Class 7 Mathematics Exercise 3 Full Solution

(1) Solution:- 1sq.foot = (c) 929sqcm.

(2) Solution:- If the length of one edge of a cube is 3m then total surface area of the cube – (a) 5sqm.

Therefore, Area of one surface= (3×3) = 9sqm.

Therefore, Total area= (6×9) = 9sqm.

(3) Solution:- Let, breadth of garden = x m.

Therefore, length of garden= 3x m.

Therefore, 2 (3x+x) = 400

=> 4x = 200

=> x = 200/4

= 50

Therefore, Breadth = 50m.

Length= (3×50)m

= 150m (c)

(4) Solution:- Area = (150×50)sqm

(d) = 7500sqm.

(5) Solution:- meaning of deci in Latin – (b) One tenth.

(6) solution:- Perimeter of land = 2 (20+15)

= (b) 70m.

(7) solution:– Without work way, length of land

= {20 – (2+2)}m

= 16m.

Breadth of land

= {15- (2+2)}

= 11m.

Therefore, Area= (16×11) sqm

= 176sqm.

(8) (a) Solution:- 40390cm= 0.4039km.

(8) (b) Solution:- 75m 250mm

= 75m+ 250mm

= 75m+ 250/100m [Because, 1m= 1000mm]

= 75+0.5

= 75025m

= 75.25/1000

= 0.07525km.

(9) Solution:- 5.37 decimetre = (5.37x 10)m

= 53.4m

Therefore, 1m= 10decm.

Therefore, 53.7m = 537decm.

(10) (a) Solution:- Area of triangle

= ½ x .10x 6

= 30sqm.

(10) (b) Solution:- Area of triangle

= ½ x 25x14cm.

= 175sqcm.

(11) Solution:- Let, breadth of a rectangular plot x m. and length = 3x m.

Therefore, perimeter= 2 (3x+x)

= 2x4x

= 8x m.

Therefore, 8x= 1000

=> x= 1000/8

X = 125

Therefore, breadth= 125m.

Therefore, length= (3×125)m.

= 375m.

Therefore, length 375m and breadth 125m.

(12) Solution:– Perimeter of rectangular park

= 2 ( l+ b)

= 2 (100+50)m.

= (2×150)m.

= 300m.

Therefore, 1m cost to fencing is 100TK.

Therefore, 300m cost to fencing in (300×100) TK.

= 30000TK.

(13) Solution:- Given, base of parallelogram = 40m

Therefore, height of parallelogram = 50m.

Therefore, Area= (40×50) = 2000sqm.

(14) Solution:- Area of cube = (4x40sqm

= 16sqm.

Therefore, Total surface area of cube

= (16×6)

= 96sqm.

(15) Solution:- Produces potatoes

= 500kg 700gm

= (500×1000) + 700gm

= 5007000gm.

Joseph, 1 piece of land produces 500 700gm potatoes.

Therefore, 11 piece of land produces (11×500 700) Gm potatoes

= 5507700gm

= 5507000+7.00gm

= 5507000/1000+ 7000gm

= (5507kg+700gm)

(16) Solution:- In 16 area land produces 28 metric ton paddy

Therefore, 1 area land produces 28/16 metric ton paddy.

= 7/4 metric ton paddy.

= 1 – ¾ metric ton.

(17) Solution:- 1 month = 30days.

In 30 days produces rod = 200000m.t.

Therefore, 1 days produces rod = 200000/30 m.t

= 666 – 2/3m.t

= 666m.ton + 2/3x 1000kg

= 666m.ton +666- 2/3kg

= 666m.ton+666kg +2/3x 1000gm

= 666m.ton+ 666kg+666-2/3gm.

(18) Solution:- In 1 day sell 20kg 400gm.

= 20 400gm.

Therefore, in 30 day sell = (20400×30) gm

= 612 000gm

= 612kg.

(19) Solution:- in 1 piece of land produces 20 kg 850gm mustard = 20850gmmustard.

Therefore, in 7 piece of land produces = (20850×7) mustard.

= 145950gm.

= (145000+950)gm.

= 145kg 950gm.

(20) Solution:- Volume of mug = 1500cu.cm.

= 1500/1000 list

= 1.5 lit.

Therefore, 1.5 lit in 1 mug

Therefore, 1 lit in 1/1.5 mug

Therefore, 270 lit in 270/1.5 mug

= 180mug.

(22) Solution:- In 1 day requires milk 125 lit

Therefore, 30 day requires mil (30×1.255) lit

= 37.50lit.

Now, 1 lit of milk cost 52TK.

Therefore, 37.50 lit of milk cost (37.50×52) TK.

= 1650TK.

(24) Solution:- Let, the breadth of room X m.

Therefore, length = (3xx) 3xm.

Therefore, area = (3Xx) sqm.

= 3x²sqm.

Therefore, 7.50TK – 1 sqm.

1TK. — 1/7.50 sqm.

Therefore, 1102.50TK. – 1×1102.50/7.50 sqm

= 147sqm.

Therefore, Area of room 147sqm

Therefore, 3x² = 147

=> x² = 147/3

= 49

=> x = √49

= 7

Therefore, Breadth is 7m out length (7×3) = 21m.

Class 7 Mathematics Exercise 2.3 Full Solution

(1) (a) Solution:- Twice divided ratio of 4:9

= √4: √9

(a) = 2:3

(3) Solution:- 4:3= 4×5: 3×5= 20:15

5:6= 5×3: 6×3= 15:18

Therefore, 20:15:18.

Second quantity is = 15 (b).

(4) Solution:- Ratio = 5:3:2

Therefore, (5+3+2)

= 10

Therefore, Maisha get = (5/10×30)

= 15m.

(5) Solution:- Tania get = 30x 2/10

= 6m.

Maria get = 30x 3/10

= 9m.

Therefore, (9-6) = 3m.

Maria get 3m more than Tania.

(7) Solution:- Let, 4^th proportional is ‘x’

Therefore, 3:5 : : 15: x

=> 3/5 = 15/x

=> x = 5×15/3

(b) = 25

(8) Solution:- C P= 1.50 TK. S P= 2.00

Profit = (2-1.50)

= 0.50TK.

Percentage profit= (0.5/1.50x 100)%

= 33 1/3%

(9) Solution:- C P of banana= 25TK. 1 bunch= 4pc

SP of banana = 27TK.

2TK profit in 1 bunch

Therefore, 1TK profit in ½ bunch

Therefore, 50TK profit in 50×1/2 bunch

= 25 bunch

(10) (a) Solution:- If the cost price is more than the selling price — (d) Loss oriented.

(10) (b) Solution:- If the cost price is less than the selling price— (b) Profitable.

(10) (c) Solution:- Time along with current—(a) Less time.

(10) (d) Solution:- Time against the current—(c) More time.

(11) Solution:– 5 workers can reap crops land 8 bigha in 6 days.

Therefore, 1 workers can reap crops land 8 bigha in 6×5/8 days.

Therefore, 25 workers can reap crops land 20 bigha in 6x5x20/25×8 days.

= 3 days.

(12) Solution:- Swapan 1 days work = 1/24

Ratan 1 days work = 1/16

Together in 1 day total = (1/24+1/10)

= 2+3/48

= 5/48

Therefore, They work 5/48 in 1 day.

They work 1 in 48/5 day.

= 3 3/5 day.

(13) Solution:- Hibiba 8 Aalima together can do,

20 day—- 1 part

1 day = 1/20part

Therefore, 8 day = 8/20 part.

Rest part = (1- 8/20)

= 20-8/20

= 12/20 = 6/10= 3/5 part

Therefore, Halima do, 3/5 part in 21 day

Therefore, 1 part in 21×5/3 day

= 35 days.

(16) Solution:- In 1^st pipe, 12hr. fill 1 tank

Therefore, 1hr fill 1/12 part.

And in 2^nd pipe,

18hr fill 1 tank.

