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Page No 77:

Question 1:

Each Equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.
(1) x − 4 = 3, x =1, 7,−7
(2) 9m = 81, m = 3, 9, −3
(3) 2a + 4 = 0, a = 2, −2, 1
(4) 3 y = 4, y = 1, 1, 2

Answer:

(1) x − 4 = 3, x =1, 7,7

x-4=3x-4+4=3+4x=7

Hence, x = 7 is the solution of the equation.

(2) 9m = 81, m = 3, 9, −3

9m=81Dividing both sides by 99m9=819m=9

Hence, m = 9 is the solution of the equation.

(3) 2a + 4 = 0, a = 2, −2, 1

2a+4=02a+4-4=-42a=-4Dividing both sides by 22a2=-42a=-2

Hence, a = 2 is the solution of the equation.

(4) 3 y = 4, y = 1, 1, 2

3-y=43-y+y=4+y3=4+y3-4=4+y-4-1=y

Hence, y = 1 is the solution of the equation.

Page No 77:

Question 2:

Solve the following equations.
(1) 17p − 2 = 49
(2) 2m + 7 = 9
(3) 3x + 12 = 2x − 4
(4) 5(x 3) = 3(x + 2)

(5) 9x8+ 1 = 10

(6) y7 + y-43 = 2

(7) 13x − 5 = 32

(8) 3(y + 8) = 10(y − 4) + 8

(9) x-9x-5 = 57

(10) y-43 + 3y = 4

(11) b+(b+1)+(b+2)4 = 21

Answer:

(1) 17p − 2 = 49

17p-2=49Adding 2 to both sides17p-2+2=49+217p=51Dividing both sides by 1717p17=5117p=3

Hence, p = 3 is the solution of the equation.

(2) 2m + 7 = 9

2m+7=9Subtracting 7 from both sides2m+7-7=9-72m=2Dividing both sides by 22m2=22m=1

Hence, m = 1 is the solution of the equation.

(3) 3x + 12 = 2x − 4

3x+12=2x-4Subtracting 12 from both sides3x+12-12=2x-4-123x=2x-16Subtracting 2x from both sides3x-2x=2x-16-2xx=-16

Hence, x = −16 is the solution of the equation.

(4) 5(x 3) = 3(x + 2)

5x-3=3x+25x-15=3x+6Adding 15 to both sides5x-15+15=3x+6+155x=3x+21Subtracting 3x from both sides5x-3x=3x+21-3x2x=21Dividing both sides by 22x2=212x=212

Hence, x212 is the solution of the equation.

(5) 9x8+ 1 = 10

9x8+1=10Multiplying both sides by 89x8×8+1×8=10×89x+8=80Subtracting 8 from both sides9x+8-8=80-89x=72Dividing both sides by 99x9=729x=8

Hence, x = 8 is the solution of the equation.

(6) y7 + y-43 = 2

y7+y-43=2Multiplying both sides by 2121y7+21y-43=2×213y+7y-4=423y+7y-28=4210y-28=42Adding 28 to both sides10y-28+28=42+2810y=70Dividing both sides by 1010y10=7010y=7

Hence, y = 7 is the solution of the equation.

(7) 13x − 5 = 32

13x-5=32Multiplying both sides by 2213x-5=2×3226x-10=3Adding 10 to both sides26x-10+10=3+1026x=13Dividing both sides by 2626x26=1326x=12

Hence, x12 is the solution of the equation.

(8) 3(y + 8) = 10(y − 4) + 8

3y+8=10y-4+83y+24=10y-40+83y+24=10y-32Adding 32 to both sides3y+24+32=10y-32+323y+56=10ySubtracting 3y from both sides3y+56-3y=10y-3y56=7yDividing both sides by 77y7=567y=8

Hence, y = 8 is the solution of the equation.

(9) x-9x-5 = 57

x-9x-5=57Multiplying both sides by 77x-9x-5=57×77x-63x-5=5Multiplying both sides by x-57x-63x-5x-5=5x-57x-63=5x-25Adding 63 to both sides7x-63+63=5x-25+637x=5x+38Subtracting 5x from both sides7x-5x=5x+38-5x2x=38Dividing both sides by 22x2=382x=19

Hence, x = 19 is the solution of the equation.

(10) y-43 + 3y = 4

y-43+3y=4Multiplying both sides by 33y-43+33y=4×3y-4+9y=1210y-4=12Adding 4 to both sides10y-4+4=12+410y=16Dividing both sides by 1010y10=1610y=85

Hence, y 85 is the solution of the equation.

(11) b+(b+1)+(b+2)4 = 21

b+b+1+b+24=21Multiplying both sides by 44×b+b+1+b+24=4×21b+b+1+b+2=843b+3=84Subtracting 3 from both sides3b+3-3=84-33b=81Dividing both sides by 33b3=813b=27

Hence, b = 27 is the solution of the equation.



Page No 79:

Question 1:

Mother is 25 year older than her son. Find son's age if after 8 years ratio of son's age to mother's age will be 49.

Answer:

Let the present age of the son be x years.

