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Exercise 7.2 | Cambridge Math | Class 8 | Linear Equations in One Variable

Cambridge Math|Class 8|Exercise 7.2 1. Solve the following:

(a)  15y=10y5
 15y10y=5
 5y=5
 y=55y=1


(b)  8a=5a3
 8+3=5a+a
 11=6a
 a=116


(c)  15x1=2(5x2)+8
 15x1=10x4+8
 15x10x=4+15x=5
 x=55x=1


(d)  14p53=3p8+730
 14p53p8=730+3
 112p15p40=7+9030
 97p40=9730
 97p×30=40×97
 p=40×9730×97
 p=43


(e)  18q3=15+9q
 189q=15+3
 9q=18
 q=189
 q=2


(f)  45y1=25y+415
 45y25y=1+415
 4y2y5=15+415
 2y5=1915
 2y×15=5×19
 y=5×192×15
 y=196


(g) 0.5(2+x)=0.1(6x)
 1.0+0.5x=0.60.1x
 1.00.6=0.5x0.1x
 0.4=0.6x
 0.40.6=x
 x=23


(h) 2(x1)+3(x2)=4(x3)
2x2+3x6=4x12
5x4x8=12
x=812x=4


(i) x53=3x15+2
x533x15=2
5(x5)3(3x1)15=2
5x259x+3=2×15
4x22=30
x=524
x=13


(j)  x+14+0.25=x
 x+14x=0.25
 (x+1)4x4=25100
 x+14=25×4100
 x4x=11
 3x=2
 x=32


(k) x+22x34=5x12
x+22x34+x12=5
4(x+2)3(x3)+6(x1)12=5
4x+83x+9+6x6=5×12
10x3x+11=60
7x=6011
x=497x=7


(l) (0.5y2)1=1(0.5y2)
0.5y21=10.5y+2)
0.5y+0.5y=3+3=6
y=6


(m) 5x+72=32x14
72+14=32x5x
7+282=3x10x2
7+28=3x10x
35=7x
357=x
x=5


(o) 34y+2y=12+4y3
34y+2y4y=123
3y+8y16y4=162
2(11y16y)=4(5)
2(5y)=20
10y=20
y=2010y=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
 3xx=28122x=16
 x=162x=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
 2112=2xx
 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  7x [⸪ the difference between ones and tens digit is 7]
⸫ The number =  10×(7x)+x=7010x+x=709x
But when the digits are interchanged,
The number becomes  10x+(7x)=10x+7x=9x+7
According to question,
 (9x+7)=(709x)+27
 9x+7=709x+27
 9x+9x=977
 x=9018
 x=5
⸫ The required number =  709×5=7045=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 =  x15
  and the second number =  2x15
According to question,
 3(x15)=2x153x45=2x5
 3x2x=4515x=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,
 x2+10000=2(x3+3000)
 x2+10000=2x3+6000
 100006000=2x3x2
 4000=4x3x6
 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  12x
⸫ The number =  10(12x)+x=12010x+x=1209x
When the digits are interchanged,
The number becomes  10(x)+(12x)=10x+12x=12+9x
According to question,
 1209x=(12+9x)+36
 1209x=48+9x
 12048=9x+9x72=18x
 x=7218x=4
⸫ The required number is  1209(4)=12036=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,
 3x44x4=57
 7(3x4)=5(4x4)
 21x28=20x20
 21x20x=2820
 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=1433322=11022x=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(x5)=2x53x15=2x5
 3x2x=155x=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=13×6xx+5=2x
 2xx=5x=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=750003000x=75000
 x=750003000x=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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