Business Mathematics and Statistics IMPORTANT QUESTION

Business Mathematics and Statistics

IMPORTANT QUESTION

Prime number check krne ke liye 2, 3, 5, 7 jese small prime numbers se divide krke pata chal jata h. Agar koi number in chhote prime numbers se divide nahi hota, toh vo prime hota hai.

Example:

Agar tumhe check karna hai ki 29 prime hai ya nahi:

1. 2 se check karo – 29 odd hai, toh 2 se nahi divide hota.
2. 3 se check karo – 29 ko 3 se divide karo, remainder bachta hai, toh 3 se divide nahi hota.
3. 5 se check karo – 29 ka last digit 0 ya 5 nahi hai, toh 5 se nahi hota.
4. 7 se check karo – 29 ko 7 se divide karo, remainder bachta hai, toh 7 se bhi nahi hota.

Jab 29 kisi bhi number se divide nahi hota, toh ye prime number hai.

General Rule:

Agar tumhara number $\sqrt{N}$ (square root) se chhoti prime numbers se divide nahi hota, toh vo prime hai. For example, 40 ke liye $\sqrt{\mathrm{40}}\mathrm{<span\; class="mspace"></span><span\; class="mrel">\approx 6.3,toh\; tumhe\; sirf\; 2,\; 3,\; 5\; se\; check\; karna\; hoga.</span>}$

 FIND THE AVERAHGE OF APP PRIME NUMBER BETWEEN 30 TO 60

To find the average of all prime numbers between 30 and 60, let's follow these steps:

Step 1: Identify the Prime Numbers Between 30 and 60

Prime numbers are numbers greater than 1 that have no divisors other than 1 and themselves.

The prime numbers between 30 and 60 are:

• 31
• 37
• 41
• 43
• 47
• 53
• 59

Step 2: Sum the Prime Numbers

Now, let's add these prime numbers together:

$$31 + 37 + 41 + 43 + 47 + 53 + 59$$

Calculating the sum step-by-step:

• $31+37=68$
• $68+41=109$
• $109+43=152$
• $152+47=199$
• $199+53=252$
• $252+59=311$

So, the total sum is 311.

Step 3: Count the Prime Numbers

Next, we need to count how many prime numbers we have:

• The prime numbers we listed are 7 in total.

Step 4: Calculate the Average

The average is calculated by dividing the total sum by the number of prime numbers:

$$\text{Average} = \frac{\text{Total Sum}}{\text{Number of Prime Numbers}} = \frac{311}{7} \approx 44.43$$

Final Answer:

The average of all prime numbers between 30 and 60 is approximately 44.43.

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Practice Questions

1. Find the average of all prime numbers between 20 and 40.

2. Find the average of all prime numbers between 10 and 30.

3. Find the average of all prime numbers between 50 and 80.

Q2 THE AVERAGE AGE OF CLASS OF 39 STUDENT IS 15 YEAR IF THE AGE OF TEACHER INCLUDED THEN THE AVERAGE INCREASED 3 MONTHS FIND THE AGE OF TEACHER

Let's solve the problem step by step.

Step 1: Understand the Given Information

• Average age of 39 students = 15 years
• Average age increases by 3 months when the teacher is included.

Note: 3 months is equal to $\frac{3}{12} = \frac{1}{4}$ years. Therefore, the new average age with the teacher included is:

$$15 + \frac{1}{4} = 15.25 \text{ years}$$

Step 2: Calculate the Total Age of the Students

To find the total age of the 39 students, we can use the formula for average:

$$\text{Average} = \frac{\text{Total Age}}{\text{Number of Students}}$$

Rearranging the formula gives us:

$$\text{Total Age of Students} = \text{Average} \times \text{Number of Students}$$

Substituting in the values:

$$\text{Total Age of Students} = 15 \times 39 = 585 \text{ years}$$

Step 3: Calculate the Total Age Including the Teacher

Now, we need to calculate the total age when the teacher is included:

• Number of individuals (students + teacher) = $39 + 1 = 40$
• New average age = 15.25 years

Using the average to find the new total age:

$$\text{Total Age with Teacher} = \text{New Average} \times \text{Total Individuals}$$

Substituting in the values:

$$\text{Total Age with Teacher} = 15.25 \times 40 = 610 \text{ years}$$

Step 4: Find the Teacher's Age

Now we can find the age of the teacher by subtracting the total age of the students from the total age including the teacher:

$$\text{Age of Teacher} = \text{Total Age with Teacher} - \text{Total Age of Students}$$

Substituting in the values:

$$\text{Age of Teacher} = 610 - 585 = 25 \text{ years}$$

Final Answer:

The age of the teacher is 25 years.

