Quantitative Aptitude PDF for Engineering Students

QUANTITATIVE APTITUDE For CIVIL ENGINEERING COMPUTER SCIENCE ENGINEERING ELECTRICAL ENGINEERING INSTRUMENTATION ENGINEERING ELECTRONICS & COMMUNICATION ENGINEERING MECHANICAL ENGINEERING  QUANTITATIVE APTITUDE...

QUANTITATIVE APTITUDE For CIVIL ENGINEERING COMPUTER SCIENCE ENGINEERING ELECTRICAL ENGINEERING INSTRUMENTATION ENGINEERING ELECTRONICS & COMMUNICATION ENGINEERING MECHANICAL ENGINEERING  QUANTITATIVE APTITUDE SYLLABUS Numbers, Linear Equations, Ratio, Proportion & Variation, Percentage, Profit–Loss and Partnership, Simple & Compound Interest, Average-Mixture-Alligation, Time and Work, Pipes & Cisterns, Time, Speed & Distance, Permutation and Combination, Probability, Geometry & Mensuration ANALYSIS OF GATE PAPERS Exam Year 1 Mark Ques. 2 Mark Ques. Total 2010 0 3 6 2011 1 2 5 2012 1 3 7 2013 0 4 8 2014 Set-1 2 2 6 2014 Set-2 2 2 6 2014 Set-3 2 2 6 2015 Set-1 1 2 5 2015 Set-2 2 1 4 2015 Set-3 3 2 7 2016 Set-1 1 2 5 2016 Set-2 1 2 5 2016 Set-3 0 3 6 2017 Set-1 1 2 5 2017 Set-2 1 3 7 2018 Set-1 1 3 7 2018 Set-2 1 3 7 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  CONTENT Topics Page No 1. NUMBERS 01 2. LINEAR EQUATIONS 10 3. RATIO, PROPORTION & VARIATION 12 4. PERCENTAGE 16 5. PROFIT, LOSS AND PARTNERSHIP 19 6. SIMPLE & COMPOUND INTEREST 22 7. AVERAGE-MIXTURE-ALLIGATION 24 8. TIME AND WORK, PIPES & CISTERNS 27 9. TIME, SPEED & DISTANCE 30 10. PERMUTATION AND COMBINATION 39 11. PROBABILITY 44 12. GEOMETRY & MENSURATION 47 13. DATA INTERPRETATION 56 14 GATE QUESTIONS 59 15 Assignment 118 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  1 NUMBERS CLASSIFICATION OF NUMBERS All the counting numbers (positive and negative) including zero are called integers. All the numbers that we see or use on a For example regular basis can be classified as given in -100, -99, -50, -40, 0, 13, 17 are all integers. the chart below. I / Z  0, 1, 2, 3 RATIONAL NUMBERS This is the set of real numbers that can be a written in the form b, where a and b are integers and b is not equal to zero b ≠ 0. All integers and all fractions are rational numbers including the finite decimal numbers (i.e. terminating). The numbers - 2 50 10 1 4, , ,  ,  , 0, 145 3 7 3 4  145  15  also represented as  and are REAL NUMBERS  1  1 examples of rational numbers. Real numbers represent actual physical a  quantities in a meaningful way for e. g. Q   : a, b  I &b  0   b  length, height, density etc. Real numbers can be further divided into subgroups. For IRRATIONAL NUMBERS example, rational/irrational, odd/even, prime/composite etc. The numbers which are not rational are NATURAL OR COUNTING NUMBERS called irrational numbers, such as 2, π. These numbers give an approximate To count objects we use counting numbers answer in terms of decimals. Also the digits like 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The set of after the decimal are non-terminating and positive counting numbers is called natural non-recurring. Thus, numbers. These are also at times called 2  1.4142135, π  3.141592 etc. positive integers. N  1, 2,3, 4.. EVEN NUMBERS WHOLE NUMBERS The set of Even Numbers is the set of integers which are divisible by 2. E.g. 2, 4, 6, The set of natural numbers taken along 8, 10… Even numbers are expressed in the with 0, gives us the set of Whole Numbers. form 2n, where n is an integer. Thus 0, -2, - Thus, the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 ….. 6 etc. are also even numbers. represent the set of whole numbers. W  0,1, 2,3, 4 ODD NUMBERS The set of odd Numbers is the set of INTEGERS integers which are not divisible by 2. E.g. 1, © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  3, 5, 7, 9… Odd numbers are expressed in right of the origin (see the figure), while the the form (2n-1), where n is an integer (not negative integers -1, -2, -3, … are associated necessarily prime). Thus, -1, -3, -9 etc. are with points on the line at distances 1, 2, 3, all odd numbers. … units respectively to the left of the origin. Note: 1) The smallest natural number is 1. 2) The smallest whole number is 0. PRIME NUMBERS The rational number 1/2 is represented on this scale by a point P halfway between 0 A natural number which does not have any 1 and +1. The negative number -3/2 or 1 other factor besides itself and unity is a 2 prime number. For example 2, 3, 5, 7, 11, 1 is represented by a point R, 1 units to the 13 etc. The set of such numbers is the set of 2 prime numbers. left of the origin. There is one and only one point on the number line corresponding to Note: each real number and conversely, to every 1) 1 is neither prime nor composite. point on the line, there corresponds one 2) The only even prime number is 2. and only one real number. 3) Two numbers are said to be The position of real numbers on a line relatively prime to each other or co- establishes an order to the real number prime when their HCF is 1 e.g. (i) 9 system. If a point A lies to the right of and 28, (ii) 3 and 5, (iii) 14 and 29 another point B on the line we say that the etc. numbers corresponding to A is greater than 4) If a number has no prime factor the number corresponding to B or that the equal to or less than its square root, number corresponding to B is less or then the number is prime. This is a greater than the number corresponding to test to judge whether a number is A. The symbols for “greater than” and “less prime or not. than” are > and < respectively. These symbols are called “inequality signs”. COMPOSITE NUMBERS Thus since 5 is to the right of 3, 5 is greater than 3(5 > 3), we may also say 3 is less The set of Composite Numbers is the set than 5(3 < 5). Similarly, since -6 is to the of natural numbers which have other left of -4, -6 is smaller than -4, (−6 < −4), factors also, besides itself and unity. E.g. 8, we may also write (−4 > −6). 