Champter-1

NUMBER SYSTEMS

In Hindu Arabic System, we use ten symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 called digits to represent any number. This is the decimal system where we use the numbers 0 to 9. 0 is called insignificant digit.

A group of figures, denoting a number is called a numeral. For a given numeral, we start from extreme right as Units place, Tens place, Hundreds place and so on.

Illustration 1 We represent the number 309872546 as shown below:

Ten Crore 108Crores 107Ten Lacs

(million) 106Lacs 105Ten Thousand 104Thousand 103Hundred 102Tens 101Units 100

309872546

We read it as

Thirty crores, ninety- eight lacs, seventy-two thousands five hundred and forty-six.

In this numeral:

The place value of 6 is 6 1 = 6

The place value of 4 is 4 10 = 40

The place value of 5 is 5 100 = 500

The place value of 2 is2 1000 = 2000 and so on.

The face value of a digit in a numbers is the value itself wherever it may be.

Thus, the face value of 7 in the above numeral is 7. The face value of 6 in the above numeral is 6 and in the above numeral is 6 and so on.

NUMBER SYSTEM

Natural numbers

Counting numbers 1, 2, 3, 4, 5,... are know as natural numbers.

The set of all natural numbers, can be represented by

N= {1, 2, 3, 4, 5,.}

Whole numbers

If we include 0 among the natural numbers, then the numbers 0, 1, 2, 3, 4, 5, are called whole numbers.

The set of whole number can be represented by

W= {0, 1, 2, 3, 4, 5}

Clearly, every natural number is a whole number but 0 is a whole number which is not a natural number.

INTEGERS

All counting numbers and their negatives including zero are know as integers.

The set of integers can be represented by

Z or I = {-4, -3, -2, -1, 0, 1, 2, 3, 4, }

Positive Integers

The set I+ ={1, 2, 3, 4,} is the set of all positive integers. Clearly, positive integers and natural numbers are synonyms.

Negative Integers

The set I- = {-1, -2, -3} is the set of all negative integers. 0 is neither positive nor negative.

Non-negative Integers

The set {0, 1, 2, 3,} is the set all non-negative integers.

Rational Numbers

The numbers of the form p/q, where p and q are integers and q 0, are known as rational numbers, e.g. 4/7, 3/2, -5/8, 0/1, -2/3, etc.

The set of all rational numbers is denoted by Q. i.e. Q ={x:x =p/q; p,q belong to I, q0}. Since every natural number a can be written as a/1, every natural number is a rational number. Since 0 can be written as 0/1 and every non-zero integer a can be written as a/1, every integer is a rational number.

Every rational number has a peculiar characteristic that when expressed in decimal form is expressible rather in terminating decimals or in non-terminating repeating decimals.

For example, 1/5 =0.2, 1/3 = 0.33322/7 = 3.1428704287, 8/44 = 0.181818., etc.

The recurring decimals have been given a short notation as

0.333. = 0.(3

4.1555 = 4.0(5

0.323232= 0.(32.

Irrational Numbers

Those numbers which when expressed in decimal from are neither terminating nor repeating decimals are known as irrational number, e.g. 2, 3, 5, , etc.

Note that the exact value of ( is not 22/7. 22/7 is rational while irrational number. 22/7 is approximate value of . Similarly, 3.14 is not an exact value of it.

Real Numbers

The rational and irrational numbers combined together are called real numbers, e.g.13/21, 2/5, -3/7, (3, 4 + (2, etc. are real numbers.

The set of all real numbers is denote by R.

Note that the sum, difference or product of a rational and irrational number is irrational, e.g. 3+ 2, 4-3, 2/3-5, 43, -75 are all irrational.

Even Numbers

All those numbers which are exactly divisible by 2 are called even numbers, e.g.2, 6, 8, 10, etc., are even numbers.

Odd Numbers

All those numbers which are not exactly divisible by 2 are called odd numbers, e.g. 1, 3, 5, 7 etc., are odd numbers.

Prime Numbers

A natural number other than 1 is a prime number if it is divisible by 1 and itself only.

For example, each of the numbers 2, 3, 5, 7 etc., are prime numbers.

Composite Numbers

Natural numbers greater than 1which are not prime, are known as composite numbers.

For example, each of the numbers 4, 6, 8, 9, 12, etc., are composite numbers.

Note:

1. The number 1 is neither a prime number nor composite number.

2. 2 is the only even number which is prime

3. Prime numbers up to 100 are:

2, 3, 5, 7, 11, 13, 17, 19,23, 29, 31, 37, 41,43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, i.e. 25 prime numbers between 1 and 100.

4. two numbers which have only 1 as the common factor are called co-primes or relatively prime to each other, e.g. 3 and 5 are co-primes.

Note that the numbers which are relatively prime need not necessarily be prime numbers, e.g. 16 and 17 are relatively prime although 16 is not a prime number.

\ADDITION AND SUBTRACTION (SHORT-CUT METHODS)

The method is best illustrated with the help of following example:

Illustration 2 54321 (9876+8976+7689) = ?

Step 1 Add 1st column:

54321

9876

8967

7689

27789

6+7+9 = 22

To obtain 1 at units place add 9 to make 31. In the answer, write 9 at units place and carry over 3.

Step 2 Add 2nd column:

3+7+6+8=24

To obtain 2 at tens place add 8 to make 32. In the answer, write 8 at tens place and carry over 3.

Step 3 Add 3rd column:

3 + 8 + 9 + 6 = 26

To obtain 3 at hundreds place, add 7 to make 33. In the answer, write 7 at hundreds place and carry over 3.

Step 4 Add 4th column:

3 + 9 + 8 + 7 = 27

To obtain 4 at thousands place add 7 to make 34. In the answer, write 7 at thousands place and over 3.

Step 5 5th column:

To obtain 5 at ten-thousands place add 2 to it to make 5. In the answer, write 2 at the ten-thousands place.

( 54321 (9876 + 8967 + 7689) = 27789.

Common Factor

A common factor of two or more numbers is a number which divides each of them exactly.

For example, 4 is a common factor of 8 and 12.

Highest common factor

Highest common factor of two or more numbers is the greatest number that divides each one of them exactly. For example, 6 is the highest common factor of 12, 18 and 24. Highest Common Factor is also called Greatest Common Divisor or Greatest Common Measure.

Symbolically, these can be written as H.C.F. or G.C.D. or G.C.M., respectively.

Methods of Finding H.C.F.

I. Method of Prime Factors

Step 1 Express each one of the given numbers as the product of prime factors.

