Class 8: Linear Equations -Exercise 22B

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

Answer:

Let the number $=x$

$4x-17=11$

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

$\\$

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

Answer:

Let the number $=x$

$4x+10=5x-5$

$x=15$

$\\$

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

Answer:

Let the original number $= x$

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

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

$x=60$

$\\$

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

Answer:

Let the number $= x$

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

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

$x=150$

$\\$

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

Answer:

Let the number $= x$

$x-21=71-x$

$2x=92$

$x=46$

$\\$

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

Answer:

Let the number $= x$

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

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

$x=40$

$\\$

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

Answer:

Let the number $= x$

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

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

$\\$

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

Answer:

Let the number $= x$

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

$x=105$

$\\$

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

Answer:

Let the number $=x$

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

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

$x=48$

$\\$

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

Answer:

Let the number $=x$

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

or $\frac{2}{15} x=10$

or $x=75$

$\\$

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

Answer:

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

Therefore

$3x=4y$

$x+y=84$

Solving

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

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

$y=3\times 12=36$

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

$\\$

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

Answer:

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

Therefore,

$4x:5x:6x$

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

Solving,

$15x=135$

$x=9$

Therefore

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

$\\$

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

Answer:

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

Given,

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

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

solving,

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

Substituting in ii)

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

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

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

or $y=25$

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

Two numbers are $15$ and $25$

$\\$

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

Answer:

Let the three consecutive numbers be

$x, x+2, x+4$

therefore,

$x+x+2+x+4=75$

$3x+6=75$

$3x=69$

$x=23$

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

$\\$

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

Answer:

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

Therefore

$x+y=25$

$7x+5y+139$

Solving we get

$x=25-y$

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

$175-2y=139$

$2y=175-139=36$

or $y=18$

The other part $=7$

$\\$

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

Answer:

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

Therefore

$x+y=180$

$x+12=2y$

Solving

$y=180-x$

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

$3x=360-12=348$

$x=116$

Therefore $y=180-116=64$

$\\$

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

Answer:

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

Given

$y=x+4$

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

Given,

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

$2x-2=x+3$

$x=5 \ and\ y=9$

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

$\\$

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

Answer:

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

Given

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

$x+2=2x-2$

$x=4$

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

$\\$

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

Answer:

Let the two digit number be $xy$

Given

$x+y=5 ...i)$

$xy+27=yx$

$10x+y+27=10y+x$

$9x+27=9y$

or $x+3=y ...ii)$

Solving i) and ii) together.

$x+3=(5-x)$

$2x=2$

$x=1$

$y=4$

Hence the number $= 14$

$\\$

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

Answer:

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

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

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

$7x=7$

$x=1$

$\\$

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

Answer:

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

Given

$x+y 110$

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

Solving

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

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

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

$y=10-65=45$

Two numbers are $45$ and $65$

$\\$

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

Answer:

Let the number subtracted from the numerator $= x$

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

$132-11x=13+6x$

$17x=119$

or $x=7$

$\\$

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

Answer:

Perimeter of right angled triangle $= 120$

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

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

Therefore

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

$30x=120$

$x=4$

Therefore, length of side $= 20, 48, 52$

$\\$

Question 24: The sum of the digits of a two-digit number is $9$ . If $9$ 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 $= xy$

$x+y=9 ...i)$

$yx+9=3(xy)$

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

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

$y+9=29x ...ii)$

Solving i) and ii)

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

$63-7x+9=29x$

$72x=36$

Or $x=2$

$y=9-2=7$

Therefore, the number $= 27$

$\\$

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

Answer:

Find the length and the breadth of the plot.

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

$l= 23 +b$

Given

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

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

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

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

$416=8b$

or $b=52m$

Therefore

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

$\\$

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 $= l$ and breadth $= b$

$l=2b$

$2l+2b=186$

$4b+2b=186$

$6b=186$

or $b=31$

$l=62$

$\\$

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

Answer:

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

$l= b+7$

Given

$21l +2b = 90$

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

$4b = 76$

Or $b= 19 cm$

$l= 19+7 = 26 cm$

$\\$

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

Answer:

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

$l+ 7 =2b$

Given,

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

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

$2l+2b=64$

Solving,

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

$6b=78$

$b=13,$

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

breadth $=13 cm$

length $=19cm$

$\\$

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 $= 5x$

If son’s age $= x$

Two years letter

Man’s age $= 5x +2$

Son’s age $= x+ 2$

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

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

Man’s age $= 30 yrs.$

$\\$

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 $= x$

Man’s age $= 2x$

$12 \ years \ ago$

Son’s age $= x-12$

Man’s age $= 2x-12$

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

$2x-12 = 3x-36$

$x= 24 = son's \ age$

Man’s age $= 48 \ years$

$\\$

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

Answer:

Let Rekha’s age $= x$

Seema’s age $= x + 10$

given

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

$3x+30=5x$

$2x=30$

or $x = 15$

Rekha’s sage $=15 yrs.$

Seema’s sage $=25 yrs.$

$\\$

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

Answer:

Let the present age of Parvati $= x yrs$

age of son $= y yrs.$

$x+y=55 ...i)$

Five years before

Parvati $=x-5 yrs.$

son $=y-5 yrs.$

Given,

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

$x-4y= -15 ...ii)$

solving i) and ii)

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

$5x=205$

or $x = 44 =$ Parvati’s age

Son’s age $= 55-44 = 11 years$

$\\$

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

Answer:

Man’s age $= 56 years$

Son’s age $= 24 years$

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

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

$56+x=48+2x$

or $x=8 years$

$\\$

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

Answer:

Let the current age of the girl $= x$

Given,

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

$x+9=3x-27$

$2x=36$

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

$\\$

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

Answer:

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

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

Total distance $= 480 km.$

$x+y=480$

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

Solving

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

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

$\\$

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

Answer:

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

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

or $x=270 km.$

$\\$

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

Answer:

Distance $= 300 km$

Let the speed of 1^st cyclist $= x$

Then speed of 2^nd cyclist $= x + 7$

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

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

Therefore

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

$4x + 48 = 300$

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

Speed of 1st cyclist $=63km/hr$

Speed of 2nd cyclist $=63+7=70 km/hr$

$\\$

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

Answer:

Let the speed of boat $= x km/hr$

Speed of stream $= 5 km/hr$

Speed of boat upstream $= x-5 km/hr$

Speed of the boat downstream $= x + 5 km/ hr$

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

$30x + 150 = 50x-250$

$400 = 20 x$

Or $x = 20 km/hr$

$\\$

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

Answer:

Let the base $= x cm$

Sides $= (x+ 4)$

Perimeter $= 62 cm$

Therefore

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

$3x+8=62$

$3x=54$

or $x=18$

Base $=18 cm$ , Sides $=22cm$

$\\$

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

Answer:

Let the number of candidates $= x$

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

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

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

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

No. of candidates $=220$

$\\$

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

Answer:

Let money with Kamal $= x$

Then money with Raman $= 3x$

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

Or $x= 450 Rs.$

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

$\\$

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

Answer:

Ratio of angle $= 2: 3: 4$

Therefore
$2x+3x+4x=180$

$9x=180$

$x=20$

Therefore, angles are $40, 60, 80$ degrees.

$\\$

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

Answer:

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

Total work $= 50x \ man \ days$

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

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

Therefore

$50x=40 (x+7)$

$5x=4x+28$

$x=28$

Original no of men $=28$

$\\$

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

Answer:

Let the two parts $= x \ and\ y$

$x + y = 600$

$0.4x-0.6y = 120$

Solving we get

$x= 480$

$y = 120$

$\\$

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

Answer:

Salary $= 150 \ Rs./day$

Fine $= 50 \ Rs./day$

Let $x$ be the number of days worked

Therefore

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

Or $x = 18$ days.

$\\$

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

aquacompass7 says:

August 9, 2016 at 5:12 pm

Thank you for following my blog.

LikeLike

abhi says:

December 17, 2017 at 6:21 pm

useful

LikeLike

	ICSE Mathematics Stu… on Class 8: Square Roots & Cu…
	Chaitanya Jha on Class 8: Square Roots & Cu…
	ICSE Student-Assista… on Class 8: Simple Interest and C…
	Aryan Panicker on Class 8: Simple Interest and C…
	Class 9: Rectilinear… on Class 8: Polygons (Lecture…

ICSE / ISC Mathematics Portal for K12 (ICSE / ISC Board) Students

Mathematics Made Easy for School Students

Class 8: Linear Equations -Exercise 22B

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

Leave a Reply Cancel reply

Share this:

Like this:

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

Leave a Reply Cancel reply