Category: Mathematical Example

# Compound Interest Mathematical Problem & Solution

## Compound Interest Mathematical Problem

Question: A Person has invested Rs. 10000 in a bank at an interest of 10% per annum. How much amount will receive After 2 years if the compounding is done?

a) Annually.
b) Semi-Annually.
c)Quarterly.
d)Monthly.
e)Daily

FVn = PV ( 1+ r/m )mn

Where, FVn = Future value after n years

n = Number of years [In the above math, n = 2]

m = Number of times compounding is done in a year

a) Annually.

Here, r = 0.10 (10%), m = 1.

FV = PV ( 1+ 0.10/1 )1×2 = 10000(1+0.10)2= 10000 × (1.10)2 = 12100 Rs.

b) Semi-Annually.

Here, r = 0.10 (10%), m = 2.

FV = PV ( 1+ 0.10/2 )2×2 = 10000(1+0.05)4= 10000 × (1.05)4 = 12155 Rs.

c) Quarterly.

Here, r = 0.10 (10%), m = 4.

FV = PV ( 1+ 0.10/4 )4×2 = 10000(1+0.025)8= 10000 × (1.025)8 = 12184 Rs.

d) Monthly.

Here, r = 0.10 (10%), m = 12.

FV = PV ( 1+ 0.10/12 )12×2 = 10000(1+0.0083)24= 10000 × (1.0083)24 = 12194 Rs.

e) Daily

Here, r = 0.10 (10%), m = 365.

FV = PV ( 1+ 0.10/365 )365×2 = 10000(1+0.000274)730= 10000 × (1.000274)730 = 12214 Rs.

How to Determine Runoff Coefficient – Mathematical Problem & Solution

How To Calculate True Bearing From Magnetic Bearing

# How to Determine the Frequency of Irrigation – Problem & Solution

## How to Determine the Frequency of Irrigation

Mathematical Problem

The following data are given to determine the frequency of irrigation

• Field capacity of soil = 45 %
• Permanent wilting point = 19 %
• Density of soil = 1.5 g/cm3
• Depth of root zone = 65 cm
• Daily consumptive use of water = 18 mm

Solution:

Available moisture = Field Capacity – Permanent wilting point = 45 – 19 = 26

Let readily available moisture = is 75 % of the available moisture,

So, readily available moisture is = 26 × 0.75 = 19.5 %

Optimum moisture content = 45 – 19.5 = 25.5 %

Now, by applying irrigation water the moisture content is to be raised from 25.5 % to 45 %.

Calculate the depth of water to be applied in each watering, by using the following equation:

$D_{w} = \frac{W_{s}\times d}{W_{w}}\times [F_{c}-M_{0}]$

Where,

• Ws = Unit Weight of soil or density of soil = 1.5 g/cm3
• Ww = Density of water = 1g/cm3
• d = Depth of root zone = 65 cm
• Fc = Field capacity = 45 %
• M0 = Optimum moisture content = 25.5 %

After putting the all value,

$D_{w} = \frac{1.5\times 65}{1}\times\left [ 0.45 – 0.255 \right ]$

$Or, D_{w} = 97.5 \times 0.195 = 19.01$ cm

Now, we can calculate the frequency of irrigation by using the following equation

$f_{w} = \frac{D_{w}}{C_{u}}$

Where,

• fw = Frequency of watering or frequency of irrigation
• Cu = Daily consumptive use of water =
• Dw = Depth of water to be applied in each watering

After putting all values, the frequency of irrigation

$f_{w} = \frac{19.01}{1.8} = 10.56 = 11 days (say)$

Find the Intensity of Rainfall

Determine Runoff by using Rational Method

Types Of Irrigation

# Find the Intensity of Rainfall When the duration of Rainfall is Given

## Find the intensity of Rainfall When the duration of Rainfall is Given

First of all, we need to know the formula used to determine rainfall intensity. There are three types of empirical formulas –

a) For storm duration of 5 to 20 minutes

$i = \frac{672}{t+10}$

b) For storm duration of 20 to 100 minutes

$i = \frac{1020}{t+10}$

c) Where rainfall is frequent

$i = \frac{3430}{t+18}$

Here, i = Intensity of rainfall in mm/hr, t = duration of the storm in minutes.

Following are the two mathematical examples for the determination of rainfall intensity

### Mathematical Problem: 1

Find the intensity of rainfall when the storm continues for a period of 75 minutes.

