Class 8: Linear Equations -Exercise 22B

Question 1: $\displaystyle 17$ less than four times a number is $\displaystyle 11$ . Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \Rightarrow 4x-17 = 11$

$\displaystyle \Rightarrow x = \frac{28}{4} = 7$

$\displaystyle \\$

Question 2: If $\displaystyle 10$ be added to four times a certain number the result is $\displaystyle 5$ less than five times the number. Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \Rightarrow 4x+10 = 5x-5$

$\displaystyle \Rightarrow x = 15$

$\displaystyle \\$

Question 3: $\displaystyle \frac{2}{3}$ Of a number is $\displaystyle 20$ less than the original number. Find the original number.

Answer:

$\displaystyle \text{Let the original number } = x$

$\displaystyle \frac{2}{3} x = x-20$

$\displaystyle \Rightarrow \frac{1}{3} x = 20$

$\displaystyle \Rightarrow x = 60$

$\displaystyle \\$

Question 4: A number is $\displaystyle 25$ more than its part. Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle x = 25+ \frac{5}{6} x$

$\displaystyle \Rightarrow \frac{1}{6} x = 25$

$\displaystyle \Rightarrow x = 150$

$\displaystyle \\$

Question 5: A number is as much greater than $\displaystyle 21$ as is less than $\displaystyle 71$ . Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle x-21 = 71-x$

$\displaystyle \Rightarrow 2x = 92$

$\displaystyle \Rightarrow x = 46$

$\displaystyle \\$

Question 6: 6 more than one-fourth of the number is two-fifth of the number. Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \frac{1}{4} x+6 = \frac{2}{5} x$

$\displaystyle \Rightarrow 6 = ( \frac{2}{5} - \frac{1}{4} )x = \frac{30}{20} x$

$\displaystyle \Rightarrow x = 40$

$\displaystyle \\$

Question 7: One-third of a number exceeds one-fourth of the number by $\displaystyle 15$ . Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \frac{1}{3} x- \frac{1}{4} x = 15$

$\displaystyle \Rightarrow \frac{1}{12} x = 15 \ or \ x = 180$

$\displaystyle \\$

Question 8: If one-fifth of a number decreased by $\displaystyle 5$ is $\displaystyle 16$ find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \frac{1}{5} x-5 = 16$

$\displaystyle \Rightarrow x = 105$

$\displaystyle \\$

Question 9: A number when divided by $\displaystyle 6$ is diminished by $\displaystyle 40$ . Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \frac{x}{6} = x-40$

$\displaystyle \Rightarrow \frac{5}{6} x = 40$

$\displaystyle \Rightarrow x = 48$

$\displaystyle \\$

Question 10: Four-fifths of a number exceeds two-third of the number by $\displaystyle 10$ . Find the number.

Answer:

$\displaystyle \text{Let the number } = x$

$\displaystyle \frac{4}{3} x = \frac{2}{3} x+10$

$\displaystyle \Rightarrow \frac{2}{15} x = 10$

$\displaystyle \Rightarrow x = 75$

$\displaystyle \\$

Question 11: Two numbers are in the ratio $\displaystyle 3:4$ and their sum is $\displaystyle 84$ . Find the number.

Answer:

Let the two numbers be $\displaystyle x$ and $\displaystyle y$

Therefore

$\displaystyle 3x = 4y$

$\displaystyle x+y = 84$

Solving

$\displaystyle \frac{4}{3} y+y = 84$

$\displaystyle \frac{7y}{3} = 84$

$\displaystyle y = 3 \times 12 = 36$

$\displaystyle \text{Hence } x = \frac{4}{3} \times 36 = 48$

$\displaystyle \\$

Question 12: Three numbers are in ratio $\displaystyle 4:5:6$ and their sum is $\displaystyle 135$ . Find the numbers.

