Algebra 1 Assignment: Comprehensive Solutions for Sections 1-4

This document presents solutions to an Algebra 1 assignment, covering multiple-choice questions and problem-solving exercises across four sections. The solutions include calculations for volume, speed, and cost analysis, as well as applications of the Pythagorean theorem. The assignment addresses various algebraic concepts, including formulas, equations, and practical applications. The document provides step-by-step solutions, making it a valuable resource for students studying Algebra 1. It includes calculations for distance, time, and speed, as well as problems involving percentages and geometric shapes. The solutions provide a comprehensive overview of the topics covered in the assignment.

Algebra
1

Table of Contents
SECTION 1......................................................................................................................................3
SECTION 2......................................................................................................................................3
SECTION 3......................................................................................................................................4
SECTION 4......................................................................................................................................7
2

SECTION 1
Multiple choice
1) The number of cans required to make this display which has three rows of cans is 6.
1 D
2 C
3 D
4 A
5 C
6 C
7 D
8 A
SECTION 2
9)
a) T = s+d
b) d= T-s
d= 34.848 – 29.837
= 5.011
10)
a) C= $3.60 + d*$2.19
where d is distance travelled
b) x= 15 km
C= $3.60+ 15*$2.19
C= $36.45
3

11)
a) 1. l= 2*π* r
b) l= 2*3.14*3
l= 18.84
SECTION 3
12)
a) s = d/t
b) Given d= 748km , t= 11.5 hours
s = 748/11.5
s= 65.043 km/h
13)
a) Volume of rabbit hutch = B * H
where, H= 2.5 m
For B, calculate area of triangle,
area of triangle = ½ * b * h
= ½ * .5m * .5m
= .125 m
thus, the volume = B* H
= .125* 2.5
= .3125 m
b) Volume of orange, v= 4/3* π*r^3
given r= 4 cm
v= 4/3*3.14*4^3
v= 267.946 cm^3
c) volume of rectangular prism = ½ * L * W *
4

H
= ½ * 15 *
60 * 55
= 24750 m
d) Volume of cylinder,v= π*r^2*h
given r=36 mm , h=72mm
v= 3.14*36^2
v= 292999.68 mm^3
14)
a) volume of cuboid =
length^2*width^2*height^2
b) volume of cylinder, v= π*r^2*h
15)
a) Given BAC = (10N-7.5H)/6.8M
N=4, H=2, M=80
BAC= (10*4-7.5*2)/80
= 0.3125
b) Given BAC=0.18, N=12, H=4
(a). 0.18= (10*12-7.5*4)/6.8M
(b). M= (10*12-7.5*4)/6.8*0.18
M= 73.529 kg
16)
a) Given t= √(2d/9.8)
1. d/4.9
5

b) d= 44 cm
or d= 0.44 m
t= √(2d/9.8)
t= √(2*0.44)/9.8
t= 0.299
c) t^2=d/4.9
d=(t^2)*4.9
d) here t=0.299/2
t=0.1495
d=(0.1495^2)*4.9
d=0.0045m
17)
a) Given SP=1.1*CP
CP= SP/1.1
b) Given SP= $65.45
CP= $65.45/1.1
CP=$59.5
c) GST of $ 65.45, Where GST = 10 %
= $ 65.45 + 10 %
= $ 71.995
18)
a) According to Pythagoras theorem
a^2+b^2= c^2
(8.1^2)+b^2 = 10.15^2
b^2= 6.1
Hence given width is 6.3 so it is not a
rectangle.
6

b) (14.4^2)+b^2=15.31^2
b^2= 5.2
Hence given width is also 5.2 so it is a
rectangle.
SECTION 4
19)
a)
b) profit = $ 28616.19
c) Profit= Income- expenses
20)
4.35+4.20+6.10+5.50 /4
=5.0375
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