Question 1: Frame a formula for each of the following statements:

(i) The area \displaystyle (A) of a rectangle is equal to the product of its length \displaystyle (l) and breadth \displaystyle (b)

\displaystyle A = l \times b  

(ii) The area \displaystyle (A) of a triangle is half the product of its base \displaystyle (b) and height \displaystyle (h)

\displaystyle A = \frac{1}{2} \times b \times h  

(iii) The volume \displaystyle (V) of the cone is one third the product of \displaystyle \pi , square of the radius \displaystyle (r) and height \displaystyle (h)

\displaystyle V = \frac{1}{3} \pi r^2 h  

(iv) The perimeter \displaystyle (P) of the rectangle is twice the sum of its length \displaystyle (l) and breadth \displaystyle (b)

\displaystyle P = 2(l+b)  

(v) The perimeter \displaystyle (P) of a square is four times it size \displaystyle (s)

\displaystyle P = 4s  

(vi) The distance \displaystyle (s) through which the body falls freely under gravity is \displaystyle 4.81 times the square of the time \displaystyle (t)

\displaystyle S = 4.81 t^2  

(vii) The reciprocal of focal length \displaystyle (f) is equal to the sum of reciprocals of the object distance \displaystyle (u) and the image distance \displaystyle (v)

\displaystyle \frac{1}{f} = \frac{1}{u} + \frac{1}{v}  

(viii) \displaystyle 2 years ago, a man whose present age is \displaystyle x years was three times as old as his son, whose present age is \displaystyle y years

\displaystyle (x-2) = 3(y -2)  

(ix) Two digit number having \displaystyle x as ten’s digit and \displaystyle y as units’s digit is \displaystyle 5 times the sum of the digits

\displaystyle 10x + y = 5(x+y)  

(x) The number of diagonals \displaystyle (d) that can be drawn from one vertex of an \displaystyle n -side polygon to all other vertices is \displaystyle 3 less than \displaystyle n

\displaystyle d = (n-3)  

\displaystyle \\

Question 2: A workman is paid \displaystyle Rs. \ x for each day he works and fined \displaystyle Rs. \ y for each day he is absent. If he works for \displaystyle N days in a month of \displaystyle 30 days, find the expression of his total earnings \displaystyle (E) in rupees.

Answer:

\displaystyle E = N \times x -(30-N) \times y

\displaystyle \\

Question 3: A purse contains \displaystyle x notes of \displaystyle Rs. 10 each, \displaystyle y notes of \displaystyle Rs. 5 each, \displaystyle z coins of \displaystyle 50 paisa each and \displaystyle t coins of \displaystyle 5 paisa each. Find the total money \displaystyle M in \displaystyle Rs .

Answer:

\displaystyle M = 10x + 5y + \frac{1}{2} z +  \frac{1}{20} t

\displaystyle \\

Question 4: A shopkeeper buys \displaystyle m kg. of rice at \displaystyle Rs. \ x per kg and another \displaystyle n kg of rice at \displaystyle Rs. \ y per kg. He mixes the two quantities and sells the mixture at \displaystyle Rs. \ z per kg. Find the expression of his (i) total profit and (ii) profit per cent.

Answer:

Total cost \displaystyle = (mx + ny)

Total Sale \displaystyle = (m+n)z

\displaystyle \text{(i) Total profit  } = (m+n)z - (mx + ny)

\displaystyle \text{ (ii) Profit percentage  } =  \{ \frac{(m+n)z - (mx + ny)}{mx + ny} \times 100  \}  

\displaystyle \\

Question 5: A man cycles for \displaystyle p hours at \displaystyle x km per hour and for another \displaystyle q hours at \displaystyle y km per hour. Find his average speed \displaystyle (A) for the whole journey.

Answer:

\displaystyle A =  \frac{px+qy}{p+q}  

\displaystyle \\

Question 6: The average age of \displaystyle x boys in a class is \displaystyle y years. A new boy of age \displaystyle z years joins the class. Find the present average age \displaystyle (A) .

Answer:

\displaystyle A =  \frac{xy+z}{x+1}  

\displaystyle \\

Question 7: A cricketer has an average score of \displaystyle 85 runs per innings in \displaystyle x innings and an average of \displaystyle 63 runs per innings in \displaystyle y innings. Find the average score \displaystyle (A) per innings.

Answer:

\displaystyle A =  \frac{85x+63y}{x+y}  

\displaystyle \\

Question 8: In a class of \displaystyle x children, each one of y children pays \displaystyle Rs. \ 10 and each of the remaining pays \displaystyle Rs.\ 6 for a charity show. Find the total amount \displaystyle (C) in rupees.