Answer:

Let the three numbers be $\displaystyle x \ y \ z$

Therefore

$\displaystyle 4x:5x:6x$

$\displaystyle 4x+5x+6x = 135$

Solving

$\displaystyle 15x = 135$

$\displaystyle x = 9$

Therefore

The three numbers are $\displaystyle 36 \ 45 \ 54$

$\displaystyle \\$

Question 13: Two numbers are in the ratio $\displaystyle 3:5$ . If each is increased by $\displaystyle 10$ then ratio between the new numbers so formed is $\displaystyle 5:7$ Find the original numbers.

Answer:

Let the two numbers be $\displaystyle x$ and $\displaystyle y$

Given

$\displaystyle \frac{x}{y} = \frac{3}{5} \ldots \ldots \ldots i)$

$\displaystyle \frac{x+10}{y+10} = \frac{5}{7} \ldots \ldots \ldots ii)$

solving

From $\displaystyle i) x = \frac{3}{5} y$

Substituting in ii)

$\displaystyle \frac{\frac{3}{5} y+10}{y+10} = \frac{5}{7}$

$\displaystyle \frac{21}{5} y+70 = 5y+50$

$\displaystyle 20 = \frac{4}{5} y$

or $\displaystyle y = 25$

$\displaystyle x = \frac{3}{5} \times 25 = 15$

Two numbers are $\displaystyle 15$ and $\displaystyle 25$

$\displaystyle \\$

Question 14: The sum of three consecutive odd numbers is $\displaystyle 75$ . Find the numbers.

Answer:

Let the three consecutive numbers be

$\displaystyle x x+2 x+4$

therefore

$\displaystyle x+x+2+x+4 = 75$

$\displaystyle 3x+6 = 75$

$\displaystyle 3x = 69$

$\displaystyle x = 23$

Therefore the three numbers are $\displaystyle 23 \ 25 \ and \ 27$

$\displaystyle \\$

Question 15: Divide $\displaystyle 25$ into two parts such that $\displaystyle 7$ times the first part added to $\displaystyle 5$ times the second part makes $\displaystyle 139$ .

Answer:

Let the two parts be $\displaystyle x$ and $\displaystyle y$

Therefore

$\displaystyle x+y = 25$

$\displaystyle 7x+5y+139$

Solving we get

$\displaystyle x = 25-y$

$\displaystyle 7(25-y)+5y = 139$

$\displaystyle 175-2y = 139$

$\displaystyle 2y = 175-139 = 36$

or $\displaystyle y = 18$

The other part $\displaystyle = 7$

$\displaystyle \\$

Question 16: Divide $\displaystyle 180$ into two parts such that the first part is $\displaystyle 12$ less than twice the second part.

Answer:

Let the two parts be $\displaystyle x$ and $\displaystyle 2y$

Therefore

$\displaystyle x+y = 180$

$\displaystyle x+12 = 2y$

Solving

$\displaystyle y = 180-x$

$\displaystyle x+12 = 2(180-x)$

$\displaystyle 3x = 360-12 = 348$

$\displaystyle x = 116$

$\displaystyle \text{Therefore } y = 180-116 = 64$

$\displaystyle \\$

Question 17: The denominator of the fraction is $\displaystyle 4$ more than its numerator. On subtracting $\displaystyle 1$ from each numerator and denominator the fraction becomes. Find the original fraction.

Answer:

Let the fraction be $\displaystyle \frac{x}{y}$

Given

$\displaystyle y = x+4$

$\displaystyle \text{Therefore the fraction } = \frac{x}{x+4}$

Given

$\displaystyle \frac{x-1}{x+4-1} = \frac{1}{2}$

$\displaystyle 2x-2 = x+3$

$\displaystyle x = 5 \text{ and } y = 9$

$\displaystyle \text{Therefore fraction } = \frac{5}{9}$

$\displaystyle \\$

Question 18: The denominator of the fraction is $\displaystyle 1$ more than the double the numerator. On adding $\displaystyle 2$ to the numerator and subtracting $\displaystyle 3$ from denominator we obtain $\displaystyle 1$ . Find the original fraction.

