Q1. Make formula for each of the following statements:
(i) The reciprocal of the focal lenght f, is equal to the sum of the reciprocals of the object distance u and image distance v.
(ii) The number of diagonals d, that can be drawn from one vertex of an n-sided polygon to all the other vertices is equal to the number of sides less 3.
(iii) The distance s metres, which a falling body covers in time t seconds, is 4.8 times the square of the time t.
(iv) The mean M of the five quantities a, b, c, d and e is equal to their sum divided by the number of quantities.
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(i) Reciprocal of the focal length f = $\dfrac{1}{\text{f}}$
$\Rightarrow$ Reciprocal of object distance u = $\dfrac{1}{\text{u}}$
$\Rightarrow$ Reciprocal of image distance v = $\dfrac{1}{\text{v}}$
$\therefore \dfrac{1}{\text{f}} = \dfrac{1}{\text{u}} + \dfrac{1}{\text{v}}$ Ans
(ii) $\text{d} = \text{n} - 3$ Ans
(iii) $\text{S} = 4.8\text{t}^{2}$ Ans
(iv) $\text{M} = \dfrac{\text{a} + \text{b} + \text{c} + \text{d} + \text{e}}{5}$ Ans
$\Rightarrow$ Reciprocal of object distance u = $\dfrac{1}{\text{u}}$
$\Rightarrow$ Reciprocal of image distance v = $\dfrac{1}{\text{v}}$
$\therefore \dfrac{1}{\text{f}} = \dfrac{1}{\text{u}} + \dfrac{1}{\text{v}}$ Ans
(ii) $\text{d} = \text{n} - 3$ Ans
(iii) $\text{S} = 4.8\text{t}^{2}$ Ans
(iv) $\text{M} = \dfrac{\text{a} + \text{b} + \text{c} + \text{d} + \text{e}}{5}$ Ans
Q2. A bus is carrying x children. If each of y children pays Rs.2.50 and each of the remaining pays Rs.5.25, find the total collection "C" in rupees.
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$\Rightarrow$ Money paid by y children = 2.50y
$\Rightarrow$ Money paid by x $-$ y children = 5.25(x $-$ y)
$\Rightarrow$ 5.25(x $-$ y) = 5.25x $-$ 5.25y
$\Rightarrow$ Collection 'C' = 2.50y $+$ 5.25x $-$ 5.25y
$\Rightarrow$ 2.50y $+$ 5.25x $-$ 5.25y = 5.25x $-$ (5.25 $+$ 2.50)y
$\Rightarrow$ 5.25x $-$ (5.25 $+$ 2.50)y = 5.25x $-$ 5.25y $-$ 2.50y
$\therefore$ 5.25x $-$ 2.75y Ans
$\Rightarrow$ Money paid by x $-$ y children = 5.25(x $-$ y)
$\Rightarrow$ 5.25(x $-$ y) = 5.25x $-$ 5.25y
$\Rightarrow$ Collection 'C' = 2.50y $+$ 5.25x $-$ 5.25y
$\Rightarrow$ 2.50y $+$ 5.25x $-$ 5.25y = 5.25x $-$ (5.25 $+$ 2.50)y
$\Rightarrow$ 5.25x $-$ (5.25 $+$ 2.50)y = 5.25x $-$ 5.25y $-$ 2.50y
$\therefore$ 5.25x $-$ 2.75y Ans
Q3. A worker is engaged for 100 days, on the condition that he will be paid Rs.x per day for each day he works and will be charged Rs.y per day for each day he is absent. Frame a formula to find his earning "E" in rupees, if he works for "d" days only.