[A number is said to be a prime number if it is exactly divisible by 1 and itself but not by any other number, e.g. 2, 3, 5, 7, etc. are prime numbers]

Step 2 Choose Common Factors.

Step 3 Find the product of lowest powers of the common factors. This is the required H.C.F. of given numbers.

Illustration 1 Find the H.C.F. of 70 and 90.

Solution 70 = 2 ( 5 ( 7

90 = 2 ( 5 ( 9

Common factors are 2 and 5.

( H.C.F. = 2 ( 5 = 10.

Illustration 2 Find the H.C.F. of 3332, 3724 and 4508

Solution 3332 = 2 ( 2 ( 7 ( 7 ( 17

3724 = 2 ( 2 ( 7 ( 7 ( 19

4508 = 2 ( 2 ( 7 ( 7 ( 23

( H.C.F. = 2 ( 2 ( 7 ( 7 = 196.

Illustration 3 Find the H.C.F. of 360 and 132.

Solution 360 = 23 ( 32 ( 5

132 = 22 ( 31 ( 11

( H.C.F. = 22 ( 31 ( = 12.

Illustration 4 If x = 23 ( 35 ( 59 and y = 25 ( 37 ( 511, find H.C.F. of x and y.

Solution The factors common to both x and y are 23, 35 and 59.

( H.C.F. = 23 ( 35 ( 59.

II. Method of Division

A. For two numbers:

Step 1 Greater number is divided by the smaller one.

Step 2 Divisor of (1) is divided by its remainder.

Step 3 Divisor of (2) is divided by its remainder. This is continued until no remainder is left.

H.C.F. is the divisor of last step.

Illustration 5 Find the H.C.F. of 3556 and 3444.

3444 )3556 (1

3444

112 ) 3444 ( 30

3360

84 ) 112 ( 1

84

28 ) 84 ( 3

84

(

B. For more than two numbers:

Step 1 Any two numbers are chosen and their H.C.F. is obtained.

Step 2 H.C.F. of H.C.F. (of(1)) and any other number is obtained.

Step 3 H.C.F. of H.C.F. (of (2)) and any other number (not chosen earlier) is obtained.

This process is continued until all numbers have been chosen. H.C.F. of last step is the required H.C.F.

Illustration 6 Find the greatest possible length which can be used to measure exactly the lengths

7 m, 3 m 85 cm, 12 m 95 cm.

Solution Required length

= (H.C.F. of 700, 385, 1295) cm = 35 cm.

Common Multiple

A common multiple of two or more numbers is a number which is exactly divisible by each one of them.

For Example, 32 is a common multiple of 8 and 16.

8 ( 4 = 32

16 ( 2 = 32.

Least Common Multiple

The least common multiple of two or more given numbers is the least or lowest number which is exactly divisible by each of them.

For example, consider the two numbers 12 and 18.

Multiples of 12 are 12, 24, 36, 48, 72,

Multiple of 18 are 18, 36, 54, 72,

Common multiples are 36, 72,

(Least common multiple, i.e. L.C.M. of 12 and 18 is 36.

Methods of Finding L.C.M.

A. Method of Prime Factors

Step 1 Resolve each given number into prime factors.

Step 2 Take out all factors with highest powers that occur in given numbers.

Step 3 Find the product of these factors. This product will be the L.C.M.

Illustration 7 Find the L.C.M. of 32, 48, 60 and 320.

Solution32 = 25 ( 1

48 = 24 ( 3

60 = 22 ( 3 ( 5

320 = 26 ( 6

( L.C.M. = 26 ( 3 ( 5 = 960.

B. Method of Division

Step 1 The given numbers are written in a line separated by common.

Step 2 Divide by any one of the prime numbers 2, 3, 5, 7, 11, which will divide at least any two of the given nu8mbers exactly. The quotients and the undivided numbers are written in a line below the first.

Step 3 Step 2 is repeated until a line of numbers (prime to each other) appears.

1 Find the product of all divisors and numbers in the last line which is the required L.C.M.

Illustration 8 Find the L.C.M. of 12, 15, 20 and 54.

Solution212,15, 20,54

2 6,15,10,27

3 3,15, 5,27

5 1, 5, 5, 9

1, 1, 1, 9

L.C.M. = 2 ( 2 ( 3 ( 5 ( 1 ( 1 ( 1 ( 9 = 540.

Note: Before finding the L.C.M. or H.C.F., we must ensure that all quantities are expressed in the same unit.

Some Useful Short-Cut Methods

1. H.C.F. and L.C.M. of Decimals

Step 1 Make the same number of decimal places in all the given numbers by suffixing zero(s) if necessary.

Step 2 Find the H.C.F./L.C.M. of these numbers without decimal.

Step 3 Put the decimal point (in the H.C.F./L.C.M. of step 2) leaving as many digits on its right as there are in each of the numbers.

2. L.C.M. and H.C.F. of Fractions

L.C.M = L.C.M. of the numbers in numerators

H.C.F. of the numbers in denominators

H.C.F. = H.C.F. of the numbers in numerators

L.C.M. of the numbers in denominators

3. Product of two numbers

= L.C.M. of the numbers ( H.C.F. of the numbers

4. To find the greatest number that will exactly divide x, y and z.

Required number = H.C.F. of x, y and z.

5. To find the greatest number that will divide x, y and z leaving remainders a, b and c, respectively.

Required number = H.C.F. of (x a), (y b) and (z c).

6. To find the least number which is exactly divisible by x, y and z.

Required number = L.C.M. of x, y and z.

7. To find the least number which when divided by x, y and z leaves the remainders a, b and c, respectively. It is always observed that (x a) = (y b) = (z c) = k (say)

( Required number = (L.C.M. of x, y and z) k.

8. To find the least number which when divided by x, y and z leaves the same remainder r in each case.

Required number = (L.C.M. of x, y and z) + r.

9. To find the greatest number that will divide x, y and z leaving the same remainder in each case.

(a) When the value of remainder r is given:

Required number = H.C.F. of (x r), (y r) and (z r).

(b) When the value of remainder is not given:

Required number = H.C.F. of ((x y)(, ((y z)( and ((z x)(10. To find the n-digit greatest number which, when divided by x, y and z.

(a) leaves no remainder (i.e. exactly divisible)

Step 1 L.C.M. of x, y and z = L

L ) n digit greatest number (

Step 2 remainder = R

Step 3Required number

= n-digit greatest number R

(b) leaves remainder K in each case Required number

= (n-digit greatest number R) + K.

