myhelper: RS Aggarwal solution class 8 chapter 8 Linear Equations Exercise 8B

Exercise 8B

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Q1 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 1:

Two numbers are in the ratio 8 : 3. If the sum of the numbers is 143, find the numbers.

Answer 1:

$Let the numbers be 8 x and 3 x . 8 x + 3 x = 143$

$\Rightarrow 11 x = 143$

$\Rightarrow x = \frac{143}{11}$

$\Rightarrow x = 13$

$∴ O ne number = 8 x = 8 \times 13 = 104$

$Other number = 3 x = 3 \times 13 = 39$

Q2 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 2:

$\frac{2}{3}$ of a number is 20 less than the original number. Find the number.

Answer 2:

$Let the original number be x . \frac{2}{3} of the number is 20 less than the original number .$

$∴ \frac{2}{3} x = x - 20$

$\Rightarrow \frac{2 x}{3} = x - 20$

$\Rightarrow 2 x = 3 (x - 20) (by cross multiplication)$

$\Rightarrow 2 x = 3 x - 60$

$\Rightarrow 2 x - 3 x = - 60$

$\Rightarrow - x = - 60$

$\Rightarrow x = 60 Therefore, the original number is 60 .$

Q3 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 3:

Four-fifths of a number is 10 more than two-thirds of the number. Find the number.

Answer 3:

$Let the number be x . Four fifths of the number is 10 more than two thirds of the number .$

$∴ \frac{4}{5} x = 10 + \frac{2}{3} x$

$\Rightarrow \frac{4 x}{5} = 10 + \frac{2 x}{3}$

$\Rightarrow \frac{4 x}{5} = \frac{30 + 2 x}{3} (L . C . M . of 1 and 3 is 3)$

$\Rightarrow 3 (4 x) = 5 (30 + 2 x) (by cross multiplication)$

$\Rightarrow 12 x = 150 + 10 x$

$\Rightarrow 12 x - 10 x = 150$

$\Rightarrow 2 x = 150$

$\Rightarrow x = \frac{150}{2} = 75$

$Therefore, the number is 75 .$

Q4 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 4:

Twenty-four is divided into two parts such that 7 times the first part added to 5 times the second part makes 146. Find each part.

Answer 4:

$Let one part be x . 7 times the first part = 7 x$

$Let the other part be (24 - x) . 5 times the second part = 5 (24 - x)$

$∴ 7 x + 5 (24 - x) = 146$

$\Rightarrow 7 x + 120 - 5 x = 146$

$\Rightarrow 7 x - 5 x = 146 - 120$

$\Rightarrow 2 x = 26$

$\Rightarrow x = \frac{26}{2} = 13$

$Therefore, one part is 13 .$

$Other part = (24 - x) = (24 - 13) = 11$

Q5 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 5:

Find the number whose fifth part increased by 5 is equal to its fourth part diminished by 5.

Answer 5:

$Let the number be x . Fifth part increased by 5 = \frac{x}{5} + 5$

$Fourth part diminished by 5 = \frac{x}{4} - 5$

$∴ \frac{x}{5} + 5 = \frac{x}{4} - 5$

$\Rightarrow 5 + 5 = \frac{x}{4} - \frac{x}{5}$

$\Rightarrow 10 = \frac{5 x - 4 x}{20}$

$\Rightarrow 10 = \frac{x}{20}$

$\Rightarrow 200 = x$

$\Rightarrow x = 200$

$Therefore, the number is 200 .$

Q7 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 6:

Three numbers are in the ratio of 4 : 5 : 6. If the sum of the largest and the smallest equals the sum of the third and 55, find the numbers.

Answer 6:

$Let the common multiple for the given three numbers be x .$

Then, the three numbers would be 4x, 5x and 6x.

∴4x + 6x = 5x + 55

⇒10x = 5x + 55

⇒ 10x - 5x = 55

⇒ 5x = 55

⇒ x = 555= 11

∴ Smallest number = 4x = 4(11) = 44

Largest number is = 6x = 6(11) = 66

Third number =5x = 5(11) = 55

Therefore, the three numbers are 44, 55 and 66.

Q7 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 7:

If 10 be added to four times a certain number, the result is 5 less than five times the number. Find the number.