72, 39 etc. Alternatively, we might say that a natural OTHER SPECIAL NUMBERS number (except 1) which is not prime is a composite number. PERFECT NUMBERS GRAPHICAL REPRESENTATION OF REAL If the sum of the divisors of N excluding N NUMBERS itself but including unity is equal to N, then N is called a perfect number. e.g. 6, 28 etc. It is often useful to represent real numbers 6=1+2+3, where 1,2 & 3 are divisors of 6 by points on a line. To do this, we choose a 28 = 1+2+4+7+14 point on the line to represent the real number zero and call this point the origin. FIBONACCI NUMBERS The positive integers +1, +2, +3,… are then associated with points on the line at Fibonacci numbers form a sequence in distance 1, 2, 3, … units respectively to the which each term is the sum of the two © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  terms immediately preceding it. It is named 21964 for its discoverer, Leonardo Fibonacci 64  4  16 (Leonardo Pisano). The Fibonacci sequence 4. Divisible by : 5 that has 1 as its first term is 1, 1, 2, 3, 5, 8, Test : The units’ digit should be 0 or 5. 13, 21, 34, 55 … These numbers are Examples : referred to as Fibonacci numbers. The 1835, 15440 defining property can be symbolically Last digits are 5 and 0 respectively represented as t r 2  t r  t r 1. 5. Divisible by : 6 Example: Test : The sum of the digits of the If X1=1 and Xn+1=2X  n+5 number should be divisible by 3 and the Where n=1,2…, then what is the value of number should be even. X100 ? Examples : 1)  5  299  6  1272 2)  5  299  6  1  2  7  2  12 3)  6  299  5 12  3  4 , Number is even 6. Divisible by : 8 4)  6  299  5 Test : The number formed by the last Solution: three digits (units’, tens’ and hundreds’) X11=1, X2=2X1+5=7, X3=2X2+5=19 of the given number should be divisible Sequence is 1,7,19,... It satisfies Option 4. by 8.  X100  6  299  5. Examples : 52672 Hence Ans. (4) 672  8  84 DIVISIBILITY TESTS 7. Divisible by : 9 Following are important divisibility tests. Test : The sum of the digits of the 1. Divisible by : 2 number should be divisible by 9. Test : The unit’s digit should be even or Examples : 0 (i.e. in the given number at the units 127296 place we should have 2, 4, 6, 8, 0) Examples : 1  2  7  2  9  6  27  26, 48 etc. 27  9  3 6  2  3,8  2  4 8. Divisible by : 10 2. Divisible by : 3 Test : The units’ digit should be 0. Test : The sum of the digits of the Examples : number should be divisible by 3. 3220 Examples : Units’ digit is zero. 12729 1  2  7  2  9  21 9. Divisible by : 11 Test : The difference between the sums 21  3  7 of the digits in the even and odd places 3. Divisible by : 4 Test : The number formed by the last should be zero or a multiple of 11. two digits (units’ and tens’) of the given Examples : number should be divisible by 4.  5  0  3  4  4  8 , Hence difference Examples : is zero. 10. Divisible by : 12 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Test : The sum of the digits of the 1) Group the numbers in sets of three from number should be divisible by 3 and the the units’ digit. number should also be divisible by 4. 2) Add the odd groups and the even groups separately. Examples : 3) The difference of the odd and the even 1728 groups should be either 0 or divisible 1  7  2  8  18 by 7. 18  3  6 Example: Is 85437954 divisible by 7? also 28  4  7 Solution: 11. Divisible by : 15 85437954 Test : The sum of the digits of the Adding up the first and the third sets, we number should be divisible by 3 & units’ get 85  954  1039 digit of the number should be 0 or 5. Now their difference is 1039  437  602 Examples : Since, 602  7  86 , hence the number is 810645 divisible by 7. 8  1  0  6  4  5  24 DIVISIBLE TEST FOR 13 24  3  8 The test holds good only for numbers with also last digit is 5. more than three digits. 12. Divisible by : 16 The test to be applied is as follows Test : The number formed by the last 1) Group the numbers in sets of three from Four digits (units’, tens’, hundreds’ and the unit’s digit. thousands’) of the given number should 2) Add the odd groups and the even groups be divisible by 16. separately. Examples : 3) The difference of the odd and the even groups should be either 0 or divisible 8320 16  520 by 13. 13. Divisible by : 25 Test : The last two digits of the number Example: Is 136999005 divisible by 13? should be 25, 50, 75 or 00. Solution: Examples : 136999005 1125, 975, 15500, 50 Adding up the first and the third sets, we The last two digits are as required. get 14. Divisible by : 125 136  5  141 Test : The last three digits of the Now their difference is 999 141  858. number should be 125, 250, 375, 500, Since, 858 13  66 , so the number is 625, 750, 875 or 000. divisible by 13. Examples : 1125, 1875, 15500, 35625, 76375, Example: Find X & Y when 22250, 49750, 50000 1) 15X8351Y is divisible by 72. The last three digits are as required. 2) 2856354XY is divisible by 99. Solution: SOME OTHER IMPORTANT TESTS 1) Since 72  8  9 , so the number must be divisible both 8 and 9. DIVISIBILITY TEST FOR 7 ⇒ The last three digits of the number The test holds good only for numbers with should be divisible by 8. Hence 51Y/8 must more than three digits and is applied as be an integer (last 3 digits), i.e. Y=2. Now, follows. © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  the given number should also be divisible 4) 0.12345  (12345  123) / 99000 by 9. 