Therefore, 1hr fill 1/18 part

If two pipes are open then in 1hr fill

= (1/12+ 1/18)

= 3+2/36

= 5/36

Therefore, 2 pipe fill 5/36 part in 1 hr.

Therefore, 2 pipe fill 1 part in 1/5/36 hr.

= 7 1/5hr.

(17) Solution:- Along with current,

Boat cross, 4hr-36km.

Therefore, 1hr= 36/4 km.

= 9km.

Therefore, Speed of beat +speed of current = 9km/hr.

Where, speed of current = 3km/hr.

Therefore, (9-3)= 6km/hr.

Speed of boat in still water is 6km/hr.

(18) Solution:- A ship travel 11 hrs — 77km against the current

Therefore, a ship travel 1 hrs — 77/11 km against the current.

= 7 km.

In still water, speed of ship 9km/hr.

Therefore, against the stream,

Speed of ship = (9-7) km/hr.

= 2km/hr.

(20) Solution:- Farmers, cultivate,

With 5 pair cows in 8 days cultivate 40 hectors land

Therefore, with 1 pair cows in 8 days cultivate 40/5 hectors land.

Therefore, with 7 pair cows in 1 days cultivate 40×7/5×8 hectors land.

Therefore, with 7 pair cows in 12 days cultivate 40x7x12/8×5 hectors land

= 84 hectors land.

(21) Solution:- Lily can do, a work in 10 hr.

Therefore, 10hr— 1 part

Therefore, 1hr — 1/10 part.

And, mily can do that work in 8hr.

Therefore, 8hr — 1 part.

Therefore, 1 hr —- 1/8 part.

They work together totally

= (1/10+ 1/8)

= 4+5/40

= 9/40

Therefore, Together, 9/40 part do in 1hr.

Therefore, 1 part do in 40/9 hr.

= 4 4/9 hr.

(22) Solution:- By 2^nd pipe,

In 30 min fill 1 tank 1/30 part tank full.

Therefore, 18 min fill 1×18/30 part tank full.

= 3/5 Part.

Therefore, Rest part = (1- 3/5)= 2/5 part

And, by 1^st pipe, 1 part tank full in 20 min

Therefore, 2/5 part tank full in 20x 2/5 min

= 8min.

(23) Solution:- 1hr= 3600 sec.

Therefore, In 3600 sec, train cover 48/3600 km.

Therefore 30 sec train 48×30/3600 km

= 2/5km.

= 2/5x 1000

= 400m.

Therefore, length of the bridge

= (400-100)m.

= 300m.

(24) Solution:- If train cover total bridge, then total distance cover by train

= (120+330)m

= 450m, ; 30km= 30,000m.

1hr= 3600sec.

30,000m distance cover in 36000sec

Therefore, 1m distance cover in 3600/30,000 sec.

Therefore, 450m distance cover in 3600×450/30,000 sec

= 54sec.

(25) (b) Solution:- Ratio given= 3:6:10

Total= 19

Given weight = 190gm.

Therefore, weight of bronze = (3/19×190)

= 30gm.

Therefore, weight of zinc = (6/19×190)

= 60gm.

Therefore, weight of silver= (190×10/19) gm

= 100 gm.

(25) (c) Solution:- Weight bronze in ornament = 30gm

Therefore, weight of zinc in ornament = 60gm

Let, x gm zinc to be added,

Therefore, 30: (60+x) = 1:3

=> 30/60+x = 1/3

=> 60+x = 90

= x= 90-60

= 30gm.

Therefore, 30 gm to be added to get ratio 1:3.

(26) (a) Solution:- Let, C P = 100TK.

(a) 10% loss, S P= (100-10) = 90TK.

Therefore, if S P 90, than C P = 100TK.

Therefore, 1 TK then C P 100/90 TK.

Therefore, 625 TK. Then C P 100×625/90

= 69 -4 -4/9 TK

Therefore, Cost price of watch 69 – 4 -4/9 TK

Therefore, Loss (C P- S P)

= (694- 4/9 – 625)

= (6250/9- 625)

= 6250-625/9

= 625/9 TK.

(26) (b) Solution:- buying price of watch 694- 4/9.TK.

(26) (c) Solution:- 10% profit, S P of watch (100+10)

= 110TK.

Therefore,. If C P 100 then S P = 110

Therefore, C P 1 than S P = 110/6250/9×10

= 6875/9 TK.

= 763- 8/9.

Class 7 Mathematics Exercise 4.1 Full Solution

(1) Solution:- (3abx 4³a)

= 12a⁴ b

(2) Solution:- (5xyx 6az)

= 30axyz.

(3) Solution:- (5ax ²X 3ax² y)

= 15a³ x⁷ y

(4) Solution:- 8a² b X (-2b²)

= -16a² b³

(5) Solution:- (-2abx²) x (10bx³ y z)

= -20ab⁴ x³ y z

(6) Solution:- (-3p² q³)x (-6p⁵ q⁴ )

= 18 p⁷q⁷

(7) Solution:- (-12m² a² x ²) x (-2ma² x²)

= 24 m³ a⁴ x⁵

(8) Solution:- (7a³ bx⁵ y²) x (-3x⁵ y⁵ a² b² )

= – 21a⁵ b³ x¹⁰y⁵

(9) Solution:- (2x+3y) (5xy)

= (10x² y +15xy²)

(10) Solution:- (5x- 4xy) (9x y)

= 45x y – 36 x y

(11) Solution:- (2a² – 3b²+c²) (a³ b²)

= 2a⁵ b² – 3a³ b⁴+ a³ b² c²

(12) Solution:- (x³ – y² + 3xyz) (x⁴ y)

= x⁷ y- x⁴ y³ + 3x⁵ y z

(13) Solution:- (2a-3b) (3a+2b)

= 6a² + 4ab- 9ab- 6b²

= 6a² – 5ab- 6b²

(14) Solution:- (a+ b) (a-b)

= a² –b²

(15) solution:- (x²+1) (x²-1)

= (x²)² – (11)²

= (x⁴-1)

(16) Solution:- (a+ b) (a+ b)

= a³+ a² b+ ab² +b³

(17) Solution:- (a² –a b +b²) (a +b)

= a³+ a² b – a² b – a b²+ a b²+ b³

= a³+b³

(18) Solution:- (x+2xy+y) 9x+y)

= x³+yx²+2x² y+2x² y+ xy² +y³

= x³+ 3x² y+ 3xy²+ y³

(19) Solution:- (x²-2xy+y²) (x-y)

= x²-x² y – 2x² y+2x y²+xy²-y³

= x³– 3xy²+ 3xy²– y³

(20) Solution:- (x+2x-3) (x+3)

= x³ + 3x² + 2x² +6x- 3x-9

= x³ + 5x²+ 3x-9

(21) Solution:- (a²+ab+ b²) (b²– ab+ a²)

= a² b² –a³b+a⁴ +ab³– a² b² +a³b+ b⁴-ab³+a² b²

= a⁴+b⁴+a² b²

(22) Solution:- (a+ b +c) (a+ b +c)

= (a + b + c)²

= a²+b²+c²+2ab+2bc+2ca

(23) Solution:- (x²-x³y +x² y²+ x³y- x² y² + xy³+ x² y² – xy³+y⁴)

= x⁴+ x² y² +y⁴

(24) Solution:- (y²-y+1) (1+y+y²)

= (y²+ y³+ y⁴-y- y^2- y³+1+y+ y²)

= y⁴+ y²+1

(25) Solution:-

A= x²+xy+y²

B= x-y

Therefore, L.H.S, AB

= (x²+xy+y²) (x-y)

= x³ – y3

= R.H.S

L.H.S=R.H.SS (proved).

(26) Solution:- AB.