11. To find the n-digit smallest number which when divided by x, y and z(a) leaves no remainder (i.e. exactly divisible)

Step 1 L.C.M. of x, y and z = L

L )n-digit smallest number(

Step 2 remainder = R

Step 3 Required number

= n-digit smallest number + (L R).

(b) leaves remainder K in each case.

Required number

= n-digit smallest number + (L R) + k.

RATIOS, PROPORTIONS AND VARIATION

A ratio is a comparison of two quantities by division. It is a relation that one quantity bears to another with respect to magnitude. In other words, ratio means what part one quantity is of another. The quantities may be of same kind or different kinds. For example, when we consider the ratio of the weight 45 kg of a bag of rice to the weight 29kg of a bag of sugar we are considering the quantities of same kind but when we talk of allotting 2 cricket bats to 5 sportsmen, we are considering quantities of different kinds. Normally, we consider the ratio between quantities of the same kind.

If a and b are two numbers, the ratio of a to b is a/b or a +b and is denoted by a : b. The two quantities that are being compared are called terms. The first is called antecedent and the second term is called consequent.

For example, the ratio 3 : 5 represents 3/5 with antecedent 3 and consequent 5.

Note:

1. A ratio is a number, so to find the ratio of two quantities, they must be expressed in the same units.2. A ratio does not change if both of is terms are multiplied or divided by the same number. Thus, 2/3= 4/6 = 6/9 etc.TYPES OF RATIOS

1. Duplicate Ratio The ratio of the squares of two numbers is called the duplicate ratio of the two numbers.For example, 32/42 or 9/16 is called the duplicate ratio of .2. Triplicate Ratio The ratio of the cubes of two numbers is called the triplicate ratio of the two numbers. For example, 33/43 or 27/64 is triplicate ratio of .3. Sub-duplicate Ratio The ratio of the square roots of two numbers is called the sub-duplicate ratio of two numbers.For example, 3/4 is the sub- duplicate ratio of 9/16.4. Sub-duplicate Ratio The ratio of the cube roots of two numbers is called the sub-triplicate ratio of two numbers.For example, 2/3 is the sub-triplicate ratio of 8/27.5. Inverse Ratio or Reciprocal Ratio If the antecedent and consequent of a ratio interchange their places, the new ratio is called the inverse ratio of the first.Thus, if a : b be the given ratio, then 1/a : 1/b or b : a is its inverse ratio.

For example, 3/5 is the inverse ratio of 5/3.

6. Compound Ratio The ratio of the product of the antecedents to that of the consequents of two or more given ratios is called the compound ratio. Thus, if a :b and c:d are two given rations, then ac : bd is the compound ratio of the given ratios, For example, if , 4/5 and 5/7 be the given ratios, then their compound ratios is 3(4(5/ 4(5(7, that is, 3/7.

PROPORTION

The equality of two ratios is called proportion.

If a/b = c/d, then a, b, c and d are said to be in proportion and we write a : b: : c: d. This is read as a is to b as c is to d.

For example, since = 6/8, we write 3; 4: : 6: 8 and say 3, 4, 6 and 8 are in proportion.

Each term of the ratio a/b and c/d is called a proportional. a, b, c, and d are respectively the first, second, third and fourth proportionals

Here, a, d are known as extremes and b, c are known as means.

SOME BASIC FORMULAE

1. If four quantities are in proportion, then product of Means = product of Extremes For example, in the proportion a : b: : c: d, we have bc = ad.

From this relation we see that if any three of the four quantities are given, the fourth can be determined.

2. Fourth proportional If a: b: :c :x, x is called the fourth proportional of a, b, c.

We have, a/b = c/x or, x = bc/a

Thus, fourth proportional of a, b, c is b c / a.

Illustrational 1 Find a fourth proportional to the numbers 2, 5, 4.

Solution Let x be the fourth proportional, then

2 : 5 : : 4 : x or 2/5 = 4/x.

(x = 5 4 /2 = 10.

3. Third proportional If a: b: : c: x, x is called the third proportional of a, b.

We have, a/b= b/x or x= b2/a.

Thus, third proportional of a. b is b2/a

Illustration 2 Find a third proportional to the numbers 2.5, 1.5

Solution Let x be the third proportional, then

2.5 : 1.5 : :1.5 : x or 2.5/1.5= 1.5/x.

(x = 1.5 1.5/2.5 = 0.9

4. Mean Proportional If a: x: : x: b, x is called the mean or second proportional of a, b.

We have, a/x =x/b or x2 = ab or x = (ab

(Mean proportional of a and b is (ab.

We also say that a, x, b are in continued proportion

Illustration 3 Find the mean proportional between 48 and 12.

Solution Let x be the mean proportional. Then, 48 : x : : x : 12 or, 48/x = x/12 or, x2= 576 or, x=24.

5. If a/b = c/d, then

(i) (a + b)/b = (c +d)/d (Componendo)

(ii) (a b)/b = (c-d)/d (Dividendo)

(iii) (a + b)/a-b = c +d/c-d = (Componendo and Dividendo)

(iv) a/b = a + c/b+d = (a c)/b-d

Illustration 4 The sum of two umber is c and their quotient is p/q. Find the numbers.

Solution Let the numbers be x, y.

Given: x + y = c(1)

and, x/y = p/q

(2)

(x/ x+y = p/p+q ( x/c = p/p+q [Using (1)]

x = pc/p +q.

SOME USEFUL SHORT-CUT METHODS

1. (a) If two numbers are in the ratio of a: b and the sum of these numbers is x, then these numbers will be ax/ a + b and bx/ a+b, respectively. or

If in a mixture of x liters of, two liquids A and B in the ratio of a: b, then the quantities of liquids A and B in the mixture will be ax / a + b litres and bx/ a + b litres, respectively.

(b) If three numbers are in the ratio a : b: c and the sum of these numbers is x, then these numbers will be ax / a + b + c , bx / a + b + c and cx / a + b + c, respectively.

Explanation

Len the three numbers in the ratio a: b: c be A, B and C. Then,

A = ka, B = kb, C =kc and, A + B + C = ka + kb + kc = x

( k(a+b+c) = x ( k = x / a + b+ c.

(A = ka = ax / a + b+ c.

B = kb = bx / a + b+ c.

C = kc = cx / a + b+ c.

Illustration 5 Two numbers are in the ratio of 4 : 5 and the sum of these numbers is 27. Find the two numbers.

Solution Here, a = 4, b = 5, and x = 27.

The first number = ax / a + b = 4 ( 27 / 4+5 = 12.

and, the second number = bx / a + b = 5 ( 27 / 4+5 = 15.

