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#### 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

(1) x − 4 = 3, x =1, 7,7

$x-4=3\phantom{\rule{0ex}{0ex}}⇒x-4+4=3+4\phantom{\rule{0ex}{0ex}}⇒x=7$

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

(2) 9m = 81, m = 3, 9, −3

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

(3) 2a + 4 = 0, a = 2, −2, 1

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

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

$3-y=4\phantom{\rule{0ex}{0ex}}⇒3-y+y=4+y\phantom{\rule{0ex}{0ex}}⇒3=4+y\phantom{\rule{0ex}{0ex}}⇒3-4=4+y-4\phantom{\rule{0ex}{0ex}}⇒-1=y$

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

#### 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) $\frac{9x}{8}$+ 1 = 10

(6) $\frac{y}{7}$ + $\frac{y-4}{3}$ = 2

(7) 13x − 5 = $\frac{3}{2}$

(8) 3(y + 8) = 10(y − 4) + 8

(9) $\frac{x-9}{x-5}$ = $\frac{5}{7}$

(10) $\frac{y-4}{3}$ + 3y = 4

(11) $\frac{b+\left(b+1\right)+\left(b+2\right)}{4}$ = 21

(1) 17p − 2 = 49

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

(2) 2m + 7 = 9

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

(3) 3x + 12 = 2x − 4

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

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

Hence, x$\frac{21}{2}$ is the solution of the equation.

(5) $\frac{9x}{8}$+ 1 = 10

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

(6) $\frac{y}{7}$ + $\frac{y-4}{3}$ = 2

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

(7) 13x − 5 = $\frac{3}{2}$

Hence, x$\frac{1}{2}$ is the solution of the equation.

(8) 3(y + 8) = 10(y − 4) + 8

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

(9) $\frac{x-9}{x-5}$ = $\frac{5}{7}$

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

(10) $\frac{y-4}{3}$ + 3y = 4

Hence, y $\frac{8}{5}$ is the solution of the equation.

(11) $\frac{b+\left(b+1\right)+\left(b+2\right)}{4}$ = 21

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

#### 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 $\frac{4}{9}$.

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,
$\frac{x+8}{x+33}=\frac{4}{9}\phantom{\rule{0ex}{0ex}}\therefore 9\left(x+8\right)=4\left(x+33\right)\phantom{\rule{0ex}{0ex}}\therefore 9x+72=4x+132\phantom{\rule{0ex}{0ex}}\therefore 9x-4x=132-72\phantom{\rule{0ex}{0ex}}\therefore 5x=60\phantom{\rule{0ex}{0ex}}\therefore x=12$

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

#### 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 $\frac{1}{2}$. Find the fraction.

Let the numerator of the fraction be x.

∴ Denominator = x + 12

∴ The fraction = $\frac{x}{x+12}$

If the numerator is decreased by 2 and the denominator is increased by 7, the new fraction obtained is $\frac{1}{2}$.
From the given information,

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

Hence, the fraction is $\frac{23}{35}$.

#### 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.

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

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

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

#### Question 4:

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

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

From the given information,

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.

#### 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.

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,

∴ The unit's digit = 2.
Tens Digit = 2 × 2 = 4.

Hence, the number is 42.

#### 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.

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

From the given information,

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

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

#### 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

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

From the given information,

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

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

#### 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.

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

Selling price of the bicycle =
= $\frac{108}{100}×x$

Amit repaired it spending ₹ 54.

Cost price of the bicycle for which Nikhil purchased it = $\frac{108}{100}x+54$
$⇒1134=\frac{108}{100}x+54\phantom{\rule{0ex}{0ex}}⇒1134-54=\frac{27}{25}x\phantom{\rule{0ex}{0ex}}⇒1080=\frac{27}{25}x\phantom{\rule{0ex}{0ex}}⇒1080×25=27x\phantom{\rule{0ex}{0ex}}⇒x=\frac{1080×25}{27}\phantom{\rule{0ex}{0ex}}⇒x=1000$

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

#### 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.

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

Total runs scored in all three 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.

#### 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.

Anil's present age = $\left(\frac{5+3x}{2}\right)$ years
$\left(5+3x\right)+\left(x\right):3\frac{\left(5+3x\right)}{2}=5:6\phantom{\rule{0ex}{0ex}}⇒\frac{5+4x}{\frac{3\left(5+3x\right)}{2}}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{2\left(5+4x\right)}{15+9x}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}⇒\frac{10+8x}{15+9x}=\frac{5}{6}\phantom{\rule{0ex}{0ex}}⇒6\left(10+8x\right)=5\left(15+9x\right)\phantom{\rule{0ex}{0ex}}⇒60+48x=75+45x\phantom{\rule{0ex}{0ex}}⇒48x-45x=75-60\phantom{\rule{0ex}{0ex}}⇒3x=15\phantom{\rule{0ex}{0ex}}⇒x=5\phantom{\rule{0ex}{0ex}}$