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Practice Questions

1. The average weight of 25 students is 50 kg. If the weight of a new student is included and the average increases to 52 kg, find the weight of the new student.

2. In a group of 15 friends, the average score in a game is 20 points. If a new friend joins and the average becomes 21 points, find the score of the new friend.

3. The average salary of 30 employees in a company is $3000. If a new employee joins and the average salary increases to $3100, find the salary of the new employee.

 

Q3 EVALUATE 28% OF 450 + 45% OF 280 - 16 2/3 OF 600

To evaluate the expression 

$28\% \text{ of } 450 + 45\% \text{ of } 280 - 16 \frac{2}{3}\% \text{ of } 600$, let's break it down step by step.

Step 1: Calculate Each Percentage

1. Calculate $28\% \text{ of } 450$:

   $$28\% \text{ of } 450 = \frac{28}{100} \times 450 = 0.28 \times 450 = 126$$

2. Calculate $45\% \text{ of } 280$:

   $$45\% \text{ of } 280 = \frac{45}{100} \times 280 = 0.45 \times 280 = 126$$

3. Calculate $16 \frac{2}{3}\% \text{ of } 600$:

   • First, convert $16 \frac{2}{3}\%$ to an improper fraction:
   $$16 \frac{2}{3} = \frac{50}{3}$$
   • Now, calculate $\frac{50}{3}\% \text{ of } 600$:
   $$\frac{50}{3}\% \text{ of } 600 = \frac{50}{3 \times 100} \times 600 = \frac{50}{300} \times 600 = \frac{50 \times 600}{300} = 100$$

Step 2: Substitute the Calculated Values into the Expression

Now we can substitute these values into the expression:

$$126 + 126 - 100$$

Step 3: Evaluate the Expression

Now, we can simplify:

$$126 + 126 = 252$$ $$252 - 100 = 152$$

Final Answer:

The final result of the expression $28\% \text{ of } 450 + 45\% \text{ of } 280 - 16 \frac{2}{3}\% \text{ of } 600$ is 152.

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Practice Questions

1. Evaluate $25\% \text{ of } 800 + 30\% \text{ of } 600 - 20\% \text{ of } 1500$.

2. Calculate $10\% \text{ of } 250 + 15\% \text{ of } 400 - 5\% \text{ of } 1200$.

3. Find $18\% \text{ of } 350 + 25\% \text{ of } 240 - 33\frac{1}{3}\% \text{ of } 900$.

Q4 THE AVERAGE AGE OF 6 CONSICUTIVE ODD NUMBER IS 30 FIND THWE LARGEST NUMBER

To find the largest of 6 consecutive odd numbers given that their average age is 30, we can follow these steps:

Step 1: Understand the Average

The average of a set of numbers is calculated by dividing the sum of the numbers by the count of the numbers. Given that the average age of the 6 consecutive odd numbers is 30, we can set up the equation:

$$\text{Average} = \frac{\text{Sum of Numbers}}{\text{Number of Numbers}}$$

So, we have:

$$30 = \frac{\text{Sum of 6 Odd Numbers}}{6}$$

Step 2: Calculate the Sum of the Odd Numbers

To find the sum, we can rearrange the equation:

$$\text{Sum of 6 Odd Numbers} = 30 \times 6 = 180$$

Step 3: Represent the Odd Numbers

Let’s represent the 6 consecutive odd numbers. If we let the smallest odd number be $x$, the next consecutive odd numbers can be represented as:

• $x$ (1st)
• $x + 2$ (2nd)
• $x + 4$ (3rd)
• $x + 6$ (4th)
• $x + 8$ (5th)
• $x + 10$ (6th)

Step 4: Set Up the Equation for Their Sum

Now, we can express the sum of these numbers:

$$x + (x + 2) + (x + 4) + (x + 6) + (x + 8) + (x + 10) = 180$$

Combining like terms gives us:

$$6x + (2 + 4 + 6 + 8 + 10) = 180$$

Calculating the sum of the constants:

$$2 + 4 + 6 + 8 + 10 = 30$$

So now we have:

$$6x + 30 = 180$$

Step 5: Solve for $x$

Now, subtract 30 from both sides:

$$6x = 180 - 30$$ $$6x = 150$$

Now, divide by 6:

$$x = \frac{150}{6} = 25$$

Step 6: Find the Largest Odd Number

Now that we have $x$, which is the smallest odd number (25), we can find the largest odd number:

$$\text{Largest Odd Number} = x + 10 = 25 + 10 = 35$$

Final Answer:

The largest of the 6 consecutive odd numbers is 35.

