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## Tamilnadu Samacheer Kalvi 9th Maths Solutions Chapter 3 Algebra Ex 3.14

Solve by any one of the methods

Question 1.

The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.

Solution:

Let the two digit number be x y

Its place value = 10x + y

After interchanging the digits the number will be y x

Its place value = 10y + x

Their sum = 10x + y + 10y + x = 110

11x + 11y = 110

x + y = 10 ………….. (1)

If 10 is subtracted from the first number, the new number is 10x + y – 10

The sums of the digits of the first number is x + y

Its 4 more than 5 times is = 5(x + y) + 4

∴ 10x + y – 10 = 5x + 5y + 4

10x + y – 5x – 5y = 4 + 10

5x – 4y = 14 ………. (2)

Substitute y = 4 in (1)

x + 4 = 10

x = 10 – 4

x = 6

The first number is 64

Question 2.

The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \(\frac{1}{2}\). Find the fraction.

Solution:

Let the numerator be x

Denominator be y

x + y = 12 ………. (1)

\(\frac{x}{y+3}=\frac{1}{2}\)

2x = y + 3

2x – y = 3 …………. (2)

Substitute y = 7 in (1)

x + 7 = 12

x = 12 – 7

x = 5

∴ The fraction is \(\frac{x}{y}=\frac{5}{7}\)

Question 3.

ABCD is a cyclic quadrilateral such that ∠A = (4y + 20)°, ∠B = (3y – 5)°, ∠C =(4x)° and ∠D = (7x + 5)°. Find the four angles.

Solution:

In a cyclic quadrilateral the sum of the four angles is 360° and

the sum of the opposite angles is 180°.

∠A + ∠C = 180°

4y + 20 + 4x = 180°

4x + 4y = 180 – 20

x + y = \(\frac{160}{4}\)

x + y = 40 ………….. (1)

∠B + ∠D = 180°

3y – 5 + 7x + 5 = 180

7x + 3y = 180 …………. (2)

Substitute y = 25° in (1)

x + 25 = 40

x = 40 – 25

x = 15°

∠A = (4y + 20)° = (4 × 25 + 20) = 100 + 20 = 120°

∠B = (3y – 5)° = (3 × 25° – 5) = 75° – 5° = 70°

∠C = (4x)° = 4 × 15° = 60°

∠D = (7x + 5)° = (7 × 15 + 5) = 105° + 5° = 110°

∴ ∠A = 120°, ∠B = 70°, ∠C = 60°, ∠D = 110°

Question 4.

On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains ₹ 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains ₹ 1500 on the transaction. Find the actual price of the T.V. and the fridge.

Solution:

Let the actual price of a T.V. = x

Let the actual price of a Fridge = y

\(\frac{5}{100}\)x + \(\frac{10}{100}\)y = 2000

\(\frac{5}{100}\) (x + 2y) = 2000

x + 2y = 2000 × \(\frac{100}{5}\)

x + 2y = 40000 ………. (1)

\(\frac{10}{100}\) x – \(\frac{5}{100}\)y = 1500

2x – y = 1500 × \(\frac{100}{5}\)

2x – y = 30000 ……….. (2)

Substitute y = 10000 in (1)

x + 2(10000) = 40000

x + 20000 = 40000

x = 40000 – 20000

x = 20000

∴ Actual price of T.V. = ₹ 20000

Actual price of Fridge = ₹ 10000

Question 5.

Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.

Solution:

Let the two numbers be x, y

\(\frac{x}{y}=\frac{5}{6}\)

⇒ 6x = 5y

6x – 5y = 0 ………… (1)

\(\frac{x-8}{y-8}=\frac{4}{5}\)

⇒ 5(x – 8) = 4(y – 8)

5x – 40 = 4y – 32

5x – 4y = 40 – 32

5x – 4y =8 ………….. (2)

Substitute y = 48 in (1)

6x – 5(48) = 0

6x – 240 = 0

6x = 240

x = \(\frac{240}{6}\) = 40

\(\frac{x}{y}=\frac{40}{48}=\frac{5}{6}\)

∴ The numbers are in the ratio 5 : 6

Question 6.

4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indians and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it take for 1 Chinese to do it?

Solution:

Let for one Indian the rate of working be \(\frac{1}{x}\)

Let for one Chinese the rate of working be \(\frac{1}{y}\)

For cross multiplication method, we write the co-efficients as

∴ 1 Chinese can do the same piece of work = \(\frac{1}{y}\) = b = \(\frac{1}{36}\) = 36 days

1 Indian can do the piece of work = \(\frac{1}{x}\) = a = \(\frac{1}{18}\) = 18 days