5) 0.12345  (12345  1234) / 90000 ⇒ 1  5  X  8  3  5  1  2  25  X should be divisible by 9. 6) 0.12345 12345 /100000 Thus X=2. Hence the number is 15283512. Example: 2) 99  9 11. Hence the number should be Let D be a decimal of the form, divisible by 9 and 11 both. D  0.a1a 2a1a 2a1a 2. ,where digits a1 & a 2 ⇒ 33  X  Y (sum of the digits) should be lie between 0 and 9. Then which of the divisible by 9. following numbers necessarily produces an ⇒  2  5  3  4  Y   8  6  5  X  integer, when multiplied by D? 1) 18 2) 108  0 or  11 or  22 3) 198 4) 208 ⇒ 14  Y  19  X  0 or  11 Solution: Or … Solving the equations Y - X – 5 = -11 It is recurring decimal and can be written and 33 + X + Y = 45, we get X = 9 and Y = 3. as D  0.a1a 2. To convert this to fraction, we Hence the number is 285635493. can write it as a1a 2 / 99. Thus when the CONVERSION OF A PURE RECURRING number is multiplied by 99 or a multiple of DECIMAL INTO FRACTION it, we shall necessarily get an integer. Of the given options, only (3) is a multiple of 99, Rule: Write the recurring figures only once hence Ans. (3) in the numerator and take as many nines in the denominator as the number of EXPONENTS repeating figures. 1) 0.6  6 / 9  2 / 3. Laws 2) 16.  16  0.6  16  6 / 9  16  2 / 3  50 / 3 1) a m  a n  a mn TO CONVERT A MIXED RECURRING For example, 23  24  234  27 DECIMAL INTO FRACTION 2) a m / a n  a mn  1/ a n m (if a  0) 35 5 2 3 3 4 1 1 Rule: In the numerator, write the For example, 2  3  3 , 6  64  2 difference between the number formed by 3 3 3 3 3)  a m   a mn n all the digits after decimal point (taking repeated digits only once) and that formed For example,  42   46 ,  34   38 3 2 by the digits which are not repeated. In the denominator, write the number 4) a  m  1/ a m formed by as many nines as there are 1 repeating digits followed by as many For example, 2  32 3 zeroes as in the number of non repeating 5) a 0  1 digits. (Any number with zero exponents is equal to 1) 0.17  (17 1) / 90  16 / 90  8 / 45 1) 2) 0.1254  (1254 12) / 9900  69 / 550 6)  a  b   a m  bm m 161 3) 2.536  2  (536  13) / 900  2 For example,  4  5  42  52 2 300 7)  a  b   a m  bm (ifb  0) m A quick summary 1) 012345.  12345 / 99999 3 3 5 5 2) 0.12345  (12345  1) / 99990 For example,    3 2 2 3) 0.12345  (12345  12) / 99900 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  8) m a  a1/m (Total 20 in number) For example, 3 27  271/3  3 SUM OF FACTORS q 9) a p/q  a p The sum of factors of the number N (as For example, 3 8 8 2 2/3 defined above) is given by the formula  a m1  1 bn 1  1 cp1  1 Example: Simplify 3   3  4 3 2 4  a  1 b  1 c  1  3   3 15 4 Where a, b, c…., m, n, p… retain the same Solution: meaning. 3   3  4 3 2 4  312  38  320  31  3 Example: Find the sum of all the factors of 240.  3   3 3  3 15 4 15 4 19 3 Solution:The sum by the above formula PRIME FACTORS   25  1 32  1 52  1 31 8  24   744 A composite number can be uniquely  2  1 3  1 5  1 1 2  4 expressed as a product of prime factors. Sum will be For example, 1 + 2 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 + 16 12  2  6  2  2  3  22  31 + 20 + 24 + 30 + 40 +48+60+80+120+240 = 20  4  5  2  2  5  22  51 744 124  2  62  2  2  31  22  31 etc. The number of ways in which a composite Every composite number can be expressed number N may be resolved into two factors in a similar manner in terms of its prime 1   p  1 q  1 r  1ifN  a p bq cr is not a factors. 2 NUMBER OF FACTORS perfect square and 1   p  1 q  1 r  1 1 The number of factors of a given composite 2 number N (including 1 and the number if N is a perfect square. itself) which can be resolved into its prime The number of ways in which a composite factors as number can be resolved into two factors N  a m  bn  cp..... which are prime to each other. Where a, b, c are prime numbers, are If N  a p bq cr  then the number of ways of 1  m1  n 1  p ..... resolving N into two factors prime to each 1 other is  1  11  11  1  2n 1 Example: Find the total number of factors 2 of 240. where n is the number of different prime Solution: factors of N. 240  2  2  2  2  3  5  24  31  51 If P is a prime number, the coefficient of Comparing with the standard format for every term in the expansion of 𝑎 + 𝑏 𝑝 the number N, we obtain except the first and the last is divisible by P. a  2, b  3,c  5, m  4, n  1, p  1 Example: In how many ways can the ∴ the total number of factors of this number 7056 be resolved into two factors? number including 1 and itself are Solution:  1  m 1  n 1  p ..... N  7056  32  24  72  3p  2q  7r  1  4   1  1  1  1  5  2  2  20 Note: N is perfect square. Number of ways Factors of 240 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, in which it can be resolved into two factors 16, 20, 24, 30, 40, 48, 60, 80, 120, 240  1/ 2 p  1 q  1 r  1  1 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  1 1 10)25(2  2  2  1 4  1 2  1  1   46  23 2 2 20 5)10(2 Example: Find the number of ways in which N=2778300 can be resolved into the 10 factors prime to each other. 