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Practice Questions

1. The average of 5 consecutive even numbers is 24. Find the largest number.

2. If the average of 7 consecutive odd numbers is 40, what is the largest number?

3. The average age of 8 consecutive even numbers is 22. What is the largest number?

Q5 THE TRADE MAN SOLD AN ARTICLE AT A LESS OF 20% IF THE SELLING PRICE HAD BEEN INCREASED BY RS 100 WHAT WAS THE LAST PRICE OF AN ARTICLE

To solve this problem, let's break it down step by step.

Step 1: Understand the Given Information

• The article is sold at a loss of 20%.
• If the selling price (SP) had been increased by Rs. 100, it would equal the cost price (CP).

Step 2: Set Up the Equations

Let:

• $CP$ = Cost Price of the article
• The selling price (SP) can be expressed as:

$$SP = CP - (20\% \text{ of } CP) = CP - 0.2 \cdot CP = 0.8 \cdot CP$$

Step 3: Set Up the Selling Price with the Increase

According to the problem, if the selling price had been increased by Rs. 100, it equals the cost price:

$$SP + 100 = CP$$

Substituting the expression for SP:

$$0.8 \cdot CP + 100 = CP$$

Step 4: Solve for Cost Price (CP)

Now, let's rearrange the equation:

$$100 = CP - 0.8 \cdot CP$$ $$100 = 0.2 \cdot CP$$

Now divide both sides by 0.2 to find CP:

$$CP = \frac{100}{0.2} = 500$$

Step 5: Final Calculation

Now that we have found the cost price of the article, we can conclude:

• Cost Price (CP) = Rs. 500

Final Answer:

The cost price of the article is Rs. 500.

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Practice Questions

1. A trader sold a watch at a loss of 15%. If the selling price had been increased by Rs. 50, it would equal the cost price. Find the cost price of the watch.

2. A book is sold at a loss of 10%. If the selling price were increased by Rs. 30, it would equal the cost price. What is the cost price of the book?

3. A merchant sells a bag at a loss of 25%. If the selling price were increased by Rs. 80, it would equal the cost price. Find the cost price of the bag.

Q6 LET 

Matrix $A$ is a 3x2 matrix:

$$A=\left(\begin{array}{cc}3& 2\\ -1& 0\\ 2& 1\end{array}\right)$$

Matrix $B$ is a 2x3 matrix:

$$B=(\begin{array}{ccc}2& 1& -1\\ 4& 0& 5\end{array}$$) FIND BA AND AB

Let's solve the multiplication of matrices $BA$ and $AB$ step by step.

Given Matrices:

Matrix $A$ is a 3x2 matrix:

$$A = \begin{pmatrix} 3 & 2 \\ -1 & 0 \\ 2 & 1 \end{pmatrix}$$

Matrix $B$ is a 2x3 matrix:

$$B = \begin{pmatrix} 2 & 1 & -1 \\ 4 & 0 & 5 \end{pmatrix}$$

We will now find both $BA$ and $AB$.

---

Step 1: Find $BA$ (Multiplying $B \times A$)

For $BA$, we need to multiply a 2x3 matrix ($B$) by a 3x2 matrix ($A$).

Matrix $B$:

$$\begin{pmatrix} 2 & 1 & -1 \\ 4 & 0 & 5 \end{pmatrix}$$

Matrix $A$:

$$\begin{pmatrix} 3 & 2 \\ -1 & 0 \\ 2 & 1 \end{pmatrix}$$

The product $BA$ will be a 2x2 matrix.

Let's calculate:

Element at (1,1) in $BA$:

$$(2 \times 3) + (1 \times -1) + (-1 \times 2) = 6 - 1 - 2 = 3$$

Element at (1,2) in $BA$:

$$(2 \times 2) + (1 \times 0) + (-1 \times 1) = 4 + 0 - 1 = 3$$

Element at (2,1) in $BA$:

$$(4 \times 3) + (0 \times -1) + (5 \times 2) = 12 + 0 + 10 = 22$$

Element at (2,2) in $BA$:

$$(4 \times 2) + (0 \times 0) + (5 \times 1) = 8 + 0 + 5 = 13$$

So, the matrix $BA$ is:

$$BA = \begin{pmatrix} 3 & 3 \\ 22 & 13 \end{pmatrix}$$

---

Step 2: Find $AB$ (Multiplying $A \times B$)

For $AB$, we need to multiply a 3x2 matrix ($A$) by a 2x3 matrix ($B$).

