Upload
trinhanh
View
214
Download
0
Embed Size (px)
Citation preview
P-80 M A T H E M A T I C S – VIII
1. (A) Rs. 100 1
2. (C) 12 years 1
3. 72% of 25 students are good in Mathematics.
(100 – 72)% of 25 students are not good in Mathematics. 12
or, 28% of 25 students are not good in Mathematics.
or, 28
25100
´ = 7 students are not good in Mathematics. 12
4. Increase in price this year = 20% [the last year’s price]
= 20% of Rs. 45,000
= ´20
45,000100
12
= Rs. 20 × 450= Rs. 9,000
This year’s price = [Last year’s price] + [Increase in price] 12
= Rs. 45,000 + Rs. 9,000= Rs. 54,000.
5. Let the original salary = Rs. xIncrease in salary = 10% of Rs. x
=10
Rs.100
x = Rs. 10
x1
New salary = Rs. Rs.10
xx+ =
11Rs
10x
11
Rs.10
x = Rs. 154000 (given)
or x =´154000 10
Rs.11
= Rs. 140000
Thus, the original salary = Rs. 140000 1
6. We have P = Rs. 8000, R = 5% p.a., T = 2 years
A = 1100
nR
Pé ù+ê úë û
A =25
8000 1100
=221
800020
1
=21 21
800020 20
= (20 × 21 × 21) = Rs. 8820 ½Now, compound interset
= A – P= Rs. 8820 – Rs. 8000= Rs. 820 ½
7. Here P = Rs. 10800, n = 3 years,
R = 1
122
% p.a. = 25
%2
p.a.
8 COMPARING QUANTITIES WORKSHEET-39
P-81C O M P A R I N G Q U A T I T I E SN
We have, A = 1100
nR
Pé ù+ê úë û
=3
25Rs. 10800 1
2 100
é ù+ê úë û´
[ Interest compounded annually, n = 3]
=3
225Rs. 10800
200
é ùê úë û 1
=225 225 225
Rs. 10800200 200 200
´ ´ ´
=675 9 9 9
Rs.4 8
´ ´ ´
´=
492075Rs.
32
Amount = Rs. 15377·34
= Rs. 15377·34 1Now, Compound interest
= Rs. 15377·34 – Rs. 10800= Rs. 4577·34 1
8. Here, we shall calculate the amount for 2 years using the C.I. formula. Then this amount
will become the principal for next 4 months, i.e., 4
12 years.
Here, P = Rs. 26400, n = 2 years
R = 15% p.a.
A = 1100
nR
Pé ù+ê úë û
= Rs. 26400 2
151
100
é ù+ê úë û
= Rs. 26400 × 2
23
20
é ùê úë û 1
= Rs. 26400 × 23 23
20 20´
= Rs. 66 × 23 × 23= Rs. 34914
Again, P = 34914, T = 4 months
=4
12years, R = 15% p.a. 1
Using S.I. =100
P R T´ ´, we have
S.I. =15 4
Rs. 34914100 12
´ ´
=5819 3
Rs.10
´= Rs.
17457
10= Rs. 1745·70
Amount = S.I. + P= Rs. (1745·70 + 34914)= Rs 36659·70 1
Thus, the required amount to be paid to the bank after 2 years 4 months = Rs 36659·70.
P-82 M A T H E M A T I C S – VIII
1. (C) 40% 1
2. (B) 10 1
3. Sale price of the pair of shoes = Rs. 900
Sales tax = 5% × Rs. 900
= ´5
(Rs. 900)100
12
= Rs. 45
Bill amount = Sale price + Sales tax
= Rs. 900 + Rs. 45
= Rs. 945. 12
4. Cost of the two soap bars
= (Rs. 35) × 2 = Rs. 70
Sales tax = 5% of Rs. 70
=5
Rs 70100
´ = Rs. 3·50 ½
Thus, the buying price
= Rs.70 + Rs. 3·50
= Rs. 73·50 12
5. The price of the calculator = Rs. 750
Sales tax = 5% of Rs. 750
=5
Rs. 750100
´
=75
Rs.2
= Rs. 37·50 ½
Bill amount = Sales Price + Sales Tax
= Rs. 750 + Rs. 37·50
= Rs. 787·50 ½
6. Total Cost Price of the two-wheeler (C.P.)
= Rs. 15000 + overhead expenses
= Rs. 15000 + Rs. 500
= Rs. 15500 1
Selling Price (S.P.)
= Rs. 18600 S.P. > C.P. She got profit
Profit = Rs. 18600 – Rs. 15500
= Rs. 3100 ½
8 COMPARING QUANTITIES WORKSHEET-40
P-83C O M P A R I N G Q U A T I T I E SN
Now, Profit percent =3100
100%15500
´
= 20% ½
Thus, she earned a profit of 20%.
7. At simple interest
SI on ` 12,000 at 6% per annum for 2 years
= 12,000 6 2
100 = ` 1,440
At compound interest
P = ` 12,000
R = 6% per annum
n = 2 years
A = R
P 1100
n
1
=
26
12,000 1100
=
23
12,000 150
=
253
12,00050
= 53 53
12,00050 50
= ` 13,483.20 1
CI = A – P
= ` 13,483.20 – ` 12,000
= ` 1,483.20
Excess amount = ` 1,483.20 – ` 1,440
= ` 43.20
Hence, I would have to pay to him as excess amount of ` 43.20. 1
8. Principal = Rs 10,000 ½
Time =1
12
years
Rate = 10% p.a.
Case I. Interest is compounded half yearly
We have R = 10% p.a. = 5% per half yearly
T = 12
1 years n = 3
A = 1100
nR
Pé ù+ê úë û
= Rs. 10000 3
51
100
é ù+ê úë û
= Rs. 10000 3
21
20
é ùê úë û
P-84 M A T H E M A T I C S – VIII
= Rs. 10,000 21 21 21
20 20 20´ ´ ´
=5 21 21 21
Rs.4
´ ´ ´
=46305
Rs.4
= Rs. 11576·25 1
Amount = Rs. 11576·25
Now, C.I. = Amount – Principal
= Rs. 11576·25 – Rs. 10000
= Rs. 1576·25 ½
Case II. Interest is compounded annually
We have R = 10% p.a.
T =1
12
year
Amount for 1 year
= 1100
nR
Pé ù+ê úë û
= Rs. 10000 1
101
100
é ù+ê úë û
=11
Rs. 1000010
´
= Rs. 11000 Interest for 1st year
= Rs. 11000 – Rs. 10000
= Rs 1000 1
Interest for next 1
2year on Rs. 11000
=100
P R T´ ´
=10 1
Rs. 11000100 2
´ ´
= Rs. 55 × 10 = Rs. 550
Total interest = Rs. 1000 + Rs. 550
= Rs. 1550 1
Since Rs. 1576.25 > Rs. 1550
Interest would be more in case of it is compounded half yearly. ½
P-85C O M P A R I N G Q U A T I T I E SN
1. (A) Rs. 51,000 1
2. (B) 4 years 1
3. Number of visitors on Sunday = 845
Number of visitors on Monday = 169
Decrease in the number of visitors
= 845 – 169 = 676 12
Percent decrease
=676
100%845
´ = (4 × 20%)
= 80% 12
4. Cost price= Rs. 2400
Profit = 16% of Rs. 2400
=16
Rs 2400100
´
= Rs.16 × 24 = Rs. 384
Selling price = Rs. 2400 + Rs. 384
= Rs. 2784
Now, selling price per article = Rs 2784 80 [ Number of articles = 80]
=348
Rs Rs.34·8010
= 1
5. (a) Cost of TV including VAT = Rs. 13500
Rate of VAT = 8%
Let the original price = Rs. x 12
x + 8% of x = Rs. 13,500
or x + 8
100x = Rs. 13500
or x8
1100
é ù+ê úë û= Rs. 13500
or x × 108
100= Rs. 13500
or x =13500 100
Rs.108
´
= Rs. 12500
Thus, the original price
= Rs 12500 1
(b) Let original price = Rs. x
Price including VAT
= Rs. x + 8% of x
8 COMPARING QUANTITIES WORKSHEET-41
P-86 M A T H E M A T I C S – VIII
=8
Rs. 1100
é ù+ê úë ûx
=108
Rs.100
x
But the original price + VAT
= Rs. 180
108
100x = Rs. 180 1
2
or x =100
Rs. 180108
´
=500
Rs.3
= 2
Rs. 1663
Thus, the original price
=2
Rs. 1663
1
6 People who like cricket = 60%
People who like footbal = 30%
People who like other games
= [100 – (60 + 30)%
= [100 – 90]%
= 10%
Now, Total number of people = 50,00,000 1
60% of 50,00,000
=60
50,00,000100
´
= 6 × 5,00,000
= 30,00,000 ½
30% of 50,00,000
=30
100× 50,00,000 1
= 3 × 5,00,000
= 15,00,000 ½
10% of 50,00,000
=10
100× 50,00,000
= 1 × 5,00,000
= 5,00,000 ½
Thus, number of people who like
Cricket 30,00,000Football 15,00,000
Other games 5,00,000
ü= ï= ý
ï= þ½
P-87C O M P A R I N G Q U A T I T I E SN
7. Total S.P. of the shoapkeeper =Rs. 16000/-S.P. of VCR = Rs. 8000/-
Let C.P. of the shopkeeper = Rs. xLoss = 4%
S.P. of VCR = C.P. of VCR – Loss
8000 = 4
100x x
x = 100 4
8000100
x ½
x = 8000 100 25000
96 3
½
C.P. of VCR = Rs. 25000
3
S.P. of TV = Rs. 8000/-
Let shopkeeper cost price by Rs. y
Profit = 8%
S.P. of TV = C.P. of TV + 8
100 of C.P. of TV ½
8000 = 8
100y y
y 8
1100
= 8000 ½
y = 8000 100
74074.40108
½
Total C.P. = x + y = Rs. 25000
0.7407.403
= 15740.740
Also total S.P. = Rs. 16000/- 1
S.P. > C.P.
Overall profit = S.P. – C.P.
= 259.26
Profit % = 259.26
10015740.740
= 1.65% 1
P-88 M A T H E M A T I C S – VIII
8 COMPARING QUANTITIES WORKSHEET-42
[A] Lab Activity
Object :
To verify experimentally that a single discount of some percentage say x% provided onan article is greater than the sum of two successive discounts of any percentage say y%and z% with x = y + z.
Let’s start :
The discount is a reduction given on the marked price. If the discount percentage isd%, then
Discount = ´Marked Price100
d½
EFGH is a rectangle (Fig. (i). Its opposite sides are equal and each angle is of measure90°.
Area (EFGH) = l × b sq. units.
H G
E F
b
l
½
(i)
Meterial Required :
(1) Thick white paper(2) Sketch pens(3) Geometry box(4) A pair of scissors(5) Fevicol ½
Procedure :
(1) Take the MP of an article as ` 100
(2) Discount (in `) on the article by a single discount of x%
= 100100
xx´ = ½
(3) Using scissors, cut a rectangle of length x and breadth unity. Name it as ABCD andcolour it (Fig. (ii))
D C
A Bx
1
1
(ii)
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-89C O M P A R I N G Q U A T I T I E SN
(4) Discounts on the article by two successive discounts of y% and z% are respectively
100100
y´ ` = y and ` (100 – y)
100
z´ = `
100
yzzæ öç ÷-è ø. ½
Sum of these discounts (in `) = y + z 100
yz-
(5) Using scissors, cut another rectangle of length y + z and breadth unity. Colour it with adifferent colour and name it as PQRS (Fig. (iii). It is obvious that y + z = x. ½
S R
P Q
N
M
(y + z)
yz100
(iii)
(6) Cut a rectangle PMNS from the rectangle PQRS obtained in step (4) along the dotted
line such that PM = 100
yz (Fig. (iii))
(7) Now, paste the remaining rectangle MQRN on the rectangle ABCD such that the pointQ concides with B and R with C (Fig. (iv)). ½
D C, R
A B, Q
N
M
(iv)
Observation :
We observe that region AMND is clearly free from overlapping on the rectangle ABCD.
