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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Problems Based on Ages

Question 24: If twice the son’s age is added to father’s age, the sum is 70 . But if twice the father’s age is added to the son’s age, the sum is 95 . Find the ages of father and son.

Answer:

Let Son’s age be x and Father’s age by y years.

Therefore 2x+ y = 70 … … … … … (i)

and x + 2y = 95 … … … … … (ii)

Multiplying (i) by 2 and subtracting (ii) from (i) we get

4x+2y = 140

\underline{ (-) \hspace{0.5cm} x+2y = 95}  

3x = 45

\Rightarrow x = 15 years.

Therefore y = 70 - 2(15) = 40 years.

Hence the age of son is 15 \ years and that of father is 40 \ years .

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Question 25: Ten year ago, father was twelve times as old as his son and ten years hence, he will be twice as old as his son will be. Find their present ages.

Answer:

Present Age Age 10 years ago Age 10 years hence
Son’s Age x x-10 x+10
Father’s Age y y-10 y+10

Given:

y - 10 = 12(x-10) \Rightarrow y - 10 = 12x - 120  \Rightarrow 12x - y = 110 … … … … … (i)

and y + 10 = 2(x+10) \Rightarrow y + 10 = 2x+20  \Rightarrow 2x - y = -10 … … … … … (ii)

Subtracting (ii) from (i)

12x - y = 110

\underline{ (-) \hspace{0.5cm} 2x - y = -10}  

10x = 120

\Rightarrow x = 12 years.

Therefore y = 2(12) + 10 = 34  years.

Hence the age of son is 12 \ years and that of father is 34 \ years .

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Question 26: Ten years later, A will be twice as old as B and five year ago, A was three times as old as B . Find the present ages of A and B ?

Answer:

Present Age Age 5 years ago Age 10 years hence
A x x-5 x+10
B y y-5 y+10

Given:

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

and x-5 = 3(y-5) \Rightarrow  x-3y= -10 … … … … … (ii)

Subtracting (ii) from (i)

x-2y = 10

\underline{ (-) \hspace{0.5cm} x-3y= -10}  

y = 20

\Rightarrow y = 20 years.

Therefore x = 2(20) + 10 = 50  years.

Hence the age of A is 50 \ years and that of B is 20 \ years .

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Question 27: Ten years ago, a father was twelve times as old as his son and ten year hence, he will be twice as old as his son will be then. Find their present ages.

Answer:

Present Age Age 10 years ago Age 10 years hence
Father x x-10 x+10
Son y y-10 y+10

Given:

x-10 = 12(y - 10) \Rightarrow  x-12y = -110 … … … … … (i)

and x+5 = 2(y+10) \Rightarrow  x-2y= 10 … … … … … (ii)

Subtracting (ii) from (i)

x-12y = -110

\underline{ (-) \hspace{0.5cm} x-2y= 10}  

-10y = -120

\Rightarrow y = 12 years.

Therefore x = 12 (12) - 110 = 144 - 110 = 34  years.

Hence the age of Father is 34 \ years and that of Son is 12 \ years .

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Question 28: The present age of father is three years more then three times that age of the son. Three years hence father’s age will be 10 years more than twice the age of the son. Determine their present ages.

Answer:

Let Father’s age be x and Son’s age by y years.

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

and x+3 = 10 + 2(y + 3) \Rightarrow x - 2y = 13 … … … … … (ii)

Subtracting (ii) from (i) we get

x - 3y = 3

\underline{ (-) \hspace{0.5cm} x - 2y = 13}  

-y = -10

\Rightarrow y = 10 years.

Therefore x = 2(10) + 13 = 33 years.

Hence the age of son is 10 \ years and that of father is 33 \ years .

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Question 29: Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of the two children. Find the age of father.

Answer:

Let Father’s age be x and Son’s age by y \& z years.

Therefore x = 3(y + z) \Rightarrow y + z = \frac{x}{3} … … … … … (i)

and x+5 = 2(y+5+z+5) \Rightarrow x+5 = 2(y+z) + 20

\Rightarrow x + 5= \frac{2}{3} x + 20 … … … … … (ii)

Substituting (i) into (ii) we get

\frac{1}{3} x = 15 \Rightarrow x = 45 \ years

Hence  Father’s age is 33 \ years .