Answer 7:

$Let the number be x .$

$∴ 10 + 4 x = 5 x - 5$

$\Rightarrow 10 + 5 = 5 x - 4 x$

⇒ 15 = x

⇒ x = 15 (by transposition)

Therefore, the number is 15.

Q8 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 8:

Two numbers are such that the ratio between them is 3 : 5. If each is increased by 10, the ratio between the new numbers so formed is 5 : 7. Find the original numbers.

Answer 8:

Let us consider x as the common multiple of both the number.
Then, first number = 3x
Second number = 5x

$∴ \frac{3 x + 10}{5 x + 10} = \frac{5}{7}$

$\Rightarrow 7 (3 x + 10) = 5 (5 x + 10) (by cross multiplication)$

$\Rightarrow 21 x + 70 = 25 x + 50$

$\Rightarrow 21 x - 25 x = 50 - 70$

$\Rightarrow - 4 x = - 20$

$\Rightarrow x = \frac{- 20}{- 4} = 5$

Therefore, the common multiple of both the numbers is 5.
First number = $3 x = 3 \times 5 = 15$
Second number = $5 x = 5 \times 5 = 25$

Q9 | Ex-8B | Class 8 | RS AGGARWAL | Chapter 8 | Linear Equations | myhelper

Question 9:

Find three consecutive odd numbers whose sum is 147.

Answer 9:

Let the first odd number be x.

Let the second odd number be ( x+2 ).

Let the third odd number be ( x+4 ).

∴ x + ( x+2 ) + ( x+4 ) = 147

⇒ x + x + 2 + x + 4 = 147

⇒ 3x + 6 = 147

⇒ 3x = 147 -6

⇒ 3x = 141

⇒ x = 1413= 47

Therefore, the first odd number is 47.

Second odd number = ( x+2 ) = ( 47+2 ) = 49

Third odd number = ( x+4 ) = ( 47+4 ) = 51

ALTERNATE METHOD

Let the first odd number be 2x+1

Let the second odd number be 2x+3

Let the third odd number be 2x+5

∴ (2x+1) + (2x+3) + (2x+5) = 147

⇒2x+1+2x+3+2x+5=147

⇒6x+9=147

⇒6x=138

⇒$x=\frac{138}{6}=23$

Therefore, the first odd number is 2x+1=2(23)+1=46+1=47

Second odd number is 2x+3=2(23)+3=46+3=49

Third odd number is 2x+5=2(23)+5=46+5=51

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Question 10:

Find three consecutive even numbers whose sum is 234.

Answer 10:

$Let the first even number b e x .$

$Let the second even number b e x + 2 .$

$Let the third even number be x + 4 .$

$∴ x + x + 2 + x + 4 = 234$

$\Rightarrow x + x + 2 + x + 4 = 234$

$\Rightarrow 3 x + 6 = 234$

$\Rightarrow 3 x = 234 - 6$

$\Rightarrow 3 x = 228$

$\Rightarrow x = \frac{228}{3} = 76$

$∴ F irst even number \Rightarrow x = 76$

$S econd even number$ ⇒ x+2 = 76+2 = 78

$T hird even number \Rightarrow x + 4 = 80$

ALTERNATE METHOD

Let the first odd number be 2(x)=2x

Let the second odd number be 2(x+1)=2x+2

Let the third odd number be 2(x+2)=2x+4

∴ (2x) + (2x+2) + (2x+4) = 234

⇒2x+2x+2+2x+4=234

⇒6x+6=234

⇒6x=228

⇒$x=\frac{228}{6}=38$

Therefore, the first odd number is 2x=2(38)=76

Second odd number is 2x+2=2(38)+2=76+2=78

Third odd number is 2x+4=2(38)+4=76+4=80

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Question 11:

The sum of the digits of a two-digit number is 12. If the new number formed by reversing the digits is greater than the original number by 54, find the original number. Check your solution.

