Universal Active Math Math Delight - 7 - Navnirmiti Learning

Conceived and Written by

Dr. Vivek Monteiro, Geeta MahashabdeUniversal Active Math team

Universal Active Math

Math Delight - 7

Supported By

Published by : Navnirmiti Learning Foundation

All rights reserved with Navnirmiti Learning Foundation

Contents developed between 2000 - 2021English Marathi Bilingual Version - 2021English Mizo Bilingual Version - 2021

Graphics and Design :

Geeta Mahashabde, Laxman Vyavahare, Lokesh Daga, Dibyajyoti Changkakoti

Universal Active Math - Math Delight VII ( Of a series of Math Delight I to VIII)

Supported By : Priyanvada Barbhai, Swati More, Varsha Khanvelkar, Vathsala Sharma,

Fundilal Mali, Swati Joshi, Khushboo Lahoti, Tanima Paul

Cover Design : Sahil Kalloli

Vivek Monteiro, Geeta Mahashabde, Navnirmiti Trust, Navnirmiti Learning Foundationã

Authors’ and Contributors’ team for the series :

Vivek Monteiro, Geeta Mahashabde, Neelima Deshpande, Dinesh Lahoti, Sushma

Dhumal, Teachers of Goa, Vathsala Sharma, Sushma Bakshi, Soni Yadav, Lokesh Daga

and Navnirmiti and Edugenie’s material generation teams

Navnirmiti Learning Foundation, Pune

9850303396 e [email protected]

w www.navnirmitilearning.org

Edugenie, Guwahati

8486020288/388 e [email protected]

w www.myedugenie.com

Navnirmiti Trust, Mumbai

022-25773215 e [email protected] Eduquality, Mumbai

Mizo Translation : Quality Education Project Team

A to F are six bank accounts. Write the values of these accounts.

+ 1

- 1

+ 1

- 1

+ 1

- 1

+ 1 + 1

+ 2

A

- 1 - 1- 1 - 1

- 1 - 1 - 1B

+ 1 + 1 + 1

- 1 - 1 - 1 - 1

+ 1 + 1

- 1 - 1 - 1

+ 1

C D

+ 1G H+ 4 - 1 - 2- 6 - 7

E F+ 5 + 5 - 3 - 2

1C VCM-GM-NLF

One Rupee in bank account.

Bank account ah chengkhat a awm.

Loan of one rupee in bank account.

Bank account ah chengkhat a awm.

Heng A atanga F bank account paruk te hi a hlutna theuh zawng chhuak rawh?

Which account is the richest?

Which account is the Poorest?

Which account has zero value?

Who is richer, A or B?

Whose account has less value, D or B?

Compare using the signs <, =, >

0 -3

- 2 +10

Write appropriate numbers considering the signs <, =, >

- 4 < + 2 < <

- 4 > + 2 > >

+2 0 +5

- 6 -8 -10

<

2C VCM-GM-NLF

Khawi account hi nge tam ber?

Khawi account hi nge thlem ber?

Khawi account hian nge zero balance nei?

A emaw B ah hian khawinge tam zawk?

D emaw B ah hian khawi hi nge tlem zawk?

A sign <, =, > hmang la, tehkhin rawh

A sign <, =,> a rawn pek sa te hi rinchhan in a number tur awm ang ziak rawh.

Here is a number line. We place 0 at a point. +1 is on its right and -1 is on its left, both

at the same distance from 0. Where will you place +2, -2, +3, -3, +4, -4?

Represent the given numbers on the number line :

+20, +30, -10, -20, -30, -40

0

-7, -3, 0, +2, +7

0 +1-1

+100, +150, -50, -150, - 300

500

- 5

123+10

1)

2)

3)

3C VCM-GM-NLF

Number line ah hian 0 chu a lai tak ah awm in +1 chu dinglam ah a awm a, -1 a vei lam ah

awm in 0 atangin an awmna hmun a in ang ve ve a, +2, -2,+3, -3, +4, -4 te hi khawi ah nge

i dah ve tak ang?

Number line ah hian heng a hnuai a number te hi a awmna hmun tur ah dah rawh :

+200, +300, +400, -100, -200, -300

0 +100

4)

-100, +50, +150, -200, -250, +300

0 +100

5)

+2a, +3a, +4a, -2a, -3a, -4a

0 a

6)

7)

10 12

+ 2

34

+42

+12

-22

-32

-42

-

4C VCM-GM-NLF

0+1

-1

+10

On a day in December 2020, at 1 pm, following were the temperatures of different locations

in India. Place those numbers on the given number line. Write the name of each place

next to its temperature.

Temperature(ْC)Place

Aizawl (Mizoram)

Trivandrum (Kerala)

Pune (Maharashtra)

Kolkata (W. Bengal)

Kashmir

Shimla (Himachal)

Leh

Khardungla (world’shighest motorable road)

+19

+32

+29

+24

- 4

+ 3

- 11

- 24

5C VCM-GM-NLF

Heng a hnuai a temperature tarlan te hi December 2020, chawhnu dar 1 a India ram hmun

hrang hrang a khawchin te an ni a, number line ah hian a number theuh pe la, an hmun

hming theuh tarlang tel bawk rawh.

A hmun Khawchin

(Khawvel a motor kawngpui sang ber)

Represent the given numbers on the number line :

1)12

+12

-14

+14

-12

112

-1 14

214

-2

0 1 2 3-1-2-3-4

0 1 2 3-1-2-3-4

2) 2.2, 0.5, -0.5, 1.5, -1.5, -3.2, 2.4, -3.4

0 1

3) -5, -4.5, -3.2, -2.1, 0, 2.1, 3.2, 4.5, 5

Compare using the signs <, =, >

(- 15) (- 32)

(3.4) (- 3.4)

>

(- 0.05) (- 0.5)

( )43

- ( )13

( 3 )12

- (- 3 )14 ( 1.25 ) ( 1.5 )

Did you discover that1. Any positive number is greater than any negative number.2. The bigger number becomes the lesser number when we attach negative sign to both numbers.

6C VCM-GM-NLF

2.2

Heng a hnuai a number te hi number line ah hian dah rawh.

Heng a sign <, =, > hmang te hian khaikhin rawh

1. Positive number chu negative number aiin a tam zawk.2. Negative number reng rengah chuan a number pangai kha lian mahse a sign kha

negative a nih avangin a te zawk ani tih kan hre thin ani.

I lo hre chhuak ve tawh hem

In class 6, you had discovered rules of adding positive and negative numbers. and subtracting

Revise it from Math Delight 6 (Page numbers 95 to 103).

Use appropriate rule and solve.

1) + and + + Total

2) – and – - Total

3) + and – (sign of the bigger

number) and (difference)

1) + 5 + 7

3) – 7 + 5

5) – 200 + 10

7) +3a – 10a

9) + 5 + 3 – 2

11) – 3 - 2 - 1

13) – 4 + 4 – 2 – 1

2) - 5 – 7

4) + 5 – 10

6) – 300 - 10

8) – 5x - 3x

10) – 6 – 2 + 4

12) – 5a + 5a – 2a

14) + 3 – 2 – 3 + 2

-1 +1 = 0

-4 +4 = 0}

7C VCM-GM-NLF

Pawl 6 I zir lai khan positive leh negative number belh leh paih dan kha i hre tawh a, Maths

delight-6 hmang khan en nawn rawh. (Page numbers 95 to 103).

Number tam zawk sign leh a inkar

1) + and + + Total

2) – and – - Total

3) + and –

-

- 4 + 4 = 0

- 1 + 1 = 0We show this as cancellation in addition.

Kan belh ah khan a thaichhiat dan kan hmu

A rules hmang la, chawk chhuak rawh.

While adding, when you remove the brackets, signs of insiders remain unchanged.

1] ( + 2 ) + ( + 3) = + 2 + 3 = + 5

2] ( – 5 ) + ( – 1) = – 5 – 1 =

3] ( – 3 ) + ( + 2 ) = ......................................... =

4] ( + 6 ) + ( – 2 ) = ......................................... =

Subtracting a positive number is like adding a negative number.

Subtracting a negative number is like adding a positive number.

Therefore while removing a bracket after a minus sign, signs of all insiders get reversed.

1] ( + 3 ) – ( + 1 ) = + 3 – 1 = + 2

2] ( – 5 ) – ( - 2 ) = – 5 + 2 =

3] ( + 6 ) – ( + 6 ) = .................................. =

5] ( – 8 ) – ( – 8 ) = ................................... =

4] ( + 10 ) – ( + 2 ) = ................................. =

6] ( + 2 - 3 ) – ( – 2 - 3 ) = ........................................... =

7] ( – 3 + 5) – ( - 5 – 2 ) = ............................................ =

Note that we are using two rules. Rule for removing the brackets, the rule for adding positive and negative numbers.

5] ( + 2 - 3 ) + ( – 2 - 3 ) = ......................................... =

6] ( – 3 + 5) + ( - 5 – 2 ) = .......................................... =

8C VCM-GM-NLF

Belh i hman laiin bracket i laksawn pawn , a sign chhuak a chu a danglam tur a ni lo.

Minus sign awm hnu a bracket kan paih khan, a chhunga a sign kha a letling.Negative number kan paih khan positive number kan neih belh. Positive number kan paih khan negative number kan neih belh.

Dan pahnih kan hmang a, chung te chu bracket paih dan leh positive leh negative number

belh dan.

Fill in the blanks.

( + 2 ) + ( ) = + 6 ( + 2 ) + ( ) = - 4 6 - ( ) = 4

6 - ( ) = 8 ( - 4 ) + ( ) = - 6 - 4 + ( ) = - 2

( - 4 ) - ( ) = - 6 ( - 4 ) - ( ) = - 2 ( ) + ( - 2) = 0

( ) - ( - 2 ) = 0 ( ) - ( + 2) = 0 ( ) + ( - 3) = - 6

( ) + ( + 3 ) = - 3 ( ) + ( ) = 0 ( ) + ( ) = + 5

( ) + ( ) = - 3 ( ) + ( ) = - 10 ( ) + ( ) = -100

Use one positive and one negative number :

( ) + ( ) = + 6

Use two negative numbers :

( ) + ( ) = - 6

Use two negative numbers :

( ) - ( ) = + 4

Use one positive and one negative number :

( ) - ( ) = - 6

Use two numbers with the same sign :

( ) + ( ) = 2

( ) - ( ) = 2

( 5a ) + ( - 2a ) = ........... ( 5a ) - ( - 2a ) = ............

( 10 x ) - ( x ) = ........... ( - 3 xy ) - ( - 5xy ) = ............

( 3a ) + ( - 4a ) - ( - a) = ...............

( - 7p ) - ( + 3p ) - ( - p) = ...............

9C VCM-GM-NLF

A kar awl dahkhat rawh.

Positive number pakhat leh negative number pakhat hmang rawh:

Negative number pahnih hmang rawh:

Negative number pahnih hmang rawh:

Positive number pakhat leh negative number pakhat hmang rawh:

Number pahnih sign in ang ve ve hmang rawh.

Multiplication of Integers

Multiplying by a positive number = Putting into your bank account those many times.

Multiplication Picture Answer

+ 1

3 x 23 taken 2 times + 1 + 1 + 1 + 1 + 1 + 6

- 1

(-3) x 2(-3) taken 2 times - 1 - 1 - 1 - 1 - 1 - 6

2 x 3

( -2 ) x 3

2 x 5

( -2 ) x 5

( -5 ) x 2

5 x 2

10C VCM-GM-NLF

Integer hmang a puntir

Positive number hmang a puntir = I bank account ah vawi tam tak i dah belh.

Puntir A lan dan A chhanna

Pa 3 atanga pa 2 puntir

(-3) atanga pa 2 puntir

Multiplication Picture Answer

(-2) x 4

+1 +1 +1 +1 +1 +1 +1 +1

x 312

x 3-12

x 3-15

x 5-15

x 2-8

x 2+8

11C VCM-GM-NLF

Puntir A lan dan A chhanna

Observe the pattern of multiplying by a positive integer on previous two pages.

Positive numbers x Positive number = (product of two numbers)

Negative numbers x Positive number = (product of two numbers) +-

Use the rule and solve.

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

( -3 ) x ( +6 ) = ( -10 ) x ( +2 ) =

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

( 1 ) x ( 1 ) = ( -1 ) x ( 1 ) =

( 1 ) x ( 0 ) = ( -1 ) x ( 0 ) =

( 0 ) x ( 0 ) =

( ) x ( +4 ) = ( ) x ( +4 ) = +16 -16

( ) x ( ) = ( ) x ( ) = +8 -8

Write three different multiplications which give the answer as -12.

( ) x ( ) = -12

( ) x ( ) = -12

( ) x ( ) = -1212C VCM-GM-NLF

A phek hmasa pahnih ami ang khan, positive = + (number pahnihn a puntir chhuak)A phek hmasa pahnih ami ang khan, positive integer hmang in puntir rawh.

Negative number leh positive number puntir = - (number pahnih a puntir chhuak)

Dan hmangin chawk rawh.

Puntir a -12 tichhuak thei turin a hnuai a pa 3 ah te hian zawng rawh.

Multiplying by a negative number - Read carefully, try out and understand.

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

Account

Multiplication What is to be done and process Answer

Take away (+2) three times

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

-6

Take away (+3) two times

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

-6

Take away (-2) three times

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+6

13C VCM-GM-NLF

An empty account has zero value, but the

account having zero value may not be

empty. e.g. an account shown in this figure.

It has +6 and -6

(+1 and -1) pairs. This ensures that we start

with an account of 0 value.

Multiplying by a positive number is putting

into your account those many times.

Multiplying by a negative number is taking

away from your account those many times.

For being able to take away, we start with

an account having zeros made of

Negative number puntir chuan = uluk takin chhiar la, hriatthiam tum in zir rawh.

I account ah zero value atangin i tan ani tih

ti chiang tu chu zero (+1 leh -1) inkawp

atanga i account i tan hmasak vang a ni.

Negative number hmang a I puntir chuan i

account nasa takin a ti phai.

A tirah chuan i account ah engmah a awm

lo.

Positive number hmang a I puntir chuan i

account nasa takin a ti pung.

Puntir A chawh dan leh a kalhmang A chhanna

2 x (-3)

2 taken (-3) times

2 hmun (-3)

(+2) vawi thum lak bo

3 x (-2)

3 taken (-2) times

3 hmun (-2)

(+3) hi vawithum kan lak hran chuan

Account a pawisa awmlo chuan hlutna a

neilo a, chutih rual chuan account hlutna

neilo dang chu a ruak a ni vek kherlo ani.

Entirnan; +6 leh - 6 te account a a landan

ang hian.

