(a) \(\ \frac{5x-7}{3x}=2 \)
⇒ \(\ 5x-7=2×3x\)
⇒ \(\ -7 = 6x-5x \)
⇒ \(\ x=-7 \)
(b) \(\ \frac{5-7y}{2+4y}=\frac{-8}{7} \)
⇒ \(\ 7(5-7y)=(-8)(2+4y) \)
⇒ \(\ 35-49y=-16-32y \)
⇒ \(\ 35+16=49y-32y\)
⇒ \(\ 51=17y ⇒ y=\frac{51}{17}⇒y=3 \)
(c) \(\ \frac{9y}{7-6y}=15 \)
⇒ \(\ 9y=15(7-6y) \)
⇒ \(\ 9y = 105-90y\)
⇒ \(\ 9y+90y = 105\)
⇒ \(\ 99y=105 ⇒ 99y=105 \)
⇒ \(\ y=\frac{105}{99} ⇒ y=\frac{35}{33} \)
(d) \(\ \frac{2p-4}{3p+2}=\frac{-2}{3} \)
⇒ \(\ 3(2p-4)=(-2)(3p+2) \)
⇒ \(\ 6p-12=-6p-4 \)
⇒ \(\ 6p+6p=12-4 \)
⇒ \(\ 12p=8 ⇒ p=\frac{8}{12} ⇒ p=\frac{2}{3}\)
(e) \(\ \frac{2a-\frac{3}{4}}{9a+\frac{4}{7}}=\frac{1}{4} \)
⇒ \(\ 4\left(2a-\frac{3}{4}\right) = 1\left(9a+\frac{4}{7}\right) \)
⇒ \(\ 4(2a)-4\left(\frac{3}{4}\right)=9a+\frac{4}{7} \)
⇒ \(\ 8a-3=9a+\frac{4}{7} \)
⇒ \(\ 8a-9a=\frac{4}{7}+3 \)
⇒ \(\ -a=\frac{21+4}{7}=\frac{25}{7} \)
⇒ \(\ a=-\frac{25}{7} \)
(f) \(\ \frac{5x-11}{3x+7}=\frac{-2}{1} \)
⇒ \(\ 1(5x-11)=(-2)(3x+7) \)
⇒ \(\ 5x-11=-6p-14 \)
⇒ \(\ 5x+6x=11-14 \)
⇒ \(\ 11x=-3 ⇒ x=-\frac{3}{11} \)
(g) \(\ \frac{5x+2}{2x-5}=\frac{22}{5} \)
⇒ \(\ 5(5x+2)=22(2x-5) \)
⇒ \(\ 25x+10=44x-110 \)
⇒ \(\ 10+110=44x-25x\)
⇒ \(\ 120=19x ⇒ x=\frac{120}{19} \)
(h) \(\ \frac{\frac{2}{5}x-4}{\frac{3}{4}x+7}=\frac{4}{5} \)
⇒ \(\ 5\left(\frac{2}{5}x-4\right) = 4\left(\frac{3}{4}x+7\right) \)
⇒ \(\ \frac{10x}{5}-20=4×\frac{3x}{4}+4×7 \)
⇒ \(\ \frac{10x}{5}-20=3x+28 \)
⇒ \(\ \frac{10x}{5}-3x=28+20 \)
⇒ \(\ \frac{10x-15x}{5}-3x=48 ⇒ -5x=48×5 \)
⇒ \(\ x=-\frac{48×5}{5} ⇒ x=-48 \)
(i) \(\ \frac{2-b}{b+16}=\frac{3}{5} \)
⇒ \(\ 5(2-b)=3(b+16) \)
⇒ \(\ 10-5b=3b+48 \)
⇒ \(\ 10-48=3b+5b⇒ -38=8b \)
⇒ \(\ b=-\frac{38}{8}⇒b=-\frac{19}{4} \)
(j) \(\ \frac{(5x-7)-(2x+3)}{6x+11}=\frac{8}{3} \)
⇒ \(\ 3[(5x-7)-(2x+3)]=8(6x+11) \)
⇒ \(\ 3(5x-7-2x-3)=48x+88 \)
⇒ \(\ 15x-21-6x-9=48x+88 \)
⇒ \(\ 9x-30=48x+88 \)
⇒ \(\ 9x-48x=88+30 \)
⇒ \(\ -39x=118 ⇒ x=-\frac{118}{39} \)
(k) \(\ \frac{ 0.3+0.7x}{2}=0.85x \)
⇒ \(\ 0.3+0.7x=2×0.85x \)
⇒ \(\ 0.3=1.70x-0.7x \)
⇒ \(\ 0.3 =x \)
⇒ \(\ x=0.3 \)
(l) \(\ \frac{1}{9x-(5x-1)}=\frac{7}{6x-(2x-1)} \)
⇒ \(\ 1[6x-(2x-1)]=7[6x-(2x-1)] \)
⇒ \(\ 6x-(2x-1)=7[6x-2x+1)] \)
⇒ \(\ 6x-2x+1=42x-14x+7 \)
⇒ \(\ 6x-2x+1=42x-14x+7 \)
⇒ \(\ 4x+1=28x+7 \)
⇒ \(\ 1-7=28x-4x ⇒-6=24x \)
⇒ \(\ x=-\frac{6}{24}⇒ x=-\frac{1}{4} \)
2. The difference between two positive integers is 51. The quotient when one integer is divided by the other is 4. Find the two integers.
Solution:
Let one of the positive integers be \(\ x\) and the other be \(\ x-51 \)
According to question,
\(\ \frac{x}{x-51}=4 \)
\(\ x=4(x-51) ⇒x=4x-204 \)
⇒ \(\ 204=4x-x⇒ 204=3x\)
⇒ \(\ \frac{204}{3}=x ⇒x=68 \)
⸫ The required positive integers are \(\ 68 \) and \(\ 68-51=17 \).
