Question 1

The degree f polynomial p(x) = x 3 + 1 + 2x = 6x + 1/x is?

Accepted Answer

x3 + 1 + 2x =6x + 1/x x3 + 1 + 2x = (6x2 + 1)/x (x3 + 1 + 2x)x = 6x2 + 1 x4 - 4x2 - 6x2 -1 =0 x4 - 4x2 + x - 1 =0 Degree of polynomial is highest exponent degree term i.e.,4.

Question 2

If x = 1 + √ 2, then what is the value of x 4 - 4x 3 + 4x 2?

Accepted Answer

x = 1 + √2 ∴ x4 - 4x3 + 4x2 = x2(x2 - 4x + 4) = x2(x - 2)2 = (1 + √2)2(1 + √2 - 2)2 =(√2 + 1)2 (√2 - 1)2 =[(√2)2 - (1)2]2 =(2 - 1)2 =1

Question 3

What is the solution of the equations x - y = 0.9 and 11(x + y) -1 = 2?

Accepted Answer

x - y = 0.9 ...(i) and 11(x + y)-1=2 ⇒ 11/ (x + y) = 2 ⇒ 2(x + y) =11 ⇒ x + y = 11/2 ...(ii) On solving Eqs.(i) and (ii),we get x = 3.2 and y = 2.3

Question 4

The product of two alternate odd integers exceeds three times the smaller by 12. What is the larger integer?

Accepted Answer

Let two alternate odd integers odd integers be (2x+1) and (2x+5). Then according to the question, (2x + 1) (2x + 5) = 3(2x + 1) + 12 ⇒ (2x + 1) (2x + 5 - 3) = 12 ⇒ 2x2 + 3x - 5 = 0 On solving this quadratic equation,we get x = 1 and x = -5/2 x = -5/2 is not a integer ∴ x = 1 Then, larger integer = 2x + 5 = 2 x 1 + 5 = 7

Question 5

The sum of two numbers is 24 and their product is 143. The sum of their squares is

Accepted Answer

Let the two numbers be P and Q. According to given question, Sum of two numbers = 24 ∴ P + Q = 24 ...................... (1) Product of two numbers = 143 and, PQ = 143 .......................... (2) As we know the formula, ∴ P 2 + Q2 = (P + Q)2 – 2PQ Put the value from the equation (1) and (2), We will get ∴ P 2 + Q2 = (24)2 – 2 × 143 ∴ P 2 + Q2 = 576 – 286 = 290

Question 6

If P = (0.08) 2, Q = 1/(0.08) 2 and R = (1 – 0.08) 2 – 1, then out of the following the true relation is

Accepted Answer

Question 7

If (5x2 – 3y2) : xy = 11 : 2, then the positive value of x is : y

Accepted Answer

Question 8

If a3 – b3 = 56 and a – b = 2, then the value of ( a2 + b2 ) is :

Accepted Answer

As we know from the algebra formula, (a – b)3 = a3 – b 3 – 3ab (a – b) .......................(1) As per given question, a3 – b 3 = 56 and a – b = 2 Put these value in above equation (1) . We will get, ⇒ 23 = 56 – 3ab x 2 ⇒ 8 = 56 – 3ab x 2 ⇒ 8 = 56 – 6ab ⇒ 6ab = 56 – 8 = 48 ⇒ ab = 8 ...................... (i) ∴ a2 + b2 = (a – b)2 + 2ab ∴ a2 + b2 = (a – b)2 + 2ab Put the value of ∴ a2 + b2 = 22 + 2 x 8 ∴ a2 + b2 = 4 + 16 = 20

Question 9

If x = 2 + √3, then the value of √x + 1 is : √x

Accepted Answer

Question 10

If 5a + 1 = 5, the value of 9a2 + 1 is 3a 25a2

Accepted Answer

Question 11

If a, b, c are real and a3 + b3 + c3 = 3abc and a + b + c ≠ 0 , then the relation between a, b, c will be

Accepted Answer

Given that :- Now ,a3 + b3 + c3 = 3abc ( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) + 3abc = 3abc ( a+b+c )( a2 + b2 + c2 - ab - bc - ca ) = 0 ∴ a² + b² + c² - ab - bc - ca = 0 or a + b + c ≠ 0 Where , a = b = c

Question 12

The line passing through the points (– 2, 8) and (5, 7)

Accepted Answer

As indicated in the graph, the line passing through the points cuts Y-axis only.