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Question 30: The ages of two friends A and B differ by 3 years. A’s father D is twice as old as A and B is twice as old as here sister C . The ages of C and D differ by 30 years. Find the ages of A and B .

Answer:

Let the age of A, B, C \ and \ D be x, y, z_1 \ and \ z_2 respectively.

Therefore

x-y = 3 … … … … … (i)

y = 2 z_2

z_1-z_2 = 30

2x- \frac{1}{2} y = 30 \Rightarrow 4x - y = 60   … … … … … (ii)

Subtracting (ii) from (i) we get

x-y = 3

\underline{ (-) \hspace{0.5cm} 4x - y = 60}  

-3x=-57

\Rightarrow x = 19 years.

Therefore y = 19 - 3 = 16 years.

Hence the age of A is 16 \ years and that of B is 19 \ years .

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Problems based on Time, Distance and Speed

Question 31: Point A and B are 90 km apart from each other on a road. A car starts from point A and another from point B at the same time. If they go in the same direction they meet in 9 hours and if they go in opposite direction, they meet in \frac{9}{7} hours. Find their speeds.

Answer:

Let the speed of the car A be x km/hr and car B be y km/ hr.

Let the cars meet at point C .

Case 1: The cars are travelling in the same direction

Distance covered by car A = 9x

Distance covered by car B = 9y

Hence 9x-9y = 90 \Rightarrow x - y = 10 … … … … … (i)

Case 2: The cars are travelling in the same direction

Distance covered by car A = \frac{9}{7} x

Distance covered by car B = \frac{9}{7} y

Hence \frac{9}{7} x+ \frac{9}{7} y = 90 \Rightarrow x + y = 70 … … … … … (ii)

Solving (i) and (ii) we get x = 40 km/hr and y = 30 km/hr

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Question 32: Point A and B are 70 km apart from each other on a road. A car starts from point A and another from point B at the same time. If they go in the same direction they meet in 7 hours and if they go in opposite direction, they meet in 1 hour. Find their speeds.

Answer:

Let the speed of the car A be x km/hr and car B be y km/ hr.

Let the cars meet at point C .

Case 1: The cars are travelling in the same direction

Distance covered by car A = 7x

Distance covered by car B = 7y

Hence 7x-7y = 70 \Rightarrow x - y = 10 … … … … … (i)

Case 2: The cars are travelling in the same direction

Distance covered by car A = 1x

Distance covered by car B = 1y

Hence 1x+1y = 70 \Rightarrow x + y = 70 … … … … … (ii)

Solving (i) and (ii) we get x = 40 km/hr and y = 30 km/hr

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Question 33: A train covers a certain distance with uniform speed. If the train would have been 6 km/hr faster, it would have taken 4 hours less than the scheduled time. And if the train was slower by 6 km/hr it would have taken 6 more hours than the scheduled time. Find the length of the journey.

Answer:

Let the actual speed of the train be x km/hr and the actual time taken by the train be y hours.

Therefore the distance covered by the train = xy

If the speed was 6 km/hr more:

xy = (x+6)(y-4) \Rightarrow -2x+3y = 12 … … … … … (i)

If the speed was 6 km/hr slower:

xy = (x-6)(y + 6) \Rightarrow x-y =6 … … … … … (ii)

Solving (i) and (ii) we get x = 30 km/hr and y = 24 hours

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Question 34: A man travels 370 km partly by train and partly by car. If he covers 250 km by train and rest by car, it takes him 4 hours. But if he travels 130 km by train and rest by car, it takes him 18 minutes longer. Find the speed of the train and the car.

Answer:

Let the speed of the train be x km/hr and the speed of the car by y km/hr

When he covers 250 km by train and rest by car, it takes him 4 hours.