Illustration 6 Three numbers are in the ratio of 3: 4 : 8: and the sum of these numbers is 975. Find the three numbers.

Solution Here, a = 3, b = 4, c = 8 and x = 975

( The first number = ax / a + b+ c = (3 (975)/ 3 + 4 + 8 = 195.

The second number = bx / a + b+ c = (4 (975)/ 3 + 4 + 8 = 260.

and, the third number = cx / a + b+ c = (8 (975)/ 3 + 4 + 8 = 520.

2. If two numbers are in the ratio of a : b and difference between these is x, then these numbers will be

a) ax/ a-b and bx/ a-b, respectively (where a > b).

b) ax/ a-b and bx/ a-b, respectively (where a < b).

Explanation

Let the two numbers be ak and bk.

Let a > b.

Given : ak bk = x

( (a b)k = x or k = x / (a-b).

Therefore, the two numbers are ax / a-b and bx/ a-b.

Illustration 7 Two numbers are in the ratio of 4 : 5. If the difference between these numbers is 24, then find the numbers.

Solution Here, a = 4, b = 5 and x = 24.

(The first number = ax/ b-a = 4 (24/5- 4 = 96 and, the second number = bx/ b-a = 5( 24 / 5-4 = 120.

3. (a). If a : b = n1 : d1 and b : c = n2 : d2, then

a : b : c = (n1(n2) : (d1( n2) : (d1 ( d2).

(b). If a : b = n1 : d1, b : c = n2 : d2, and c : d = n3 : d3 then

a : b : c : d= (n1( n2( n3) : (d1( n2( n3 ) : (d1( d2( n3 ) : (d1( d2( d3 ).

Illustration 8 If A : B = 3 : 4 and B : C = 8 : 9, find A : B : C.

Solution Here, n1 = 3, n2 =8, d1 =4 and d2 = 9.

(a : b : c = (n1(n2) : (d1(n2) : (d1(d2)

= (3(8) : (4(8) : (4(9)

= 24 : 32 : 36 or, 6: 8 : 9.

Illustration 9 If A : B = 2 : 3, B:C = 4 : 5 and C : D = 6 : 7, find A :D.

Solution Here, n1 = 2, n2 = 4, n3 = 6, d1 = 3, d2 = 5 and d3 = 7.

(A : B : C : D = (n1(n2(n3) : (d1(n2(n3) : (d1 ( d2 ( n3) : (d1 ( d2 ( d3)

= (2 ( 4 ( 6) : (3 ( 4 ( 6) : (3 ( 5 ( 6) : (3 (5 (7)

= 48 : 72 : 90: 105: or, 16: 24 : 30 ; 35. Thus, A : D = 16 : 35.

4. (a) The ratio between two numbers is a : b. If x is added to each of these numbers, the ratio becomes c : d. The two numbers are given as:

ax(c d) / ad bc and bx(c- d) / ad bc.

Explanation

Let two number be ak and bk.

Given : ak +x / bk+x = c/d (akd +dx= cbk + cx

( k(ad bc) = x(c d)

(k =x(c-d)/ ad bc.

Therefore, the two numbers are ax(c-d) / ad-bc and bx(c-d) / ad- bc

(b) The ratio between two numbers is a : b. if x is subtracted from each of these numbers, the ratio becomes c : d.

The two numbers are given as:

ax(d-c) / ad-bc and bx(d-c) / ad- bc

Explanation

Let the two numbers be ak and bk.

Given : ak-x/bk-x = c/d ( akd-xd = bck-xc

( k(ad-bc) = x(d-c)

( k = x(d-c)/ad-bc.

Therefore, the two numbers are ax(d-c)/ad-bc and bx(d-c)/ad-bc

Illustration 10 Given two numbers which are in the ratio of 3 : 4, If 8 is added to each of them, their ratio is

changed to 5 : 6. Find two numbers.

Solution We have,

a : b = 3 : 4, c : d = 5 : 6 and x = 8.

(The first number = ax(c d)/ ad bc = 3 ( 8((5-6) / (3( 6- 4 (5) = 12.

and, the second number = bx(c d)/ ad bc

= 4 ( 8((5-6) / (3( 6- 4 (5) = 16.

Illustration 11 The ratio of two numbers is 5 : 9. If each number is decreased by 5, the ratio becomes 5: 11.

Find the numbers.

Solution We have,

a : b = 5: 9, c: d = 5: 11 and x =5.

( The first number = ax (d c)/ ad bc

= 5 ( 5((11-5)/ (5(11- 9 (5) = 15.

and, the second number= bx(d-c)/ad-bc

9(5((11-5)/(5(11-9(5)= 27.

5. (a) If the ratio of two numbers is a: b, then the numbers that should be added to each of the numbers in order to make this ratio c : d is given by ad-bc/ c-d.

Explanation

Let the required number be x.

Given: a+x/ b+x = c/d ( ad+ xd = bc + xc

( x(d-c) =bc ad or, x =ad-bc/c-d.

(b)If the ratio of two numbers is a : b, then the number that should be subtracted from each of the numbers in order to make this ratio c : d is given by bc-ad/c-d.

Explanation

Let the required number be x.

Given: a-x/ b-x = c/d ( ad- xd = bc - xc

( x(c-d) =bc ad or, x = bc-ad/c-d.

Illustration 12 Find the number that must be subtracted from the terms of the ratio 5 : 6 to make it equal to 2 : 3.

Solution We have a : b= 5 : 6 and c: d =2 : 3.

( The required number = bc-ad/ c-d

= 6 ( 2-5(3/ 2-3 =3.

Illustration 13 Find the number that must be added to the terms of the ratio 11 : 29 to make it equal to 11 : 20.

Solution We have, a : b= 11 : 29 and c: d =11: 20.

( The required number = ad-bc/ c-d

= 11 ( 20-29(11/ 11-20 = 11.

PERCENTAGES

Introduction

The term per cent means per hundreds or for every hundred. It is the abbreviation of the Latin phrase per centum. Scoring 60 per cent marks means out of every 100 marks the candidate scored 60 marks.

The term per cent is sometimes abbreviated as p.c. The symbol % is often used for the term per cent.

Thus, 40 per cent will be written as 40%.

A fraction whose denominator is 100 is called a percentage and the numerator of the fraction is called rate per cent,

e.g. 5/100 and 5 per cent means the same thing, i.e. 5 parts out of every hundred parts.