Solution:

As the duration of rainfall is in between 20 to 100 minutes, so, the formula will be
$i = \frac{1020}{t+10}$
The intensity of rainfall,

$i = \frac{1020}{75+10}$

$Or, i = \frac{1020}{85} = 12 mm/hr$

### Mathematical Problem: 2

Find the intensity of rainfall when the rainfall occurs frequently and the total duration of rainfall is 115 minutes:

Solution:

As the rainfall occurs frequently, we will use 3rd formula, $i = \frac{3430}{t+18}$
So, the intensity of rainfall is

$i = \frac{3430}{115+18}$

$Or, i = \frac{3430}{133} = 25.78 mm/hr$

Determine the Frequency of Irrigation

Determine Runoff Coefficient

Types of Rainfall

# How to Determine Runoff by using Rational Method – Mathematical Problem & Solution

## How to Determine Runoff by using Rational Method

First of all, we need to understand which formula or expression is used for the determination of runoff. In the rational method, the runoff is given by the following expression:

$Q = \frac{KiA}{36}$

Where,

Q = Runoff in cumec
K = Coefficient of runoff
i = Intensity of rainfall in cm/hr
A = Catchment area in hectares

The above formula may also be expressed as follows,

$Q = \frac{KiA}{360}$

This formula applies when the intensity of rainfall is given in mm/hr. Otherwise, all units are the same as above.

### Mathematical Problem

A catchment area of 35 sq km consists of two-thirds rural area and one-third urban area. The rainfall in the whole catchment area is recorded as 30 mm/hr. Find the total runoff from the catchment area.

Solution:

We know the total are = 35 sq km = 3500 hectares ( 1 sq km = 100 hectares)

Now,

Rural area = $3500 \times \frac{2}{3}$ = 2333.33 hectares

Urban area = $3500 \times \frac{1}{3}$ = 1166.66 hectares.

Rainfall intensity i = 30 mm/hr.

Assuming, runoff coefficient for urban area = 0.50 and for rural area = 0.30

In rational method, the runoff is given by,

$Q = \frac{KiA}{360}$

Or, $\large Q = \frac{(0.30\times 25\times 2333.33) + (0.50\times 25\times 1166.66)}{ 360 }$

Or, $\large Q = \frac{17499.97+14583.25}{360}$

Or, Q = 89.12 cumec.

Empirical Formula Used For Determination of Runoff

How to Calculate Annual Runoff

Burkli Ziegler Formula – To Determine the Peak Runoff Rate

How to Determine Runoff Coefficient

# How to Determine Runoff Coefficient – Mathematical Problem & Solution

## How to Determine Runoff Coefficient

When the nature of the surface of different areas and the impermeability factors of the areas are known or given, then the value of the runoff coefficient(K) may be determined by the following expression:

$K= \frac{a_{1}p_{1}+a_{2}p_{2}+\cdot \cdot \cdot \cdot \cdot }{A}$

Where,

a1, a2 = Area of different surfaces.
p1, p2 = Impermeability factors of those surfaces.
A = Total surface area.

### Mathematical Problem

A catchment area consists of the following surfaces:

1. Cultivated area = 50 hectares (p = 0.20)
2. Forest area = 30 hectares (p = 0.10)
3. Garden = 5 hectares (p = 0.05)
4. Residential = 15 hectares (p = 0.50).

Find the runoff coefficient of that catchment area.

Solution

Here, the total area will be = 50 + 30 + 5 + 15 = 100 hectares.

From the expression,

$K= \frac{a_{1}p_{1}+a_{2}p_{2}+\cdot \cdot \cdot \cdot \cdot }{A}$

Here,

$a_{1}$ = 50 and $p_{1}$ = 0.20

$a_{2}$ = 30 and $p_{2}$ = 0.10

$a_{3}$ = 5 and $p_{3}$ = 0.05

$a_{4}$ = 15 and $p_{4}$ = 0.50

We get the runoff coefficient,

Determine Runoff by using Rational Method

Empirical Formula Used For Determination of Runoff

How to Calculate Annual Runoff || Problem & Solution ||

Burkli Ziegler Formula – To Determine the Peak Runoff Rate

# Find the Correction for Curvature & Refraction for a Given Distance

## Find the Correction for Curvature

Mathematic Problem

Find the correction for curvature for a distance of 900 m and 2.5 km.

Solution

1. When the distance is 900 m 0r 0.9 km

Correction for curvature (in meters) = 0.0785 D2

After Putting the value of D in the above equation [D is in Km]

Correction for curvature (in meters) = 0.0785 × 0.92 = 0.063 m

2. When the distance is 2.5 km

Correction for curvature (in meters) = 0.0785 × 2.52 = 0.49 m

## Find the Correction for Refraction

Mathematical Problem

Find the correction for refraction for a distance of 700 m and 1.5 km

Solution

1. When the distance is 700 m 0r 0.7 km

Correction for refraction (in meters) = 0.0112 D2 = 0.0112 × 0.7 2 = 0.0055 m.

2. When the distance is 1.5 km

Correction for refraction (in meters) = 0.0112 D2 = 0.0112 × 1.5 2 = 0.0252 m.

Convert Quadrantal Bearings to Whole Circle Bearing

# Calculate the Horizontal Length After Slope Correction of Tape

## Calculate the Horizontal Length After Slope Correction of the Tape

Mathematical Problem:

The downhill end of the 30m tape is held 70cm too low. What is the horizontal length?