½
Result :
A single discount of x% is greater than the two successive discounts of y% and z% suchthat x = y + z. ½
[B] Fill in the Blanks 5 × ½ = 2½
1. 90
2. 120
3. Marked
4. Cost price
5. R
P-90 M A T H E M A T I C S – VIII
[C] True/False 4 × ½ = 2
1. False
2. False
3. False
4. True
[D] Viva-Voce
1. The price reduces . 1
2. The first discount is applicable on the marked price, the second discount is applicableon the reduced price after the first discount and so on. 1
3. Rs. 250. 1
4. A single discount of 10%. 1
5. Discount = Marked Price – Sale Price. 1
[E] Quiz
1. Value Added Tax. 1
2. Discount percent = Discount
MarkedPrice× 100 1
3. Increase or decreases in population. 1
4. Cost Price. 1
5. Two successive discounts of 12% and 8%. 1
P-91ALGEBRAIC EXPRESSIONS AND IDENTITIES
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
1. (D) 2 2 2 11p q pq 1
2. (A) 2 7 5 10xy yz zx xyz 1
3. (a) 25 3xyz zy
25 3
5 3
xyz zy-
-
Terms
Their coefficients12
(b) 1 + x + x2
2 1
1 1 1
x xTerms
Their coefficients12
4. (a) ab – bc
+ bc – ca
– ab + ca
0 + 0 + 0 = 0 12
(b) a – b + ab 12
b – c + bc – a c + ac
0 + 0 + ab + 0 + bc + ac = ab + bc + ca 12
5.æ ö æ ö- ´ç ÷ ç ÷è ø è ø
3 310 6
3 5pq p q = 3 310 6
3 5pq p q
é ùæ ö æ ö- ´ ´ ´ç ÷ ç ÷ê úè ø è øë û
1
=4 410 6
3 5p q
æ ö æ ö- ´ ´ç ÷ ç ÷è ø è ø
= 4 44p q . 1
6. 2( 1) 5a a a = 2 1 5a a a a a
= 3 2 5a a a 12
(i) For a = 0,
a3 + a2 + a + 5 = (0)3 + (0)2 + 0 + 5 = 5 12
(ii) For a = 1,
3 2 5a a a = 3 2(1) (1) 1 5 = 1 + 1 + 1 + 5
= 8 12
(iii) For a = – 1
3 2 5a a a = 3 2( 1) ( 1) 1 5
= – 1 + 1 – 1 + 5
= 4. 12
9ALGEBRAIC EXPRESSIONSAND IDENTITIES WORKSHEET-43
P-92 M A T H E M A T I C S – VIII
7. (a + b)(a – 2b + 3c) = a × (a – 2b + 3c) + b(a – 2b + 3c)
= 2 22 3 2 3a ab ac ab b bc
=2 2( 2 ) 3 2 3a ab ab ac b bc 1
= 2 22 3 3a b ab bc ac
and (2a – 3b) × c = 2ac – 3bc 1
+ - + - - ´( )( 2 3 ) (2 3 )a b a b c a b c = 2 2( 2 3 3 ) (2 3 )a b ab bc ac ac bc- - + + - -
= 2 22 3 3 2 3a b ab bc ac ac bc- - + + - +
= 2 22 (3 3 ) (3 2 )a b ab bc bc ac ac
[Combining the like terms bc and ac]
= 2 22 6a b ab bc ac . 1
8. (1.5x – 4y)(1·5x + 4y + 3) – 4·5x + 12y
= + + - + +1·5 (1·5 4 3) 4 (1·5 4 3)x x y y x y – 4·5x 12y
= (1·5 1·5 ) (1·5 4 ) (1·5 3)x x x y x - ´ - ´ - ´(4 1·5 ) (4 4 ) (4 3)y x y y y - +4·5 12x y 1
= + + - - -2 22·25 6 4·5 6 16 12x xy x xy y y - +4·5 12x y
= 2 22·25 (6 6) (4·5 4·5) 16x xy x y (12 12)y 1
= 2 22·25 (0) (0) 16 (0)x xy x y y
= 2 22·25 0 0 16 0x y
= 2 22·25 16x y . 1
Another Method :
(1·5x – 4y)(1·5x + 4y + 3) – 4·5x + 12y
= (1·5x – 4y)(1·5x + 4y) + (1·5x – 4y) × 3 – 4·5x + 12y 2
= (1·5x)2 – (4y)2 + (4·5x – 12y) – (4·5x + 12y)
= 2·25x2 – 16y2. 1
P-93ALGEBRAIC EXPRESSIONS AND IDENTITIES
1. (A) 2xy 1
2. (B) a6 1
3. 12a – 9ab + 5b – 3
– 4a – 7ab + 3b + 12 12
(–) (+) (–) (–) Subtract
8a – 2ab + 2b – 15 12
4. We have,
(– 4p) × (7pq) = (– 4 × 7) × p × pq
= – 28p2q 1
5. 2 3( 5)( 3) 5a b = 2 3 3( 3) 5( 3) 5a b b 1
= 2 3 2 3( ) ( 3) (5 )a b a b (5 3) 5
= 2 3 2 33 5 15 5a b a b
= 2 3 2 33 5 20a b a b . 1
6. 2 2(2 5) (2 5)x x
= 2 2[(2 ) (5) 2 2 5]x x 2 2[(2 ) (5) 2 2 5]x x 1
= 2 2[4 25 20 ] [4 25 20 ]x x x x
= 2 24 25 20 4 25 20x x x x
= 20x + 20x
= 40x. 1
Alternative Method :
Since, a2 – b2 = (a + b)(a – b)
(2x + 5)2 – (2x – 5)2
= [(2x + 5) + (2x – 5)][(2x + 5) – (2x – 5)] 1
= (4x)[2x + 5 – 2x + 5]
= (4x)(10) = 40x. 1
7. L.H.S. =
24 3
23 4
m n mn
=
2 24 3 4 3
2 23 4 3 4
m n m n mn
1
=2 216 9
2 29 16
m mn n mn
9ALGEBRAIC EXPRESSIONSAND IDENTITIES WORKSHEET-44
P-94 M A T H E M A T I C S – VIII
=2 216 9
( 2 2)9 16
m mn n 1
=2 216 9
(0)9 16
m mn n
=2 216 9
9 16m n 1
= R.H.S. Hence Proved.
8. L.H.S. = 2 2(4 3 ) (4 3 )pq q pq q
= [(4pq + 3q) + (4pq – 3q)] × [(4pq + 3q) – (4pq – 3q)]
(... x2 – y2 = (x + y) (x – y)) 1
= (8pq) (4pq + 3q – 4pq + 3q) 1
= (8pq) (6q)
= 248 = R.H.S.pq Hence Proved. 1
P-95ALGEBRAIC EXPRESSIONS AND IDENTITIES
1. (B) x + 1 1
2. (C) 2 2a b 1
3. 3x(4x – 5) + 3 = (3x)(4x) – (3x)(5) + 3
= (3 × 4) × (x × x) – 15x + 3
= 12x2 – 15x + 3 ½
4. 71 = 70 + 1
2(71) = 2(70 1)
= (70)2 + 2 × 70 × 1 + (1)2 1
[·.· (a + b)2 = a2 + 2ab + b2]
= 4900 + 140 + 1
= 5041. 12
5. (i) Using the Identity,
(a + b)(a – b) = 2 2a b
We have,
(6x + 7)(6x – 7) = 2 2(6 ) (7)x
= 236 49.x 1
(ii) Using the identity, 2( )a b = a2 + b2 + 2ab, we have
3 3
2 4 2 4
x y x y
=
23
2 4
x y
=
2 23 3
22 4 2 4
x y x yæ ö æ ö æ ö æ ö+ + ´ ´ç ÷ ç ÷ ç ÷ ç ÷
è ø è ø è ø è ø
=2 29 3
4 16 4
x y xy . 1
6. L.H.S. = 2(3 7) 84x x
= + + ´ ´ -2 2(3 ) (7) 2 3 7 84x x x
= 29 49 42 84x x x
= 29 49 42x x 1
R.H.S. = 2(3 7)x
= 2 2(3 ) (7) 2 3 7x x
= 29 49 42x x 1
9ALGEBRAIC EXPRESSIONSAND IDENTITIES WORKSHEET-45
P-96 M A T H E M A T I C S – VIII
Since L.H.S. = R.H.S.
2(3 7) 84x x = 2(3 7)x
7. 1·05 × 9·95 = (1 + 0·05)(1 – 0·05) 1
= (1)2 – (0·05)2 1
= 1 – 0·0025
= 1·0000 – 0·0025
= 0·9975. 1
8. We have
- + - - +2 25 2 5 11 3 28p q pq pq q p
2 24 5 3 7 8 10p q pq pq q p 1
(–) (–) (+) (–) (+) (+)
2 27 8 18 5 38p q pq pq q p 2
P-97ALGEBRAIC EXPRESSIONS AND IDENTITIES
9ALGEBRAIC EXPRESSIONSAND IDENTITIES WORKSHEET-46
1. (A) A binomial 1
2. (B) 225 30 9c c 1
3. 2 251 49 = (51 + 49) (51 – 49)
= (100) × (2)
= 200. 1
4. (x2 – 5)(x + 5) + 25 = x2(x + 5) – 5(x + 5) + 25
= 2 2( ) ( 5) (5 ) (5 5) 25x x x x 1
= 3 25 5 25 25x x x
= + -3 25 5 .x x x 1
5. (x + 7y) × (7x – y) = x(7x – y) + 7y(7x – y)
= ´ + ´- + ´ + ´-( 7 ) ( ) (7 7 ) (7 )x x x y y x y y 1
= - + -2 27 49 7x xy xy y
= + -2 27 48 7x xy y . 1
6. First expression
= 3a(a + b + c) – 2b(a – b + c)
= (3a) × (a) + (3a) × (b) + (3a) × (c) – (2b) × (a)
+ (2b) × (b) – (2b) × (c)
= 3a2 + 3ab + 3ac – 2ab + 2b2 – 2bc
= 3a2 + 2b2 + 3ab – 2ab – 2bc + 3ac
= 3a2 + 2b2 + ab – 2bc + 3ac 12
Second expression
= 4c (– a + b + c)
= 4c × (– a) + 4c × b + 4c × c
= – 4ac + 4bc + 4c2 12
P-98 M A T H E M A T I C S – VIII
Subtraction
– 4ac + 4bc + 4c2
3a2 + 2b2 + ab + 3ac – 2bc
– – – – +————————————————
– 3a2 – 2b2 – ab – 7ac + 6bc + 4c2 1
7. Area of the new rectangle = (l + 5) × (b – 3) 2
8. Volume of a cuboid = Length × Breadth × Height
= (x2 – 2) [(2x + 2) (x – 1)]
= (x2 – 2) [2x2 – 2x + 2x – 2] 1
= (x2 – 2) [2x2 – 2]
= 2(x2 – 2) (x2 – 1)
= 2[x4 – x2 – 2x2 + 2] 1
= 2[x4 – 3x2 + 2]
= 2x4 – 6x2 + 4. 1
P-99ALGEBRAIC EXPRESSIONS AND IDENTITIES
9 WORKSHEET-47
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
ALGEBRAIC EXPRESSIONSAND IDENTITIES
[A] Lab Activity
Object :
To verify the following identity by paper cutting and pasting.
(a + b)2 = a2 + 2ab + b2
Pre-requisite Knowledge :
(i) Area of a square
(ii) Area of a rectangle
(iii) Addition of algebraic expressions
Materials Required :
(i) Thick chart paper
(ii) A pair of scissors
(iii) White papers
(iv) Sketch pens
(v) Fevicol. 1
Procedure :
(1) Draw a square of side ‘a’ units on a thick sheet of paper.
(2) Draw another square of side ‘b’ units on the thick sheet of paper.
a2 b2
a
a b
b
½
(3) Draw two rectangles each of length ‘a’ units and breadth ‘b’ units as shownbelow :
ab ab
a a
b b½
P-100 M A T H E M A T I C S – VIII
(4) Draw a square of side (a + b) units.
(a + b)2 (a + b)
(a + b)
½
(5) Take the cutouts of all the above figures.
(6) Arrange and paste the cutouts 2 2, ,a b ab and ab on the cutout of the square (a + b)2.
ab
abb2
a2
½
Observations :
The square of side (a + b) units is completely covered by the cutouts of a2, b2, ab and
ab. ½
Conclusion :2( )a b = 2 2a ab ab b
Or2( )a b = 2 22a ab b . ½
[B] Fill in the Blanks 5 × ½ = 2½
1. 2
2. Coefficient
3. 7
4. Monomial
5. Trinomial
[C] True/False 5 × ½ = 2½
1. True
2. False
P-101ALGEBRAIC EXPRESSIONS AND IDENTITIES
3. True
4. False
5. True
[D] Viva-Voce ½ × 10 = 5
1. (b) 1
2. 2
3. Yes
4. Two
5. Binomial
6. An expression is a group of terms, separated by + or – signs whereas a polynomial is a
group of terms separated by + or – or × signs.
7. (c) – 3
8. ‘Variable’ is a symbol which can have different values whereas a ‘constant’ has a fixed
value.
9. If an equality is true for all values of variables in it, then it is called an identity.
10. 0.