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$\Rightarrow$ Money earned for 'd' days = dx
$\Rightarrow$ No. of days absent = (100 - d)
$\Rightarrow$ Money given for (100 - d) days = y(100 - d)
$\therefore$ E = dx - y(100 - d) Ans
$\Rightarrow$ No. of days absent = (100 - d)
$\Rightarrow$ Money given for (100 - d) days = y(100 - d)
$\therefore$ E = dx - y(100 - d) Ans
Q4. A shopkeeper buys x kg sugar for Rs.y and sells it at Rs.z per kg. Find the expression for his:
(i) Profit
(ii) Profit percent
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$\Rightarrow$ C.P. = xy
$\Rightarrow$ S.P. = xz
(i) Profit = xz $-$ xy Ans
(ii) Profit p.c. = $\dfrac{xz - xy}{xy} \times 100$ Ans
$\Rightarrow$ S.P. = xz
(i) Profit = xz $-$ xy Ans
(ii) Profit p.c. = $\dfrac{xz - xy}{xy} \times 100$ Ans
Q5. A shopkeeper buys 'a' kg of rice at Rs.x per kg and another 'b' kg of rice at Rs.y per kg. If he sells the mixture at Rs.z per kg, find the expression for his:
(i) Total profit
(ii) Profit percent
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$\Rightarrow$ C.P. = ax $+$ by
$\Rightarrow$ S.P. = (a $+$ b)z
(i) Total profit = (a $+$ b)z $-$ (ax $+$ by) Ans
(ii) Profit p.c. = $\dfrac{(\text{a} + \text{b})z - (\text{ax} + \text{by})}{\text{ax} + \text{by}} \times 100$ Ans
$\Rightarrow$ S.P. = (a $+$ b)z
(i) Total profit = (a $+$ b)z $-$ (ax $+$ by) Ans
(ii) Profit p.c. = $\dfrac{(\text{a} + \text{b})z - (\text{ax} + \text{by})}{\text{ax} + \text{by}} \times 100$ Ans
Q6. A man buys 'm' articles at Rs.x each and another n articles for Rs.y. If he sells all the articles at Rs.z per article, frame an equation to find his total profit.
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$\Rightarrow$ C.P. = mx $+$ ny
$\Rightarrow$ S.P. = (m $+$ n)z
$\therefore$ Profit = (m $+$ n)z $-$ (mx $+$ ny) Ans
$\Rightarrow$ S.P. = (m $+$ n)z
$\therefore$ Profit = (m $+$ n)z $-$ (mx $+$ ny) Ans
Q7. During a certain month, a firm posted x letters with Rs.5 stamps on each and another y letters with Rs.3.50 stamps on each. Obtain an expression to find the total money, Rs.M, spent on postage.
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$\Rightarrow$ Total price of x letters = 5x
$\Rightarrow$ Total price of y letters = 3.50y
$\therefore$ M = (5x $+$ 3.50y) Ans
$\Rightarrow$ Total price of y letters = 3.50y
$\therefore$ M = (5x $+$ 3.50y) Ans
Q8. Find the total money M in rupees, in a purse, if the purse contains 2x notes of Rs.5 each, 3y notes of Rs.2 each, 6z coins of Rs.1 each and 8r coins of 50 paise each.
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$\Rightarrow$ Total amount 2x notes = (2x $\times$ 5) = 10x
$\Rightarrow$ Total amount 3y notes = (3y $\times$ 2) = 6y
$\Rightarrow$ Total amount 6z coins = (6z $\times$ 1) = 6z
$\Rightarrow$ Total amount 8r coins = (8r $\times \dfrac{50}{100}$) = 6z
$\Rightarrow \text{8r} \times \dfrac{50}{100} = \text{8r} \times \dfrac{1}{2} = \text{4r}$
$\therefore \text{M} = 10\text{x} + 6\text{y} + 6\text{z} + 4\text{r}$ Ans
$\Rightarrow$ Total amount 3y notes = (3y $\times$ 2) = 6y
$\Rightarrow$ Total amount 6z coins = (6z $\times$ 1) = 6z
$\Rightarrow$ Total amount 8r coins = (8r $\times \dfrac{50}{100}$) = 6z
$\Rightarrow \text{8r} \times \dfrac{50}{100} = \text{8r} \times \dfrac{1}{2} = \text{4r}$
$\therefore \text{M} = 10\text{x} + 6\text{y} + 6\text{z} + 4\text{r}$ Ans
Q9. A worker in a factory is paid Rs.x per hour for the normal work and double this rate for the overtime work. Find the total earning "E", in rupees of a worker who works for 20 hours out of which y hours is the overtime.
