{"id":585562,"date":"2022-03-16T09:27:13","date_gmt":"2022-03-16T09:27:13","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=585562"},"modified":"2022-03-17T06:17:16","modified_gmt":"2022-03-17T06:17:16","slug":"maharashtra-board-solutions-class-8-maths-practice-set-12-2-chapter-12-equations-in-one-variable","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/maharashtra-board-solutions-class-8-maths-practice-set-12-2-chapter-12-equations-in-one-variable\/","title":{"rendered":"Maharashtra Board Solutions Class 8-Maths (Practice Set 12.2): Chapter 12- Equations in One Variable"},"content":{"rendered":"\n

Class 8: Maths Chapter 12 solutions. Complete Class 8 Maths Chapter 12 Notes.<\/p>\n\n\n\n

Maharashtra Board Solutions Class 8-Maths (Practice Set 12.2): Chapter 12- Equations in One Variable<\/h2>\n\n\n\n

Maharashtra Board 8th Maths Chapter 12, Class 8 Maths Chapter 12 solutions<\/p>\n\n\n\n

Question 1.
Mother is 25 years older than her son. Find son\u2019s age, if after 8 years ratio of son\u2019s age to mother\u2019s age will be \\frac { 4 }{ 9 }49.
Solution:<\/strong>
Let the son\u2019s present age be x years.
\u2234 Mother\u2019s present age = (x + 25) years
After 8 years,
Son\u2019s age = (x + 8) years
Mother\u2019s age = (x + 25 + 8) = (x + 33) years
Since, the ratio of the son\u2019s age to mother\u2019s age after 8 years is \\frac { 4 }{ 9 }49.
\u2234 \\frac{x+8}{x+33}=\\frac{4}{9}x+8x+33=49
\u2234 9 (x + 8) = 4 (x + 33)
\u2234 9x + 72 = 4x + 132
\u2234 9x \u2013 4x = 132 \u2013 72
\u2234 5x = 60
\u2234 x = \\frac { 60 }{ 5 }605
\u2234 x = 12
\u2234 Son\u2019s present age is 12 years.<\/p>\n\n\n\n

Question 2.
The denominator of a fraction is greater than its numerator by 12. If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction is equivalent to \\frac { 1 }{ 2 }12 . Find the fraction.
Solution:<\/strong>
Let the numerator of the fraction be x.
The denominator of a fraction is greater than its numerator by 12.
\u2234 Denominator of the fraction = (x + 12)
\u2234 The required fraction = \\frac { x }{ x+12 }xx+12
For the new fraction,
numerator is decreased by 2.
\u2234 The new numerator = (x \u2013 2)
Also, denominator is increased by 7.
\u2234 The new denominator = (x + 12) + 7
= (x + 19)
Since, the new fraction is equivalent to \\frac { 1 }{ 2 }12.
\u2234 \\frac{x-2}{x+19}=\\frac{1}{2}x\u22122x+19=12
\u2234 2(x \u2013 2) = 1(x + 19)
\u2234 2x \u2013 4 = x + 19
\u2234 2x \u2013 x = 19 + 4
\u2234 x = 23
\u2234 The required fraction = \\frac{x}{x+12}=\\frac{23}{23+12}=\\frac{23}{35}xx+12=2323+12=2335
\u2234 The required fraction is \\frac { 23 }{ 35 }2335<\/p>\n\n\n\n

Question 3.
The ratio of the weights of copper and zinc in brass is 13:7. Find the weight of zinc in a brass utensil weighing 700 gm.
Solution:<\/strong>
Let the weight of zinc in the brass utensil be x gm.
Since, the ratio of the weights of copper to zinc in brass is 13:7.
\"Maharashtra
\u2234 Weight of copper in the brass utensil = \\left(\\frac{13}{7} x\\right)(137x) gm
The weight of the brass utensil = 700 gm
\u2234 \\frac { 13 }{ 7 }x+x=700137x+x=700
\u2234 \\frac { 13 }{ 7 }x137x x \u00d7 7 + x \u00d7 7 = 700 \u00d7 7
\u2234 13x + 7x = 4900
\u2234 20x = 4900
\u2234 x=\\frac { 4900 }{ 20 }x=490020
\u2234 x = 245
\u2234 The weight of zinc in the brass utensil is 245 gm.<\/p>\n\n\n\n