Therefore:

\frac{250}{x} + \frac{370-250}{y} = 4 \Rightarrow \frac{125}{x} + \frac{60}{y} = 2 … … … … … (i)

When he travels 130 km by train and rest by car, it takes him 18 minutes longer

\frac{130}{x} + \frac{370-130}{y} = 4 \frac{18}{60}   \Rightarrow \frac{130}{x} + \frac{240}{y} = \frac{43}{10} … … … … … (i)

Now substituting \frac{1}{x} =u and \frac{1}{y} = v we get

125 u + 60 v = 2 … … … … … (iii)

130 u + 240 v =  \frac{43}{10}  … … … … … (iv)

Solving (iii) and (iv) we get u = \frac{1}{100}  and v = \frac{1}{80}

Therefore x = 100 km/hr and y = 80 km/hr

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Question 35: A man travels 760 km partly by train and partly by car. If he covers 160 km by train and rest by car, it takes him 8 hours. But if he travels 240 km by train and rest by car, it takes him 12 minutes longer. Find the speed of the train and the car.

Answer:

Let the speed of the train be x km/hr and the speed of the car by y km/hr

When he covers 160 km by train and rest by car, it takes him 8 hours.

Therefore:

\frac{160}{x} + \frac{760-160}{y} = 8 \Rightarrow \frac{20}{x} + \frac{75}{y} = 1 … … … … … (i)

When he travels 240 km by train and rest by car, it takes him 12 minutes longer

\frac{240}{x} + \frac{760-240}{y} = 8 \frac{12}{60}   \Rightarrow \frac{240}{x} + \frac{520}{y} = \frac{41}{5} … … … … … (i)

Now substituting \frac{1}{x} =u and \frac{1}{y} = v we get

20 u + 75 v = 1 … … … … … (iii)

240 u + 520 v =  \frac{41}{5} … … … … … (iv)

Solving (iii) and (iv) we get v = \frac{1}{100} and u = \frac{1}{80}

Therefore x = 100 km/hr and y = 60 km/hr

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Question 36: A girl travels 300 km partly by train and partly by car. If she covers 60 km by train and rest by car, it takes her 4 hours. But if she travels 100 km by train and rest by car, it takes her 10 minutes longer. Find the speed of the train and the car.

Answer:

Let the speed of the train be x km/hr and the speed of the car by y km/hr

When she covers 60 km by train and rest by car, it takes her 4 hours.

Therefore:

\frac{60}{x} + \frac{300-60}{y} = 4 \Rightarrow \frac{15}{x} + \frac{60}{y} = 1 … … … … … (i)

When she travels 100 km by train and rest by car, it takes her 10 minutes longer

\frac{100}{x} + \frac{300-100}{y} = 4 \frac{10}{60}   \Rightarrow \frac{100}{x} + \frac{200}{y} = \frac{25}{6} … … … … … (ii)

Now substituting \frac{1}{x} =u and \frac{1}{y} = v we get

15 u + 60 v = 1 … … … … … (iii)

100 u + 200 v =  \frac{25}{6} … … … … … (iv)

Solving (iii) and (iv) we get v = \frac{1}{80} and u = \frac{1}{60}

Therefore x = 60 km/hr and y = 80 km/hr

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Question 37: A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.

Answer:

Let the speed of the boat be x km/hr and that of the stream is y km/hr.

Therefore speed of boat up stream = x - y km/hr

and Speed of boat downstream = x + y km/hr

Therefore we have \frac{32}{x-y} + \frac{36}{x+y} = 7 … … … … … (i1)

Similarly, \frac{40}{x-y} + \frac{48}{x+y} = 9 … … … … … (ii)

Let \frac{1}{x-y} = u and \frac{1}{x+y} = v

Therefore we get

32 u + 36 v =7 … … … … … (iii)

and 40 u + 48 v = 9 … … … … … (vi)

Solving (iii) and (iv) we get u = \frac{1}{8} and v = \frac{1}{12}

Therefore x-y = 8 km/hr and x+y = 12 km/hr. Solving these two equations we get x = 10 km/hr and y = 2 km/hr

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Question 38: A sailor goes 8 km downstream in 40 minutes and returns in 1 hours. Determine the speed of the sailor in still water and the speed of the current.

Answer:

Let the speed of the sailor be x km/hr and that of the stream is y km/hr.