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$\Rightarrow$ Normal time = 20 $-$ y
$\Rightarrow$ Earning for normal time = x(20 $-$ y)
$\Rightarrow$ Earning for overtime = 2xy
$\therefore$ E = x(20 $-$ y) $+$ 2xy Ans
$\Rightarrow$ Earning for normal time = x(20 $-$ y)
$\Rightarrow$ Earning for overtime = 2xy
$\therefore$ E = x(20 $-$ y) $+$ 2xy Ans
Q10. Frame a formula for each of the following:
(i) Find the number of hours h in x days and y hours.
(ii) Frame a formula for finding the number of students n that may be seated in a class room in which there are S single seats and D double seats.
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(i) h = 24x $+$ y Ans
(ii) n = S $+$ 2D Ans
(ii) n = S $+$ 2D Ans
Q11. If Rs.2b are equally distributed among 3n boys, how many paise did each get?
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$\Rightarrow$ Given, total money = Rs. 2b
$\Rightarrow$ Given, total no. of days = $\dfrac{2\text{b}}{3\text{n}}$
$\therefore$ Share of 1 boy in paise = $\dfrac{2\text{b}}{3\text{n}} \times 100 = \dfrac{200\text{b}}{3\text{n}}$ Ans
$\Rightarrow$ Given, total no. of days = $\dfrac{2\text{b}}{3\text{n}}$
$\therefore$ Share of 1 boy in paise = $\dfrac{2\text{b}}{3\text{n}} \times 100 = \dfrac{200\text{b}}{3\text{n}}$ Ans
Q12. A man bough 2a articles at 3p paise each and sold them at 2q paise each. Find his profit P in rupees.
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$\Rightarrow$ C.P. of 2a articles = 3p
$\therefore$ Total C.P. = 2a $\times$ 3p = 6ap
$\Rightarrow$ S.P. of 1 article = 2q
$\therefore$ Total S.P. = 2a $\times$ 2q = 4aq
As such, profit = $\dfrac{4\text{aq} - 6\text{ap}}{100}$ Ans
$\therefore$ Total C.P. = 2a $\times$ 3p = 6ap
$\Rightarrow$ S.P. of 1 article = 2q
$\therefore$ Total S.P. = 2a $\times$ 2q = 4aq
As such, profit = $\dfrac{4\text{aq} - 6\text{ap}}{100}$ Ans
Q13, Find the number of kilometers k, walked by a man in 75 minutes at the rate of x. per hour.
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$\Rightarrow$ Given time = 75 minutes
$\Rightarrow$ 75 minutes = $\dfrac{75}{60} = \dfrac{5}{4} = 1.25$ hrs.
$\Rightarrow$ Given speed = x km/hr
$\therefore$ Required distance = 1.25 $\times$ x = 1.25x Ans
$\Rightarrow$ 75 minutes = $\dfrac{75}{60} = \dfrac{5}{4} = 1.25$ hrs.
$\Rightarrow$ Given speed = x km/hr
$\therefore$ Required distance = 1.25 $\times$ x = 1.25x Ans
Q14. A boy has p rupees. If he buys a articles of b rupees each, how much is left with him?
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$\Rightarrow$ Given, total money with the boy = Rs. p
$\Rightarrow$ Price of 1 article = Rs. b
$\Rightarrow$ Price of a articles = (b $\times$ a) = Rs.ab
$\therefore$ Money left with boy = Rs. (p $-$ ab) Ans
$\Rightarrow$ Price of 1 article = Rs. b
$\Rightarrow$ Price of a articles = (b $\times$ a) = Rs.ab
$\therefore$ Money left with boy = Rs. (p $-$ ab) Ans
Q15. A car can go k kms on l litres of petrol. How far can the car go on p litres?
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$\Rightarrow$ Kilometers covered by car on 'l' litres petrol = k
$\Rightarrow$ Kilometers covered by car on 1 litre petrol = $\dfrac{\text{k}}{l}$
$\Rightarrow$ Kilometers covered by car on 1 litre petrol = $\dfrac{\text{k}}{l} \times \text{p} = \dfrac{\text{kp}}{l}$ Ans
$\Rightarrow$ Kilometers covered by car on 1 litre petrol = $\dfrac{\text{k}}{l}$
$\Rightarrow$ Kilometers covered by car on 1 litre petrol = $\dfrac{\text{k}}{l} \times \text{p} = \dfrac{\text{kp}}{l}$ Ans