= (a²-ab+b²) (a+ b)

= a³+b³

(27) Solution:- L.H.S,

(a²+1) (a – 1) (a²+1)

= (a²-1) (a²+1)

= (a²)² – (1)²

= a⁴ – 1

= R.H.S

(28) Solution:- L.H.S,

(x +y) (x-y) (x +y)

= (x²-y²) (x²+y²)

= (x²)² – (y²)²

= x⁴-y⁴

= R.H.S

Class 7 Mathematics Exercise 4.2 Full Solution

(1) Solution:-

45a⁴÷ 9a²

= 45a⁴/9a²

= 5a²

(2) Solution:-

-24a⁵÷ 3a²

= – 24a⁵/3a²

= -8a³

(3) Solution:-

30a⁴ x³ ÷ 6a² x

= – 30a⁴ x³ / 6a² x

= -5 a² x²

(4) Solution:- – 28 x⁴ y³ z² ÷ 4x y z

= – 28 x⁴ y³ z² /4x y² z

= – 7 x³ y z

(5) Solution:-

-36a³ z³ y² ÷ (-4ayz)

= – 36a³ z³ y²/- 4ayz

= 9a²y z²

(6) Solution:-

– 22 x³ y²z÷ – 2xyz

= – 22 x³ y² z/ – 2xy z

= 11 x² y

(7) Solution:-

(3a³ b³– 2a² b³) ÷ a²b²

= 3a³b² – 2a² b² /a² b²

= 3 a³ b² /a² b² – 2a² b³ /a² b²

= 3a- 2b

(8) Solution:-

(36 x⁴ y³ + 9x⁵ y²) ÷ 9xy

= 36 x⁴ y³ /9 x y+ 9x⁵ y² /9xy

= 4 x³ y² + x⁴ y

(9) Solution:-

(a³ b⁴– 3a⁷ b⁷) ÷ – a³ b³

= a³ b⁴ / a³ b³ – 3a⁷ b⁷/ a³ b³

= b- 3a⁴ b⁴

(10) Solution:-

(6a b -9a b) V 3a b

= 6a b/3a b – 9a b/3a b

= 2a b- 3a b

(11) Solution:-

(15x³ y³ + 12x³ y²– 1 x⁵ y³ ) ÷ 3x² y²

= 15x³ y³/ 3x² y² + 12x³ y² / 3x² y² – 12x³ y²/ 3x² y²

= 5xy+ 4x – 4x³ y

(12) Solution:-

(6x⁸ y⁶ z- 4x⁴ y³ z² + 2x²y²z²) ÷ 2 x² y² z

= 6x⁸ y⁶ z/ 2 x² y² z – 4x⁴ y³ z²/ 2 x² y² z+2x²y²z² /2 x² y² z

= 3 x⁶ y⁴ – 2 x²z + z

(13) Solution:-

(24a² b² c- 15a⁴ b⁴ c⁴ – 9a² b⁶ c²) ÷ (-3ab²)

= 24a² b² c /-3ab²– 15a⁴ b⁴ c⁴ /-3ab²– 9a² b⁶ c²/ -3ab²

= – 8a c + 5a³ b² c⁴+ 3ab⁴ c²

(14) Solution:-

(a³ b³ + 2a² b³) ÷ (a+2b)

= a³ b³/ a+2b + 2a² b³/ a+2b

= a² b²

(15) Solution:-

(6x² +x-2) ÷ (2x -1)

= (6x² +x-2)/(2x-1)

= 3x+2

Class 7 Mathematics Exercise 4.3 Full Solution

(1) Solution:- 3a² b x (-4ab²)

= – 12 a³ b³ (d)

(2) Solution:- 20 a⁶b³/4a³ b

= 5 a³ b³(c)

(3) Solution:- – 25x³ y/ 5xy³

= -5x²/y²

(4) Solution:- (8a-2b) + (7a+4b)

= (8×3-2×2) + (-7×3+4×2)

= (24-4) + (-21+8)

= 20-13

= 7.

(5) Solution:- x³+2x²-1

= (-1)³ +{2x (-1)²}

= -1+2-1

= 0

(6) Solution:- 10x⁶ y⁵ z⁴/-5×2 y² z²

= -2x⁴ y³ z²

(7) Solution:- (a) (i) & (ii) are correct.

(8) Solution:- ( m^x)^y

= (m³)²

= m⁶ Answer (d)

(9) Solution:- a⁰= 1 (c)

(10) Solution:- x⁷/x-²

= x⁹ (a)

(11) Solution:- x- {x- (x-y)}

= x- (x-x +y)

= x-y (c)

(12) Solution:- (x +y) x [x-{x-9x-y)}]

= (x +y) x (x-y)

= x²– y² (d)

(13) Solution:- a⁵ x (-a³)⁰ x a-⁵

= -a^5+3-5

= -a³ (d)

(14) Solution:- [2-{(1+1)-2}]

= 2-{2-2}

= 2

(15) Solution:- 7+2[-8-{-3-(-2-3)}-4]

= 7+2[-8-{-3+5}-4]

= 7+2[-8-2-4]

= 7+2[-14]

= 7-28

= -21

(16) Solution:- -5-[-8-{-4-(-2-3)}+13

= -5-[-8-{-4+5}+13

= -5-[-8-1+13]

=-5-4

= -9

(17) Solution:- 7-2 [-6+3(-5+2(4-3)}]

= 7-2 [-6+3{-5+2×1}]

= 7-2[-6+3x-3]

= 7-2 [-6-9]

= 7-2 (-15)

= 7+30

= 37

(18) Solution:- x-{a+ (y-b)}

= x-(a +y-b)

= x-a-y +b

(19) Solution:- 3x+(4y-z) – {a-b- (2c-4a) -5a}

= 3x+4y-z[a-b-2c+4a-5a}

= 3x+4y-z-[-b-2c]

= 3x+4y-z+b+2c

(20) solution:- -a-[-3b-{-2a-(-a-4b)}]

= -a-[-3b-{-2a+a+4b}]

= -a-[-3b-{-2a+a+4b}]

= -a-{-3b+a-4b}

= -a-{-7b+a}

= 7b-2a

(22) Solution:- {2a-(3b-5c)}-[a-{2b-(c-4a)}-7c]

= {2a-3b+5c}- [a-[2b-c+4a}-7c]

= 2a-3b+5c- [a-2b+c-4a-7c]

= 2a-3b+5c-[-3a-2b-6c]

= 2a-3b+5c+3a+2b+6c

= 5a-b+11c

(23) Solution:- -a+[-6b-{-15c+(-3a-9b-13c)}]

= -a+[-6b-{-15c-3a-9b-13c}]

= -a+[-6b-{-28c-3a-9b}]

= -a+[6b+28c+3a+3b]

= -a+3b+28c+3a

= 2a+3b+28c

(24) Solution:- -2x-[-4y-{-6z-(8x-10y+12z)}]

= -2x-[-4y-{-6z-8x+10y-12z}]

= -2x-[-4y-{-18z-8z-8z-8x+10y}]

= -2x-[-4y+18z+8x-10y]

= -2x-[-14+18z+8x]

= -2x+14y+18z+8x

= -10x+14y-10z

(27) Solution:- 20 –[{( 6a+3b) – (5a-2b)}+6]

= 20-[{(6a+3b-5a+2b}+6]

= 20-[{a+5b}+6]

= 20-[a+5b+6]

= 20-a-5b-6

= 14-a-5b

(29) Solution:- 15a+2[3b+3(2a+2(2a+b)}]

= 15a+2[3b+3 (2a+4 a-2b)}]

= 15a + 2[3b+3 (2a-4 a-2b)}]

= 15a+2[3b-6a-6b]

= 15a+2[-3b-6a]

= 15a-6b-12a

= 3a-6b

(29) Solution:- [8b-3{2a-3(2b+5)-5(b-3)}]-3b

= [8b-3{2a-6b-15-5b+15}]-3b

= [8b-3{2a-11b}]-3b

= [8b-6a+33b]-3b

= 41b-6a3b

= 38b-6a

(30) Solution:- a-b +c-d

= a- (b-c+ d)

(31) Solution:- a-b-c +d-m +n-x +y

= a-(b +c-d)-m+ (n-x) +y

(32) Solution:- 7x-5y+8z-9

1^st, 7x-5y- (-8z+9)