(-2) vawi thum kan lak hran chuan

(-2) x (-3)

(-2) taken(-3) times

3 vawi (-2)

zat kan lak in

Account

Multiplication What is to be done and process Answer

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

+1

-1

(+2) x (-4)

(-4) x (-2)

(+4) x (-2)

(-1) x (-8)

(+1) x (-8)

(-2) x (-4)

14C VCM-GM-NLF

Puntir A chawh dan leh a kalhmang A chhanna


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

( -3 ) x ( - 4) = ( 3 ) x ( - 4) =

( - 6) x ( ) = ( ) x ( - 2 ) =

( ) x ( ) = ( ) x ( ) =

( 1 ) x ( - 1 ) = ( -1 ) x ( - 1 ) =

( 10 ) x ( - 2) = (- 10) x ( - 2) =

( ) x ( - 2 ) = ( ) x ( +2 ) = - 10 + 10

Write six different multiplications which give the answer as + 15

( ) x ( ) = +15

( ) x ( ) = +15

( ) x ( ) = +15

Observe the pattern of multiplying by a negative integer on previous two pages.

Positive numbers x Negative number = (product of two numbers)

Negative numbers x Negative number = (product of two numbers) -

+

+ 12 - 12

+ 15 - 15

( ) x ( ) = +15

( ) x ( ) = +15

( ) x ( ) = +15

15C VCM-GM-NLF

Negative number leh negative number puntir chuan = + (number pahnih a puntir chhuak)

A phek hmasa lam a kan lo tih tawh ang khan negative integer hmang in puntir rawh. Positive number leh negative number puntir chuan = - (number pahnih a puntir chhuak)

A rule hmang la, chawk chhuak rawh.

Puntir a +15 tichhuak thei tu tur a hnuai ami pa 6 ah te hian zawng chhuak rawh.

Rules for multiplication of integers :

( + ) x ( + ) = + product

( - ) x ( + ) = - product

( + ) x ( - ) = - product

( - ) x ( - ) = + product


( +6 ) x ( +3 ) = ( +4 ) x ( - 4) =

( - 6 ) x ( +3 ) = ( - 4 ) x ( - 4) =

(+7) x ( ) = ( ) x ( ) =

( - 7 ) x ( ) = ( ) x ( ) =

( + 6 ) x ( - 3) = (- 4) x ( + 4) =

( - 6 ) x ( - 3 ) = ( +4 ) x ( +4) =

+21 - 20

+ 21 +20

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

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

( + 2 ) x ( - 3 ) x ( 0 ) =

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

( - 2 ) x ( - 3 ) x ( - 4) =

16C VCM-GM-NLF

Multiplication of integer chawh dan tur:

A rule hmang la , zawng chhuak rawh

( - 1) x ( - 1) =

( - 1) x ( - 1) x ( - 1) =

( - 1) x ( - 1) x ( - 1) x ( - 1) =

( - 1) x ( - 1) x ( - 1) x ( - 1) x ( - 1) =

( - 1) x ( - 1) x ( - 1) x ( - 1) x ( - 1) x ( - 1) =

Write the missing numbers and signs.

( ) X ( ) = ( )

( ) X ( ) = ( )

( ) X ( ) = ( )

( ) X ( ) = ( )

( ) X ( ) = ( )

( ) X ( ) = ( )

( ) X ( ) = ( )X ( )

( ) X ( ) = ( )X ( )

( ) X ( ) = ( )X ( )

( ) X ( ) = ( )X ( )

- +3 8 24

25- +

8-222

2 3 6---

2052 +++

1 6 0

6 6- -

14+ +

- 273

549+

17C VCM-GM-NLF

A sign leh number awmlo te hi zawng chhuak rawh.

Let’s write the numbers to be multiplied along the sides of a rectangle and product inside

the rectangle. Write the missing numbers and products.

+5

+2

+6

+3+10

-5

+2

- 6

-3-10

x

+3 - 3

y

x

- x - x

-y

ab ab

Write two different pairs which have ab as their product.

18C VCM-GM-NLF

Rectangle hmang a puntir in :

rectangle chhung a factor leh product te hi zawng chhuak rawh.

Number pahnih in ang lo inkawp ab a product nei

+5

+3 +15

+2

Distributive property

+6

[ (+5) + (+2) ] x (+3) = [ (+5) x (+3) ] + [ (+2) x (+3) ]

= (+5+2) x (+3) = 7 x 3 = 21

= Sum of areas of two small rectangles= Rectangle pahnih te belh khawm.

= [(+5)x(+3)] + [(+2)x(+3)]

= 15 + 6 = 21

b

a ab

c

ac

[ b + c ] x a = a x [ b + c ] = [ a x b ] + [ a x c ]

-5

-3

-2

[ (-5) + (-2) ] x (-3) = [ ( ) x ( ) ] + [ ( ) x ( ) ]

= [ ] + [ ]

=

19C VCM-GM-NLF

Number puntir chhuah tawh, a hnu a belh leh chu distributive property ani.

= Rectangle zawng zawng zau zawng.

Area of big rectangle

Factors Expansion / Expression Value

3 x 102

3 x (100 + 2)

= (3 x 100) + (3 x 2)

= 300 + 6

306

5 x (100 + 1)

2 x (3a - 5a)

(x + y) x z

a x (x + y - z)

3 x (200 + 50 + 3)

3 x 198

= 3 x (200 - 2)

20C VCM-GM-NLF

A puntirna Ti zau A hlutna chhuak

Write division statements and discover the rules for division.

+5

+2 +10

-5

+2 -10

+10

+2= +5

+10

+5

- 10

+2

- 10

- 5

-5

-2 +10

+10

- 2

+10

- 5

=

= =

= =

Rules for division of integers

++

= + sem

-+

= - sem

+-

= - sem

--

= + sem

21C VCM-GM-NLF

A sem dan kalhmang ziak lang la, sem nan a dan kan hman te kha zawng chhuak rawh.

Integer hmang a sem dan tur

++

= + division

-+

= - division

+-

= - division

--

= + division

Use the rules and solve.

+14

+7=

- 27

+9=

+36

- 6=

- 50

- 10=

+ 13

- 1=

0

- 4=

- ab

- a=

2+ x

- x=

- 5

0= Not

Defined

( ab - ac )

a=

a x ( b - c)

a= b - c

2( x - xy )

x=

abc

- bc=

a

- 1=

- x

- 1= =

xy - xyz

- xy=

22C VCM-GM-NLF

A rule hmang la, chawk chhuak rawh

Draw your own rough sketch for solving each problem. Draw, think, count, use formula and

get the answer.

Minimum temperature in Shimla was (- 2) degree Celsius and maximum temperature was 11 degree Celsius. What is the difference between minimum to maximum?

Students get (+3) marks for each cooperative act and get (-2) marks for each competitive act. Find the marks obtained by each student.

Name Number of acts and procesure Total marks

A

B

C

D

2 cooperative and 4 competive

[ 2 x (+3) + 4 x (-2)] = +6 - 8 = - 2 -2

3 cooperative and 3 competitive



23C VCM-GM-NLF

Chhutchhuah tur te chhut turin nangmahin siam rawh. A chhanna te i hmuh theihnan siam

la, ngaihtuah la, chhut la, formula te hmang ang che.

Shimla a an khawchin hniam lai ber chu (-2) degree celsius ani a, asan lai ber chu 11

degree Celsius ani. A hniam lai ber leh a san lai ber danglamna eng nge ni?

Hming An thil tih zat leh an tih dan Marks belhkhawm

A khawm a an tih 2 leh intihsiakna 4




School naupang te chuan a khawm a hna an thawhna theuh atangin (+3) marks an hmu a, (-2) marks chu intihsiakna an neih atangin a ni. School naupang tin te marks hmuhzat zawng chhuak rawh.

How many groups of 2

can be made from 6? 3

=6

2

3

1=

Draw the missing pebbles.

Write the missing number.

=4

1 2=

=8

4

4=

Equivalent Fractions

Paruk atang hian pahnih zel awm turin hlawm

engzat nge kan lak theih ang? 3

Lung kimlo te ziah belh rawh. Number kimlo

ziah belh rawh.

Number chung leh hnuai hlutna inang

24C VCM-GM-NLF

Draw the missing pebbles.

Write the missing number.

= =3

1

= =

= =

= =

=3

1=

2

=1

4=

8

=6

3=

4

=2

3=

9

Lung kimlo te ziah belh rawh. Number kimlo

ziah belh rawh.

25C VCM-GM-NLF

=63

105=

64

128=

64

32

=46 9=

46

8=

46

2

=13

3=

13

4=

13

2

=14

3=

14

4=

14

2

=52 8=

52

15=

52 4

=21 2

=3

=4

=5

=10

=100

=2

= = = = =1 2 3 4 5 10 100

=32 4

=6

=8

=10

=20

=100

=2

= = = = =3 15 60 18 21 100 90

Write equivalent fractions in numbers.

Use pebbles if required.

But think for yourself.

A tul chuan lung nawi-te hmang la.Mahse nangmah ngeiin i ngaihtuah dawn nia.

Then hran inangte hi number in ziak rawh.

26C VCM-GM-NLF

The length of one foot was measured.

It is shown on the grid below as 1 whole length.

Draw line segments showing all other lengths.

1

2

1

2

1

2

1

2

11

3

Draw line segments of the following lengths in cm.

2

1

1

1

2

1

2

21

2

1

4

41

4

61

4

Kephah pakhat sei zawng chu teh a ni a. Kephah 1 sei

zawng chu grid ah hian tarlan a ni. A dang sei zawng hi

line hmangin tarlang rawh.

A hnuai ami te hi line hmangin a sei zawng (cm) tarlang rawh.

27C VCM-GM-NLF

A segment of 2 cm

2 x 3 = 2 cm taken 3 times = 6 cm

1

2 x 2

(2 cm) taken 3 times a segment of 6 cm

1 x 4

x 4

Show the given multiplication in Length.

1 cm x 4 = cm

cm x 2 = cm

cm x 6 = cm

1

2 x 6

1

21

cm x 4 = cm

A hnuai ami te hi puntir in length ah chantir rawh. 2 x 3 = 2 cm hmun 3 = 6 cm

Line 2 cm a rinCentimeter hnih hmun 3 a lakin

Centimeter 6 a sei line a siam

28C VCM-GM-NLF

This

circl

e is

1 w

hole

Roti

/ B

haka

ri / P

izza

/ D

osa

/ P

uri e

tc.

Multi

plic

atio

nM

eanin

gP

ictu

reP

roce

ssA

nsw

er

3 x

2

2 x

1

1 x

2

3 x

2

x 2

1 2

x 4

1 4

x 8

1 8

x

21 2

6 1

He c

ircl

e p

um

pui h

i Roti

/ B

haka

ri / P

izza

/D

osa

/ P

uri e

tc. ang a

ni a

.

Puntir

Aw

mzi

aA

lem

A c

haw

h d

an

Chhanna

3 take

n 2

tim

es.

Pa 3

hm

un 2

take

n 2

tim

es.

1 2

A c

hanve

( )

hm

un 2

1 2

29C VCM-GM-NLF

Meaning of cancellation in multiplication

x 2 =1

2half taken 2 times.

Two halves make a whole.

Thus x 2 = 11

2One whole is made.

x 4 =1

4taken 4 times.

x 4 =1

4

x 4 =1

2taken 4 times.

= halves

= wholes

x 4 =1

2 x 2 x 2 = 21

2

Semna a thaichiat awmzia

A chanve hmun 2

A chanve pahnih in a pumpui siam

A pumpui a insiam.Chutiang chuan

Hmun 4 a lakin

Hmun li a lakin

A chanve

A pumpui

30C VCM-GM-NLF

This

circl

e is

1 w

hole

Roti

/ B

haka

ri / P

izza

/ D

osa

/ P

uri e

tc.

Multi

plic

atio

nM

eanin

gP

ictu

reA

nsw

er

Pro

cess

2 x

8

1 4 x

8

1 4

take

n 8

tim

es.

1 4

x 6

1 3

x 12

1 4

He r

inbia

l pum

pui h

i Roti

/ B

haka

ri / P

izza

/D

osa

/P

uri e

tc. ang

a n

i a.

Puntir

Aw

mzi

aA

lem

A c

haw

h d

an

Chhanna

Hm

un 4

a then a

hm

un

1 v

aw

i 8 a

lak

in

31C VCM-GM-NLF

This

circl

e is

1 w

hole

Roti

/ B

haka

ri / P

izza

/ D

osa

/ P

uri e

tc.

Multi

plic

atio

nP

ictu

reP

roce

ss a

nd A

nsw

er

x

4

=

22 4

x 4

2 4

x 4

3 4

x 4

3 2

He r

inbia

l pum

pui h

i Roti

/ B

haka

ri / P

izza

/D

osa

/P

uri e

tc. ang

a n

i a.

Puntir

A le

mA

chaw

h d

an le

h C

hhanna

32C VCM-GM-NLF

Colour the given multiplications.

Write the length (l) breadth (b) and area of each rectangle (answer).

4 x 3

l = 4b = 3 A = 4 x 3 = 12

4 x 2 4 x 1

4 x x 2 x 1

x 2

1

2

1

2

1

2

x 1

2

1

2

3

2 x

1

2

5

2

l = b = A =

l = b = A =

l = b = A =

l = b = A =

l = b = A =

l = b = A =

l = b = A =

l = b = A =

Puntir tur te hi chei la, a dung, a vang leh a area zau zawng te hi chawk chhuak bawk rawh.

33C VCM-GM-NLF

1 x 1

1/2 x 1/4

1

1

5/6 x 1/2

1=

2/9 x 3/4

1/4

2 divisions in length, 4 divisions in breadthTotal 8 parts in the whole,1 part of that type

1/8=

1/2

=

....... divisions in length, ..... divisions in breadthTotal ........parts in the whole,..... parts of that type

=


2/9

3/4

5/6

1/2

Hlawm khatah hian then sawm 8 awmin then

sawm 1 chu.

A dung pahnih thenin, a vang 4 a then sawm in.

Length (a dung) ____ thenin, a breadth

(a vang) _____a then sawm in.

Hlawm khatah hian then sawm

___awmin then sawm ____ chu.

Length (a dung) ____ thenin, a breadth

(a vang) _____a then sawm in.

Hlawm khatah hian then sawm

___awmin then sawm ____ chu.

34C VCM-GM-NLF

Make the required number of divisions in length and breadth and find the answers-

5/6 x 5/6

=

....... divisions in length, ..... divisions in breadth

Total ........parts in the whole,..... parts of that type

Can you guess the answer of 3/6 x 2/6 ?

=

2/4 x 3/4


Can you guess the answer of 2/4 x 2/4?

=

A dung leh a vang te hi a then sawm dan turin then sawm la, a chhanna ziak bawk rawh.

Hlawm khatah hian then sawm ___awmin then

sawm ____ chu.

Length (a dung)____ thenin, a breadth (a vang)

_____a then sawm in.

3/6 x 2/6 hi a chhanna i hre thei em?

=

Length (a dung)____ thenin, a breadth (a vang)

_____a then sawm in.

Hlawm khatah hian then sawm ___awmin then

sawm ____ chu.

2/4 x 2/4 hi a chhanna i hre thei em?

Have you discovered the rule,

product of two fractions =

product of numerators

product of denominators

Fraction pahnih puntir chhuah = A chung zawk puntir chhuah

A hnuai zawk puntir chhuah

A tih dan tur i hre chhuak em.