Answer:

\displaystyle C = 10y +( x-y)\times 6 = 4y+6x

\displaystyle \\

Question 9: A shopkeeper marks each article at \displaystyle Rs.\ m and gives \displaystyle 20\% discount on the marked price. If the cost price of an article \displaystyle Rs. \ C , find the formula for profit \displaystyle (P) . Find P when \displaystyle m = 300 \text{ and } C = 200 .

Answer:

\displaystyle P = 0.8m - C

\displaystyle P = 0.8 \times 300 - 200 = 40

\displaystyle \\

Question 10: The hiring charges \displaystyle (h) of a taxi were \displaystyle Rs. \ 100 plus \displaystyle Rs.\ 8 per km for distances traveled beyond \displaystyle 25 km. If the distance traveled is \displaystyle x km, write a formula for the hiring charges.

Answer:

\displaystyle h = 100 + (x - 25) \times 8

\displaystyle \text{If }  h = 140 \Rightarrow x = 25 +  \frac{140 - 100}{8} = 30

\displaystyle \\

Question 11: Make \displaystyle b as a subject in the formula, \displaystyle P = 2 (l + b) . Find \displaystyle b , when \displaystyle P = 66 \text{ and } l = 18 .

Answer:

\displaystyle P = 2 (l + b)

\displaystyle \Rightarrow b = (  \frac{P}{2} - l)

\displaystyle b = (  \frac{66}{2} - 18) = 15

\displaystyle \\

Question 12: Given: \displaystyle s = ut -  \frac{1}{2} gt^2

(i) Make \displaystyle g , the subject of the formula

(ii) Find \displaystyle g , when \displaystyle t = 20, s = 10 \text{ and } u = 50 .

Answer:

\displaystyle \text{(i) } s = ut -  \frac{1}{2} gt^2

\displaystyle \Rightarrow  \frac{1}{2} gt^2 = ut - S

\displaystyle \Rightarrow g =  \frac{2}{t^2} (ut-S)

\displaystyle \text{(ii) } t = 20, s = 10 \text{ and } u = 50

\displaystyle g =  \frac{2}{10^2} (50 \times 10 -10) = 9.8

\displaystyle \\

Question 13: \displaystyle \text{ Given:  } T = 2 \pi  \sqrt{\frac{l}{g}}  

(i) Make \displaystyle l as the subject of the formula

(ii) Find \displaystyle l , when \displaystyle T = 2, g = 9.8 \text{ and } \pi = \sqrt{10}

Answer:

\displaystyle \text{(i) } T = 2 \pi  \sqrt{\frac{l}{g}}  

\displaystyle \Rightarrow T^2 = 4 \pi^2  \frac{l}{g}  

\displaystyle l =  \frac{gT^2}{4 \pi^2}  

\displaystyle \text{(ii) } \text{When   } T = 2, g = 9.8 \text{ and } \pi = \sqrt{10}

\displaystyle l =  \frac{9.8 \times 2^2}{4 (\sqrt{10})^2} = 0.98

\displaystyle \\

Question 14: Let \displaystyle S =  \frac{n}{2} \{2a + (n-1) d \}

(i) Find \displaystyle a when \displaystyle S = 185, n = 10 \text{ and } d = 3

(ii) Find \displaystyle d when \displaystyle a = 11, n = 10 \text{ and } S = 380

Answer:

\displaystyle \text{(i) } S =  \frac{n}{2} \{2a + (n-1) d \}

\displaystyle \Rightarrow  \frac{2S}{n} = 2a + (n-1)d

\displaystyle \Rightarrow a =  \frac{1}{2} \{ \frac{2S}{n} - (n-1)d  \}  

\displaystyle \Rightarrow a =  \frac{1}{2} \{ \frac{2 \times 185}{10} - (10-1) \times 3  \} = 5

\displaystyle \text{(ii) } S =  \frac{n}{2} \{2a + (n-1) d \}

\displaystyle \Rightarrow  \frac{2S}{n} = 2a + (n-1)d

\displaystyle d =  \frac{1}{n-1} \{ \frac{2S}{n} - 2a  \}  

\displaystyle d =  \frac{1}{10-1} \{ \frac{2 \times 380}{10} - 2 \times 11  \} = 6

\displaystyle \\

Question 15: Let \displaystyle v = \sqrt{u^2+2ax}

(i) Make \displaystyle x , the subject of the formula

(ii) Find \displaystyle x , when \displaystyle v = 35, u = 25 \text{ and } a = 50

Answer:

\displaystyle \text{(i) } v = \sqrt{u^2+2ax}

\displaystyle v^2 = u^2 + 2ax

\displaystyle \Rightarrow x =  \frac{1}{2a} (v^2 - u^2)

\displaystyle \text{(ii) } x =  \frac{1}{2 \times 50} (35^2 - 25^2) = 6

\displaystyle \\

Question 16: The volume \displaystyle (V) of a hollow cylindrical pipe with outer radius \displaystyle (R) , inner radius \displaystyle (r) and length \displaystyle (h) is given by the formula, \displaystyle V = \pi (R^2 - r^2)h .