2^nd, 7x+{-5y-(-8z+9)}

Therefore, 7x+{-5y-(-8z+9)}

(33) Solution:- (a) (15x²+7x-2)- (5x-1)

= 15x²+7x-2-5x+1

= 15x²+2x-1

(b) (15x²+7x-2) x (5x-1)

= 75x³+35x²-10x-15x²-7x+2

= 75x³+20x²-17x+2

(34) Solution:- (a) A+B

= (x²-xy+y²) + (x²+xy+y²)

= (x²-xy+y²+x²+xy+y²)

= (2x²+2y²)

= 2 (x²+y²)

(b) Solution:- A x B

= (x²-xy+y²) x (x²+xy+y²)

= x⁴-x³ y+x² y²+x³y-x² y²+xy³+ x² y²– xy³+y⁴

= (x⁴+ x² y²+y⁴)

Class 7 Mathematics Exercise 5.1 Full Solution

(1) Solution:- (a+5)²

= a²+2x z x 5+5²

= a²+10a+25

(2) Solution:- (5x-7)²

= (5x)² – 2x5xX7+ (7)²

= 25x²-70x+49

(3) Solution:- (3a-11xy)²

= (3a)2 – 2x3ax11xy+ (11xy)2

= 9a²– 66axy+ 121x² y²

(4) Solution:- (5a²+9m²)²

= (5a²)²+2x5a².9m²+ (9m²)²

= 25a⁴+ 90a² m²+ 81 m⁴

(5) Solution:- (55)²

= (50+5)²

= (50)2+ 2x50x5 +5²

= 2500+500+25

= 3025

(6) Solution:- (990)²

= (1000-10)²

= (1000)² – 2x1000x10+(10)²

= 1000000-20000+100

= 980100

(7) Solution:- (xy-6y)²

= (x y)²– 2.xy.6y+ (6y)²

= x² y²– 12xy² +30y²

(8) Solution:- (a x-by)²

= (a x)²– 2.ax.by + (by)²

= a² x²– 2axby + b² y²

(9) Solution:- (97)2

= (100-3)²

= (100)²– 2.100.3+ 3²

= 10000- 600+9

= 9409

(10) Solution:- (2x+y-z)²

= {(2x+y)-z}²

= (2x+y)² – 2. (2x+y).z+ z²

= (2x)2 + 2X2xXy+y²– 4zx-2yz+z²

= 4x²+4xy+y²-4zx-2yz+x²

(11) Solution:- (2a-b+3c)²

= {(2a-b)+ 3c}²

= (2a-b)²+2. (2a-b).3c+(3c)²

= (2a)²– 2.2a.b.+b²+12ac-6bc+9c²

= 4a²-4ab+b²+12ac-6bc+9c²

= 4a²+b²+9c²-4ab+12ac-6bc

(13) Solution:- (x²+y²-z²)²

= {(x²+y²)² – (z²)²}

= (x2+y²)²– 2. (x²+y²). Z² + (z²)²

= (x²)²+ 2x²y²+ (y²)²– 2x² z² – 2y² z²+ z⁴

= x⁴+y⁴+z⁴+2x² y²+ 2y² z²– 2z² x²

(14) Solution:- (3x-2y+z)²

= {(3x-2y) + z}²

= (3x-2y)²+ 2. (3x-2y) . z +z²}

= 9x²-12xy+4y²+6xz-4yz+z²

= 9x²+4y²+z²+12xy+6xz-4yz

(15) Solution:- (b c +ca +a b)²

= {(b c +c a) + a b}²

= (b c+ c a)² + 2. (b c +c a). a b+ (a b)²}

= (b c)2+2.bc.ca+ (ca)2 + 2ab2c +2a2bc+ a2 b2

=b²c²+ c²a²+a²b²+ 2ac²+2ab²c+2a²bc

(17) Solution:- (2a+1)² – 4a(2a+1)+ 4a²

= (2a1)²– 2. (2a+1). 2a + (2a)²

= {(2a+1) – 2a}²

= (2a+1-2a)²

= (1)² = 1

(18) Solution:- Let, 5a+3b= x ; 4a-3b= y

Given, x²+2xy+y²

= (x +y)²

= (5a+3b+4a-3b)²

= (9a)²

= 81a²

(19) Solution:- Let, 7a+b= x ; 7a-b= y

Therefore, given, x²-2xy+y²

= (x-y)²

= (7a+b-7a+b)²

= (2b)²

= 4b²

(20) Solution:- Let, (2x+3y)= a ; 2x-3y= b

Given, a²+ 2ab+b²

= (a +b)²

= (2x+3y+2x-3y)²

= (4x)²

= 16x²

(21) Solution:- (5x-2) = a ; 5x+7= b

Given, a2+b2- 2ab

= (a-b)2

= (5x-2-5x-7)²

= (-9)²

= 81

(22) Solution:- (3ab-cd)² + 9 (c d-a b) + 6 (3ab-cd) (c d-a b)

= (3ab-cd)² + 2. (3ab-cd) x 3 (c d-a b) + [3 (c d- a b)]²

Let, 3ab-cd= x ; 3 (c d-a b) = y

Therefore, x²+ 2xy +y²

= (x +y)²

= {(3ab-cd) +3 (c d-a b)}²

= (3ab-cd+3cd-3ab)²

= (2cd)²

= 4c²d²

(23) Solution:- (2x+5y+3z)2 + (5y+3z-x)2- 2 (5y+3z-x) (2x+5y+3z)

Let, (2x+5y+3z) = a ; 5y+3z-x = b

Therefore, a²+b²-2a b

= (a-b)²

= {2x+5y+3z-5y-3z+x}²

= (3x)²

= 9x²

(24) Solution:- (2a-3b+4c)2 + (2a+3b-4c)2 + 2 (2a-3b+4c) (2a+3b-4c)

Therefore, Let, 2a-3b+4c= x and, (2a+3b-4c) = y

Therefore, x²+ y²+2xy

= (x +y)²

= (2a-3b+4c+2a+3b-4c)²

= (4a)²

= 16a²

(25) Solution:- 25x²+36y²– 60xy ; when x= -4, y= -5

Therefore, 25x² + 36y²– 60xy

= (5x)² + 2.5x.6y+ (6y)²

= (5x-6y)²

= {5(-4) -6 (-5)}²

= (-20+30)²

= (10)² = 100

(26) Solution:- 16a²– 24ab- 9b²

= (4a)² – 2.4a.3b+ (3b)²

= (4a-3b)²

= (4×7-3×6)²

= (28-18)²

= (10)²

= 100

(27) Solution:- (9x²+30x+25)

= (3x)²+2.3x.5+5²

= (3x+5)²

= {3x (-2) + 5}²

= (-6+5)²

= (-1)²

=1

(28) Solution:- 81a³+18ac+c²

= (9a)²+ 2.9a.1 +c²

= (9a+c)²

= ({9×7-67}²

= (63-67)²

= (-4)²

= 16

(29) Solution:- a²+ b²= (a +b)²– 2ab

= (5)²– 2x 12

= 25-24

= 1

(31) Solution:- x+ 1/x = 5 (given)

=> (x+1/x)²= (5)²

=> x²+ 1/x² = 25-2

= x²+ 1/x² = 23

Both side square,

(x²+1/x²)² = (23)²

=> (x²-1/x²)² + 4.x². 1/x²= 529

=> (x²– 1/x²)

= 529-4

= 525

(32) Solution:- a +b= 8: a-b- 4

We know that,

4ab= (a +b)² – (a -b)²

= 8²– 4²

= 64-16

= 48

Therefore, a b= 12

Class 7 Mathematics Exercise 5.2 Full Solution

(1) Solution:- (4x+3) (4x-3)

= (4x)² – (3)²

= 16x² -9

(2) Solution:- (13-12p) (13+12p)

= (13)² – (12p)²

= 169 – 144p²

(3) Solution:- (ab+3) (ab-3)

= (a b)² – (3)²

= a² b² – 9

(4) Solution:- (10-xy) (10+xy)

