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

**Exercise
3.10**

**Miscellaneous
Practice Problems**

** **

**1. The sum of three numbers is 58. The
second number is three times of two-fifth of the first number and the third number
is 6 less than the first number. Find the three numbers.**

**Solution:**

Here what we know

*a + b + c* = 58 (sum of three numbers is 58)

Let the first number be ‘*x*’

*b* = *a* + 3 (the second number is three times of 2/5 of the first number)

*b *= 3 × (2/5) *x* = (6/5) *x*

Third number = *x* − 6

Sum of the numbers is given as 58.

∴ *x* + 6/5 *x* + (*x* − 6) =
58

Multiplying by 5 throughout, we get

5 × *x* + 6*x* + 5 × (*x* – 6) = 58 × 5

5*x* + 6*x* + 5*x* – 30 = 290

∴ 16*x* = 290 + 30

∴ 16*x* = 320

∴* x* = (320/16) *x* = 20

**Answer:**

1^{st} number = 20

2^{nd} number = 3 × 2/5 × 20 = 24

3^{rd} number = 24 – 6 = 14

** **

**2. In triangle ABC, the measure of ****∠****B is two-third of the measure of ****∠****A. The measure of ****∠****C is 20****°**** more
than the measure of ****∠****A. Find
the measures of the three angles.**

**Solution:**

Let angle ∠A be *a*°

Given that ∠B = 2/3 × ∠A = 2/3 *a*

& given ∠C = ∠A + 20 = *a* + 20

Since A, B & C are angles of a triangle, they add up to 180°
(Δ property)

∴ ∠A +∠B + ∠C = 180°

⇒ *a *+ 2/3 *a* + *a* + 20 =
180°

{[3*a* + 2*a* + 3*a*] / 3} + 20 = 180°

8*a* / 3 = 180 – 20
= 160

∴ *a* = [160 × 3] / 8 = 60°

∠B = 2/3 × ∠A = 2/3 × 60 = 40°

∠C = 80°

** **

**3. Two equal sides of an isosceles triangle
are 5y−2 and 4y+9 units. The third side is 2y+5 units. Find ῾y᾿ and the perimeter
of the triangle.**

**Solution:**

Given that 5*y* − 2 & 4*y* + 9 are the equal sides
of an isosceles triangle.

∴ The 2 sides are equal

=> 5*y* − 2 = 4*y* + 9

∴ 5*y* − 4*y* =
9 + 2 (by transposing)

∴ *y* = 11

∴ 1^{st} side = 5*y*
− 2 = 5 × 11 − 2 = 55 − 2 = 53

2^{nd} side = 53

3^{rd} side = 2*y* + 5 = 2 × 11 + 5 = 22 + 5 = 27

Perimeter is the sum of all 3 sides

∴ P = 53 + 53 + 27 = 133
units

** **

**4. In the given figure, angle XOZ and
angle ZOY form a linear pair. Find the value of x.**

**Solution:**

Since ∠XOZ & ∠ZOY form a linear pair,

by property, we have their sum to be 180°

∴ ∠XOZ + ∠ZOY = 180°

∴ 3*x* − 2 + 5*x* + 6 = 180°

8*x* + 4 = 180 = 8*x* = 180 − 4

∴ 8*x* =
176 ⇒ *x* = 176 / 8 ⇒ *x* = 22°

XOZ = 3*x* − 2 = 3 × 22 − 2 = 66 − 2 = 64°

YOZ = 5*x* + 6 = 5 × 22 + 6

= 110 + 6 = 116

** **

**5. Draw a graph for the following data:**

**Does the graph represent a linear relation?**

**Solution:**

Graph between side of square & area

When we plot the graph,
we observe that it is not a linear relation.