Answer 11:

$Let the digit in t h e units place be x .$

$D igit in t h e tens place = (12 - x)$

$∴ Original number = 10 (12 - x) + x = 120 - 9 x$

$On reversing the digits, we have x at the tens place and (12 - x) at the units place .$

$∴ New number = 10 x + 12 - x = 9 x + 12 New number - Original number = 54$

$\Rightarrow 9 x + 12 - (120 - 9 x) = 54$

$\Rightarrow 9 x + 12 - 120 + 9 x = 54$

$\Rightarrow 18 x - 108 = 54$

$\Rightarrow 18 x = 54 + 108$

$\Rightarrow 18 x = 162$

$\Rightarrow x = \frac{162}{18} = 9$

$Therefore, the digit in t h e units place is 9 .$

$D igit in tens place = (12 - x) = (12 - 9) = 3$

$Therefore, the original number is 39 .$

$Check : The original number is 39 .$

$S um of the digits in the original number = (3 + 9) = 12$

$N ew number obtained on reversing the digit s = 93$

$New number - Original number = (93 - 39) = 54$

$Thus, both the given conditions are satisfied by 39 .$

$Hence, the original number i s 39 .$

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Question 12:

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number. Check your solution.

Answer 12:

$Let the digit in t h e units place be x .$

$D igit in t h e tens place = 3 x O riginal number$

$= 10 (3 x) + x = 30 x + x$

$On reversing the digits, we have x at the tens place and (3 x) at the units place .$

$∴ N ew number = 10 (x) + 3 x = 10 x + 3 x$

$New number = Original number - 36$

$\Rightarrow 10 x + 3 x = 30 x + x - 36$

$\Rightarrow 13 x = 31 x - 36$

$\Rightarrow 36 = 31 x - 13 x$

$\Rightarrow 36 = 18 x$

$\Rightarrow 18 x = 36$

$\Rightarrow x = \frac{36}{18} = 2$

$Therefore, the digit in t h e units place is 2 .$

$D igit in t h e tens place = (3 x) = 3 \times 2 = 6$

$Therefore, the original number is 62 .$

$Check : New number + 36 = Original Number 26 + 36 = 62$

$Henc e, both the conditions are satisfied .$

$Therefore, the original number i s 62 .$

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Question 13:

The denominator of a rational number is greater than its numerator by 7. If the numerator is increased by 17 and the denominator decreased by 6, the new number becomes 2. Find the original number.

Answer 13:

$Let the numerator be x .$

$The denominator is greater than the numerator by 7 .$

$∴ (x + 7)$

$∴ \frac{x + 17}{(x + 7) - 6} = 2$

$\Rightarrow \frac{x + 17}{x + 1} = 2$

$\Rightarrow x + 17 = 2 (x + 1) (by cross multiplication)$

$\Rightarrow x + 17 = 2 x + 2$

$\Rightarrow x - 2 x = 2 - 17$

$\Rightarrow - x = - 15$

$\Rightarrow x = 15$

$Therefore, the numerator is 15 .$

$Denominator = (x + 7) = (15 + 7) = 22$

$∴ Original number = \frac{15}{22}$

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Question 14:

In a fraction, twice the numerator is 2 more than the denominator. If 3 is added to the numerator and to the denominator, the new fraction is $\frac{2}{3}$ . Find the original fraction.

Answer 14:

$Denominator, d = x$

$It is given that twice the numerator is equal to two more than the denominator . ∴ Twice of numerator, 2 n = x + 2 ∴ Numerator, n = \frac{x + 2}{2}$

$∴ \frac{n + 3}{d + 3} = \frac{2}{3}$

$\Rightarrow 3 (n + 3) = 2 (d + 3) (b y c r o s s m u l t i p l i c a t i o n)$

$\Rightarrow 3 n + 9 = 2 d + 6$

$\Rightarrow 3 n - 2 d = 6 - 9$

$\Rightarrow 3 n - 2 d = - 3$

$On replace d by x and n by \frac{x + 2}{2} :$

$\Rightarrow 3 (\frac{x + 2}{2}) - 2 x = - 3$

$\Rightarrow \frac{3 x + 6 - 4 x}{2} = - 3 (taking the L . C . M . of 2 and 1 as 2)$

$\Rightarrow 6 - x = - 6 (by cross multiplication)$

$\Rightarrow - x = - 6 - 6$

$\Rightarrow x = 12$

$The denominator is 12 .$

$∴ Numerator = \frac{x + 2}{2} = \frac{12 + 2}{2} = \frac{14}{2} = 7$

$∴ O riginal fraction = \frac{7}{12}$

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Question 15:

The length of a rectangle exceeds its breadth by 7 cm. If the length is decreased by 4 cm and the breadth is increased by 3 cm, the area of the new rectangle is the same as the area of the original rectangle. Find the length and the breadth of the original rectangle.

