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

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

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

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

$\displaystyle \\$

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.

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

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

$\displaystyle \Rightarrow x = 15$

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow x = 60$

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow x = 150$

$\displaystyle \\$

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

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

$\displaystyle x-21 = 71-x$

$\displaystyle \Rightarrow 2x = 92$

$\displaystyle \Rightarrow x = 46$

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow x = 40$

$\displaystyle \\$

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

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

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

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

$\displaystyle \\$

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

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

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

$\displaystyle \Rightarrow x = 105$

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow x = 48$

$\displaystyle \\$

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

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

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

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

$\displaystyle \Rightarrow x = 75$

$\displaystyle \\$

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

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.

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.

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.

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 $\displaystyle 139$ .

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.

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.

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.

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.

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?

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.

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.

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.

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.

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 $\displaystyle 360 m^2$ .

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.

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.

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.

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.

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.

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.

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.

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?

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?

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 $\displaystyle 60\ \text{ km/hr }$ .

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.

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

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.

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.

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.

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?

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.

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?

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 $\displaystyle 120$ .

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?

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

$\displaystyle \\$