35C VCM-GM-NLF

Now, apply the rule and find answers. Write the answer in its lowest form.

x 1

4

1

4 = x

1

4

1

2 =

x 4

5

2

3 = x

3

7

1

4 =

x 2

5

7

8 = x

8

5

3

4 =

x 3

2

2

3 = x

4

5

5

4 =

x 6

8

4

3 = x

6

7

5

2 =

x 43

8 = x

1

1

1 =

9

1

Have you discovered the rule,

product of two fractions =

product of numerators

product of denominators

Fraction pahnih puntir chhuah = A chung zawk puntir chhuah

A hnuai zawk puntir chhuah

A tih dan tur i hre chhuak em.

A tih dan tur hmangin a chhanna ziak rawh. A chhanna hi a te thei ang berin dah ang che.

36C VCM-GM-NLF

Commutativity

Solve the problem. Do the same operation by changing the order of numbers. Do you get the

same answer? If yes, that opearation is commutative, else, not commutative.

3 + 2 = 5 2 + 3 = 5

In addition, the answer remains the same even if you change the sequence.

This is called commutativity.

Problem

in reverse order Operation

Commutative / not commutative

5 - 2 = 3 2 - 5 = -3 Not Commutative

10 + 2 =

3 x 2 =

6 ÷ 2 =

Try with other numbers.

Belh reng reng ah hian number awmna te thlak thleng pawh ni ila, a chhanna ngai-tho chuan commutativity a ni.

Number in dawt dan te thlak thleng in chawk la, a chhanna a danglam em? A danglam loh

chuan commutative a ni anga a danglam chuan a ni lo tihna a ni mai.

Thalk thleng in A tih dan

Number dang hmangin chawk ve rawh

Subtraction

Paih

37C VCM-GM-NLF

3 x 2 2 x 3

3 taken 2 times. 2 taken 3 times.= 6 =

Fill in the blanks.

4 x 3 = 3 x

4 x b = b x a

4 x = 2 x 4

3 x 10 = x 3 5 x 4 = 4 x

1

4 x 4 = 4 x 1

2 x 2 = 2 x

1

4

1

2 x = x 1

4

5

6

6

5 x = x 5

6

xx y = x x

2a x b = b x 3a x 2b = 2b x

Commutativity

3 hmun 2 2 hmun 3

A kar awl dah khat rawh

38C VCM-GM-NLF

1This is one whole chocolate.

Its length = 1 and breadth = 1 .

Area = length x breadth = 1 x 1 = 1 square

Some fractions of this chocolate are coloured here. Write the area and multiplication related to

coloured portion in comparison with this 1 whole. One example is solved.

Picture of coloured portion

1

12=

1

4x

1

3

1/4

1/3

= x

= x

= x

A dung = 1, a vang = 1Hei hi chocolate hlawm pakhat a ni.

Zau lam = a dung x a vang = 1x1 = Kil li intiat 1

Hetah hian chocolate then hran thenkhat te chei ani a. A hnuai a chocolate chei dum lai ang

zel hian fraction in a zau lam ziak la. A dung leh a vang puntir chhuak tu ziak rawh. Entirna

pakhat tarlan ani.

1

1 1

A inchei dum zat

Coloured fraction(Area) = Length x Breadth

Chei dumna fraction a dahin = (A zau zawng) = A Dung x A Vang

39C VCM-GM-NLF

Multiplication Picture Procedure Answer

1

2x

1

4

1/2

1/41 x 1

2 x 4

1

8

1

2x

1

3

1

4x

2

3

1

4x

2

32

1

2

5x

3

6

1 x3

4

Pun tir A lem A kal hmang Chhanna

40C VCM-GM-NLF

2

1x

1

2

2

1/2 = 1

2

1x

1

2= 1

3

2x

2

3

3

2x

2

3=

3

4x

4

3

3

4x

4

3=

41C VCM-GM-NLF

x = 3

21

2

3x =

5

3

3

5

x = 2

51 x = 1

1

8

x = 15 x = 11

Here, the product of two numbers is 1. Such numbers are called as multiplicative inverse or

reciprocal of each other.

4

5x

5

4

4/5

5/4

= 1

4

5x

5

4= 1

Number

Reciprocal

6

1

6

3

5

1

10

5

3100 1 0

Not defined

Hetah hian number pahnihin a hring chhuah te chu 1 ani a. Chung number te chu

multiplicative inverse emaw reciprocal (inlet thei) an ni.

In let thei Sawi fiah loh

42C VCM-GM-NLF

Write additive and multiplicative inverse :

Additive Inverse Multiplicative InverseNumber

14

+ 3

34

-

- 3.5

x

- 2a

0Does not exist.

काढता येत नाह�.

(+2) + (-2) = 0

Therefore (+2) and

(-2) are

additive inverse

of each other.

(5) x ( ) = 1

Therefore 5 and

are multiplicative

inverse of each other.

15

15

( -5 ) x ( - ) = 1

Therefore (-5) and (- )

are multiplicative

inverse of each other.

15

15

Chuvang chuan (+2)

leh (-2) te chu a let

zawng a in belh ve ve

thei an ni ta ani.

te chu a let zawng a puntir theih ve ve an ni.

Chuvang chuan 5 leh 15

Chuvang chuan (-5) leh

a puntir theih ve ve an ni.

( - ) te chu a let zawng

15

A let zawng a belh theih leh puntir theih te ziak rawh.

43C VCM-GM-NLF

Observe the equality on slates. Draw appropriate number of roties on empty slates.

= =

= =

= =

= =

1

2

3

1

Bawm (Slates) a mi te hi an in tluktlan dan en la.

Chhang (roties) in mil zat tur a sir a bawm ah hian ziak rawh.

44C VCM-GM-NLF

1

2

Draw appropriate number of roties on empty slates. Write the missing numbers.

=

4

2

= =

4

2

= =2 2

2

1

How many groups of 2 can be made from 4? 2

=

1

= == =1

How many groups of 1/2 can be made from 1?

=

1

= =

1

= =1

How many groups of 1/4 can be made from 1?

1

1

2

1

4

1

4

Chhang (roties) in mil tur bawm ruak ah hian ziak rawh. Number kimlo ziak bawk la.

4 atang hian 2 theuh awmin group engzat nge kan siam theih ang?

1 atang hian ½ theuh awmin group engzat nge kan siam theih ang?

1 atang hian ¼ theuh awmin group engzat nge kan siam theih ang?

45C VCM-GM-NLF

= =

2

1

How many groups of 1/2 can be made from 1? 2

Make the required parts of the number in numerator and find the answer.

= =1

= =1

=1

1

2

1

1

2

2

1

4

1

1

2

3 =

1 atang hian ½ theuh awmin group engzat nge kan siam theih ang?

A tul angin number a chung zawk a awm tur hi siam la a chhanna ziak rawh.

46C VCM-GM-NLF

Study carefully the answers on previous page and fill in the blanks in the following questions.

=

1

1

2

2 =

1

1

3

=

1

4 =1

5

5

=

2

=1

5

1

3

3

= =1

3

4

9 12

Phek hmasa zawk a chhanna kha uluk takin zir la, a hnuai a kar awlah hian hnawh khat rawh.

Dividing by a fraction = Multiplying by its reciprocal

A them then hran (fraction) hmang a sem

= a inlet (reciprocal) hmang a puntir

47C VCM-GM-NLF

1

2

2

Division By making portions Procedure Answer

4=

1

2

1x

2

1=

4

14

1

4

1

1

4

3

1

4

1

2

1

8

1

4

1

4

3

2

Sem A hrang a siam in A kal hmang Chhanna

48C VCM-GM-NLF

Division By making portions Procedure Answer

1=

2

1

4x

2

1=

1

21

2

1

4

1

3

1

3

1

2

1

2

1

8

1

4

1

2

3

4

3

4

3

2

1

8

2

4

Sem A hrang a siam in A kal hmang Chhanna

49C VCM-GM-NLF

1

2x

3

1=

3

21

3

1

2 3

2

1

2

2

3

2

3

2

3

5

6

1

3

2

3

4

6

=3

2

This is one whole chocolate bar. Colour both fractions in

each division problem and write the answer by observation.

Also solve by procedure.

=

=

=

=

Hei hi chocolate tlawn khat a ni. Uluk taka enin then hrang la, i sem anga, chung te chu i chei

dawn nia. Kalhmang pangai tak hmangin a dikna zawng rawh

Division

Sem

By making portions

A hrang a siam in

Procedure

A kal hmang

Answer

Chhanna

50C VCM-GM-NLF

110

= 1

Fraction

Number written in housesDecimal

form

0.1

1210

Colour the squares to show fraction and write it in houses

and using decimal point. Fill all blank columns.

4100

35100

1.2

170100

0 1 0

U 110

1

100

1 2 0

U 110

1

100

U 110

1

100

U 110

1

100

U 110

1

100

Decimal point hmangin squares (kil li nei) te hi fractions ti

lang turin chei la, an mahni bawm theuhah ziak rawh.A

bawm ruak ah te i ziak dawn nia.

Picture

A lem Number an bawmah theuh ziah

51C VCM-GM-NLF

Writing expanded form into short form

300 + 20 + 4

3 x 100 + 2 x 10 + 4 x 1

300 + 20 + 4 +

110

610

3 x 100 + 2 x 10 + 4 x 1 + 6 x 110

110

110

5 x

324

324.6

324

400.5

21.65

1.52

0.5

11.58

205.67

250.09

324.6

2 x 10 + 5 x 1 +

300 +

3 x 100 + 5 x 1 + 6 x + 4 x

5 x 100 + 2 x 10 + 0 x 1 + 3 x + 1 x

Writing short form as expanded form

510

1100

1100

Expanded form te a tawi zawng a ziah.

A tawi a awm te hi expanded form (a sei zawngin) ziak rawh

52C VCM-GM-NLF

Dividing a number by 10, 100, 1000

number number x 10 number x 100 number x 1000

number number / 10 number / 100 number / 1000

Multiplying a number by 10, 100, 1000

125 12.5 1.25 0.125

42

0.234

1000

500.05

25

17.231

0.5463

0.0001

999.009

10, 100, 1000 hmanga number sem.

10, 100, 1000 hmanga number puntir.

53C VCM-GM-NLF

510

3 +

Different ways of converting Fractions to Decimals :

1. Making denominator as 10, 100, 1000 etc. (when denominator is a factor of 10,100,1000)

72

=7 x 52 x 5

=3510

= 3.550

=15

8125

=125

=20

2. Converting into improper fraction and then converting the remaining part into decimal.

72

= = = 3.550

=115

10520

=25

=28

12

3 +

3. Direct division including decimal point (Easier when denominator is a single digit number)

15 5 ) 1.0

0.2

01 0

- 1 0 0

15

= 0.2

128

Observe the numbers carefully and decide which method to use. Don’t jump to procedure.

Number chung leh hnuai awm point a chawh chhuah dan chi hrang hrang te:

1. A hnuaia number te hi 10, 100, 1000 in dah la ( a hnuai a number te hi 10,100, 1000 te in puntir rawh)

2. A chung number lian zawk, a hnuai number te zawk ah chantir la, a bak la bang zawng chu point ah chantir rawh.

3. Number lian zawk atanga number te zawk kan sem in point a chhuak tur ani( A hnuaia number pakhat a awm in a awlsam lehzual)

A number hi ngun tak a thlir chung in eng dan hmang a chawh tur nge i ngaihtuah ang. A chawh dan kal hmang en lo in.

54C VCM-GM-NLF

kilometer (km)millimeter (mm) centimeter (cm) meter (m)

Units of distance measurement and their conversions

110001000001000000

10 mm = 1 cm 100 cm = 1 m 1000 m = 1 km

11001000 0.001

1

1

Diameter of earth

12742 km

Thickness of paper

0.1 mm

Thil inkar hlat zawng tehna leh an inlehlin kual dan.

Khawvel zau zawng

Height of MountEverest

8848.86 m

Mount Everest san zawng

Diameter of 10-rupee coin

2.7 cm

Cheng 10 pawisa thir zau zawng

Lehkha puan chhah zawng

55C VCM-GM-NLF

kilogram (kg)

Units of mass (weight) measurement and their conversions

1000 mg = 1 g 1000 g = 1 kg

milligram (mg) gram (g)

110001000000

11000

0.001 0.000001

One liter of water weighs

1 kg

1

0.001

A house-mouse weighs

An ant weighs

3 mg

8818 kg

Indian elephant weighs

19 g

Thil inhlawm khawm tehna leh an inlehlin kual dan

Thil inhlawm khawm tehna Bukna te tak te Thil rih zawng tehna

Tui liter 1 rih zawng

In a awm thin sazu rih zawng

Fanghmir rih zawng

India sai rih zawng

56C VCM-GM-NLF

Plot the numbers on number lines.

0 1-1

2, - 2, 0.5, - 0.5, 2.5, -2.5, 8.1, - 7.1

0 1 m0.1m

0.01m

0.1 m, 0.01m, 0.35m, 50 cm, 1.5 m, 0.7 m, 0.06 m, 161 cm

0 0.1- 0.1

0.2, - 0.2, 0.35, - 0.35, 0.83, -0.54, 0.54

0.1

0.01, 0.001, 0.15, 0.05, 0.083, 0.061, 0.005

0

A hnuaia number te hi a rin ah khuan dah la, chhinchhiah tel zel rawh.

57C VCM-GM-NLF

Write the numbers in houses and do the addition or subtraction.

T U 110

1

100

U 110

5.2 + 4

Answer Answer

63.5 + 9.75

U 110

T U 110

1

100

6.5 - 0.5 40 - 25.4

6 + 2.5 = 7 - 1.5 =

12.5 + 3 = 50 - 10.5 =

T U 110

Answer

60 + 0.5

T U 110

Answer

25 - 4.5

Answer Answer

Number te hi an in theuh ah dah la, belh emaw, paih hmang in chawk chhuak rawh.

58C VCM-GM-NLF

Chhanna Chhanna Chhanna

Chhanna Chhanna Chhanna

Write the numbers in houses and do the addition or subtraction.

H T U 110

1

100

H T U 110

1

100

324.25 + 28.05

Answer

उ�र

Answer

उ�र

63.5 + 9.75

H T U 110

1

100

H T U 110

1

100

629.75 - 58.9

Answer

उ�र

Answer

उ�र

40 - 25.4

5.5 + 4.5 = 7.5 - 1.5 =

10.48 + 3.05 = 50.0 - 8.05 =

Number te hi an in theuh ah dah la, belh emaw, paih hmang in chawk chhuak rawh.

59C VCM-GM-NLF

Fraction multiplicationEquivalent of answer having 10, 100 etc in denominator

Decimal form of answer

510

x 310

= 15100

15100

0.15

32

x 12

=34

75100

0.75

12

x 52

=

32

x 35

=

54

x 25

=

710

x 35

=

13100

x 110

=

950

x 910

=

125

x 925

=

Number chung leh hnuai awm puntir

A hnuai a number te chhanna intluk tlang 10,100 etc.