1. To Convert a fraction into a per cent:

to convert any fraction l/m to rate per cent, multiply it by 100 and put % sign, i.e. l/m ( 100%

2. To Convert a Percent into a Fraction:

To convert a per cent into a fraction , drop the per cent sign and divide the number by 100.

3. To find a percentage of a given number: x % of given number (N) = x/100 ( N.

Some useful shortcut methods

1. (a) if A is x% more than that of B, then B is less than that of A by

(b) If a is x% less than that of B, then B is more than that of A by

2. If a is x% of C and B is y% of C, then A = x/y ( 100% of B.

3. (a) If two numbers are respectively x% and y% more than a third number, then the first number is of the second and the second is of the first.

(b) If two numbers are respectively x% and y% less than a third number, then the first number is of the first.

4. (a) If the price of a commodity increases by P%, then the reduction in consumption so as not to increase the expenditure is .

(b) If the price of a commodity decreases by p%, then the increase in consumption so as not to decrease the expenditure is .

5. If a number is changed (increased/decreased) successively by x% and y%, then net% change is given by (x+y+(xy/100))% which represents increase or decrease in value according as the sign is +ve or ve.

If x or y indicates decrease in percentage, then put ve sign before x or y, otherwise +ve sign.

6. If two parameters A and B are multiplied to get a product and if A is changed (increased/decreased) by x% and another parameter B is changed (increased/decreased) by y%, then the net% change in the product (A ( B) is given (x+y+(xy/100))% which represents increase or decrease in value according as the sign in +ve or ve.

If x or y indicates decrease in percentage, then put ve sign before x or y, otherwise +ve sign.

7. If the present population of a town (or value of an item) be P and the population (or value of item) changes at r% per annum, then

(a) Population (or value of item) after n years =

(b) Population (or value of item) n years ago =

where r is +ve or ve according as the population (or value of item) increase or decreases.

8. If a number A is increased successively by x% followed by y% and then by z%, then the final value of A will be

In case a given value decreases by any percentage, we will use a negative sign before that.

9. In an examination, the minimum pass percentage is x%. If a student secures y marks and fails by z marks, then the maximum marks in the examination is .

10. In an examination x% and y% students respectively fail in two different subjects while z% students fail in both the subjects, then the percentage of students who pass in both the subjects will be (100-(x+y-z))%.

EXERCISE

1. What is 20% of 50% of 75% of 70?

1. 5.25

2. 6.75

3. 7.25

4. 5.50

2. Ram sells his goods 25% cheaper than Shyam and 25% dearer than Bram. How much percentage is Brams goods cheaper than Shyams?

1. 33.33%

. 50%

3. 66.66%4. 40%

3. In an election between 2 candidates, Bhiku gets 65% of the total valid votes. If the total votes were 6000. What is the number of valid votes that the other candidate Mhatre gets if 25% of the total votes were declared invalid?

1. 1625

2. 1575

3. 1675

4. 1525

4. In a medical certificate, by mistake a candidate gave his height as 25% more than normal. In the interview panel, he clarified that his height was 5 feet 5 inches. Find the percentage correction made by the candidate from his stated height to his actual height.

1. 20

2. 28.56

. 25

4. None

5. Arjit Sharma generally wears his fathers coat. Unfortunately, his cousin Shaurya poked him one day that he was wearing a coat of length more than his height by 15%. If the length of Arjits fathers coat is 120 cm then find the actual length of his coat.

1. 105

2. 108

3. 104.344. 102.72

6. ***In a mixture of 80 liters of milk and water, 25% of the mixture is milk. How much water should be added to the mixture so that milk becomes 20% of the mixture?

1. 20 liters

2. 15 liters3. 25 liters4. None

7. 50% of a% of b is 75% of b% of c. Which of the following is c?

1. 1.5a

2. 0.667a3. 0.5a

4. 1.25a

8. ***A landowner increased the length and the breadth of a rectangular plot by 10% and 20% respectively. Find the percentage change in the cost of the plot assuming land prices are uniform throughout his plot.

1. 33%

2. 35%

3. 22.22%4. None

9. The height of a triangle is increased by 40%. What can be the maximum percentage increase in length of the base so that the increase in area is restricted to a maximum of 60%?

1. 50%

2. 20%

3. 14.28%4. 25%

10. The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboids.

1. 77%

2. 75%

3. 88%

4. 98%

11. The salary of Amit is 30% more than that of Varun. Find by what percentage is the salary of Varun less than that of Amit?

1. 26.12%

2. 23.07%3. 21.23%4. None

12. ***The price of sugar is reduced by 25% but in spite of the decrease, Aayush ends up increasing his expenditure on sugar by 20%. What is the percentage change in his monthly consumption of sugar?

1. +60%

2. 10%

3. +33.33%4. 50%

13. The price of rice falls by 20%. How much rice can be bought now with the money that was sufficient to buy 20 kg of rice previously?

1. 5kg

2. 15 kg

3. 25 kg

4. 30 kg

14. 30% of a number when subtracted from 91, gives the number itself. Find the number.

1. 60

2. 65

3. 70

4. None

15. ***At an election, the candidate who got 56% of the votes cast won by 144 votes. Find the total number of voters on the voting list if 80% people cast their vote and there were no invalid votes.

1. 360

2. 720

3. 1800

4. 1500

16. The population of a village is 1,00,000. The rate of increase is 10% per annum. Find the population at the start of the third year?

1. 1, 33,100

2. 1, 21, 0003. 1, 20, 0004. None

17. the population of the village of Gavas Is 10, 000 at this moment. It increases by 10% in the first year. However, in the second year, due to immigration, the population drops by 5%. Find the population at the end of the third year if in the third year the population increases by 20%.

1. 12, 340

2. 12, 5403. 1, 27, 5404. 12, 340

18. A man invests Rs.10,000 in some shares in the ratio 2:3:5 which pay dividends of 10%, 25% and 20% (on his investment) for that year respectively. Find his dividend income.

1. 1900

2. 2000

3. 2050

4. 1950

Averages & MixturesWhenever we are asked the marks scored by us in any examination, we usually tell the marks in percentage, taking the percentage of total marks of all subjects. This percentage is called average percentage. Also, in a class, If there are 100 students, instead of knowing the age of individual student, we usually talk about average age.

The average or mean or arithmetic of a number of quantities of the same kind is equal to their sum divided by the number of those quantities. For example, the average of 3, 11, 15, 18,19, and 23 is

3 + 9 +11+ 15+ 18+ 19+ 23+ /7 = 98/7 = 14.