Therefore speed of sailor up stream = x - y km/hr

and Speed of boat downstream = x + y km/hr

Therefore we have \frac{8}{x+y} = \frac{40}{60} \Rightarrow x+y = 12 … … … … … (i1)

Similarly, \frac{8}{x-y} = 1 \Rightarrow x-y = 8 … … … … … (ii)

solving (i) and (ii) we get x = 10 km/hr and y = 2 km/hr

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Question 39: A boat covers 24 km upstream and 28 km downstream in 6 hours. Also, it covers 30 km upstream and 21 km downstream in 6.5 hours. Find the speed of the boat in still water and that of the stream.

Answer:

Let the speed of the boat be x km/hr and that of the stream is y km/hr.

Therefore speed of boat up stream = x - y km/hr

and Speed of boat downstream = x + y km/hr

Therefore we have \frac{24}{x-y} + \frac{28}{x+y} = 6 … … … … … (i1)

Similarly, \frac{30}{x-y} + \frac{21}{x+y} = 6.5 … … … … … (ii)

Let \frac{1}{x-y} = u and \frac{1}{x+y} = v

Therefore we get

24 u + 28 v =6 … … … … … (iii)

and 30 u + 21 v = 6.5 … … … … … (vi)

Solving (iii) and (iv) we get u = \frac{1}{6} and v = \frac{1}{14}

Therefore x-y = 6 km/hr and x+y = 14 km/hr. Solving these two equations we get x = 10 km/hr and y = 4 km/hr

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Question 40: A boat covers 12 km upstream and 40 km downstream in 8 hours. Also, it covers 16 km upstream and 32 km downstream in 8 hours. Find the speed of the boat in still water and that of the stream.

Answer:

Let the speed of the boat be x km/hr and that of the stream is y km/hr.

Therefore speed of boat up stream = x - y km/hr

and Speed of boat downstream = x + y km/hr

Therefore we have \frac{12}{x-y} + \frac{40}{x+y} = 8 … … … … … (i1)

Similarly, \frac{16}{x-y} + \frac{32}{x+y} = 8 … … … … … (ii)

Let \frac{1}{x-y} = u and \frac{1}{x+y} = v

Therefore we get

12 u + 40 v =8 \Rightarrow 3u+10v=2 … … … … … (iii)

and 16 u + 32 v = 8 \Rightarrow 2u+4v=1 … … … … … (vi)

Solving (iii) and (iv) we get u = \frac{1}{4} and v = \frac{1}{8}

Therefore x-y = 4 km/hr and x+y = 8 km/hr. Solving these two equations we get x = 6 km/hr and y = 2 km/hr

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Question 41: X takes 3 hours more than Y to walk 30 km. But if X doubles his pace, he is ahead of Y by 1 \frac{1}{2} hours. Find their speed of walking.

Answer:

Let the speed of the X be x km/hr and that of Y is y km/hr.

Therefore \frac{30}{x} = \frac{30}{y} +3 \Rightarrow \frac{10}{x} - \frac{10}{y} = 1

and \frac{30}{2x} + 1 =  \frac{30}{y} \Rightarrow \frac{15}{x} - \frac{30}{y} = -1

Now substituting \frac{1}{x} =u and \frac{1}{y} = v we get

10u - 10v = 1 … … … … … (iii)

15u - 30v = -1 … … … … … (iv)

Solving (iii) and (iv) we get u = \frac{3}{10} \Rightarrow x = \frac{10}{3} km/hr

and v = \frac{1}{5} \Rightarrow y = 5 km/hr

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Question 42: A man walks a certain distance with a certain speed. If he walks 0.5 km/hr faster he would take 1 hours lesser. But if he walks 1 km/hr slower, he takes 3 more hours. Find the distance covered by the man and his speed of walking.

Answer:

Let the actual speed of the man be x km/hr and the actual time taken by the train be y hours.

Therefore the distance covered by the man = xy

If the speed was 0.5 km/hr more:

xy = (x+0.5)(y-1) \Rightarrow x-0.5y=-0.5 … … … … … (i)

If the speed was 6 km/hr slower:

xy = (x-1)(y + 3) \Rightarrow 3x-y = 3 … … … … … (ii)

Solving (i) and (ii) we get x = 4 km/hr and y = 9 hours

Distance covered = 9 \times 4 = 36 km/hr

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Question 43: A train covered distance at a uniform speed. If the train could have been 10 km/hr faster, it would have taken two hours lesser then the scheduled time. But if it were slower by 10 km/hr, it would have taken 3 more hours than the scheduled time. Find the distance covered by the train.

