2

1.

When a = 4, b8, c = 10, d = 0.

Find the value of

5a+6b+2c+d+4a+7b+3c+5a+6b+2d =(5x4)+(6x8) + (2 × 10)+0+(4×4)+(7x8)+ (3x10)+(5x4)+(6x8)+(2×0)

1

20+48+20+0+16+56 +30+20 +48+0=258

2. Find the value of

=

6a+226 +29c+19d

(6 × 4)+(22 × 8) + (29 × 10) +(19×0)
= 24+176+290+0=490

3. Find the value of

=

47a+246 +16c+12d

(47 × 4)+(24×8) + (16 × 10)+(12×0)
= 188+192+160+0= 540

4. Find the value of

21a+27b+51c+31d

= (21x4)+(27 x 8)+(51 x 10)+(31 × 0)
= 84+216+510+0 =810

When a 8, b = 10, c = 12, d = 4, e = 0. 5. Find the value of

ab+5ac-2ad +7e+6de +56

= (8×10)+5(8×12) — 2(8×4)+(7×0)+6(4×0)

+(5×10)

=80+(5 × 96)-(2×32)+0+(6x0)+50
= 610-64546

6. Find the value of

4ac+15bd-7bce+6cd-4ab-7ae

=4(8×12)+15 (10 × 4)-7(10x 12 × 0)+6(12×4) -4(8x10)-7(8x0)

= 384+600-0+288-320-0=952

7. Find the value of

7abd+15b-8bd-7de+14ab-8bcd+4ac

=7(8×10 × 4)+(15 × 10) - 8 (10 × 4) -7(4x0)+ 14(8x10)-8(10x12x4)+4(8×12) =2240+150—320—0+1120-3840+384 =3894-4160=-266

=

8. Find the value of

14ad-27bcd-17ae +21bd

14(8×4)-27(10 × 12 × 4)-17(8×0)+21(10 × 4) = (14×32)-(27 × 480) — (17 × 0)+(21 × 40) = = 1288-1296011672

1, b = 2, c = 3, d = 5, e = 0.

Find the values of

(1) 2a+3b-2c+3d-4c+4a.
(2) 3a-2b+4d+4c-3d-+8.
(3) 6a-3b+4c-3d+4a-3b+7.
(4) 6c-3b+4a-2a+6b-3d.
(5) 7a+4d-3c+4a-3e.
(6) 7a-2b+4c-3d+4a.
(7) 8a-4d--4a-3b+bc-4d.
(8) 7e-6c+4b−2abc÷3cde.

Ans. 21.

Ans. 24.

Ans. 2.

Ans. 11.

Ans. 22.

Ans. 4.

Ans. -36
Ans.-22.

When a quantity is squared or multiplied by itself, this is expressed by a small figure called an index placed to the right of the quantity, and a little above it.

Thus 22 means 2 x 2; a2 means a Xa, and so on. When a number is cubed or multiplied by itself so that there are 3 terms, as 2 × 2 × 2, this is expressed by a small figure called an index placed as before, but the figure this time is 3. Thus 23 is 2 cubed, or the third power of 2; a3 is a xaxa, or the third power of a.

Similarly a1 means a raised to the fourth power, or axaxaxa, and so on.

When a 4, b = 6, c = 8, d= 10, e = 0 9. Find the value of

15a2-12bc-3b2c+5c3d—d2c

= 15(4×4)

12 (6×8) — 3 [(6×6) ×8]+

5 [(8x8x8) x 10]-[(10x 10) × 8]
=240-576-864+25600-800
- 25840-2240 = 23600

10. Find the value of

3a2-7ab2+6a2b2 — d2

=3(4×4)—7[4(6 × 6)]+6[(4×4) (6 × 6)]—(10 × 10) =(3x16)-7(4×36)+6(16 x 36)-100 =48-1008+3456—100 = 2396

11. Find the value of

14a2d-7a2b2+7c2d2-8e

= 14 [(4×4) 10] 7 [(4×4) (6×6)] +
7[(8×8) (10×10)] — 8×0
(14×160)-(7x576)+(7 × 6400)-0
=2240-4032+44800 43008

=

12. Find the value of

3abcde+4a e-3a-ed-7de

= 3(4x6x8x10x0) + 4 [(4×4) 0] —
3 [(4×4) 0× 101-7 (10×0)
=0+(4×0) - (3x0)-0=0

13. Find the value of

a2-b2-c2-d2-e2+ab-bc+de

= (4×4)-(6×6) — (8×8)-(10×10)-(0x0)+

==

(4x6)-(6x8)+(10x0)
=16-36-64-100-0+24-48+0
=40-248=-208

When a quantity is enclosed within brackets ( or [ ], it means that all the quantities are to be considered as one: thus, (8+12) means 20, (12-8) means 4, (8-12) means-4, (a+2a+c) means 3a +c, (3a-6a+c) means -3a+c, -(6a+3a-b) means (9a-b), which is -9a+b.