= (10)² – (x y)²

= 100 – x²y²

(5) Solution:- (4x²+3y²) (4x²-3y²)

= (4x²)² – (3y²)²

= 16x⁴– 9y⁴

(6) Solution:- (a-b-c) (a+ b+ c)

= a- (b +c)}{a+ (b +c)}

= {a +(b +c)} {a- (b +c)}

= a2- (b +c)²

= a²– (b²+2bc+c²)

= a²-b² -2bc-c²

(8) Solution:- (x-1/2a) (x-5/2a)

= x2+ (-1/2a – 5/2a) x + (-1/2 a) (-5/2a)

= x2+ (-a-5a/2)x + 5a2/4

= x2+ (-6a/2)x + 4a2+4

= x2- 3ax + 5/4a2

(9) Solution:- (1/4x – 1/3y) (1/4x + 1/3y)

= (1/4x)2 – (1/3y)2

= x2/16 – y2/9

(10) Solution:- (a⁴+3a² x²+ 9x⁴) (9x⁴– 3a² x² + a⁴)

= {(a⁴+9x⁴) + 3a² x²} {a⁴+ 9x⁴) – 3a² x²}

= (a⁴+9x⁴)² – (3a²n²)²

= (a⁴)²+ 2x2a⁴x9x⁴+ (9x⁴)² – 9a⁴x⁴

= a⁸+ 18a⁴ x⁴ + 81x⁸– 9 a⁴ x 4⁴

(11) Solution:- (x+1) (x-1) (x²+1)

= (x²-1) (x²+1)

= (x2)² – (1)²

= x⁴-1

(12) Solution:- (9a²+b²) (3a+b) (3a-b)

= (9a²+b²) (9a² – b²)

= (9a²)² – (b²) ²

= 81a⁴ – b⁴

Class 7 Mathematics Exercise 5.3 Full Solution

(1) Solution:- x²+x y +z x +y z

= x( x +y) + z (x +y)

= (x +y) (x +z)

(2) Solution:- a²+bc+ ca+ a b

= a²+ a b +ca +b c

= a (a +b) + c(a +b)

= (a +b) (a + c)

(3) Solution:- a b(p x+ q y)+a²q x +b²x + b² p y

= a b p x + a b q y + a² q x+ b² p y

= a b p x+ a² q x+ b² p y + a b q y

= a x (b p +a q) +by (b p +a q)

=  (b p +a q) (a x +by)

(4) Solution:- 4x²-y²

= (2x)² – (y)²

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

(5) Solution:- 9a² – 4b²

= (3b)² – (2b)²

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

(6) Solution:- a² b² – 49y²

= (a b)² – (7y)²

= (ab+7y) (ab-7y)

(7) Solution:- 16x⁴ – 81y⁴

= (4x²)² – (9y²)²

= (4x²+9y²) (4x²-9y²)

= (4x²+9y²) {(2x)² – (3y)²}

= (4x²+9y²) (2x+3y) (2x- 3y)

(8) Solution:- a² – (x +y)²

= (a+ x+ y) (a-x-y)

(9) Solution:- (2x-3y+5z)² – (x-2y+3z)²

= (2x-3y+5z+x-2y+3z) (2x-3y+5z-x+2y-3z)

= (3x-5y+8z) (x-y+2z)

(10) Solution:- (4a+ 8a²+9a⁴)

= (2)²+2x 2x 3a²+ (3a²)² – 4a²

= (2+3a²)² – (2a)²

= (2+ 3a²+2a) (2+ 3a²-2a)

(11) Solution:- 2a²+6a-80

= 2 (a²+3a-40)

= 2 (a²+8a-5a-40)

= 2{a (a+8) – 5 (a+8)}

= 2 (a+8) (a-5)

(12) Solution:- y²-6y- 91

= y² – 13y+7y- 91

= y(y-13) + 7(y-13)

= (y-13) (y+7)

(13) Solution:- p²– 15p+ 56

= p²– 8p-7p+56

= p (p-8) -7(p-8)

= (p-8) (p-7)

(14) Solution:- 45a⁸ – 5a⁴x⁴

= 5a⁴ (9a⁴-x⁴)

= 5a⁴ {(3a²)² – (x²)²}

= 5a⁴ (3a²+x²) (3a²-x²)

(15) Solution:- a²+3a-40

= a²+8a-5a-40

= a (a+8) -5 (a+8)

= (a+8) (a-5)

(16) Solution:– (x²+1)² – (y²+1)²

= {(x²+1) + (y²+1) }{(x²+1) – (y²+1)}

= (x²+1+y²+1) (x²+1-y²-1)

= (x²+y²+2) (x +y) (x-y)

= (x²+y²+2) (x +y) (x-y)

(17) Solution:- x²+11x+30

= x²+5x+6x+30

= x (x+5) + 6 (x+5)

= (x+5) (x+6)

(18) Solution:- a²-b²+2bc-c²

= a²-(b²-2bc+c²)

= (a)² – (b-c)²

= (a+ b-c) (a-b +c)

(19) Solution:- 144x⁷– 25x³ a⁴

= x³ (144x⁴– 25a⁴)

= x³{(12x²)² – (5a²)²}

= x³ (12x²+5a²) (12x²-5a²)

(20) Solution:- 4x²+12xy+9y²-16a²

= (2x)² + 2X 2 x X 3y+ (3y)² – 16a²

= (2x+3y)² – (4a)²

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

Class 7 Mathematics Exercise 7.2 Full Solution

(1) Solution:- Let, the number is = x

Therefore, 2x +5 = 25

=> 2x= 25-5

=> x= 20/2

= 10 Therefore, The number is 10.

(2) Solution:- Let, the number ix ‘x’

Therefore, x- 27= -21

=> x-27= -21

=> x= -21+27

= 6

Therefore, Number is 6.

(3) Solution:- Let, the number is ‘x’

Therefore, x/3 = 4

=> x= 12

Therefore, The number is 12.

(4) Solution:- Let, the number is ‘x’

Therefore, (x-5) x5 = 20

=> 5x- 25 – 20

=> 5x-25 = 20

=> 5x = 20+25

=> x= 45/5

= 9

Therefore, Number is 9.

(5) Solution:- Let, the number is = x

Therefore, x/2 – x/3 = 6

=> 3x -2x/6 = 6

=> x= 36

Therefore, the number is 36.

(6) Solution:- Let, the numbers are, x, x+1, x+2

Therefore, x+x+1x+2 = 63

=> 3x = 63-3

=> x= 60/3

= 20

Therefore, 1^st number is = 20

2^nd, number is = 20+1= 21

3^rd number is = 20+2= 22

(7) Solution:- Let, the smaller no. is = x

Bigger No. is = (55-x)

Therefore, 5(55-x) = 6x

=> 275-5x= 6x

=> 11x= 275

=> x= 275/11

= 25

Therefore, smaller number is = 25

Bigger number is = (55-25)

= 30.

(8) Solution:- Let, Rita has x Rs.

Gita has (x-6) TK.

Mita has (x+12) TK.

Therefore, x +x-6+x +12 = 180

=> 3x +6= 180

=> 3x= 180-6

=> x= 174/3

= 58 TK.

Therefore, Rita has 58TK.

Gita has (58-6) = 48 TK.

Mita has (58+12) = 70TK.

(9) Solution:- Let, the price of khata= x TK.

Therefore, price of pen = (75-x) TK.

Therefore, x-5= 2(75-x+2)

=> x-5 = 2(77-x)

=> x-5= 154-2x

=> x-5 = 154-2x

=> x+2x= 154+5

=> x= 159/3

= 53

Therefore, price of khata 53TK.

Therefore, price of pen (75-53)TK.

= 22TK.

(10) Solution:- Let, the total No. of fret = x

Therefore, Apple = x X ½ = x/2

Orange = x X1/3 = x/3

Orange= 40piece

Therefore, x/2 + x/3 +40= x

=> 3x+2x+240/6 = x

=> 5x+240 = 6x

=> x= 240

Therefore, Total No. 240.