A chhanna a point chawh chhuah

60C VCM-GM-NLF

Multiplication of decimal fractions

0.5 x 0.3 = 5

10x 3

10=

15100

= 0.15

0.5 x 0.03 =

0.5 x 0.03 =

0.05 x 0.03 =

Observe the above answers. Did you discover the rule?

Multiply the two numbers as if there is no decimal point. Count the total number of digits on

the right of decimal point in two numbers. Put the decimal point in answer such that it will

have these many digits on the right of decimal point.

Solve using the rule.

15.03

x 5.7

36.24

x 10.02

0.009

x 0.9

Decimal fraction a puntirna

A chung a chhanna hi en la, a kalhmang i hrethiam em? Decimal point awm lo ang maiin

number pahnih puntir la,decimal point ding lam a digit awm zawng zawng kha number

pahnihin belh rawh. A chhanna ah decimal point kan dah ang a, tichuan digit tam tak kha

decimal point ding lam ah a awm ang.

A chawh dan tur diktak in chawk chhuak rawh.

61C VCM-GM-NLF

10.8

0.03=

10.8 x 100

0.03 x 100

Division of decimal fractions

Method 1 - Convert into fractions

=25

5= 5

2.5 x 10

0.5 x 10

2.5

0.5=

Method 2 - Multiply numerator and

denominator by an appropriate number

between 10, 100, etc to convert them into

integers.

2.5

0.5=

25

10

5

10

=25

10

10

5x = 5

10.8

0.03=

==

0.5

0.005=

0.5

0.005=

Decimal fraction semdan

Fraction a lehlin

A number chung leh hnuai te hi 10, 100

inkar hmang hian puntir tur, tichuan a

number pumpui( integer) a kan leh theihna

turin

62C VCM-GM-NLF

Find :

3.6 ÷ 0.2 7.75 ÷ 0.25

3.25 ÷ 0.5 30.94 ÷ 0.7

0.35 ÷ 5 7 ÷ 3.5

0.25 ÷ 0.5 0.7 ÷ 35

Zawng chhuak rawh.

63C VCM-GM-NLF

Data Handling

The first step is to collect data. This chart

has the data of number of persons in the

families of some of our teachers.

Let’s understand basic concepts of data

handling through one simple and real

example.

Name Number of members in the family

Baby Lalthakimi Chinzah

Zoramthanga

7

4

T.Vanlalhruaia 5

R.Lalzuimawia 2

Andrew 3C.Lalrotluangi 9

C. Zirthangzela 2

C.Lianzela 8Zorengpuii 3

Vanlalsiami 7

C.Lalthawmmawia 6

Lalchhanhima hnamte 9Lalsangliana 5

Ramtharnghaki 3Zonunhrangi 2

R. Vanlalhluna 3

Lalhrinsaka Pachuau 5

V Lalremruata 6

6Michael Lalkrosvuana

Thil hmuhchhuah te dah khawmna/ fawm khawmna

He data ah hian kan zirtirtute

chhungkaw zat tarlan ani.

A hmasa ber hi data lakkhawm ani a.

Data handling tobul hi entirna hote

hmang in hriatthiam tum ila.

Hming Chhungkua a member awm zat

64C VCM-GM-NLF

Observe the chart on previous page and answer the following questions :

Ÿ What is the number of members in the

largest family?..........................

Ÿ Can the minimum number of family

members be zero?........................

Ÿ How many families have 3

members?..........................

Ÿ How many families’ information do we

have? (N).................................

Ÿ What is the total number of persons in all

families together? (M)............................

Ÿ How many families that have 2

members?......................

Ÿ How many members are there in Baby’s

family?....................

Ÿ Whose family has 7

members?..........................................

Ÿ What is the number of members in the

smallest family?....................

This is called as MEAN.

Ÿ Can the maximum number of family

members be 100?...........................

Ÿ The number of members that has come

more often is ...........................

Ÿ What is the average of number of family

members? (M/N)..................................

This is called as MODE.

65C VCM-GM-NLF

Ÿ Chhungkaw engzah chanchin nge kan

hriat theih? (N)..................................

Ÿ Chhungkaw member pathum nei te kha

engzah nge an nih?.............................

Ÿ Member zat lang zing ber chu ................

Ÿ Tu te chhungkua ber ah nge member

pasarih awm?...................................

Ÿ Chhungkaw member pahnih nei te kha

engzah nge an nih?.............................

Ÿ Chawhrualin chhungkaw member

engzah nge an nih? (M/N) ...................

Ÿ Chhungkaw tam ber kha member

engzah nge an nih?...........................

Ÿ A vai a belh khawm in chhungkaw

member engzat nge awm? (M) ...............

Ÿ A tam thei ang ber a chhutin chhungkaw

member zat hi za a ni thei angem?.........

He thil te hi number chawhrual tihna ani.

Ÿ Chhungkaw tlem ber te kha member

engzah nge an nih?...........................

Ÿ A tlem thei ang ber a lak in chhungkaw

member awm miah lohna a awm thei

ngem?..................

Ÿ Baby-I te chhungkua hi engzah nge an

nih?..................

Page hmasa ami chart en chungin a hnuaia zawhna te hi chhang rawh:

He thil hi number lang tam ber (Mode)

tihna ani.

2, 2, 2, ...................................................................

This is called as MEDIAN.

Arrange the data of number of family

members in ascending order (from smallest

to biggest). Also include the repeated

numbers.

Check whether you have got 19 numbers in

the above list.

The middle value of this list is ....................Number i hmuh chhuah hrang hrang atanga

a lai ber chu ........................

A chunga i chhanna hi en la, number

sawmpakua i hmuchhuak em en rawh.

For our data of families :

Mean =

Mode =

Median =

Mean = Average

Median (The middle number when you

arrange the data in ascending order)

Mode = The number that comes more

often

Chawhrual = Chawhruala chhuak

Number lang tam ber = Number

lang tam ber

Number te ber atanga number lian

ber i rem chhoh a number a lai tak

a awm.

66C VCM-GM-NLF

Number te ber (number tlem atanga tam

ber) atanga tan in chhungkaw member zat

hi rem tur ani. Number inzia knawn a awm

pawn ziak tel zel rawh.

He thil hi number a laia awm lak chhuah tihna ani.

Mode = Number lang tam ber =

Mean = Chawhrual =

In chhungkaw zat :

Median = A lai tak =

Daily income of 11 families is given here :

500, 940, 650, 500, 470, 2330, 570, 540, 600, 500, 3400

Mean =

Mode =

Median=

From the average income we may feel that the income of families is ok.

A chawhrual a lak in chhungkaw sum lakluh zat hi a awm tawk kan ti mai thei.

Mode gives us an idea of what more families are earning.

Mode chuan chhungkaw tam zawkin an thawh chhuah zat min kawhhmuh.

Median tells us the middle number.

Observe mean, mode and median carefully. What can you say about the incomes of these families?

67C VCM-GM-NLF

Chhungkaw sawmpakhat te nitin a sum lakluh zat chu:

Chawhrual

Number lang tam ber

A lai tak

Median chuan number a laia awm min kawhhmuh .

Mean, mode leh median te hi uluk tak a en chung in heng chhungkaw ho sum lakluh ah hian enge sawi tur i neih?

Measu

re the follo

win

g a

ngle

s usi

ng p

rotr

act

or

an

d fi

ll in

the c

hart

.

Angle

Short

Nam

eLong n

am

e 2

ways

Measu

reTyp

e o

f angle

ÐB

ÐA

BC

040

Acu

te a

ngle

ÐC

BA

Q

ZX Y

N

L

M

GE

F

A

BC

P

R

68C VCM-GM-NLF

A h

nuaia

tarlan te h

i Pro

tract

or

hm

angin

teh la

, a k

ara

wl t

e k

hu h

naw

hkh

at ra

wh.

Kil

A h

min

g la

m taw

iA

sei z

aw

ng a

a h

min

g c

hi h

nih

inT

he

Eng a

ng k

il nge a

n n

ih

Figure Information

ÐABD = .........................................

ÐDBC = .........................................

ÐABD + ÐDBC = ...........................

ÐABC = .........................................

ÐRMQ = .........................................

ÐPMS = .........................................

ÐRMP = ........................................

ÐQMS = .......................................

ÐRMQ = Ð ...................................

ÐRMP = Ð ...................................

ÐMOQ = .........................................

ÐNOQ = .........................................

ÐMOQ + ÐNOQ = ..........................

ÐMOP = .........................................

ÐPON = .........................................

ÐMOP + ÐPON = ..........................

Fill in the blanks without actually measuring the angles.

ÐEBD = ...............................

ÐABD = ...............................

ÐCBE = ...............................

ÐABC + ÐCBE + ÐEBD + ÐABD = ...........

A

B

D

C

M

P

R

Q

S

035

055

050 0130

A

080

O

Q

N

M

P

C

B

E

D

0140040

69C VCM-GM-NLF

A lem An chanchin

A kil teh kher lovin a karawl dah khat rawh.

Triangle

Sid

es

Angle

sTyp

e b

ase

d o

n s

ides

Typ

e b

ase

d o

n a

ngle

s

AB

= 4

cm

AC

= 4

cm

BC

= 4

cm

0Ð

A =

60

0Ð

B =

60

0Ð

C =

60

Acu

te a

ngle

d triangle

Equila

tera

l triangle

A

BC

5

060

060

060

B

030

C0

30

0120

3

5.2 5

054

080

046

4.2

DA

BC

A

P

Q

R

3.6

70C VCM-GM-NLF

Kil

thum

nei

A s

ir (

a p

ang)

A k

ilA

sir a

wm

dan

ata

ng a

an n

ihna

A k

il aw

mdan

ata

nga a

n n

ihna

Triangle

Sid

es

Angle

s

4.5 5

2.7

080

067

033

045

045

090

L

M

N

PQ

R

030

030

012

0

X

YZ

5

3.5

3.5

Measu

re the s

ides

71C VCM-GM-NLF

Kil

thum

nei

A s

ir (

a p

ang)

A k

il

Typ

e b

ase

d o

n s

ides

A s

ir a

wm

dan

ata

ng a

an n

ihna

Typ

e b

ase

d o

n a

ngle

sA

kil

aw

mdan

ata

nga a

n n

ihna

Match the given triangles with their types.

Type based on angles Type based on sides

Acutetriangle

Equilateral triangle

Righttriangle

Isoscelestriangle

Obtusetriangle

Scalenetriangle

Can a right angled

triangle be equilateral ? _______________

Can equilateral triangle be

obtuse angled triangle ? ______________

72C VCM-GM-NLF

An nihna milin heng kil thum nei te hi thai zawm rawh.

A kil awmdan a zir a an nihna A sir awmdan atanga a nihna

Right angle triangle hi a sir a inchen vek

thei em? _______________________

Triangle sir inchen vek equilateral hi

obtuse angled triangle a ni thei

angem?_______________________

AÐA = .........................................

ÐB = .........................................

ÐC = .........................................

ÐA + ÐB + ÐC = .........................B C

P Q

R

ÐP = .........................................

ÐQ = .........................................

ÐR = .........................................

ÐP + ÐQ + ÐR = .........................

ÐU = .........................................

ÐV = .........................................

ÐW = .........................................

ÐU + ÐV + ÐW = .........................

U

V W

ÐL = .........................................

ÐM = .........................................

ÐN = .........................................

ÐL + ÐM + ÐN = .........................

L

M N

ÐX = .........................................

ÐY = .........................................

ÐZ = .........................................

ÐX + ÐY + ÐZ = .........................

X

Y Z

For each triangle, measure the angles and find the sum.

73C VCM-GM-NLF

Triangle (kil thum nei) theuh te hi teh la, belh khawm rawh.

Angles of a triangle add up to

Kil thum nei te kil zawng zawng zat chu

Figure Observe and Find

ÐA = .........................................

ÐQ = .........................................

ÐC = .........................................

ÐM = .........................................

ÐDEF + ÐEFG = .........................

ÐD = .........................................

L

M N

CB

A

090 025

030

030

060

060

P

QR

A

C

B

050

D E

F G

040070

ÐACB + ÐACD = 180

ÐACB + .......... = 180

ÐACB = .................

ÐA = ......................

A

050 0120

B C D

Sum of the angles of a triangle is 180. Use this to find the missing angles.

74C VCM-GM-NLF

Kil li nei kil belhkhawm zat chu 180 a ni a. A karawl dahkhah nan hmang rawh.

A lem Uluk takin thlir la zawng chhuak rawh.

A

060 0120

B C D

060

060

P

030

Q R S

060

090

0ÐA = ............... 0ÐB = ...............

0ÐACB = ............... 0ÐACD = ...............

0ÐP = ............... 0ÐQ = ...............

0ÐPRQ = ............... 0ÐPRS = ...............

0ÐACD = Exterior angle = ..........

Sum of its interior

opposite angles = ÐA + ÐB0

= ..........

0ÐPRS = Exterior angle = ..........

Sum of its interior

opposite angles = ÐP + ÐQ0

= ..........

75C VCM-GM-NLF

Protractor hmangin heng kil thum nei pathum te hi teh la, chutiang ang bawkin a kil pawn

ami te thleng hian i teh dawn nia. Eng nge i hmuhchhuah?

Measure all three angles of the given triangles and the given exterior angle using a

protractor. What do you observe?

0ÐACD = A kil pawnlam = ..........

A chhunglam a kil lehlam belhkhawm0

= ÐA + ÐB = ..........

0ÐPRS = A kil pawnlam = ..........

A chhunglam a kil lehlam belhkhawm0

= ÐP + ÐQ = ..........

An exterior angle of a triangle is equal to sum of its interior opposite angles.

Triangle a a pawnlam kil te hi a chhunglam kil belhkhawm nen a intluk ani.

Kil thum nei lem eng ang pawh siam la.

A pawnlam ah a kil te siam tel bawk la.

Chumi ah chuan kil ruk (six angles) i

siam thei angem? A chung ami triangle

(kil thum nei) hi i en tel dawn nia.

Draw any triangle. Draw all exterior

angles. Could you draw 6 angles?

Check the above property for your

triangle.

P

030 025

R

Q

S

A

070 0130B

C D

ÐACD = ........................

ÐACB = ........................

ÐA + ÐB = ÐA CD

ÐA + 70 = ........................

ÐA = ...............................

Find the missing angles in the figure by using properties of triangles.

Use two properties.0

1) Angles of a triangle add up to 180 .

2) Exterior angle is equal to the sum

of interior opposite angles.

P M N Q

L

070 070

0160

0100

ÐPQR = ........................

ÐRQS = ........................

ÐL = ........................

ÐLMP = ........................

ÐLNQ = ........................

ÐZ = ........................

ÐXYZ = ........................x

Y z

76C VCM-GM-NLF

Triangle (kil thum nei) nihna hmangin a kil awmlo te khu zawng chhuak rawh.

1)Triangle kil zawng zawng belhkhawm hi

180' ani.

Heng an nihna pahnih te hi hmang rawh.

2)A pawnlam a kil awm te hi a chhunglam a

kil awm belhkhawm nen an intluk ani.

Figure

A lem

FindZawngchhuak rawh

Revision of basic constructions

Example : Draw a perpendicular bisector RS of line segment PQ using compass :

P Q

R

S

RS PQ

PA = AQ

A L M

=

........................

