k + 2k + 8k = 156 WILL BE WRONG since it is the total amount, not the number of coins.

The left hand side of this equation is the total number of coins and the right hand side, 156, is 
NOT the total number of coins but is the value of the number of coins. 

Thus, the number of coins will have to be transferred to value of coins, in Rs.
3k Re.1 coins will amount to 3k × 1 = Rs. 3k
2k 50-paise coins will amount to 2k × 0.5 = Rs. k
8k 25-paise  coins will amount to 8k × 0.25 = Rs. 2k.

Thus, total amount in bag will be 3k + k + 2k i.e. 6k and this will be equal to Rs. 156. 
Thus, 6k = 156 i.e. k = 26
We want to find the number of 50-paise coins and the required answer will be 2k i.e. 2 × 26 = 52 coins.


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Quantitative Aptitude: Partnership concepts & problems

Today, I am going to explain the concept of Partnership in Quantitative aptitude.
Partnership
Partnership is when two or more people pool in money as capital for a common venture. The profit of the venture is then divided among the people depending on the amount of money that each has invested. 

Different Investments, Same Time period of Investing
If the amount invested by the partners are C1 , C2 , C3  then the profit is distributed in the ratio C1  : C2  : C3

E.g. Rahul and Rohit get in Rs. 4000 and Rs. 5000 to fund a new venture. In what ratio should they divide the profit of Rs. 1,80,000 earned at the end of the year?

Profit has to be divided among Rahul and Rohit in the ratio of their investments i.e. 4 : 5
Let Rahul’s and Rohit’s share of profit be 4k and 5k respectively. The two share together would be the entire profit i.e. 4k + 5k = 1,80,000
i.e. 9k = 1,80,000 i.e. k = 20,000
Rahul’s share = 4 × 20,000 = 80,000
Rohit’s share = 5 × 20,000 = 1,00,000.

Same Investments, different Time periods

If the investments made by the partners are same, but the time period is different, the profit is divided in the ratio of the time periods.

When the investment and also the time period is different

Let there be three partners, one invests C1 for t1 time, second invests C2 for t2 time and third invests C3 for t3 time. The profit is shared in the ratio C1 × t1 : C2 × t2 : C3 × t3

E.g. A and B enter into a partnership with Rs. 8,000 and 15,000 respectively. After 3 months C joins them by investing Rs. 10,000. 4 months before the first year is completed, B quits, taking his invested amount back with him. In what ratio should the profit of Rs. 2,55,000 earned in the first year be distributed among the three?

A has invested 8,000 for 12 months.
B has invested 15,000 for 8 months.
C has invested 10,000 for 9 months. 
Thus the profit has to be distributed in the ratio of 8 × 12 : 15 × 8 : 10 × 9
i.e. 16 : 20 : 15  

A’s share = 16 × 2,55,000= 80,000
                     51

B’s share = 20 × 2,55,000= 1,00,000
                     51 

C’s share =  15 × 2,55,000= 75,000
                      51

When one partner has different amounts invested in different time periods

Let’s say in a partnership between A and B, A invests Rs. Ia for a time period of ta. But B invests Rs. Ib1 for a period of tb1 time and Rs. Ib2 for a period of tb2 time. In this case the profit will be divided between A and B in the ratio Ia × ta : (Ib1 × tb1 + Ib2 × tb2)

E.g. A, B and C enter into a partnership. They invest Rs. 40,000, Rs. 80,000 and Rs. 1,20,000 respectively. At the end of the first year, B withdraws Rs. 40,000 while at the end of second year, C withdraws Rs. 80,000. In what ratio will the profit be shared at the end of three years?

The profit has to be shared in the ratio of (40 × 3) : (80 × 1) + (40 × 2) : (120 × 2) + (40 × 1) i.e. 120 : 160 : 280 i.e. 3 : 4 : 7

Working Partner drawing a salary

In this case, first the salary of the partner is deducted from the profit & after that, whatever profit is left, is distributed among partners as we did in previous methods.

E.g. Two partners, A and B invest Rs. 10,000 and Rs. 15,000 in  a partnership firm which makes a profit of Rs. 50,000 at the end of the year. But since A is a working partner, he is entitled to a salary of Rs. 20,000 for the year. What is the amount that B receives?

The profit left after deducting A's salary = Rs. 50,000 – Rs. 20,000 = Rs. 30,000 
This amount will now be divided among A and B in the ratio of their investments i.e. in the ratio 10,000: 15,000 i.e 2: 3 
Thus, amount received by B =3/5th of Rs. 30,000 i.e. Rs. 18,000



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Tips to improve performance in IBPS Online Quantitative Aptitude tests

Friends, there are many of us who are smart enough to solve Quantitative Aptitude Section or say Numerical Ability sections of various papers with 100% accuracy [Given the test is pen-paper based & no time constraints]..BUT the main purpose of the Online Aptitude test is to check your ability to solve those questions within the given time limit with maximum accuracy.