** **

**Challenging
Problems**

** **

**6. Three consecutive integers, when taken
in increasing order and multiplied by 2, 3 and 4 respectively, total up to 74. Find
the three numbers.**

**Solution:**

Let the 3 consecutive integers be ‘*x*’, ‘*x* + 1’
& ‘*x* + 2’

Given that when multiplied by 2, 3 & 4 respectively &
added up, we get 74

i.e [ 2 × *x* ] +
[ 3 × (*x* + 1) ] + [ 4 (*x* + 2) ] = 74

Simplifying the equation, we get

2*x* + 3*x* + 3 + 4*x* + 8 = 74

9*x *+ 11 = 74

9*x *= 63 ⇒ *x* = 63 / 9 = 7

First number = 7

Second numbers = *x* + 1 ⇒ 7 + 1 = 8

Third numbers = *x* + 2 ⇒ 7 + 2 = 9

∴ The numbers are 7, 8 & 9

** **

**7. 331 students went on a field trip.
Six buses were filled to capacity and 7 students had to travel in a van. How many
students were there in each bus?**

**Solution:**

Let the number of students in each bus be ‘*x*’

∴ number of students in 6 buses = 6 × *x* = 6*x*

Apart from 6 buses, 7 students went in van

A total number of students is 331

∴ 6*x* + 7 = 331

∴ 6*x* = 331 − 7 = 324

∴ *x* = 324 / 6 = 54

∴ There are 54 students in each bus.

** **

**8. A mobile vendor has 22 items, some
which are pencils and others are ball pens. On a particular day, he is able to sell
the pencils and ball pens. Pencils are sold for ****₹****15 each and ball pens are sold at ****₹****20 each.
If the total sale amount with the vendor is ****₹****380, how many pencils did he sell?**

**Solution:**

Let vendor have ‘*p*’ number of pencils & ‘*b*’
number of ball pens

Given that total number of items is 22

∴ *p* + *b* = 22 ……..(I)

Pencils are sold for ₹ 15
each & ball pens for ₹ 20 each

total sale amount = 15 × *p* + 20 × *b*

= 15*p* + 20*b* which is given to be 380.

∴ 15*p* + 20*b* =
380

Dividing by 5 throughout,

15*p*/5 + 20*b*/5 = 380 / 5 ⇒ 3*p* + 4*b* =
76 ………..(2)

Multiplying equation (1) by 3 we get

3 × *p* + 3 × *b* = 22 × 3

⇒ 3*p* + 3b = 66
………..(3)

Equation (2) − (3) gives

3*p* + 4*b*
= 76

(−) 3*p* + 3*b*
= 66

__0 + b = 10__

∴ *b* = 10

∴ *p* = 12

He sold 12 pencils

** **

**9. Draw the graph of the lines y = x,
y = 2x, y = 3x and y = 5x on the same graph sheet.
Is there anything special that you find in these graphs?**

**Solution:**

(i) *y* = *x*, (ii) *y* = 2*x*, (iii) *y*
= 3*x* (iv) *y* = 5*x*

**(i) y = x**

When *x *= 1, *y*
= 1

*x* = 2, *y* = 2

*x* = 3, *y* = 2

**(ii) y = 2x**

When *x* = 1, *y*
= 2

*x* = 2, *y* =
4

*x* = 3, *y* =
6

**(iii) y = 3x**

when *x* = 1, *y*
= 3

*x* = 2, *y* =
6

*x* = 3, *y* =
9

**(iv) y = 5x**

When *x* = 1, *y*
= 5

*x* = 2, *y* = 10

*x* = 3, *y* = 15

When we plot the above points & join the points to form
line, we notice that the lines become progressively steeper. In other words,
the slope keeps increasing.

** **

**10. Consider the number of angles of
a convex polygon and the number of sides of that polygon. Tabulate as follows:**

**Use this to draw a graph illustrating
the relationship between the number of angles and the number of sides of a polygon.**

**Solution:**

Shapes: Trianlge, Rectangle, Pentagon, Hexagon

Angles :

** **

**Answer:**

**Exercise 3.10 **

**Miscellaneous Practice
Problems **

**1. x = 20 **

**2. 60°, 40°, 80° **

**3. y = 11 units p=133 units **

**4. 116°,64°**

**Challenging Problems **

**6. 7,8,9 **

**7. 54 **

**8. 12 pencils**

** **

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

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