........................

Draw a perpendicular bisector CD of line segment AB using compass :

Example : Draw a perpendicular from an external point P to segment RS.

PA RS0Ð PAR = Ð PAS = 90

P

R SA

Draw a perpendicular from an external point A to segment BC.

B C

A

0Ð ............. = Ð ............... = .........

............. .............

Draw 3 different line segments, Draw perpendicular bisectors of each of them.Draw 3 different line segments. Take point A such that it is outside of each of these segments.Draw a perpendicular to each of the line segments from point A.

77C VCM-GM-NLF

Remkhawm dan ennawnna

Entirnan : Compass hmangin line chu a ding ngil in hmun hnihah tan tlangin siam la, a tan tlangtu pakhat zawk chu RS ani ang a, pakhat zawk chu PQ.

Compass hmangin line ngil intantlang CD leh AB siam rawh.

Entirna: Line ngil chu P point a hmangin siam la, a tantlang tu chu RS tiin i vuah ang.

'A' point a hmangin line pawnlam atanga tantlang turin BC siam rawh.

Line bung hrang hrang pathum siam la, a ngil (ding ngil) a intantlang theuh thei turin siam rawh. Line bung hrang hrang pathum siam rawh le. Point A chu i la hrang anga, line bung hrang hrang pawna awm turin. Point A atang chuan line bung hrang hrang te chu a ding ngil a awmin i siam ang.

Median : A line segment joining a

vertex to the midpoint of opposite side.

Draw all three medians of each triangle.

Is AD a median ? Why?

Is QS a median ? Why?

Q

S

R

P

B D C

A

78C VCM-GM-NLF

A lai a awm: Line bung hrangin a chhip a zawm

anga a lehlam pang point lai tak thlengin.

Triangle tinah hian a lai tak i siam ang.

AD hi a lai tak a ni em? Eng nge a chhan?

QS hi median a ni em? Eng nge a chhan?

Altitude : Perpendicular from vertex of a triangle to the opposite side.

A

B P C

C

Q A B

B

C A R

Table

Plum line

AP, CQ and BR are altitudes of DABC.

Draw all three altitudes of each triangle using set square.

A

B C

R

Q

D

P

RQ

S

U

T

L

NM

79C VCM-GM-NLF

A san lam: Triangle chhip lehlam atanga ngil (ding ngil) at.

AP, CQ leh BR te hi ABC te san zawng a ni.

Set square hmangin triangle pathum san lam siam rawh.

Draw all three medians and all three altitudes of DABC.

A

B C

Is AP an altitude of DABC? Why?

B P C

A

Is QR an altitude of DPQS? Why?P

Q

R S

Is LM an altitude of DLMN? Why?L

M N

80C VCM-GM-NLF

AP hi ABC te san zawng a ni em? Eng nge a chhan?

QR hi PQS te san zawng a ni em? Eng nge a chhan?

LM hi LMN te san zawng a ni em? Eng nge a chhan?

ABC te atang hian a lai tak pathum leh a san zawng pathum siam rawh.

Pythagoras Theorem.

Measure all three sides and verify. Use calculator.

Triangle Side 1 Side 2 Hypotenuse Verification (Use calculator)

a

b

c

xz

y

p

qr

e

fd

l

n

m

a=........... b=........... c=............

2 2a =............ b =............

2 2a + b =................

2c =...............

x=........... y=........... z=............

2 2x =............ y =............

2 2x + y =................2z =.................

p=...........

d=...........

q=...........

e=...........

r=............

f=............

2 2p =............ q =............

2 2p + q =................2r =.................

2 2d =............ e =............

2 2d + e =................2

f =.................

l=........... m=.......... n=............

2 2l =............ m =............

2 2l + m =................2

n =.................

81C VCM-GM-NLF

A sir leh a pang pathum te hi teh la, finfiah rawh. Calculator i hmang dawn nia.

Kil li nei A sir 1 A sir 2 Finfiahna (Calculator hmang rawh)

From the above examples is the sum of

squares of right-angle making sides of a

triangle equal to the square of hypotenuse?

A chunga entirna in a tih lan chu right-angle squares belhkhawm in triangle sides a siam a, chu chu hypotenuse square nen a in tluk ani.

Find a Find b

Find x Find m

3

4

a

610

b

5

5x 7

9.9

m

82C VCM-GM-NLF

In any right triangle, the sum of squares of right-angle making sides is equal to the square of hypotenuse.

Pythagoras Theorem

Find the missing side using Pythagoras theorem.Pythagoras theorem hmangin a sir a awl awm te hi zawng chhuak rawh.

a zawng rawh.

Right triangle ah chuan right angle pang a square belhkhawm chuan square eptu sir a tluk ani.

b zawng rawh.

m zawng rawh.x zawng rawh.

Draw and cut this triangleEveryone got the same triangle?

Test of congruency

Side 1 = 6 cmSide 2 = 8 cmSide 3 = 5 cm

SSS is the test of congruency

0Angle 1 = 80

0Angle 2 = 70

0Angle 3 = 30

Side 1 = 8 cmSide 2 = 8 cmSide 3 = 11 cm

Side 1 = 7 cmSide 2 = 9 cm

0Angle between these two sides = 50

Side 1 = 6 cmSide 2 = 8 cm

0Angle between these two sides = 90

0One angle = 90Side 1 = 8 cmHypotenuse = 13 cm

Side BC = 7 cm0Ð B = 600Ð C = 40

In triangle ABC,

Side QR = 11 cmSide PQ = 6 cm

0Ð R = 30

In triangle PQR,

0Ð L = 800Ð M = 600Ð N = 40

In triangle LMN,

83C VCM-GM-NLF

Triangle chu i siam anga i chepthla bawk ang

You will get two different triangles

Triangles inanglo pahnih i nei ang

No

Aih

AAA is NOT thetest of congruency

AAA chu test of congruency A NILO.

SSS chu test ofcongruency a ni.

Use scale, protractor, compass and pencil to draw the following triangles on loose papers. Write the names of vertices inside the triangle. Cut out the triangle. For each triangle check whether everyone’s triangle is congruent (exactly identical)

Lehkha puan awl ah heng a hnuaia triangle te hi scale, protractor, compass leh pencil hmangin siam la, triangle te chu i chep thla dawn nia. Triangle tinah chuan triangle dang nen an in ang em tih i endik zel ang.

Mi zawng zawngin triangle inang an nei vek em?

A inang em tih endik na

Yes

Aw

See the markings on two triangles.

By which test are they congruent?

Write the congruency of sides and angles

Have you discovered that

- if SSS, SAS, ASA, AAS and Hypotenuse-Side (HS) are the same in two triangles,

you get congruent trangles.

- You may not get congruent triangles if SSA and AAA are the same in two triangles.

B C

A

Q R

P

ABC @ PQR

Ð A @ Ð P

Ð B @ Ð Q

Ð C @ Ð R

AB @ PQ

AC @ PR

BC @ QR

By SSS test

R

S

T

D

E

F

D

E

F

U

V

W

X

Y Z

A

B C

090 090

K

L M

S

T U

84C VCM-GM-NLF

nih chuan, triangle thuhmun pahnih in nei dawn ani.

I hmuchhuak em; -SSS, SAS, ASA, AAS leh a pang hypotenuse (HS) te hi triangle inang pahnih an

-SSA leh AAA te hi triangle inang an nih chuan triangle thuhmun pahnih in nei lo ang.

Triangles pahnih a in thai te hi i hmu em? Eng

endikna hmangin nge thuhmun an nih? huhmun

leh inang an nihna kil leh a sir te ziak rawh.

Sanika has 10 pencils. Sarang has 5 pencils.

Compare in two ways :

Sanika has 2 times that of Sarang. This kind

of comparison by division (or by mutiple) is

called as ratio.

2

1

10

5= =

Sanika’s pencils

Sarang’s pencils

2

1

10

5= =

सा�नका�या पेि�सल�

सारंग�या पेि�सल�

1)

2)

1)

2)

The ratio is also written as 2 : 1

We read it as ‘Two is to One’

= =Sarang’s pencils

Sanika’s pencils

ratio == :

= =Sanika’s pencils

Total pencils

ratio == :

= =Sarang’s pencils

Total pencils

ratio == :

Ratio

Inkungkaihna

Sanika-i chuan pencil 10 a nei a.

Sarang'an pencil 5 a nei a. Pencil an neih

zat hi chi hnih in han khaikhin teh le.

Sanika-i chuan Sarang-a pencil neih let

hnih a nei a. Hetianga sem (emaw puntir)

tura an inkungkaihna kan khaikhinna dan

hi ratio chu a ni.

Hmun hnih a then a hmun khat' tiin

kan chhiar thin.

Ratio hi 2:1 angin kan ziak thin ani.

85C VCM-GM-NLF

Of the 24 hours day, Mangesh sleeps for 8

hours.

He is awake for ____________ hours.

= =Hours of sleep

Hours of being awake= =

A teacher has to check total 60 notebooks.

She has finished checking 36 notebooks.

Find the ratio of work done to the total work.

There are 30 girls and 24 boys in a class.

Find the ratio of girls to boys.

Nikhat a darkar 24 chhung hian Mangesh-a

chu darkar 8 a mu a.

Darkar___chhung a harh a.

A mut chhung darkar

A harh chhung darkar

Zirtirtu chuan notebook 60 endik tur a nei a.

Chung zingah chuan notebook 36 a endik zo

tawh a. A endik zawh tawh zat ratio zawng

chhuak rawh.

Class khatah chuan hmeichhia 30 leh mipa

24 an awm a. Hmeichhia leh mipa te

inkungkaihna (ratio) zat hi zawng chhuak

rawh.

86C VCM-GM-NLF

Mother took 4 katories of rice and 2 katories

of dal to make Khichadi. Write the following

numbers and ratios.

The number of people is doubled because of

guests. Now she needs total 12 katories of

grains. The ratio of rice : dal has to remain

the same. Find the quantity of rice and dal in

it.

=Rice : Dal :

=Dal : Rice :

=Rice : Total grains :

= :Dal : Total grains

Rice = katories Dal = katories

Total grains = katories

=Rice : Dal :


2 1

12

2

1=

We find the

equivalent number

such that numerator

and denominator

add to 12.

2

1=

=Rice : Dal :

=Dale : Rice :

=Rice : Total grains :

= :Dal : Total grains

Ka nu in Khichadi siam nan buh katories 4

leh dal katories 2 a hmang a. A hnuaiah

hian a number leh ratio te ziak rawh.

Mipui lokal khawm te chu mikhualte avang

chuan a letin an pung a. Chuvang chuan buh

leh dal chu Katories 12 a mamawh belh ta a.

Buh leh dal ratio awmsa ti danglam lovin buh

leh dal a hman ngai zat zawng chhuak rawh.

Number intluk

chhut chhuak tur

hian a chung leh a

hnuai ah 12 belh

ve ve rawh.

87C VCM-GM-NLF

=Rice : Dal :


2 1

12

Quantities Process Ratio

Find the ratio of given quantities. (Note : To compare two quanitities the units must be the same.)

16, 2016

20

4

5

4 : 5

100, 60

1 hour, 30 min

1 km, 1 m

150 cm, 100 cm

12, 6, 60 12, 6, 60

2 1 10

2 : 1 : 10

100, 75, 50

A hnuai ami te hi an inkungkaihna zawng chhuak rawh. (Thubelh: Number pahnih te

inkungkaihna khaikhin tur hian number kan lak zat hi a in ang tur ani)

88C VCM-GM-NLF

Nora

and G

ozo

love

carr

ots

.

Let’s

div

ide the c

arr

ots

am

ong them

as

per

a d

iffere

nt ru

le e

ach

tim

e.

Nora

Gozo

Tota

l Carr

ots

Logic

of

the r

ule

Ratio

(N

ora

: G

ozo

)E

quiv

ale

nt R

atio

sN

ora

gets

G

ozo

gets

12

Equal t

o b

oth

1 : 1

1 1

6 6=

66

12

Nora

’s w

eig

ht is

dou

ble

that of G

ozo

.S

he s

hould

get

double

carr

ots

.

2 : 1

12

1 : 3

Gozo

is 3

tim

es

more

act

ive.

So h

e s

hould

get 3 tim

es.

Dis

trib

uti

ng

a q

uan

tity

in

a g

iven

rati

o.

Rati

o a

wm

hm

an

g a

sem

Nora

leh G

azo

te c

huan c

arr

ot an n

gain

a e

m e

m a

, a

hnuaia

tih

dan tur

hm

ang h

ian I h

an s

em

daw

n the a

ng.

Carr

ot

belh

kha

wm

Dan k

alh

mang

Inku

ngka

ihna

(Nora

: G

ozo

)In

kungka

ihna in

zat

Nora

-I h

ang z

at

Gozo

- a c

hang z

at

A in

zat ve

ve in

Nora

chu G

ozo

aia

a le

t a a

rih

ava

ngin

a le

tin a

chang tam

ang.

Gozo

chu N

ora

aiin

a le

t th

um

in a

harh

vang z

aw

k a. C

huva

ng c

huan a

let th

um

in a

chang tam

ang.

89C VCM-GM-NLF

Total Carrots

20

20

20

24

Nora gets Gozo gets

15 5

Ratio (Nora : Gozo)

3 : 1

1 : 1

Equivalent Ratios

15

5

3

1=

3 : 2

16 4

24

24

24

15 9

7 : 5

16 8

Carrot

belhkhawm

Nora

-I chang zat

Gazo

-a chang zatInkungkaihna inzat

Inkungkaihna

(Nora : Gozo)

90C VCM-GM-NLF

Ketaki and Pranav started a stall of lemon juice. Ketaki spent Rs. 100 and Pranav spent Rs. 80

to buy materials. They decided to share the profit in the ratio of their investment (money spent )

as both were putting in equal efforts to run the stall.

Total Profit

18

36

45

Ketaki’s investment

Pranav’s investment= = Ratio :=

Equivalent Ratios Ketaki gets Pranav gets

5

4

10

8= 10 8

= =

100

80

5 4

:5 4

100

80

90

Day

1

2

3

4

Ketaki leh Pranav te chuan Nimbu tui zawrhna dawr an hawng a. An bungrua hman ho lei nan

Ketaki chuan cheng 100 a thawh a, Pranav chuan cheng 80 a thawh ve bawk a. an dawr siam

atanga an hlawkna hmuh te chu an pawisa thawh dan a zir a insem an tum a.

Ketaki pawisa peipun

Pranav-a pawisa peipun

NiHlawkna

belhkhawmInkungkaihna inzat Ketaki chang zat

Pranav

-a chang zat

Inkungkaihna =

91C VCM-GM-NLF

Two friends A and B, earned a profit of Rs. 800/- by selling vegetables. They decided to sharethe profit in the ratio of number of members in their families as 3 : 5. How will the share Rs. 800?

Three friends A, B and C earned 2500 rupees by selling vegetables. How will you distribute it

as per the ratio of number of members in the family which is 2 : 3 : 5?