00 Solution: HCF =5. N  22  34  52  73 If there are more than two numbers, we The required number is the same as the will repeat the whole process with the HCF number of ways of resolving 2  3  5  7 obtained from two numbers as the divisor into two factors which is equal to and so on. The last divisor will then be the 1/ 2 1  11  11  11  1  23  8. required HCF of the numbers. HCF OF NUMBERS Example: Find the HCF of 10, 25 and 30. Solution: It is the highest common factor of two or We can find the HCF of 10 and 25 i.e. 5. more given numbers. It is also called GCF Now we have to find the HCF of 5 and 30 (greatest common factor). which is 5. So, the HCF of 10, 25 and 30 is 5. e.g. HCF of 10 and 15=5, HCF of 55 and Note: If we have to find the greatest 200=5, HCF of 64 and 36=4 etc. number that will exactly divide p, q and r, then required number = HCF of p, q and r. Example: Find the HCF of 88, 24 and 124. Example: Find the greatest number that Solution: will exactly divide 65, 52 and 78. 88  2  44  2  2  22  2  2  2 11  23 111 Solution: Required number = HCF of 65, 52 and 78 = 24  2 12  2  2  6  2  2  2  3  23  31 13 124  2  62  2  2  31  23  311 If we have to find the greatest number that  HCF  22  4 will divide p, q and r leaving remainders a, b and c respectively, then the required DIVISION METHOD TO FIND HCF number By division method, we start with the two = HCF of (p - a), (q - b) and (r - c) numbers and proceed as shown below, till the remainder becomes zero. Example: Find the greatest number that will exactly divide 65, 52 and 78 leaving Example: Find the HCF of 12 and 48. remainders 5, 2 and 8 respectively. Solution: Solution: Required number 12)48(4 = HCF of (65 - 5), (52 - 2) and (78 - 8) = HCF of 60,50 and 70 =10. 48 00 If we have to find the greatest number that HCF =12. will divide p, q and r leaving remainders Here, when the remainder is not zero, the same remainder in each case, then divide the previous divisor with that required number = HCF of the absolute remainder and proceed in the same way values of until you get the remainder as zero. The (p - q), (q - r) and (r - p). last divisor is the required HCF. Example: Find the greatest number that Example: Find the HCF of 10 and 25. will divide 65, 81 and 145 leaving the same Solution: remainder in each case. © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Solution: Required number LCM OF NUMBERS = HCF of (81-65),(145-81) and (145-65) Lowest common multiple of two or more = HCF of 16,64 and 80=16 numbers is the smallest number which is Example: How many numbers below 90 exactly divisible by all of them. and other than unity exist, such that the e.g.LCMof 5,7,10  70, HCF of that number and 90 is unity? LCMof 2, 4,5  20, Solution: LCMof 11,10,3  330. 90  32  2  5 FACTORIZATION METHOD TO FIND LCM To find the LCM of the given numbers, first resolve all the numbers into their prime factors and then the LCM is the product of highest powers of all the prime factors. Example: Find the LCM 40, 120 and 380. Solution: Numbers of multiples of 2 = 45 40  4 10  2  2  2  5  23  51 ; Numbers of multiples of 3 = 30 120  4  30  2  2  2  5  3  23  51  31 ; Numbers of multiples of 5 = 8 Numbers of multiples of 2 & 5 = 9 380  2 190  2  2  95  2  2  5 19  22  51 191 Numbers of multiples of 2 & 3 = 15 Numbers of multiples of 3 & 5 = 6  Required LCM  23  51  31 191  2280 Numbers of multiples of 2,3 & 5 = 3 DIVISION METHOD TO FIND LCM  Total   24  6  12   12  3  6  3 Write the given numbers separately. Then  42  24  66 divide by 2 and write the result below the ∴ 24 numbers (including unity) compared numbers divisible by 2. If it is not divisible with 90 have only ‘1’ as common factor. by 2 then try with 3, 5, 7…. Etc. Leave the Hence required result is 24-1=23. others (those not divisible) untouched. Do the same for all steps till you get 1 as the Example: A riot hit state was deployed remainder in each column. with 3 different regiments of black commandos, each having 115, 161 and 253 Example: Find the LCM 6, 10, 15, 24 and commandos respectively. Each regiment 39. has commandos domicile of one particular Solution: state only and each regiment came from 2 6 10 15 24 39 different states. The Commander-in-Charge 2 3 5 15 12 39 further plans to split the regiments into 2 3 5 15 6 39 smaller groups but in such a way that all 3 3 5 15 3 39 groups have same number of commandos 5 1 5 5 1 13 and each group has commandos belonging 13 1 1 1 1 13 to a particular state only. What is the 1 minimum number of groups that can be LCM  2  2  2  3 5 13  1560 formed? Note: If we have to find the least number Solution: which is exactly divisible by p, q and r, then 115  23  5; 161  23  7; 233  23 11 the required number =LCM of p, q and r.  HCF  23 Hence minimum number of groups that can Example: Find the least number that is be formed = 23. exactly divisible by 6, 5 and 7. © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Solution: Required number = LCM of 6, 5 Solution: The numbers can be written as and 7 = 210. 0.6, 0.9, 1.5, 1.2 and 3.0 If we have to find the least number which Consider them as 6,9,15,12,30 ⟹ HCF=3 when divided by p, q and r leaves the ⟹Required HCF=0.3 and LCM=180. remainders a, b and c respectively, then if it ⟹Required LCM=18.0 is observed that (p-a)=(q-b)=(r-c)=K(say), Note: If the first number in the above then the required number =(LCM of p,q and example had been 0.61, then the equivalent r)-(K) integers would have been 61, 90, 150, 120 Example: Find the least number which and 300 etc. when divided by 6, 7 and 9 leaves the remainders 1, 2 and 4 respectively. HCF & LCM OF FRACTIONS Solution: Here, (6-1) = (7-2) = (9-4) = 5 HCF of fractions = HCF of numerators ÷ Required number =(LCM of 6,7 and 9)-5 LCM of denominators = 126 – 5 = 121. LCM of fractions = LCM of numerators ÷ If we have to find the least number which HCF of denominators when divided by p, q and r leaves the same Example: Find the HCF and LCM of remainder ‘a’ each case, then required 5 3 7 : , and. number 16 4 15 =(LCM of p,q and r) + a. Solution: Example: Find the least number which HCF   HCFof 5,3, 7   1 when divided by 15, 20 and 30 leaves the remainders 5 in each case.  