(11) Solution:- Let, the age of son = x yr.

Father’s present age = 6x yr.

Therefore, 5yrs later,

Age of son = (x+5) yr.

11 of father = (6x+5) yr.

Therefore, (x+5) + (6x+5) = 45

=> 7x+10 = 45

=> 7x = 45-10

=> x= 35/7

= 5

Therefore, son’s present age 5yr.

Father present age = (5×6) = 30yr.

(12) Solution:- Liza: sikha age = 2:3

Therefore, Let, age = 2x & sikha’s age = 3x yr.

Therefore, 2x+3x= 30

=> 5x= 30

=> x= 30/5

= 6yr.

Therefore, Liza’s age = 2×6 = 12yr.

Sikha’s age= 3×6 = 18 yr.

(13) Solution:- Let, suman’s run = x

And, Iman’s man = 2x-5

Therefore, x+2x-5 = 58

=> 3x= 58+5

=> x= 63/3

=> 21

Iman’s run = (21×2-5)

= 37.

(14) Solution:- Let, the distance = ‘x’ km.

10min = 10/60

= 1/6

Therefore, to cover x km in 30km/hr take time x/30hr,

And, to cover 25km taking time x/25hr.

Therefore, x/25 – x/30 = 1/6

=> 6x-5x/150= 1/6

=> x = 150/6

= 25km.

(15) Solution:- Let, the breadth= ‘x’ m

Length= 3x m

Therefore, 2(x+3x) = 40

=> 8x= 40

=> x= 40/8

= 5m

Therefore, Breadth is = 5m

Length is = (3×5) = 15m.

Class 7 Mathematics Exercise 7.1 Full Solution

(1) Solution:- 4x+1 = 2x+7

=> 4x-2x = 7-1

=> x= 6/2

= 3

(2) Solution:- 5x-3 = 2x+3

=> 5x-2x = 3+3

=> x = 6/3

= 2

(3) Solution:- 3y+1 = 7y-1

=> 3y-7y= -1-1

=> – 4y = -2

=> y= 2/4

= ½

(4) Solution:- 7y-5 = y-1

=> 7y-y = -1 +5

=> y= 4/6

= 2/3

(5) Solution:- 17-2z = 3z+2

=> -2z-3z = 2-17

=> – 5z = -15

=> z = 15/5

= 3

(6) Solution:- 13z-5 = 3-2z

=> 13z +2z = 3+5

=> z =8/15

(7) Solution:- x/4 = 1/3

=> x= 4/3

(8) Solution:- x/2+1= 3

=> x/2 = 3-1

=> x= 4

(9) Solution:- x/3+5 = x/2+7

=> x/3 – x/2 = 7-5

=> 2x-3x/6 = 2

=> -x= 12

=> x= -12

(10) Solution:- y/2 – y/3 = y/5 – 1/6

=> y/2 – y/3- y/5 = -1/6

=> 15y-10y-6y/30 = -1/6

=> -y/30 = -1/6

=> y = 30/6

=> y = 5

(11) Solution:- y/5- 2/7 = 5y/7 – 4/5

=> y/5 – 5y/7 = -4/5+27

=> 7y-25y/35 = 10-28/35

=> -18y/35 = -18/35

=> y= 18/18

=1

(12) Solution:- 2x-1/3 = 5

=> 2x-1 = 15

=> 2x= 15+1

=> x= 16/2

= 8

(13) Solution:- 5x/7 + 4/5 = x/5+ 2/7

=> 5x/7 – x/5 = 2/7-4/5

=> 25x-7x/35 = 10-28/35

=> 18x/35= -18/35

Therefore, x= -18/18

= -1

(14) Solution:- y-2/4 +2y-1/3 = y- 1/3

=> y-2/4 + 1/3 = y – 2y-1/3

=> 3(y-2) +4= 12y-4 (2y-1) [Both side x by 12]

=> 3y-6 + 4= 12y – 8y +4

=> 3y-12y+8y= 4+6-4

=> -y= 6

=> y= -6

(15) Solution:- 3y+1/5 = 3y-7/3

=> 15y-35 = 9y +3

=> 15y-9y= 3+35

=> 6y= 38

=> y= 38/6

= 19/3

(16) Solution:- x +1/2 – x-2/3 – x-3/5 = 2

=> 15x +15 – 10x +20 – 6x +18/30 = 2

=> -x +53/30 = 2

=> -x +53 = 60

=> -x = 60- 53

=> x= -7

(17) Solution:- 2(x+3) = 10

=>2x +6=10

=> x= 4/2

= 2

(18) Solution:- 5(x-2) = (x-4)

=> 5x-10 = 3x-12

=> 5x-3x = -12+10

=> 2x= -2

=> x = -2/2

= -1

(19) Solution:- 7(3-2y) + 5(y-1) = 34

=> 21-14y+5y-5 = 34

=> -9y = 34-21+5

=> -y= 18/9

=> y= -2

(20) Solution:- (z-1) (z+2) = (z+4) (z-2)

=> z²+2- z-2= z2-2z-2z+4z-8

=> z²+z-2= z²+2z-8

=> z²+z-z²= -8+2

=> -z= 6

Therefore, z= 6

Class 7 Mathematics Exercise 6.2 Full Solution

(1) Solution:- 2/3a, 3/5ab

=>2/3a= 2x5b/3ax5b = 10b/15ab

= 3/5ab = 3×3/5abx3 = 9/15ab

(2) Solution:- x/y z, y/z x

Therefore, x/y z= x Xx/y z x x = x²/x y z

y/z x = y x y/z x X y = y²/x y z

(3) Solution:- 1/a +b + 1/a-b

= a-b +a +b/(a-b) (a-b)

(c) = 2a/a²-b²

(4) Solution:- x/2+1= 3

=> x/2 = 3-1

=> x= 4

(5) Solution:- a/b = a x c/b x c = ac/b c

(6) Solution:- 4a² b- 9b³/4a² b +6ab²

= b (2a=3b) (2a-3b)/2ab (2a+3b)

= b(2a-3b)/2ab

(7) Solution:- a/x+ b/x – c/x

= a +b-c/x

(8) Solution:- (a) (x+2) (x-2)

(9) Solution:- x²+4x+4/x²-4

= x²+2. X.2+4/(x+2) (x-2)

= (x+2)²/(x+2) (x-2)

= (x+2)²/(x+2) (x-2)

= X+2/x-2

(10) Solution:- 3a/5 + 2b/5 = 3a+2b/5

(11) Solution:- 1/5x+ 2/5x

= 1+2/5x

= 3/5x

(12) Solution:- x/2z + y/3b

= 3bx+2ay/6ab

(13) Solution:- 2a/x+1 + 2a/x-2

= 2a(x-2) + 2a(x+1)/(x+1) (x-2)

= 4ax-4a+2a/(x+) (x-2)

= 4ax-2a/(x+1) (x-2)

= 2a(x-1)/(x+1) (x-2)

(14) Solution:- a/a+2 + 2/a-2

= a (a-2) + 2(a+2)/(a+2) (a-2)

= a²-2a+2a+4/a²-4

= a²+4/a² – 4

(15) Solution:- 3/x²-4x-5 + 4/x+1

= 3/x²-5x+x-5 + 4/x+1

= 3/x(x-5) (x+1) + 4/x+1

= 3+ 49x-5)/(x-1) (x-5)

= 3+ 4x-20/(x+1) (x-5)

= 4x-17/(x+1) (x-5)

(16) Solution:- 2a/7 – 4b/7

= 2a-4b/7

(17) Solution:- 2x/5a – 4y/5a

= 2x-4y/ 5a

(18) Solution:- a/8x- b/4y

= ay-2bx/8xy

(19) Solution:- 3/x+3- 2/x+2

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

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

= x/(x+3) (x+2)

(20) Solution:- p + q/p q – q + r/q r

= p r + q r-p q-p r/p q r

= q(r-p)/p q r

= (r-p)/p r

(22) Solution:- 5/a²+6a+5 = 1/(a-1)

