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

Given

$\displaystyle 168=n(n+2)$

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

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

Given

$\displaystyle 420 = 16t^2+4t$

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

Hence $\displaystyle t = 5 \text{ sec }$

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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 $\displaystyle xy$

Given

$\displaystyle x \times y = 24$

$\displaystyle y = x+2$

Therefore

$\displaystyle x(x+2) = 24$

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

Hence $\displaystyle y = 6$

Therefore the number is $\displaystyle 46$

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Question 4: The age of the 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 $\displaystyle = n$

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

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

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

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 $\displaystyle = x \text{ years }$

Let the age of son $\displaystyle = y \text{ years }$

Given

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

$\displaystyle x=y^2$

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

$\displaystyle y^2-8y+7 = 0 \Rightarrow y = 7 \text{ or } 1$ (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 $\displaystyle = x \text{ years }$

Let the age of son $\displaystyle = y \text{ years }$

Given

$\displaystyle x=2y^2$

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

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

$\displaystyle 2y^2-3y-20=0 \Rightarrow y = 4 \text{ or } -2.5$ (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 $\displaystyle = 15 \text{ km/hr }$

Let the speed of the stream $\displaystyle = x \text{ km/hr }$

Therefore,

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

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

$\displaystyle x^2 = 25 \Rightarrow x = 5 \text{ or } -5$ (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 $\displaystyle = x \text{ Rs. }$

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

But the servant eats 3 oranges

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

Therefore

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

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

$\displaystyle x^2-3x-180=0 \Rightarrow x = 15 \text{ or } -12$ (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 $\displaystyle = x$

Therefore

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

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

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

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 $\displaystyle = x$

Let the weekly wage of the worker $\displaystyle = y \text{ Rs. }$

Given

$\displaystyle xy = 3150 ... ... ... ... i)$

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

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

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

Solving i) and ii)

$\displaystyle y(y+5)=3150$

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

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 $\displaystyle \text{ } = x$

$\displaystyle \text{Total cost price } = 1200 \text{ Rs. }$

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

$\displaystyle \text{Number of articles sold } = (x-10)$

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

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

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

$\displaystyle \text{ } (x-10)(1200+2x) = 1260x$

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

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

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

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