[E] Quiz
1. 2 2 sq. cmx 1
2. 2 2 2( ) 2a b a ab b 1
3. 2( )( ) ( )x a x b x a b x ab 1
4. ( )( )a b a b 1
5. No 1
6. An equation is not true for every value of variable in it, but an identity is. 1
P-102 M A T H E M A T I C S – VIII
1. (B) 6 1
2. (A) 3 faces 1
3. A polyhedron has a minimum number of your faces. 1
4. No, not always, because it can be a cuboid also. 1
5. Since, a prism is a polyhedron having two of its faces congruent and parallel, where as
other faces are parallelogram.
(i) No, a nail is not a prism. 12
(ii) Yes, unsharpened pencil is a prism. 12
(iii) No, table weight shown is not a prism. 12
(iv) Yes, box is a prism. 12
6. A solid is a polyhedron if Eulers formula F + V – E = 2 is satisfied. ½
Here, F = 10, E = 20 and V = 15
10 + 15 – 20 = 2
or 25 – 20 = 2 ½
or 5 = 2, which is not true
i.e. F + V – E 2 1
Thus, such a polyhedron is not possible.
7. Here :
Number of vertices (V) = 20
Number of edges (E) = 30
Let the number of faces = F. If do decahedron is to exist.
Using Euler’s formula, we have
F + V = E + 2 ....(1) 1
Substituting the values of V and E in (1), we get
F + 20 = 30 + 2 1
F + 20 = 32
F = 32 – 20
F = 12
Thus the required number of faces = 12. 1
8. The (bases) of the given solid are congruent polygons each of six sides.
It is a hexagonal prism. ½
In a hexagonal prism, we have :
The number of faces = 8 ½
The number of edges = 18 ½
The number of vertices = 12 ½
10 VISUALISING SOLID SHAPES WORKSHEET-48
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-103V I S U A L I S I N G S O L I D S H A P E S
1. (A) triangles 1
2. (D) 12 1
3. The plural of a polyhedron is polyhedra. 1
4. If the base of a pyramid is a triangle, then we call it as a tetrahedron. 1
5. A polyhedron is bounded by four or more than four polygonal faces. 12
(i) No, it is not possible that a polyhedron has 3 triangles for its faces.
(ii) Yes, 4 triangles can be the faces of a polyhedron. 12
(iii)Yes, a square and 4 triangles can be the faces of a polyhedron. 1
6. (i)If we look at the given solid structure from the top, we would see just a square. 1
(ii)If we look at it from a side, i.e. left or right, then we would se a figure as shown here.
1
7. (i) A cylinder is not a polyhedron. 12
(ii) A cuboid is a polyhedron. 12
(iii) A cube is a polyhedron. 12
(iv) A cone is not a polyhedron. 12
(v) A sphere is not a polyhedron. 12
(vi) A pyramid is a polyhedron. 12
8. (i) Polyhedron : A solid shape bound by polygons which are called its faces is called a
polyhedron. The faces meet at line segments called edges and edjes meet at points
called vertices. For example, cube, cuboid, etc. 1
(ii) Prism : A prism is a solid, whose faces are parallelograms and whose ends (or
bases) are congruent parallel rectilinear figures. 1
(iii) Pyramid : A pyramid is a polyhedron whose base is a polygon of any number of
sides and whose other faces are triangles with a common vertex. 1
10 VISUALISING SOLID SHAPES WORKSHEET-49
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-104 M A T H E M A T I C S – VIII
1. (C) 5 1
2. (B) 6 1
3. When the bases of a prism are parallelograms, then it is called a parallelopiped. It is
formed by six parallelogram. 1
4. Both of the prisms and cylinders have their base and top as congruent faces and parallel
to each other. Also, a prism becomes a cylinder as the number of sides of its base
becomes larger and larger. i.e. A cylinder is a limiting case of a prism as the number of
sides of the base of a regular prism approach infinity. 1
5. The front view and the top view of the given solid are given below : 1
(i) (ii)
Front view Top view
2
6. Here, number of faces (F) = 20
Number of vertices (V) = 12
Let the number of edges be E.
Using Euler’s formula, we have 1
F + V = E + 2
20 + 12 = E + 2
32 = E + 2
E = 32 – 2 = 30
Thus, the required number of edges = 30. 1
7. If a polyhedron is having number of faces as F, number of edges as E and the number of
vertices as V, then the relationship
F + V = E + 2
is known as Euler’s formula. 1
Following figure is a solid pentagonal prism.
10 VISUALISING SOLID SHAPES WORKSHEET-50
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-105V I S U A L I S I N G S O L I D S H A P E S
It has : Number of faces (F) = 7
Number of edges (E) = 15 1
Number of vertices (V) = 10
Substituting the values of F, E and V in the relation,
F + V = E + 2
we have 7 + 10 = 15 + 2
17 = 17
Which is true, the Euler’s formula is verified. 1
8. (a)1 1
12 3
æ öç ÷+ +è økm 1
(b) Meena’s. 1
(c) Letter box, Petrol pump, Hospital. 1
P-106 M A T H E M A T I C S – VIII
1. (C) Cuboid 1
2. (C) Sphere 1
3. The pyramid and cones are alike because their lateral faces meet at a vertex. Also a
pyramid becomes a cone as the number of sides of its base becomes larger and larger. 1
4. 6 vertices 1
5. (i) F + V = E + 2
F + 6 = 12 + 2
F + 6 = 14
F = 14 – 6 = 8 1
(ii) F + V = E + 2
20 + 12 = E + 2
32 = E + 2
E = 32 – 2 1
E = 30
6. In figure (i), we have
F = 7, V = 10 and E = 15 1
F + V – E = 7 + 10 – 15
= 17 – 15 = 2, 1
which is true.
Thus, Euler’s formula is verified.
(ii) In figure (ii), we have
F = 9, V = 9 and E = 16 1
F + V – E = 9 + 9 – 16
= 18 – 16 = 2,
which is true.
Thus, Euler’s formula is verified. 1
7. At least 4 planes can enclose a solid. Tetrahedron is the simple polyhedron. Following
figure represent a simplest regular polyhedron called tetrahedron. 2
10 VISUALISING SOLID SHAPES WORKSHEET-51
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-107V I S U A L I S I N G S O L I D S H A P E S
A tetrahedron has :
4 triangular faces, i.e. F = 4 1
4 vertices, i.e., V = 4
6 edges, i.e., E = 6
Now, substituting the values of F, V and E in Euler’s formula, i.e.,
F + V = E + 2
we have, 4 + 4 = 6 + 2
8 = 8, which is true.
Thus, Euler’s formula is verified for a tetrahedron. 1
P-108 M A T H E M A T I C S – VIII
10 VISUALISING SOLID SHAPES WORKSHEET-52
[A] Lab Activity
Make a prism and verify the Euler’s formula for it.
Object :
To make a prism say triangular prism and to verify the Euler’s formula for it.
Materials Required
(1) Cardboard (2) Ruler (3) A pair of scissors (4) Cellophane tape (5) Fevicol (6) Glazedpapers ½
Procedure :
(1) Cut two dimensional equilateral triangles of sides 15 cm each from the cardboard as
shown in fig. (i).
(i)
(2) Cut out the three identical rectangular surfaces each of length 20 cm and breadth 15 cm
leaving tab, on sides as shown in fig. (ii).
(3) Put one of the triangles horizontally. Then put the three rectangular faces vertically on
the three edges of the triangular surface such that breadth of one rectangular face lies
exactly on each edge of the triangle.
(4) Now, apply fevicol on the tabs and use the cellophane tape to fix these pairs of edges as
shown in fig. (iii).
(ii) (iii) (iv)
(5) In the same way, you have to fix the other triangular surfaces on the top of the figure
obtained in step 4.
(6) Paste Red glazed paper on the triangular faces and three other different colours on
rectangular faces.
(7) The three dimensional figure thus obtained is a triangular prism as shown in the fig.
(iv). 1½
Observation :
We observe that
Number of faces, F = 5
Number of vertices, V = 6
Number of edges, E = 9 ½
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-109V I S U A L I S I N G S O L I D S H A P E S
Result :
Verification of Euler’s formula :
F + V – E = 2
Here, F + V – E = 5 + 6 – 9
= 11 – 9 = 2
Hence, Euler’s formula is verified. ½
[B] Fill in the Blanks 4 × ½ = 2
1. 8, 12, 6
2. Vertex
3. Prism
4. Tetrahedron
[C] True/False 4 × ½ = 2
1. True
2. False
3. True
4. False
[D] Match the column 1 × 5 = 5
(1) (C)
(2) (D)
(3) (E)
(4) (B)
(5) (A)
[E] Viva-Voice
1. F + V – E = 2 1
2. 6 1
3. Parallelogram 1
4. Hexagonal prism 1
5. 6 1
[F] Quiz 6 × ½ = 3
1. 4
2. 4
3. 8
4. 12
5. No
6. Dice
P-110 M A T H E M A T I C S – VIII
1. (C) 7 cm 12. (D) 15 cm 1
3. Area of a rhombus =1
2 × Product of diagonals
Area of given rhombus=1
7·5 cm 12 cm2 1
2
=1 75 12
2 10 1 cm2
= 15 × 3 cm2
= 45 cm2.12
4. Area of a quadrilateral =1
2 × Diagonal × [Sum of the perpendiculars
on the diagonal from opposite vertices] 12
=21
24 (8 13) m2
= ´ ´ 2124 21 m
2
= 212 21 m
= 2252 m . 12
5. Area of a parallelogram = Base × Corresponding height
Area of a tile =224 10
m100 100
(converting cm into m)
=2240
m10000
1
= 20·024 m
Now area of the floor = 1080 m2
Number of tiles =Total area
Area of one tile
=1080
0·024
= 45000 tiles. 1
6. A rhombus is a parallelogram
Area of a parallelogram = Base × Height
Area of the rhombus = Base × Height
= 6 × 4 cm2
= 24 cm2 1
Let the other diagonal be d.
Also area of the rhombus =1
8 42
d d [ One of the diagonals = 8 cm]
11 MENSURATION WORKSHEET-53
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-111M E N S U R A T I O N
So, 4d = 24
or d =24
4 = 6 cm
Thus, the required other diagonal is 6 cm. 1
7. (a) Side of the square = 60 m
Its perimeter = 4 × side
= 4 × 60 m = 240 m ½
Area of the square = Side × Side
= 60 m × 60 m
= 3600 m2 12
(b) Perimeter of the rectangle = Perimeter of the given square
Perimeter of the rectangle = 240 m
But the perimeter of a rectangle = 2 × [Length + Breadth] = 240 m
or 2 × [80 m + Breadth] = 240 m
or 80 m + Breadth =240
2 m
= 120 m
Breadth = (120 – 80) m 1
= 40 m
Now, Area of the rectangle
= Length × Breadth
= 80 × 40 m2
= 3200 m2
Since, 3600 m2 > 3200 m2 ½
Area of the square field (a) is greater. ½
8. Perimeter of food-piece (a)
= (r + 1·5 + 2·8 + 1·5) cm
=22 2·8
5·8 cm7 2
æ ö´ +ç ÷
è ø
= (22 × 0·2 + 5·8) cm
= (4·4 + 5·8) cm = 10·2 cm 1
Perimeter of food-piece (b)
= (r + 2 + 2) cm
=22 2·8
4 cm7 2
æ ö´ +ç ÷
è ø
= (22 × 0·2 + 4) cm
= (4·4 + 4) cm = 8·4 cm 1
Since 10·2 > 8·4 > 7·2
The ant will take a longer round in case of food-piece (b). 1
P-112 M A T H E M A T I C S – VIII
1. (C) 2rh 1
2. (B) 7 cm 1
3. Let the height of the cuboid = h cm
Now base area × Height = Volume
or 180 × h = 900 12
or h =900
180 = 5 cm
Hence, the required height of the cuboid = 5 cm. 12
4. (i) Area of quadrilateral ABCD
= 1
2AC [Sum of perpendiculars on AC from opposite vertices] ½
= 1
6 cm [3 cm + 5 cm]2
= 1
6 cm 8 cm = 3 cm 8 cm2
= 24 cm2 ½
5. Tiles are rhombus shaped, having d1 = 45 cm and d2 = 30 cm.
Area of a tile (rhombus) = 1 2
1
2d d
=21
45 30 cm2
= 45 × 15 cm2 = 675 cm2 12
Total number of tiles = 3000
Area of floor = 675 × 3000 cm2
= 2025000 cm2
=22025000
m100 100´
2 2[ 1 m = 100 100 cm ]
=22025
m10
12
Cost of polishing the floor= Rs. 4 per sq. m
Cost of polishing the floor = Rs. 4 × 2025
10
= Rs. 2 × 405
= Rs. 810. 1
11 MENSURATION WORKSHEET-54
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-113M E N S U R A T I O N
6. Let the length of the side along the river
= 2x cm
Other side (parallel to the road) = x cm 12
Area of the trapezium shaped field
= 212 100 m
2x x
= 2 21 3003 100 m m
2 2x 1
2
But area of the field = 10500 m2
300
2x = 10500 1
2
or x =10500 2
300
=105 2
35 23
= 70 m.
