Module 2: Mathematics

Part A: Ratio And Proportion

Let N be the given number and it is to be divided into two

parts in the ratio a : b. Then,

First part =

aN/(a+b)

Second

part=bN/(a+b)

In case, N is to be

divided into three

parts in the ratio a

: b : c. Then,

First

part=aN/(a+b+c)

second part=bN/(a+b+c)

Third part=cN/(a+b+c)

Example 1: If x : y = 3 : 8 and y : z = 4 : 9, find (i) x : z and

(ii) x : y : z.

Solution 1:

(i) x : y=3 : 8 x/y=3/8 eqn(1)

(ii) y : z=4 : 9 y/z=4/9 eqn(2)

Multiplying (1) and (2), we have

(x / y ) .( y / z) = (3 / 8) . (4/9) = 1 / 6

x : z = 1 : 6 = 3 : 18

x : y = 3 : 8

and y : z = 4 : 9 gives

y : z = 8 : 18

x : y : z = 3 : 8 : 18.

Example 2: Two numbers are in the ratio 3:5. If 8 is added

to each number, the ratio becomes 2:3, find the numbers.

Solution2: Since the numbers are in the ratio 3:5, let the

required numbers be 3x and 5x.

now , (3x + 8) / (5x + 8) = 2 / 3

9x + 24 = 10x + 16

x=8

The required numbers are 3(8) = 24 and 5(8) = 40.

Example 3: In a one-day

cricket match between

India and South Africa,

Kumble conceded 27 runs

in 9 overs and 2 balls,

while Pathan conceded 31

runs in 10 overs. Who had

a better performance?

Solution 3: Kumble

conceded 27 runs in 9 overs and 2 balls,

Runs conceded by Kumble in 1 over = 27 * (3/8) = 81 / 28

i.e., in 9 2/3 = 9 1/3 = 28 / 3

Similarly, runs conceded by Pathan in 1 over = 31 / 10

Since, 81 / 28 < 31 / 10

Kumble's performance was better.

Evaluate Your Grasp:

1. What should be added to each term of the ratio 7:13 so

that the ratio becomes 2:3?

2. Find the greatest ratio among the following: 14:23, 5:12,

61:92

3.The ages of two brothers are in the ratio 5:7. If the

elder is 12 years 3 months old, what is the age of the

younger brother?

4.The monthly salaries of two

persons are in the ratio 3:5. If

each receives a raise of Rs 200 in

the salary, the ratio is altered to

13:21. Find their respective

increased salaries.

5. The ratio of the present ages

of the

father

and the son is 5:2. After 8

years, the ratio of their ages

would be 2:1. Find their present

ages.

Answers

1. 5

2. 61:92

3. 8 years 9 months

4. Rs 2,600; Rs 4,200

5. 40 years; 16 years.

Example 4: Divide Rs 1,870 into

three parts such that half of the

first part, one-third of the second

part and one-sixth of the third

part are equal

Solution 4: Let the three parts of

Rs 1,870 be Rs x, Rs y & Rs z.

Then, x + y + z = 1,870 ..(1)

And, x/2 = y/3 = z/6

Let, x/2 = y/3 = z/6 = K(Say)

then x=2K , y=3K , z = 6K..(2)

Substituting these values in (1), we have

2K + 3K + 6K = 1,870

11K = 1,870

K = 170

The three parts are x= 2 x 170 = 340, y= 3 x 170= 510 and

z= 6 x 170= 1,020

Evaluate Your Grasp

Divide Rs 2,911 among A, B and C

such that A receives 30% more

than B and B receives 20% less

than C.

[Hint: Here, A : B : C = 26 : 20 :

25.] Answer

Rs 1,066; Rs 820; Rs 1,025

Example 5: A bag contains Rs 147 in the form of 1-rupee,

50 paise and 25 paise coins in the ratio 1 : 3 : 4. Find the

number of each type of coins.

