The taxi charges in a city consist of a fixed charge

Given: The charge for 7 km is Rs. 27 The charge for 5 km is Rs. 21 FormulaUsed: Total Charge = Fixed Charge + Rate per kilometer×Distance Calculation: Let fixed price be Rs. p and Ratebe Rs. rper kilometer 7× r + Rs. p = Rs. 27 ⇒ 7r +Rs. p = Rs. 27 ----(i) 5× r + Rs. p = Rs. 21 ⇒ 5r + Rs. p = Rs. 21 ----(ii) Subtracting (ii) from (i) ⇒ 2r = Rs. 6 ⇒ r = Rs. 6/2 ⇒ r = Rs. 3 ∴ The charge per kilometer is Rs. 3

The taxi charges in a city consist of a fixed 2 M charge together with the charge for the distance covered. For a distance of 1 0 k m, the charge paid is R s 1 0 5 & for a journey of 1 5 k m the charge paid is R s 1 5 5. What w the fixed charges and the charge per k m? How much does a person have to pay for travelling a distance of 2 5 k m? The taxi charge in Hyderabad are fixed , along with the charge for the distance covered. Upto first 3 km you will be charged a certain minimum amount. from tbere onward you have to pay addionally for every kilimeter travelled. For the first 10 km , the charge paid is Rs. 166. For a journey 15 km. The charge oaid is Rs. 256

Let fixed charge = R s . x and charge per kilometer = R s . y According to the question, x + 1 1 0 y = 6 9 0 … ( i ) and x + 2 0 0 y = 1 0 5 0 … (ii) On subtracting Eq.(i) from Eq. (ii), we get x + 2 0 0 y − x − 1 1 0 y = 1 0 5 0 − 6 9 0 ⇒ 9 0 y = 3 6 0 ⇒ y = 4 0 On putting the value of y = 4 0 in Eq. (i), we get x + 1 1 0 ( 4 0 ) = 6 9 0 ⇒ x + 4 4 0 = 6 9 0 ⇒ x = 6 9 0 − 4 4 0 = 2 5 0 Hence, monthly fixed charges is R s . 2 5 0 and charge per kilometer is R s . 4 0

The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 315 and for a journey of 15 km, the charge paid is Rs. 465. What are the fixed charges and the charge per kilometer ? How much does a person have to pay for travelling a distance of 32 km ? Let the fixed charge be Rs. x and the charge per kilometer be Rs. y. The charges for 10 km = Rs. 10y The charges for 15 km = Rs. 15y According to the question, x+ 10y = 315 ...(1) x+ 15y = 465 ....(2) Solving the equations, we get - 5y = - 150 ⇒ y = 30 and x = 315 - 10y = 315 - 10(30) = 15 So, the fixed charges is Rs. 15 and the charges per kilometer is Rs. 30. To travel 32 km, a personal has to pay = Rs. 15 + Rs. 30( 32 ) = Rs. 15 + Rs. 960 = Rs. 975.

Form the pair of linear equations for the following problems and find their solution by substitution method (i) The coach of a cricket team buys 7 bats and 6 balls for Rs.3800. Later, she buys 3 bats and 5 balls for Rs.1750. Find the cost of each bat and each ball. (ii) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km? Maths Q&A Form the pair of linear equations for the following problem and find their solution by substitution method: The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs . 105 and for a journey of 15 km, the charge paid is Rs . 155. What are the fixed charges and the charge per km? How much does a person have to pay for traveling a distance of 25 km? Step 1: Assume variables and form one equation Let, the fixed charge (in rupees) = x And, the per kilometer charge (in rupees) = y Now, according to the question, fixed charge + charge for 10 km = 105 ⇒ x + 10 × y = 105 ⇒ x + 10 y = 105 ⇒ x = 105 - 10 y …(i) Step 2: Form the second equation Again, according to the question, fixed charge + charge for 15 km = 155 ⇒ x + 15 × y = 155 ⇒ x + 15 y = 155 …(ii) Step 3: Calculate y b...

Let fixed charges be Rs.x and rate per km be Rs.y. Then as per the question x + 80y = 1330 ………(i) x + 90y = 1490 ……..(ii) Subtracting (i) from (ii), we get 10y = 160 ⇒ y = 160/10 = 16 Now, putting y = 16, we have x + 80 × 16 = 1330 ⇒x = 1330 – 1280 = 50 Hence, the fixed charges be Rs.50 and the rate per km is Rs.16.

2022-23 54 (iv) The taxi charges in a city consist of a fixed charge together with the charge fix the distance covered. For a distance of 10 km, the charge paid is ₹100 and fix a journey of 15 km, the charge paid is ₹155. What are the fixed charges and the charge per km ? How much does a person have to pay for traveling a dotrance odif (v) A fraction becomes 11 9 ​ , if 2 is added to both the numeranor and the dengeminaise: If, 3 is added to both the numerator and the denominator it becoemes 6 5 ​ , Pind the fraction. (vi) Five years hence, the age of Jacob will be three times that of his ove the yeas? ago. Jacob's age was seven times that of his son. What are their pereenk skes? 3.4.2 Elimination Method Now let us consider another method of eliminating (i.e. removing) whe rakthbl this is sometimes more convenient tham the substitution method. I er us wee how this mhehi works. Views: 5,747 P on the ground the angle of elevation of the top of a 10 m tall building is 3 0 ∘. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 4 5 ∘. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 ​ = 1.732 ) Views: 5,926 P = and B = { Positive od 2022-23 54 (iv) The taxi charges in a city consist of a fixed charge together with the charge fix the distance covered. For a distance of 10 km, the charge paid is ₹100 and fix a journey of 15 km, the charge paid is ₹155. What are the fixed ...