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

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 ten’s digit be x and the unit digit be y.

The number is 10x + y

If the digits are interchanged

The new number is 10y + x

By the given first condition

10x + y + 10y + x = 110

11x + 11y = 110

x + y = 10 → (1) (Divided by 11)

Again by the given second condition

10x + y – 10 = 5(x + y ) + 4

10x + y – 10 = 5x + 5y + 4

5x – 4y = 14 → (2)

(1) × 5 ⇒ 5x + 5y = 50 → (3)

(2) × 1 ⇒ 5x – 4y = 14 → (2)

(3) – (2) ⇒ 9y = 36

y = 36/9

= 4

Substitute the value of y = 4 in (1)

x + y = 10

x + 4 = 10

x = 10 – 4

= 6

∴ The number is (10 × 6 + 4) = 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” and the denominator be “y”

∴ The fraction is \(\frac{x}{y}\)

By the given first condition

x + y = 12 → (1)

Again by the second condition

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

2x = y + 3

2x – y = 3 → (2)

(1) + (2) ⇒ 3x = 15

x = \(\frac{15}{3}\) = 5

Substitute the value of x = 5 in (1)

5 + y = 12

y = 12 – 5

= 7

∴ The fraction is \(\frac{5}{7}\)

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:

ABCD is a cyclic quadrilateral ∠A + ∠C = 180°

(Sum of the opposite angles of a cyclic quadrilateral is 180°)

(4y + 20)° + (4x)° = 180°

4y + 20 + 4x = 180

4x + 4y = 180 – 20

4x + 4y = 160

x + y = 40 → (1) (divided by 4)

∠B + ∠D = 180° (Sum of the opposite angles of a cyclic quadrilateral)

(3y – 5)° + (7x + 5)° = 180°

3y – 5 + 7x + 5 = 180

7x + 3y = 180 → (2)

(1) × 3 ⇒ 3x + 3y = 120 → (3)

(3) – (2) ⇒ -4x = – 60

4x = 60

x = \(\frac{60}{4}\)

Substitute the value of x = 15 in (1)

15 + y = 40

y = 40 – 15 = 25

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

∴ ∠A = 120°

∠B = 3y – 5 = 3(25) – 5 = 75 – 5 = 70

∴ ∠B = 70°

∠C = 4x = 4(15) = 60

∴ ∠C = 60°

∠D = 7x + 5 = 7(15) + 5

∠D = 105 + 5 = 110°

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

Question 4.

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

Solution:

Let the cost price of the TV be Rs “x” and the cost price of the fridge be Rs “y”.

By the given condition

Multiply by 20

x + 2y = 40000 → (1)

Again by the given second condition

Multiply by 20

2x – y = 30000 → (2)

(2) × 2 ⇒ 4x – 2y = 60000 → (3)

(1) + (3) ⇒ 5x + 0 = 100000

x = \(\frac{100000}{5}\)

= 20000

Substitute the value of x = 20000 in (1)

20000 + 2y = 40000

2y = 40000 – 20000

= 20000

y = \(\frac{20000}{2}\)

= 10000

Cost price of a TV = Rs 20,000

Cost price of a fridge = Rs 10,000

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 and y.

By the given first condition

x : y = 5 : 6

6x = 5y (Product of the extreme is equal to the product of the means)

6x – 5y = 0 → (1)

Again by the given second condition

x – 8 : y – 8 = 4 : 5

5(x – 8) = 4(y – 8)

5x – 40 = 4y – 32

5x – 4y = – 32 + 40

5x – 4y = 8 → (2)

(1) × 4 ⇒ 24x – 20y = 0 → (3)

(2) × 5 ⇒ 25x – 20y = 40 → (4)

(3) – (4) ⇒ – x + 0 = -40

∴ x = 40

Substitute the value of x = 40 in (1)

6(40) – 5y = 0

240 – 5y = 0 ⇒ – 5y = -240

5y = 240

y = \(\frac{240}{5}\)

= 48

The two numbers are 40 and 48 [∴ The ratio of the number = 40 : 48 are 5 : 6]

Question 6.

4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indian 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 the time taken by a Indian be “x”

Time taken by a Chinese be “y”

Work done by a Indian in one day = \(\frac{1}{x}\)

Work done by a Chinese in one day = \(\frac{1}{y}\)

By the given first condition

(4 Indian + 4 Chinese) finish the work in 3 days

\(\frac{4}{x}\) + \(\frac{4}{y}\) = \(\frac{1}{3}\) → (1)

Again by the given second condition

(2 Indian + 5 Chinese) finish the work in 4 days

\(\frac{2}{x}\) + \(\frac{5}{y}\) = \(\frac{1}{4}\) → (2)

Solve the equation (1) and (2)

Let \(\frac{1}{x}\) = a; \(\frac{1}{y}\) = b

4a + 4b = \(\frac{1}{3}\)

12a + 12b = 1 → (3) (Multiply by 3)

2a + 5b = \(\frac{1}{4}\)

8a + 20b = 1 → (4) (Multiply by 4)

(3) × (2) ⇒ 24a + 24b = 2 → (5)

(4) × (3) ⇒ 24a + 60b = 3 → (6)

(5) – (6) ⇒ -36b = -1

b = \(\frac{1}{36}\)

Substitute the value of b = \(\frac{1}{36}\) in (3)

12a + 12(\(\frac{1}{36}\)) = 1

12a + \(\frac{1}{3}\) = 1

36a + 1 = 3

36a = 2

a = \(\frac{2}{36}\) = \(\frac{1}{18}\)

But \(\frac{1}{x}\) = a ⇒ \(\frac{1}{x}\) = \(\frac{1}{18}\)

x = 18

\(\frac{1}{y}\) = b ⇒ \(\frac{1}{y}\) = \(\frac{1}{36}\)

y = 36

∴ Time taken by a 1 Indian is 18 days

Time taken by a 1 Chinese is 36 days