Thus, the length of the side along the river
= 2x = 140 m. 12
7. Area of trapezium I :
Parallel sides are 24 cm and 16 cm.
Height =28 20
4 cm2
Area = ´ + ´ 21[16 24] 4 cm
2
= 40 × 2 cm2
= 80 cm2 1
Area of trapezium II :
Parallel sides are 20 cm and 28 cm.
Height =24 16
2
= 4 cm
Area =21
[20 28] 4 cm2 1
= 48 × 2 cm2 = 96 cm2
Area of trapezium III :
Area of trapezium III = Area of trapezium I
= 80 cm2 12
P-114 M A T H E M A T I C S – VIII
Area of trapezium IV :
Area of trapezium IV = Area of trapezium II
= 96 cm2. 12
8. Total surface area of the cuboid (a)
= 2[lb + bh + hl]
= 22[(60 40) (40 50) (50 60)] cm
= 22[2400 2000 3000] cm 1
= 2 22[7400 cm ] 14800 cm
Total surface area of the cube (b)
= 6l2 = 6 (50 × 50)
= 6(2500)
= 15000 cm2 1
Since the total surface area of the second (b) is more. 1
Cuboid (a) will require lesser material.
P-115M E N S U R A T I O N
1. (B) 13860 litres ( 1 l = 0.001 m3) 1
2. (C) 1 litre ( 1 l = 1000 cm3) 1
3. Let d1 and d2 be the diagonals of the rhombus.
d1 = 12 cm and d2 = 9·2 cm
Area of rhombus = 1 2
1
2d d 1
2
Area of given rhombus =21
12 9·2 cm2
= 6 × 9·2 cm2 = 55·2 cm2. 12
4. (i) l = 6 cm, b = 4 cm and h = 2 cm
Total surface area of the cuboid
= 2( )lb bh hl
= 2(6 × 4 + 4 × 2 + 2 × 6) cm2
= 2(24 + 8 + 12) cm2 1
= 2 × 44 cm2 = 88 cm2
5. Here, r = 7 m and h = 3 m
Total surface area of the cylindrical tank
= 2r(r + h)
= 2222 7 (7 3) m
7´ ´ ´ + 1
= 2 × 22 × 10 m2
= 440 m2
Thus, 440 m2 metal is required. 1
6. Lateral surface area of a cylinder
= 4224 cm2 = 2rh
4224 = (2r) (33)
and length of the rectangular sheet = l cm = circumference of circular base = 2r
33 × l = 4224
l =4224
33 = 128 cm 1
Now perimeter of the rectangular sheet
= 2[Length + Breadth]
= 2[128 cm + 33 cm]
= 2 × 161 cm
= 322 cm 1
11 MENSURATION WORKSHEET-55
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-116 M A T H E M A T I C S – VIII
7. Total surface area of a suitcase
= 2(lb + bh + hl)
= 2(80 48 48 24 24 80) 1
= 22(3840 1152 1920) cm
= 22(6912) cm
= 213824 cm
Total S. A. of 100 suitcases
= 2 2100 13824 cm 1382400 cm 1
Surface area of 1 m of tarpaulin
= 100 cm 96 cm´
= 9600 cm2
i.e. 9600 cm2 surface covered by 1 m of tarpaulin.
1382400 cm2 surface covered by tarpaulin 1382400
9600 metre or 144 m.
Thus, 144 metres of tarpaulin will be requird to cover 100 suitcases. 1
8. Let the side be x cm.
Total S. A. of the cube = 2 26 cmx 1
But the S. A. of the cube = 2600 cm
6x2 = 600 1
or x2 =600
6 = 100
or x2 = 102 x = 10 1
The required side of the cube = 10 cm.
P-117M E N S U R A T I O N
1. (D) 4·62 litres 1
2. ... 1 cubic cm = 0.001 1
(b) Side (edge) of the cube = 1·5 cm
Volume of the cube = (edge)3
= 3(1·5 m)
=315 15 15
m10 10 10
=3 33375
m 3·375 m1000
. 1
3. (i) Radius (r) = 7 cm
Height (h) = 10 cm
Volume of the cylinder = r2h
=2 322
7 10 cm7
=322
7 7 10 cm7
= 322 7 10 cm
= 1540 cm3 12
(ii) Base area = 250 m2
Height = 2 m
Volume of the cylinder = Base area × Height
= 250 m2 × 2 m
= 500 m3. 12
4. The road roller is a cylinder
Radius =84
cm = 42 cm2
Length (h) = 1 m = 100 cm
Lateral surface area = 2rh
= 2222 42 100 cm
7
= 2 × 22 × 6 × 100 cm2
= 26400 cm2 1
Area levelled by the roller in 1 revolution = 26400 cm2
or Area levelled in 750 revolutions
= 2750 26400 cm
= 2750 26400 m
100 100
= 15 × 132 m2 = 1980 m2 1Thus, the required area of the road
= 1980 m2.
11 MENSURATION WORKSHEET-56
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-118 M A T H E M A T I C S – VIII
5. Volume of the cuboid = 60 cm × 54 cm × 30 cm ½= (60 × 54 × 30) cm3
Volume of the small cube = (6 × 6 × 6) cm3 ½
Required Number of cubes =Volume of the cuboid
Volume of the small cube½
=3
3
(60 54 30) cm
(6 6 6) cm
=60 54 30
6 6 6
= 10 × 9 × 5 = 450 ½
Thus, 450 small cubes can be placed in the given cuboid.
6. Volume of the reservoir = 108 m3
1 m 3 = 1000 litres
Capacity of the reservoir = 108 × 1000 litres
= 108000 litres 1
Amount of water poured in 1 minute = 60 litres
Amount of water to be poured in 1 hour
= 60 × 60 litres 1
Thus, number of hours required to fill the reservoir = 108000
60 60 = 30
The required number of hours = 30. 1
7. (i) Total length of the fencing surrounding it
= Perimeter of the park
= 30 m + 20 m + 30 m + 20 m
= 100 m 1
(b) Land occupied by the park = Area of the park
= 30 × 20 m2
= 600 m2 1
(c) Area of cemented path = Area of park – Area of park left after cementing the path.
Now, since path is 1 m wide, so, the rectangular area left after cementing the path
= {(30 – 2) × (20 – 2)} m2
= (28 × 18) m2 = 504 m2 12
Area of cemented path= 600 m2 – 504 m2 = 96 m2
Number of cement bags used
=area of the path
area cemented by 1 bagIf 1 bag of cement is required to cement 4 m2 area, then the number of cement bagsused
=96
4 = 24 1
2
(d) Area of rectangular beds = 2 × (1·5 × 2) m2 = 6 m2
Area of the park left after cementing the path= 504 m2
Area covered by the grass
= 504 m2 – 6 m2 = 498 m2 . 1
P-119M E N S U R A T I O N
11 MENSURATION WORKSHEET-57
[A] Lab Activity
Objective :To derive the formula for the total surface area of a right circular closed cylinder.
Materials Required :1. A closed cylindrical box2. A pair of scissors3. A pen4. Graph papers ½
Procedure :
(1) Take the cylindrical box. Let its radius be r and height be h.(2) Trace one of its bases on a graph paper and cut it (Fig. (i)). ½
r
h
½
(i)(3) You get a circular strip of radius r (Fig. (ii))(4) The circumference of this strip is 2r and area is r2 (Fig. (ii)).(5) Thus, the surface area of each base of the cylindrical box is r2. ½(6) Take another graph paper in such a way that its width is equal to the height of the box,
i.e. h.(7) Wrap this paper around the box such that it just fits around the box (cut out the excess
paper) (Fig. (iii)). ½
(ii)
½
(iii)
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-120 M A T H E M A T I C S – VIII
(8) The shape of this paper obtained is a rectangular strip as shown in Fig. (iv).(9) The length of this strip is same as the circumference of the circular strip, obtained in the
step 3, i.e., 2r, and height h.(10) Area of this (rectangular) strip = 2rh. 1
This represent the lateral surface area of the cylindrical box, i.e., lateral surface area of thebox = 2rh.
h
2r
1
(iv)
Result :
Total surface area of the cylindrical box= 2 × surface area of each base + lateral surface area= 2 × r2 + 2rh= 2r (r + h).
Hence, we obtain that total surface area of a right circular closed cylinder = 2r(r + h)sq. units. 1
[B] Fill in the blanks ½ × 4 = 2
1. Base, height
2. Base, height
3. Sum of parallel sides, distance between them
4. 128
[C] True/False : ½ × 4 = 2
1. True
2. False
3. True
4. False
[D] Viva-Voce 1 × 5 = 5
1. 4 cm
2. 2rh
3. No vertex
4. 3, one curved face and two circular faces
5. They are identical.
[E] Quiz 1 × 5 = 5
1. a3 cubic units
2. 1000
3. 24 cm2
4. 1 litre
5. 1 : 1
P-121E X P O N E N T A N D P O W E R SS
1. (B) 1
101
2. (D) 16 1
3. The multiplicative inverse of 2– 4 is 24. 1
4. 22
1 1 1( 4) ·
( 4) ( 4) 16( 4)
1
5. (i) 1 = 13
3
1
2 =
33
3
1 1
22
Now
2
3
1
2
=
23 3 21 1
2 2
[Using (am)n = amn]
=
6
6
1 1
2 2
1
(ii)4
4 5( 3)
3
=
45
( 3)3
= 4[( 1) 5]- ´
= 4 4[( 1) (5) ]- ´
= 4 41 (5) (5) . 1
6. (i) 0 1 2(3 4 ) 2 0 1 1
1 and a aa
0 1 2(3 4 ) 2 =21
1 24
=5
4 54 1
(ii)
2 2 21 1 1
2 3 4
= (2)2 + (3)2 + (4)2
= 4 + 9 + 16
= 29 1
7. (i) m ma b = ( )mab
1 1 2(2 4 ) 2 = 1 2(2 4) 2
= 1 2 1 2(2 2 ) 2 1
12 EXPONENTS AND POWERS WORKSHEET-58
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-122 M A T H E M A T I C S – VIII
= 1 2 1 2(2 ) 2
= 3 1 2(2 ) 2
= 3 22 2
= 3 ( 2) 1 12 (2)
2 1
2
(ii) 13 = 1 11 1 1, 4 and 5
3 4 5- -= =
1 1 1 0[3 4 5 ] =
01 1 1
3 4 5
1
=
020 15 12
60
=
047
160
12
8. m na a = m na
35 5m = 55
( 3)5m- - = 55 1
35m = 55
Since, bases 5 are equal, therefore exponents are also equal. 1
i.e., m + 3 = 5
or m = 5 – 3 = 2
Thus, the required value of m is 2. 1
P-123E X P O N E N T A N D P O W E R SS
1. (D) 1 1
2. (D) 7
31
3.
5
5
1 1 1 1
1 1 1 1 1 12 12 2 2 2 2 322
-æ ö= = =ç ÷
æ ö æ öè ø æ ö ´ ´ ´ ´ç ÷ ç ÷ç ÷ è ø è øè ø
= 32 1
4. (i) 0·000000564 = 564
1000000000
= 2
9 7
5·64 5·6410
10 10 = 5·64 × 10– 7 1
0·000000564 = 5·64 × 10– 7
(ii) 0·0000021 = 21
10000000 =
2·1 10
10000000
= 6
2·1
10 = 2·1 × 10– 6
0·0000021 = 2·1 × 10– 6 1
5.1 3
4
8 5
2
=
1
4
8 (5 5 5)
2
=41
2 1258 1
=1
2 2 2 2 1258
= 2 × 125 = 250. 1
6. 1 1 1(5 2 ) 6 = 1 1(5 2) 6 1
= 1 1(10) 6
= 1 1(10 6) (60)- -´ =
=1
601
7. (i) (a)– 1 =1
a
11 11 1
3 4
=1
1 1
1 1
3 4
= 1 1(3 4) ( 1)
=1
11
1
12 EXPONENTS AND POWERS WORKSHEET-59
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-124 M A T H E M A T I C S – VIII
(ii) a– m =1ma
48
5
=
45
8
Now,
7 45 8
8 5
=
7 45 5
8 8
=
7 45
8
- +æ öç ÷è ø
=
35
8
-æ öç ÷è ø
[ ]m n m na a a
=
38
5
æ öç ÷è ø
=8 8 8
5 5 5
´ ´
´ ´
=512
125. 1
8. 125 = 5 × 5 × 5 = 53
10– 5 = (2 × 5)– 5 = 2– 5 × 5– 5
6– 5 = 5 5 5(2 3) 2 3- - -´ = ´ 1
5 5
7 5
3 10 125
5 6
=
5 5 5 3
7 5 5
3 2 5 5
5 2 3
- - -
- - -
´ ´ ´
´ ´
=
5 5 5 3
5 5 7
3 2 5 5
3 2 5
1
= 5 ( 5) 5 ( 5) 5 3 ( 7)3 2 5- -- - -- - + - -´ ´
= 5 5 5 5 5 103 2 5
= 0 0 53 2 5
= 51 1 5
= 55. 1
P-125E X P O N E N T A N D P O W E R SS
1. (C) – 1 1
2. (C) 3 1
3. (i) 63·02 10 = 2 6302 10 10- -´ ´
= 8302 10-´
=302
100000000
= 0·00000302 12
(ii) 4·5 × 104 =45
1000010
= 45 × 1000
= 45000 12
4. (i) 1
1000000 =
66
11 10
10
1 micron = 1 × 10– 6 m12
(ii) 0·07 =2
2
7 77 10
100 10
Thickness of a thick paper is 7 × 10– 2 mm. 12
5. ( )m n mna a=
222
3
=
( 2) 22
3
1
=
4 4
4
( 2)2
3 (3)
m m
m
a a
b b
=4
4
3
( 2)
=3 3 3 3
( 2) ( 2) ( 2) ( 2)
=81
.16
1
6. Since, 25 = 52
4
3 8
25
5 10
t
t
=
2 ( 4) ( 8)
3
5
5 5 2
t - - -
-
´
´ ´
=2 3 1 4 85
2
t 1
=4 45
2
t´
12 EXPONENTS AND POWERS WORKSHEET-60
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-126 M A T H E M A T I C S – VIII
= 45 5 5 5
2t
´ ´ ´´
=4625.
