Q16. A man can run at the rate of m metres per second.
(i) How many minutes will he take to go 4 kms.
(ii) How many metres will he run in 3 hours?
(iii) How many kms will he run in h hours?
Show Answer
(i) Given, speed = $\dfrac{\text{m}}{\text{sec}}$
$\because$ Time = $\dfrac{\text{Distance}}{\text{Speed}}$
$\Rightarrow$ Here, 4 kms = 4000 metres
$\Rightarrow$ And, 1 minute = 60 seconds
$\therefore$ Times taken to cover 4 kms = $\dfrac{4000}{60} = \dfrac{200}{3}$ Minutes Ans
(ii) New distance = 4000 metres
$\Rightarrow$ New time = $\dfrac{200}{3}$
$\therefore$ New speed = $\dfrac{4000 \times 3}{200} = 20 \times 3 = 60$ metres/minutes
$\Rightarrow$ As such, metres covered in 3 hrs = $60(3 \times 60) = 60 \times 180 = 10800$ metres Ans
(iii) New distance = 4000 metres
$\Rightarrow$ Converting 10800 metres into kilometres = $\dfrac{10800}{1000} = \dfrac{54}{5}$ kms
$\Rightarrow$ Now, in 3 hrs a man can walk = $\dfrac{54}{5}$ kms
$\Rightarrow$ Now, in 1 hr a man can walk = $\dfrac{54}{5 \times 3} = \dfrac{18}{5}$ kms
$\therefore$ In 'h' hrs a man can walk = $\dfrac{18}{5} \times h = 3.6$h kms Ans
$\because$ Time = $\dfrac{\text{Distance}}{\text{Speed}}$
$\Rightarrow$ Here, 4 kms = 4000 metres
$\Rightarrow$ And, 1 minute = 60 seconds
$\therefore$ Times taken to cover 4 kms = $\dfrac{4000}{60} = \dfrac{200}{3}$ Minutes Ans
(ii) New distance = 4000 metres
$\Rightarrow$ New time = $\dfrac{200}{3}$
$\therefore$ New speed = $\dfrac{4000 \times 3}{200} = 20 \times 3 = 60$ metres/minutes
$\Rightarrow$ As such, metres covered in 3 hrs = $60(3 \times 60) = 60 \times 180 = 10800$ metres Ans
(iii) New distance = 4000 metres
$\Rightarrow$ Converting 10800 metres into kilometres = $\dfrac{10800}{1000} = \dfrac{54}{5}$ kms
$\Rightarrow$ Now, in 3 hrs a man can walk = $\dfrac{54}{5}$ kms
$\Rightarrow$ Now, in 1 hr a man can walk = $\dfrac{54}{5 \times 3} = \dfrac{18}{5}$ kms
$\therefore$ In 'h' hrs a man can walk = $\dfrac{18}{5} \times h = 3.6$h kms Ans
Q17. A man can take f sheep to their market and return with g sheep. How many sheep has he sold?
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$\Rightarrow$ No. of sheep sold = (f $-$ g) Ans
Q18. How old will a boy be in 15 years, if he was x years old 4 years ago?
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$\Rightarrow$ Required age = $15 + (x + 4) = x + 19$ Ans
Q19. A man earns Rs.32 per day and boy earns Rs.15 per day. At this rate, find the weekly earning of:
(i) 3 men and 5 boys
(ii) $x$ men and $y$ boys
Show Answer
$\Rightarrow$ Earning of 1 man per day = Rs.32
$\Rightarrow$ Earning of 3 men per day = $32 \times 3 = \text{Rs.}96$
$\Rightarrow$ Earning of 1 boy per day = Rs.15
$\Rightarrow$ Earning of 5 boys per day = $15 \times 5 = \text{Rs.}75$
(i) 1 week's earning of 3 men and 5 boys = $7(96 + 75)$
$\Rightarrow 672 + 525 = \text{Rs.}1197$ Ans
(ii) 1 week's earning of x men and y boys = $7(32x + 75y)$ Ans
$\Rightarrow$ Earning of 3 men per day = $32 \times 3 = \text{Rs.}96$
$\Rightarrow$ Earning of 1 boy per day = Rs.15
$\Rightarrow$ Earning of 5 boys per day = $15 \times 5 = \text{Rs.}75$
(i) 1 week's earning of 3 men and 5 boys = $7(96 + 75)$
$\Rightarrow 672 + 525 = \text{Rs.}1197$ Ans
(ii) 1 week's earning of x men and y boys = $7(32x + 75y)$ Ans
Q20. Out of m matches played by a team 20 were, 10 were draw and remaining were lost. If the team gets two points for a win nothing for a draw and minus two for a loss: express the result fo the team.
Show Answer
$\Rightarrow$ No. of matches won = 20
$\therefore$ Points earned = $20 \times 2 = 40$
$\Rightarrow$ No. of matches not lost = $20 + 10 = 30$ (Wins $+$ Draws)
$\therefore$ No. of matches lost = $\text{m} - 30$
$\Rightarrow$ As such, points lost = $-2(\text{m} - 30) = -2\text{m} + 60$
$\therefore$ Match result = $40 - 2\text{m} + 60$
$\Rightarrow (60 + 40) - 2\text{m} = 100 - 2\text{m}$ Ans
$\therefore$ Points earned = $20 \times 2 = 40$
$\Rightarrow$ No. of matches not lost = $20 + 10 = 30$ (Wins $+$ Draws)
$\therefore$ No. of matches lost = $\text{m} - 30$
$\Rightarrow$ As such, points lost = $-2(\text{m} - 30) = -2\text{m} + 60$
$\therefore$ Match result = $40 - 2\text{m} + 60$
$\Rightarrow (60 + 40) - 2\text{m} = 100 - 2\text{m}$ Ans
Q21. Out of a collection of 6a stamps, 2b stamps are lost. The remaining stamps are divided equally among 10 boys. Find the share of each.