Answer:

Let the fraction be $\displaystyle \frac{x}{2x+1}$

Given

$\displaystyle \frac{x+2}{2x+1-3} = 1$

$\displaystyle x+2 = 2x-2$

$\displaystyle x = 4$

Fraction $\displaystyle = \frac{4}{9}$

$\displaystyle \\$

Question 19: The sum of the digits of a two-digit number is $\displaystyle 5$ . On adding $\displaystyle 27$ to the number its digits are reversed. Find the original number.

Answer:

Let the two digit number be $\displaystyle xy$

Given

$\displaystyle x+y = 5 \ldots \ldots \ldots i)$

$\displaystyle xy+27 = yx$

$\displaystyle 10x+y+27 = 10y+x$

$\displaystyle 9x+27 = 9y$

or $\displaystyle x+3 = y \ldots \ldots \ldots ii)$

Solving i) and ii) together.

$\displaystyle x+3 = (5-x)$

$\displaystyle 2x = 2$

$\displaystyle x = 1$

$\displaystyle y = 4$

Hence the number $\displaystyle = 14$

$\displaystyle \\$

Question 20: What same numbers should be added to each one of the number $\displaystyle 15 23 29 44$ to obtain numbers which are in proportion?

Answer:

Let the number added to each one of $\displaystyle 15 23 29 44$ be $\displaystyle x$

$\displaystyle \frac{15+x}{23+x} = \frac{29+x}{44+x}$

$\displaystyle 660+59x+ x^2 = 667+52x+ x^2$

$\displaystyle 7x = 7$

$\displaystyle x = 1$

$\displaystyle \\$

Question 21: The sum of two numbers is $\displaystyle 110$ . One-fifth of the larger number is $\displaystyle 8$ more than one-ninth of the smaller number. Find the numbers.

Answer:

Let the two numbers be $\displaystyle x$ and $\displaystyle y$

Given

$\displaystyle x+y 110$

$\displaystyle \frac{1}{5} x = \frac{1}{9} y+8$

Solving

$\displaystyle \frac{1}{5} x = \frac{1}{9} (110-x)+ 8$

$\displaystyle (\frac{1}{5} + \frac{1}{9} )x = \frac{110}{9} +8 = \frac{182}{9}$

$\displaystyle x = \frac{182 \times 45}{9 \times 14} = 65$

$\displaystyle y = 10-65 = 45$

Two numbers are $\displaystyle 45$ and $\displaystyle 65$

$\displaystyle \\$

Question 22: A number is subtracted from the numerator of the fraction $\displaystyle \frac{12}{13}$ and six times that number is added to the denominator. If the new fraction is $\displaystyle \frac{1}{11}$ then find the number.

Answer:

Let the number subtracted from the numerator $\displaystyle = x$

$\displaystyle \frac{12-x}{13+6x} = \frac{1}{11}$

$\displaystyle 132-11x = 13+6x$

$\displaystyle 17x = 119$

or $\displaystyle x = 7$

$\displaystyle \\$

Question 23: A right angled triangle having perimeter $\displaystyle 120\ \text{ cm }$ has its two-side perpendicular side in the ratio $\displaystyle 5:12$ . Find the lengths of its sides.

Answer:

Perimeter of right angled triangle $\displaystyle = 120$

Perpendicular sides $\displaystyle = 5x \ and \ 12x$

Hypotenuse $\displaystyle = \sqrt{(5x)^2+(12x)^2} = 13x$

Therefore

$\displaystyle 5x+12x+13x = 120$

$\displaystyle 30x = 120$

$\displaystyle x = 4$

Therefore length of side $\displaystyle = 20 48 52$

$\displaystyle \\$

Question 24: The sum of the digits of a two-digit number is $\displaystyle 9$ . If $\displaystyle 9$ is added to the number formed by reversing the digits then the result is thrice the original number. Find the original number.