2t 1
7. Since
21
4 =
24
1 = (4)2 = 16
31
3 = (3)3 = 3 × 3 × 3 = 27 1
41
2= (2)4 = 2 × 2 × 2 × 2 = 16
2 3 41 1 1
4 3 2 = 16 + 27 + 16 = 59 1
8. (i) 0·00000000000942 =942
100000000000000
=2
14
9·42 10
10
= 9·42 × 102 – 14
= 9·42 × 10– 12 1
(ii) 6020000000000000 = 602 × 10000000000000
= 13602 10
= 2 136·02 10 10
= 156·02 10 . 1
9. Thickness of one book = 20 mm
Thickness of 5 books = 5 × 20 mm
= 100 mm
Again,
Thickness of 1 paper sheet = 0·016 mm
Thickness of 5 paper sheets
= 5 × 0·016 mm
= 0·080 mm 1
Total thickness = 100 mm + 0·080 mm
= 100·08 mm
= 1·0008 × 102 mm. 1
P-127E X P O N E N T A N D P O W E R SS
12 EXPONENTS AND POWERS WORKSHEET-61
1. (B) 6 1
2. (A) (ab)m 1
3. (i) 3 10p p = 3 ( 10) 7( ) ( )p p [... am × an = am+n] 12
= 7
1
p
(ii) 2 5 63 3 3 = 2 ( 5) 63 + - +
= 8 5 33 3- = . 12
4. 33
15
5
--´ = 3 3 05 5 1.- + = = 1
5. 0·000,000,000,000,000,000,16
=16
100000000000000000000
= 20 19
1·6 10 1·6
10 10
= 191·6 10
Charge on an electron is 1·6 × 10– 19 coulomb. 1
6. 34500000 = 345 × 100000
= 3.45 × 100 × 100000
= 3.45 × 102 × 105 12
= 3.45 × 102+5
= 3.45 × 107
Thus, 34500000 = 3.45 × 107 12
7. We have (80 + 5– 1)– 2 × 32
=
221
1 35
0 1 1
1 and x xx
=
226
35
=
225
36
1
=5 5
3 36 6
=5 5
2 2
=25
4. 1
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-128 M A T H E M A T I C S – VIII
8. Since 125 = 5 × 5 × 5 = 53
6– 5 = (2 × 3)– 5 = 5 52 3
5 5
4 5
2 3 125
5 6
=
5 5 3
4 5 5
2 3 5
5 2 3
1
=5 5
3 45 5
2 35
2 3
1
= 1 × 1 × 57
= 5 × 5 × 5 × 5 × 5 × 5 × 5
= 78125. 1
9. Distance between Sun and Moon
= 1·496 × 1011 – 3·84 × 108 1
= 1·496 × 1000 × 108 – 3·84 × 108
= (1496 – 3·84) × 108 m 1
= 1492·16 × 108 m
= 1·49216 × 1011 m. 1
P-129E X P O N E N T A N D P O W E R SS
12 EXPONENTS AND POWERS WORKSHEET-62
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
[A] Lab Activity
Object :
To verify law of exponents using paper folding.
Let’s Start :
We know that, 5 2 32 2 2 8. Let us verify it by paper folding.
Materials Required :
A butter paper. 1
Procedure :
(1) Take a butter paper or a very thin paper. Fold it as many number of times as power ofdividend. Here, we have folded it 5 times (see Fig. i). 1
1
(i)
(2) Open up the fold and count the number of boxes you get.
(3) Then make groups of boxes equal to divisor, that is, 4 in this case (see Fig. ii). 1
1
(ii)
(4) Now, count the number of groups you get.
(5) The number of groups you get will be equal to the quotient.
Observation
We observe that as we go on increasing the number of folds the number of boxes multiplies.
Conclusion
We conclude that the total boxes we obtain is equal to dividend and the number ofgroups gives quotient. 1
Result
25 22 = 8, i.e., 23
Hence, am an = am – n 1
[B] Fill in the Blanks ½ × 5 = 2½
1. 3·6 × 105
2. 1·23 × 10– 5
P-130 M A T H E M A T I C S – VIII
3.9
4
4. 0·003
5. 0·000532
[C] True/False ½ × 5 = 2½
1. True
2. False
3. True
4. True
5. False
[D] Viva-Voce ½ × 6 = 3
1. First number is 5 times the second number.
2. 1·275 × 10–5
3. (– 5)–9
4. 1/729
5. 24
6. 22
[E] Quiz 1 × 5 = 5
1. 24
2. No
3. x = 2
4. – 1
5. 7·25 × 10– 4
P-131D I R E C T A N D I N V E R S E P R O P O R T I O N S
1. (D) 720 1
2. (A) 17·6 1
3. Clearly, 3 5 7 9 12 1
6 10 14 18 24 2
x
y= = = = = = 1
(Constant)
x, y are directly proportional.
4. Let the required distance be x km. Then; we have
36 25
432 x
Quantity of petrol (in litres)
Distance (in km)
Clearly, less is the quantity of petrol consumed, less is the distance covered. 12
So, it is a case of direct proportion
36 25 1 25
432 12x x x = 12 × 25 = 300
required distance is 300 km. 12
5. We have :
60 154 hours Rs. 60
4 1100 25
8 hours Rs. 1008 2
140 3512 hours Rs. 140
12 3180 15
24 hours Rs. 18024 2
Parking Time Parking Charges Parking charges/Parking time
1
Since15
1
15 35 15
2 3 2
The parking charges are not in direct proportion with the parking time. 1
6. Let the required number of sugar crystals be x in 5 kg. of sugar. We have :
(i) 62 9 10
5 y
Weight of sugar Number of sugar
(in kg) crystals
13 DIRECT AND INVERSE PROPORTIONS WORKSHEET-63
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-132 M A T H E M A T I C S – VIII
Since, more the amount of sugar, more would be the number of sugar crystals. ½
it is a case of direct proportion.
2
5 =
69 10
x
½
or 2 × x = 6[9 10 ] 5
or x =6 69 10 5 45 10
2 2
½
= 22·5 × 106 = 2·25 × 107 [Standard form]
Thus, the required number of sugar crystals = 2·25 × 107. ½
7. Let the red pigment be represented by x1, x2, x3, ..... and the base represented by y1, y2,
y3, ..... .
As the base increases, the required number of red pigments will also increase.
The quantities are in direct proportion. ½
i.e. 1
1
x
y = 32
2 3
.....xx
y y ½
For x1 = 1, y1 = 8
1
1
x
y =
1
8
Now, 2
2
x
y =
2
1 4 1
8 8y ½
or y2 = 4 8 32
3
3
x
y =
3
1 7 1
8 8yÞ = ½
or y3 = 7 × 8 = 56
4
4
x
y =
4
1 12 1
8 8y ½
or y4 = 12 × 8 = 96
5
5
x
y =
5
1 20 1
8 8y
or y5 = 20 × 8 = 160
Thus, the required parts of base are : 32, 56, 96 and 160. ½
P-133D I R E C T A N D I N V E R S E P R O P O R T I O N S
8. Let the required distance covered in the map be x cm. We have :
18 1
72 x
Actual distance Distance
covered on the road represented
(in km) on the map (in cm)1
Since, it is a case of direct proportion,
18
72 =
1
x1
or 18 × x = 72
or x =72
418
Thus, the required distance on the map is 4 cm. 1
P-134 M A T H E M A T I C S – VIII
1. (A) 2700 12. (C) 29·5 kg 13. Let the required number of days be x. Then we have
45 35
49 x
Number of men
Number of days
Clearly, less men will take more days to finish the work. 12
So, it is a case of inverse proportion. 45 × 49 = 35 × x
x =45 49
6335
The required number of days = 63. 12
4. Let the required number of workers be x. Then we have
14
45 35
xNumber of workers
Number of hours
For finishing the work in less hours, more workers will be needed. 12
So, it is a case of inverse proportion. 14 × 45 = x × 35
x =14 45
1835
Hence, the required number of workers = 18. 12
5. Since the speed is constant.
for longer distance, more time will be required.
So, it is a case of direct proportion.
Let the required distance to be travelled in 5 hours (= 300 minutes) be x km. 1We have
14 25
300x
Distance (in km.) Time in (minutes)
12
14
x =
25
300or 25 × x = 14 × 300
or x =14 300
25
´
= 14 × 12 = 168
The required distance = 168 km. 12
6. Number of animals added = 10
now, the total number of animals
= 10 +20 = 30
For more number of animals, the food will last less number of days. 1
It is a case of inverse proportion.
13 DIRECT AND INVERSE PROPORTIONS WORKSHEET-64
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-135D I R E C T A N D I N V E R S E P R O P O R T I O N S
Thus, we have 30 × x = 20 × 6
or x =20 6
430
Therefore, the food will now last for 4 days. 17. (i) Obviously, more the number of spokes, less is the measure of angle between a pair of
consecuitve spokes. It is a case of inverse proportion. 1
Thus, 4 × 90° = 8 × x1 x1 = 4 90
458
´ °= °
4 × 90° = 10 × x2 x2 = 4 90
3610
´ °= °
4 × 90° = 12 × x3 x3 = 4 90
3012
´ °= °
Thus, the table is completed as below :
4 6 8 10 12
90 60 45 36 30° ° ° ° °
Number of spokes
Angle beween a pair of
consecutive spokes1
(ii) Let required number of spokes be n.
n × 40° = 4 × 90°
or n =4 90
40
= 9
The required number of spokes = 9. 1
8. (i) Let the time taken by the remaining persons to complete the job be x. 2 persons – 1 person = 1 personand lesser the number of person, more will be the number of days to complete the job.
it is a case of inverse proportion. ½
We have
2 3
1 x
Number of persons Number of days
2 × 3 = 1 × x
or x =2 3
61
1
1 person will complete the job in 6 days.(ii) Let the number of persons required to finish the job in 1 day be y.
2 3
1y
Number of persons Number of days
2 × 3 = 1 × y ½
or y =2 3
61
6 persons will be required to complete the job in 1 day. 1
P-136 M A T H E M A T I C S – VIII
1. (B) 5 : 3 12. (B) xy = k, being a constant 13. Let x bottles be filled in 5 hours.
840 6
5x
Number of bottles filled Number of hours
For more number of hours, more number of bottles would be filled. Thus given quantities
vary directly. 12
840
x =
6
5
or 6x = 5 × 840
or x =5 840
6
= 5 × 140 = 700 12
Thus, the required number of bottles = 700.
4. Let the number of workers employed to build the wall in 30 hours be y. We have thefollowing table.
48 30
15 y
Number of hours
Number of workers
Obviously more the number of workers, faster will they build the wall. So, the number
of hours and number of workers vary in inverse proportion. 12
So 48 × 15 = 30 × y
Therefore,48 15
30
= y
or y = 24
i.e., to finish the work in 30 hours, 24 workers will be required. 12
5. More is the number of persons, less is the time to complete job.
it is a case of inverse proportion. ½
3 4
4 x
Number of days
Number of persons to Complete the
wiring job½
3 × 4 = 4 × x
or x =3 4
34
the required number of days = 3. 1
13 DIRECT AND INVERSE PROPORTIONS WORKSHEET-65
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-137D I R E C T A N D I N V E R S E P R O P O R T I O N S
6. Suppose that x machines would be required. We have the following table.
Number of machines 42
Number of days 63 54
x
Lesser the number of machines, more will be the number of days to produce the samenumber of articles. 1
So, this is a case of inverse proportion.
