Exercise 7.1 | Cambridge Math | Class 8 | Linear Equations in One Variable

1. Solve the following:

(a) $\ x+8=2$
⇒ $\ x=2-8$
⇒ $\ x=-6$


(b) $\ 2z-15=30$
⇒ $\ 2z=30+15$
⇒ $\ z=\frac{45}{2}$
⇒ $\ z=22.5$



(c) $\ \frac{p}{2}=12$
⇒ $\ p=12×2$
⇒ $\ p=24$


(d) $\ 8x-(2x+4)=20$
⇒ $\ 8x-2x-4=20$
⇒ $\ 6x=20+4$
⇒ $\ x=\frac{24}{6}$
⇒ $\ x=4$


(e) $\ \frac{1}{2}x-5=7$
⇒ $\ \frac{1}{2}x=7+7=12$
⇒ $\ x=12×2 = 24$


(f) $\ \frac{x}{3}+1=\frac{2}{3}$
⇒ $\ \frac{x}{3}=\frac{2}{3}-1$
⇒ $\ \frac{x}{3}=\frac{2-3}{3}=-\frac{1}{3}$
⇒ $\ x=\frac{-1×3}{3} ⇒ x=-1$


(g) $\ 3(y-2)=27$
⇒ $\ y-2=\frac{27}{3}=9$
⇒ $\ y=9+2 ⇒ y=11$


(h) $\ \frac{2}{3}(p+2)+\frac{1}{5}(p-4)=\frac{7}{5}$
⇒ $\ \frac{2p+4}{3}+\frac{p-4}{5}=\frac{7}{5}$
⇒ $\ \frac{5(2p+4)+3(p-4)}{15}=\frac{7}{5}$
⇒ $\ 10p+20+3p-12=\frac{7×15}{5}$
⇒ $\ 13p+8=7×3$
⇒ $\ 13p= 21-8=13$
⇒ $\ x=\frac{13}{13}⇒ x=1$


(i) $\ \frac{z}{2}-5=\frac{6}{3}$
⇒ $\ \frac{z}{2}-5=2$
⇒ $\ \frac{z}{2}=2+5=7$
⇒ $\ z=7×2 ⇒ z=14$


(j) $(5y+2)-\left(\frac{16}{3}y+1\right)=8$
⇒ $5y+2-\frac{16}{3}y-1=8$
⇒ $5y-\frac{16}{3}y+1= 8$
⇒ $\frac{15y-16y}{3}= 8-1$
⇒ $-y=7×3 ⇒ y=-21$


(k) $\ 3(x-5)-7=14$
⇒ $\ 3x-15-7=14$
⇒ $\ 3x-22=14 ⇒3x=14+22$
⇒ $\ x=\frac{36}{3} ⇒x=12$


(l) $\ \frac{1-x}{0.3}=3$
⇒ $\ 1-x=3×0.3=0.9$
⇒ $\ 1-0.9=x ⇒ x=0.1$


(m) $\ \frac{x}{3}+\frac{5}{2}=-\frac{3}{2}$
⇒ $\ \frac{x}{3}=-\frac{3}{2} -\frac{5}{2}$
⇒ $\ \frac{x}{3}=\frac{-3-5}{2}=-\frac{8}{2}=-4$
⇒ $\ x=(-4)×3 ⇒ x=-12$


(n) $\ \frac{15}{4}-7x=9$
⇒ $\ \frac{15}{4}-9=7x$
⇒ $\ \frac{15-36}{4}=7x$
⇒ $\ \frac{-21}{4×7}=x$
⇒ $\ x=\frac{-3}{4}$


(o) $\ \frac{x}{3}+1=\frac{7}{15}$
⇒ $\ \frac{x}{3}=\frac{7-15}{15}= \frac{-8}{15}$
⇒ $\ x=\frac{-8×3}{15}$
⇒ $\ x=-\frac{8}{5}$


(p) $1.6=\frac{y}{1.5}$
⇒ $\ y=1.5×1.6$
⇒ $\ y=2.40$




2. The sum of three consecutive multiples of 6 is 126. Find these three multiples.
Solution:
Let the three consecutive multiples of 6 be $\ 6x, 6(x+1), 6(x+2)$
According to question,
$\ 6x+6(x+1)+6(x+2)=126$
⇒ $\ 6x+6x+6+6x+12=126$
⇒ $\ 18x=126-18$
⇒ $\ x=\frac{108}{18}=6$
⸫ The required three consecutive multiples of 6 are- $\ 36, 42, 48$