Most of the bank exams mainly IBPS & other Competitive exams are conduct online these days. You may have the ability to solve the numerical questions but this 'Online' thing may become a hurdle in your success unless you make a strategy to face it.

So here are some tips that will help you improve your performance in IBPS Quantitative Aptitude (or Numerical Ability) tests:

  • Practice: Practice as much online tests as you can. Choose the same format of online test that you may be facing in the real exam. The more you practice, the better your score will be.
  • There are a lot of Numerical ability tests available online. You just have to google about it.
  • A good book: Buy a good book of quantitative Aptitude to improve your basics. Most of the questions asked in these exams are of not more than 10th class level. I suggest R.S. Agarwal Quantitative Aptitude book & M. Tyra Quicker Maths which contains some Vedic Mathematics tricks.
  • Basic Calculations: Practice basic calculations involving addition, subtraction, multiplication, division, percentage increase, fractions etc on daily basis. I know it sounds silly for a pro(in school time) in Mathematics but it is a necessary thing to do to improve your score.
  • Try mental maths. Numerical Ability section in IBPS exams is designed to make you rely on quicker mental calculations otherwise its hard to get a very good score. The less paper you use for calculations, the better you could score.
  • Prepare questions of every type. Example- Just because there was no question on Boat & streams last year, you should NOT leave the topic. 
  • Do not pay attention to anything else while attempting the test. Just Concentrate on the test. [For boys- ladkiyaan paper ke baad bhi dekhi ja sakti hain]
  • Leave the questions for reviewing later which you feel could take more time to solve than estimated but you know how to solve them..
  • Don't panic if you can't solve the question. Just move on & try to solve another one.
  • DO NOT try attempt the questions about which you have no idea how to solve. After completely attempting the paper, if time permits (that too after revision), you could give these questions a try.
  • Allot time to numerical Ability section. IBPS exams are basically of 2 hours. So 40 minutes to solve Numerical Ability section of IBPS Papers should be the maximum limit. 
  • Get comfortable with the negative marking concept. Do not mark the answer unless you are very sure about its correctness. Most of the candidates get over-confident & mark too many answers that reduces their score.
  • You should keep track of your performance in these Online Practice Aptitude tests. Make a diary & write down the marks you get in every test. You will eventually notice improved performance if you have taken the steps mentioned above.
  • Read the instructions carefully before attempting the test. You may have given the test many times but you never know when they are going to change the rules.
  • Get 7 hour sleep daily. Keep a good mood while attempting test.
I hope these tips help you to score more in the Numerical ability section of IBPS & other exams.


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Quantitative Aptitude Tips & Tricks-1

Quantitative Aptitude is one of the things that becomes a head ache for students from non-technical background. However, the aptitude questions asked in any of the exams like CSAT, SSC, IBPS etc are of very moderate level. They require only practice & with a good strategy anyone can master this section.

Here I am posting some tricks. Mix them with your practice & try to acquaint with them to solve the questions in less time with same accuracy

Finding number of Factors
To find the number of factors of a given number, express the number as a product of powers of prime numbers.

In this case, 48 can be written as 16 * 3 = (24 * 3)
Now, increment the power of each of the prime numbers by 1 and multiply the result.
In this case it will be (4 + 1)*(1 + 1) = 5 * 2 = 10 (the power of 2 is 4 and the power of 3 is 1)
Therefore, there will 10 factors including 1 and 48. Excluding, these two numbers, you will have 10 – 2 = 8 factors.

Sum of n natural numbers
-> The sum of first n natural numbers = n (n+1)/2
-> The sum of squares of first n natural numbers is n (n+1)(2n+1)/6
-> The sum of first n even numbers= n (n+1)
-> The sum of first n odd numbers= n^2

Finding Squares of numbers
To find the squares of numbers near numbers of which squares are known
To find 41^2 , Add 40+41 to 1600 =1681
To find 59^2 , Subtract 60^2-(60+59) =3481

Finding number of Positive Roots
If an equation (i:e f(x)=0 ) contains all positive co-efficient of any powers of x , it has no positive roots then.
Eg: x^4+3x^2+2x+6=0 has no positive roots .

Finding number of Imaginary Roots
For an equation f(x)=0 , the maximum number of positive roots it can have is the number of sign changes in f(x) ; and the maximum number of negative roots it can have is the number of sign changes in f(-x) .
Hence the remaining are the minimum number of imaginary roots of the equation(Since we also know that the index of the maximum power of x is the number of roots of an equation.)

Reciprocal Roots
The equation whose roots are the reciprocal of the roots of the equation ax^2+bx+c is cx^2+bx+a

Roots
Roots of x^2+x+1=0 are 1,w,w^2 where 1+w+w^2=0 and w^3=1

Finding Sum of the roots
For a cubic equation ax^3+bx^2+cx+d=o sum of the roots = – b/a sum of the product of the roots taken two at a time = c/a product of the roots = -d/a
For a biquadratic equation ax^4+bx^3+cx^2+dx+e = 0 sum of the roots = – b/a sum of the product of the roots taken three at a time = c/a sum of the product of the roots taken two at a time = -d/a product of the roots = e/a

Maximum/Minimum
-> If for two numbers x+y=k(=constant), then their PRODUCT is MAXIMUM if x=y(=k/2). The maximum product is then (k^2)/4
-> If for two numbers x*y=k(=constant), then their SUM is MINIMUM if x=y(=root(k)). The minimum sum is then 2*root(k) .