Answer:

Let the two-digit number $\displaystyle = xy$

$\displaystyle x+y = 9 \ldots \ldots \ldots i)$

$\displaystyle yx+9 = 3(xy)$

$\displaystyle 10y+x+9 = 3(10x+y)$

$\displaystyle 10y+x+9 = 30x+3y$

$\displaystyle y+9 = 29x \ldots \ldots \ldots ii)$

Solving i) and ii)

$\displaystyle 7(9-x)+9 = 29x$

$\displaystyle 63-7x+9 = 29x$

$\displaystyle 72x = 36$

Or $\displaystyle x = 2$

$\displaystyle y = 9-2 = 7$

Therefore the number $\displaystyle = 27$

$\displaystyle \\$

Question 25: The lengths of a rectangle plot of land exceeds its breadth by $\displaystyle 23 m$ if the length is decreased by $\displaystyle 15 m$ . and the breadth is increased by $\displaystyle 7 m$ . the area is reduced by $\displaystyle 360 m^2$ .

Answer:

Find the length and the breadth of the plot.

Let the length $\displaystyle = l$ and breadth $\displaystyle = b$

$\displaystyle l = 23 +b$

Given

$\displaystyle (l-15)(b+7) = lb-360$

$\displaystyle (23+b-15)(b+7) = (23+b)b-360$

$\displaystyle (b+8)(b+7) = 23b+b^2-360$

$\displaystyle b^2+15b+56 = 23b+b^2-360$

$\displaystyle 416 = 8b$

or $\displaystyle b = 52m$

Therefore

$\displaystyle l = b+23 = 52+23 = 75m$

$\displaystyle \\$

Question 26: The length of the rectangular park is twice its breadth. If the perimeter of the park is 186 m find its length and breadth.

Answer:

Let the length $\displaystyle = l$ and breadth $\displaystyle = b$

$\displaystyle l = 2b$

$\displaystyle 2l+2b = 186$

$\displaystyle 4b+2b = 186$

$\displaystyle 6b = 186$

or $\displaystyle b = 31$

$\displaystyle l = 62$

$\displaystyle \\$

Question 27: The length of the rectangle is $\displaystyle 7 \text{ cm }$ more than its breadth. If the perimeter of the rectangle is $\displaystyle 90\ \text{ cm }$ find its length and breadth.

Answer:

Let the length $\displaystyle = l$ breadth $\displaystyle = b$

$\displaystyle l = b+7$

Given

$\displaystyle 21l +2b = 90$

$\displaystyle 2(b+7) +2b = 90$

$\displaystyle 4b = 76$

Or $\displaystyle b = 19 \text{ cm }$

$\displaystyle l = 19+7 = 26 \text{ cm }$

$\displaystyle \\$

Question 28: The length of a rectangle is $\displaystyle 7 \text{ cm }$ less than twice its breadth. If the length is decreased by $\displaystyle 2\ \text{ cm }$ and breadth increased by $\displaystyle 3\ \text{ cm }$ the perimeter of the resulting rectangle is $\displaystyle 66 \text{ cm }$ . find the length and the breadth of the original rectangle.

Answer:

Let the length $\displaystyle = l$ and breadth $\displaystyle = b$

$\displaystyle l+ 7 = 2b$

Given

$\displaystyle 2(l-2)+ 2(b+3) = 66$

$\displaystyle 2l-4+2b+6 = 66$

$\displaystyle 2l+2b = 64$

Solving

$\displaystyle 2(2b-7) + 2b = 64$

$\displaystyle 6b = 78$

$\displaystyle b = 13$

$\displaystyle l = 2 \times 13-7 = 19$

breadth $\displaystyle = 13 \text{ cm }$

length $\displaystyle = 19cm$

$\displaystyle \\$

Question 29: A man is five times as old as his son. In two years’ time he will be four times as old as his son. Find their present ages.