SOME BASIC FORMULAE

1. Average = sum of quantities/ Number of quantities

2. Sum of quantities = Average ( Number of quantities

3. Number of quantities = Sum of quantities/ Average

Illustration 1 A man purchased 5 toys at the rate of Rs 200each, 6 toys at the rate of Rs 250each and 9 toys at the rate of Rs 300 each. Calculate the average cost of one toy.

Solution Price of 5 toys = 200 ( 5 = Rs 1000

Price of 6 toys = 250 ( 6 = Rs 1500

Price of 9 toys = 300 ( 9 = Rs 2700

Average price of 1 toy = 1000 + 1500 + 2700/ 20

= 5200/20 = Rs 260.Illustration 2 The average marks obtained by 200 students in a certain examination are 45. Find the total marks.

Solution Total marks

= Average marks ( Number of students

= 200 ( 45 = 900.

Illustration 3 Total temperatures for the month of September is 8400C, If the average temperature of that month is 280C, find of how many days is the month of September.

Solution Number of days in the month of September

= Total temperature/ Average temperature = 840/28

= 30days.

SOME USEFUL SHORTCUT METHODS

1. Average of two or more groups taken together

a) If the number of quantities in two groups be n1 and n2 and their average is x and y, respectively, the combined average (average of all of then put together) is

n1x +n2y / n1 + n2

Explanation

No. of quantities in fist group = n1 Their average = x

Sum = n1 ( xNo. of quantities in second group = n2

Their average = y Sum = n2 ( yNo. of quantities in the combined group = n1+n2Total sum (sum of quantities of first group and second group) = n1x+n2y

Combined Average = n1x+n2y./ n1 +n2.

b). If the average of n1 quantities is x and the average of n2 quantities out of them is y, the average of remaining group (rest of the quantities) is

n2x n2y/ n1- n2.

Explanation

No. of quantities = n1Their average = x

( Sum = n1x

No of quantities taken out n2Their average = y

( Sum = n2y

Sum of remaining quantities = n1x n2y

No. of remaining quantities = n1 n2( Average of remaining group = n1x n2y/ n1 n2Illustration 4 The average weight of 24 students of section A of a class is 58 kg whereas the average weight of 26 students of section B of the same class is 60. 5 kg. Find the average weight of all the 50 students of the class.

Solution Here, n1 = 24, n2 = 26, x = 58 and y = 60.5.

( Average weight of all the 50 students

= n1x+n2y/ n1 +n2

= 24 58 + 2660.5 / 24+26

= 1392 +1573/ 50 = 2965/ 50 =59.3kg.

Illustration 5 Average salary of all the 50 employees including 5 officers of a company is Rs 850. If the average salary of the officers is Rs 2500, find of the class.

Solution Here, n1 = 50, n2 =5, x = 850and y = 2500.

( Average salary of the remaining staff

= n1x-n2y/ n1-n2 = 50(850 -5(2500 / 50-5

= 42500-12500/ 45 = 30000/ 45

= Rs 667(approx)

2. If (x is the average of x1, x2, , xn, then

a) The average of x1 + a, x2 + a, ., xn + a is (x +a.

b) The average of x1 - a, x2 - a, ., xn - a is (x -a.

c) The average of ax1, ax2,.,axn is ax, provided a 0.

d) The average of x1 / a, x2 / a, ., xn / a is(x /a, provided a 0.

Illustration 6 The average value of six numbers 7, 12, 17, 24, 26 and 28 is 19. If 8 is added to each number, what will be the new average?

Solution The new average = (x +a.

= 19+8 = 27.

Illustration 7 The average value of x numbers is 5x. If x 2 is subtracted from each given number, what will be the new average?

Solution The new average =(x -a.

= 5x- (x-2) = 4x +2.

Illustration 8 The average of 8 numbers is 21.If each of the numbers multiplied by 8, find the average of a new set of numbers.

Solution The average of a new set of numbers

= a(x = 8( 21 = 168.

3. The average of n quantities is equal to x. If one of the given quantities whose value is p, is replaced by a new quantity having value q, the average becomes y, then q = p+n(y-x)

Illustration 9 The average weight of 25 persons is increased by 2 kg when one of them whose weight is 60kg, is replaced by a new person. What is the weight of the new person?

Solution The weight of the new person

= p + n(y-x)

= 60 + 25(2)= 110kg

4. a). The average of n quantities is equal to x. When a quantity is removed, the average becomes y. The value of the removed quantity is n(x- y)+y.

b) The average of n quantities is equal to x. When a quantity is added, the average becomes y. The value of the new quantity is n(y-x)+y

Illustration10 The average are of 24 students and class teacher is16 years, If the class teachers age is excluded, the average age reduces by 1 year. What is the age of the class teacher?

Solution The age of class teacher

= n(x- y) + y

= 25(16 15) + 15 = 40 years.

Illustration 11 The average age of 30 children in a class is 9 years. If the teachers age be included, the average age becomes 10years. Find the teachers age.

Solution The teachers age

= n(y- x) + y

= 30(10 9) +100 = 40 years.

5. a).The average of first n natural numbers is (n +1) /2

b). The average of square of natural numbers till n is (n +1)(2n+1)/6.

c). The average of cubes of natural numbers till n is n(n +1)2/4

d). The average of odd numbers from 1 to n is (last odd number +1) / 2

e). The average of even numbers from 1 to n is (last even number + 2) / 2.

Illustration 12 Find the average of first 81natural number.

Solution The required average

= n + 1/ 2 = 81 + 1 /2 = 41.

Illustration 13 What is the average of squares of the natural numbers from 1 to 41?

Solution The required average

= (n+1)(2n+1)/ 6 = (41+1)(2(41+1)/ 6

= 42 ( 83/ 6 = 3486/ 6 = 581Illustration 14 Find the average of cubes of natural numbers from 1 to 27.

Solution The required average

= n(n +1)2 / 4 = 27((27+1)2 / 4

27 ( 28 ( 28 / 4 = 21168 / 4 = 5292.

Illustration 15 What is the average of odd numbers from 1 to 40?

Solution The required average

= last odd number + 1/ 2 = 39 +1/ 4 =20.

Illustration 16 What is the average of even numbers from 1 to 81?

Solution The required average

= last even number + 2/ 2 = 80+2 = 41.

