Comprehensive Math Assignment: Supermarket Analysis & Applications

This mathematics assignment presents detailed solutions to a variety of problems. It includes statistical analysis of supermarket shopping times, including calculating mean, median, quartiles, and outliers, along with box plot representations. The assignment also covers mixture problems, determining the correct ratios for saline solutions, and geometric problems involving the dimensions of squares. Furthermore, it addresses linear programming concepts related to animal feed mixtures and break-even point calculations for a manufacturing business. Finally, the assignment solves a problem involving exponential decay and carbon dating to determine the age of burned trees. Desklib offers many such solved assignments and past papers for students to enhance their learning.

MATHEMATICS QUESTIONS
TABLE OF CONTENTS
QUESTION 1..................................................................................................................................2
QUESTION 2..................................................................................................................................6
QUESTION 3..................................................................................................................................7
QUESTION 4..................................................................................................................................7
QUESTION 5..................................................................................................................................7
QUESTION 6................................................................................................................................10

QUESTION 1
Supermarket 1
Shopping
time (in
minutes)
10
12
15
25
30
1
94
21
8
Supermarket 2
Shopping
time (in
minutes) Reason to go shopping
22 Weekly groceries
12 Buy lunch
16 Buy lunch
18 Buy lunch
19 Buy lunch
25 Weekly groceries
18 Buy lunch
39
Prepare for a flat warming
party
15 Buy lunch
37
Prepare for a flat warming
party
33 Weekly groceries
15 Buy lunch
36
Prepare for a flat warming
party
18 Buy lunch

38
Prepare for a flat warming
party
39 Weekly groceries
16 Buy lunch
39
Prepare for a flat warming
party
14 Buy lunch
11
Prepare for a flat warming
party
1.
Combined mean: Supermarket A + B
S. No.
Supermarket
A + B
(Shopping
time: in
minutes)
1 10
2 12
3 15
4 25
5 30
6 1
7 94
8 21
9 8
10 22
11 12
12 16
13 18
14 19
15 25
16 18
17 39
18 15
19 37
20 33
21 15

22 36
23 18
24 38
25 39
26 16
27 39
28 14
29 11
Mean = ∑x / n
= 696 / 29
= 24 minutes
2.
Shopping
time (in
minutes)
1
8
10
12
15
21
25
30
94
Arranging data of first supermarket in ascending manner:
Median = (N + 1) / 2
= (9 + 1)/ 2
= 5
According to the 5th item median shopping time is 15 minutes.
Computation of quartile is enumerated below:
Particulars Figures
Quartile 1 10
Quartile 2 15

Quartile 3 25
3.
Lower outlier = Q1 – (1.5 * IQR)
= 10 – (1.5 * 15)
= 10 – 22.5
= -12.5
Higher outlier = Q3 + (1.5 * IQR)
= 25 + (1.5 * 15)
= 25 + 22.5
= 47.5
4.
Box plot pertaining to supermarket 1 and 2 is enumerated below:
Labels
Sample 1
(Supermarket 1)
Sample 2
(Supermarket 2)
Min 1 11
Q1 10 15.25
Median 15 18.5
Q3 25 36.85
Max 94 39
IQR 15 20.5
Upper Outliers 1 0
Lower Outliers 0 0

QUESTION 2
Solution
Solution 1: Saline 1% and water 99%
Solution 2: Saline 3% and water 97%
Solution 3: Saline 2.5% and water 97.5%
The ratio in which first and second mixture must be mixed to obtain 2.5% saline solution is:
=(2.5-1) : (3-2.5)
=0.5:1.5
= 1:3
Total quantity = 30 gram
Dividing 30 gm in ratio 1:3 we get:
(30/4)*1 and (30/4)*3 which is equals to 7.5 gm and 22.5 gm.
Thus for first type of saline solution 7.5 gm and for second type of saline solution 22.5 gm
solution must be taken.

QUESTION 3
Solution
Let the side of both squares is A and B respectively.
Square A is 9m2 smaller than B it means its side is 3 m less than that of B. So
A = B-3
Change in side A: (B-3+5) = B+2
(B+2)2 = 2B2
B2 + 2B + 4 = 2B2
B2 - 2B - 4 = 0
B = 4 and -1
Considering only positive value:
Side of square B = 4m
Side of square A = 4-3 = 1 m
QUESTION 4
Solution
Let two types of feeds are x and y
One gm of x = 20E1 + 5 E2
One gm of y = 10E1 + 15 E2
If equation 2 is multiplied with 3 then:
3y = 30 E1 + 45 E2
1x = 20E1 + 5E2
On adding both equations amount of each E1 and E2 equates to 50 gm.
1x + 3y = 50E1 +50E2
Thus 1 gm of first type of feed and 3 gm of second type of feed is required to make element
E1 and E2 in desired quantity.
QUESTION 5
Shopping trends of customers in second supermarket is presented below:

Bar graph clearly exhibits that, in the context of second supermarket, customers take
more time for buying lunch. Through graphical presentation it can be stated that time taken
6.
Bin Frequency
11 1
18 9
25 3
32 0
More 7
1. Price function
P = 400 x +10000

2. Determining weekly revenue and cost function
Cost function is:
C(x) = 600x+10000
Manufacture is selling computers at different price level:
Revenue function
R(x) = 1000x
R(x) = 800x
3. Assessing weekly profit function of manufacturer
Profit function of manufacturer can be determined in the following manner:
P (x) = 1000x – (600x+10000)
= 400 x – 10000
P (x) = 800x – (600x+10000)
= 200 x - 10000
4. Calculating break-even point
Particulars Formula
Figures
(in £)
Figures
(in £)
Selling price per unit 1000 800
Units 30 50
Variable cost per unit 600 600
Contribution per unit 400 200
Fixed cost 10000
Break-even point (in
units)
Fixed cost / contribution per
unit 25 50
Break-even point (in £) 25000 40000
The above depicted table clearly shows that BEP will be different at varied price level.
Accordingly, in the case of 30 units and 1000 SPU, manufacturer is required to sell 25 computers
for attaining the level of no profit & loss. On the other side, business entity needs to sell 50
computers for getting BEP level when price per unit accounts for £800 respectively.
5. Presenting relationship between cost and revenue graphically
When units are 30

Y axis (in £)
At the level of 50 units
Y axis (in £)
QUESTION 6
Solution
1. Half-life of carbon 14 = T1/2 = 5600 years
Relation between half-life and decay constant k is: T1/2 = 0.693/k
So k = 0.693/5600
K = 1.38 * 10-4
2.
Amount of carbon found = 10% of original amount
N /N0 = 0.1
N = N0 e-kt
0.1 = e-kt
Taking log both sides
ln 0.92 = -kt
-2.30 = -kt
From above k = 1.38 * 10-4
Time t = -2.30 / (1.38 * 10-4)
t = 1.769 * 104 years

Thus trees were burned and felled nearly 1.769 * 104 years ago.