(i) Make \displaystyle h the subject of the formula

(ii) Find \displaystyle h , when \displaystyle R = 2.6, r = 2.3, \pi =  \frac{22}{7} \text{ and } V = 115.5

Answer:

\displaystyle \text{(i) } V = \pi (R^2 - r^2)h

\displaystyle h =  \frac{V}{\pi (R^2 - r^2)}  

\displaystyle \text{(ii) } h =  \frac{115.5 \times 7}{22 \times (2.6^2 - 2.3^2)} = 25

\displaystyle \\

Question 17: Let \displaystyle x =  \frac{m+n}{m-n} ,

(i) Make \displaystyle n the subject of the formula

(ii) Find \displaystyle n , when \displaystyle m = 36 \text{ and } x = 2

Answer:

\displaystyle \text{(i) } x =  \frac{m+n}{m-n}  

\displaystyle mx-nx = m+ n

\displaystyle \Rightarrow m(x-1) = (x+1) n

\displaystyle \Rightarrow n =  (\frac{x-1}{x+1}) m

\displaystyle \text{(ii) } \Rightarrow n =  (\frac{2-1}{2+1}) \times 36 = 12

\displaystyle \\

Question 18: Let \displaystyle x =  \frac{3-4p}{p+2q} ,

(i) Make \displaystyle p the subject of the formula

(ii) Find \displaystyle p , when \displaystyle q =  \frac{1}{2} \text{ and } x =  \frac{1}{5}  

Answer:

\displaystyle \text{(i) } x =  \frac{3-4p}{p+2q}  

\displaystyle px + 2qx = 3 - 4p

\displaystyle px + 4p = 3 - 2qx

\displaystyle p =  \frac{3 -2qx}{x+4}  

\displaystyle \text{(ii) } p =  \frac{3 -2 \times \frac{1}{2} \times \frac{1}{5}}{\frac{1}{5}+4} =  \frac{2}{3}  

\displaystyle \\

Question 19: Let \displaystyle R =  \sqrt{\frac{3V}{\pi h}} ,

(i) Make \displaystyle h the subject formula

(ii) Find \displaystyle h , when \displaystyle V = 13.5, R = 2.5 \text{ and } \pi=\frac{22}{7}

Answer:

\displaystyle \text{(i) } R =  \sqrt{\frac{3V}{\pi h}}  

\displaystyle \pi R^2 h = 3V

\displaystyle \Rightarrow h =  \frac{3V}{\pi R^2}  

\displaystyle \text{(ii) } \Rightarrow h =  \frac{3 \times 13.5}{\pi 2.5^2} = 2.06

\displaystyle \\

Question 20: If \displaystyle \frac{1}{f} = \frac{1}{u} + \frac{1}{v} , find \displaystyle u \text{ in terms of } v \text{ and } f . Find \displaystyle u when \displaystyle v = 32 \text{ and } f = 24 .

Answer:

\displaystyle \frac{1}{f} = \frac{1}{u} + \frac{1}{v}  

\displaystyle \Rightarrow \frac{1}{u} = \frac{1}{f} - \frac{1}{v}  

\displaystyle \Rightarrow \frac{1}{u} = \frac{v-f}{fv}  

\displaystyle \Rightarrow u =  \frac{fv}{v-f}  

\displaystyle \Rightarrow u =  \frac{24 \times 32}{32-24} =96

\displaystyle \\

Question 21: If \displaystyle r = \sqrt{x^2 +y^2} , express \displaystyle y \text{ in terms of } r \text{ and } x . Find \displaystyle y , when \displaystyle r = 17 \text{ and } x = 8 .

Answer:

\displaystyle r = \sqrt{x^2 +y^2}

\displaystyle y^2 = r^2 - x^2

\displaystyle y = \sqrt{r^2 - x^2}

\displaystyle y = \sqrt{17^2 - 8^2} = 15

\displaystyle \\

Question 22: Make \displaystyle b the subject of \displaystyle x =  \sqrt{\frac{a-b}{a+b}}  

Answer:

\displaystyle x =  \sqrt{\frac{a-b}{a+b}}  

\displaystyle x^2 =  \frac{a-b}{a+b}  

\displaystyle ax^2+bx^2 = a- b

\displaystyle b(x^2 + 1) = a (1 - x^2)

\displaystyle \Rightarrow b = \frac{a(1-x^2)}{1+ x^2}