**The sum of two numbers is 40. If the smaller number is doubled, it becomes 14 more than the larger number. Find the numbers.**

Solution:

Let x and y be two numbers then it is given that, x + y = 40

Let x be a smaller number, if it is doubled i.e, 2x it becomes 14 more than the larger number i.e, 2x = y + 14

x + y = 40 …………..(1)

2x – y = 14 ………….(2)

———————————

3x = 54

x = ^{54}/_{3} = 18

Substitute the value of x in equation(1)

18 + y = 40

y = 40 – 18 = 22

Therefore, the smaller number x = 18 and the larger number y = 22.

**Two numbers are such that twice the smaller number added to 2 gives larger number. Also, double of the larger number is 1 less than five times the smaller number. Find the numbers.**

Solution:

Let x and y be two number. Let x be the smaller number and y be the larger number. Then, it is given that 2x+2 = y and 2y = 5x – 1

2x + 2 = y ………..(1)

5x – 1 = 2y ………(2)

—————————–

(1)x5⇒ 10x + 10 = 5y

(2)x2⇒ 10x – 2 = 4y

(-) (+) (-)

—————————–

12 = y

Substitute the value of y in (1)

2x + 2 = 12

2x = 12 – 2 = 10

x = ^{10}/_{2} = 5

Therefore, the smaller number x = 5 and the larger number y = 12.

**Find the fraction which becomes**^{1}/_{2}when the denominator is increased by 4 and^{1}/_{8}when the numerator is diminished by 5?

Solution:

Let x and y be two number. Let ^{x}/_{y }be the fraction.

the fraction which becomes ^{1}/_{2} when the denominator is increased by 4

⇒ ^{1}/_{2} = ^{x}/_{y+4} i.e, 2x = y + 4

^{1}/_{8} when the numerator is diminished by 5

⇒ ^{1}/_{8} = ^{x-5}/_{y} i.e., 8(x-5) = y ⇒8x – 40 = y

2x – y = 4 ………..(1)

8x – y = 40 …………(2)

(-) (+) (-)

—————————-

-6x = -36

x = 6

Substitute the value of x in (1),

2×6 – y = 4

12 – y = 4

-y = 4 – 12 = – 8

y = 8

Therefore, x = 6 and y = 8

**A father is 25 years older than his son. After 8 years the ratio of their ages**

**Will be 13: 8 Find their present ages.**

Solution:

Let x be the age of son then x+25 = y is the age of the father.

After 8 years, ratio of their ages, F : S = 13 : 8

x+25+8 : x+8 :: 8 : 13

8y – 8 = x ⇒ 8y – x = 8

13y – 8 = x + 25 ⇒13y – x = 32

(-) (+) (-)

———————————————

-5y = -25

y = 5

Thus, 8y – x = 8

8×5 – x = 8

-x = 8 – 40 = – 32

x = 32

Therefore, x = 32 and x + 25 = 32 + 25 = 57

**The measures of the sides (in cm) of a triangle are**^{5}/_{3}x + y +^{1}/_{2}, 2x +^{1}/_{2 }y and^{2}/_{3}x + 2y +^{5}/_{2}. For what vales of x and y the triangle is equilateral? Also find the measures of the sides of the triangle.

Solution:

Sides of the triangle are ^{5}/_{3} x + y + ^{1}/_{2} , 2x + ^{1}/_{2 }y and ^{2}/_{3} x + 2y + ^{5}/_{2}.

If it is equilateral triangle then

^{5}/_{3} x + y + ^{1}/_{2} = 2x + ^{1}/_{2 }y

^{5}/_{3} x + y – 2x – ^{1}/_{2 }y = –^{1}/_{2}

^{5x-6x}/_{3} + ^{2y-y}/_{2} = –^{1}/_{2}

–^{x}/_{3} + ^{y}/_{2} = –^{1}/_{2}

-2x + 3y = – 3

2x – 3y = 3 ……………(1)

2x + ^{1}/_{2 }y = ^{2}/_{3} x + 2y + ^{5}/_{2}

2x + ^{1}/_{2 }y – ^{2}/_{3} x – 2y = ^{5}/_{2}

^{6x-2x}/_{3} + ^{y-4y}/_{2} = ^{5}/_{2}

12x – 4x + 3y – 12y = 15

8x – 9y = 15 ………….(2)