Answer:

Let the actual speed of the train be x km/hr and the actual time taken by the train be y hours.

Therefore the distance covered by the train = xy

If the speed was 10 km/hr more:

xy = (x+10)(y-2) \Rightarrow2x-10y = -20 … … … … … (i)

If the speed was 10 km/hr slower:

xy = (x-10)(y + 3) \Rightarrow 3x-10y=30 … … … … … (ii)

Solving (i) and (ii) we get x = 50 km/hr and y = 12 hours

Distance covered = 50 \times 12 = 60 km

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Miscellaneous Problems

Question 44: A taxi charge comprises of a fixed charge and a variable charge per km. For a journey of 10 km the charge paid is Rs. \ 75 and for a journey of 15 km, the charge paid is Rs. \ 110 . What will the person pay for a journey of 25 km.

Answer:

Let the fixed charges = x Rs. and the Variable charges = y Rs. / km

Therefore: x + 10y = 75 … … … … … (i)

and x + 15y = 110 … … … … … (ii)

Subtracting (i) from (ii) we get 5y = 35 \Rightarrow y = 7 Rs. / km

Therefore x = 75 - 10(7) = 5 Rs.

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Question 45: A part of the monthly hostel charges is fixed and the other part depends on the number of days that you have stayed. When a student A stays for 20 days, he pays Rs. 1000 as the hostel charge where as the student B who stays for 26 days pays Rs. 1180 for the hostel charges. Find the fixed charge and the daily charge for the stay.

Answer:

Let the fixed charges = x Rs. and the Variable charges = y Rs. / day

Therefore: x+20y = 1000 … … … … … (i)

and x+15y = 110 … … … … … (ii)

Subtracting (i) from (ii) we get 6y=180 \Rightarrow y = 30 Rs. / day

Therefore x = 1000 - 20 \times 30 = 400 Rs.

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Question 46: A man starts his job with a certain monthly salary and earns a fixed increment every year. If his slary was Rs. 1500 after 4 years, and Rs. 1800 after 10 years, what was his starting salary and his annual increment.

Answer:

Let the starting salary = x Rs. and the increment = y Rs. / year

Therefore: x+4y = 1500 … … … … … (i)

and x+10y = 1800 … … … … … (ii)

Subtracting (i) from (ii) we get 6y=300 \Rightarrow y = 50 Rs. / year

Therefore starting salary x = 1500 - 4 \times 50 = 1300 Rs.

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Question 47: Students of a class are made to stand in rows. If there is one student extra in a row, there would be 2 rows less. If one student is less in a row, there would be 3 rows more. Find the number of students in the class.

Answer:

Let the number of students in a row be x   and the number of rows be  y .

Therefore the total number of students = xy

If there was one student more in each row:

xy = (x+1)(y-2) \Rightarrow 2x -y = -2 … … … … … (i)

If there was one student less in a row:

xy = (x-1)(y + 3) \Rightarrow 3x-y = 3 … … … … … (ii)

Solving (i) and (ii) we get x = 5   and y = 12 rows

Total number of students = 5 \times 12 = 60

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Question 48: 8 men and 12 boys can finish the work in 10 days while 6 men and 8 boys can finish the work in 14 days. Find the time taken by one men alone and that by one boy alone to finish the work.

Answer:

Let one man can complete the work in x days and one boy can finish the work in y days alone.

Therefore one man’s work = \frac{1}{x}

and one boys work = \frac{1}{y}

Since 8 men and 12 boys can finish the work in 10 days, we get

10( \frac{8}{x} + \frac{12}{y} ) = 1 \Rightarrow  \frac{80}{x} + \frac{120}{y} = 1 … … … … … (i)

Similarly 6 men and 8 boys can finish the work in 14 days , we get

14( \frac{6}{x} + \frac{8}{y} ) = 1 \Rightarrow  \frac{84}{x} + \frac{112}{y} = 1 … … … … … (ii)

Let \frac{1}{x} =u and \frac{1}{y} = v we get

80u + 120 v = 1 … … … … … (iii)

84u + 112 v = 1 … … … … … (iv)

Solving (iii) and (iv) we get u = \frac{1}{140} and v = \frac{1}{280}

Hence x = 140 days  and y = 280 days

Therefore we can say that a man alone will take 140 days while a boy working alone will finish the work in 280 days.