Matrix $A$:

$$\begin{pmatrix} 3 & 2 \\ -1 & 0 \\ 2 & 1 \end{pmatrix}$$

Matrix $B$:

$$\begin{pmatrix} 2 & 1 & -1 \\ 4 & 0 & 5 \end{pmatrix}$$

The product $AB$ will be a 3x3 matrix.

Let's calculate:

Element at (1,1) in $AB$:

$$(3 \times 2) + (2 \times 4) = 6 + 8 = 14$$

Element at (1,2) in $AB$:

$$(3 \times 1) + (2 \times 0) = 3 + 0 = 3$$

Element at (1,3) in $AB$:

$$(3 \times -1) + (2 \times 5) = -3 + 10 = 7$$

Element at (2,1) in $AB$:

$$(-1 \times 2) + (0 \times 4) = -2 + 0 = -2$$

Element at (2,2) in $AB$:

$$(-1 \times 1) + (0 \times 0) = -1 + 0 = -1$$

Element at (2,3) in $AB$:

$$(-1 \times -1) + (0 \times 5) = 1 + 0 = 1$$

Element at (3,1) in $AB$:

$$(2 \times 2) + (1 \times 4) = 4 + 4 = 8$$

Element at (3,2) in $AB$:

$$(2 \times 1) + (1 \times 0) = 2 + 0 = 2$$

Element at (3,3) in $AB$:

$$(2 \times -1) + (1 \times 5) = -2 + 5 = 3$$

So, the matrix $AB$ is:

$$AB = \begin{pmatrix} 14 & 3 & 7 \\ -2 & -1 & 1 \\ 8 & 2 & 3 \end{pmatrix}$$

---

Final Answer:

• Matrix $BA$ is:

$$BA = \begin{pmatrix} 3 & 3 \\ 22 & 13 \end{pmatrix}$$

• Matrix $AB$ is:

$$AB = \begin{pmatrix} 14 & 3 & 7 \\ -2 & -1 & 1 \\ 8 & 2 & 3 \end{pmatrix}$$

Practice Questions:

1. Question 1:
   Let $C = \begin{pmatrix} 1 & 4 \\ 0 & 2 \\ 3 & 1 \end{pmatrix}$ (3x2 matrix)
   Let $D = \begin{pmatrix} 2 & 1 & 3 \\ 4 & 0 & 5 \end{pmatrix}$(2x3 matrix)
   Find the product $DC$ and $CD$.

2. Question 2:
   Let $E = \begin{pmatrix} 5 & 2 \\ 3 & 1 \\ 4 & 0 \end{pmatrix}$ (3x2 matrix)
   Let $F = \begin{pmatrix} 1 & 0 & 2 \\ 3 & 4 & 1 \end{pmatrix}$ (2x3 matrix)
   Calculate $FE$ and $EF$.

3. Question 3:
   Let $G = \begin{pmatrix} 6 & 1 \\ 2 & 3 \\ 5 & 4 \end{pmatrix}$ (3x2 matrix)
   Let $H = \begin{pmatrix} 0 & 2 & 1 \\ 3 & 1 & 4 \end{pmatrix}$ (2x3 matrix)
   Find the products $HG$ and $GH$.

4. Question 4:
   Let $I = \begin{pmatrix} 4 & -2 \\ 3 & 1 \\ 0 & 5 \end{pmatrix}$ (3x2 matrix)
   Let $J = \begin{pmatrix} 1 & 3 & 2 \\ -1 & 0 & 4 \end{pmatrix}$ (2x3 matrix)
   Calculate $IJ$ and $JI$.

Instructions:

1. Try to calculate the products step by step as we did previously.
2. Write down each element of the resulting matrix.
3. Check your answers with the results from multiplying the matrices.\

UTSAV

Q7 FIND THE INVERSE OF MATRIX 3*3 A= 1 2 -1 , 2 4 7 , 3 2 6

$$To find the inverse of aTo find the inverse of aTO FIND 3×33 \times 3 matrix, we can use the formula:

$$A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A)$$

where $\text{det}(A)$ is the determinant of matrix $A$ and $\text{adj}(A)$ is the adjugate of $A$.