Solution:

Given,

l = Length of the tape = 30 m.
h = 70 cm = 0.70 m

Correction for slope = $\large \frac{h^{2}}{2l}$ = $\large \frac{0.7^{2}}{2\times 30}$ = 0.00817 m.

Hence, the horizontal length = 30 – 0.00817 = 29.99183 m

Calculate the Correct Length of the Tape

# Find out the Discharge of a Siphon Spillway – Problem & Solution

## Find out the Discharge of a Siphon Spillway

Mathematical Problem:

Find out the discharge of a siphon spillway from following data:

• Number of siphon units = 4.
• Area at throat in m2 = 3.
• Full reservoir level = 150 m.
• R.L of centre of outlet = 128.
• Tailwater level on D/S side during rains = 130.
• Tailwater level during winter = 125.
• Discharge coefficient = 0.60.

Solution:

Case 1

In rainy days outlet remains submerged and hence discharge depends upon the tailwater level.

Working head = R.L of reservoir – R.L. or T.W.L. = 150 – 130 = 20 m.

$Q = CA \sqrt{2gh}$

Or, $Q = 0.60\times 3 \sqrt{2g\times 20}$

Or, Q = 35.75 cumecs.

So, the discharge of four units = 4 × 35.75 = 143 cumecs.

Case II

In winter T.W.L falls down and spillways discharge free in the air.

Available head = Reservoir level – R.L of center of outlet = 150 – 128 = 22 m.

$Q = CA \sqrt{2gh}$

Or, $Q = 0.60\times 3 \sqrt{2 \times 9.81 \times 22}$

Or, Q = 37.5 cumecs.

Discharge of four units = 4 × 37.5 = 152 cumecs.

Standard Irrigation Water

How to Calculate Annual Runoff – Problem & Solution

How to Calculate Discharge of the Water if the Depth and Crop Period are Given

# Determine the Discharge of the Channel for which it is to be Designed

## Determine the Discharge of the Channel for which it is to be Designed

Mathematical Problem

A channel is to be designed for irrigating 6000 hectares of Kharif crop and 5500 hectares of Rabi crop. The water requirement for the Kharif crop is 70 cm and the water requirement for Rabi is 35 cm. The Kor period for Kharif is 2 weeks and for Rabi is 3 weeks and 4 days. Determine the discharge of the channel for which it is to be designed.

Solution:

Using the relation

∇ = (8.64 × B)/ D

Discharge for Kharif crop

Here,

• ∇ = 70 cm = 0.70 m
• B = 2 weeks = 14 days

Duty (D) = (8.64 × 21)/ 0.70 = 259.2 hectares/cumec.

Area to be irrigated = 6000 hectares.

Required discharge of channel for Kharif crop = (6000/259.2) = 23.15 cumec.

Discharge for Rabi crop

Here,

• ∇ = 35 cm = 0.35 m.
• B = 3 weeks and 4 day = (21 + 4) = 25 days.

Duty (D) = (8.64 × 25)/ 0.35 = 617.14 hectares/cumec.

Area to be irrigated = 5500 hectares.

Required discharge of channel for Kharif crop = (5500/617.14) = 8.91 cumec.

Therefore, the channel is to be designed for the maximum discharge of 23.15 cumec, because this discharge capacity of the channel will be able to supply water to both seasons.

Relation Between Base, Delta, and Duty

Factors Affecting the Duty of Water

How to Calculate Discharge of The Water If The Depth & Crop Period are Given

# How to Calculate Discharge of The Water If The Depth & Crop Period are Given

## How to Calculate Discharge of The Water If The Depth & Crop Period are Given

Mathematical Problem

A Waterhouse has a culturable commanded area of 2200 hectares. The intensity of irrigation for crop A is 50% and for B is 35%, both the crops being Rabi crops. Crop A and B have a crop period of 25 days and 20 days respectively. Calculate the discharge of the water if the depth for crops A and B is 12 cm and 18 cm respectively.

Solution:

Culturable land for irrigation = 2200 hectares.

Area of irrigation for Crop A = 2200 × 50% = (2200 × 50)/100 = 1100 hectares.

Area of irrigation for Crop B = 2200 × 35% = (2200 × 35)/100 = 770 hectares.

Using the relation.

D = (8.64 × B)/ ∇

Discharge for crop A

Here,

• ∇ = 12 cm = 0.12 m.
• B = 25 days

Duty (D) = (8.64 × 25)/0.12 = 1800 hectares/cumec.

Required discharge for crop A = (1100/1800) = 0.61 cumec.

Discharge for crop B

Here,

• ∇ = 18 cm = 0.18 m
• B = 20 days

Duty (D) = (8.64 × 20)/0.18 = 960 hectares/cumec.

Required discharge for crop B = ( 770/960) = 0.80 cumec.

S0, the total discharge of water course = (0.61 + 0.80) = 1.41 cumec.