Question 13

If the difference of two numbers is 3 and the difference of their squares is 39; then the larger number is:

Accepted Answer

Let the numbers are p and q. p – q = 3 … (1) p2 – q2 = 39 (p – q) (p + q) = 39 ⇒ x + y = 13 … (2) Adding eqn (1) and (2) p + q + p – q = 16 ⇒ p = 8 ∴ q = 3 Hence, 8 is the larger number.

Question 14

Let a = √6 - √5, b = √5 - 2, c = 2 - √3 Then point out the correct alternative among the four alternatives given below.

Accepted Answer

We know that , √6 = 2.44, √5 = 2.23, √3 = 1.73 a = √6 - √5 = 2.44 - 2.23 = 0.21 b = √5 - 2 = 2.23 - 2 = 0.23 c = 2 - √3 = 2 - 1.73 = 0.27

Question 15

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

Accepted Answer

Given that :- a2 + b2 + c2 = 2 (a – b – c) – 3 ⇒ a2 + b2 + c2 - 2 (a – b – c) + 3 = 0 ⇒ a2 + b2 + c2 - 2a + 2b + 2c + 3 = 0 ⇒ ( a2 + 1 - 2a ) + ( b2 + 1 + 2b ) + ( c2 + 1 + 2c ) = 0 ⇒ ( a - 1 )2 + ( b + 1 )2 + ( c + 1 )2 = 0 This is possible when ( a - 1 )2 = 0, ( b + 1 )2 = 0, and ( c + 1 )2 = 0 ⇒ a = 1, b = –1, c = –1 Thus, 2a – 3b + 4c = 2 (1) – 3 (–1) + 4 (–1) 2a – 3b + 4c = 2 + 3 – 4 = 1.

Question 16

If the square of the sum of two numbers is equal to 4 times of their product. then the ratio of these numbers is:

Accepted Answer

Let the number be p and q. According to question, (p + q)2 = 4pq ⇒ p2 + q2 + 2pq – 4pq = 0 ⇒ (p – q)2 = 0 ⇒ p = q Hence , required ratio is 1 : 1 .

Question 17

A man buys 3 cows and 18 goats in ₹ 47,200. Instead if he would have bought 8 cows and 3 goats, he had to pay ₹ 53,000 more. Cost of one cow is:

Accepted Answer

Question 18

In a two-digit number, the digit at the unit's place is 1 less than twice the digit at the ten's place. If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number…

Accepted Answer

Since two digit number = 10p + q According to question , → q = 2p – 1.. (i) When digits are interchanged then new number = 10q + p then original number – [new number – original number] = 20 → 10p + q – [10q + p – (10p + q)] = 20 → 10p + q – 10q – p + 10p + q = 20 19p – 8q = 20 19p – 8 (2p – 1) = 20 (Using eq. (i)) 19p – 16p + 8 = 20 3p = 12 → p = 4 From (i) q = 2 × 4 – 1 ⇒ q = 7 ∴ original number = 10p + q = 10 × 4 + 7 = 47

Question 19

The linear equation such that each point on its graph has an ordinate four times its abscissa is:

Accepted Answer

The linear equation is y = 4x, When, x = 1, y = 4

Question 20

A sum of Rs. 13.50 is made by 38 coins which are either 50 paise or 25 paise. Find the number of 25 paise coins?

Accepted Answer

Let the number of 25 paise coins and 50 paise cons be x and y,respectively. According to the question, x + y = 38 Also, 25x + 50y = 1350 On multiplying Eq.(i) by 25 and substracting from Eq.(ii), we get y =16 On substituting the value of y in Eq.(i),we get y=22 So,the number of 25 paise coins in 22.

Elementary Algebra Questions