Given Matrix:

$$A = \begin{pmatrix} 1 & 2 & -1 \\ 2 & 4 & 7 \\ 3 & 2 & 6 \end{pmatrix}$$

Step 1: Calculate the Determinant of $A$

The determinant of a $3 \times 3$ matrix is calculated as follows:

$$\text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg)$$

For our matrix $A$:

$$a = 1, \quad b = 2, \quad c = -1 \\ d = 2, \quad e = 4, \quad f = 7 \\ g = 3, \quad h = 2, \quad i = 6$$

So, we can compute:

$$\text{det}(A) = 1(4 \cdot 6 - 7 \cdot 2) - 2(2 \cdot 6 - 7 \cdot 3) + (-1)(2 \cdot 2 - 4 \cdot 3)$$

Calculating each term step by step:

1. $4 \cdot 6 = 24$
2. $7 \cdot 2 = 14$
3. $2 \cdot 6 = 12$
4. $7 \cdot 3 = 21$
5. $2 \cdot 2 = 4$
6. $4 \cdot 3 = 12$

Now substitute these into the determinant formula:

$$\text{det}(A) = 1(24 - 14) - 2(12 - 21) - (4 - 12)$$

Simplifying:

$$= 1(10) - 2(-9) - (-8)$$ $$= 10 + 18 + 8$$ $$= 36$$

Step 2: Calculate the Adjugate of $A$

The adjugate of a $3 \times 3$ matrix is the transpose of the cofactor matrix. We need to calculate the cofactor for each element.

$$\text{Cofactor}(A) = \begin{pmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{pmatrix}$$

Where each cofactor $C_{ij}$ is computed as:

$$C_{ij} = (-1)^{i+j} \cdot \text{det}(M_{ij})$$

$M_{ij}$ is the minor of element $a_{ij}$, found by deleting the $i$-th row and $j$-th column.

Let's calculate the cofactors:

1. Cofactor $C_{11}$:

   • Minor $M_{11} = \begin{pmatrix} 4 & 7 \\ 2 & 6 \end{pmatrix}$
   • $\text{det}(M_{11}) = (4 \cdot 6) - (7 \cdot 2) = 24 - 14 = 10$
   • $C_{11} = 1 \cdot 10 = 10$

2. Cofactor $C_{12}$:

   • Minor $M_{12} = \begin{pmatrix} 2 & 7 \\ 3 & 6 \end{pmatrix}$
   • $\text{det}(M_{12}) = (2 \cdot 6) - (7 \cdot 3) = 12 - 21 = -9$
   • $C_{12} = -1 \cdot -9 = 9$

3. Cofactor $C_{13}$:

   • Minor $M_{13} = \begin{pmatrix} 2 & 4 \\ 3 & 2 \end{pmatrix}$
   • $\text{det}(M_{13}) = (2 \cdot 2) - (4 \cdot 3) = 4 - 12 = -8$
   • $C_{13} = 1 \cdot -8 = -8$

4. Cofactor $C_{21}$:

   • Minor $M_{21} = \begin{pmatrix} 2 & -1 \\ 3 & 6 \end{pmatrix}$
   • $\text{det}(M_{21}) = (2 \cdot 6) - (-1 \cdot 3) = 12 + 3 = 15$
   • $C_{21} = -1 \cdot 15 = -15$

5. Cofactor $C_{22}$:

   • Minor $M_{22} = \begin{pmatrix} 1 & -1 \\ 3 & 6 \end{pmatrix}$
   • $\text{det}(M_{22}) = (1 \cdot 6) - (-1 \cdot 3) = 6 + 3 = 9$
   • $C_{22} = 1 \cdot 9 = 9$

6. Cofactor $C_{23}$:

   • Minor $M_{23} = \begin{pmatrix} 1 & 2 \\ 3 & 2 \end{pmatrix}$
   • $\text{det}(M_{23}) = (1 \cdot 2) - (2 \cdot 3) = 2 - 6 = -4$
   • $C_{23} = -1 \cdot -4 = 4$

7. Cofactor $C_{31}$:

   • Minor $M_{31} = \begin{pmatrix} 2 & -1 \\ 4 & 7 \end{pmatrix}$
   • $\text{det}(M_{31}) = (2 \cdot 7) - (-1 \cdot 4) = 14 + 4 = 18$
   • $C_{31} = 1 \cdot 18 = 18$

