Q: If a2 + b2 + c2 = 2 (a b c) 3 then the value of 2a 3b + 4c is

As per the given question , we have a2 + b2 + c2 = 2 (a b c) 3 ⇒ a2 + b2 + c2 = 2a + 2b + 2c + 3 = 0 ⇒ a2 2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0 ⇒ (a 1)2 + (b + 1)2 + (c + 1)2 = 0 [If x2 + y2 + z2 = 0 ⇒ x = 0; y = 0; z = 0] ∴ a 1 = 0 ⇒ a = 1 b + 1 = 0 ⇒ b = 1 c + 1 = 0 ⇒ c = 1 ∴ 2a 3b + 4c = 2 + 3 4 = 1

Question 1

If the cost of 2 apples, 3 oranges and a pear is Rs.62 and the cost of 5 apples, 6 oranges and 4 pears is Rs. 120, then find the cost of 3 apples, 3 oranges and 3 pears?

Accepted Answer

Let the cost of each apple,orange nd pear be Rs.x,y and z, respectively.Then, 2x + 3y + z = 62 ...(i) 5x + 6y + 4z = 20 ...(ii) On subtracting Eq.(i)from Eq.(ii), we get 3x + 3y + 3z = 58 So, the cost of 3 apple,3 oranges and 3 pears is Rs. 58.

Question 2

Two candidates attempt to solved a quadratic equation of from x 2 + px + q = 0. One starts with a wrong value of p and finds the roots to be 2 and 6. Other starts with a wrong value of q and find the root to be 2 and -9. Find the correct roots and e…

Accepted Answer

When p is wrong i.e., -b/a = (α + β) is wrong but c/a =(αβ) is correct. Then αβ = c/a = 2 x 6 = 12 ....(i) Again, when q is wrong i.e.,c/a = (α.β ) is wrong but -b/a = α+β is correct. Then, -b/a = α+β= 2 + (-9) = -7 So, the required correct quadratic equations is x2 -(α + β)x + α.β=0 ⇒ x2 - (-7)x + 12 = 0 ⇒ x2 + 7x + 12 = 0 and correct roots this equations are -3, -4.

Question 3

If a + b + 1 = 0, then the value of (a 3 + b 3 + 1 – 3ab) is

Accepted Answer

As we know from the formula, A3 + B3 + C3 = ( A + B + C) (A2 + B2 + C2 – AB – BC – CA) + 3ABC If A + B + C = 0 then, A3 + B3 + C3 = 0 x (A2 + B2 + B2 – AB – BC – CA) + 3ABC A3 + B3 + C3 = 0 + 3ABC A3 + B3 + C3 - 3ABC= 0 As we know that from given question, A = a, B = b and C = 1, Put these value in above equation, we will get a3 + b3 + 13 - 3 x a x b x 1 = 0 a3 + b3 + 1 - 3ab = 0

Question 4

If a + 1 = 1, then the value of a3 is : a

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Question 5

If ax + by = 6, bx – ay = 2 and x 2 + y 2 = 4, then the value of ( a 2 + b 2) would be :

Accepted Answer

As we know that given in question, ax + by = 6 ........................ (i) bx – ay = 2 ........................ (ii) On squaring the both equation and adding, a2 x2 + b2 y2 + 2abxy + b2 x2 + a2 y2 – 2abxy = 36 + 4 ⇒ x2 (a2 + b2) + y2 (a2 + b2) = 40 ⇒ (a2 + b2) (x2 + y2) = 40 As per given question, x2 + y2 = 40 , Put this value in above equation, we will get ⇒ (a2 + b2) × 4 = 40 ⇒ a2 + b2 = 10

Question 6

In xy–plane, P and Q are two points having co– ordinates (2, 0) and (5, 4) respectively. Then the numerical value of the area of the circle with radius PQ is

Accepted Answer

According to question , PQ = √( 5 - 2 )2 + ( 4 - 0 )2 PQ = √9 + 16 = 5 ∴ Area of circle = πr2 Area of circle = 25 π sq. units

Question 7

In the xy–coordinate system, if ( a, b) and ( a + 3, b + k) are two points on the line defined by the equation x = 3y – 7, then k =?

