Question 1

Find the remainder when f(x) = x 4 - 6x 3 + 2x 2 + 3x + 1 is divided by (x - 2)?

Accepted Answer

When f(x) = x4 - 6x3 + 2x2 + 3x + 1 is divided by (x - 2), then remainder can be calculated by putting x - 2 = 0 i.e. x = 2 Now, put x = 2 in f(x) f(2) = (2)4 - 6(2)3 + 2(2)2 + 3 x 2 + 1 = 16 - 48 + 8 + 6 + 1 = -17 ∴ The remainder will be - 17.

Question 2

If the polynomials ax 3 + 4x 2 -3x - 4 and x 3 - 4x + a leave the same remainder when divided by (x - 3). Then, what is the value of a?

Accepted Answer

Since, p(x) = ax3 + 4x2 + 3x - 4 and q(x) = x3 -4x + a divided by x - 3, hence p(3) = q(3) ⇒ a x 33 + 4 x 32 + 3 x 3 - 4 = 33 - 4 x 3 + a ∴ a = -1

Question 3

A student was asked to find 2/5th of a number and he instead multiplied it by 5/2. As a result he got an answer which was more than the correct answer by 5208. What was the number?

Accepted Answer

Let the number be x. Then, 5x/2 - 2x/5 = 5208 ⇒ (25x - 4x)/10 = 5208 ⇒ 21x/10 = 5208 ⇒ 21x = 5208 x 10 ⇒ x = (5208 x 10)/21 ⇒ x = 2480 So, the original numbers was 2480.

Question 4

A man covers a distance of 18 km in 5 h party by walking and partly by running. if he walks at 2 km/h and runs at 6km/h, then find the distance he covers by running?

Accepted Answer

Let the distance covered by walking be x km. Speed while walking = 2 km/h Time spent on walking = x/2 h Now, distance covered by runnig = (18 - x) km Speed while running = 6 km/h Time spent on running = (18 - x)/6 h ∴x/2 + (18 -x)/6 = 5 ⇒ (3x + 18x - x)/6 = 5 ⇒ 2x + 18 = 30 ⇒ 2x = 12 ⇒ x = 6 ∴ The distance covered by running = 18 - 6 = 12 km

Question 5

Ramesh and Ganesh had some oranges initially. If Ramesh gave 5 oranges to Ganesh, then Ganesh will have thrice as many oranges as Ramesh. Instead of that, if Ganesh was to give 5 oranges to Ramesh, then they both will have the same number of oranges…

Accepted Answer

Let the number of oranges Ramesh has be x and the number of oranges Ganesh has be y. According to the first condition, y + 5 = 3(x - 5) ⇒ 3x - y + 10 = 0 ...(i) According to the second condition, x + 5 = y - 5 ⇒ x - y + 10 = 0 ...(ii) On subtracting Eq. (ii) from Eq. (i), we get x = 15 On substituting the value of x in Eq.(ii), we get y = 25 ∴ The ratio of oranges to be distributed Ramesh and Ganesh = 15:25 = 3:5

Question 6

A two-digit number is such that the ten's digit exceeds twice the unit's digit by 2 and the number obtained by interchanging the digits is 5 more than three times the sum of the digits. Find the two-digits number?

Accepted Answer

Let the digit in the unit's place be y and digit in the ten's place be x. ∴ Number = 10x + y Then, x = 2y + 2 ⇒ x - 2y - 2 = 0 ...(i) Number obtained by reversing the digits = 10y + x Then, 10y + x = 5 + 3(x + y) ⇒ 2x - 7y + 5 = 0 ...(ii) On multiplying Eq.(i) by 2 and then subtracting it from Eq. (ii), we get y = 3 On substituting the value of y in Eq.(i), we get x = 8 So, the two-digit number is 83.

Question 7

In an examination consisting of 60 questions, three marks are given for every correct answer and one mark is deducted for every wrong answer. A student attempts all the questions and gets 120 marks. How many questions did he mark correct?

Accepted Answer

Let the number of correct answers marked be c and the number of wrong answers marked be w. According to the question, c + w = 60 ...(i) Also, 3c - w = 120 ...(ii) On adding Eqs.(i) and (ii), we get c = 45 So, the questions marked correct are 45.