Solution 5: As 1-rupee, 50 paise and

25 paise coins are in the ratio 1 : 3 : 4,

let the number of these coins be x, 3x

and 4x respectively. According to the

question, x + 3x*1/2 + 4x*1/4 = 147

Therefore, x=42, and

The number of 1-rupee coins = 42

The number of 50 paise coins = 3 x 42 = 126

The number of 25 paise coins = 4 x 42 = 168.

Example 6: The work done by (x - 3) men in (2x + 1) days

and the work done by (2x + 1) men in (x + 4) days is in the

ratio 3 : 10, find x.

Solution 6:

Let K units be the work done by 1 man in 1 day. Then,

( x - 3)(2x + 1)K / (2x + 1)(x + 4)K = 3/10

10(x - 3)(2x + 1) = 3(2x + 1)(x + 4)

10(2x^2 - 5x - 3) = 3(2x^2 + 9x + 4)

20x^2 - 50x - 30 = 6x^2 + 27 x + 12

14x^2 -77x- 42 = 0

2x^2 - 11x - 6 = 0

(x - 6)(2x + 1) = 0

x - 6 = 0 or 2x + 1 = 0

x = 6

because, x= -1/2 is an inadmissible value

Evaluate Your Grasp

1. In a certain

examination the

number of those who

passed was 4 times

the number of those

who failed . If there

had been 35 fewer

candidates and 9

more had failed, the

number would have

been in the ratio 2 : 1 find the total number of candidates.

[ Hint : Let the total number of candidates be x According to the question, Number of candidates passed = 4x/5 Number of candidates failed = x/5 in the second situation, Total number of canditates = x-35 Number of candites failed = (x/5) + 9 Number of candidates passed =(x - 35 - x/5 - 9) = (4x/5) - 44 Since the ratio is given to be 2 : 1 ,we have (4x/5) - 44 = 2(x/5 + 9) Find the value of x.]

2. A man can do as much work in 4 hours as a woman can do

in 6 hours. If they work together at a job, and receive Rs

300 for it, how should they share this amount?

[ Hint: Let the work done by a man in 4 hours be x units. Then, the work done by a man in 1 hour = x/4 units work done by a woman in 1 hour = x/6 units The ratio of wages = The ratio of work done = 1/4 : 1/6 ]

3 . A wall of bricks

tumbled down,

leaving a height of

60 cm standing. You

were told that 5/8

of the wall had

fallen. How many

centimetres of

bricks would you

have to add to

rebuild the wall?

[Hint: Standing part of the wall = (1-5/8) = 3 / 8 Ratio of the fallen part to the standing part = 5/8 : 3/8 i.e 5:3 Fallen part : 60 cm = 5:3 Fallen part = 100 cm ] Answers

1. 155 candidates

2. Rs 180; Rs 120

3. 100 cm

Proportion If four quantities a, b, c and d are such that the ratio a : b

is equal to the ratio c : d, then we say that a, b, c and d are

in proportion.

Continued Proportion

Three quantities a, band c are said to be in continued

proportion if a : b = b : c i.e.

a/b = b/c

Here, b is called the mean proportional of a and c. also , b=

+(ac)

Moreover, c is called the third proportional to a and b.

Similaly, four quantities a, b, c and d are said to be in

continued proportion if

a:b=b:c=c:d

a/b = b/c =c/d

Addition equivalence property

If a/b = c/d = e/f = .. then each ratio is equal to

(a+c+e+.) / (b+d+f+.)

Example 7: If x, 5, 10, yare in continued proportion, then

find x & y

Solution 7: Since x, 5, 10 and y are in continued proportion,

then

x / 5 = 5 / 10 = 10 / y

x / 5 = 1 / 2 = 10 / y

x = 5/2 and y = 20

Example 8: What number should be subtracted from each

of the numbers 21,38,55 and 106 so that they become in

proportion?

Solution 8: Let x be subtracted from each number so that

they may be in proportion

(21 - x) / (38 - x) =

(55 - x) / (106 - x)

(21 - x)(106 - x) =

(38 - x) (55 - x)

2,226 - 21x - 106x +

x2 = 2,090 - 38x -

55x + x2

2,226 -127x = 2,090

- 93x

34x = 136 x = 4

Hence, the required number is 4.

