Problems Based on Articles and their cost

Question 4: 5 pens and 6 pencils together cost Rs. \ 9 and 3 pens and 2 pencils cost Rs. \ 5 . Find the cost of one pen and one pencil.

Answer:

Let the cost of a pen = x \ Rs . and the cost of a pencil = y \ Rs .

Therefore given:

5x+6y = 9 … … … … … (i)

3x+2y = 5   … … … … … (ii)

Now multiplying equation( (i) by 3 and equation (ii) by 5   we get

15x+18y=27   … … … … … (iii)

15x+10y = 25   … … … … … (iv)

Subtracting (iv) from (iii) we get

15x+18y=27

\underline{ (-) \hspace{0.5cm} 15x+10y = 25}  

8y = 2

\Rightarrow y = 0.25 \ Rs.

Substituting y = 0.25 Rs. in (i) we get

5x+6(0.25) = 9 \Rightarrow 5x = 7.50 \Rightarrow x = 1.50 \ Rs.

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Question 5: A person has pens and pencils with together are 40  in number.If she had 5 more pencils and 5 less pens, then the number of pencils would become 4 times the number of pens. Find the number of pens and pencils that the person had.

Answer:

Let the number of pens = x \ Rs . and the number of pencil = y \ Rs .

Therefore given:

x+y = 40 … … … … … (i)

4(x-5) = y+5 \Rightarrow 4x-y = 25   … … … … … (ii)

Now multiplying equation( (i) by 4    we get

4x+4y=160   … … … … … (iii)

Subtracting (i1) from (iii) we get

4x+4y=160

\underline{ (-) \hspace{0.5cm} 4x-y = 25}  

5y = 135

\Rightarrow y = 27 

Substituting y = 27  in (i) we get

x = 40 -(27) = 13

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Question 6: 5 books and 7 pens together cost Rs. \ 79 whereas 7 books and 5 pens together cost Rs. \ 77 . Find the cost of one books and two pens.

Answer:

Let the cost of a book = x \ Rs . and the cost of a pen = y \ Rs .

Therefore given:

5x+7y = 79 … … … … … (i)

7x+5y = 77   … … … … … (ii)

Now multiplying equation( (i) by 7 and equation (ii) by 5   we get

35x+49y=553   … … … … … (iii)

35x+25y = 385   … … … … … (iv)

Subtracting (iv) from (iii) we get

35x+49y=553

\underline{ (-) \hspace{0.5cm} 35x+25y = 385}  

24y = 168

\Rightarrow y = 7 \ Rs.

Substituting y = 7 Rs. in (i) we get

5x+7(7) = 79 \Rightarrow 5x = 30 \Rightarrow x = 6 \ Rs.

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Question 7: On selling a TV at 5\% gain and a fridge at 10\% gain, a shopkeeper gains Rs. \ 2000 . But if he sells the TV at 10\% gain and the fridge at 5\% loss, he gains Rs. \ 1500 on the transaction. Find the actual prices of the TV and the fridge.

Answer:

Let the cost of TV = x \ Rs . and the cost of Fridge = y \ Rs .

Therefore given:

0.05x+0.10y=2000 … … … … … (i)

0.10x-0.05y=1500   … … … … … (ii)

Now multiplying equation( (i) by 2    we get

0.20x-0.10y=3000    … … … … … (iii)

Adding (iii) from (i) we get

0.05x+0.10y=2000

\underline{ (-) \hspace{0.5cm} 0.20x-0.10y=3000}  

0.25x=5000

\Rightarrow x = 20000 \ Rs.

Substituting y = 20000 Rs. in (i) we get

0.05(20000)+0.10y=2000 \Rightarrow y = 10000 \ Rs.

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Question 8: A lending library has a fixed charge for the first three days and an additional charge for each additional day. Person A paid Rs. \ 27 for a book kept for 7 days  while Person B paid Rs. \ 21 for the book  kept for 5 days. Find the fixed charge and the charge for each additional day.

Answer:

Let the fixed cost = x \ Rs . and the cost per day = y \ Rs .

Therefore given:

x+(7-3)y = 27 \Rightarrow x+4y=27 … … … … … (i)

x+(5-3)y = 21 \Rightarrow x+2y=21   … … … … … (ii)

Subtracting (ii) from (i) we get

x+4y=27

\underline{ (-) \hspace{0.5cm} x+2y=21}  

2y=6

\Rightarrow y = 3 \ Rs.

Substituting y = 3 Rs. in (i) we get

x+4(3)=27 \Rightarrow x = 15 \ Rs.

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Problems Based on Numbers

Question 9: The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digits is 18 . Find the number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given: x+y = 8 … … … … … (i)

10y + x - (10x+y) = 18 \Rightarrow 9y - 9x = 18 \Rightarrow y-x = 2 … … … … … (ii)

Adding (i) and (ii) we get y = 5  \Rightarrow x = 8-5 = 3

Hence x = 3, y = 5 and the number is 53 .