Answer 15:

$Let the breadth of the original rectangle be x cm .$

$Then, its length will be (x + 7) cm .$

$The area of the rectangle will be (x) (x + 7) {cm}^{2} .$

$∴ (x + 3) (x + 7 - 4) = (x) (x + 7)$

$\Rightarrow (x + 3) (x + 3) = x^{2} + 7 x$

$\Rightarrow x^{2} + 3 x + 3 x + 9 = x^{2} + 7 x$

$\Rightarrow x^{2} + 6 x + 9 = x^{2} + 7 x$

$\Rightarrow 9 = x^{2} - x^{2} + 7 x - 6 x$

$\Rightarrow 9 = x$

$\Rightarrow x = 9 (by transposition) Breadth of the original rectangle = 9 cm$

$Length of the original rectangle = (x + 7) = (9 + 7) = 16 cm$

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Question 16:

The width of a rectangle is two-thirds its length. If the perimeter is 180 metres, find the dimensions of the rectangle.

Answer 16:

$Let the width of the rectangle be x cm .$

$It is \frac{2}{3} of the length of the rectangle .$

$This means that the length of the rectangle will be \frac{3}{2} x .$

$Perimeter of the rectangle = 2 (x) + 2 (\frac{3}{2}) x = 180 m$

$∴ 2 x + \frac{6 x}{2} = 180$

$\Rightarrow \frac{4 x + 6 x}{2} = 180 (taking the L . C . M . of 1 on the L . H . S . of the equation)$

$\Rightarrow 10 x = 2 \times 180 (by cross multiplication)$

$\Rightarrow 10 x = 360$

$\Rightarrow x = \frac{360}{10} = 36$

$Therefore, the width of the rectangle is 36 m .$

$Length of the rectangle will be = \frac{3}{2} x = \frac{3}{2} (36) = 54 m$

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Question 17:

An altitude of a triangle is five-thirds the length of its corresponding base. If the altitude be increased by 4 cm and the base decreased by 2 cm, the area of the triangle remains the same. Find the base and the altitude of the triangle.

Answer 17:

$Let the length of the base of the triangle be x cm .$

$Then, its altitude will be \frac{5}{3} x cm . Area of the triangle = \frac{1}{2} (x) (\frac{5}{3} x) = \frac{5}{6} x^{2}$

$∴ \frac{1}{2} (x - 2) (\frac{5}{3} x + 4) = \frac{5}{6} x^{2}$

$\Rightarrow (\frac{x - 2}{2}) (\frac{5 x + 12}{3}) = \frac{5 x^{2}}{6}$

$\Rightarrow \frac{(x - 2) (5 x + 12)}{6} = \frac{5 x^{2}}{6}$

$\Rightarrow \frac{5 x^{2} + 12 x - 10 x - 24}{6} = \frac{5 x^{2}}{6}$

$\Rightarrow 5 x^{2} + 2 x - 24 = 5 x^{2} (cancelling the denominators from both the sides since they are same)$

$\Rightarrow 5 x^{2} - 5 x^{2} + 2 x = 24$

$\Rightarrow 2 x = 24$

$\Rightarrow x = \frac{24}{2} = 12 m$

$Therefore, the bas e of the triangle is 12 m .$

$Altitude of the triangle = \frac{5}{3} x = \frac{5}{3} (12) = 20 m$

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Question 18:

Two angles of a triangle are in the ratio 4 : 5. If the sum of these angles is equal to the third angle, find the angles of the triangle.

Answer 18:

$Let the common multiple of all the three angles be x .$

$Then, the first angle will be 4 x . And the second angle w i l l be 5 x .$

$In a triangle, sum of all the three angles will be equal to 180 ° .$

$∴ T hird angle = 180 - (4 x + 5 x) = 180 - 9 x$

$∴ 4 x + 5 x = 180 - 9 x$

$\Rightarrow 9 x = 180 - 9 x$

$\Rightarrow 9 x + 9 x = 180$

$\Rightarrow 18 x = 180$

$\Rightarrow x = \frac{180}{18} = 10$

$First angle = 4 x = 4 \times 10 = 40^{°}$

$Second angle = 5 x = 5 \times 10 = 50 °$

$Third angle = 4 x + 5 x = 9 x = 9 \times 10 = 90 °$

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Question 19:

A steamer goes downstream from one port to another in 9 hours. It covers the same distance upstream in 10 hours. If the speed of the stream be 1 km/h, find the speed of the steamer in still water and the distance between the ports.

