7th Maths : Term 1 Unit 3 : Algebra : Exercise 3.4 : Miscellaneous Practice problems, Challenge Problems, Text Book Back Exercises Questions with Answers, Solution

**Exercise
3.4**

** **

__Miscellaneous
Practice problems__

** **

**1. Subtract −3***ab*** −8 from 3***ab*** + 8. Also, subtract
3***ab*** + 8 from −3***ab*** −8.**

**Subtract –3 ab –8 from 3 ab + 8**

3*ab *+ 8 – (–3 ab – 8)

3*ab* + 8 + 3 *ab* + 8

3*ab* + 3 *ab* + 8 + 8

(3 + 3)* ab *+16

6* ab *+ 16

**Subtract 3 ab + 8 from –3ab –8**

–3*ab* – 8 – (3* ab* + 8)

– 3*ab* –8 – 3* ab* – 8

– 3*ab* – 3* ab *– 8 – 8

(–3 –3)* ab *–16

– 6* ab *–16

** **

**2. Find the perimeter of a triangle whose sides
are ***x*** + 3***y***, 2***x*** + ***y***, ***x*** − ***y***.**

Sides of the triangle *a*
= *x* + 3*y*

b = 2*x* + *y*

c = *x – y*

Perimeter of a triangle = a + b + c

= *x *+ 3*y* + 2*x *+ y + x – y

= *x* + 2*x* + *x* + 3*y*
+ *y* – *y*

= (l +2 + l)*x* + (3 + l– l) *y*

= 4*x* + 3*y*

** **

**3. Thrice a number when increased by 5 gives 44.
Find the number.**

Thrice a number when increased by 5 gives 44.

Let the number be *x*

3*x* + 5 = 44

3*x* + 5 – 5 = 44 – 5

3*x* = 39

3/3 *x =* 39/3

* x* = 13

The number *x* =13

** **

**4. How much smaller is 2***ab*** + 4***b*** − ***c*** than 5***ab*** − 3***b*** + 2***c*

5*ab* – 3*b* + 2*c
*– (2*ab* + 4*b *– *c*)

5*ab* – 3*b* + 2*c
*– 2*ab* – 4*b *+ *c*

5*ab* – 2*ab *– 3*b* – 4*b* + 2*c *+ *c*

(5–2) *ab* + (– 3 – 4)*b* + (2+l) *c *

3*ab* – 7*b *+ 3*c*

** **

**5.
Six times a number subtracted from 40 gives −8. Find the number.**

Six times a number subtracted from 40 gives –8

Let the number be *x*

40 – 6*x* = –8

40 – 40 – 6x = –8 – 40

– 6*x *= –48

–6*x* / –6 = –48 / –6

* x *= 8

**The number x = 8**

** **

__Challenge
Problems__

** **

**6. From the sum of 5***x*** + 7***y*** − 12 and 3***x*** − 5***y*** + 2, subtract
the sum of 2***x*** − 7***y*** − 1 and −6***x*** + 3***y*** + 9.**

The sum 5*x* + 7*y* –12 + 3*x *– 5*y* + 2

5*x* + 3*x* + 7*y*
–5*y* –12 + 2

(5 + 3)*x* + (7 – 5) *y* –10

8*x* + 2*y *– 10

The sum 2*x *– 7*y* –1 – 6*x* + 3*y* + 9

2*x *– 6*x* –7*y*
+ 3*y *–1 + 9

(2 – 6)*x* + (–7 +3)*y *+8

– 4*x *– 4*y *+ 8

(8*x* + 2*y *–10) – (–4*x* – 4*y* + 8)

8*x* + 2*y* – 10 + 4*x* + 4*y *– 8

8*x* + 4*x* + 2*y*
+ 4*y* – 10 – 8

(8 + 4)*x* + (2 + 4)*y* – 18

**12 x + 6y
–18**

** **

**7. Find the expression to be added with 5***a*** − 3***b*** + 2***c*** to get ***a*** − 4***b*** − 2***c***?**

*a* – 4*b* – 2c – (5*a* –3*b*
+ 2*c*)

*a* – 4*b* –2*c* –5*a*
+3*b* –2*c*

*a* – 5*a* – 4*b* + 3*b
*– 2*c* – 2*c*

(l – 5)*a* + (–4 +3)*b* + (–2 –2)*c*

– 4*a* – *b* – 4*c*

– 4*a* – *b* – 4*c*
to be added with 5*a* – 3*b* + 2*c*
to get *a* – 4*b* – 2*c*

** **

**8. What should be subtracted from 2***m*** + 8***n*** + 10 to get**** ****−3***m*** + 7***n*** + 16?**

2*m* + 8*n* + 10 – (–3*m* + 7*n* + 16)

2*m* + 8*n* + 10 + 3*m* – 7*n* – 16

2*m* + 3*m* + 8*n*
– 7*n* + 10 –16

(2+3) *m* + (8 – 7) *n* – 6

5*m* + *n* – 6

5*m* + *n* – 6 should be subtracted from

2*m* + 8*n* + 10 to get –3*m* + 7*n* + 16

** **

**9. Give an algebraic equation for the following
statement:**

*“ The difference between the area
and perimeter of a rectangle is 20”*

Area of a rectangle = *lb*

Perimeter of a rectangle = 2(*l+b*)

The different between the area and Perimeter of a rectangle is
20.

* lb* – 2 (*l + b*) = 20

** **

**10.
Add: 2***a*** + ***b*** + 3***c*** and ***a*** + 1/3 ***b*** + 2/5 ***c*

2*a* + b + 3*c* +*
(a* + 1/3 *b* + 2/5 *c)*

2*a* + *a* + b + 1/3 *b* + 3*c* + 2/5 *c*

(2+1) *a* + (1 + 1/3) *b* + (3 + 2/5) *c*

3*a* + 4/3 *b* + 17/5 *c*

** **

Puzzles are instrumental for the origin of Algebra. **Leelavathi** is the first puzzle book in India
written by Baskaracharya (Baskara II) who lived in Maharashtra during 12th century.
The modified version of the interesting puzzle is for you.

“The beads from a pearl necklace fell
down. One third of the pearls fell on the ground. One fifth of the pearls rolled
under cot. Two persons started collecting the pearls. One person was able to collect
one sixth and the other person collected one tenth of the pearls. If only 6 pearls
are left in the necklace, find the total number of pearls in the necklace?”

Let the total number of pearls be
*x*.

Using the given details, we can create an equation.

*x *= 6 × 5 = 30 pearls

** **

**ANSWERS**

**EXERCISE-3.4 **

1. 6 *ab* + 16; − 6*ab* − 16

2. 4* x *+ 3y

3.* x *= 13

4. 3*ab* − 7b + 3c

5.* x *= 8

** Challenge Problems **

6. 12*x* + 6* y *− 18

7. −4*a* − *b* − 4*c*

8. 5*m* + *n* − 6

9. *lb* − 2 ( *l* + *b*) = 20

10. 3a + 4/3 b + 17/5 c

Tags : Questions with Answers, Solution | Algebra | Term 1 Chapter 3 | 7th Maths , 7th Maths : Term 1 Unit 3 : Algebra

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7th Maths : Term 1 Unit 3 : Algebra : Exercise 3.4 | Questions with Answers, Solution | Algebra | Term 1 Chapter 3 | 7th Maths

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