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Question 10: The sum of two digit number and the number obtained by reversing the order of its digits is 121 , and the two digits differ by 3 . Find the number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given:

10y + x +  (10x+y) = 121 \Rightarrow 11x+11y = 121 \Rightarrow x+y = 11 … … … … … (i)

Also  x-y = \pm 3 … … … … … (ii)

Consider +ve sign

Solving x+y = 11 and x-y = 3 \Rightarrow x = 7 and y =4 and the number is 74

Consider -ve sign

Solving x+y = 11 and x-y = -3 \Rightarrow x = 4 and y =7 and the number is 47

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Question 11: The sum of two digit number and the number formed by interchanging its digits is 110 . If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the first number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given:

10y + x +  (10x+y) = 110 \Rightarrow 11x+11y = 110 \Rightarrow x+y = 10 … … … … … (i)

Also  (10y+x-10) = 4 + 5(x+y) \Rightarrow 5y-4x=14 … … … … … (ii)

Solving x+y = 10 and 5y-4x=14 \Rightarrow x = 4 and y =6 and the number is 64

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Question 12: The sum of two numbers is 8 . If their sum is 4 times their difference, find the numbers.

Answer:

Let the two numbers be x and y

Therefore x+y = 8 … … … … … (i)

x+y = 4(x-y) \Rightarrow 3x - 5y = 0 … … … … … (ii)

Now multiplying equation (i) by 5 and adding  (i) and (ii)  we get

5x+5y = 40

\underline{ (+) \hspace{0.5cm} 3x - 5y = 0}  

8x=40

Therefore we get  x = 5 and y = (8-5) = 3

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Question 13: The sum of the digits of a two digit number is 15 . The number obtained by reversing the order of the digits of the given number exceeds the given number by 9 . Find the given number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given:

10y + x + 9 = 10x + y \Rightarrow 9x-9y = 9 \Rightarrow x-y = 9 … … … … … (i)

Also  x+y = 15 … … … … … (ii)

Solving x+y = 15 and x-y = 9 we get  x =8 \ and \ y = 7 and the number is 78

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Question 14: The sum of two numbers is 1000 and the difference between their squares is 256000 . Find the numbers.

Answer:

Let the two numbers be x and y

Therefore x+y = 1000 … … … … … (i)

x^2-y^2  256000  \Rightarrow (x-y)(x+y) = 256000 \Rightarrow x-y = 256  … … … … … (ii)

Adding  (i) and (ii)  we get

x+y = 1000

\underline{ (+) \hspace{0.5cm} x-y = 256}  

2x=1256

Therefore we get  x = 628 and y = (1000-628) = 372

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Question 15: A 2 digit number is four times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given:

10y + x +  18 = 10x + y \Rightarrow 9x-9y = 18 \Rightarrow x-y = 2 … … … … … (i)

Also  10y + x  = 4(x+y) \Rightarrow x=2y \Rightarrow x-2y=0 … … … … … (ii)

Solving x-2y=0 and x-y = 2 we get  x =4 \ and \ y = 2 and the number is 24

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Question 16: A 2 digit number is such that the product of its digits is 20 . If 9 is added to the number, the digits interchange their places. Find the number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given:

10y + x +9 = 10x + y \Rightarrow 9x-9y = 9 \Rightarrow x-y = 1 … … … … … (i)

Also  xy  = 20 \Rightarrow y= \frac{20}{x} … … … … … (ii)

Substituting (ii) in (i) we get

x^2 - \frac{20}{x} = 1

\Rightarrow x^2 - x - 20 = 0 

\Rightarrow (x-5)(x+4) = 0 \Rightarrow x = 5 \ or \  x = -4 (not possible)

For x = 5 we get y = \frac{20}{5} = 4

We get  x =5 \ and \ y = 4 and the number is 45

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Question 17: 7 times two digit number is equal to 4 times the number obtained by reversing the digits. If the difference between the digits is 3 . Find the number.

Answer:

Let the digit in unit place is x and that in tenth place is y .

Therefore Number = 10y + x

Number formed by reversing the digits = 10x+ y

Given:

7(10y + x) = 4(10x + y) \Rightarrow 33x = 66y \Rightarrow x= 2y … … … … … (i)

Also  x-y=3 … … … … … (ii)

Solving x= 2y and x-y=3 we get  x =6 \ and \ y = 3 and the number is 36

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Problems Based on Fractions

Question 18: A fraction become \frac{4}{5} if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and denominator, the fraction becomes \frac{1}{2} . What is the fraction?

