8th Maths : Chapter 3 : Algebra : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise
3.5**

**Miscellaneous
Practice Problems**

** **

**1. Subtract: ****−****2( xy)^{2} ( y ^{3} **

**Solution:**

5*y*^{2} (*x*^{2} *y*^{3}
− 2*x*^{4}*y* + 10*x*^{2}) − [(−2) (*xy*)^{2}
(*y*^{3} + 7*x*^{2}*y* + 5)]

= [5*y*^{2} (*x*^{2} *y*^{3})
− 5*y*^{2} (2*x*^{4} *y*) + 5*y*^{2}
(10*x*^{2})] − [(−2) *x*^{2}*y*^{2} (*y*^{3}
+ 7*x*^{2} *y* + 5)]

= (5*y*^{5}*x*^{2} − 10 *x*^{4}*y*^{3}
+ 50 *x*^{2}*y*^{2}) − [(−2*x*^{2}*y*^{2})
(*y*^{3}) + (−2*x*^{2}*y*^{2}) (7*x*^{2}*y*)
+ (−2*x*^{2}*y*^{2}) (5)]

= 5*x*^{2}*y*^{5} − 10*x*^{4}*y*^{3}
+ 50*x*^{2}*y*^{2} − [−2*x*^{2} *y*^{5}
− 14*x*^{4}*y*^{3} − 10*x*^{2}*y*^{2}]

= 5*x*^{2}*y*^{5} − 10*x*^{4}*y*^{3}
+ 50*x*^{2}*y*^{2} + 2*x*^{2} *y*^{5}
+ 14*x*^{4}*y*^{3} + 10*x*^{2}*y*^{2}

= (5 + 2) *x*^{2}*y*^{5} + (−10 + 14) *x*^{4}*y*^{3}
+ (+50 + 10)*x*^{2}*y*^{2}

= 7*x*^{2}*y*^{5} + 4*x*^{4}*y*^{3}
+ 60*x*^{2}*y*^{2}

** **

**2. Multiply ****(**** 4 x^{2} **

**Solution:**

(4*x*^{2} + 9) (3*x* − 2) = 4*x*^{2}
(3*x* − 2) + 9 (3*x* − 2)

= (4*x*^{2}) (3*x*) − (4*x*^{2})
(2) + 9 (3*x*) − 9(2) = (4 × 3 × *x* × *x*^{2}) − (4 × 2
× *x*^{2}) + (9 × 3 × *x*) − 18

= 12*x*^{3} − 8*x*^{2} + 27*x* − 18
(4*x*^{3} + 9) (3*x*− 2) = 12*x*^{3} − 8*x*^{2}
+ 27*x* − 18

** **

**3. Find the simple interest on Rs. 5 a ^{2}b^{2} for 4ab years
at 7b% per annum.**

**Solution:**

Simple interest = [ Principal
× years × rate of interest ] / 100

= [ 5*a*^{2}*b*^{2} × 4*ab* × 7*b*
] / 100 = [ (5 × 4 × 7) (*a*^{2} × *b*^{2} × *a *×
*b* × *b*) ] / 100

= (140/100) (*a*^{3} *b*^{4}) = (14/10)
*a*^{3} *b*^{4}

Simple Interest = (7/5) *a*^{3} *b*^{4}

** **

**4. The cost of a note book is Rs. 10 ab. If Babu has Rs. **

**Solution:**

For ₹10 *ab* the number of note books can buy = 1.

For ₹ (5*a*^{2}*b* + 20*ab*^{2} +
40 *ab*) the number of note books can buy = Total amount
/ cost of 1 note book

= [ 5*a*^{2}*b* + 20*ab*^{2} + 40*ab*
] / 10 *ab* = [ 5*a*^{2}*b* / 10*ab* ] + [ 20*ab*^{2}
/ 10*ab* ] + [ 40*ab* / 10*ab* ]

= 1/2 [ *a*^{2−1}
*b*^{1−1 }] + [ 2 *a*^{1−1} *b*^{2−1} ] +
[ 4 *a*^{1−1} *b*^{1−1} ] = (1/2)*a*^{1}
*b*^{0 }+ 2 *a*^{0} *b*^{1} + 4 *a*^{0}
*b*^{0}

Number of note books he can buy = (½)*a* + 2*b* + 4

** **

**5. Factorise : ****(**** 7 y^{2} **

**Solution:**

7*y*^{2} − 19*y* – 6 is of the form *ax*^{2}
+ *bx* + *c* where *a* = 7; *b* = −19; *c* = − 6

**Product = −42 : Sum = −19**

1 × −42 = −42 : 1 + (−42) = −41

2 × −21 = −42 : 2 + (−21) = −19

The product *a* × *c* = 7*x* − 6 = − 42

sum *b* = −19

The middle term − 19*y* can be written as − 21*y* + 2*y*

7*y*^{2} − 19*y* − 6 = 7*y*^{2} −
21*y* + 2*y* − 6

= 7*y* (*y* − 3) + 2 (*y* − 3) = (*y* − 3) (7*y*
+ 2)

7*y*^{2 − }19*y* − 6 = (*y* − 3) (7*y*
+ 2)

** **

**Challenging
problems**

** **

**6. A contractor uses the expression 4 x^{2} **

**Solution:**

Given Number of rooms = *x* + 2

Number of rooms × Number of outlets = amount of wire.