= 5/(a-1) (a-b) + 1/(a-1)

= 5+11 (a-5)/(a-1) (a-5)

= 5+ a-5/(a-1) (a-5)

= a/(a-1) (a-5)

(23) Solution:- 1/x+2 – 1/x²-4

= 1/x+2 – 1/(x+2) (x-2)

= x-2 – 1/(x+2) (x-2)

= x-3/x²-4

(24) Solution:- a/3 + a/6 -3a/8

= 8a+4a-9a/24

= 3a/24

= a/8

(25) Solution:- a/b – 3a/2b+ 2a/3b

= 6a-9a+4a/6b

= a/6b

(26) Solution:- x/y z – y/z x +z/x y

= x²-y²+z²/x y z

(27) Solution:- x-y/x y + y-z/y z + z-x/z x

= z x-y z+ x y-x z +y z-x y/ x y z

= 2/ x y z

= 0

(28) Solution:- (a) x²– 3xy – 4y²

= x²– 4xy + x y – 4y²

= x(x-4y) +y(x-4y)

= (x +y) (x-4y)

(b) Solution:- x/x +y , x/x-4y

= L.C.M of denominator= (x+ y) (x-4y)

=> x/x +y = x X(x-4y)/ (x+ y) (x-4y) = x(x-4y)/ (x-y) (x-4y)

=> x/x-4y = x X(x +y) /(x +y) (x-4y) = x(x +y)/(x +y) (x-4y)

(c) Solution:- x/x +y + x/x-4y + y/x²-3xy-4y²

= x/ x +y + x/x-4y + y/(x +y) (x-4y)

= x²– 4xy + x²+ x y +y/ (x +y) (x- 4y)

= 2x² – 3x y +y/ (x +y) (x-4y)

(29) Solution:- A= 1/x²+3x , B= 2/x²+5x+6

C= 3/x²– x-12

(a) Solution:- x²=5x+6

= x²+3x+2x+6

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

= (x+3) (x+2)

(b) Solution:- x² + 3x = X(x+3)

X²+ 5x+6 = (x=3) (x+2)

X²– x-12 = x²-4x+3x-12

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

= (x+3) (x-4)

Therefore, L.C.M of denominator= x(x+2) (x+3) (x-4)

Let, A= 1/x²+3x

= 1x (x+2) (x-2)/x((x+3) x (x+2) (c-4)

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

B= 2/x²+5x+6

=2/(x+2) (x+3)

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

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

C= 3/x²-x-12

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

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

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

(c) Solution:-A+B-c

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

= x² -4x+2x-8 +2×2-8+2x²-8x-3x²-6x/x(x+2) (x+3) (x-4)

= 3x²-3x²-18x+2x-8/x(x+2) (x+3) (x-4)

= -8-16x/x(x+2) (x+3) (x-4)

Class 7 Mathematics Exercise 6.1 Full solution

(1) Solution:- a² b/a³ c

= b/ac

(2) Solution:- a² b c/ab² c

= a/b

(3) Solution:- x³y³z³/x²y²z²

= x y z

(4) Solution:- x²+x/x y +y

= x(x+1)/y(x+1)

= X/y

(5) Solution:- 4a² b/6a³ b

= 4/6x a² b/a³ b

= 2/3×1/a

= 2/3a

(6) Solution:- 2a-4ab/1-4b²

= 2a (1-2b)/(1)²– (2b)²

= 2a (1-2b)/(1+2b) (1-2b)

= 2a/(1+2b)

(7) Solution:- 2a+3b/4a²-9b²

= 2a+3b/(2a)² – (3b)²

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

= 1/2a-3b

(8) Solution:- a²+4a+4

= a²+2a+2a+4/(a)² – (2)²

= a(a+2)+2 (a+2)/(a+2) (a-2)

= (a+2) (a+2)/(a+2) (a-2)

= (a+2)/(a-2)

(9) Solution:- x2-y2/(x +y)2

= (x +y) (x-y)/ (x +y) (x +y)

= X-y/x +y

(10) Solution:- x²+2x-15/x² +9x+20

= x²+5x-3×15/x²+5x+4x+20

= X(x+5) -3 (x+5)/x (x-5) +4 (x+5)

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

(11) Solution:- a/b c, d/a c

Denominator b c 8 ac, L.C.M= a b c

Therefore, a/ac = a x a/b c x a

= a²/a b c

And a/ac = a x b/a c x b

= a b/a b c

(12) Solution:- x/p q, y/p r

L.C.M of denominator = p q r

Therefore, x/p q = x X r/p q x r = x r/p q r

y/p r= y X x/p r x q = y q/p q r

(13) Solution:- 2x/3m, 3y/2n

L.C.M of denominator= 6mn

Therefore, 2x/3m = 2xX2n/3mx2n= 4xn/6n

= 3y/2n = 3yx3m/2mx3m= 9my/6mn

(14) Solution:- a/a-b, b/a +b

Therefore, L.C.M of denominator (a +b) (a-b)

Therefore, a/a-b= a x(a +b)/(a-b) (a +b) = a(a +b)/a² – b²

=> b /a +b = b x(a-b)/(a +b) (a-b) = b(a-b)/a² – b²

(15) Solution:- x²/a²-2ab, y²/a+2b

Therefore, L.C.M of denominator= (a+2b) (a²-2ab)

= (a+2b) (a-2b)a

Therefore, x²/a²-2ab= x²X(a²+2ab)/(a²-2ab) (a²+2ab)= x² (a²+2ab)/a (a²-4b²)

=> y²/a+2b= y²xa (a-2b)/(a+2b) (a-2b)a = ay² (a-2b)/a (a²-4b²)

(16) Solution:- 3/a²-4, 2/a (a+2)

L.C.M, of denomination, = (a²-4) a (a+2)

= a (a+2) (a-2)

3/a²-4 = 3xa/(a+2) (a-2)a

= 3a/a(a2-4)

2/a(a+2) = 2(a-2)/a(a+2) (a-2)

= 2 (a-2)/a(a²-4)

(17) Solution:- a/a²-9, b/a+3

Therefore, a²-9, (a)² – (3)² = (a+3) (a-3)

L.C.M of denominator= a (a+3) (a-3)

a/a²-9 = a. 1/(a+3) (a-3) = a/a² – 9

= b/(a+3) = b(a-3)/(a+3) (a-3) = b(a-3)/a² -9

(19) Solution:- a/a-b, b/a +b, c/a(a +b)

L.C.M, of denominator = a (a +b) (a-b)

a/ a-b = a x a(a +b) /(a-b) a(a +b)= a²(a +b)/a(a²-b²)

b /a +b= b x a (a-b)/(a +b) x a (a-b) = a b(a-b)/a (a²-b²)

c/a (a +b) = c x(a-b)/a (a +b) (a-b) = c(a-b)/a (a²-b²)

(20) Solution:- 2/x²– x- 2, 3/x²+x-6

= x²-x-2 = (x+1) (x-2)

X²+x-6= (x-3) (x-2)

Therefore, (x+1) (x-2) (x+3)

Therefore, 2/x²-x-2= 2x(x+3)/(x+1) (x-2) x (x+3)

3/x²+x-6 = 3x(x+1)/(x+3) (x-2) (x-1)

Class 7 Mathematics Exercise 5.4 Full Solution

(1) Solution:- (a-5)² = a²– 2xax5 + (5)²

(b) = a²-10a+25

(2) Solution:- (x +y)²+ 2(x +y) (x-y) + (x-y)²

= (x +y + x-y)²

= (2x)²

= 4x²

(3) Solution:- We know, 4ab = (a +b)² – (a-b)²

=> 4ab = 4²-2²

= 16-4

= 12

=> a b = 12/4 = 3 (a)

(4) Solution:- If a quantity is divisible without remainder another quantity is called – (c) Multiple

(5) Solution:- Least common multiple of a, a², a (a +b) is a² (a +b) (d) Answer.