Accepted Answer

Given in the question, Points (a, b) and [(a + 3), (b + k)] will satisfy the equation x – 3y + 7 = 0. Points (a, b) in the equation x – 3y + 7 = 0, We will get ∴ a – 3b + 7 = 0 ...................... (i) Points (a + 3), (b + k) in the equation x – 3y + 7 = 0, We will get a + 3 – 3 (b + k) + 7 = 0 ⇒ a + 3 – 3b – 3 k + 7 = 0 (Rearrange the equation) ⇒ a – 3b + 7 + 3 – 3 k = 0 (a – 3b + 7 = 0 , Put the value from Equation (1) ), We will get ⇒ 0 + 3 – 3k = 0 ⇒ 3k = 3 ⇒ k = 1

Question 8

If a = √ 3 - √ 2 , b = √ 3 + √ 2 , then the value of √3 + √2 √3 - √2 a 2 + b 2 is : b a

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Question 9

Area of the trapezium formed by x-axis; y-axis and the lines 3x + 4y = 12 and 6x + 8y = 60 is:

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Question 10

If x = 1 + √2 + √3, then the value of ( 2x4 – 8x3 – 5x2 + 26x – 28) is

Accepted Answer

Given in the question , x - 1 = √2 + √3 On squaring both sides , x 2 – 2x + 1 = 2 + 3 + 2√6 ⇒ x 2 - 2x - 4 = 2 √6 On squaring again, ⇒ x4 + 4x2 + 16 – 4x3 – 8x2 + 16x = 24 ⇒ x4 – 4x3 – 4x2 + 16x – 8 = 0 ⇒ 2x4 – 8x3 – 8x2 + 32x – 16 = 0 ⇒ 2x4 – 8x3 – 5x2 + 26x – 28 – 3x2 + 6x + 12 = 0 ⇒ 2x4 – 8x3 – 5x2 + 26x – 28 = 3x2 – 6x – 12 ⇒ 2x4 – 8x3 – 5x2 + 26x – 28 = 3 (x2 – 2x – 4) ⇒ 2x4 – 8x3 – 5x2 + 26x – 28 = 3 x 2√6 = 6 √6

Question 11

If 4x = 18y, then the value of ( x/y - 1) is

Accepted Answer

7 2

Question 12

If a2 + b2 + c2 = 2 (a  b  c)  3 then the value of 2a  3b + 4c is

Accepted Answer

As per the given question , we have a2 + b2 + c2 = 2 (a  b c) 3 ⇒ a2 + b2 + c2 = 2a + 2b + 2c + 3 = 0 ⇒ a2  2a + 1 + b2 + 2b + 1 + c2 + 2c + 1 = 0 ⇒ (a 1)2 + (b + 1)2 + (c + 1)2 = 0 [If x2 + y2 + z2 = 0 ⇒ x = 0; y = 0; z = 0] ∴ a  1 = 0 ⇒ a = 1 b + 1 = 0 ⇒ b =  1 c + 1 = 0 ⇒ c =  1 ∴ 2a  3b + 4c = 2 + 3  4 = 1

Question 13

If x = 2 + √3, y = 2 - √3, then the value of x 2 + y 2 is x3 + y3

Accepted Answer

Question 14

If 2x - 1 = 6, then the value of x2 + 1 is 2x 16x2

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Question 15

If x = 4ab , then the value of x + 2a + x + 2b is a + b x - 2a x - 2b

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Question 16

The area of the triangle formed by the lines 5x + 7y = 35, 4x + 3y = 12 and x-axis is

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Question 17

The area (in sq. unit) of the triangle formed by the three graphs of the equations x = 4, y = 3, and 3x + 4y = 12, is

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Question 18

If x = √a + 1 , y = √a - 1 , then the value of √a √a x4 + y4 - 2x2 y2 is

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Question 19

If a + b + c = 8, then the value of (a – 4)3 + (b – 3)3 + (c – 1)3 – 3 (a – 4) (b – 3) (c – 1) is

Accepted Answer

We have p3 + q3 + r3 – 3pqr = (p + q + r) (p2 + q2 + r2 –pq – qr - rp) Here p = a – 4, q = b – 3, r = c –1 So, given expression is (p + q + r) (p2 + q2 + r2 – pq – qr – rp) = (a – 4 + b – 3 + c – 1) (p2 + q2 + r2 – pq – qr – rp) = (a + b + c – 8) (p2 + q2 + r2 – pq – qr – rp) = (8 – 8) (p2 + q2 + r2 – pq – qr – rp) ∴ (a – 4)3 + (b – 3)3 + (c – 1)3 – 3 (a – 4) (b – 3) (c – 1) = 0

Question 20

Minimum value of x2 + 1 - 3 is x2 + 1

Accepted Answer

Elementary Algebra Questions