8. Cofactor $C_{32}$:

   • Minor $M_{32} = \begin{pmatrix} 1 & -1 \\ 2 & 7 \end{pmatrix}$
   • $\text{det}(M_{32}) = (1 \cdot 7) - (-1 \cdot 2) = 7 + 2 = 9$
   • $C_{32} = -1 \cdot 9 = -9$

9. Cofactor $C_{33}$:

   • Minor $M_{33} = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$
   • $\text{det}(M_{33}) = (1 \cdot 4) - (2 \cdot 2) = 4 - 4 = 0$
   • $C_{33} = 1 \cdot 0 = 0$

Cofactor Matrix:

$$\text{Cof}(A) = \begin{pmatrix} 10 & 9 & -8 \\ -15 & 9 & 4 \\ 18 & -9 & 0 \end{pmatrix}$$

Adjugate Matrix:

The adjugate is the transpose of the cofactor matrix:

$$\text{adj}(A) = \begin{pmatrix} 10 & -15 & 18 \\ 9 & 9 & -9 \\ -8 & 4 & 0 \end{pmatrix}$$

Step 3: Calculate the Inverse of $A$

Now we can find the inverse:

$$A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) = \frac{1}{36} \cdot \begin{pmatrix} 10 & -15 & 18 \\ 9 & 9 & -9 \\ -8 & 4 & 0 \end{pmatrix}$$

Final Inverse Matrix:

$$A^{-1} = \begin{pmatrix} \frac{10}{36} & \frac{-15}{36} & \frac{18}{36} \\ \frac{9}{36} & \frac{9}{36} & \frac{-9}{36} \\ \frac{-8}{36} & \frac{4}{36} & 0 \end{pmatrix}$$

This simplifies to:

$$A^{-1} = \begin{pmatrix} \frac{5}{18} & \frac{-5}{12} & \frac{1}{2} \\ \frac{1}{4} & \frac{1}{4} & \frac{-1}{4} \\ \frac{-2}{9} & \frac{1}{9} & 0 \end{pmatrix}$$

Q 8 HOW MANY MINUTES DOES A TAKE TO COVER A DISTANCE 40M IF HE RUNS AT A SPED OF 40KM/H

To find out how many minutes it takes for a person (A) to cover a distance of 40 meters while running at a speed of 40 km/h, we can use the formula:

$$\text{Time} = \frac{\text{Distance}}{\text{Speed}}$$

Step 1: Convert the Speed to Meters per Second

First, we need to convert the speed from kilometers per hour (km/h) to meters per second (m/s):

$$\text{Speed (m/s)} = \frac{\text{Speed (km/h)} \times 1000}{3600}$$

For $40$ km/h:

$$\text{Speed} = \frac{40 \times 1000}{3600} = \frac{40000}{3600} \approx 11.11 \text{ m/s}$$

Step 2: Calculate the Time in Seconds

Now we can calculate the time it takes to cover 40 meters:

$$\text{Time (seconds)} = \frac{\text{Distance (m)}}{\text{Speed (m/s)}} = \frac{40}{11.11} \approx 3.6 \text{ seconds}$$

Step 3: Convert Time to Minutes

To convert seconds to minutes, we divide by 60:

$$\text{Time (minutes)} = \frac{3.6 \text{ seconds}}{60} \approx 0.06 \text{ minutes}$$

Conclusion

So, it takes approximately 0.06 minutes for A to cover a distance of 40 meters at a speed of 40 km/h.

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Practice Questions

1. If A runs at a speed of 30 km/h, how long would it take him to cover the same distance of 40 meters?

2. How many minutes does A take to cover 100 meters at a speed of 50 km/h?

3. If A increases his speed to 60 km/h, how long will it take him to cover a distance of 80 meters?

$$Q9\; WORKER\; A\&B\; TAKES\; 12\; HR\; AND\; 8\; HR\; TO\; DO\; A\; SAME\; WORK\; HOW\; LONG\; SHOULD\; IT\; TAKE\; IF\; BOTH\; A\; ANB\; WORK\; TOGETHER$$

$$\text{Work Rate of A} = \frac{1}{\text{Time taken by A}} = \frac{1}{12} \text{ work/hour}$$

$$Q10\; FIND\; CUBE\; ROOT\; OF\; 2744$$

$$2744$$

$$<br>$$

$$Q11\; A\; CYCLER\; COVER\; A\; DISTANCE\; OF\; 750\; M\; IN\; 20\; MINUTES\; AND\; 30\; SEC\; WHAYT\; IS\; THE\; SPEED\; IN\; KMH$$

$$750$$