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Question 49: 2 men and 7 boys can finish the work in 4 days while 4 men and 4 boys can finish the work in 3 days. Find the time taken by one men and one boy to finish the work.

Answer:

Let one man can complete the work in x days and one boy can finish the work in y days alone.

Therefore one man’s work = \frac{1}{x}

and one boys work = \frac{1}{y}

Since 2 men and 7 boys can finish the work in 4 days, we get

4( \frac{2}{x} + \frac{7}{y} ) = 1 \Rightarrow  \frac{8}{x} + \frac{28}{y} = 1 … … … … … (i)

Similarly 4 men and 4 boys can finish the work in 3 days , we get

3( \frac{4}{x} + \frac{4}{y} ) = 1 \Rightarrow  \frac{12}{x} + \frac{12}{y} = 1 … … … … … (ii)

Let \frac{1}{x} =u and \frac{1}{y} = v we get

8u + 28 v = 1 … … … … … (iii)

12u + 12 v = 1 … … … … … (iv)

Solving (iii) and (iv) we get u = \frac{1}{15} and v = \frac{1}{60}

Hence x = 15 days  and y = 60 days

Therefore we can say that a man alone will take 15 days while a boy working alone will finish the work in 60 days.

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Question 50: 2 women and 5 men can finish the work in 4 days while 3 women and 6 men can finish the work in 3 days. Find the time taken by one woman alone and that by one man alone to finish the work.

Answer:

Let one woman can complete the work in x days and one men can finish the work in y days alone.

Therefore one man’s work = \frac{1}{x}

and one boys work = \frac{1}{y}

Since 2 women and 5 men can finish the work in 4 days, we get

4( \frac{2}{x} + \frac{5}{y} ) = 1 \Rightarrow  \frac{8}{x} + \frac{20}{y} = 1 … … … … … (i)

Similarly 3 women and 6 men can finish the work in 3 days , we get

3( \frac{3}{x} + \frac{6}{y} ) = 1 \Rightarrow  \frac{9}{x} + \frac{18}{y} = 1 … … … … … (ii)

Let \frac{1}{x} =u and \frac{1}{y} = v we get

8u + 20 v = 1 … … … … … (iii)

9u + 18 v = 1 … … … … … (iv)

Solving (iii) and (iv) we get u = \frac{1}{18} and v = \frac{1}{36}

Hence x = 18 days  and y = 36 days

Therefore we can say that a woman alone will take 18 days while a boy working alone will finish the work in 36 days.

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Question 51: The ratio of income of two people is 9:7 and the ratio of their expenditure is 4:3 . If each one of them saves Rs. \ 200 per month, find the monthly incomes.

Answer:

Let the income of the people be 9x and 7x respectively. Similarly their expenses will be 4y and 3y respectively.

Therefore given based on savings

9x-4y = 200 … … … … … (i)

7x-3y = 200 … … … … … (ii)

Solving (i) and (ii) we get x = 200 \ Rs.   and y = 400 \ Rs.

Therefore the incomes are Rs. \ 1800 and Rs. \ 1400 respectively.

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Question 52: The income of X and Y are in the ratio of 8:7 and their expenditures are in the ratio of 19:16 . If each save Rs. \ 1250 , find their income.

Answer:

Let the income of the people be 8x and 7x respectively. Similarly their expenses will be 19y and 16y respectively.

Therefore given based on savings

8x-19y = 1250 … … … … … (i)

7x-16y = 1250 … … … … … (ii)

Solving (i) and (ii) we get x = 750 \ Rs.   and y = 250 \ Rs.

Therefore the incomes are Rs. \ 6000 and Rs. \ 5250 respectively.

