(a) \(\ 15y=10y-5 \)
⇒ \(\ 15y-10y=-5 \)
⇒ \(\ 5y=-5 \)
⇒ \(\ y = \frac{-5}{5}⇒y=-1 \)
(b) \(\ 8-a=5a-3 \)
⇒ \(\ 8+3=5a+a \)
⇒ \(\ 11 = 6a \)
⇒ \(\ a = \frac{11}{6} \)
(c) \(\ 15x-1=2(5x-2)+8 \)
⇒ \(\ 15x-1=10x-4+8 \)
⇒ \(\ 15x-10x=4+1 ⇒ 5x=5 \)
⇒ \(\ x= \frac{5}{5} ⇒ x=1 \)
(d) \(\ \frac{14p}{5}-3=\frac{3p}{8}+\frac{7}{30} \)
⇒ \(\ \frac{14p}{5}-\frac{3p}{8}=\frac{7}{30}+3 \)
⇒ \(\ \frac{112p-15p}{40}=\frac{7+90}{30} \)
⇒ \(\ \frac{97p}{40}=\frac{97}{30} \)
⇒ \(\ 97p×30=40×97 \)
⇒ \(\ p=\frac{40×97}{30×97} \)
⇒ \(\ p=\frac{4}{3} \)
(e) \(\ 18q-3=15+9q\)
⇒ \(\ 18-9q=15+3 \)
⇒ \(\ 9q = 18 \)
⇒ \(\ q= \frac{18}{9} \)
⇒ \(\ q= 2 \)
(f) \(\ \frac{4}{5}y-1=\frac{2}{5}y+\frac{4}{15} \)
⇒ \(\ \frac{4}{5}y-\frac{2}{5}y=1+\frac{4}{15} \)
⇒ \(\ \frac{4y-2y}{5}=\frac{15+4}{15} \)
⇒ \(\ \frac{2y}{5}=\frac{19}{15} \)
⇒ \(\ 2y×15=5×19 \)
⇒ \(\ y=\frac{5×19}{2×15} \)
⇒ \(\ y=\frac{19}{6} \)
(g) \( 0.5(2+x)=0.1(6-x) \)
⇒ \(\ 1.0+0.5x=0.6-0.1x \)
⇒ \(\ 1.0-0.6=-0.5x-0.1x \)
⇒ \(\ 0.4=-0.6x \)
⇒ \(\ \frac{0.4}{0.6}=-x \)
⇒ \(\ x=-\frac{2}{3}\)
(h) \(2(x-1)+3(x-2)=4(x-3)\)
⇒ \(2x-2+3x-6=4x-12 \)
⇒ \(5x-4x-8=-12\)
⇒ \(x=8-12 ⇒ x=-4\)
(i) \(\frac{x-5}{3}=\frac{3x-1}{5}+2 \)
⇒ \( \frac{x-5}{3}-\frac{3x-1}{5}=2 \)
⇒ \( \frac{5(x-5)-3(3x-1)}{15}=2 \)
⇒ \( 5x-25-9x+3=2×15 \)
⇒ \( -4x-22=30 \)
⇒ \( -x=\frac{52}{4}\)
⇒ \( x=13\)
(j) \(\ \frac{x+1}{4}+0.25=x \)
⇒ \(\ \frac{x+1}{4}-x=-0.25 \)
⇒ \(\ \frac{(x+1)-4x}{4}=-\frac{25}{100} \)
⇒ \(\ x+1-4= -\frac{25×4}{100} \)
⇒ \(\ x-4x=-1-1 \)
⇒ \(\ -3x=-2 \)
⇒ \(\ x=\frac{3}{2} \)
(k) \(\frac{x+2}{2}-\frac{x-3}{4}=5-\frac{x-1}{2}\)
⇒ \(\frac{x+2}{2}-\frac{x-3}{4}+\frac{x-1}{2}=5\)
⇒ \(\frac{4(x+2)-3(x-3)+6(x-1)}{12}=5\)
⇒ \(4x+8-3x+9+6x-6=5×12\)
⇒ \(10x-3x+11=60\)
⇒ \(7x=60-11\)
⇒ \(x=\frac{49}{7} ⇒ x=7\)
(l) \((0.5y-2)-1=1-(0.5y-2)\)
⇒ \(0.5y-2-1=1-0.5y+2)\)
⇒ \(0.5y+0.5y=3+3=6\)
⇒ \( y=6\)
(m) \(5x+\frac{7}{2}=\frac{3}{2}x-14\)
⇒ \(\frac{7}{2}+14=\frac{3}{2}x-5x\)
⇒ \(\frac{7+28}{2}=\frac{3x-10x}{2}\)
⇒ \(7+28=3x-10x\)
⇒ \(35=-7x\)
⇒ \(\frac{35}{7}=-x\)
⇒ \(x=-5\)
(o) \(\frac{3}{4}y+2y=\frac{1}{2}+4y-3 \)
⇒ \(\frac{3}{4}y+2y-4y=\frac{1}{2}-3 \)
⇒ \(\frac{3y+8y-16y}{4}=\frac{1-6}{2}\)
⇒ \(2(11y-16y)=4(-5)\)
⇒ \(2(-5y)=-20\)
⇒ \(-10y=-20\)
⇒ \(y=\frac{20}{10}⇒ y=2 \)
2. A mother is 24 years older than her son. After 4 years, the age of the mother will be three times the age of the son. Find the present ages of mother and son.