Question 4.
Find three consecutive whole numbers whose sum is more than 45 but less than 54.
Solution:<\/strong>
Let the three consecutive whole numbers be (x \u2013 1), x and (x + 1).
\u2234 Sum of the three numbers
= (x \u2013 1) + x + (x + 1)
= 3x
Given that, the sum of the three numbers is greater than 45 and less than 54.
When the sum of the three numbers is 45,
3x = 45
\u2234 x = \\frac { 45 }{ 3 }453
\u2234 x = 15
When the sum of the three numbers is 54,
\u2234 3x = 54
\u2234 x = \\frac { 54 }{ 3 }543
\u2234 x = 18
\u2234 the value of x is greater than 15 and less than 18.
\u2234 the value of x is either 16 or 17<\/p>\n\n\n\n

Case I:
If the value of x is 16, then the three consecutive whole numbers are
(16 \u2013 1), 16,(16 + 1)i.e., 15, 16, 17<\/p>\n\n\n\n

Case II:
If the value of x is 17, then the three consecutive whole numbers are (17 \u2013 1), 17, (17 + 1) i.e., 16, 17, 18.
\u2234 The three consecutive whole numbers are 15, 16, 17 or 16, 17, 18.<\/p>\n\n\n\n

Question 5.
In a two-digit number, digit at the ten\u2019s place is twice the digit at unit\u2019s place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. Find the number.
Solution:<\/strong>
Let the digit at unit\u2019s place be x.
The digit at the ten\u2019s place is twice the digit at unit\u2019s place.
\u2234 The digit at ten\u2019s place = 2x<\/p>\n\n\n\n

<\/td>Digit in units place<\/td>Digit in tens place<\/td>Number<\/td><\/tr>
Original Number<\/td>x<\/td>2x<\/td>(2x \u00d7 10) + x = 20x + x = 21x<\/td><\/tr>
New Number<\/td>2x<\/td>x<\/td>(x \u00d7 10) + 2x = 10x + 2x = 12x<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Since, the sum of the original number and the new number is 66.
\u2234 21x + 12x = 66
\u2234 33x = 66
\u2234 x = \\frac { 66 }{ 33 }6633
\u2234 x = 2
\u2234 Original number = 21x = 21 \u00d7 2 = 42
\u2234 the original number is 42.<\/p>\n\n\n\n

Question 6.
Some tickets of Rs 200 and some of Rs 100, of a drama in a theatre were sold. The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100. The total amount received by the theatre by sale of tickets was Rs 37000. Find the number of Rs 100 tickets sold.
Solution:<\/strong>
Let the number of tickets sold of Rs 100 be x.
The number of tickets of Rs 200 sold was 20 more than the number of tickets of Rs 100.
\u2234 Number of tickets sold of Rs 200 = (x + 20)
\u2234 Total amount received by the theatre through the sale of tickets = 100 \u00d7 x + 200 \u00d7 (x + 20)
= 100x + 200x + 4000
= 300x + 4000
Since, the total amount received by the theatre through the sale of tickets = Rs 37000
\u2234 300x + 4000 = 37000
\u2234 300x = 37000 \u2013 4000
\u2234 300x = 33000
\u2234 x=\\frac { 33000 }{ 300 }x=33000300
\u2234 x = 110
\u2234 110 tickets of Rs 100 were sold.<\/p>\n\n\n\n

Question 7.
Of the three consecutive natural numbers, five times the smallest number is 9 more than four times the greatest number, find the numbers.
Solution:<\/strong>
Let the three consecutive natural numbers be (x \u2013 1), x and (x + 1).
Here, the smallest number is (x \u2013 1) and the greatest number is (x + 1).
Since, five times the smallest number is 9 more than four times the greatest number.
\u2234 5 \u00d7 (x \u2013 1) = [4 \u00d7 (x + 1)] + 9
\u2234 5x \u2013 5 = 4x + 4 + 9
\u2234 5x \u2013 5 = 4x + 13
\u2234 5x \u2013 4x = 13 + 5
\u2234 x = 18 .
\u2234 the three numbers are (18 \u2013 1), 18, (18 + 1)
i. e., 17, 18, 19
\u2234 The three consecutive natural numbers are 17,18 and 19.<\/p>\n\n\n\n