Answer:

Let the fraction = \frac{x}{y}

Therefore based on the given conditions:

\frac{x+1}{y+1} = \frac{4}{5} \Rightarrow 5x + 5 = 4y + 4 \Rightarrow 5x - 4y = -1 … … … … … (i)

and \frac{x-5}{y-5} = \frac{1}{2} \Rightarrow 2x-10 = y - 5 \Rightarrow 2x - y = 5   … … … … … (ii)

Multiplying (ii) by 4 and subtracting it from (i)  we get

5x-4y = -1

\underline{ (-) \hspace{0.5cm} 8x-4y=20}  

-3x = -21

Therefore x = 7 .

Substituting x = 7 in (i) we get 4y = 5(7)+1 \Rightarrow y = 9

Therefore the fraction is \frac{7}{9}

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Question 19: A denominator of a fraction is 4 more than twice the numerator. When both the numerator and denominator are decreased by 6 , the denominator becomes 12 times the numerator. Determine the fraction.

Answer:

Let the fraction = \frac{x}{y}

Therefore based on the given conditions:

y = 2x+ 4 \Rightarrow 2x-y = -4 … … … … … (i)

and y-6 = 12(x-6) \Rightarrow 12x - y = 66   … … … … … (ii)

Subtracting (ii) from (i) we get

2x-y = -4

\underline{ (-) \hspace{0.5cm} 12x - y = 66}  

-10x = -70

Therefore x = 7 .

Substituting x = 7 in (i) we get y = 2(7) + 4 = 18

Therefore the fraction is \frac{7}{18}

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Question 20: A fraction becomes \frac{1}{3} is 1 is subtracted from both its numerator and denominator. If 1 is added to both numerator and denominator, it becomes \frac{1}{2} . Find the fraction.

Answer:

Let the fraction = \frac{x}{y}

Therefore based on the given conditions:

\frac{x-1}{y-1} = \frac{1}{3} \Rightarrow 3x-3 = y-1 \Rightarrow 3x-y=2 … … … … … (i)

and \frac{x+1}{y+1} = \frac{1}{2} \Rightarrow 2x+2 = y +1 \Rightarrow 2x - y = -1   … … … … … (ii)

Subtracting (ii) from (i) we get

3x-y=2

\underline{ (-) \hspace{0.5cm} 2x - y = -1}  

x = 3

Therefore x = 3 .

Substituting x = 3 in (i) we get y = 3(3) -2 = 7

Therefore the fraction is \frac{3}{7}

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Question 21: If the numerator of the fraction is multiplied by 2 and the denominator is reduced by 5 the fraction becomes \frac{6}{5} . And if the denominator is doubled and the numerator is increased by 8 , the fraction becomes \frac{2}{5} . Determine the fraction.

Answer:

Let the fraction = \frac{x}{y}

Therefore based on the given conditions:

\frac{2x}{y-5} = \frac{6}{5} \Rightarrow 10x = 6y - 30 \Rightarrow 5x-3y=-15 … … … … … (i)

and \frac{x+8}{2y} = \frac{2}{5} \Rightarrow 5x+40=4y \Rightarrow 5x-4y=-40   … … … … … (ii)

Subtracting (ii) from (i) we get

5x-3y=-15

\underline{ (-) \hspace{0.5cm} 5x-4y=-40}  

y=25

Therefore y = 25 .

Substituting y=25 in (i) we get 5x = 3(25) - 15 = 60 \Rightarrow x = 12

Therefore the fraction is \frac{12}{25}

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Question 22: The sum of the numerator and denominator of a fraction is 18 . If the denominator is increased by 2 , the fraction reduces to \frac{1}{3} . Find the fraction.

Answer:

Let the fraction = \frac{x}{y}

Therefore based on the given conditions:

x+y = 18 … … … … … (i)

and \frac{x}{y+2} = \frac{1}{2} \Rightarrow 3x=y-2 \Rightarrow 3x-y = 2   … … … … … (ii)

Adding (i) and (i) we get

x+y = 18

\underline{ (-) \hspace{0.5cm} 3x-y = 2}  

4x = 20

Therefore x = 5 .

Substituting x = 5 in (i) we get y = 18 - 5 = 13

Therefore the fraction is \frac{5}{18}

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Question 23: The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator  are decreased by 1 , the numerator becomes half the denominator. Determine the fraction.

Answer:

Let the fraction = \frac{x}{y}

Therefore based on the given conditions:

x+y + 3= 2y \Rightarrow x - y = -3 … … … … … (i)

and x-1 = \frac{1}{2} (y-1) \Rightarrow 2x - 2 = y - 1 \Rightarrow 2x - y = 1   … … … … … (ii)

Subtracting (ii) from (i) we get

x - y = -3

\underline{ (-) \hspace{0.5cm} 2x - y = 1}  

-x = -4

Therefore x = 4 .

Substituting x = 4 in (i) we get y = 4+3 = 7

Therefore the fraction is \frac{4}{7}

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Exercise 7(b) continued …

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