Hence, 42 × 63 = x × 54 1
x =42 63
54
x = 49
Hence, 49 machines would be required. 1
7. Let x1 = p1, x2 = 200, x3 = 300
are y1 = 60, y2 = 30, y3 = p2
Since x and y are in inverse variation.
(i) x1y1 = x2y2
p1 × 60 = 200 × 30 1
p1 =200 30
60
´ = 100. 1
(ii) Also x2y2 = x3y3
200 × 30 = 300 × p2
p2 =200 30
300
´ = 20. 1
8. Actual length of the bacteria
=5
50000 cm
=1
10000 cm = 10–4 cm 1
More the number of times a photograph of a bacteria is enlarged, more the lengthattained. So, the number of times a photograph of a bacteria is enlarged and the lengthattained are directly proportional to each other.
So,1
1
x
y =2
2
x
y
50000
5 =
2
20000
y 1
50000y2 = 5 × 20000
y2 =5 20000
50000
y2 = 2
Hence, its enlarged length would be 2 cm. 1
P-138 M A T H E M A T I C S – VIII
1. (A) Inverse proportion 1
2. (C) 20 days 1
3. Reduced number of children = 24 – 4 = 20
Since, more the number of children, less is the quantity of sweets, that each child gets.
It is a case of inverse variation.
24 × 5 = 20 × x 12
or x =24 5
20
= 6
Each student will get 6 sweets. 12
4. Let the required length of the model of the ship = x m.
We have :
28 12
9x
Length of the ship Height of the mast
Since, more the length of the ship, more would be the height of its mast.
It is a case of direct proportion.
28
x =
12
91
or x =28 9
12
= 7 × 3 = 21
Thus, the required length of the model ship = 21 m. 2
5. Present duration of each period = 50 minutes
= 45 minutes 1
Duration of each period after reduction.
Let the present number of periods after reduction in time be ‘x’.
Since, more the number of periods, less is the duration of a period.
It is a case of inverse variation. 1
we have
9 50
45x
Duration of ech periodNumber of periods
(in minutes)
1
9 × 50 = x × 45
x =9 50
45
= 10
Thus, the required number of periods = 10. 1
13 DIRECT AND INVERSE PROPORTIONS WORKSHEET-66
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-139D I R E C T A N D I N V E R S E P R O P O R T I O N S
6. ·.· Quantity of water required for 2 persons
= 300 mL
Quantity of water required for 5 persons
=300
52 mL = 750 mL 1
Quantity of sugar required for 2 persons
= 2 spoons
Quantity of sugar required for 5 persons
=25
2 = 5 spoons 1
Quantity of tea leaves required for 2 persons
= 1 spoon
Quantity of tea leaves required for 5 persons
=15
2 =
122
spoons 1
Quantity of milk required for 2 persons
= 50 mL
Quantity of milk required for 5 persons
=50
52 mL = 125 mL 1
Thus, Mohan will require 750 mL of water, 5 spoons of suagar, 122
spoons of tea leaves
and 125 mL of milk of 5 persons. 1
P-140 M A T H E M A T I C S – VIII
13 DIRECT AND INVERSE PROPORTIONS WORKSHEET-67
[A] Lab Activity
Objective :
To show relation between two things in direct and indirect proportion by plotting the
data of both on a graph paper.
Let’s Start :
Observe the following tables :
(A)
1 2 3 4
25 50 75 100
No. of items
purchased
Total Cost
(in Rs.)
½
(B)
5 10 15 20
300 150 100 75
No. of children
No. of days food
will last in pantry½
Materials Required :
(i) Data (ii) Graph papers (iii) Pencil (iv) Ruler.
Procedure :
(1) Taking suitable scale plot the points of Data 1 (Table A) on the graph paper and join the
points (see Fig. (i)).
(2) Take suitable scale plot the points of Data 2 (Table B) on a different graph paper and
join the points (see Fig. (ii)).
(3) Observe and compare both graphs by moving from left to right.
1
(i)
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-141D I R E C T A N D I N V E R S E P R O P O R T I O N S
1
(ii)
Observations :
For 1st graph : We observe that as we go from left to right the number of items
purchased increases, so the total cost also increases.
For 2nd graph : We observe that as number of children increases the number of days
for which the food will last decreases.
Conclusion :
We conclude that 1st graph is of direct proportion and 2nd graph is of indirect proportion.
2
Result :
In direct proportion, the two things given are in direct relation, i.e., if one increases the
other also increases and if one decreases the other also decreases. In indicrect proportion,
the result is inverse that is the two given things are in inverse relation, i.e., if one
increases other will decrease and vice-versa. ½
[B] Fill in the Blanks ½ × 5 = 2½
1. 10– 4 cm
2. 3
3. 120
4. 2
5. 13
[C] True/False ½ × 4 = 2
1. True
2. True
3. False
4. False
P-142 M A T H E M A T I C S – VIII
[D] Viva-Voce
1. If the number of items purchased will be reduced, then the total cost will also reduce.1
2. This happens beause items purchsed and total cost are in direct proportion. 1
3. In graph (ii), we observe downward sloping curve because as the number of children is
increasing, the no. of days the food will last is decreasing. 1
4. This happens because number of children and the number of days food will last are in
indirect proportion. 1
5. For a given job, more the number of workers, less will be the time taken to complete thework. 1
[E] Quiz
1. Rs. 80 1
2. For the given table 1
·3
x
y 1
3. k is constant of proportionality. 1
4. Simple interest. 1
5. The required number of sheets is 375. 1
P-143F A C T O R I S A T I O N
1. (A) 10xy(4x + 3y)(4x– 3y) 1
2. (D) 5(z + 4)(z – 4) 1
3. (a) 14pq = 2 7 p q = (2 7 )p q
28p2q2 = 2 2 7 p p q q´ ´ ´ ´ ´ ´
= (2 7 ) 2p q p q
The common factor = (2 7 )p q
= 14pq 12
(b) 2x = 1 2 1 2x x
3x2 = 1 3 1 3x x x x´ ´ ´ = ´ ´ ´
and 4 = 1 2 2 1 2 2´ ´ = ´ ´
The common factor = 1. 12
4. (a) 7x = 7 (7)x x
42 = 2 7 3 (7) 2 3
7x – 42 = 7[( ) (2 3)]x
= 7[ 6]x 12
(b) 6p = 2 3 p
= (2 3) p´ ´
12q = 2 × 2 × 3 × q
= (2 × 3) × 2 × q
6p – 12q = (2 3)[( ) (2 )]p q
= 6[ 2 ]p q . 12
5. 2 33x y = 3 x x y y y
= ( ) 3x x y y y
3 210x y = 2 5 x x x y y
= ( ) 2 5x x y y x 1
2 26x y z = 2 3 x x y y z
= ( ) 2 3x x y y z
The common factor = (x × x × y × y)
= 2 2x y . 1
6. 2 249 84 36y yz z+ + = 2 2(7 ) 2(7 )(6 ) (6 )y y z z 1
= 2(7 6 )y z
= (7 6 )(7 6 )y z y z 1
14 FACTORISATION WORKSHEET-68
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-144 M A T H E M A T I C S – VIII
7. 2 7 10y y = 2 2 5 10y y y [Splitting 7y in 2y + 5y such that
2y × 5y = 10y2] 1
= ( 2) 5( 2)y y y
= ( 2)( 5)y y 1
2 7 10
( 5)
y y
y
=
( 2)( 5)( 2)
( 5)
y yy
y
+ += +
+
2( 7 10) ( 5)y y y = (y + 2). 1
8. 25 25 20p p = 25( 5 4)p p
= 5(p2 – p – 4p + 4)
= 5[ ( 1) 4( 1)]p p p- - - 1
= 5[( 1)( 4)]p p
25 25 20
1
p p
p
=
5( 1)( 4)
( 1)
p p
p
- -
-1
=5 ( 4)
1
p´ -
Thus, 2(5 25 20) ( 1) 5( 4)p p p p 1
P-145F A C T O R I S A T I O N
1. (D) q3 – p3 1
2. (C) 162 1
3. 2 8 8x xy x y
= ( ) 8( )x x y x y+ + + 12
= ( )( 8)x y x . 12
4. 2 24 9p q = 2 2(2 ) (3 )p q 12
= (2 3 )(2 3 )p q p q . 12
[Using a2 – b2 = (a + b)(a – b)]
5. (i) 4 4a b = 2 2 2 2( ) ( )a b
= 2 2 2 2( )( )a b a b 12
2 2[using ( )( )]a b a b a b
= 2 2( )( )( )a b a b a b 12
(ii) p4 – 81 = 2 2 2( ) (9)p
= 2 2( 9)( 9)p p 12
= 2 2 2( 9)[( ) (3) ]p p
= 2( 9)( 3)( 3)p p p+ + - 12
6. 2 4 6 2 2 363 7a b c a b c =2 4 6
2 2 3
63
7
a b c
a b c
=2 4 6
2 2 3
3 3 7
7
a b c
a b c
=2 4 6
2 2 33 3
a b c
a b c
= 2 2 4 2 6 39 a b c 1
= 0 2 39 a b c
= 2 3 2 39 1 9b c b c . 1
7. 2 29 16x y = 2 2(3 ) (4 )x y
= (3 4 )(3 4 )x y x y 1
14 FACTORISATION WORKSHEET-69
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-146 M A T H E M A T I C S – VIII
2 212 (9 16 )
4 (3 4 )
xy x y
xy x y
=
2 2 3 (3 4 ) (3 4 )
2 2 (3 4 )
x y x y x y
x y x y
1
=3 (3 4 )
1
x y
2 212 (9 16 ) 4 (3 4 )xy x y xy x y- ¸ + = 3(3x – 4y) 1
8. 3a – 12 = 3(a – 4)
5b – 30 = 5(b – 6)
96 = 2 × 2 × 2 × 2 × 2 × 3
and 144 = 2 × 2 × 2 × 2 × 3 × 3 1
96 (3 12)(5 30)
144( 4)( 6)
abc a b
a b
- -
- -
=2 2 2 2 2 3 3 ( 4) 5 ( 6)
2 2 2 2 3 3 ( 4) ( 6)
a b c a b
a b
´ ´ ´ ´ ´ ´ ´ ´ ´ ´ - ´ ´ -
´ ´ ´ ´ ´ ´ - ´ -1
= 2 × 5 × a × b × c = 10abc
Thus, 96abc(3a – 12)(5b – 36) 144(a – 4)(b – 6) = 10abc. 1
P-147F A C T O R I S A T I O N
1. (C) (x + 2) 1
2. (C) (x – 1)(x2 + 1) 1
3. 428 56x x =428
56
x
x =
42 2 7
2 2 2 7
x
x
12
=4 1 31 1
2 2x x . 1
2
4. 2(5 6 ) 3x x x =25 6 (5 6)
3 3
x x x x
x x
- -=
=5 6
3
x-. 1
5. Taking out 2x as common from each term, we have
3 2 22 2 2x xy xz = 2 2 22 ( )x x y z . 1
6. 3z – 24 = 3(z – 8)
2 29 (3 24)
27 ( 8)
x y z
xy z
=
2 23 3 3 ( 8)
3 3 3 ( 8)
x y z
x y z
´ ´ ´ ´ ´ -
´ ´ ´ ´ ´ -
=
2 22 1 2 1x y
x yx y
= xy. 1
7. 10x – 25 = 5(2x – 5)
10 25
5
x =
5(2 5)(2 5)
5
xx
1
8. 4 4( )x y z = 2 2 2 2[ ] [( ) ]x y z
= 2 2[( ) ( ) ]x y z 2 2[( ) ( ) ]x y z 1
2 2[Using ( )( )]a b a b a b
We can factorise [x2 – (y + z)2] further as
2 2( ) ( )x y z = [( ) ( )][( ) ( )]x y z x y z
= ( )( )x y z x y z 1
4 4( )x y z = ( )( )x y z x y z 2 2[ ( ) ]x y z 1
14 FACTORISATION WORKSHEET-70
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-148 M A T H E M A T I C S – VIII
9. We have
4 4( )x x z = 2 2 2 2[ ] [( ) ]x x z- -
= 2 2 2 2[ ( ) ][ ( ) ]x x z x x z 1
Now, factorising x2 – (x – z)2 further, we have
2 2( )x x z = [ ( )][ ( )]x x z x x z
= ( )( )x x z x x z
= (2 )( )x z z 1
4 4( )x y z = 2 2(2 )[ ( ) ]z x z x x z
= 2 2 2(2 )[ ( 2 )]z x z x x xz z 1
= 2 2 2(2 )[ 2 ]z x z x x xz z
= 2 2(2 )(2 2 )z x z x xz z 1
P-149F A C T O R I S A T I O N
14 FACTORISATION WORKSHEET-71
1. (A) 12x 1
2. (B) a + b and a + c 1
3. 5 316 144x x = 3 216 ( 9)x x 12
= 3 2 216 [( ) (3) ]x x
= 316 ( 3)( 3)x x x . 12
[Using a2 – b2 = (a – b)(a + b)]
4. 2 27 21p q = 7 7 3p p q q
= 7[ 3 ]p p q q 12
= 2 27( 3 )p q .