Question 8.
Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending Rs 54. Then he sold the bicycle to Nikhil for Rs 1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.
Solution:<\/strong>
Let the cost price at which Raju purchased the bicycle be Rs x.
Since, Raju sold the bicycle at 8% profit to Amit.
\u2234 Selling price of bicycle for Raju = x + 8% of x
\"Maharashtra
Since, Amit spent Rs 54 on repairing the bicycle and then sold it to Nikhil for Rs 1134, at no loss and no profit.
\u2234 Selling price of bicycle + repairing cost = Rs 1134
\"Maharashtra
\u2234 The cost price of the bicycle at which Raju purchased it is Rs 1000.<\/p>\n\n\n\n

Question 9.
A cricket player scored 180 runs in the first match and 257 runs in the second match. Find the number of runs he should score in the third match so that the average of runs in the three matches be 230.
Solution:<\/strong>
Let the number of runs required by the cricket player to score in the third match be x.
Number of runs scored by the player in first match = 180
Number of runs scored in second match = 257
\u2234 Total runs scored by the player = 180 + 257 + x = 437 + x
Average of runs in the three matches = \\frac { 437+x }{ 3 }437+x3
Since, the average of runs should be 230.
\\frac { 437+x }{ 3 }=230437+x3=230
\u2234 437 + x = 230 \u00d7 3
\u2234 437 + x = 690
\u2234 x = 690 \u2013 437
\u2234 x = 253
\u2234 The cricket player should score 253 runs in the third match.<\/p>\n\n\n\n

Question 10.
Sudhir\u2019s present age is 5 more than three times the age of Viru. Anil\u2019s age is half the age of Sudhir. If the ratio of the sum of Sudhir\u2019s and Viru\u2019s age to three times Anil\u2019s age is 5:6, then find Viru\u2019s age.
Solution:<\/strong>
Let Viru\u2019s present age be x years.
Sudhir\u2019s present age is 5 more than three times the age of Viru.
\u2234 Sudhir\u2019s present age = (3x + 5) years
Anil\u2019s age is half the age of Sudhir.
\u2234 Anil\u2019s present age = \\left(\\frac{3 x+5}{2}\\right)(3x+52) years
Since, the ratio of the sum of Sudhir\u2019s and Viru\u2019s age to three times Anil\u2019s age is 5:6.
\"Maharashtra
\u2234 2 \u00d7 (24x + 30) = 45x + 75
\u2234 48x + 60 = 45x + 75
\u2234 48x \u2013 45x = 75 \u2013 60
\u2234 3x = 15
\u2234 x = \\frac { 15 }{ 3 }153
\u2234 x = 5
\u2234 Viru\u2019s present age is 5 years.<\/p>\n\n\n\n

Intext Questions and Activities<\/strong><\/p>\n\n\n\n

Question 1.
Write correct numbers in the boxes given. (Textbook pg. no. 78)
length is 3 times the breadth<\/strong>
\"Maharashtra
Perimeter of the rectangle = 40
2(__x + __x) = 40
2 \u00d7 __ x = 40
__ x = 40
x = __
\u2234 Breadth of rectangle = __ cm and Length of rectangle = __ cm
Solution:
<\/strong>length is 3 times the breadth
\"Maharashtra
Perimeter of the rectangle = 40
\u2234 2(3x + 1x) = 40
\u2234 2 \u00d7 4x = 40
\u2234 8x = 40
\u2234 x = 5
\u2234 Breadth of rectangle = 5 cm and Length of rectangle = 15 cm<\/p>\n\n\n\n

Download PDF<\/strong><\/h2>\n\n\n\n

Maharashtra Board Solutions Class 8-Maths (Practice Set 12.2): Chapter 12- Equations in One Variable<\/p>\n\n\n\n

Download PDF<\/strong>: Maharashtra Board Solutions Class 8-Maths (Practice Set 12.2): Chapter 12- Equations in One Variable PDF<\/a><\/p>\n\n\n\n

Chapterwise Maharashtra Board Solutions Class 8 Maths :<\/strong><\/h2>\n\n\n\n