Question

7, Q.3 Solve any two subquestions from the following: 1) Sachin invested in a national saving certificcate scheme. In the first year he invested Rs. 5,000 in the second year Rs. 7,000, in the third year Rs. 9,000 and so on. Find the total amount that he invested in 12 years. 2) How many two digit numbers are divisible by 4 ? 3) Solve the following simultaneous equations using cramer's rule. x + 2y = -1; 2x - 3y = 12​

Answer 1

Answer:-

Giving Answer of 2nd and 3rd Question

Question:-2} How many two digit numbers are divisible by 4?

Answer:-2} 22 two

Therefore, there are a total 22 two-digit numbers which are divisible by 4.

Question:-3}Solve the following simultaneous equations using cramer's rule. x + 2y = -1; 2x - 3y = 12

Answer:- 3}The given equations are

he given equations are⇒ x + 2y = – 1 … i

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …ii

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iii

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get⇒ x = – 1 – 2y

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get⇒ x = – 1 – 2y⇒ x = – 1 – 2 × −2

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get⇒ x = – 1 – 2y⇒ x = – 1 – 2 × −2⇒ x = – 1 + 4

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get⇒ x = – 1 – 2y⇒ x = – 1 – 2 × −2⇒ x = – 1 + 4⇒ x = 3

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get⇒ x = – 1 – 2y⇒ x = – 1 – 2 × −2⇒ x = – 1 + 4⇒ x = 3Here, x = 3 and y = – 2.

he given equations are⇒ x + 2y = – 1 … i⇒ 2x – 3y = 12 …iiFrom equation i, we get⇒ x = – 1 – 2y … iiiPutting x’s value in Equation ii, we get⇒ 2−1–2y – 3y = 12⇒ – 2 – 4y – 3y = 12⇒ – 7y = 14⇒ y = 14/- 7⇒ y = – 2Putting y’s value in equation iii, we get⇒ x = – 1 – 2y⇒ x = – 1 – 2 × −2⇒ x = – 1 + 4⇒ x = 3Here, x = 3 and y = – 2.Hence, the solution of the given equation is x = 3, y = – 2.