(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 \(\ 10\)
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