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.

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Let the number of articles \displaystyle  = x

Therefore

\displaystyle  x\times 100 - 4800 = \frac{4800}{x} \times 15

\displaystyle  100x^2-4800x-72000=0 \Rightarrow x = 60 or -12 \text{ (not possible) }

Therefore the number of articles bought is 60