Question 1: The sum $S \ of \ n$ successive odd numbers starting from 3 is given by the relation: $S=n(n+2)$. Determine $n$, if the sum is 168.

Given

$168=n(n+2)$

$n^2+2n-168=0 \Rightarrow n = 12 or -14 \ (not \ possible)$

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Question 2: A stone is thrown vertically downwards and the formula $d=16t^2+4t$ gives the distance, d meters, that is falls in t seconds. How long does it take to fall 420 meters?

Given

$420 = 16t^2+4t$

$16t^2+4t-420 = 0 \Rightarrow t = 5 or - 5.25 \ (not \ possible)$

Hence $t = 5 \ sec$

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Question 3: The product of the digits of a two-digit number is 24. If its unit’s digit exceeds twice its ten’s digit by 2; find the number.

Let the number be $xy$

Given

$x \times y = 24$

$y = x+2$

Therefore

$x(x+2) = 24$

$x^2+2x-24 = 0 \Rightarrow x = 4 or -6 \ (not \ possible)$

Hence $y = 6$

Therefore the number is $46$

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Question 4: The age of two sisters are 11 years and 14 years. In how many years’ time will the product of their ages be 304?

Let the number of years $= n$

Therefore $(11+n)(14+n) = 304$

$154+25n+n^2=304$

$n^2+25n-150=0 \Rightarrow n= 5, -30 \ (not \ possible)$

Hence the product of their ages be 304 in 5 years.

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Question 5: One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present ages.

Let the age of father $= x \ years$

Let the age of son $= y \ years$

Given

$(x-1) = 8(y-1)$

$x=y^2$

Therefore $(y^2-1) = 8(y-1)$

$y^2-8y+7 = 0 \Rightarrow y = 7 or 1 \ (not \ possible)$

Therefore son’s age is 7 years and father’s age is 49 years

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Question 6: The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

Let the age of father $= x \ years$

Let the age of son $= y \ years$

Given

$x=2y^2$

$x+8 = 3(y+8) + 4$

$2y^2+8 = 3y+28$

$2y^2-3y-20=0 \Rightarrow y = 4 or -2.5 \ (not \ possible)$

Therefore Son’s age is 4 years and Father’s ages is 32 years.

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Question 7: The speed of a boat in still water is 15 km/hr. it can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.

Speed of the boat $= 15 km/hr$

Let the speed of the stream $= x km/hr$

Therefore,

$\frac{30}{15-x}+\frac{30}{15+x}= \frac{9}{2}$

$10(15+x)+10(15-x) = \frac{3}{2}(225-x^2)$

$x^2 = 25 \Rightarrow x = 5 \ or \ -5 \ (not \ possible)$

Hence the speed of the steam is 5 \ km/hr \$

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Question 8: A person sends his servant to the market to buy oranges worth Rs.15. The servant having eaten three oranges on the way, the person pays 25 paisa per orange more than a market price. Find the number of oranges received back by the person.

Let the market price of the orange  $= x Rs.$

Number of oranges bought by the servant $= \frac{15}{x}$

But the servant eats 3 oranges

Therefore the real cost of each orange for this person $= \frac{15}{x-3}$

Therefore

$\frac{15}{x-3} -\frac{15}{x}= \frac{1}{4}$

$15x-15(x-3) = \frac{1}{4}v(x^2-3x)$

$x^2-3x-180=0 \Rightarrow x = 15 or -12 \ (not \ possible)$

Hence the person receives 12 oranges

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Question 9: 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paisa less. Find the number of children.

Let the number of children $= x$

Therefore

$\frac{250}{x} - \frac{250}{x+25} = \frac{1}{2}$

$250x+6250-250x=\frac{1}{2} (x^2+25x)$

$x^2+25x-12500=0 \Rightarrow x = 100 or -125 \ (not \ possible)$

Hence the number of children is 100

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Question 10: An employer finds that if he increases the weekly wages of each worker more by Rs. 5 and employs five workers less, he increases his weekly wage bill from Rs. 3,150 to Rs. 3,250. Find the weekly wages of the workers.

Let the number of workers $= x$

Let the weekly wage of the worker $= y Rs.$

Given

$xy = 3150 ... ... ... ... i)$

$(x-5)(y+5) = 3250$

$xy-5y+5x-25=3250$

or $x-y = 25 ... ... ... ... ii)$

Solving i) and ii)

$y(y+5)=3150$

$y^2+25y-3150 = 0 \Rightarrow 45 or -70 \ (not \ possible)$

Hence the weekly wage of the workers is 45 Rs.

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Question 11: A trader bought a number of articles for Rs.1,200. Ten were damaged and he sold each of the remaining articles at Rs.2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction.

Let the number of articles bought $= x$

Total cost price $= 1200 Rs.$

Cost of one article $= \frac{1200}{x}$

Number of articles sold $= (x-10)$

Selling price of one article $= \frac{1200}{x} + 2$

Total Selling Price $= (x-10) \times (\frac{1200}{x} + 2)$

Therefore $(x-10) \times (\frac{1200}{x} + 2) - 1200 = 60$

$(x-10)(1200+2x) = 1260x$

$1200x-12000+2x^2-20x=1260x$

$2x^2-80x-12000=0 \Rightarrow x = 100 or -60 \ (not \ possible)$

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Question 12: The total cost price of a certain number of identical article is Rs.4800. by selling the article Rs.100 each, a profit equal to the cost price of 15 article is made. Find the number of articles bought.

Let the number of articles $= x$
$x\times 100 - 4800 = \frac{4800}{x} \times 15$
$100x^2-4800x-72000=0 \Rightarrow x = 60 or -12 \ (not \ possible)$