Age of the mother = (x + 25) years

After 8 years,
Age of son = (x + 8) years
Age of mother = (x + 25+ 8) years
                        = (x + 33) years

From the given information,
x+8x+33=499x+8=4x+339x+72=4x+1329x-4x=132-725x=60x=12

Hence, the present age of the son is 12 years.

Page No 79:

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 with 12. Find the fraction.

Answer:

Let the numerator of the fraction be x.

∴ Denominator = x + 12

∴ The fraction = xx+12

If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction obtained is 12.
From the given information,
x-2x+12+7=12 2x-2=1x+12+7 2x-4=x+19 2x-x=19+4 x=23

∴ Numerator of the fraction is 23.
∴ Denominator of the fraction is 23 + 12 = 35.

Hence, the fraction is 2335.



Page No 80:

Question 3:

The ratio of weight of copper and zinc in brass is 13 : 7. find the weight of zinc in a brass utensil weighing 700 gm.

Answer:

The ratio of weight of copper and zinc in brass is 13 : 7.

Let the weight of the copper be 13x.
Weight of zinc be 7x.

From the given information,

Given weight of brass = 700 gm

Weight of copper in brass + weight of zinc in brass = 700 gm

13x+7x=700 20x=700 x=35

∴ Weight of zinc in brass = 7 × 35 = 245 gm

Hence, the weight of zinc in a brass utensil is 245 gm.

Page No 80:

Question 4:

Find three conescutive whole numbers whose sum is more than 45 but less than 54.

Answer:

Let the three conescutive whole number be x, x + 1, x + 2.

From the given information,

45<x+x+1+x+2 and x+x+1+x+2<54 45<3x+3 and 3x+3<54 42<3x and 3x<51 14<x and x<17 x=15,16

If x = 15,
∴ The other two numbers are 16 and 17.

If x = 16,
∴ The other two numbers are 17 and 18.

Hence, three conescutive whole numbers are 15, 16, 17 or 16, 17, 18.

Page No 80:

Question 5:

In a two digit number, digit at the ten's place is twice the digit at units's place. If the number obtained by interchanging the digits is added to the original number, the sum is 66. find the number.

Answer:

Let the digit at the unit's place be x.

Digit at tens place = 2x

Original number = (2x)×10 + x = 21x

Number obtained by interchanging the digits = (x)×10 + 2x = 12x

From the given information,

12x+21x=66 33x=66 x=2
∴ The unit's digit = 2.
Tens Digit = 2 × 2 = 4.

Hence, the number is 42.

Page No 80:

Question 6:

Some tickets of ₹ 200 and some of ₹ 100, of a drama in a theatre were sold. The number of tickets of ₹ 200 sold was 20 more than the number of tickets of ₹ 100. The total amount received by the theatre by sale of tickets was ₹ 37000. Find the number of ₹ 100 tickets sold.

Answer:

Let the number of tickets of ₹ 100 be x.
Number of tickets of ₹ 200 = x + 20

From the given information,

100x+200x+20=37000 100x+200x+4000=37000 300x=37000-4000 300x=33000 x=110

∴ The number of tickets of ₹ 100 = 110.
Number of tickets of ₹ 200 = 110 + 20 = 130.

Hence, the number of tickets of ₹ 100 is 110.

Page No 80:

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

Answer:

Let the three conescutive natural number be x, x + 1, x + 2.

From the given information,

5x=4x+2+9 5x=4x+8+9 5x-4x=17 x=17

∴ The three consecutive natural numbers are 17, 17 + 1, 17 + 2.

Hence, the three consecutive natural numbers are 17, 18, 19.

Page No 80:

Question 8:

Raju sold a bicycle to Amit at 8% profit. Amit repaired it spending ₹ 54. Then he sold the bicycle to Nikhil for ₹ 1134 with no loss and no profit. Find the cost price of the bicycle for which Raju purchased it.

Answer:

Let the cost price of the bicycle for which Raju purchased it be x.

Selling price of the bicycle = 108% of cost price
                                            = 108100×x

Amit repaired it spending ₹ 54.

Cost price of the bicycle for which Nikhil purchased it = 108100x+54
1134=108100x+541134-54=2725x1080=2725x1080×25=27xx=1080×2527x=1000

Hence, the cost price of the bicycle for which Raju purchased it is ₹ 1000.

Page No 80:

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.

Answer:

Runs scored in First match = 180
Runs scored in Second match = 257

Total runs scored in all three matches = Average × Number of matches
                                                             = 230×3
                                                             = 690

Runs scored in Third match = 690 − (180 + 257)
                                             = 690 − 437
                                             = 253

Hence, runs he should score in the third match is 253.

Page No 80:

Question 10:

Sudhir's present age is 5 more than three times the age of Viru. Anil's age is half the age of Sudhir. If the ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5:6, then find Viru's age.

Answer:

Let Viru's present age be x years.
Sudhir's present age = (5 + 3x) years
Anil's present age = 5+3x2 years

∴ From the given information,
5+3x+x:35+3x2=5:65+4x35+3x2=5625+4x15+9x=5610+8x15+9x=56610+8x=515+9x60+48x=75+45x48x-45x=75-603x=15x=5

Hence, Viru's age is 5 years.



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