3. Karan has three gift boxes of different sizes. The first box is 10 cm longer than the third box and the second box is 5 cm longer than the third box. If the total length of the three boxes is 75 cm, then find the length of each box.
Solution:
Let the length of the third box be $\ x$ cm
and so the length the first box = $\ x+10$ cm
and the length of the second box = $\ x+5$ cm
According to question,
$\ x+(x+10)+(x+5)=75$
⇒ $\ 3x=75-15$
⇒ $\ x=\frac{60}{3}$
⇒ $\ x=20$
⸫ The required length of the third box = $\ 20$ cm
and so the length the first box = $\ 20+10=30$ cm
and the length of the econd box = $\ 20+5=25$ cm



4. Two numbers are in the ratio 10: 3. If their difference is 35, find the numbers.
Solution:
Let $\ x$ be the common ratio between the numbers.
and so the first number = $\ 10x$
and the second number = $\ 3x$
According to question,
$\ 10x-3x=35 ⇒ 7x=35$
$\ ⇒ x=\frac{35}{7}⇒x=5$
⸫ The required first number = $\ 10\times 5=50$
and the second number = $\ 3\times 5=15$



5. The sum of four consecutive multiples of 7 is 294. Find these multiples.
Solution:
Let the four consecutive multiples of 7 be $\ 7x, 7(x+1), 7(x+2), 7(x+3)$
According to question,
$\ 7x+7(x+1)+7(x+2)+7(x+3)=294$
⇒ $\ 7x+7x+7+7x+14+7x+21=294$
⇒ $\ 28x=294-42$
⇒ $\ x=\frac{252}{28}=9$
⸫ The required four consecutive multiples of 7 are $\ 63, 70, 77, 84$



6. The perimeter of a rectangle is 72 m. Its breadth is 10 m less than its length. Find the dimensions of the rectangle.
Solution:
Let the length of the rectangle be $\ x$ m and so the breadth be $\ x-10$ m
According to question,
Perimeter of the rectangle = $\ 72$
$\ ⇒ 2[x+(x-10)]=72$
$\ ⇒ 2(2x-10)=72$
$\ ⇒4x-20=72$
$\ ⇒4x=72+20$
$\ ⇒x=\frac{94}{4}$
$\ ⇒x=23$
⸫ The length = $\ 23$ m
and breadth = $\ 23-10=13$ m
⸫ The required dimention of the plot is $\ 23~m×13~m$.



7. What should be added to twice the rational number $\ \frac{-7}{3}$ to get $\ \frac{3}{7}$ ?
Solution:
Let the number be $\ x$
According to question,
$\ 2\left(\frac{-7}{3}\right)+x=\frac{3}{7}$
$\ ⇒x-\frac{14}{3}=\frac{3}{7}$
$\ ⇒x=\frac{3}{7}+\frac{14}{3}$
$\ ⇒x=\frac{9+98}{21}$
$\ ⇒x=\frac{107}{21}$
⸫ The required number is $\ \frac{107}{21}$



8. The perimeter of a rectangle is 13 cm and its width is $\ 2\frac{3}{4}$ cm. Find its length.
Solution:
Let the length of the rectangle be $\ x$ cm
Given, the breadth = $\ 2\frac{3}{4}=\frac{11}{4}$
According to question,
Perimeter of the rectangle = $\ 13$
$\ ⇒ 2\left(\ x+\frac{11}{4}\right)=13$
$\ ⇒ x+\frac{11}{4}=\frac{13}{2}$
$\ ⇒ x=\frac{13}{2}-\frac{11}{4}$
$\ ⇒ x=\frac{26-11}{4}$
$\ ⇒ x=\frac{15}{4}$
$\ ⇒ x=3\frac{3}{4}$
⸫ The required length of the rectangle is $\ 3\frac{3}{4}$ cm



9. The present age of Raman's mother is three times the present age of Raman. After 5 years, their ages will add to 66 years. Find their present ages.
Solution:
Let the present age of Raman be $\ x$ years and that of his mother be $\ 3x$ years.
According to question,
$\ (x+5)+(3x+5)=66$
$\ ⇒ x+5+3x+5=66$
$\ ⇒4x= 66-10$
$\ ⇒ x=\frac{56}{4}$
$\ ⇒ x= 14$
⸫ The required present age of Raman is $\ 14$ years
and that of his mother is $\ 3\times 14=42$ years.