Thian pahnih A leh B te chuan thlai anhralhna atangin cheng 800 an help a. an thlai hralhna chu

an chhungkaw zat ratio 3:5 angin insem an tum a. Cheng 800 hi engtiangin nge an insem ang?

Thian pathum A, B leh C te chuan thlai an hralhna atangin cheng 2500 an hlawh a. Engtiangin

nge chhungkaw member ratio 2:3:5 ang zelin I sem ang?

92C VCM-GM-NLF

Number

10

100

1000

5000

2500

2400

ProcessRatio

3 : 2

3 : 2

7 : 5

Distribution

6, 4

Divide the given number in the given ratio.

3 and 2 given once, 5 are over. We can give two times. So 6 and 4

3 : 2

1 : 4

1 : 4

A hnuaia number te hi an ratio in ziak ang zelin sem rawh.

Inkungkaihna A tih dan tur Sem

3 leh 2 te chu vawikhat kan pe a,

5 a zo a. Vawi hnih kan pe leh a,

chutichuan 6 leh 4 angin.

93C VCM-GM-NLF

Indian flag has the ratio of length : breadth as 3 : 2. The flag of Bangladesh has the ratio 5 : 3.Which flag looks longer, which flag has more ratio?

=3

2 6=

5

3 6

3

2 6

5<

To compare 3/2 and 5/3 let’s find their equivalent fractions having the same denominator, 6.

One glass of lemonade has lemon : water in the ratio 1 : 4. The other glass has ratio 2 : 7

Which lemonade will taste more sour?

flag looks longer.

A division has 20 girls and 18 boys. B division has 20 girls and 20 boys. C division has 18 girls

and 20 boys. Which class has the highest ratio of girls to boys?

Indian flag len zawng a dung:a vang ratio chu 3:2 a ni a, Bangladesh flag ratio ve thung chu

5:3 a ni. Khawi flag nge sei hmel zawk a, khawi flag ratio hi nge lian zawk ang?

3/2 leh 5/3 khaikhin tur hian a hnuai ah number 6 dah ve ve in a fraction inang kan zawng

chhuak dawn nia.

Flag lang sei zawk.

No pakhat a nimbu tui a tak zawng nimbu tui: Tui ratio chu 1:4 a ni a. No dang leh ami chu 2:7

ani ve thung a. Khawi no zawk khi nge thur zawk ang?

Division A ah chuan hmeichhia 20 leh mipa 18 an awm a, division B ah chuan hmeichhia 20

leh mipa 20, division C ah chuan hmeichhia 18 leh mipa 20 an awm bawk a. Khawi division

ah khian nge a ratio a lakin hmeichhia leh mipa sang ber ang.

94C VCM-GM-NLF

Percentage

Find equivalent fraction having 100 in denominator. Write using % sign.

FractionEquivalent having 100 in denominator Percentage

1

250%

50

100

1 x 50

2 x 50=

1

4

1

10

3

4

2

5

20%

90%

3

2

Za zelah

Fraction in ang a hnuai a 100 dahin zawng chhuak rawh. % sign hmang ang che.

100 a hnuaia awm inang Za zelah

95C VCM-GM-NLF

Colour the given portion.

15%

30%

40%

1

5

2%

3

10

4

25

9

20

A hnuaia in ziak ang zelin chei rawh.

96C VCM-GM-NLF

Join each figure with fraction or percentage of coloured portion.

15%

100%

40%

1

2

1

100

5%

1

4

4

5

A fraction emaw percentage in ziak ang zel in chei rawh.

97C VCM-GM-NLF

Mangoes Cost (Rs.)

1 10

5 ? 5

1

50

10=

We say, 1, 5, 10 and 50 are in proportion.

One mango costs 10 rupees. What is the cost of 5 mangoes?

Mangoes Cost (Rs.)

?

=

_____, ______, _____, _____ are in proportion.

_____, _____, _____, _____

One mango costs 8 rupees. What is the cost of 4 mangoes?

We write mangoes below mangoes, rupees below rupees. To find the cost of many from 1, we multiply the number by the cost of 1.

Equivalent ratios and four numbers in proportion

Theihai pakhat chu cheng 10 man a ni a. Theihai panga chu engzat man nge ni ang?

1, 5, 10 leh 50 te chu a inlen hleih dan

angin an awm kan ti a.

Theihai pakhat chu cheng 8 man a ni a. Theihai pali chu engzat man nge ni ang?

te chu an inlen hleih a.

Theihai hnuaiah theihai kan ziak a, pawisa hnuaiah pawisa. A hlawm a an hlutna pakhat

zawng chhuak turin a number chhuak tura chu a hlawma awm zat mal pakhat man zat nen

kan puntir thin.

98C VCM-GM-NLF

Ratios intluktlang leh number pali te an inlen hleih

Mangoes Cost (Rs.)

5 60

1 ? 1

5

12

60=

We say, 5, 1, 60 and 12 are in proportion.

5 mangoes cost 60 rupees. What is the cost of 1 mango?

Mangoes Cost (Rs.)

?

=


_____, _____, _____, _____

10 mangoes cost 70 rupees. What is the cost of 1 mango?

We write mangoes below mangoes, rupees below rupees. To find the cost of 1 from many, we divide the cost of many by the number.

Theihai panga chu cheng 60 man a ni a. Theihai pakhat chu engzat man nge ni ang?

5, 1, 60 leh 12 te chu a inlen

hleih dan angin an awm kan ti a.

Theihai sawm chu cheng 70 man a ni a. Theihai pakhat chu engzat man nge ni ang?


Theihai hnuaiah theihai kan ziak a, pawisa hnuaiah pawisa. A hlawm a an hlutna pakhat

zawng chhuak turin a number chhuak tura chu a hlawma awm zat mal pakhat man zat nen

kan sem thin.

99C VCM-GM-NLF

5 45

1

5 mangoes cost 45 rupees. What is the cost of 1 mango? From that find the cost of 7 mangoes.

Hours km

=


_____, _____, _____, _____

240 km in 4 hours. How many km in 5 hours?

7

45

5

45

5

7 x

5 45

745

5

7 x

=

Unitary Method

Theihai panga man chu cheng 45 a ni a. Theihai pakhat chu engzat man nge ni ang?

Chumi atang chuan theihai pasarih man zawng chhuak rawh.

Mangoes

Theihai

Cost (Rs.)

A man zat (Cheng)

Mangoes

Theihai

Cost (Rs.)

A man zat (Cheng)

Skipping this step

Hemi step hi kan tih hmaih ang.

Darkar 4 ah 240 Km. Darkar 5 ah Km engzat nge ni ang?

Darkar


100C VCM-GM-NLF

Read the table below. Write the question in words. Find the answer.

=


_____, _____, _____, _____

10 rupees off on purchase of 100 rupees. How many rupees off on purchase of 60 rupees?

Petrol in lit km

=3 60

5 ?

Cheng 100 man hu thil lei in cheng10 in a tlawm a.

Cheng 60 man leiin cheng engzatin nge tlawm thei ang?


A hnuaia table hi chhiar la. Zawhna te hi a thu in ziak rawh.

A chhanna zawng chhuak bawk rawh.

Petrol bur a awm


_____, _____, _____, _____ te chu an inlen hleih a.

101C VCM-GM-NLF

=

A car goes 20 km in 1 liter petrol. How many liters of petrol will be needed to go 100 km?

liter km

4 notebooks cost Rs. 80/-. How many notebooks can you get in 200 rupees?

Manas walked 10 km in 2 hours. How much distance will he walk in 3 hours?

Car chuan petrol liter khat in 20 km a tlan thei a. 100 km tlan turin petrol liter engzat nge a

hman ngai ang?

Notebook pali chu cheng 80 man a ni a. Cheng 200 in notebook engzat nge I lei theih ang?

Darkar hnih chhungin Manas-a chu 10 km a hla a kal a.

Darkar thum chhungin engtia hla nge a kal theih ang?

C VCM-GM-NLF 102

=

Meena has got 40 marks out of 50. How many marks has she got out of 100?

Mangesh got 180 out of 200 in mathematics. How many percent marks has he got?

50

40

100

When we make the denominator of fraction as 100,

we call the number in numerator as ‘per cent’.(per hundred).

Meena has got 80 per cent marks. We write it as 80%.

80

30 marks out of 50 means how many per cent?

=200 100

_____ %

=100

_____ %

= _____

90% means how many out of 50?

Percentage

Meena chuan Mark 50 ah mark 40 a hmu a. 100 zelah mark engzat nge a hmuh ang?

Fraction hnuaia 100 kan dah hian a chung zawk a

number chu 'Per cent' kan ti thin. Meena-I chuan 80

percent a hmu a, 80% tiin kan ziak thin.

Mangesh-a chuan mark 200 ah 180 a hmu a, engzat percent nge a hmuh?

Mark 50 a 30 hi engzat percent nge?

90% hi 50 ah engzat nge?

103C VCM-GM-NLF

Za zelah

20 percent discount means a discount of how many rupees on 90 rupees of purchase?

100 20

90100

90 x 20

A discount of 100 rupees on purchase of 400 rupees. What is the percentage discount?

%

20% students are absent in a school having 500 students. How many students are absent?

Cheng 90 man ah 20 percent min tlawm hian cheng engzat man-in nge kan thil lei tur kha

kan lei theih ang?

Purchase Discount

Thil lei Tlawm

Cheng 400 man a cheng 100 min tlawm hian engzat nge min tlawmna percent zat?

Purchase Discount

Thil lei Tlawm

Zirlai naupang 500 zing a 20% te chu an kal lo a, naupang engzat nge kal lo?

104C VCM-GM-NLF

Four quantities in each of the following

situations are in proportion to each other.

We can write them as a pair of equivalent

fractions. Find the answers considering this

equivalence:

1. One mango costs 10 rupees. What is the

cost of 3 mangoes?

10. There are 15 girls and 10 boys in a

class. What is the percentage of girls in the

class? What is the percentage of boys in the

class?

9. 4 mangoes out of 400 were rotten.

What was the percentage of rotten

mangoes?

3 Mangoes 30 rupees

4. 15 rupees of discount on purchase of 100

rupees. How much

5. A car goes 20 km in 1 lit of petrol. How

much petrol would be needed to go 100

km?

7. 1/2 means what per cent? (How many out

of 100?)

1 Mango 10 rupees

discount on purchase of 60 rupees?

8. 90 per cent is how many out of 10

and how many out of 50?

1/3 = 10/30

2. Five mangoes cost 60 rupees, what is the

cost of 1 mango?

3. A car goes 240 km in 4 hours. How many

km will it go in 5 hours?

6. 40 marks out of 50 is equivalent to how

many marks out of 100?

4. Cheng 100 man thil kan lei khan cheng

15 min tlawm a. Cheng 60 man thil lei ta ila

cheng engzatnge min tlawm ang?

7. 1/2 hi za zelah engzat nge ni ang? (100

ah engzat nge?)

1 Theihai Cheng 10

2. Theihai panga chu 60 man ani a, theihai

pakhat hi engzat man nge ni ang le?

3 Theihai Cheng 30

3. Car chu darkar li chhungin 240 km lai a

tlan a. Darkar nga tlan dawn ta se km

engzat nge a tlan ngai ang?

5. Petrol litre 1 hmangin car chu 20 km a

tlan a. 100 km a hla tlan dawn ta ila petrol

litre engzat nge ka mamawh ang?

1. Theihai pakhat chu cheng 10 man ani a.

Theihai 3 man engzat nge ni ang?

Heng a hnuai a thil thleng atanga hmun li te

hi an in inkungkaih theuh a. A inkawp in

emaw a hlutna inangin emaw kan ziak thei

ang. A hlutna te ngaihtuah chungin a

chhanna te zawng rawh.

1/3 =10/30

6. Mark 50 ah mark 40 hi mark 100 a mark

engzat nen nge intluk thei ang?

8. Za ah 90 hi sawm (10) ah leh

sawmnga(50) ah engzat zel nge ni ve ang

le?

9. Theihai 400 zing a 4 te chu a tawih a. Za

ah engzat zel nge tawih ang a kan chhut

theih ang?

10. Class ah chuan Hmeichhia 15 leh mipa

10 an awm a. Za ah engzat nge Hmeichhia

zawng hi ni ang? Za ah engzat nge mipa

awm ang?

105C VCM-GM-NLF

A number that can be expressed in the form , where p and q are integers and q ¹ 0 is

called a Rational number.

pq

Rational numbers

31

32

– 36

–24

3

– 15

Numberpq

in form

Instandard form

331

62

93

3010

1

1.2

1.25

0

106C VCM-GM-NLF

Number awmze nei

ni ta chu Rational number a ni.

hmanga number kan sawifiah theih, chumi a p leh q te chu number pum ni si leh q ¹ 0 pq

Some equivalent rational numbers

Rational number intluk te Pawm tlan dan

Revise perimeter and area from class 6 - Pages 154 to 161.

Perimeter and Area

Find the area of right angled triangle by observation. Note that it is half of dotted rectangle.

Verify the area of each triangle using formula ( 1/2 x base x height )

Area of parallelogram = base x height. Find area of each parallelogram.

107C VCM-GM-NLF

A tlang leh a zau lam

Perimeter leh area class 6 zirlai phek 154 atanga 161 ennawnna.

Uluk taka en chungin right angled triangle zau zawng hi zawng chhuak rawh. Rectangle

chanve atanga chhunhan ani tih i hre dawn nia.

Formula hmangin triangle zau zawng te hi i finfiah dawn nia. (1/2 x base x height)

Parallelogram zau zawng = base x height.

Parallelogram zau zawng I zawng chhuak theuh dawn nia.

Circumference and diameter of circle

Measure the circumference and diameter of three flat circular objects like plate, bangle,

table top, etc. using a thread or using tailor’s tape.

For each circle find (circumference / Diameter). Use calculator to find this value.

Object Circumference (C) Diameter (d) C/d

Plate

Did you get C/d almost same for all circles?

Is it somewhere near 3.14? Or 22/7?

From the definition of pi, C = p x d

Diameter = 2 x radius.

C = p x 2 x r = 2 x p x r

Object Diameter (d) Radius( r = d/2) Circumference

C = 2 p r

Cycle rim

Plate

63 cm

28 cm

Earth 12742 km

The ratio of circumference of a circle to its

diameter is called as pi. It is written as p

Find circumference of the following objects. Use pi = 22/7 or 3.14. (Use calculator).

108C VCM-GM-NLF

Rinbial vettu leh rinbial tan tlangtu

Plate, ngun, dawhkan bial khuhna leh a dang; tlang leh a in tantlang na te khawl-la emaw

tailor's tape emaw hmangin the rawh. Rinbial tinah hian heng circumference (a tlang) leh

diameter (a intantlang) te hi i awng chhuak dawn nia. Calculator hmangin a hlutna te hi i

zawng chhuak dawn nia.

Anih loh pawn 22/7? Rinbial leh a tlang (a vettu) te a tan tlanga tehna (diameter) te nen a an inlaichin dan hi pi ani. Hetiang

hian an ziak thin p

C/d te hi rinbial tin deuh thaw ah khan a inang in i hmuchhuak em? 3.14 vel a ni em?