Solution:
Let the present age of son be \(\ x\) years
and that of the mother be \(\ x+24\) years
According to question,
\(\ 3(x+4)=(x+24+4) \)
⇒ \(\ 3x+12=x+28\)
⇒ \(\ 3x-x=28-12 ⇒ 2x=16 \)
⇒ \(\ x=\frac{16}{2} ⇒x=8 \)
⸫ The required present age of son is \(\ 8\) years
and that of the mother is \(\ 8+24=32\) years.
3. Kashyap is 21 years old and Kinshuk is 15 years younger than Kashyap. After how many years will Kashyap be twice as old as Kinshuk?
Solution:
Let the number of years after which Kashyap will be twice as old as Kinshuk be \(\ x \)
Accoeding to question,
\(\ x+21=2(x+6) \)
⇒ \(\ x+21=2x+12 \)
⇒ \(\ 21-12=2x-x \)
⇒ \(\ 9=x \)
⸫ After 9 years Kashyap will be twice as old as Kinshuk.
4. The sum of the digits of a 2-digit number is 7. The number obtained by interchanging the digits exceeds the original number by 27. Find the number.
Solution:
Let the ones digit of the number be \(\ x\)
and so the tens digit be \(\ 7-x \) [⸪ the difference between ones and tens digit is 7]
⸫ The number = \(\ 10×(7-x)+x = 70-10x+x=70-9x \)
But when the digits are interchanged,
The number becomes \(\ 10x+(7-x) = 10x+7-x=9x+7 \)
According to question,
\(\ (9x+7)=(70-9x)+27 \)
⇒ \(\ 9x+7=70-9x+27 \)
⇒ \(\ 9x+9x=97-7 \)
⇒ \(\ x=\frac{90}{18} \)
⇒ \(\ x=5\)
⸫ The required number = \(\ 70-9×5=70-45=25 \)
5. One number is twice another number. If 15 is subtracted from both the numbers, then one of the new numbers becomes thrice that of the other new number. Find the numbers.
Solution:
Let the first number be \(\ x\) and the second be \(\ 2x \).
When 15 is subtracted,
the first number = \(\ x-15 \)
and the second number = \(\ 2x-15 \)
According to question,
\(\ 3(x-15)=2x-15 ⇒ 3x-45=2x-5 \)
⇒ \(\ 3x-2x=45-15⇒x=30 \)
⸫ The required first number is \(\ 30 \)
and the second number is \(\ 2×30=60\).
6. Mr Khan won a cash prize of a certain amount in the jackpot of a carnival. He gave half of this money and an additional amount of 10,000 to his wife. He also gave one third of the money received and an additional amount of 3000 to his son. If his wife got double the amount his son got, find the amount of money Mr Khan won in the jackpot.
Solution:
Let the amount of money Mr. Khan won in tha jackpot be Rs. \(\ x\).
According to question,
\(\ \frac{x}{2}+10000=2\left(\frac{x}{3}+3000\right) \)
⇒ \(\ \frac{x}{2}+10000=\frac{2x}{3}+6000 \)
⇒ \(\ 10000-6000 =\frac{2x}{3}-\frac{x}{2}\)
⇒ \(\ 4000 =\frac{4x-3x}{6}\)
⇒ \(\ 4000×6 =x\)
⇒ \(\ x=24000 \)
⸫ The required amount of money Mr. Khan won in the jackpot is Rs. \(\ 24000 \).