Answer 19:

$Let the speed of the steamer in still water be x km / h .$

$Speed (downstream) = (x + 1) km / h$

$Speed (upstream) = (x - 1) km / h$

$Distance covered in 9 hours while going downstream = 9 (x + 1) km$

$Distance covered in 10 hours while going upstream = 10 (x - 1) km$

$But both of these distances will be same . 9 (x + 1) = 10 (x - 1)$

$\Rightarrow 9 x + 9 = 10 x - 10$

$\Rightarrow 9 + 10 = 10 x - 9 x$

$\Rightarrow 19 = x$

$\Rightarrow x = 19$

$Therefore, the speed of the steamer in still water is 19 km / h .$

$Distance between the ports = 9 (x + 1) = 9 (19 + 1) = 9 \times 20 = 180 km$

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Question 20:

The distance between two stations is 300 km. Two motorcyclists start simultaneously from these stations and move towards each other. The speed of one of them is 7 km/h more than that of the other. If the distance between them after 2 hours of their start is 34 km, find the speed of each motorcyclist. Check your solution.

Answer 20:

$Let the speed of one motorcyclist be x km / h .$

$So, the speed of the other motorcyclist will be (x + 7) km / h .$

$D istance travelled by the first motorcyclist in 2 hours = 2 x km$

$D istance travelled by the second motorcyclist in 2 hours = 2 (x + 7) km$

$Therefore, 300 - (2 x + (2 x + 14)) = 34$

$\Rightarrow 300 - (2 x + 2 x + 14) = 34$

$\Rightarrow 300 - 4 x - 14 = 34$

$\Rightarrow 286 - 4 x = 34$

$\Rightarrow 286 - 34 = 4 x$

$\Rightarrow 252 = 4 x$

$\Rightarrow x = \frac{252}{4} = 63$

$Therefore, the speed of the first motorcyclist is 63 km / h .$

$The speed of the second motorcyclist is (x + 7) = (63 + 7) = 70 km / h .$

$Check : The distance covered by the first motorcyclist in 2 hours = 63 \times 2 = 126 km$

$The distance covered by the second motorcyclist in 2 hours = 70 \times 2 = 140 km$

$The distance between the motorcyclists after 2 hours = 300 - (126 + 140) = 34 km (which is the same as given)$

$Therefore, the speeds of the motorcyclists are 63 km / h and 70 km / h, respectively .$

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Question 21:

Divide 150 into three parts such that the second number is five-sixths the first and the third number is four-fifths the second.

Answer 21:

$Let the first number be x . Then, the second number will be \frac{5}{6} x .$

$Third numbe r = \frac{4}{5} (\frac{5}{6} x) = \frac{2}{3} x$

$∴ x + \frac{5 x}{6} + \frac{2 x}{3} = 150$

$\Rightarrow \frac{6 x + 5 x + 4 x}{6} = 150 (multiplying the L . H . S . by 6, which is the L . C . M . of 1, 6 and 3)$

$\Rightarrow 15 x = 150 \times 6 (by cross multiplication)$

$\Rightarrow 15 x = 900$

$\Rightarrow x = \frac{900}{15} = 60$

$Therefore, the first number is 60 . Second number = \frac{5}{6} x = \frac{5}{6} (60) = 50$

$Third number = \frac{2}{3} x = \frac{2}{3} (60) = 40$

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Question 22:

Divide 4500 into two parts such that 5% of the first part is equal to 10% of the second part.