3. The numerator of a rational number is less than the denominator by 8. If the denominator is decreased by 1 and numerator is increased by 17, then the number obtained is \(\ \frac{3}{2} \). Find the rational number.
Solution:
Let the denominator of the rational number be \(\ x\) and so the numerator be \(\ x-8 \)
⸫ The rational number = \(\ \frac{x-8}{x} \)
According to question,
\(\ \frac{(x-8)+17}{x-1}=\frac{3}{2} \)
⇒ \(\ 2(x-8+17)=3(x-1) \)
⇒ \(\ 2x-16+34=3x-3 \)
⇒ \(\ 18+3=3x-2x \)
⇒ \(\ x=21 \)
⸫ The required rational number is \(\ \frac{13}{21} \)
4. The denominator of a rational number is greater than its numerator by 3. If numerator and denominator are increased by 1 and 4, respectively, the number obtained is \(\ \frac{1}{2} \). Find the rational number.
Solution:
Let the numerator of the rational number be \(\ x\) and so the denominator be \(\ x+3 \)
⸫ The rational number =\(\ \frac{x-8}{x} \)
According to question,
\(\ \frac{x+1}{x+3+4}=\frac{1}{2} \)
⇒ \(\ 2(x+1)=1(x+3+4) \)
⇒ \(\ 2x+2=x+7 ⇒2x-x=7-2\)
⇒ \(\ x=5 \)
⸫ The required rational number is \(\ \frac{5}{8} \)
5. The ratio of two numbers is 3: 5. If each number is increased by 10, the ratio between the numbers so formed is 5: 7. Find the two original numbers.
Solution:
Let the common ratio between the numbers be \(\ x \)
So the numbers are \(\ 3x \) and \(\ 5x \)
According to question,
\(\ \frac{3x+10}{5x+10}=\frac{5}{7} \)
⇒ \(\ 21x+70=25x+50 \)
⇒ \(\ 70-50=25x-21x ⇒20=4x \)
⇒ \(\ \frac{20}{4} ⇒ x= 5 \)
⸫ The required numbers are \(\ 3×5=15 \) and \(\ 5×5=25 \)
6. A 2-digit number is such that ones digit is 1 more than the tens digit. The ratio of this number and the number formed by reversing its digits is 5: 6. Find the number.
Solution:
Let the tens digit of the number be \(\ x \) and so the ones digit be \(\ x+1 \)
⸫ The number is \(\ 10x+(x+1)=10x+x+1=11x+1 \)
But, when the digits are reversed,
The number becomes \(\ 10(x+1)+x=10x+10+x=11x+10 \)
According to question,
\(\ \frac{11x+1}{11x+10}=\frac{5}{6} \)
⇒ \(\ 5(11x+10)=6(11x+1) \)
⇒ \(\ 55x+50=66x+6 \)
⇒ \(\ 50-6=66x-55x ⇒ 44=11x \)
⇒ \(\ \frac{44}{11} =x ⇒ x=4 \)
⸫ The required number is \(\ =11×4+1=44+1=45 \)
7. A rational number is such that its denominator is twice its numerator. If both the numerator and the denominator are increased by 13, the rational number becomes \(\ \frac{2}{3} \) in the simplest form. Find the rational number.
Solution:
Let the numerator of the rational number be \(\ x\) and so the denominator be \(\ 2x \)
⸫ The rational number = \(\ \frac{x}{2x} \)
According to question,
\(\ \frac{x+13}{2x+13}=\frac{2}{3} \)
⇒ \(\ 3(x+13)=2(2x+13) \)
⇒ \(\ 3x+13=4x+26 \)
⇒ \(\ 39-26=4x-3x \)
⇒ \(\ x=13 \)
⸫ The required rational number is \(\ \frac{13}{26} \)
8. The length of a rectangle is twice its breadth. If the ratio of the perimeter of the rectangle and its area is1:2, find the dimensions of the rectangle.
Solution:
Let the breadth of the rectangle be \(\ x \) and the length be \(\ 2x \)
Perimeter of the rectangle = \(\ 2(x+2x) \) unit
Area of the rectangle = \(\ x×2x \) sq. unit
According to question,
\(\ \frac{2(x+2x)}{x×2x}=\frac{1}{2} \)
⇒ \(\ 2[2(x+2x)]=1(x×2x) \)
⇒ \(\ 2[2x+4x]=x×2x \)
⇒ \(\ 4x+8x=x×2x ⇒12x=x×2x \)
⇒ \(\ \frac{12x}{2x}=x ⇒ x=6 \)
The length= \(\ 6 \) unit
The length= \(\ 2×6=12 \) unit
⸫ The required dimension of the rectangle is \(\ 12 \) unit × \(\ 6 \) unit.
9. A fraction is such that the sum of its numerator and denominator is 15. If 8 is added to the numerator and12 is added to the denominator, the fraction becomes \(\ \frac{3}{4} \). Find the original fraction.
Solution:
Let the numerator of the fraction be \(\ x\) and the denominator be \(\ 15-x \).
⸫ The fraction = \(\ \frac{x}{15-x} \)
According to question,
\(\ \frac{x+8}{15-x+12}=\frac{3}{4} \)
⇒ \(\ 4(x+8)=3(15-x+12) \)
⇒ \(\ 4x+32=45-3x+36 \)
⇒ \(\ 4x+3x=81-32 ⇒ 7x=49 \)
⇒ \(\ x=\frac{49}{7} ⇒ x=7 \)
⸫ The required original fraction is \(\ \frac{7}{15} \)
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