7. The sum of the digits of a 2-digit number is 12. The given number exceeds the number obtained by interchanging the digits by 36. Find the given number.
Solution:
Let the ones digit of the number be \(\ x \) and so tens digit be \(\ 12-x \)
⸫ The number = \(\ 10(12-x)+x=120-10x+x=120-9x \)
When the digits are interchanged,
The number becomes \(\ 10(x)+(12-x)=10x+12-x=12+9x \)
According to question,
\(\ 120-9x=(12+9x)+36 \)
⇒ \(\ 120-9x=48+9x \)
⇒ \(\ 120-48=9x+9x ⇒ 72=18x \)
⇒ \(\ x=\frac{72}{18}⇒ x=4 \)
⸫ The required number is \(\ 120-9(4)=120-36=84 \)
8. The ages of Suman and Neelam are in the ratio 3: 4. 4 years ago, their ages were in the ratio 5:7. Find their ages.
Solution:
Let the common ratio between the ages of Suman and Neelam be \(\ x \).
⸫ The present age of Suman be \(\ 3x \) years and that of the Nelaam be \(\ 4x \) years
According to question,
\(\ \frac{3x-4}{4x-4}=\frac{5}{7} \)
⇒ \(\ 7(3x-4)=5(4x-4) \)
⇒ \(\ 21x-28=20x-20 \)
⇒ \(\ 21x-20x=28-20 \)
⇒ \(\ x=8 \)
⸫ The required present age of Suman is \(\ 3×8=24 \) years and that of Neelam is \(\ 4×8=32\) years.
9. The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can the original number be?
Solution:
Let the ones digit of the number be \(\ x \) and so tens digit be \(\ x+3 \)
⸫ The number = \(\ 10(3+x)+x=30+10x+x=30+11x \)
When the digits are interchanged,
The number becomes \(\ 10(x)+(x+3)=10x+x+3=11x+3 \)
According to question,
\(\ 11x+30+11x+3=143 \)
⇒ \(\ 22x+33=143 \)
⇒ \(\ x=\frac{143-33}{22}=\frac{110}{22}⇒ x=5 \)
⸫ The required number is \(\ 11(5)+30=55+30=85 \)
10. Shoaib is twice as old as Shehzad. Five years ago his age was three times Shehzad's age. Find their present ages.
Solution:
Let the present age of Shehzad be \(\ x\) years and that of Shoaib be \(\ 2x\) years.
According to question,
\(\ 3(x-5)=2x-5 ⇒ 3x-15=2x-5 \)
⇒ \(\ 3x-2x=15-5⇒x=10 \)
⸫ The required present age of Shehzad is \(\ 10\) years and that of Shoaib is \(\ 2×10=20\) years.
11. Dinakar's mother's present age is six times Dinakar's present age. Dinakar's age five years from now will be one third of his mother's present age. What are their present ages?
Solution:
Let the present age of Dinakar be \(\ x\) years and that of his mother be \(\ 6x\) years.
According to question,
\(\ x+5=\frac{1}{3}×6x ⇒ x+5=2x \)
⇒ \(\ 2x-x=5⇒x=5 \)
⸫ The required present age of Dinakar is\(\ 5\) years and that of his mother be \(\ 6×5=30\) years.
12. There is a narrow rectangular plot reserved for a club in Malgudi. The length and breadth of the plot are in the ratio 11: 4. At the rate 100 per metre, it will cost the village panchayat 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the common ratio between the length and the breadth of the rectangle be \(\ x \).
⸫ The length = \(\ 11x \) and breadth = \(\ 4x \)
We know, the perimeter of a rectangle = \(\ 2(length+breadth) \)
= \(\ 2(11x+4x)=2×15=30x \)
Given, Cost of fencing each meter = Rs. 100
We also know,
Perimeter × Cost of fencing each meter = Total cost of fencing
⇒ \(\ 30x×100=75000 ⇒ 3000x = 75000 \)
⇒ \(\ x=\frac{75000}{3000}⇒x=25 \)
⸫ The length = \(\ 11×25=275 \) m
and breadth = \(\ 4×25=100 \) m
⸫ The required dimention of the plot is \(\ 275~m×100~m \).
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