Note here that when a minus sign precedes a bracket, it changes all the signs of the quantities, as in the last example; so with the + sign before a bracket we can remove the bracket without affecting the equation, but if we remove the bracket having a negative sign before it, we must change all the signs.

It will be better for the pupil at this early stage to keep the brackets till the end of the solution of the problem.

When α = 16, b = 12, c=8, d= 10, e 4, f=0 1. Find the value of

a, e÷b, 2÷(a+b), d÷(b+c), f÷(a+3),

(bcd+adf)÷ce+(b2—c2+d2)÷d+abc÷(a+b+c)

= [(12×8×10) + (16 × 10x0)] ÷ (8×4) + [(12×12)-(8×8)+(10 × 10)] ÷ 10+(16 × 12×8) ÷ (16+12+8)

=

(960+0)÷32+(144—64+100)÷10+(1536÷36) =30+18+128=90

3. Find the value of

3

(3c-4d)÷d+(c−e÷d)÷(a+b)−(a+b+c)
÷ (a-b-c)

= [ (3 × 8) — (4 × 10)]÷10+(8-4+10)÷(16+12)

=

-(16+12+8)÷(16-12-8)

(24-40)

10+14÷ 28-36 ÷ --4

= −18 +1 3+ 36 = −7%

4. Find the value of

(a2+b2)÷(2cd+f)+(c2+e2)÷2ab+(3b2—c)÷4 = [(16×16) + (12 × 12)] ÷ [2 (8×10) + 0] + [(8x8)+(4×4)]÷2(16×12)+[3(12×12)-8]÷4 =(256+144)÷[(2 × 80)+0]+(64+16)÷(2 × 192) +[(3x144)-8] ÷ 4

= (400÷160)+(80÷384)+(432-8)÷4

=400+8+424 10817

5. Find the value of

(10+2d2+2e3)÷ (4a-4c-2b)-(6+4d-26) +(6d-4c) = [(10+200+128)÷8]-(22÷28) = (338÷8) — 22 = 2 3 2 2

1161 = 4113

6. Find the value of

(8ab2-c2-d2)÷(3e2-d2)+12cd-8de

=

{8 [16 (12×12)]

(8×8) (10×10)}÷

[3 (4×4)-(10 × 10)] + 12 (8 × 10)-8 (10x4) (18432-64-100 48-100)+960-320 18268960-320 =

-52

28813

When we wish to take the root of a quantity, this is expressed by the sign put before the quantity. Thus 16 means square root of 16 4, or that number which multiplied by itself will make 16. So √64 is 8. means cube root; thus 3/8 = 2,

and so on.

Of course if we square a quantity first, and then take the square root of it, the quantity remains as before thus 42 is 4; a is a, and so on.

:

When a = 50, b = 32, c = 16, d= 8, e= 1. Find the value of

2, ƒ = 0.

√b2 +√a2÷√d2+2 √d2÷4√/a2+6√✅a2÷3√/b2 =b+a÷d+2d ÷ 4a+6a ÷ 3b

=32+50÷8+ (2x8)÷(4×50)+(6×50)÷(3 × 32)

= 32

1250+16+625

32180141,9

200

2. Find the value of

5a35-3c4÷6+b÷c3÷÷d+d2 ÷ b2 ×c÷8

= (5×125000) ÷ 5--(3 × 65536)÷6+32÷4096 → 8+64 1024 × 16 ÷ 8

(a+b)÷(d2+5d) × d÷2b+4d+2c÷d+d÷b = (50+32)÷[(8x8)+(5x8)] x8÷ (2×32) +

(4×8)+(2x16)+8+8÷32
= 82-X +64 +331⁄2= 8,41

104