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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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?

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.

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.

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.

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.

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.

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.

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.

 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 $B$?

 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.

 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.

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.

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 $B$.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

$\\$

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.

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

$\\$

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.

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

$\\$

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.

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.

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.

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.

$\\$

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.

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 $= 5 \times 12 = 60$

$\\$

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.

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.

$\\$

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.

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.

$\\$

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.

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.

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.

$\\$

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.

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.

$\\$

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 $\angle D = (4x-7)^o$.

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 $\angle D = 125^o$.

$\\$

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.

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.

$\\$

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.

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

$\\$

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.

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.

$\\$

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.

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 $y = 26 \ Rs.$

$\\$

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.

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 $20$.

$\\$

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.

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 $= 9 \times 4 = 36$