Evaluate Your Grasp

1. Find x , if 2.5 : 1.5 = x: 3.

2. What number should be subtracted from each of the

numbers, 23,30,57 and 78 so that the remainders are in

proportion?

3. Find the mean proportional of 3 and 27

4. Find the third proportional to 0.5, 0.25

Answers

1. x=5 2. 6

3. 9 4. 0.125

Part B: FRACTIONS

FRACTION =

Numerator/Denominator.

Example1 : Arrange 7/13,

493/971, 87/165,123/235 in

descending order

7/13=0.538

493/971=0.508

87/165=0.527

123/235=0.523

Now 0.538 > 0.527 > 0.523 > 0.508

therefore 7/13 > 87/165 > 123/235> 493/971

FRACTIONAL PART OF A NUMBER

Fractional part of a

number is simply the

product of the related

fraction and the given

number.

Example2:

2/3 rd of 60 is 40

3/4th of 36 = 27

1/9th of 36 = 4

2/3rd of 36 = 24

Example3: A man travels 1/4

th part by scooter, 3/8 th by

car and rest 48 km by bus.

Find the total distance

covered.

Solution3: Here, total

distance is to be found out.

Fraction related to rest 48

km=1-(1/4+3/8)

rest distance/its related

fraction=total distance

48/[1-(1/4+3/8)]=total

distance

total distance =48/(3/8)=128 km

Fraction related to balance part = 1 - (sum of all other

fractions)

Example4: A person spends 3/8th part of his salary on

food, 1/12th part of his salary on education, 1/4 th part of

his salary on clothing. He is now left with Rs. 550. Find his

total salary.

Solution4: Fraction related to rest part =1-

(3/8+1/12+1/4)=7/24

total salary=rest amount/fraction related to rest

part=550/(7/24)=Rs 2640

Example5: A person

spends 3/8 th part of his

salary on food, 1/12 th of

the rest part on

education and 1/4 th of

the remainder on

clothing. He is now left

with Rs 550. Find his

total salary.

Solution5: Here, spending

on the second item (i.e.

education) depends on

the amount left after

spending on the first item (i.e. food). Similarly, spending on

the third item (i.e. clothing) depends on the amount left

after spending on the first item and the second item.

Fraction for balance part=(1-first fraction)*(1- second

fraction)

So fraction for balance (rest) part=(1-3/8)(1-1/12)(1-

1/4)=5/8*11/12*3/4=55/128

Therefore total salary = Rest amount/Fraction related to

rest part=550/(55/128) =Rs 1280

Example 6: 5/12 part of what amount will be equal to 3 3/4

(3 three fourth) part of Rs 100.

Solution 6: Let the amount be Rs x

(5/12)x = 3 3/4 * 100

(5/12)x = (15/4) * 100

x = (12/5) * (15/4) * 100

=> x = Rs 900

Required amount is Rs 900.

Example7: 4/7 of a pole is in the mud. When 1/3rd of it is

pulled out, 250cm of the pole is still in the mud. Find the

full length of the pole

Solution7: Total length of pole=Length in mud/ Part in mud

= 250/[4/7-1/3] = 1050

Length of pole = 1050 cm

Example8: How much is to be added with 0.685 of 325 to

get 300?

Solution8: Let 'x' is to be added. Then,

x + (0.685 x 325) = 300

x = 300 - 222.625 = 77.375.

Example9: Which one of the following fraction is less than

1/3? (a) 22/63 (b)4/11 (c) 15/46 (d) 33/98

Solution9:Step 1 Reverse the test fraction, i.e.

1/3 becomes 3/1 = 3.