5. 2 3 266 11pq r qr¸ =2 3
2
66
11
pq r
qr1
=2 3 11
11
p q q r r r
q r r
12
=2 3
1
p q r = 6pqr. 1
2
6. Using the identity a2 – b2 = (a + b)(a – b), we have
2 2( ) ( )l m l m = [( ) ( )]l m l m [( ) ( )]l m l m
= [ ][ ]l m l m l m l m 1
= (2 )(2 ) 2 2( )l m l m
= 4lm. 1
7. 2 2 23 (50 98) 26 (5 7)ay y y y
=
2 2
2
3 (50 98)
26 (5 7)
ay y
y y
=
2 2
2
3 2 (25 49)
26 (5 7)
ay y
y y12
=
2 2 2
2
3 2 [(5 ) (7) ]
26 (5 7)
ay y
y y12
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-150 M A T H E M A T I C S – VIII
=
2
2
3 2 (5 7)(5 7)
26 (5 7)
ay y y
y y =
2
2
3 2(5 7)
26
ay y
y
=3 3
(5 7) (5 7)13 13
a ay y 1
9. (m2 – 14m – 32) (m + 2) =
2 14 32
2
m m
m
=
2 16 2 32
2
m m m
m1
=
( 16) 2( 16)
2
m m m
m
=
( 16)( 2)
2
m m
m
= (m – 16) 1
10. Joseph and Sirish are not correct while Suman is correct as she has followed the rule
mentioned. ½+½+½+½
P-151F A C T O R I S A T I O N
14 FACTORISATION WORKSHEET-72
[A] Lab Activity
Objective :
To verify experimentally that the factors of 3xy + 2y + 3x + 2 are (3x + 2)(y + 1) such that3xy + 2y + 3x + 2 = (3x + 2)(y + 1).
Let’s Start :
We know that area of a rectangle is the product of its length and breadth. ½Area of rectangle ABCD = a × b = ab (Fig. (i))We have
3xy = 3x × y2y = 2 × y3x = 3x × 1
and 2 = 2 × 1
A B
D C
a
b
½
(i)Materials Required :
(i) White thick paper sheet (ii) Pen (iii) Ruler (iv) A pair of scissors (v) Fevicol (vi)Colours 1
Procedure :(1) Take suitable values of x and y.(2) Cut out two rectangles of lengths 3x units and 2 units and breadth y units each from the
paper sheet.
D1 C1
A1 B13x
y
C2D2
B2A22
y
(ii) (iii)Colour these rectangles with different colours and name them as A1B1C1D1 and A2B2C2D2
respectively (Fig. (ii) and (iii)). ½
(3) Area of rectangle A1B1C1D1 = 3x × y
= 3xy.
(4) Area of rectangle A2B2C2D2 = 2 × y = 2y. ½
(5) Join and paste these two rectangles as shown in Fig. (iv)
D1 C2
C1D2
A1 B2B1A2
3 + 2x
y y
½
(iv)
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-152 M A T H E M A T I C S – VIII
(6) Area of rectangle 1 2 2 1A B C D
= Area of rectangle 1 1 1 1A B C D + Area of rectangle 2 2 2 2 (Fig. iv)A B C D
= 3xy + 2y (Use steps 3 and 4).
(7) Cut out two other rectangles of lengths 3x units and 2 units and breadth 1 unit eachfrom the paper sheet.Colour these rectangles with different colours and name them as
A3B3C3D3 and 4 4 4 4A B C D (Fig. (v) and (vi)). ½
D3 C3
A3 B3
1
3x(v)
D4 C4
A4 B4
1
2
½
(vi)
(8) Area of rectangle 3 3 3 3 3 1 3 .A B C D x x= ´ =
(9) Area of rectangle 4 4 4 4 2 1 2.A B C D
(10)Join and paste these two rectangles as shown in (Fig. (vii)).
D3 C4
A3 B4
C3 D4
B3 A4
3 + 2x
1 1½
(vii)
(11) Area of rectangle 3 4 4 3A B C D
= Area of rectangle 3 3 3 3A B C D + Area of rectangle 4 4 4 4 (Fig. (vii))A B C D
= 3x + 2 (Use steps 8 and 9).
(12)Now, join and paste the rectangles 1 2 2 1A B C D and 3 4 4 3A B C D as shown in (Fig. (viii)). 1
y+1
3 + 2x
D , A1 3 C , B2 4
C4D3
B2A1
(viii)
(13) Area of the rectangle 1 2 4 3A B C D
= Area of rectangle 1 2 2 1A B C D + Area of rectangle 3 4 4 3A B C D (Fig. (viii))
= 3 2 3 2.xy y x+ + + (Use steps 6 and 11)
(14)Length of the rectangle 1 2 4 3 2 3A B C D x and breadth of it = y + 1 (Fig. (viii)).
(15)Therefore, area of the rectangle 1 2 4 3A B C D
= (3 2) ( 1)x y
= (3 2)( 1)x y
Result :
We obtain that the factors of 3xy + 2y + 3x + 2 are 3x + 2 and y + 1 such that
3xy + 2y + 3x + 2 = (3x + 2)(y + 1). 1
P-153F A C T O R I S A T I O N
[B] Fill in the Blanks ½ × 5 = 2½
1. 6
2. 6(x – 7)
3. a(a2 + ab + b2)
4. (x + 4)2
5. 9801, (... 992 = (100 – 1)2 = (100)2 – 2 × 100 × 1 + 12
= 10000 – 200 + 1
= 9801)
[C] True/False ½ × 5 = 2½
1. True
2. False
3. True
4. True
5. False
[D] Viva-Voce ½ × 8 = 4
1. 1, b and c
2. 1, 2 and 4
3. 1
4. 2y
5. 1, 2, 3, 4, 6 and 12
6. No
7. (x + 3)(x – 3)
8. xy
[E] Quiz ½ × 8 = 4
1. No
2. 1
3. 2( 4)( 2)( 2)x x x
4. Every natural number
5. 5 × 3 × x × y
6. – 2x
7. 2 × x
8. No
P-154 M A T H E M A T I C S – VIII
1. (B) 2 axes 1
2. (A) 0 1
3. (i) On y-axis 1
(ii) On x-axis 1
4. In case (iii), there are an infinite number of temperatures at the same time which is not
possible. 1
Case (iii) does not represent a time-temperature graph. 1
5. (a) The patient’s temperature at 1 p.m. was 36·5º. 1
(b) The patient’s temperatre 38·5°C was at 12 noon. 1
(c) The patient’s temperature was same (i.e., 36·5°C) at 1 p.m. and 2 p.m.
(d) The patient’s temperature at 1:30 p.m. was 36·5°C [because the temperature of thepatient was constant (i.e. 36·5°C) from 1 p.m. to 2 p.m.] 1
(e) The temprature of patient showed an upward trend during 9 a.m. to 11 a.m. and 2
p.m. to 3 p.m. 1
6. From the figure, we have
The co-ordinates of P are (0, 4). 1
The co-ordinates of Q are (5, 5).
The co-ordinates of R are (10, 3). 1
The co-ordinates of S are (10, 2). 1
The co-ordinates of T are (8, 0). 1
15 INTRODUCTION TO GRAPHS WORKSHEET-73
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-155I N T R O D U C T I O N T O G R A P H S
1. (B) 1 1
2. (D) Its perpendicular distance from the y-axis. 1
3. At the point on intersection of the x and y-axis. 1
4. (i) A, x = – 3, y = 2 ½
(ii) C, x = 0, y = – 7 ½
5. (a) (i) Company’s sales in 2003 were Rs. 4 crores.
(b) Difference between the sales in 2002 and 2006 = [Rs. 8 crores] – [Rs. 4 crores]
= Rs. 4 crores 1
(c) The greatest difference between the sales of two consecutive years 2004 and 2005.
16. (i) The forecast temperature was the same as the actual temperature on Tuesday,
Friday and Sunday. 1
(ii) The maximum forecast temperature during the week was 35°C. 1
(iii) The minimum actual temperature during the week was 15°C. 1
7. (i) A (5, 100), D (12, 500). 1
(ii) The car started from the town A at 5 a.m. 1
(iii) The car stopped at town B for 1 hour (6 a.m. to 7. a.m.) 1
(iv) The car reached at town D at 12 noon.
The car look 2 hours to reach from town C to town D. 1
15 INTRODUCTION TO GRAPHS WORKSHEET-74
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-156 M A T H E M A T I C S – VIII
1. (B) A straight line 1
2. (A) (0, 0) 1
3. (i) 6 (ii) 0 (iii) 2
(iv) – 3 1
4. (a) The time is taken along the x-axis. The scale along x-axis is 4 units = 1 hour. 1
(b) Total travel time = 8 a.m. to 11·30 a.m. = 1
32
hours. 1
(c) Distance of the merchant from the town = 22 km. 1
(d) Yes, the stoppage time = 10·00 a.m. to 10·30 a.m. 1
5. Linear graph to show, snow fall in different years
12
11
10
9
8
7
6
5
4
3
2
1
Da
ys
for
wh
ich
rain
fall
rec
eived
Years2003 2004 2005 2006
X
Y
3
6. From the graph, the vertices of ABC are : A, B and C. We find that
The co-ordinates of A are (2, 5). ½
The co-ordinates of B are (2, 1). ½
The co-ordinates of C are (6, 1). ½
15 INTRODUCTION TO GRAPHS WORKSHEET-75
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-157I N T R O D U C T I O N T O G R A P H S
The required triangle with vertices as A(5, 2), B(1, 2) and C(1, 5) is given in the following
graph.
1½+1
P-158 M A T H E M A T I C S – VIII
1. (B) (3, 4) 1
2. (A) (1, 0) 1
3. (i) A, x = 0, y = 5, (ii) B, x = – 6, y = – 4. 1
4. On plotting the points A(2, 3) and B(3, 2) and joining them, draw a line. On extending
the line meets the x-axis at C(5, 0)and the y-axis at D(0, 5). 1
D (0, 5)
4
3
2
1
O(0, 0) 1 2 3 4 5 6 7
A(2, 3)
B(3, 2)
C(5,0)
X
Y
1
5. (i) The co-ordinates of :
O are (0, 0)
A are (2, 0)
B are (2, 3)
C are (0, 3) 1
(ii) The co-ordinates of :
P are (4, 3)
Q are (6, 1)
R are (6, 5)
S are (4, 7) 12
(iii) The co-ordinate of :
K are (10, 5)
L are (7, 7)
M are (10, 8) 12
6. (i) Taking the side of the square along the x-axis and the perimeter along the y-axis
and plotting the points (2, 8), (3, 12), (3·5, 14), (5, 20) and (6, 24), we get the required
graph as a straight line.
15 INTRODUCTION TO GRAPHS WORKSHEET-76
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-159I N T R O D U C T I O N T O G R A P H S
This graph is a linear graph. 1
Y
24
20
16
12
8
4
Per
imet
er (
in c
m)
1 2 3 3·54 5 6
Side (in cm)
X(0, 0)
1
(b) Taking the side of the square along the x-axis and area (in cm2) along the y-axis, we
can draw the required graph by plotting the points (2, 4), (3, 9), (4, 16), (5, 25) and
(6, 36) as shown in the following figure.
Y
40
30
20
10Are
a (
in c
m)
2
(0,0) 1 2 3 4 5 6 7 8 9
Side (in cm)
1
the graph in not a straight line.
it is not a linear graph.
7. (a) The horizontal axis (or the x-axis) indicates the matches played during the
year 2007. The vertical axis (or the y-axis) shows the total runs scored in each
match. 1
(b) The dotted line shows the runs scored by Batsman A. (This is already indicated at
the top of the graph). 1
(c) During the 4th match, both have scored the same number of 60 runs. (This is
indicated by the point at which both graphs meet). 1
(d) Batsman A has one great ‘‘peak’’ but many deep ‘‘valleys’’. He does not appear to be
consistent. B, on the other hand has never scored below a total of 40 runs, even
though his highest score is only 100 in comparison to 115 of A. Also A has scored a
zero in two matches and in a total of 5 matches he has scored less than 40 runs.