Answer:

Let the man’s age $\displaystyle = 5x$

If son’s age $\displaystyle = x$

Two years letter

Man’s age $\displaystyle = 5x +2$

Son’s age $\displaystyle = x+ 2$

$\displaystyle 5x +2 = 4(x+2)$

$\displaystyle x + 6 years = son's \ age$

Man’s age $\displaystyle = 30 yrs.$

$\displaystyle \\$

Question 30: A man is twice as old as his son. Twelve years ago the man was thrice as old as his son. Find their present ages.

Answer:

Let the son’s age $\displaystyle = x$

Man’s age $\displaystyle = 2x$

$\displaystyle 12 \ years \ ago$

Son’s age $\displaystyle = x-12$

Man’s age $\displaystyle = 2x-12$

$\displaystyle 2x-12 = 3 (x-12)$

$\displaystyle 2x-12 = 3x-36$

$\displaystyle x = 24 = son's \ age$

Man’s age $\displaystyle = 48 \ years$

$\displaystyle \\$

Question 31: Seema is $\displaystyle 10 \ years$ elder than Rekha. The ratio of their ages is $\displaystyle 5:3$ . Find their ages.

Answer:

Let Rekha’s age $\displaystyle = x$

Seema’s age $\displaystyle = x + 10$

given

$\displaystyle \frac{x+10}{x} = \frac{5}{3}$

$\displaystyle 3x+30 = 5x$

$\displaystyle 2x = 30$

or $\displaystyle x = 15$

Rekha’s sage $\displaystyle = 15 yrs.$

Seema’s sage $\displaystyle = 25 yrs.$

$\displaystyle \\$

Question 32: $\displaystyle 5 \ years$ ago the age of Parvati was $\displaystyle 4$ times the age of her son. The sum of their present ages is $\displaystyle 55 years$ . Find Parvati’s age.

Answer:

Let the present age of Parvati $\displaystyle = x yrs$

age of son $\displaystyle = y yrs.$

$\displaystyle x+y = 55 \ldots \ldots \ldots i)$

Five years before

Parvati $\displaystyle = x-5 yrs.$

son $\displaystyle = y-5 yrs.$

Given

$\displaystyle (x-5) = 4(y-5)$

$\displaystyle x-4y = -15 \ldots \ldots \ldots ii)$

solving i) and ii)

$\displaystyle x-4 (55-x) = -15$

$\displaystyle 5x = 205$

or $\displaystyle x = 44 =$ Parvati’s age

Son’s age $\displaystyle = 55-44 = 11 years$

$\displaystyle \\$

Question 33: A man is $\displaystyle 56$ years old and his son is $\displaystyle 24$ years old. In how many years the father will be twice as old as his son at that time?

Answer:

Man’s age $\displaystyle = 56 years$

Son’s age $\displaystyle = 24 years$

Let in $\displaystyle x years$ man would be twice the age of son

$\displaystyle 56 +x = 2(24 = x)$

$\displaystyle 56+x = 48+2x$

or $\displaystyle x = 8 years$

$\displaystyle \\$

Question 34: 9 years hence a girl will be $\displaystyle 3$ times as old as she was $\displaystyle 9$ years ago. How old is she now?

Answer:

Let the current age of the girl $\displaystyle = x$

Given

$\displaystyle x + 9 = 3(x- 9)$

$\displaystyle x+9 = 3x-27$

$\displaystyle 2x = 36$

$\displaystyle x = 18 \ years = age \ of \ the \ girl$

$\displaystyle \\$

Question 35: A man made a trip of $\displaystyle 480\ km$ in $\displaystyle 9$ hours. Some part of trip was covered at $\displaystyle 45 \text{ km/hr }$ and the remaining at $\displaystyle 60\ \text{ km/hr }$ . find the part of the trip covered by him at $\displaystyle 60\ \text{ km/hr }$ .