Inequalties
-> x + y >= x+y ( stands for absolute value or modulus ) (Useful in solving some inequations)
-> a+b=a+b if a*b>=0 else a+b >= a+b
-> 2<= (1+1/n)^n <=3 -> (1+x)^n ~ (1+nx) if x<< When you multiply each side of the inequality by -1, you have to reverse the direction of the inequality.

Product Vs HCF-LCM 
Product of any two numbers = Product of their HCF and LCM . Hence product of two numbers = LCM of the numbers if they are prime to each other.

AM GM HM
For any 2 numbers a>b a>AM>GM>HM>b (where AM, GM ,HM stand for arithmetic, geometric , harmonic menasa respectively) (GM)^2 = AM * HM

Sum of Exterior Angles
For any regular polygon , the sum of the exterior angles is equal to 360 degrees hence measure of any external angle is equal to 360/n. ( where n is the number of sides)
For any regular polygon , the sum of interior angles =(n-2)180 degrees
So measure of one angle in
Square—–=90
Pentagon–=108
Hexagon—=120
Heptagon–=128.5
Octagon—=135
Nonagon–=140
Decagon–=144

Problems on clocks
Problems on clocks can be tackled as assuming two runners going round a circle , one 12 times as fast as the other . That is , the minute hand describes 6 degrees /minute the hour hand describes 1/2 degrees /minute . Thus the minute hand describes 5(1/2) degrees more than the hour hand per minute .
The hour and the minute hand meet each other after every 65(5/11) minutes after being together at midnight. (This can be derived from the above) .

Co-ordinates
Given the coordinates (a,b) (c,d) (e,f) (g,h) of a parallelogram , the coordinates of the meeting point of the diagonals can be found out by solving for [(a+e)/2,(b+f)/2] =[ (c+g)/2 , (d+h)/2]

Ratio
If a1/b1 = a2/b2 = a3/b3 = ………….. , then each ratio is equal to (k1*a1+ k2*a2+k3*a3+…………..) / (k1*b1+ k2*b2+k3*b3+…………..) , which is also equal to (a1+a2+a3+…………./b1+b2+b3+……….)

Finding multiples
x^n -a^n = (x-a)(x^(n-1) + x^(n-2) + …….+ a^(n-1) ) ……Very useful for finding multiples .For example (17-14=3 will be a multiple of 17^3 – 14^3)

Exponents
e^x = 1 + (x)/1! + (x^2)/2! + (x^3)/3! + ……..to infinity 2 <>GP
-> In a GP the product of any two terms equidistant from a term is always constant .
-> The sum of an infinite GP = a/(1-r) , where a and r are resp. the first term and common ratio of the GP .

Mixtures
If Q be the volume of a vessel q qty of a mixture of water and wine be removed each time from a mixture n be the number of times this operation be done and A be the final qty of wine in the mixture then ,
A/Q = (1-q/Q)^n

Some Pythagorean triplets:
3,4,5———-(3^2=4+5)
5,12,13——–(5^2=12+13)
7,24,25——–(7^2=24+25)
8,15,17——–(8^2 / 2 = 15+17 )
9,40,41——–(9^2=40+41)
11,60,61——-(11^2=60+61)
12,35,37——-(12^2 / 2 = 35+37)
16,63,65——-(16^2 /2 = 63+65)
20,21,29——-(EXCEPTION)

Function
Any function of the type y=f(x)=(ax-b)/(bx-a) is always of the form x=f(y) .

Finding Squares
To find the squares of numbers from 50 to 59
For 5X^2 , use the formulae
(5X)^2 = 5^2 +X / X^2
Eg ; (55^2) = 25+5 /25 =3025
(56)^2 = 25+6/36 =3136
(59)^2 = 25+9/81 =3481

Successive Discounts
Formula for successive discounts
a+b+(ab/100)
This is used for successive discounts types of sums.like 1999 population increses by 10% and then in 2000 by 5% so the population in 2000 now is 10+5+(50/100)=+15.5% more that was in 1999 and if there is a decrease then it will be preceded by a -ve sign and likewise.

Rules of Logarithms:
-> loga(M)=y if and only if M=ay
-> loga(MN)=loga(M)+loga(N)
-> loga(M/N)=loga(M)-loga(N)
-> loga(Mp)=p*loga(M)
-> loga(1)=0-> loga(ap)=p
-> log(1+x) = x – (x^2)/2 + (x^3)/3 – (x^4)/4 ………to infinity [Note the alternating sign ...Also note that the logarithm is with respect to base e]


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