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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.

Answer:

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?

Answer:

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.

Answer:

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?

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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 &s=1$

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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  = 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.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

Let the number of articles = x  

Therefore

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

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

Therefore the number of articles bought is 60.

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