LCM of 16, 4,15 240 Solution: LCM   LCM of 5,3, 7   105  105 Required number = (LCM of 15,20 and 30)  HCFof 16, 4,15 1 + 5. = 60 + 5 = 65. Note: CYCLICITY OF UNIT DIGIT LCM × HCF=product of two numbers. (Valid only for "two") The concept of cyclicity is used to identify the unit digit of any power. The unit digit in Example: Find the LCM of 25 and 35 if any operation depends on only the unit their HCF is 5. digit of the numbers used. Solution: For instance the units digit of different Pr oduct of the number 25  35 powers LCM    175 of 2 are as given below: HCF 5 21 22 23 24 25 26 Example: By using the rule that LCM = 2 4 8 6 2 4 Product of two numbers ÷ HCF, find LCM of As can be seen from the above table the 26 and 442. units digit of 25 is same as 21 , and that of 26 Solution: HCF of the two can be found as is same as that of 22 which implies that the 26 = 13 × 2, 442 = 2 × 17 × 13 ⟹ HCF = units digit of 2 is getting repeated after a 26⟹ LCM = 26 × 442/26 = 442 cycle of 4. Similarly the units digit of 3 for different powers of 3 are 3,9,7,1,3,9,7 and HCF & LCM OF DECIMALS 1. Similarly different numbers have different cyclicity which can be used to Example: Calculate the HCF and LCM of ascertain the units digit of bigger numbers. 0.6, 0.9, 1.5, 1.2 and 3. © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  2 LINEAR EQUATIONS LINEAR EQUATIONS A pair of numbers which satisfies both equations is called a simultaneous solution An equation where the maximum power of of the given equations or a solution of the any variable is unity (one) is a linear system of equations. There are three cases equation. A linear equation is of the form: which can be described geometrically. AX + BY + C = O (Here we assume that the coefficients of x Where A and B are co-efficient and C is a and y in each equation are not both zero.) constant 1) The system has exactly one solution: The equation is called LINEAR because the Here the lines corresponding to the graph of the equation on the X-Y Cartesian linear plane is a straight line. The sets of values of equations intersect in one point as X & Y satisfying any equation(s) are called shown in fig. below. its solution(s). Consider the equation 2x + y = 4. Now, if we substitute x = -2 in the equation, we obtain 2. (-2) + y = 4 or-4 + y = 4 or y = 8. Hence (- 2, 8) is a solution. If we substitute x = 3 in the equation, we obtain 2.3 + y = 4 or 6 + y = 4 or y =-2. Hence (3, -2) is a solution. The following table lists six possible values for x and the corresponding values for y, i.e. six The system a1X  b1Y  c1 and solutions of the equation. X -2 -1 0 1 2 3 a 2 X  b2 Y  c2 has a unique solution, if y 8 6 4 2 0 -2 a1 b1 . If we plot the solutions of the equation 2x + a 2 b2 y = 4 which appear in the table above then 2) The system has no solution: Here the we see that they all lie on the same line. lines corresponding to the linear We call this line the graph of the equation equations are parallel as shown in fig. since it corresponds precisely to the below. solution set of the equation. TWO LINEAR EQUATIONS IN TWO UNKNOWNS The system a1X  b1Y  c1 & a 2 X  b2 Y  c2 We now consider a system of two linear a b c equations, in two unknowns x and y: has no solution if 1  1  1. a 2 b2 c2 a1x  b1y  c1 3) The system has an infinite number of a 2 x  b2 y  c2 solution: Here the lines corresponding © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  to the linear equations coiincide as Method of solving: Suppose we are given shown in fig. below. three equations containing three unknown quantities X, Y and Z. Mark these equations (1), (2) and (3). Now, from any two equations say (1) and (2), try to eliminate (or get rid of ) one unknown quantity, say Z, i.e., to say, multiply (1) and (2) by such suitable number which will make the coefficients of Z in (1) and (2) equal; then The system a1X  b1Y  c1 & add or subtract to cancel Z. Mark this a 2 X  b2 Y  c2 has infinitely many equation (4). After this, take any two a b c equations, say (2) and (3), and from this, solution if 1  1  1 eliminate Z in a similar way. Mark this a 2 b2 c2 equation (5). Now, solving (4) and (5) by any of the methods given previously, we SYSTEM OF LINEAR EQUATIONS will find X and Y. Substituting the value of X and Y in any of the equations (1), (2) and Consistent System: A system (of 2 or 3 or (3), we will get the value of Z. more equations taken together) of linear Example: Solve X+Y+Z=6, 2X+2Y+3Z=13, equations is said to be consistent, if it has at 3X+4Y+5Z=22 least one solution. Solution: Inconsistent System: A system of We have X + Y + Z = 6 … (1) simultaneous linear equations is said to be 2X + 2Y + 3Z = 13 … (2) inconsistent, if it has no solutions at all. 3X + 4Y + 5Z = 22 … (3) e.g. X + Y = 9; 3X + 3Y = 8 Multiply (1) by 3 and subtract (2) and we Clearly there are no values of X & Y which have simultaneously satisfy the given equations. 3(X + Y + Z) - (2X + 2Y + 3Z) = 18 - 13 So the system is inconsistent. or X + Y = 5 … (4) Again, multiply (1) by 5 and subtract (3), Example: Find k for which the system 6x - then 2y = 3, kx – y = 2 has a unique solution. 