6. a).If n is odd: The average of n consecutive numbers, consecutive even numbers or

consecutive odd numbers is always the middle number.

b). If n is even: The average of n consecutive numbers, consecutive even numbers or consecutive odd numbers is always the average of the middle two numbers.

c). The average of first n consecutive numbers is (n+1).

d). The average of first n consecutive odd numbers is n.

e). The average of squares of first n consecutive even number is2 (n+1)(2n+1) / 3.

f). The average of squares of consecutive even number till n is (n+)(n+2) / 3.

g). The average of squares of squares of consecutive odd numbers till n is n(n+2)/ 3.

h). If the average of n consecutive numbers is m, then the difference between the smallest and the largest number is 2(n-1).

Illustration 17 Find the average of 7 consecutive numbers 3, 4, 5, 6, 7, 8, 9.

Solution The required average= middle number=6.

Illustration 18 Find the average of consecutive odd numbers 21, 23, 25, 27, 29, 31, 33, 35.

Solution The required average

= average of middle two numbers

= average of 27 and 29

= 27+29 / 2 = 28.

Illustration 19 Find the average of first 31 consecutive even numbers.

Solution The required average = (n+1) = 31+ 1= 32.

Illustration 20 Find the average of first 50 consecutive odd numbers.

Solution The required average = n = 50.EXERCISE

1. The average of 13 papers is 40. The average of the first 7 papers is 42 and of the last seven papers is 35. Find the marks obtained in the 7th paper?

(A) 23

(B) 38

(C) 19

(D) None of these

2. The average age of the Indian cricket team playing the Nagpur test is 30. The average age of 5 of the players is 27 and that of another set of 5 players, totally different from the first five, is 29. If it is the captain who was not included in either of these two groups, then find the age of the captain.

(A) 75

(B) 55

(C) 50

(D) Cannot be determined

3. A bus goes to Ranchi from Patna at the rate of 60 km per hour. Another bus leaves Ranchi for Patna at the same times as the first bus at the rate of 70 km per hour. Find the average speed for the journeys of the two buses combined if it is known that the distance from Ranchi to Patna is 420 kilometers.

(A) 64.615 kmph

(B) 64.5 kmph

(C) 63.823 kmph

(D) 64.82 kmph

4. A train travels 8 km in the first quarter of an hour, 6 km in the second quarter and 40 km in the third quarter. Find the average speed of the train per hour over the entire journey.

(A) 72 km/h

(B) 18 km/h

(C) 77.33 km/h

(D) 78.5 km/h

5. The average weight of 6 men is 68.5 kg. If I is known that Ram and Tram weigh 60 kg each, find the average weight of the others.

(A) 72.75 kg

(B) 75 kg

(C) 78 kg

(D) None of these

6. The average score of a class of 40 students is 52. What will be the average score of the rest of the students if the average score of 10 of the students is 61.

(A) 50

(B) 47

(C) 48

(D) 49

7. The average age of 80 students of IIM, Bangalore of the 1995 batch is 22 years. What will be the new average if we include the 20 faculty members whose average age is 37 years?

(A) 32 years

(B) 24 years

(C) 25 years

(D) None of these

8. Out of the three numbers, the first is twice the second and three times the third. The average of the three numbers is 88. The smallest number is

(A) 72

(B) 36

(C) 42

(D) 48

9. The sum of three numbers is 98. If the ratio between the first and second is 2 : 3 and that between the second and the third is 5 : 8, then the second number is

(A) 30

(B) 20

(C) 58

(D) 48

10. The average height of 30 girls out of a class of 40 is 160 cm and that of the remaining girls is 156 cm. The average height of the whole class is

(A) 158 cm

(B) 158.5 cm

(C) 159 cm

(D) 157 cm

11. The average weight of 6 persons is increased by 2.5 kg when one of them whose weight is 50 kg is replaced by a new man. The weight of the new man is

(A) 65 kg

(B) 75 kg

(C) 76 kg

(D) 60 kg

12. The average age of A, B C and D five years ago was 45 years. By including X, the present average age of all the five is 49 years. The present age of X is

(A) 64 years

(B) 48 years

(C) 45 years

(D) 40 years

13. The average salary of 20 workers in an office is Rs. 1900 per month. If the managers salary is added, the average salary becomes Rs. 2000 per month. What is the managers annual salary?

(A) Rs. 24, 000

(B) Rs. 25,200

(C) Rs. 45,600

(D) None of these

14. The average weight of a class of 40 students is 40 kg. If the weight of the teacher be included, the average weight increases by 500 gm. The weight of the teacher is

(A) 40.5 kg

(B) 60 kg

(C) 62 kg

(D) 60.5 kg

15. In a Infosys test, a student scores 2 marks for every correct answer and loses 0.5 marks for every wrong answer. A student attempts all the 100 questions and scores 120 marks. The number of questions he answered correctly was

(A) 50

(B) 45

(C) 60

(D) 68

16. The average of the first ten natural numbers is

(A) 5

(B) 5.5

(C) 6.5

(D) 6

17. The average of the first ten whole numbers is

(A) ****4.5

(B) 5

(C) 5.5

(D) 4

18. The average of the first ten even numbers is

(A) 18

(B) 22

(C) 9

(D) 11

19. The average weight of a class of 30 students is 40 kg. If, however, the weight of the teacher is included, the average become 41 kg. The weight of the teacher is

(A) 31 kg

(B) 62 kg

(C) 71 kg

(D) 70 kg

20. 30 oranges and 75 apples were purchased for Rs. 510. If the price per apple was Rs. 2, then the average price of oranges was

(A) Rs. 12

(B) Rs. 14

(C) Rs. 10

(D) Rs. 15

21. A batsman made an average of 40 runs in 4 innings, but in the fifth inning, he was out on zero. What is the average after fifth innings?

(A) 32

(B) 22

(C) 38

(D) 49

22. The average weight of a school of 40 teachers is 80 kg. If, however, the weight of the principle be included, the average decreases by 1 kg. What is the weight of the principal?

(A) 109 kg

(B) 29 kg

(C) 39 kg

(D) None of these

23. The average age of Ram and Shyam is 20 years. Their average age 5 years hence will be

(A) 25 years

(B) 22 years

(C) 21 years

(D) 20 years

24. The average of 20 results is 30 and that of 30 more results is 20. For all the results taken together, the average is

(A) 25

(B) 50

(C) 12

(D) 24

25. The average of 5 consecutive numbers is 18. The highest of these numbers will be

(A) 24

(B) 18

(C) 20

(D) 22

26. Three years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average of the family is the same today. What is the age of the baby?

(A) 1 years

(B) 2 years

(C) 6 months

(D) 9 months

27. Varun average daily expenditure is Rs. 10 during May, Rs. 14 during June and Rs. 15 during July. His approximate daily expenditure for the 3 months is

(A) Rs. 13 approximately(B) Rs. 12

(C) Rs. 12 approximately

(D) Rs. 10

28. A ship sails out to a mark at the rate of 15 km per hour and sails back at the rate of 20 km/h. What is its average rate of sailing?