10. Maria has 3 times as many two-rupee coins as she has five-rupee coins. If she has, in all, a sum of 77, how many coins of each denomination does she have?
Solution:
Let the number of Rs. 5 coins be $\ x$ and so the number of Rs. 2 coins be $\ 3x$
According to question,
$\ 5(x)+2(3x)=77$
$\ ⇒ 5x+6x=77$
$\ ⇒11x= 77$
$\ ⇒ x=\frac{77}{11}$
$\ ⇒ x= 7$
⸫ The required number of Rs. 5 coin is $\ 7$
and the number of Rs. 2 coin is $\ 3\times 7=21$



11. The sum of three consecutive multiples of 11 is 363. Find these multiples.
Solution:
Let the three consecutive multiples of 11 be $\ 11x, 11(x+1), 11(x+2)$
According to question,
$\ 11x+11(x+1)+11(x+2)=363$
⇒ $\ 11x+11x+11+11x+22=363$
⇒ $\ 33x=363-33$
⇒ $\ x=\frac{330}{33}=10$
⸫ The required three consecutive multiples of 11 are $\ 110, 121, 132$



12. The difference between two whole numbers is 66. The ratio of the two numbers is 2:5. What are the two numbers?
Solution:
Let $\ x$ be the common ratio between the whole numbers.
⸫ The first number be $\ 5x$ and the second number be $\ 2x$
According to question,
$\ 5x-2x=66 ⇒ 3x=66$
$\ ⇒ x=\frac{66}{3}⇒x=22$
⸫ The required first number is $\ 5\times 22=110$
and the second number is $\ 2\times 22=44$



13. Deveshi has a total of 590 as currency notes in the denominations of 50, 20 and 10. The ratio of the number of 50 notes and 20 notes is 3: 5. If she has a total of 25 notes, how many notes of each denomination does she have?
Solution:
Let the common ratio between the number of notes each denomination be $\ x$
⸫ The number of Rs. 50 note Deeveshi has be $\ 3x$
and that of Rs. 20 note be $\ 5x$
⸫ The number of Rs. 10 note = $\ 25-(3x+5x)$
   = $\ 25-8x$   [⸪ Total number of notes= 25]
According to question,
$\ 50(3x)+20(5x)+10(25-8x)=590$
$\ ⇒ 150x+100x+250-80x=590$
$\ ⇒ 250x-80x=590-250$
$\ ⇒ 170x=340$
$\ ⇒ x = \frac{340}{170}$
$\ ⇒ x = 2$
⸫ Deeveshi has-
    Rs. 50 note $\ =3\times 2=6$ Nos.
    Rs. 20 note $\ =5\times 2=10$ Nos.
    Rs. 10 note $\ =25-(8\times 2) =25-16=9$ Nos.



14. Three consecutive integers are such that when they are taken in increasing order and multiplied by 3, 4 and 5 respectively, they add up to 386. Find these numbers.
Solution:
Let the three consecutive integer be $\ x$, $\ (x+1)$ and $\ (x+2 )$
Now, $\ (x)\times 3=3x$
$\ (x+1)\times 4=4x+4$
and $\ (x+2)\times 5=5x+10$
According to question,
$\ 3x+4x+4+5x+10 = 386$
$\ ⇒ 12x+14= 386$
$\ ⇒ 12x = 386-14$
$\ ⇒ x = \frac{372}{12}$
$\ ⇒ x = 31$
⸫ The required three consecutive integers are $\ 31, 32$ and $\ 33$.


15. One of the two digits of a two-digit number is twice the other digit. If you interchange the digit of the two-digit number and add the resulting number to the original number, you get 66. What is the original number?
Solution:
Let the ones digit of the number be $\ x$ and so the tens digit be $\ 2x$
⸫ The number is $\ 10(2x)+x=20x+x=21x$
But when the digits are interchanged,
The number becomes $\ 10(x)+2x=10x+2x=12x$
According to question,
$\ 12x+21x=66 ⇒ 33x=66$
$\ ⇒ x=\frac{66}{33} ⇒ x=2$
⸫ The required number is $\ 10(4)+2=42$



16. A school organised an essay competition and decided that a winner in the competition would get a prize of 1000 and a participant who does not win would get a prize of 200. The total prize money distributed is 16000. Find the number of winners, if the total number of participants is 40.
Solution:
Let the number of winner be $\ x$
the number of non-winner (runner) =$\ 40-x$
According to question,
$\ 1000x+200(40-x)=16000$
⇒ $\ 1000x+8000-200x=16000$
⇒ $\ 800x=16000-8000$
⇒ $\ x=\frac{8000}{800}$
⇒ $\ x=10$
⸫ The required number of winner is $\ 10$