Detailed Maths Project Solution: Algebra and Equation Analysis

This document presents a detailed solution to a math project, covering various algebraic equations and problem-solving techniques. The solution includes step-by-step explanations for problems involving significant figures, percentage calculations, ratios, linear equations, and quadratic equations. It demonstrates methods for solving for unknown variables, simplifying expressions, and applying algebraic principles to real-world scenarios, offering a comprehensive guide for students seeking to understand and master these mathematical concepts. Desklib provides this and other solved assignments to aid students in their studies.

Maths Project

Question 1
(F)
The three significant figures,
= 1.496 *1011 /3 *108
= 498.66 seconds
= 998.66 /60
= 8.311mins.
= 8.311/60
= 0.13851 hours.
No, danesh was not corrected about the answer. 1011 * 109 = 1020. While danesh multiplied the
power, he should added those values.
Question 2
(A)
3/5*240
= 144
=5/100*240
=36
240-144-36-12
= 48
(B)
200* 103.3 *101.5 = 209.7
First year = 3.3 percent
Second year = 1.5 percent
Question 3
(A)
while this have been given that,
2 pens each with black
5 pens each with red
6 pens each with green pens
1

and total number of 122 pens were sold.
so, the total number of green pens which have been sold =96
(B)
=9:4:5:2
=20:45:36:90
(C)
Ben has 20 packets,
total weight is 32 kg
32*20= 640
Lago has 25 packets,
total weight = 640/25
= 25.6
(D)
here, given that
4 bakers* 72 cakes =6*x
X=48
Question 4
(A)
p2=x-y /xy
y=8.5*109
y=4*108
= 10^8(85-4)/34*10^17
=1.54*10^-5
(B)
a) = 2*a +2*t = 5*t +7
=2*a = 3*t +7
=t = 2a -7/3
b) x = √7y +9/ y -1
(C)
i. (3x+2) (2x-1)
= 2𝑥(3𝑥+2)−1(3𝑥+2)
2

= 2𝑥(3𝑥+2)−1(3𝑥+2)
= 6𝑥2+4𝑥−3𝑥−2
= 6𝑥2+𝑥−2
ii. x2+3x - 4 /2x2 - 5x +3
= (x-1) (x+4)/ (2x+1)(x+3)
4 (D)
I. while in given linear equation in X & Y has infinite number of solution
= put x=2
= 3x-2y =7
= 3*2 -2y =7
=6 -2y =7
= 2y =6-7
= 2y -1
= y = -1/2
so the value can be, X = 2, Y= -1/2
II. h+7 /3 +2h -1/2 =5/6
= ℎ+7/3+2ℎ−1/2=5/6
= ℎ+11/6+2ℎ=5/6
= h=-1/3
4(e)
a) x2 +2x -35=0
= x² + (7 - 5)x - 35 = 0
=> x² + 7x - 5x + 35 = 0
=> x (x + 7) - 5 (x - 7) = 0
=> (x - 5) (x + 7) = 0
Therefore, Either (x - 5) = 0 or, (x + 7) = 0
=> x = 5 or, x => -7
b) 6x2-x-15=0
=2x-3=0
=3x+5=0
3

=2x=3
= 3x=-5
=x=3/2
= x=-5/3
c) 5x2+6x-2=0
5x2-6x-2/5= 0/5
= 5x2/5 -6x /5 -2/5=0
we know that,
(a-b)2=a2-2ab+b2
= -2ab = -6x /5
= -2b =-6/5
= b= 3/5
now, in given equation
x2-6x/5-2/5=0
= x2-6x/5+(3/5)2-2/5-(3/5)2=0
= (x-3/5)2= (3/5)2+2/5
= (x-3/5)2 =9/25+2/5
=(x-3/5)2 =19/25
= - 3 + √19/5 , -3 - √ 19/5
4