Answer 22:

$Let the first part be x . Let the second part be (4500 - x) .$

$∴ 5 % o f x = 10 % o f (4500 - x)$

$\Rightarrow (\frac{5}{100}) x = (\frac{10}{100}) (4500 - x)$

$\Rightarrow \frac{5 x}{100} = \frac{45000 - 10 x}{100}$

$\Rightarrow 5 x = 45000 - 10 x (by cancellation of same denominators from both the sides)$

$\Rightarrow 5 x + 10 x = 45000$

$\Rightarrow 15 x = 45000$

$\Rightarrow x = \frac{45000}{15} = 3000$

$Therefore, the first part is 3000 .$

$Second part = (4500 - x) = (4500 - 3000) = 1500$

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Question 23:

Rakhi's mother is four times as old as Rakhi. After 5 years, her mother will be three times as old as she will be then. Find their present ages.

Answer 23:

$Let the present age of Rakhi be x .$

$Then, the present age of Rakhi' s mother will be 4 x .$

$After five years, Rakhi' s age will be (x + 5) .$

$After five years, her mother' s age will be (4 x + 5) .$

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

$\Rightarrow 4 x + 5 = 3 x + 15$

$\Rightarrow 4 x - 3 x = 15 - 5$

$\Rightarrow x = 10$

$Present age of Rakhi = 10 years$

$Present age of Rakhi' s mother = 4 (x) = 4 \times 10 = 40 years$

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Question 24:

Monu's father is 26 years younger than Monu's grandfather and 29 years older than Monu. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Answer 24:

$Let the age of Monu' s father be x years .$

$The age of Monu' s grandfather will be (x + 26) .$

$Then, the age of Monu will be (x - 29) .$

$∴ x + (x + 26) + (x - 29) = 135$

$\Rightarrow x + x + 26 + x - 29 = 135$

$\Rightarrow 3 x - 3 = 135$

$\Rightarrow 3 x = 135 + 3$

$\Rightarrow 3 x = 138 \Rightarrow x = \frac{138}{3} = 46$

$∴ A ge of Monu' s father = 46 years$

$Age of Monu' s grandfather = (x + 26) = (46 + 26) = 72 years$

$Age of Monu = (x - 29) = 46 - 29 = 17 years$

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Question 25:

A man is 10 times older than his grandson. He is also 54 years older than him. Find their present ages.

Answer 25:

$Let the age of the grandson be x years .$

$Then, his grandfather' s age will be 10 x .$

$Also, the grandfather is 54 years older than his grandson .$

$∴ Age of the grandson = x + 54 10 x = x + 54$

$\Rightarrow 10 x - x = 54$

$\Rightarrow 9 x = 54$

$\Rightarrow x = \frac{54}{9} = 6$

$Therefore, the grandson' s age is 6 years .$

$Grandfather' s age = 10 (x) = 10 \times 6 = 60 years$

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Question 26:

The difference between the ages of two cousins is 10 years. 15 years ago, if the elder one was twice as old as the younger one, find their present ages.

Answer 26:

$Let the age of the younger cousin be x .$

$Then, the age of the elder cousin will be (x + 10) . 15 years ago :$

$Age of the younger cousin = (x - 15)$

$Ag e of elder cousin = (x + 10 - 15) = (x - 5)$

$∴ (x - 5) = 2 (x - 15)$

$\Rightarrow x - 5 = 2 x - 30$

$\Rightarrow x - 2 x = - 30 + 5$

$\Rightarrow - x = - 25$

$\Rightarrow x = 25$

$Therefore, the present age of the younger cousin is 25 years .$

$Present age of elder cousin = (x + 10) = (25 + 10) = 35 years$

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Question 27:

Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Answer 27:

$Let the number of deer in the herd be x .$

$The number of deer grazing in the field is (\frac{1}{2}) x .$

$Remaining deer = x - \frac{x}{2} = \frac{x}{2}$

$N umber of deer playing nearby = \frac{3}{4} (\frac{x}{2}) = \frac{3}{8} x$

$The number of deer drinking water from the pond is 9 .$

$∴ 9 + \frac{3}{8} x + \frac{1}{2} x = x$

$\Rightarrow \frac{72 + 3 x + 4 x}{8} = x (multiplying the L . H . S . by 8, which is the L . C . M . of 1, 8 and 2)$

$\Rightarrow 72 + 7 x = 8 x (by cross multiplication)$

$\Rightarrow 72 = 8 x - 7 x$

$\Rightarrow 72 = x$

$\Rightarrow x = 72$

$T otal number of deer in the herd = 72$