Since A has a lot of ups and downs, B is a more consistent and reliable batsman. 1
P-160 M A T H E M A T I C S – VIII
15 INTRODUCTION TO GRAPHS WORKSHEET-77
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
[A] Lab Activity
Objective :To prepare a graph paper and hence draw a linear graph of x + y = 3 on it.
Let’s Start :A graph paper is a plane surface consisting horizontal and vertical lines, on whichnumerical facts/change of one or more quantities are shown. A graph is a two-dimensionalfigure. A linear graph is of a line. ½
Materials Required :(i) A white paper (ii) Pencil (iii) Pen (iv) Sketch pens (v) Geometry box.
Procedure :(a) To prepare a graph paper :
(1) Take a point O in the middle of a paper [Fig. (i)](2) Draw a horizontal, semi-dark straight line passing through O. Name it as XOX [Fig. (ii)].
This line is called the x-axis. ½(3) Draw another semi-dark straight line which is perpendicular to the x-axis and passes
through O. Name it as YOY (Fig. iii). This line is called the y-axis.
O
X' X
O ½
(i) (ii)(4) The point O is the common point on both the axes, i.e., x-and y-axes. The point O is
called the origin. Its coordinates are (0, 0).(5) Draw many more light lines of equal separation, parallel to x-axis (Fig. (iv)).(6) Again draw many more light lines of equal separation parallel to the y-axis (Fig. (v)). ½
Y
X' X
Y'
O
Y
X' X
Y'
O
(iii) (iv) ½
OX' X
Y
Y'
(v) ½
P-161I N T R O D U C T I O N T O G R A P H S
ResultWe get a graph paper.
(b) To draw a linear graph of the line x + y = 3 :(1) First find out the coordinates of at least two points on the line.(2) Put x = 0 in x + y = 3 and it will give the value of y, i.e., y = 3. ½(3) Put y = 0 in x + y = 3 and it will give the value of x, i.e., x = 3.(4) Thus, you get the following table for the values of x and y.
0 3
3 0
x
y ½
(5) Plot two points A(0, 3) and B(3, 0) on the graph paper (Fig. vi)(6) Joint AB and extend to both the sides (Fig. vii). ½
Result :
AB
is the graph of the required line.
Y
A
X' X
Y'
O
(0, 3)
B
(3, 0)
Y
A
XX'
Y'
O
(0, 3)
B(3, 0)
x+y=3
½
(vi) (vii)
[B] Fill in the Blanks ½ × 1 = 2
1. y-axis2. A line parallel to the x-axis3. x = 04. Line graph
[C] True/False 1 × 3 = 3
1. True2. False3. True
[D] Viva-Voce
1. (0, 0) 12. 90° 13. Yes, x = y 14. Yes 15. A graph is a visual base to show the numerical facts/change in one or more quantities. 1
[E] Quiz
1. Second quadrant 12. 5 units 13. No 14. x-coordinate and y-coordinate 15. 2. 1
P-161P L A Y I N G W I T H N U M B E R S
1. (D) 100 × 4 + 10 × 2 + 1 × 1 12. (D) 804 1
3. The one’s digit, when divided by 5, must leave a remainder of 3 so the one’s digit mustbe either 3 or 8. 1
4. If remainder = 1, then the one’s digit of ‘N’ must be either 1 or 6. 1
5. 616
We have, 6 + 1 + 6 = 13
( A number is divisibile by 9 if sum of its digits is divisible by 9) ½
and 13 is not divisible by 9
616 is also not divisible by 9. ½
6. For A + S to be 2 or number whose one’s digit is 2 we must have A = 7. ½
For A + 8 to have one’s digit as 3 we must have A = 5
1 4 9 = B
or B = 1 4 12
4 5
9 8
1 4 3
+
Clearly C = 1 1
Thus A = 5, B = 4 and C = 1
7. We have the sum of the digits of 51x3
= 5 + 1 + x + 3 = 9 + x 1
Since, 51x3 is divisible by 9.
(9 + x) must be disible by 9.
(9 + x) must be equal to 0 or 9 or 18 or 27 or ..... 1
But x is a digit, then
9 + x = 9 x = 0
9 + x = 18 x = 9
x = 27 x = 18, which is not possible.
The required value of x = 0 or 9. 1
8. Since the ones digit of A × 3 is A.
We must have A = 0 or A = 5 ½
Now, let us look for B,
If B =1, then BA × B3 would at most be equal to 19 × 19
i.e. at the most it must be equal to 361.
But product is 57A which is more than 500.
So we cannot have B = 1 ½
If B = 3, then BA × A3 would be more than 30 × 30 i.e. more than 900.
16 PLAYING WITH NUMBERS WORKSHEET-78
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-162 M A T H E M A T I C S – VIII
But 57A is less than 600
B 3 ½
B can be only equal to 2.
It is either 20 × 23 or 25 × 23.
But 20 × 23 = 460
So not possible 25 × 23 = 575 ½
A = 5; B = 2.
9. ·.· The value of a or b cannot exceed 9.
The sum a + b cannot exceed 18. ½
Yes ! dad is a multiple of 11.
dad is less than or equal to 198. ½
All the 3-digit numbers which one multiples of 11 upto 198 are
110, 121, 132, 143, 154, 165, 176, 187 and 198. 1
Clearly dad = 121
d = 1, a = 2 1
P-163P L A Y I N G W I T H N U M B E R S
1. (A) 1 1
2. (C) 28 1
3. If remainder = 4, then the one’s digit of ‘N’ must be either 4 or 9. 1
4. (i) 108
We have 1 + 0 + 8 = 9
and 9 is divisible by 3.
108 is also divisible by 3. 1
(ii) 616
We have 6 + 1 + 6 = 13
and 13 is not divisible by 3.
Thus 616 is also not divisible by 3. 1
5. Since 24x is a multiple of 3, its sum of digits 6 + x is a multiple of 3; Since x is a digit, it
can only be that 6 + x = 6 or 9 or 12 or 15. Therefore, x = 0 or 3 or 6 or 9. Thus, x can have
any of four different values. [1+1+1]
6. We have 3 + 1 + z + 5 = 9 + z
31z5 is divisible by 3.
(9 + z) must be divisible by 3. 1
since, z is a digit.
If 9 + z = 9, then z = 0
If 9 + z = 12, then z = 3,
If 9 + z = 15, then z = 6,
If 9 + z = 18, then z = 9,
If 9 + z = 21, then z = 12, 1
Possible values of z are 0, 3, 6 or 9.
7. (i) 73 = 70 + 3
= 10 × 7 + 3 × 1 = 10 × 2 + 3 1
(ii) 129 = 100 + 29 + 9
= 100 × 1 + 10 × 2 + 1 × 9 = 100 × 1 + 10 × 20 + 9 1
(iii) 302 = 100 × 3 + 10 × 0 + 1 × 2 = 300 + 0 + 2 1
16 PLAYING WITH NUMBERS WORKSHEET-79
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-164 M A T H E M A T I C S – VIII
1. (B) 54 1
2. (D) a = 1 1
3. N is odd; so its one’s digit is odd. Therefore, the one’s digit must be 1, 3, 5, 7, or 9. 1
4. Chosen number = 27
Number with reversed digits = 72
Sum of the two numbers = 27 + 72 = 99
Now, 99 = 11[9]
= 11[2 + 7]
= 11 [Sum of the digits of the chosen number] 1
This works always as if we choose the number say ab which is actually 10a + b. In
reversing the digits we have 10b + a as the number.
Adding the two we get
10a + b + 10b + a
= 11a + 11b
= 11 (a + b)
which is a multiple of 11. 1
5. We have 2 + 1 + y + 5 = 8 + y
21y5 is multiple of 9
(8 + y) must be divisible by 9 1
(8 + y) will be divisible by 9 if 8 + y is either 0, 9, 18, 27, ....., etc. 1
Since, y is a digit
So, 8 + y = 0 is not possible. The only possibility is 8 + y = 9 1
8 + y = 9 y = 9 – 8 or y = 1.
6. We have 3 + 1 + z + 5 = 9 + z
31z5 is divisible by 9. 1
(9 + z) must be equal to 0, or 9 or 18 or 27, .....
16 PLAYING WITH NUMBERS WORKSHEET-80
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-165P L A Y I N G W I T H N U M B E R S
But z is a digit. 1
9 + z = 9 or 9 + z = 18 1
If 9 + z = 9, then z = 0 and if 9 + z = 18, then z = 9.
7. (i) 10 × 5 + 6 = 50 + 6 = 56 1
(ii) 100 × 7 + 10 × 1 + 8 = 700 + 10 + 8 = 718 1
(iii) 100 × a + 10 × c + b = 100a + 10c + b = a c b 1
P-166 M A T H E M A T I C S – VIII
1. (B) 0, 3, 6 or 9 1
2. (B) 440 1
3. For N 5, remainder = 4
One’s digit can be 4 or 9 ....(i)
Again for N 2, remainder = 1
N must be an odd number.
So, one’s digit of N must be
1, 3, 5, 7 or 9 ....(ii) 1
From (i) and (ii), the one’s digit of N must be 9.
4. (i) 25 = 20 + 1 × 5
= 10 × 2 + 5 12
(ii) 73 = 70 + 3
= 10 × 7 + 1 × 3 12
5. 329
We have 329 = 300 + 20 + 9
= 100 × 3 + 10 × 2 + 1 × 9 1
6. We have
100 × 7 + 10 × 0 + 9 × 1
= 700 + 0 + 9
= 709
7. (i) 294
We have 2 + 9 + 4 = 15 2
and 15 is divisible by 3.
Thus, 294 is also divisible by 3. 1
8. (i) B can be 2, 4, 6 or 8.
We need product 111 or 222, or 333 or 444 or 888 out of them 111 and 333 are rejected.
Possible products are 222, 444 or 888
To obtain
The possible value is B = 4
6
A B
B B B
´
4
6
4
A
‘A’ can be either 2 or 7.
16 PLAYING WITH NUMBERS WORKSHEET-81
S
U
M
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
P-167P L A Y I N G W I T H N U M B E R S
A × 6 means 2 × 6 = 1 2 or 7 × 6 = 4 2 + 2 + 2
1 4 4 4 1 A × 6 = 7 × 6 is the accepted valueNow, 2 7 4 × 6
4 4 4 12
Thus, A = 7 and B = 4(ii) 10 – 1 = 9 B = 9
Also 9 – 1 = 8 – 1 = 7 12
A = 7
We have 7 1 + 1 9 9 0 1
Thus, A = 7 and B = 9
9. (a) Yes ! it is necessary that (a – b + c) should be divisible by 11. 1
(b) [(b + d) – (a + c)] is divisible by 11. 1
(c) Yes ! we can say that a number will be divisible by 11 if the difference between thesum of digits at its odd places and that of digits at the even places is divisible by 11. 1
P-168 M A T H E M A T I C S – VIII
16 PLAYING WITH NUMBERS WORKSHEET-82
[A] Lab ActivityObject :
To test the divisibility of a number by grouping method.
Let’s Start
Recall all the divisibility rules.
Materials Required :
Math box. 1
Procedure :
(1) Throw out some match sticks from the match box on the table (Fig. (i)). (say 30 sticks)
Scatteredpencils
Pencil box
(i)(2) Make groups of two sticks each out of scattered sticks (see Fig. (ii)) and record your
observation in the observation table.
1
(ii)(3) Now make groups of three sticks each and record your observation [see Fig. iii(a)].
(a)
(b)
½
(iii)Keep repeating the grouping for different numbers (4, 5, 6, 7 and so on) and record theobservations. ½
Observations :
2 3 4 5 6 7 8 9 10 11
1. 30
S. Total No. of Grouping complete with numbers or not Number is
No. Sticks thrown divisible by
× × × × × 2, 3, 5, 6, 10
2.
3.
4.
5.
Conclusion :
From the above activity it can be concluded that if the grouping is completed then thenumber is divisible by the number used for grouping and if grouping is not completedthen the number is not divisibly by number used for grouping 1
F
O
R
M
A
T
I
V
E
-
A
S
S
E
S
S
M
E
N
T
1
P-169P L A Y I N G W I T H N U M B E R S
Result :
A given number is divisible by a number only when it completely divides the number,leaving no remainder. 1
[B] Fill in the Blanks ½ × 4 = 2
1. 10 × 5 + 2
2. 75
3. 100 × 1 + 10 × 2 + 3
4. 345
[C] True/False 1 × 4 = 4
1. True
2. False
3. True
4. True
[D] Viva-Voce 1 × 3 = 3
1. A number is divisible by 3 when sum of its digits is divisible by 3.
2. 9
3. Yes
[E] Quiz
1. No 1
2. A = 5 1
3. (i) A = 7, B = 4
(ii) A = 7, B = 9
(iii) A = 4, B = 7 1
4. 2 1
5. 4 1