Show Answer
$\Rightarrow$ Remaining no. of stamps = 6a $-$ 2b
$\Rightarrow$ No. of boys = 10
$\therefore$ Share of 1 boy = $\dfrac{6\text{a} - 2\text{b}}{10}$
$\Rightarrow \dfrac{6\text{a} - 2\text{b}}{10} = \dfrac{3\text{a} - \text{b}}{5}$ (Upon dividing by 2) Ans
$\Rightarrow$ No. of boys = 10
$\therefore$ Share of 1 boy = $\dfrac{6\text{a} - 2\text{b}}{10}$
$\Rightarrow \dfrac{6\text{a} - 2\text{b}}{10} = \dfrac{3\text{a} - \text{b}}{5}$ (Upon dividing by 2) Ans
Q22. x men can do a peice of work in 2d days. How many men will be required to complete the same work in 7 days?
Show Answer
$\Rightarrow$ No. of men required to complete the work in 2d days = x
$\Rightarrow$ No. of men required to complete the work in 1 days = 2d(x)
$\therefore$ No. of men required to complete the work in 7 days = $\dfrac{2\text{d}(\text{x})}{7}$ days Ans\\\\\
$\Rightarrow$ No. of men required to complete the work in 1 days = 2d(x)
$\therefore$ No. of men required to complete the work in 7 days = $\dfrac{2\text{d}(\text{x})}{7}$ days Ans\\\\\
Q23. How many minutes are there in x hours, y minutes and z seconds?
Show Answer
$\Rightarrow$ We have = $\dfrac{\text{x}}{60} + \dfrac{60\text{y}}{60} + \dfrac{60\text{z}}{60}$
$\therefore \dfrac{\text{x}}{60} + \text{y} + \dfrac{\text{z}}{60}$ Ans
$\therefore \dfrac{\text{x}}{60} + \text{y} + \dfrac{\text{z}}{60}$ Ans
Q24. A man in withdrawing his monthly salary gets x notes of 100 rupees each, y notes of 50 rupees each, z notes of 10 rupees and 5 notes of 5 rupees each. What is the monthly salary of the man?
Show Answer
$\Rightarrow$ Monthly salary = [100x $+$ 50y $+$ 10z $+$ 25]
$\therefore$ We have = 5[20x $+$ 10y $+$ 2z $+$ 5] Ans
$\therefore$ We have = 5[20x $+$ 10y $+$ 2z $+$ 5] Ans
Q25. The average of m numbers is P and average of n other numbers is Q, find an expression for the average of $(m + n)$ numbers.
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$\Rightarrow$ We have = $\dfrac{\text{mP + \text{nQ}}}{\text{m} + \text{n}}$ Ans
Q26. x articles are bought at 25 paise each and y articles are bought at 40 paise each. Find the total cost of articles in rupees.
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$\Rightarrow \dfrac{25\text{p} + 40\text{p}}{100} = \dfrac{5\text{p} + 8\text{p}}{20}$ (Upon dividing by 5) Ans
Q27. A labourer is engaged for 40 days on condition that he recieves Rs.25. for each day he works and loses Rs.8 for each day he is absent. How much does he recieves if he remains absent for x days?
Show Answer
$\Rightarrow$ Income earned = 25(40 $-$ x)
$\Rightarrow$ Income unearned = 8x
$\therefore [40(25) - 25\text{x} - 8\text{x}] = \text{Rs.}(1000 - 33\text{x})$ Ans
$\Rightarrow$ Income unearned = 8x
$\therefore [40(25) - 25\text{x} - 8\text{x}] = \text{Rs.}(1000 - 33\text{x})$ Ans
Q28. A man buys P stamps of denomination 10 paise each and Q stamps of denomination 25 paise each. Find an expression for the total cost of stamps in rupees.
Show Answer
$\Rightarrow \dfrac{10\text{P} + 25\text{Q}}{100} = \dfrac{2\text{P} + 5\text{Q}}{20}$ (upon dividing by 5) Ans
Change the subject of formulae for the indicated letter (Q 29-68):
Q29. F = $\dfrac{9}{5}\text{C} + 32$; for C
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$\Rightarrow \text{F} = \dfrac{9}{5}\text{C} + 32$
$\Rightarrow \text{F} - 32 = \dfrac{9}{5}\text{C}$
$\Rightarrow \dfrac{5}{9}\text{F} - 32 = \text{C}$
$\therefore \text{C} = \dfrac{5}{9}\text{F} - 32$ Ans
$\Rightarrow \text{F} - 32 = \dfrac{9}{5}\text{C}$
$\Rightarrow \dfrac{5}{9}\text{F} - 32 = \text{C}$
$\therefore \text{C} = \dfrac{5}{9}\text{F} - 32$ Ans
Q30. A = $\text{P}(1 + \text{rt})$; for t
Show Answer
$\Rightarrow$ A = $\text{P}(1 + \text{rt})$
$\Rightarrow$ A = P $+$ Prt
$\Rightarrow$ A $-$ P = Prt
$\therefore$ t = $\dfrac{\text{A} - \text{P}}{\text{Pr}}$ Ans
$\Rightarrow$ A = P $+$ Prt
$\Rightarrow$ A $-$ P = Prt
$\therefore$ t = $\dfrac{\text{A} - \text{P}}{\text{Pr}}$ Ans