Answer:

Let the distance covered at $\displaystyle 45 km/ hr = x$

Let the distance covered at $\displaystyle 60 km/ hr = y$

Total distance $\displaystyle = 480 km.$

$\displaystyle x+y = 480$

$\displaystyle \frac{x}{45} + \frac{y}{60} = 9$

Solving

$\displaystyle \frac{480-y}{45} + \frac{y}{60} = 9$

or $\displaystyle y = 300 \ km \ and \ x = 180 \ km$

$\displaystyle \\$

Question 36: A motorist traveled from town $\displaystyle A$ to town $\displaystyle B$ at an average speed of $\displaystyle 54\ \text{ km/hr }$ . on his return journey his average speed was $\displaystyle 60\ \text{ km/hr }$ . if the total time taken is $\displaystyle 9 hours$ find the distance between the two towns.

Answer:

Let the distance between town A and B $\displaystyle = x$

$\displaystyle \text{Therefore } \frac{x}{54} + \frac{x}{60} = 9.5$

or $\displaystyle x = 270 km.$

$\displaystyle \\$

Question 37: The distance between two stations is $\displaystyle 300\ km$ . two motor-cyclist start simultaneously from these stations and move towards each other. The speed of one of them is $\displaystyle 7\ \text{ km/hr }$ faster than that of other. If the distance between them after $\displaystyle 2 hours$ is $\displaystyle 34\ km$ find the speed of each motor-cycle

Answer:

Distance $\displaystyle = 300 km$

Let the speed of 1st cyclist $\displaystyle = x$

Then speed of 2nd cyclist $\displaystyle = x + 7$

Distance covered by 1s cyclist in 2hr $\displaystyle = 2x$

Distance covered by 2nd cyclist in 2 hr $\displaystyle = 2( x+7)$

Therefore

$\displaystyle 2x+34+2(x+7) = 300$

$\displaystyle 4x + 48 = 300$

$\displaystyle x = \frac{252}{4} = 63 \text{ km/hr }$

Speed of 1st cyclist $\displaystyle = 63 \text{ km/hr }$

Speed of 2nd cyclist $\displaystyle = 63+7 = 70 \text{ km/hr }$

$\displaystyle \\$

Question 38: A boat travels $\displaystyle 30\ km$ upstream in river in the same period of time as it travels $\displaystyle 50\ km$ downstream. If the ratio of stream be $\displaystyle 5 \text{ km/hr }$ find the speed of the boat in still water.

Answer:

Let the speed of boat $\displaystyle = x \text{ km/hr }$

Speed of stream $\displaystyle = 5 \text{ km/hr }$

Speed of boat upstream $\displaystyle = x-5 \text{ km/hr }$

Speed of the boat downstream $\displaystyle = x + 5 \text{ km/hr }$

$\displaystyle \text{Therefore } \frac{30}{x-5} = \frac{50}{x+5}$

$\displaystyle 30x + 150 = 50x-250$

$\displaystyle 400 = 20 x$

Or $\displaystyle x = 20 \text{ km/hr }$

$\displaystyle \\$

Question 39: The length of each of the equal sides of an isosceles triangle is $\displaystyle 4\ \text{ cm }$ longer than the base. If the perimeter of the triangle is $\displaystyle 62\ \text{ cm }$ find the lengths of the sides of the triangle.

Answer:

Let the base $\displaystyle = x \text{ cm }$

Sides $\displaystyle = (x+ 4)$

Perimeter $\displaystyle = 62 \text{ cm }$

Therefore

$\displaystyle (x+4) + (x) + (x+ 4) = 62$

$\displaystyle 3x+8 = 62$

$\displaystyle 3x = 54$

or $\displaystyle x = 18$

Base $\displaystyle = 18 \text{ cm }$ Sides $\displaystyle = 22cm$

$\displaystyle \\$

Question 40: A certain number of candidates appeared for an examination in which one-fifth of the whole plus $\displaystyle 16$ secured first division one-fourth plus $\displaystyle 15$ secured second division and one-fourth minus $\displaystyle 25$ secured third division if the remaining $\displaystyle 60$ candidates failed find the total number of candidates appeared.