(6) Solution:- H.C.F of 2a and 3b is 1 (a)

(7) Solution:- If a, b are real numbers – (d) (i), (ii), (iii) are correct.

(8) Solution:- (d) x y (x +y) (x-y)

(9) Solution:- H.C.F if two algebraic expression, (d) x (x-y)

(10) Solution:- The L.C.M is — (d) x y (x +y) (x+2y)

(11) Solution:- L.C.M of 9x²– 25y² and 15ax-25xy is (d) 5a (9x²-25y²)

(12) Solution:- H.C.F of x³ y⁵ and a²-b² is (d) 1.

(13) Solution:- x-1/x = 0; (d) (i), (ii), (iii) Correct.

(14) Solution:- a+1/a= 4 ; a²-4a+1= (d) 0.

(15) Solution:- (a+5)² = a²+2.a.5+25

= a²+10a+25

(16) Solution:- We know, 4ab= (a +b)² – (a-b)²

=> 4ab = (8)² – (4)²

= 64-16

= 48

=> a b= 48/4

= 12

(17) Solution:- 1^st, 3a² b² c= 3² x a² x a x a xb²xb²xc²

2^nd, 6ab2 c2= 2×3²x a²xb²xb²xc²xc

Therefore, H.C.F= 3a² b²c

(18)Solution:- 1^st, 5ab² x²= 5²xa²xb²xbxXxX

2^nd, 10a² by² = 2×5²xa²xaxaxb²xyxy

H.C.F = 5ab

(19) Solution:- 1^st, 3a² x² = 3²xa²x ax X x

2^nd, 6axy² = 2×3²x a²x X x y² x y

3^rd, 9 ay²= 3 x3²xa² xy² x y

H.C.F = 3a

(20) Solution:- 1^st, 16a³ x⁴ y = 2x2x2x2xa³xX⁴xy

2^nd, 40a² y³ x= 2²x 2x 2x5xa²xy³x X

3^rd, 28a x³= 2²x2 x7 xa²x X³

H.C.F= 2ax

(21) Solution:- 1^st, a²+ab = a (a +b)

2^nd, a²-b²= (a +b) (a-b)

Therefore, a2- b2= (a +b) (a-b)

Therefore, H.C.F = (a +b)

(22) Solution:- 1^st, x³y – xy³

= x y (x²-y)

= x y (x +y) (x-y)

2^nd, (x-y)² = (x-y) (x-y)

Therefore, H.C.F = (x-y)

(23) Solution:- 1^st, x² +7x+12

= x²+4x+3x+12

= x (x+4) +3(x+4)

= (x+4) (x+3)

2nd, x²+9x+20

= x² + 5x+ 4x+20

= x(x+5) +4 (x+5)

= (x+5) (x+4)

(24) Solution:- 1^st, a³-ab²= a (a²-b²)

= a (a +b) (a-b)

2^nd, a⁴+2a³b + a² b²

= a² (a²+2ab+b²)

= a² (a +b)²

= a. a (a +b) (a +b)

Therefore, H.C.F = a (a +b)

(25) Solution:- 1^st, a²-16= (a)² – (4)²

= (a+4) (a-4)

2^nd, 3a +12 = 3 (a+4)

3^rd, a²+ 5a +4

= a²+ 4a + a+ 4

= a (a +4) +1(a +4)

= (a +4) (a +1)

(27) Solution:- 1^st, 6a³ b² c

= 2x 3x a³x b²xc

2^nd, 9a⁴ bd²= 3 x3 xa⁴x b x d²

Therefore, L.C.M = 2x 3x 3xa⁴xb²xc xd²

= 18a⁴ b² cd²

(28) Solution:- 1^st, 5x² y²= 5x x²xy²

2^nd, 10xz³= 2x 5x X xz³

3^rd, 15y³ z⁴= 3x 5 x y³x z4

Therefore, L.C.M= 2x 3 x5x X²x y³xz⁴

= 30x² y³ z⁴

(29) Solution:- 1^st, 2p² xy2= 2x p²x X x y²

2^nd, 3p q² = 3x p x q²

3^rd, 6pqx² = 2 x 3 x p x q x X²

Therefore, L.C.M= 6p² q² x² y²

(30) Solution:- 1^st, (b²-c²)= (b +c) (b-c)

2^nd, (b +c)² = (b +c)

Therefore, L.C.M= (b +c)² (b-c)

(31) Solution:- 1^st, x²=2x

= x( x+2)

2^nd, x²+3x+2

= x(x+2) +1(x+2)

= (x+2) (x-1)

Therefore, L.C.M= x (x+1) (x+2)

(32) Solution:- 1^st, 9x²-25y²

= (3x)² – (5y)²

= (3x +5y) (3x -5y)

2^nd, 15ax-25ay = 5a (3x-5y)

(33) Solution:- 1^st, x²-3x-10

= x²-5x=2x-10

= (x-5) (x+2)

2^nd, x²-10x+25

= x²– 5x-5x+25

= (x-5) -5 (x-5)

= (x-5) (x-5)

= (x-5)²

Therefore, L.C.M= (x-5)² (x+2)

(34) Solution:- 1^st, a²-7a+12

= a²-3a-4a+12

= a (a-3) -4 (a-3)

= (a-3) (a-4)

2^nd, a²+a-20

= a²+5a-4a-20

= a(a+5) -4(a+5)

= (a-4) (a+5)

3^rd, a²+ 2a-15

= a²+5a-3a-15

= (a-5) -3(a+5)

= (a-3) (a+5)

Therefore, L.C.M, (a-3) (a-4) (a+5)

(35) Solution:- 1^st, x²-8x+15

= x²-5x-3x+15

= x (x-5) -3(x-5)

= (x-5) (x-3)

2^nd, x²-25= (x+5) (x-5)

3^rd, x²+2x-25

= x²-3x+5x-15

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

= (x-3) (x+5)

Therefore, L.C.M = (x-3) (x2-25)

(36) Solution:- 1^st, (x+5)

2^nd, x²+5x

= x (x+5)

3^rd, x²+7x+10

= (x²+5x+2x+10)

= (x+5) (x+2)

Therefore, L.C.M= x (x+5) (x+2)

(37) (a) Solution:- (a +b)

= 2x-3+2x+5

= (4x+2)

(b) Solution:- a²= (2x-3)²

= (2x)²– 2. 2x.3+ (3)2

= 4x²-12x +9

(c) Solution:- a b

= (2x-3) (2x=50

= 4x²+10x-6x-15

= 4x²+4x-15

If, x= 2;

4x (2)²+ 4×2-15

= 16+8-15

= 24-15

= 9

(38) (a) Solution:- x²+3x-10

= x²+5x-2x-10

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

= (x+5) (x-2)

(b) Solution:- 1^st, x²-625

= (x)²– (25)²

= (x+25) (x-25)

= (x+25){(x)² – (5)²}

= (x+25) (x+5) (x-5)

2^nd, x²+3x-10

= (x=5) (x-2)

L.C.M = (x+5) (x-2) (x-5) (x+25)

(c) Solution:- We get,

From 1^st expression, (x+5) (x-2)

2^nd, expression (x²+25) (x+5) (x-5)

L.C.M= (x+5) (x-2) (x-5) (x² +25)

(39) Solution:- (a) (3x-2y+z)²

= {(3x-2y) +z)²

= (3x-2y)²+2(3x-2y) z+ z²

= (3x)² – 2.3x.2y+ (2y)²+ 6xz- 4yz +z²

= 9x²+ 4y²+z²– 12xy +6x²– 4yz

(b) Solution:- 1^st, x²-3x-10

= (x+2)(x-3)

2rd, X³+6x²+8x

= x(x+6x+8)

= x(x²+4x+2x+8)

= x{x (x+4) +2 (x+4)}

= x(x+4) (x+2)

Therefore, H.C.F= (x+2)

(c) Solution:- 3^rd, expression, x⁴-5x³-14x²

= x²(x²-5x-14)

= x²(x²-7x+2x-14)

= x²{x(x-7) +2(x-7)}

= x² (x-7) (x+2)