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Question 53: Find the four angles of a cyclic quadrilateral ABCD in which \angle A = (2x-1)^o , \angle B = (y+5)^o , \angle C = (2y+15)^o and \angle D = (4x-7)^o .

Answer:

In a cyclic quadrilateral ABCD, A+C = 180^o and B+D = 180^o

Therefore

2x-1 + 2y+15 = 180 \Rightarrow  2x+2y = 166 \Rightarrow x + y = 83 … … … … … (i)

and 4x-7+y+5 = 180 \Rightarrow 4x+y = 182 … … … … … (ii)

Solving (i) and (ii) we get x = 33 and y = 50

Hence the angles are \angle A = 65^o , \angle B = 55^o , \angle C = 115^o and \angle D = 125^o .

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Question 54: In a \triangle ABC , \angle A = x^o , \angle B = 3x^o and \angle C = y^o . If 3y-5x = 30 , prove that the triangle is a right-angled triangle.

Answer:

In a triangle, the sum of all the three angles is 180^o .

Therefore

x + 3x + y = 180 \Rightarrow 4x + y = 180   … … … … … (i)

and given -5x+3y = 30   … … … … … (i)

Solving (i) and (ii) we get , x = 30 and y = 50

Therefore \angle A = 30^o, \angle B = 90^o and \angle y = 50 . Therefore the triangle is a right angled triangle.

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Question 55: If in a rectangle, the length is increased and the breadth is decreases each by 2 units, the area is reduced by 28 sq. units. However, if the length is reduced by 1 units and breadth is increased by 2 units, the area increases by 33 sq. units. Find the area of the rectangle.

Answer:

Let the length be x   and the breadth be  y .

Therefore the area = xy

If the length is increased and the breadth is decreases each by 2 units, the area is reduced by 28 sq. units:

xy-28 = (x+2)(y-2) \Rightarrow x-y=12 … … … … … (i)

If the length is reduced by 1 units and breadth is increased by 2 units, the area increases by 33 sq. units

xy+33 = (x-1)(y + 2) \Rightarrow 2x-y=35 … … … … … (ii)

Solving (i) and (ii) we get x = 23   and y = 11 units

Total area = 23 \times 11 = 253 sq. units

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Question 56: Half the perimeter of the garden, whose length is 4 more than its width is 36 \ m . Find the dimensions of the garden.

Answer:

Let the width = x units

Therefore the length = x+4 units

Hence \frac{1}{2} (2x+2(x+4)) = 36 \Rightarrow x + x + 4 = 36 \Rightarrow x = 16 units

Therefore length is 20   units.

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Question 57: If A gives Rs. \ 30 to B , then B will have twice the amount of money left with A . But if B gives Rs. \ 10 to A , then A will have thrise as much as as is left with B . How much money does each have.

Answer:

Let A have x \ Rs. and B have y \ Rs.

Therefore 2(x-30) = y + 30 \Rightarrow 2x-y = 90 … … … … … (i)

and (x+10) = 2(y-10) \Rightarrow x - 3y = -20   … … … … … (ii)

Solving (i) and (ii) we get x = 58 \ Rs. and y = 26 \ Rs.

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Question 58: A scored 40 marks in a test, getting 3 marks for each right answer and 1 mark for wrong answer. Had 4 marks been awarded for each right answer and 2 marks deducted for each wrong answer then A would have scored 50 marks. How many questions were in the test.

Answer:

Let the number of questions answered right are x and that answered wrong are y

Therefore we have 3x-y = 40 … … … … … (i)

and 4x-2y = 50   … … … … … (ii)

Solving (i) and (ii) we get x = 15 and y = 5

Therefore the total number of questions in the test are 20 .

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Question 59: Students of a class are made to stand in rows. If there are 3 students extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer:

Let the number of students in a row be x   and the number of rows be  y .

Therefore the total number of students = xy

If there was one student more in each row:

xy = (x+3)(y-1) \Rightarrow x-3y=-3 … … … … … (i)

If there was one student less in a row:

xy = (x-3)(y + 2) \Rightarrow 2x-3y=6 … … … … … (ii)

Solving (i) and (ii) we get x = 9   and y = 4 rows

Total number of students = 9 \times 4 = 36