He objects atang hian rin bial (a vettu) hi zawng chhuak rawh. pi=22/7 emaw 3.14 hi hmang la. (Calculator i hmang dawn nia)

5 x 1 = ..............

a x 1 = ..............

3

2x 1 = ............

1 x = ............7

4

a

bx 1 = ............

1 x = ............a

b

(-a) x 1 = ...........

5 x 0 = ...............

a x 0 = ..............

3

2x 0 = ............

0 x = ............7

4

a

bx 0 = ............

0 x = ............a

b

(-a) x 0 = ...........

Find equivalent fractions of 2/3

2

3= = = = =

Find equivalent fractions of a/b

a

b= = = = =

Find equivalent fractions of 2a/3b

2a

3b= = = = =

Revision of basics - A tir atanga ennawnna

2/3 equivalent fraction zawng chhuak rawh

a/b equivalent fraction zawk chhuak rawh

2a/3b equivalent fraction zawng chhuak rawh

109C VCM-GM-NLF

Write the fraction in its reduced form.

10

15=

5 x 2

5 x 3=

2

3

12

30=

14

42=

45

30=

One student has done this.

Discuss what is wrong, why it is wrong :

45

30= 30 + 15

30= 15

Cancellation can be done only when the number is written as multiplication of its factors.

5 x 3

5 x 2=

Find :

a x b

a x c=

a x b

c x d=

10 x 3

15 x 2=

2x y

xy=

a =

2a

2 x 3 x 4

3 x 2 x 5=

xyz

3xy=

9

23a=

120

40=

x + y

xy=

150a =

2120a

Last problem : watch carefully. We can’t cancel when there is + sign.

A hnuaia fraction te hi a te thei ang berin siam rawh.

Naupang pakhatin hei hi a ti a. Eng nge dik lo

a, eng nge a chhan sawi ho rawh u.

A hnuaia numberte hi an number puntir ang a ziah chauhin a cancel theih ani.

Zawng chhuak rawh:

Problem tawp ber : ngun takin en la, belhna chhinchhiahna (+) a awm in kan cancel thei lo a ni.

110C VCM-GM-NLF

Equations - Preparations.

Picture Expression

x

2x

3x + 4

4x + 1

is an unknownx

x

+

- ( )

- ( )

+

111C VCM-GM-NLF

Intlukpui -Inbuatsaihna

A lem hriatloh ani. awmze entirna

Write the algebraic expressions :

Description

Sum of x and y

Algebraic Expression

x + y

xy

a and b both squared and added

5 added to three times the product of m and n

10 - yz

2x + 3x + 2

2 (3a + 6b)

pq14

Your age is x. Your brother is 5 years elder than you.Sum of your and your brother's age.

112C VCM-GM-NLF

Algebraic expressions te hi ziak rawh

Sawifiahna

I kum chu x a ni a. I unaupa chu nangai in kum 5 in a upa a. I kum leh i unaupa kum belhkhawm.

+ and total

Addition

+1 + 1 =

-1 - 1 =

+1 - 1 =

-1 + 1 =

+4 + 4 =

-4 - 4 =

+4 - 4 =

-4 + 4 =

+1 and -1 cancel by making a zero.

+4 and -4 cancel by making a zero.

+ 2 + 1 = + 1 + 1 + 1 = + 3

- 2 - 1 = - 1 - 1 - 1 = - 3

+ 2 - 1 = + 1 + 1 - 1 = + 1

- 2 + 1 = - 1 - 1 + 1 = - 1

- and total

Difference with sign

of bigger number

Use these rules and solve :

+ 5 + 3 =

- 5 - 3 =

+ 20 - 5 =

- 20 + 5 =

- 100 + 50 =

- 25 - 25 =

+ 200 - 100 =

- 100 + 150 =

113C VCM-GM-NLF

Belh

+1 leh -1 te zero siam a sut.

+4 leh -4 te zero siam a sut.

+ leh a belhkhawm

- leh a belhkhawm

Number lian zawk ah chuan a sign te a danglam

Heng dan hmang hian chawk rawh

2a + 3a =

- 2a - 3a =

+ 2a - 3a =

- 2a + 3a =

100x + 50x =

- 100x - 50x =

+ 100x - 50x =

- 100x + 50x =

+ 5 - 2 - 3 + 4 + 3 = + 7

- 3 - 4 + 6 - 2 + 1 =

- 2 - 3 - 1 - 5 - 4 =

+ 7 - 3 - 4 + 5 - 1 =

3a - 5a + 2a =

-4x - 2x + 7x =

2 2 2 25x - x + 3x - 2x =

10xy - 5xy - 5xy + 2xy =

+ 1 + 1 + 1 - 1 + 1 - 1 - 1 =

+ 5 - 5 + 5 - 5 - 5 + 5 =

a - a - a - a + a + a =

2 2 2 2 22x - x + x - 2x + x =

114C VCM-GM-NLF

2 Mangoes + 3 Apples

2 आबं े + 3 सफरचंद=+

+ =

You collected like terms together

x y 2x 4y

y x x x y

-x +y -2x +4y

-y +x +x +x -y

=

=

=

-3x + 5y

+3x - 2y

+ 3y

+

-3x + 5y + 3x - 2y

= 3y

OR

115C VCM-GM-NLF

A thuhmun in i lakhawm a

Write the addition problem by collecting like terms together. Solve it.

- 2x + 3y - y + 5x

+- 5x + y + 2y + x

3x + 2y

- 4x + 3y

- x + 5y

Vertical Addition

=

3x + 2y - 4x + 3y

3x - 4x + 2y + 3y

-x + 5y

=

=

OR

5x + 2y

12x + 7y+ 4a + 3b

a + 3b+

19x - 3y

2x + 5y+ 6a - 9b

- a - 4b+

2 2- 3xy + 6x y2 2- 5xy - 2x y+

Horizontal Addition

116C VCM-GM-NLF

A thuhmun te dahkhawm in belhna tur zawhna siam la, chawkchhuak rawh.

A ding zawng a belh

A khamphei zawng a belh

2 2 2 2 2 2 2 2m n + m - n - m n - m + n

2 2- 3xy + 3x y2 2- 5xy - 2x y+

2 2 x - 3xy + 2y2 2- x - 10xy + y+

Simplify

2 2 2 2 - 4m n + 3mn + m n - 2mn

20a - 5b - 15a + 20b

117C VCM-GM-NLF

Tiawlsam

2 2x - 3x + 2x - 2x

21b - 32 + 7b - 20

2 2 2- z + 13z - 5z + 7z

2 2 2 22x - 5y + 3 + 3y - 5x - 4

2 2 2 2a - 2ab - b + 2ab - a + b

118C VCM-GM-NLF

Add the given terms.

3mn , - 5mn , + 8mn , -4 mn

- 7xy + 5 , 12xy + 2 , 9xy - 8

a + b - 3 ,

5m - 7n

b - a + 3

, 3n - 4m

119C VCM-GM-NLF

Heng a hnuaia tarlan te hi belh rawh

Subtraction

Subtracting a positive number is like adding a negative number.

Subtracting a negative number is like adding a positive number.

- (+1 ) = - 1

- 2a - 3a =

- (+a) =

- (x + 5) = - x - 5

- (- x - 2) = + x + 2

- (- 1 ) = + 1

- 2a - 3a =

- (- a) =

- (- y + 2) = + y - 2

- (y - 2) = - y + 2

While removing the bracket of expression to be subtracted, invert the signs of all terms.

(+1 ) - (+ 1) = 1 - 1 = 0

(+1 ) - (- 1) = 1 + 1 = 2

(-1 ) - (+ 1) = - 1 - 1 = -2

(-1 ) - (- 1) = - 1 + 1 = 0

120C VCM-GM-NLF

Paihna

Positive number paih chu negative number belh ang a ni. Negative number paih chu positive number belh ang a ni.

Bracket te i phelh laiin heng paih tur sign te hi i letling dawn nia.

+ 5 - (+ 3) =

- 5 - (- 3) =

+ 20 - (- 5) =

+ 20 - (+ 5) =

- 100 - (+ 50) =

- 25 - (- 25) =

+ 200 - (- 100) =

+ 200 - (+ 100) =

+ 2a - (3a) =

- 2a - (- 3a) =

+ 2a - (- 3a) =

- 2a - (+ 3a) =

100x - (+ 50x) =

- 100x - (- 50x) =

+ 100x - (- 50x) =

- 100x - (+ 50x) =

121C VCM-GM-NLF

3a + (- 5a) - (- 2a)

=

- 4x - (- 2x) + (- 7x)

=

+ 5 - (+ 2) + (- 2) - (- 4)

+ 5 - 2 - 2 + 4

+ 5 + 4 - 2 - 2

+ 9 - 4

+ 5

=

=

=

=

No change in sign while adding.

Invert the sign while subtracting.

2 2 2 2 5x - x - (- 3x ) + (- 2x )

=

- (- 10xy) - (- 5xy) + (- 5xy) - 10xy

=

122C VCM-GM-NLF

Kan belh laiin a sign kan thlak lo ang.Paih laiin a sign te kan letling ang

7x - 3y

+ 12x + 7y-

- -

- 5x - 10y

4a + 3b

a + 8b-

19x - 3y

2x + 5y- 6a - 9b

- a - 4b-

2 2- 3xy + 6x y2 2- 5xy - 2x y-

2 2- 3xy + 6x y2 2- 5xy - 2x y-

x + y + z

x - y + z-- x - y - z

- x - y - z-

123C VCM-GM-NLF

4x + 3y - (x + 2y)

4x + 3y - x - 2y

4x - x + 3y + 2y

3x + y

=

=

=

Solve

- 3a - 2b - (-2a + 3b)

9x - 3z - (2x - 6y) 2 2(3y + 5y - 4) - (8y - y - 4)

p - (p - q) - q - (q - p) 2 2 2 23ab - 2a - 2b - (5a - 7ab + 5b )

2 24m - 3mn + 8 - (- m + 5mn) a - b - (a - b)

124C VCM-GM-NLF

Chawkchhuak rawh

Mother has p chocolates.

She wants to share them equally

among 4 children.

Child will get ________ chocolates.

One notebook costs 10 rupees.

What is the cost of n notebooks?

You have m rupees.

Your sister has 5 rupees more.

Your sister has __________ rupees

Your age is 12

Next year your age will be __________

Your age is a

Next year your age will be __________

Your height is h cm.

Your friend is shorter by 10 cm.

Your friend's height is ________ cm.

A child is x years old.

His mother's age is twice his age.

Mother's age is ___________

Father's age is 5 more than

mother's age. Father's age is __________.

Uncle's age is 3 less than father.

Uncle's age is _______________

125C VCM-GM-NLF

I kum chu zat chu a.

Nakum a i kum zat tur chu_____________

I kum chu zat chu 12.

Nakum a i kum zat tur chu_____

I unaunu in nangai in cheng 5 in a nei tam a.

I unaunu chuan cheng ____________ a nei.

m zat pawisa i nei a.

I sanzawng chu h cm a ni a.

I thian chu nang aiin 10 cm in a tawi zawk a.

I thian sanzawng chu __________ a ni.

Notebook pakhat chu cheng 10 a ni a.

n notebooks chu engzat man nge?

A fa pali te hnenah chuan intluktlang

takin sem a duh a. A fate chuan

_____chocolate an dawng ang.

Ka nu chuan chocolates p a nei a.

Naupang chu kum x a ni a.

A nu kum chu naupang kum let hnih a ni a.

A patea kum chu ______________

A patea kum chu a pa aiin a let 3 in a tlem a.

A nu kum zat chu ____________

A pa kum chu a nu kum let 5 a ni a. A pa

kum chu ______________

Evaluating the expressions :

x x + 2 2x 2x+2 2x 2x +2

3 3 + 2 = 5 2 x 3 = 6(2 x 3)+2= 6 + 2 = 8

3 x 3 = 9 9+2=11

5

1

0

Find the value of given expressions if x = -2

3x - 4= 3 x (-2) - 4= -6 - 4= -10

- 4x + 6

25x 2 x + 3x - 3

A hnuaia numberte hlutna chhutna :

A hnuaia number x = -2 hi a hlutna zawng chhuak rawh

126C VCM-GM-NLF

Cost of one notebook is x rupees. We purchase y such notebooks. What is the total cost?

xy= =

x = 7, y = 8

xy = (5 + 2) x 8 = 7 x 8= 56

x = 5 + 2, y = 8

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

x = 5 + 2, y = 5 - 4

One child did 5 + 2 x 8 and got 56.

Another child did 5 + 2 x 8 and got 21.

Can you figure out what happened?

Try using different calculators.

We have to tell the calculator which

operation to do first (where to put

brackets)

Cost of one notebook is 12 rupees. We purchase 6 such notebooks. What is the total cost?

For different values of x and y, find the value of xy.

x = 3 + 4, y = 3 - 2 xy =

Notebook pakhat man chu Rs.12 a ni a. Notebook 6 kan lei a, a vaia a man zat zawng chhuak rawh?

Notebook pakhat man chu Rs. x a ni a. Notebook y kan leia, a vaia a man zat zawng chhuak rawh?

A hranga x leh y hlutna hriatnan, xy hlutna zawng chhuak rawh.

A dang leh chuan 5 +2 x 8 ah 21 a hmu chhuak a.

Naupang pakhat in 5+2 x 8 ah 56 a hmu chhuak a.

Eng nge a chhan zawng chhuak rawh?

Calculators hrang hrangte han hmang chin la .

Chu calcutator chu a tih hmasak ber tur kan hrilh

ang. (khawiah nge bracket kan dahna tur)

127C VCM-GM-NLF

Find the value of given expressions using given values of variables. Also use calculator.

x = 8, y = 4xy

=

x = 5+3, y = 3+1xy

=

One child did 5 + 3 ÷ 3 + 1 and got 7.

Another child did (5+3) ÷ (3+1) and got 2.

Can you figure out what happened?

Try using different calculators.

We have to tell the calculator which

operation to do first (where to put brackets)

Find using formula. Also solve using calculator.

14

1

14

11 x

12

31 x

32

1

ab

1

12

6

A hnuaia number hlutna te zawng chhuak la. Calculator hmang bawk rawh.

Calculators hrang hrangte han hmang chin la.

Chu calcutator chu a tih hmasak ber tur kan

hrilh ang. (khawiah nge bracket kan dahna tur)

Eng nge a chhan zawng chhuak rawh?

A dang leh chuan (5+3) (3+1) ah 2 a hmu ÷

chhuak a.

Naupang pakhat in 5 + 3 3 + 1 ah 7 a hmu ÷

chhuak a.

Formula hmangin zawng chhuak la, calculator hmangin chawk bawk rawh.

128C VCM-GM-NLF

Find :

23

53

xa

b

c

dx

a

bcx

a

b

b

ax

2a

b

2b

ax

a

1

1

ax

a = 10, b = 2, Find :

ंa = 10, b = 2, याव�न �कमती काढा :

a

ba

1

b÷

1 b

a

1

Zawng chhuak rawh.