Step 2 Reverse each alternative and find which alternative

is greater than 3/1

63/22

2 36 48 64 72

2 18 24 32 36

2 9 12 16 18

2 9 6 8 9

2 9 3 4 9

2 9 3 2 9

3 9 3 1 9

3 3 1 3

1 1

LCM=2*2*2*2*2*2*3*3=576

DIFFERENCE

BETWEEN HCF

AND LCM

HCF of x, y and z is

the Highest Divisor

which can exactly

divide x, y and z.

LCM of x, y and z is

the Least Dividend

which is exactly

divisible by x, y and z.

Solved Examples

Example 1: Find the greatest number that will exactly

divide 200 and 320.

Solution 1: Required number = HCF of 200 and 320 = 40.

Example 2: Find the least number which when divided by

27, 35, 45 and 49 leaves the remainder 6 in each cases.

Solution 2: Required number = (LCM of 27, 35, 45 and 49) +

6 i.e. 6615 + 6 = 6621.

Example 3: Find the

greatest possible length of a

scale that can be used to

measure exactly the

following lengths of cloth; 3

m, 5 m, 10 cm and 12 m 90

cm.

Solution 3: The lengths of

cloth to be measured are,

300 cm, 510 cm and 1290 cm

the required length of the

scale is HCF of 300, 510 and 1290 i.e. 30

the greatest possible length of the scale to be used =30 cm

Example 4: Find the greatest 4 digit number such that it is

exactly divisible by 12,15,20,35

Solution 4: Step 1 LCM of 12, 15, 20 and 35 = 420.

Step 2 420) 9999 ( 23

-9660

339

:. required number = 9999 - 339 = 9660.

Example 5: There are two electrical wires, one is a 9 m 60

cm long aluminum wire and the other is a 5 m 12 cm long

copper wire. Find the

(a) maximum length that can be equally cut from each wire

in such a way that the total length of each wire is exactly

divisible by it.

(b) how many such largest possible pieces are available in

each kind of wire?

Solution 5: 9 meter 60 cm = 960 cm and 5 meter 12 cm =

512 cm.

(a) The required largest piece = HCF of 960 and 512 cm, i.e.

64 cm.

(b) Number of such aluminum wire pieces=960/64=15 nos

And number of such copper wire pieces =512/64=8 nos

Example 6: HCF and LCF of two numbers are 16 and 240

respectively If one of the number is 48 find the other

number

Solution 6: We know, HCF x LCM = Number1 x Number2

therefore second number =16*240/48=80

Example 7:Among how many students 175 bananas and 105

oranges can be equally divided?

Solution 7: HCF of 175 and 105 =35

Evaluate your grasp:

1. Find the greatest number that will divide 148, 246 and

623 leaving remainders 4, 6 and 11 respectively.

2. Find the least number which when divided by 36, 48 and

64 leaves the remainders 25, 37 and 53 respectively.

3. Find the smallest number which when (a) increased by 8

(b) decreased by 8 is exactly divisible by 15, 21, 30

4. Find the smallest 4 digit number such that it is exactly

divisible by 12, 15, 20 and 35.

5. Four bells first begin to toll together and then at

intervals of 6, 7, 8 and 9 seconds respectively. Find how

many times the bells toll together in two hours and at what

interval they toll together?

6. Find the (a) the greatest 4-digit number, and (b) the

smallest 4-digit number such that when they are divided by

12, 18, 21 and 28, it leaves a remainder 3 in each case.

7. Find the numbers between 200 and 300 such that when

they are divided by 6, 8 or 9,

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

(b) it leaves in each case a remainder 5

8. Find out the HCF of 11, 0.121 and 0.1331

9. Find out the LCM of 2.2, 540 and 108

10. Find out HCF of 3^5 ,3^9,3^14

11. Find out the LCM of 4^5, 4^-81,4^-12 and 4^7

Answer Key:

1.) 12 2.) 565

3.) a) inc by 8 b) dec by 8 4.) 1260

5.) 14 times 6.) a) 9831 b) 1011

7.) a) 216,240,264& 288 b) 221,245,269& 293

8.) 0.0011 9.) 5940

10.) 3^5 11.) 4^12