Answer:

Let the number of candidates $\displaystyle = x$

$\displaystyle ( \frac{1}{5} x+6)+ ( \frac{1}{4} x+15 )+( \frac{1}{4} x-25)+ 60 = x$

$\displaystyle 6+60 = x-( \frac{1}{5} + \frac{1}{4} + \frac{1}{4} )x$

$\displaystyle x(1- 14/20) = 66$

$\displaystyle x = \frac{20 \times 66}{6} = 220$

No. of candidates $\displaystyle = 220$

$\displaystyle \\$

Question 41: Raman has $\displaystyle 3$ times as much money as Kamal. If Raman gives $\displaystyle Rs. 750$ to Kamal then Kamal will have twice as much as left with Raman. How much had each originally?

Answer:

Let money with Kamal $\displaystyle = x$

Then money with Raman $\displaystyle = 3x$

$\displaystyle 2(3x-750) = (x+750)$

Or $\displaystyle x = 450 Rs.$

Kamal has $\displaystyle 450 Rs.$ And Raman $\displaystyle 1350 Rs.$

$\displaystyle \\$

Question 42: The angles of triangle are in ratio $\displaystyle 2:3:4$ . Find the angles.

Answer:

Ratio of angle $\displaystyle = 2: 3: 4$

Therefore

$\displaystyle 2x+3x+4x = 180$

$\displaystyle 9x = 180$

$\displaystyle x = 20$

Therefore angles are $\displaystyle 40 60 80$ degrees.

$\displaystyle \\$

Question 43: A certain number man can finish a piece of work in $\displaystyle 50 \text{ days. }$ If there are $\displaystyle 7$ more men the work can be completed $\displaystyle 10$ days earlier. How many men were originally there?

Answer:

Let $\displaystyle x$ men finish work in $\displaystyle 50 days$

Total work $\displaystyle = 50x \ man \ days$

$\displaystyle x + 7 \ men \ finish \ work \ in \ 49 \ days$

Total work $\displaystyle = (x + 7) \times 40$

Therefore

$\displaystyle 50x = 40 (x+7)$

$\displaystyle 5x = 4x+28$

$\displaystyle x = 28$

Original no of men $\displaystyle = 28$

$\displaystyle \\$

Question 44: Divide $\displaystyle 600$ in two parts such that $\displaystyle 495$ of one exceeds $\displaystyle 60$ of the other by $\displaystyle 120$ .

Answer:

Let the two parts $\displaystyle = x \text{ and } y$

$\displaystyle x + y = 600$

$\displaystyle 0.4x-0.6y = 120$

Solving we get

$\displaystyle x = 480$

$\displaystyle y = 120$

$\displaystyle \\$

Question 45: A workman is paid $\displaystyle Rs. 150$ for each day he works and is fined $\displaystyle Rs. 50$ for each day he is absent. In a month of $\displaystyle 30$ days he earned $\displaystyle Rs. 2100$ . For how many days did he remain absent?

Answer:

Salary $\displaystyle = 150 \text{ Rs. / Day }$

Fine $\displaystyle = 50 \text{ Rs. / Day }$

Let $\displaystyle x$ be the number of days worked

Therefore

$\displaystyle 150x-(30-x)50 = 2100$

Or $\displaystyle x = 18 \text{ days. }$

$\displaystyle \\$

3 thoughts on “Class 8: Linear Equations -Exercise 22B”

aquacompass7 says:

August 9, 2016 at 5:12 pm

Thank you for following my blog.

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abhi says:

December 17, 2017 at 6:21 pm

useful

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Devershi Samrat (jaiswal). says:

February 11, 2020 at 10:53 am

Thanks a lot to provide a good fascility to help as a reference during printing mistakes in books . Devershi Samrat (Theologist, a Principal , a super teaching specialist and principaling ). Padrauna( UP).

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Class 8: Linear Equations -Exercise 22B

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