(*x* + 2) × Number of outlets = 4*x*^{2} + 11*x*
+ 6

Number of outlets = [ 4*x*^{2} + 11*x* + 6 ] /
[ *x *+ 2] ...(1)

Now factorising 4*x*^{2} + 11*x* + 6 which is
of the form* ax*^{2} + *bx* + *c* with *a* = 4 *b*
= 11 *c* = 6.

The product *a* × *c* = 4 × 6 = 24

sum *b* = 11

**Product = 24 : Sum = 11**

1 × 24 = 24 : 1 + 24 =
25

2 × 12 = 24 : 2 + 12 = 14

3 × 8 = 24 : 3 + 8 = 11

The middle term 11*x* can be written as 8*x* + 3*x*

∴ 4*x*^{2} + 11*x* + 6 = 4*x*^{2} +
8*x* + 3*x* + 6

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

4*x*^{2} + 11*x* + 6 = (*x* + 2) (4*x*
+ 3)

Now from (1) the number of outlets

[ 4*x*^{2} +
11*x* + 6 ] / [ *x* + 2] = [ (*x* + 2) (4*x* + 3)] / (*x*
+ 2) = 4*x* + 3

∴ Number of outlets =
4*x* + 3

** **

**7. A mason uses the expression x^{2} **

**Solution:**

Given length of the room = *x* + 4

Area of the room = *x*^{2} + 6*x* + 8 = *x*^{2}
+ 6*x* + 8

Length × breadth = *x*^{2} + 6*x* + 8

Breadth = [ *x*^{2} + 6*x* + 8 ] / [ *x*
+ 4 ] ……(1)

Factorizing *x*^{2} + 6*x* + 8, it is in the
form of *ax*^{2} + *bx* + *c*

Where *a* = 1 ; *b* = 6 ; *c* = 8;

The product *a* × *c* = 1 × 8 = 8

Sum = 6

**Product = 8 : Sum = 6**

1 × 8 = 8 : 1 + 8 = 9

2 × 4 = 8 : 2 + 4 = 6

The middle term 6*x* can be written as 2*x* + 4*x*

∴ *x*^{2} + 6*x* + 8 = *x*^{2} + 2*x* + 4*x*
+ 8

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

*x*^{2} + 6*x* + 8 = (*x* + 2) (*x* + 4)

Now from (1)

breadth = [ *x*^{2} + 6*x* + 8 ] / (*x* +
4) = [ (*x* + 2) (*x* + 4) ] / (*x* + 4) = (x + 2)

∴ Width of the room =
*x* + 2

** **

**8. Find the missing term: y^{2 }+ (…) x + 56 = (y+7)(y+..)**

**Solution:**

We have (*x* + *a*) (*x* + *b*) = *x*^{2}
+ (*a + b*)*x* + *ab*

56 = 7 × 8

∴ *y*^{2} + (7 + 8)*x* + 56 = (*y*
+ 7) (*y* + 8)

** **

**9. Factorise : 16 p^{4} **

**Solution:**

16*p*^{4 }−^{ }1** **= 2^{4} *p*^{4}
− 1 = (2^{2})^{2} (*p*^{2})^{2} − 1^{2}

= (2^{2}*p*^{2})^{2} − 1^{2}

Comparing with *a*^{2} − *b*^{2} = (*a
+ b*) (*a − b*) where *a* = 2^{2} *p*^{2} and
*b* = 1

∴ (2^{2}*p*^{2})^{2}
− 1^{2} = (2^{2}*p*^{2 }+ 1) (2^{2}*p*^{2}
− 1)

= (4*p*^{2} + 1) (4*p*^{2} − 1)

∴ 16*p*^{4} − l = (4*p*^{2} + 1) (4*p*^{2}
− 1) = (4*p*^{2} + 1) (2^{2}*p*^{2 − }1^{2})

= (4*p*^{2} + 1) [(2*p*)^{2} − 1^{2}]
= (4*p*^{2} + 1) (2*p* + l) (2*p* − 1)

[ ∵ using *a*^{2} − *b*^{2}
= (*a + b*) (*a − b*)]

∴ 16*p ^{4}* − 1 = (4

** **

**10. Factorise : 3 x^{3} – 45x^{2}y + 225xy^{2} – 375y^{3}**

= 3*x*^{3} − 45*x*^{2}*y* + 225*xy*^{2}
− 375*y*^{3 }

= 3(*x*^{3} − 15*x*^{2}*y* + 75*xy*^{2}
− 125*y*^{3})

= 3(*x*^{3} − 3*x*^{2}(5*y*) + 3*x*(5*y*)^{2}
− (5*y*)^{3})

= 3(*x* − 5*y*)^{3}

** **

**Exercise 3.5 **

**Miscellaneous Practice
Problems **

**1. 7 x^{2}y^{5} + 4 x^{4}y^{3} + 60x^{2} y^{2}
**

**2.12 x^{3} − 8x^{2} + 27x −18 **

**3. S.I = 7/5 a ^{3}b^{4} **

**4. 1/2 a + 2b
+ 4 **

**5. ( y − 3)(7y + 2)**

**Challenging Problems **

**6. 4 x + 3 **

**7. x + 2 **

**8. ( y + 7)( y + 8) **

**9. (4 p^{2} + 1)( 2p +1)(2p −1) **

**10. 3( x − 5y)^{3}**

^{}

Tags : Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

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8th Maths : Chapter 3 : Algebra : Exercise 3.5 | Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths

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