5(X + Y + Z) - (3X + 4Y + 5Z) = 30 - 22 Solution: The given system will have a or 2X + Y = 8 … (5) unique solution Subtracting (4) from (5), we have X = 3. a b 6 2 Hence, from (4), Y = 2. if 1  1 i.e.  or k  3. a 2 b2 k 1 Substituting these values of X and Y in (1), we have 3+2+Z=6 or Z=1. Example: What is the value of k for which the system x + 2y = 3, 5x + ky =-7 is consistent? Solution: The given system will be inconsistent if a1 b1 c1   i.e. a 2 b2 c2 1 2 3 if  . Hence, k  10. 5 k 7 SIMULTANEOUS EQUATIONS INVOLVING THREE UNKNOWN QUANTITIES © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  3 RATIO, PROPORTION & VARIATION RATIO It is the compounded ratio of two equal ratios. Thus the duplicate ratio of The comparison between two quantities of a : bis a 2 / b2 or a 2 : b2 the same kind of unit is the ratio of one quantity to another. Two quantities of Example: Find the duplicate ratio of 4: 5. different kinds cannot be compared. Thus, Solution: The duplicate ratio of 4 : 5 is there is no relation between 20 rupees and 16:25. 20 men. The ratio of a and b is usually a TRIPLICATE RATIO written as a : b or. b It is the compounded ratio of three equal ANTECEDENT AND CONSEQUENT ratios. Thus the triplicate ratio of In the ratio a : b, a is called the antecedent a : bis a 3 / b3 or a 3 : b3 (the first term) and b is called the consequent (the second term) Example: Find the triplicate ratio of 4: 5. Solution: The triplicate ratio of 4:5 is 64:125. Note: The ratio of two numbers a and b, written SUB-DUPLICATE RATIO a as a : b, is the fraction provided b  0. b For any ratio a : b its sub-duplicate ratio is Ratio is always reduced to its simplest defined as a : b. a form. Thus a : b  , b  0. if a  b  0, the b Example: Find the sub-duplicate ratio of16:25. ratio is 1 : 1or 1/1=1 Solution: Sub-duplicate ratio of 16:25 is 4 2 16 : 25 , i.e. 4 :5. 1) The ratio of 4 to 6  4 : 6   6 3 2 4 2/3 4 SUB-TRIPLICATE RATIO 2) :   3 5 4/5 6 3y 5x 20x For any ratio a ∶ b, its sub-triplicate ratio is 3) 5x :   4 3y / 4 3y defined as 3 a: 3 b. COMPOUNDED RATIO Example: Find the sub-triplicate ratio of 27:64. When two or more ratios are multiplied Solution: Sub-triplicate ratio of 27:64 is term wise, the ratio thus obtained is called 3 27 : 3 64 i.e. 3 :4. their compounded ratio. For the ratios a : b and c : d, the compounded ratio is ac : bd. COMPARING RATIOS Example: What is the compounded ratio, To compare two ratios, we express them as for 2 : 3 and 4 : 5? fractions and then compare. Solution: The compounded ratio is Example: Which is greater 3:4 or 4 : 5? 2×4∶3×5 or 8∶15. Solution: 3 4 DUPLICATE RATIO 3 : 4  and 4 : 5 . 4 5 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  3 3  5 15 4    528   704 4 4  5 20 3 4 4  4 16 The number of boys to be decreased   5 5  4 20 =1170-704=466. 16 15 4 3 Since   or,  , Example: If a : b = 3 : 4 and b : c = 6 : 13, 20 20 5 4 then find a : b : c. So, 4 : 5 > 3 : 4. Solution: The best way to solve such Note: A ratio is said to be in its simplest questions is to make b common in the two form if the HCF of the antecedent and the ratios. consequent is 1. Thus, we can write a : b = 9 : 12 and b : c = 12 : 26. Now that b is equal in both the Example: Divide 2400 in the ratio3 ∶ 5. ratios, we can write the same as Solution: a : b : c The first part is 3 units and the second part 9 : 12 is 5 units. The total of both the parts = 3 12 : 26 units + 5 units = 8 units. Thus, we can write 𝑎 ∶ 𝑏 ∶ 𝑐 = 9 ∶ 12 ∶ 26. Here, 8 units = 2400, Using formula directly, we can get So, 1 unit = 2400/8 = 300. a∶ b∶ c=(3×6): (4×6) ∶ (4×13)=9∶ 12∶ 26 The first part = 3 units = 3 × 300 =900. The second part = 5 units = 5 × 300 =1500. PROPORTION Example: A sum of money is divided A statement expressing the equality of two between Vinod and Lokesh in the ratio of ratios is called a proportion, i.e. if 3∶7. Vinod gets Rs. 240. What does Lokesh 𝑎∶ 𝑏=𝑐∶𝑑 get? a c Solution: Or  then a, b, c & d are said to be in b d Vinod gets 3 units = Rs. 240. proportion. Here a and d are called the So, 1 unit = 240/3 = 80. extremes and b and c are called the Therefore, 7 units = 7 × 80 =560. means. Also d is called the fourth Thus, Lokesh gets Rs. 560. proportional to a, b and c. Thus, we can write a : b = c : d, PROBLEMS LEADING TO THE a is the first proportional, APPLICATION OF RATIOS b is the second proportional, c is the third proportional, Example: The ratio of the number of boys d is the fourth proportional. to the number of girls in a school of 1638 is For example, 5:10= 22 : 44 is in proportion. 5: 2. If the number of girls increased by 60, Each quantity in a proportion is called a then what must be the decrease in the term. The first and the last terms are number of boys to make the new ratio of known as the extremes while the second boys to girls as 4 : 3? and the third term are called the means. Solution: For the four quantities to be in proportion, 5 Number of boys  1638  1170 7 2 Number of girls  1638  468 7 Number of girls after increase  468  60  528 Total number of boys as per new ratio © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Product of means = Product of extremes. Example: Find the mean proportional between 9a 2 b and 25b3. CONTINUED PROPORTION Three or more quantities are said to be in Solution: Let x be the mean proportional. continued proportion, when the ratio of the 9a 2 b x first and the second is equal to ratio of the    x 2  9a 2 b.25b3  x  15ab2 x 25b3 second and the third and so on. Thus a, b, c are in continued proportion if a b VARIATION a ∶ b ∶ ∶ b ∶ c i.e., . b c Similarly, a, b, c, d, … are in continued Most of us would still remember proportion if, statements like “The distance travelled a b c varies directly as the speed of the vehicle.” a : b :: b : c :: c : d ., i.e.,     , b c d These and similar statements have precise Proportions are equations and can be mathematical meanings and they represent transformed using procedures for a specific type of function called variation equations. Some of the transformed functions. equations are used frequently and are The three general types of variation called the laws of proportion. functions are direct variation, inverse If a : b = c : d, then variation and joint variation. ad = bc {Product of extremes = Product of means} DIRECT VARIATION b d If two quantities X & Y are related such that 1)  (Invertendo) any increase or decrease in ‘Y’ produces a a c a b proportionate increase or decrease in ‘X’ or 2)  (Alternendo) vice versa, then the two quantities are said c d to be in direct proportion. a b cd 3)  (Componendo) b d X is directly proportional to Y is written as a b cd X  Yor X  KY. 4)  (Dividendo) b d In other words X : Y  X / Y  K. Here K is a b cd a constant whose value for a particular 5)  (Componendo & Dividendo) a b cd variation is same. In the proportion a : b = b : c, c is called the Consider X1  KY1 and X2  KY2 , dividing third proportional to a and b, and b is X Y called a mean proportional between a and the two we get 1  1. X 2 Y2 c. Thus, b2  ac. This is known as Thus, the chances of your success in the continued proportion. test are directly proportional to the number of hours of sincere work devoted Example: Find the fourth proportional to every day. 12a 2 ,9a 2 b and 6ab2. Example: If X  Y and x  9 when y  30 , Solution: Let x be the fourth proportional, then find the relation between x and y. Find then 1 12a 2 6ab 2 6ab 2.9a 2 b 9 3 x when y  7 and y when x = 6.   x   ab 2 9a 2 b x 12a 2 2 Solution: © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  3 3 6 2 Let x  ky, then9  k  30  , k  ,i.e.x  y. When y  21, 21  ,x  10 10 x 7 1 3  1 1 When y  7 , x   7   2. JOINT VARIATION 2 10  2  4 3 When x  6, x  y  y  20. a) A varies jointly as B and C and is 10 denoted by A  BC Or A  kBC (where k is a constant). Example: Different sizes of the car have b) A varies directly as B and inversely as C different models. The weight of a car model B kB varies directly as the cube of its length. The and is denoted by A  Or A  weight of a car model of length 3 cm is 10 C C gm. What is the weight of a car model of (where k is a constant). length 12 cm? c) If A varies as B when C is constant, and Solution: Let W gm be the weight of a car if A varies as C when B is constant then model and L cm be its length. A varies as BC when B and C both vary.  A  BC Or A  kBC (where k is a  W  L3or W  kL3 (where k is a constant) constant). 10 10  10  k(3)3.  k  i.e.W  L3 27 27 Example Given, a varies as b when c is 10 constant, and as c 2 when b is constant. When L  12, W  (12)3  640gm. If 27 𝑎 = 770, then b=15 & c=7,and when c=3 & The required weight is 640 gm. a=132 find b. Solution: INVERSE VARIATION As a  bc2 or a  kbc2 (where k is a constant) Here two quantities X & Y are related such  770  k 15 (7)2 that, any increase in X would lead to a decrease in Y or any decrease in X would 22 22 k i.e.a  bc2.When c  3, a  132 lead to an increase in Y. Thus the quantities 21 21 X & Y are said to be inversely related and X 22 132   b  3  b  14 2 is inversely proportional to Y is written as 21 X  1/ Y or X  k / Y or XY  k (Constant) Thus, X1Y1  X2 Y2 Note: Or the product of two quantities remains 1) If A  B and B  C, then A  C constant. 2) If A  C and B  C, then  A  B  C Thus, the chances that you will be able to cheat in a test are inversely proportional to  (AB)  C. the smartness of the invigilator. 3) If A  BC, then A / C  B and A / B  C. 4) If A  B and C  D, then AC  BD. Example: If y varies inversely as x, and 5) If A  B, thenAn  Bn. 𝑦 = 3 when x  2 , then find x when y  21. 6 If A  B and A  C, then A   B  C  and Solution: 1 k A  (B  C) y  or y  (where k is a constant) 7) If A  B, then AP  BP where P is any x x k 6 quantity, constant or variable. then 3  , k  6i.e. y  2 x © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  4 PERCENTAGE PERCENTAGE EXPRESSING DECIMAL AS A PERCENTAGE “Percent” implies “for every hundred”. This Any decimal fraction can be converted into concept is developed to make the a percentage by multiplying it by 100: comparison of fractions easier by 0.5 100 50 0.5    50% equalizing the denominators of all fractions 100 100 to hundred. For example, 7/11 as percentage is Note: represented as 1) When two numbers x and y are given, 7 7 100 (7 100) /11 63.63 then one number can be expressed as a     63.63% 11 11100 100 100 percentage of the other, in the following way. x X as a percentage of y  100 Percentage can also be represented as y decimal fractions. In such a case it is y effectively equivalent to the proportion of Y as a percentage of x  100 x the original quantity 2) x % of y = y% of x 20 For example, 20 % is the same as , i.e. 100 PERCENTAGE INCREASE or DECREASE of 0.2 Since any ratio is also basically a a quantity is the ratio expressed in division, each ratio can also be expressed percentage of the actual INCREASE or as a percentage. The terms “ratio” and DECREASE of the quantity to the original percentage “can be used interchangeably amount of the quantity i.e., along with the corresponding mention of the denominator being taken as 100 PERCENTAGE INCREASE 1 Actualin crease For example: a ratio of can be converted  100 2 Original quanity to a percentage figure as 1 1 50 50    50% PERCENTAGE DECREASE 2 2  50 100 Actual decrease Any percentage can