(A) 16.85 km

(B) 17.14 km

(C) 17.85 km

(D) 18 km

29. The average temperature on Monday, Tuesday and Wednesday was 41 0C and on Tuesday, Wednesday and Thursday it was 40 0C. If on Thursday it was exactly 39 0 C, then on Monday, the temperature was

(A) 42 0C

(B) 46 0C

(C) 23 0C

(D) 26 0C

30. The average of 20 results is 30 out of which the first 10 results are having an average of 10. The average of the rest 10 results is

(A) 50(B) 40

(C) 20 (D) 25

31. A man had seven children. When their average age was 12 years a child aged 6 years died. The average age of the remaining 6 children is

(A) 6 years

(B) 13 years

(C) 17 years

(D) 15 years

32. The average weight of 35 students is 35 kg. If the teacher is also included, the average weight increases to 36 kg. The weight of the teacher is

(A) 36 kg

(B) 71 kg

(C) 70 kg

(D) 45 kg

33. The average of x, y and z is 45. x is as much more than the average as y is less than the average. Find the value of z.

(A) 45

(B) 25

(C) 35

(D) 15

34. The average salary per head of all the workers in a company is Rs. 95. The average salary of 15 officers is Rs. 525 and the average salary per head of the rest is Rs. 85. Find the total number of workers in the workshop.

(A) 660

(B) 580

(C) 650

(D) 46

PROFIT AND LOSSBusiness transactions have now-a-days become common feature of life. When a person deals in the purchase and sale of any item, he either gains or loses some amount generally. The aim of any business is to earn profit. The commonly used terms in dealing with questions involving sale and purchase are:

Cost Price The cost price of an article is the price at which an article has been purchased. It is the abbreviated as C.P.

Selling Price The selling price of an article is the price at which an article has been sold. It is abbreviated as S.P.

Profit or Gain If the selling price of an article is more that the cost price, there is a gain or profit.

Thus, Profit or Gain = S.P- C.P.

Loss If the cost price of an article is greater than the selling price, the suffers a loss.

Thus, Loss = C.P- S.P.

Note that profit and loss are always calculated with respect to the cost price of the item.

Illustration 1. (i)If C.P. = Rs. 235, S.P. = Rs. 240, then profit = ?

(ii) If C.P. = Rs. 116, S.P. = Rs. 107, then loss = ?

Solution (i) Profit = S.P.- C.P. =Rs. 240- 235 =Rs.5.

(ii) Loss = C.P.- S.P. = Rs. 116- 107 =Rs.9.SOME BASIC FORMULAE

1. Gain on Rs. 100 is Gain per cent

Gain% = (Gain ( 100)/C.P

Loss on Rs. 100 is Loss per cent

Loss% = (Loss ( 100)/C.P

Illustration 2 The cost price of a shirt is Rs. 200 and selling price is Rs. 250. Calculate the % profit.

Solution We have, C.P. = Rs. 200, S.P = Rs. 250.

Profit = S.P.- C.P. = 250- 200 =Rs.50.

( Profit% = profit( 100/ C.P = 50( 100/ 250 = 25%

Illustration 3 Anu bought a necklace for Rs. 750 and sold it for Rs. 675. Find her percentage loss.

Solution Here, C.P. = 750, S.P. = Rs. 675.

Loss= C.P- S.P. = 750-675 = Rs. 75.

( Loss% = Loss ( 100/ C.P = 75( 100/ 750 = 10%

2. When the selling price and gain% are given:

C.P = 100(S.P / (100+Gain%)

3. When the cost and gain per cent are given;

S.P = (100+Gain%)(C.P/ 100

4. When the cost and loss per cent are given:

S.P = (100-Loss%)(C.P / 100

5. When the selling price and loss per cent are given:

C.P =(100)(S.P / (100-Loss%)

Illustration 4 Mr. Sharma buys a cooler for Rs. 4500. For how much should he so that there is a gain of 8%?Solution We have, C.P. = Rs. 4500, gain% = 8%

(S.P = (100+Gain%/100)(C.P.

= (100+ 8/ 100) ( 4500

108/100 ( 4500 = Rs. 4860

Illustration 5 By selling a fridge Rs. 7200, Pankaj loses 10%. Find the cost price of the fridge.Solution We have, S.P. = Rs. 7200, loss% = 10%.

( C.P =(100/100-Loss%)(S.P.

= (100/100-10) ( 7200

100/90 ( 7200= Rs. 8000.

Illustration 6 By selling a pen for Rs. 99, Mohan gains 12 %. Find the cost price of the pen.

Solution Here, S.P. = Rs. 99, gain% = 12 % or 25/2%.

(C.P =(100/100+Gain%)(S.P.

= (100/100+25/2) (99

= (100(2/ 225) (99 =Rs. 88

SOME USEFUL SHORT-CUT METHODS

1. If a man buys x items for Rs. y and sells z items for Rs. w, then the gain or loss per cent made by him is

(xw/zy -1) ( 100%.

ExplanationS.P. of z items = Rs. w

S.P. of x items = Rs. w/z x

Net profit =w/z x-y.

(% profit = w/z x-y/y (100%

i.e.(xw/zy -1) (100,

which represent loss, if the result is negative.

Note: In the case of gain per cent the result obtained bears positive sign whereas in the case of loss per cent the result obtained bears negative sign.

How to remember:

1. Cross-multiply the numbers connected by the arrows (xw and zy)

2. Mark the directon of the arrows for crossmultiplicaton. The arrow going down forms the numerator while the arrow going up forms the denominator (xw/ zy).

Illustration 7 If 11 oranges are bought for Rs. 10 and sold at 10 for Rs. 11 what is the gain loss%?

Solution

% profit= (xw/zy -1) ( 100%

= (11(11/10(10-1) (100%

= 21/100 (100% = 21%

Illustration 8 A fruit seller buys apples at the rate of Rs 12 per dozen and sells them at the rate of 15 for Rs.12. Find his percentage gain or loss.

Solution% gain or loss = (xw/ zy -1) ( 100%

= (12( 12/15 (12 -1) (100%

= -36/144 ( 100% = -25%

Since the sign is ve, there is a loss of 25%.

2. If the cost price of m articles is equal to the selling price of n articles, then

% gain or loss = ( m-n/n) (100

[If m > n, it is gain and if m