(1)x4⇒8x – 12y = 12

(2)x1⇒8x – 9y = 15

(-) (+) (-)

————————-

-3y = -3

y = 1

Then, substituting y in (1),

8x – 9y = 15

8x = 15 + 9(1) = 15 + 9 = 24

x = ^{24}/_{8} = 3

Thus, y = 1 and x = 3

^{5}/_{3} x + y + ^{1}/_{2} = 2x + ^{1}/_{2 }y = ^{2}/_{3} x + 2y + ^{5}/_{2}

^{5}/_{3} x3 + 1 + ^{1}/_{2} = 2×3 + ^{1}/_{2 }x1 = ^{2}/_{3} x3 + 2 + ^{5}/_{2}

^{15}/_{3} + 1 + ^{1}/_{2} = 6 + ^{1}/_{2} = ^{6}/_{3} + 2 + ^{5}/_{2}

^{30+6+3}/_{6} = ^{36+1}/_{6} = ^{12+12+15}/_{6}

^{39}/_{6} = ^{39}/_{6} = ^{39}/_{6}

^{13}/_{2} = ^{13}/_{2} = ^{13}/_{2}

6.5 = 6.5 = 6.5

**A sailor goes 8 km down-stream in 40 minutes and returns back to the starting point in 1 hr. Find the speed of the sailor in still water and speed of the current.**

Solution:

Distance travelled = 8 km

Speed of stream y and speed of back x

Downstream is x + y

8 = (x + y)^{40}/_{60}

8 = (x + y)^{2}/_{3} ⇒24 = 2x + 2y

12 = x + y ——–(1)

upstream x – y

8 = (x – y) ^{60}/_{60}

⇒ 8 = (x – y)

⇒8 = x – y ———–(2)

From (1) and (2)

x + y = 12

x – y = 8

(-)(+)(-)

—————–

2y=4

y = 2

From(1), 12 = x + y ⇒12 = x + 2

⇒x = 12 – 2 = 10

Therefore, the speed of the boat is x = 10 and y = 2

**There is a number is equal to 4 times the sum of its digits. IF 27 is added to the number, the number’s digit get reversed. Find the number.**

Solution:

Let the digits be x and y

Let the original number be 10y + x .

10y + x = 4(x+y)

10y + x = 4x + 4y

10y – 4y = 4x – x

6y = 3x

x = 2y

The number be 10x + y when reversed if 27 is added to it, the digdits gets reversed.

10y + x + 27 = 10x + y

10y – y – 10x + x = -27

9y – 9x = -27

y – x = -3

y – 2y = -3

-y = -3

Substitute the value of y in (1)

x = 2y = 2(3) = 6

Therefore the number 10y + x = 10(3) + 6 = 30 + 6 = 36

**A boat goes upstream 30 km and downstream 44km in 10hrs. Its also goes up stream 40 kkm and downstream 55km in 13hrs. Find the speed of the stream and the boat.**

Solution:

Let the speed of the boat in stil

water = x km/hr

Let the speed of the stream = y km/hr

Downstream x+ y

Upstream = x – y

^{30}/_{x – y }+ ^{44}/_{x + y} = 10…………..(1)

^{40}/_{x – y }+ ^{55}/_{x + y} = 13…………….(2)

(1)x4⇒ ^{120}/_{x – y }+ ^{176}/_{x + y} = 40

(2)x3⇒^{120}/_{x – y }+ ^{165}/_{x + y} = 39

(-) (-) (-)

——————————————-

^{11}/_{x + y} = 1

11 = x + y ………………..(3)

Therefore, eq(1) becomes,

^{30}/_{x – y }+ ^{44}/_{11} = 10

^{30}/_{x – y }+ 4 = 10

^{30}/_{x – y }= 10 – 4 = 6

30 = 6(x – y)

x – y = 5…………..(4)

From (3) and (4), we have

x + y = 11

x – y = 5

—————————

2x = 16

x = 8

Substitute x = 8 in (3)

8 + y = 11

y = 11 – 8 = 3

y = 3

Thus, the speed of the boat is 8km/hr and the speed of the stream is 3km/hr.

## 1 thought on “Simultaneous Linear Equation – Exercise 3.5.5 – Class IX”

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