be expressed as a  100 Original quanity decimal fraction by dividing the percentage figure by 100 and conversely, any decimal fraction can be converted to percentage by For example: If the production of rice went multiplying it by 100. up from 2150 MT in 2005 to 300 MT in 2006 , then the percentage increase in rice Expressing a percentage value as a fraction: production from 2005 to 2006 is calculated x as follows: X% = x out 100 = Actual increase = 300-250 = 50MT 100 Percentage increase: 75 3 So, 75% =75 out of 100 =  Actualincrease from 2005 to 2006 100 4  100 Actual production of ricein 2005 Any percentage can be expressed as a decimal fraction by dividing the percentage 50  100  20% figure by 100 250 © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  1) If the increase on a value of 350 is 15% Q.1 If 45% of a certain number is 990, the new quantity is 1.15 × 350 = 402.5 then find the value of 54% of that (where 1.15=1+0.15, 0.15 being the number. decimal equivalent of 15 % ) Solution: 2) If the production in 2005 is given as Let the number be x 400 MT and the increase from 2004 to 45 Given, x  990 2005 is given to be 25% then the 100 production in 2004 will be equal to  x  2200 400/1.25=320 MT (where 1.25=1+0.25, 54 0.25 being the decimal equivalent of  required value  x  1188 100 25%) 3) Similarly, if there is a decrease of 12% Q.2 The ratio of the monthly salaries of on a quantity of 225, then the new A in December 2004 and in January quantity will be equal to 225×0.88 3 1 (where 0.88=1-0.12, 0.12 being the 2005 was 3 : 4. Find the decimal equivalent of 12%).If the 5 2 production in 2005 is given as 400MT percentage increase in his salary. and it is a decrease of 13% from 2004, Solution: then the production in 2004 will be The ratio of the monthly salaries 3 1 18 9 equal to 400/0.87 (where 0.87=1-0.13, 3 : 4  : 18 2  :  9  5   4 : 5 0.13 being the decimal equivalent of 5 2 5 2 13%). Let the monthly salary of A in December 2004 =4x On the basis of percentage increase, we can His monthly salary in January 2005 write down how many times the old value would be=5x. Required percentage gives the new value. For example, if the increase on his salary percentage increase is 100%,we can 5x  4x conclude that the new value is 2 times the  100  %  25% 4x old value and if the percentage increase is 300% then the new value is 4 times the old Q.3 There are three numbers. The first value. If the percentage increase is 450% and the second numbers are 50% then the new value is 5.5 times the old less and 60% less respectively than value. In general if the percentage increase the third. What percentage of the  p  first number is the second? is p% then the new value is   1 times  100  Solution: the old value. Let the third number be 100 Conversely, if we know how many times  50  First number  100 1    50 the old value gives the new value, we can  100  find out the percentage increase in the old  60  value to get the new value. For example, if Second number  100 1    40  100  the new value is 3 times the old value, the 40 4 percentage increase in the old value to get  it is th i.e. th or 80% of the first the new value is 200%. If the new value is 50 5 4.25 times the old value, then the number. percentage increase is 325%. In general if the new value is k times the old value, then Q.4 Ram got 30% in a test and failed by the percentage increase is (k  1) 100 % 10 marks.If the pass marks in the test was 70, find the maximum WORKED OUT EXAMPLES marks in it © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  Solution: Area of rectangle Let the maximum mark in the test  1.14l 1.08b  1.2312 lb be M. Increase in area  1.2312lb  lb 30 Ram’s mark  M i.e., 0.2312 lb 100 23.12 30  Percentage increase  lb Pass mark  M  10 100 100 i.e., 23.12% 30 M  10  70i.e., M  200 100 Q.5 The price of an article is decreased by 20%. By what percentage must the consumption of it be increased in order to retain the expenditure on it? Solution: Let the initial price be Rs. 100/gm Let the initial consumption be 1 gm Initial expenditure =Rs. 100  20  New price  100 1    100   Rs.80 / gm New expenditure =Rs 100 100 New consumption  gm 80 5  gm 4 Consumption must increase by 5    1100% 4   25% 1 Q.6 The length of a rectangle increases by 14% and the breadth by 8%. What is the consequent percentage increase in area? Solution: Let length and breadth of the rectangle be l and b Area of rectangle =lb Length is increased by 14% 14  New length  l  l    l  0.14l   1.14l 100 Breadth is increased by 8%  New breadth  b  8% b  b  0.08 b  1.08 b © Copyright Reserved by Gateflix.in No part of this material should be copied or reproduced without permission  5 PROFIT–LOSS AND PARTNERSHIP PROFIT -LOSS Loss X 2) Loss%  100%  100% Suppose a shopkeeper buys an article from CP CP a manufacturer. The price at which he buys 100  Gain% 100  Loss% 3) SP  CP  or CP  the article is called the cost price of the 100 100 article. We write C.P. for cost price. 100 100 4) CP  SP  or SP  The shopkeeper sells the article at a price 100  Gain% 100  Loss% which is generally more than its cost price. Note: If C.P. of both the items is same and The amount for which he sells the article is the percentage loss and gain are equal, then known as the selling price. We write S.P. for net loss or profit is zero. selling price. The excess of the selling price over the cost Example: Two shirts were having a cost price of an article is called the profit or the price of Rs. 200 each. One was sold at a gain. So, profit of 15% and the other was sold at a