129C VCM-GM-NLF

Find :

23

53

x

y

a = 10, b = 2, Find :

ंa = 10, b = 2, याव�न �कमती काढा :

a

b4a

5b

a÷

b a

b

1

÷

23

35x=

25

=

a

b

c

d÷

u

v

10

2

15

3

2a3b

a9b÷

5x

4y

4y

5x

1

÷b

a

1

Zawng chhuak rawh.

130C VCM-GM-NLF

Equations

We have kept an unknown number in cup.Let;s call is as x. Write the following equations in the language of x.

=

x = 3

+ =

= + =

+ = + =

= +

=

131C VCM-GM-NLF

Thil zat intlukpui sawina

A zat hriatloh number hemi no hnuaiah hian kan dah a.

Chuchu x tiin a hming kan vuah anga. x hmangin a awmzia leh a nihphung te ziak rawh le.

Description Equation

x = 5

2 subtracted from a number given 3.

A number divided by 3 is 4.

Two times a number plus 5 is 8.

Three times a number minus 2 given 10.

Half of a number plus 1 is 5.

132C VCM-GM-NLF

Sawifiahna

A number is equal to 5

Number chu 5 nen an intluk a

4 added to a number is 7

4 chu number nena belh chuan 7 ani a

3 times a number is 15

Number vawi thum lak chu 15 a ni a.

2 chu a number tarlan 3 in kan paih a.

Number 3 a kan sem chu 4 a ni a

Number vawihnih a kan lak leh 5 belh chu 8 a ni a.

Number vawi thum lak leh 2 kan paih chuan 10 a ni

Number chanve leh 1 kan belh chuan 5 alo ni a.

Description

x = 4

Equation

2x = 6

x + 2 = 6

3x + 1 = 10

4x - 2 = 2

= 5x

2

- 4 = 1x

3

133C VCM-GM-NLF

Sawifiahna

A number is called n.

What is the number after n?

What is the number before n?

Breadth of a rectangle is b.

Length is double of breadth.

Write length _________________

Side of an equilateral triangle is s.

Its perimeter (sum of all three sides)

is _________________

Riya was born when her mother was 25 years old.

Today Riya's age is r. Her mother's age is _______________.

Grandmother's age is 10 times that of her grandson's age.

If grandson's age is m, grandmother's age is ______________.

If grandmother's age is 50 years, grandson's age is _____________ years.

b

s

134C VCM-GM-NLF

n hnua number chu eng number nge?

n hma a number chu eng number nge?

Number chu n ani a.

A dung chu a vang lam vawihnih a ni.

A dung chu __________________

Rectangle vang lam chu b ani a.

A pang zawng zawng belh khawm zat

chu __________________

Equilateral triangle pang chu s ani a.

Vawiin ah chuan Riya-i kum chu r ani a. A nu kum zat chu _______________

Riya-i pian kum ah chuan a nu chu kum 25 a ni a.

Ka pi kum chu a tupa kum let 10 in a tam a.

A tupa kum chu m ni ta se, ka pi kum chu ________________

Ka pi chu kum 50 nita se, a tupa kum zat chu __________________ a ni ang.

Solving equations

Solve worksheets 174 to 179 of Math Delight 6.

Solve.

x + 5 = 8 m + 4 = 3 2p = 20

= 3 7 = y + 2 10 = 5tx

4

Equation

x + 3 = 5

x - 2 = 4

3x = 12

= 5x

2

To have only variable on left hand side Solution

x + 3 - 3 = 5 - 3

x = 2

Subtract 3 from both sides.

135C VCM-GM-NLF

Math Delight 6 Worksheets page 174 atanga 179 te hi chawk rawh.

A vei lamah chauh danglamna a awm A dikna zawn chhuahna

A lehlam lehlam atangin 3 kan paih ang

Chawk chhuak rawh

2x - 4 = 2

2x - 4 + 4 = 2 + 4

2x = 8

3x + 5 = 8

4y - 3 = 13

n - 1 = 1 2

3

- 3 = 2

m

4+ 1 = 6

x

2

x = 4

2x

2

8

2

4=

Added 4 on both sides.

Divided by 2 on both sides.

Did you multiply by in second step?3

2

3

2

136C VCM-GM-NLF

A tawn tawn ah 4 kan belh ang.

A tawn tawn ah 2 kan sem ang

hi a step pahnihna ah khan i puntir em?

5t + 28 = 10

2y + =

3l

2

2

3=

5

2

17

2

7m + = 13 19

2

+ 3 = 2

x = - 10

2b

3- 5 = 3

5

2

6z + 10 = - 2

a

5

137C VCM-GM-NLF

2 ( x + 4 ) = 12

-4 ( 2 + x ) = 8

4 = 5 ( p - 2 )

3 ( x + 1 ) = 0

3 ( n - 5 ) = 21

4 ( 2 - x ) = 8

- 4 = 5 ( p - 2 )

4 ( 5 - x ) = 0

138C VCM-GM-NLF

Factors

Factors : Making parts by multiplication till we get all prime numbers.

90 = 10 x 9

90 = 2 x 5 x 3 x 3

10 and 9 can be further factorised.

Vertical method of factorization

We want prime factors of a number. We

divide the number by the first prime

number 2, as many times possible. Then

we divide by 3 as many times as possible.

We divide by 2, 3, 5, 7, 11,13, 17, 19, 23....

sequentially as many times as we can. We

repeat this till we get 1. The product of all

prime numbers you divided by is the prime

factorization.

842

4222

213

77

1

84 = 2 x 2 x 2 x 3

Observe the number to be factorised.

If you can easily think of breaking it by

multiplication, horizontal method will be

easy. Otherwise vertical method will

certainly guide you to the answer.

Number , a chuang awm lova semral theitu number te chu factor an ni.

Prime number ( semral theitu number) zawng zawng kan hriat chhuah theih nan a puntirna hmanga thensawm.

10 leh 9 te hi an la thensawm leh theih a.

A ding ngil zawnga thensawm dan

Prime factors a numberte kan duh chuan,a

hmasa berah chuan prime number hmasa

ber 2 in kan sem phawt anga,tichuan 3 in a

sem theih chen chen kan sem leh phawt

ang.heng prime number kan tih

2,3,5,7,11,13,17,19,23…te hian a tam thei

ang berin kan sem ang a tawp a 1 a

chhuah hma loh chuan.Semna atan a

prime number I hman zawng zawng te kha

prime prime factorization chu an ni.

Number chu han en vang vang la,awlsam

taka puntirna hmanga I thiah theih

chuan,horizontal method( a khamphei

zawng a thensawm dan) hmang la I awlsam

ang,anih loh chuan vertical method ( a ding

ngil zawng a thensawm dan ) hmang la a

chhanna I hmu chhuak ngei ang.

139C VCM-GM-NLF

Find the prime factors (Use any method) :

30 45

49

60

39

66

70 65

132 114

Prime factor te zawng chhuak rawh.(I method duh zawng zawng I hmang thei ang)

140C VCM-GM-NLF

Find the prime factors.

26ab2 2x y

2aby

2 23x y

2 3p q

2 26x y

221p q3 3 340x y z

The above terms are called as Monomials.

Prime factor te zawng chhuak rawh.

A chunga a in ziah dan phung khi monomial kan ti a.

141C VCM-GM-NLF

Factorise :

26xy + 4y

210xy + 15x y 2 214p - 42pq

2 230a b - 25ab

Each of the above expressions has two terms. They are called as Binomials.

= 2 x 3 x x x y + 2 x 2 x y x y

= 2 y ( 3x + 2y)

You know that a x (b+c) = ab + ac

This equation has left hand side in the form of a product of two terms. They are factors.Right hand side of the equation is in the form of addition of two terms. It is expansion.If we want to find factors from expansion, we may take the common term as a multiplier.

29x + 3x

A chuang awm lo in semral rawh

In hria anga

He equation veilam hi thil pahnih puntirna danin a rawn dah a, chu chu factor kan ti a.Dinglama equation hi chu thil chihnih belh danin a rawn dah a. Chu chu expansion ani.Expansion atanga factor zawn chhuah kan duh chuan an pahnih atanga inang chu puntirtu ah kan la ang.

A chung ami te khi I en chuan ziahdan phung chi hnih theuh I hmu ang a,khing ang chi khi

Binomial kan ti ani.

142C VCM-GM-NLF

Words used in algebra

2x + 6x + 9

9 is a constant.

21 is a coefficient of x and 6 is a coefficient

of x.

2It has 3 terms : x , 6x and 9.

This is an expression. It is a trinomial.

x is a variable.

3a + 4b + 5

It is called ..........................................

This expression has ............. terms.

Constant is ...............................

Coefficient of a is .....................

Term 1 :

Coefficient of b is .....................

Term 2 :

Variables in this are ................................

Term 3 :

2x + 6x =

This expression has ............. terms.

Coefficient of x is .....................

Therefore it is called ..................................

x is called as ................................

2Coefficient of x is .....................

Term 1 :

The factors of this expression are :

Term 2 :

Factorise :

अवयव पाडा :

Algebra a kan thu /hawrawp hman te

x hi variable(engzat pawh ani thei)ani a

Kan hmuh ang hian ziahdan phung chi thum

in a in ziak a,hei hi trinomial kan tih chu ani.

21 hi x coefficient(number puntir puitlingtu

pakhat)ani a chutiang bawk in 6 hi x

coefficient ani.

9 hi constant(danglam thei lo)ani a.

a coefficient chu ……...............................

Term 1 :

Term 3 :

A variables chu…………..........................

A constant chu…………...........................

b coefficient chu………

A chunga mi hi ziah dan phung chi …….

ziah ani a. Chu chu ………. kan t.

Term 2 :

Term 2 :

A chunga mi hi ziah dan phung chi …….

ziah ani a. Chu chu ………. kan t.

X hi …………................ ani a

Term 1 :

2x coefficient chu ……....................

x coefficient chu………...................

Heng kan semral te ziahdan phung te chu :

143C VCM-GM-NLF

Expand :

4 (2a - 3b) 6x (2xy +3y - 5)

- 4 (- 5a - b) 13

12

a + k15( )

4 x (2 - 3 + 5)

= (4 x 2) - (4 x 3) + (4 x 5)

= 8 - 12 + 20

= 16

-3 (3 - 2 - 1)

14

43

a + 8 b( ) 15

52

x - y25( )-

Ti zau rawh.

144C VCM-GM-NLF

Indices

As Product

2 x 2 x 2

3 x 3 x 3 x 3

5 x 5 x 5

4 x 4 x 4 x 4

2 x 2 x 2 x 2 x 2

Using Powers

32

2

45

27

Number

8

64

81

81

64

16

16

9

3

8

4

2

145C VCM-GM-NLF

As Product

x x x x x

a x a x b

4y

3x

2a

Using Powers

2ab

3a

3 3a b

a x x x x

3 2x + y(x x x x x) + (y x y)

(x x x x x) – (y x y)

(x x x x x) – ( x x x)

146C VCM-GM-NLF

Mixed Form

2² x 2³

3² x 3³

54² x 4

x² x x

3x² x x

2a² x a x b + b

2 3 5 12 x 2 x 3 x 3

2 3 5 12 x 2 x 3 x 3

Expanded Form

(2 x 2) x (2 x 2 x 2)

(x x x x x) x (x x x x x x x)

Number

52

4x

10y

3a

6b

Fill in the blanks. A karawl dah khat rawh.

147C VCM-GM-NLF

A inpawlh in A zauh zawngin

1. Draw a cube of 1x1x1





1 unit

148C NLF-EG

3. Cube lem 3x3x3 in siam rawh.





149C NLF-EG

4. Draw a cuboid of 2x2x3





1. Cuboid lem 1x1x2 in siam rawh





Draw first letter of your name in different ways. (One example is shown here).

150C NLF-EG

I hming intanna letter kha i duh hrang hrang in siam rawh. (Entirna pakhat dah ani)

Draw the following 3D shapes on isometric dot paper.

Make more shapes using Jodo blocks, observe each shape carefully and draw it.

151C NLF-EG

Judo blocks hmangin a lem dang i siam bawk dawn nia, tin i lem siam te chu uluk taka

endik chungin i siam dawn nia.

Isometric dot paper ah hian heng 3D lem te hi siam rawh.

Make 12 different flat shapes using 5 Jodo Blocks (snap cubes).

Watch each of the shapes carefully and draw it on isometric dot paper.

Uluk taka endik chungin isometric dot paper ah te hian i siam dawn nia.

Judo Blocks 5 hmangin thil phek lem 12 i siam dawn nia. (snap cubes)

152C NLF-EG

Draw these shapes on isometric dot paper.

Isometric dot paper ah heng lem te hi i siam dawn nia.

153C NLF-EG

Build the structure shown in this picture using Jodo blocks.

Front view Top view Right side view Left side view Rear view

Observe the shape from all sides and compare with the views given here.

Make this shape using Jodo blocks, observe from all sides and draw the views in this table.

154C NLF-EG

A hnuaia thlalak tarlan ang hian Judo blocks hmangin han rem ve chhin teh le.

Right Side View

Top View

Rear View

Left Side View

Front View

A chung atanga thlirin

A hnung lam atanga thlirin

A dinglam atanga thlirinA vei lam atanga thlirin

A hmalam atanga thlirin

A lem a a hmelhmang te hi hnun takin endik la thlirdan rawn tarlan hmang hian i khaikhin dawn nia.

Hetiang ang pianhmang hi Judo blocks hmangin siam ve la, a pang zawng zawng te hi uluk takin i endik dawn nia.



A dinglam atanga thlirin

A vei lam atanga thlirin

A hnung lamatanga thlirin

Front view Top view Right side view Left side view Rear view



A hnung lam atanga thlirin

A vei lam atanga thlirin

A hnung lamatanga thlirin

StructureFront view Top view Right side view

Make the given structure using Jodo blocks and draw the views in this table.

155C NLF-EG

Judo blocks hmangin a hnuaia ruangam ang hian i siam ve dawn nia. A landan hemi table ah hian i siam bawk dawn nia.



A dinglam atanga thlirin

Look

at th

e fro

nt vi

ew

, to

p v

iew

and the r

ight si

de v

iew

. B

uild

the s

truct

ure

havi

ng these

fro

nt, top a

nd r

ight si

de v

iew

s usi

ng J

odo b

lock

s.

Circl

e the s

hape fro

m A

, B

and C

whic

h m

atc

hes

the s

hape y

ou m

ade.

Fro

nt vi

ew

Top v

iew

Rig

ht

side v

iew

AB

C

AB

C

AB

C

156C NLF-EG

A h

ma

lam

ata

nga th

lirin

A c

hung

ata

nga thlir

inA

din

gla

m

ata

nga t

hlir

in

A h

mala

m,

a c

hungla

m le

h a

din

gla

m a

tangin

thlir

la. Ju

do b

lock

hm

angin

a h

mala

m, a c

hungla

m

leh a

hm

ala

m r

uangam

langin

sia

m r

aw

h. A

pia

nhm

ang A

,B le

h C

te m

il ch

u i

rinbia

l daw

n n

ia.

Documents

Universal Active Math Math Delight - 7 - Navnirmiti Learning