Question 8

A person has only ₹ 1 and ₹ 2 coins with him. If the total number of coins that he has is 50 and the amount of money with him is 75, then the number of coins ₹ 1 and ₹ 2, respectively are?

Accepted Answer

Total number of coins = 50 Let ₹ 1 coins = x and ₹ 2 coins = y Now,according to the question, x + y = 50 ... (i) and x + 2y = 75 ...(ii) On solving Eqs.(i) and (ii),we get x = 25 and y = 25

Question 9

In a rare coins collection, there is one gold coins for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1:2. Based on the information, the total number of coins in the…

Accepted Answer

Let the number of gold coins initially be x and the number of non-gold coins be y. According to the question, 3x = y .....(i) When 10 more gold coins are added, then total number of gold coins becomes x + 10 and the number of non-gold coins remain the same as y. Now,we have 2(10 + x) = y ....(ii) On solving these two equations, we get x = 20 and y = 60 ∴ Total number of coins in the collection at the end is equal to x + 10 + y = 20 + 10 + 60 = 90.

Question 10

If the price of a book when reduced by Rs. 5,a person can buy 2 more books for Rs. 200. Find the original price of the book?

Accepted Answer

Let the original price of the book be Rs. x. ∴ Number of books bought for Rs.200 = 200/x and reduced price of the book=Rs.(x - 5) ∴ Number of books bought for Rs.200 = 200/(x - 5) Now, according to the condition, 200/(x - 5) - 200/x = 2 ⇒ (200x - 200x + 1000)/(x - 5)x = 2 ⇒ (200x - 200x + 1000)/x(x - 5) = 2 ⇒ 1000/x(x - 5) = 2 ⇒ 1000/x(x - 5) = 1 ⇒ 1000/x(x - 5) = 2 ⇒ 500/x(x - 5) = 1 ⇒ 500 = x(x-5) ⇒ x2 -5x - 50 = 0 ⇒ x2- 25 + 20x - 500 = 0 ⇒ x(x - 25) + 20(x - 25) = 0 ⇒ (x - 25)(x + 20) = 0 ⇒ x - 25 = 0 or x + 20 = 0 ⇒ x = 25 0r -20 ⇒ x = 25 (since,x cannot be negatives) So, the original price of the bok is Rs.25.

Question 11

The sum of the squares of two numbers is 234 and the square of their difference is 144. Find the product of the two numbers?

Accepted Answer

Let the two numbers be x and y. Then, x2 + y2 = 234 ...(i) and (x - y)2 = 144 ...(ii) Now, (x - y)2 = 144 ⇒ x2 + y2 - 2xy = 144 ⇒ 234 - 2xy = 144 [using Eq.(i)] On substituting the value of x2 + y2 from Eq.(i) in Eq.(ii), we get 2xy = 90 ⇒ xy = 45 So, the product of the two numbers is 45.

Question 12

The degree of 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 = 6x22 + 1 x4 + x + 2x2 - 6x2 - 1 = 0 x4 - 4x2 + x - 1 = 0 Degree of polynomial is highest exponent degree term i. e., 4.

Question 13

A piece of cloth cost ₹ 120. If the piece, then was 2 m longer and each metre of cloth costs ₹ 2 less than cost of the piece of cloth would have remained unchanged. What is the original rate per metre of the cloth?

Accepted Answer

Let the length of the piece be L m. Then,rate per metre = ₹ 120/L New length = (L + 2) m Since, the cost remains same. ∴ New rate per metre =Rs.120/(L + 2) According to the given condition, 120/L - 120/(L + 2) = 2 ⇒ [120(L + 2) - 120L]/L(L + 2) = 2 ⇒ [120L + 240 - 120L] / L(L + 2) = 2 ⇒ [120L + 240 -120L] / L(L + 2) = 2 ⇒ 240/L(L + 2) = 2 ⇒ 240 = 2L(L + 2) ⇒ 120 = L(L + 2) = L2 + 2L ⇒ L2 + 2L - 120 = 0 ⇒ L2 + 12L - 10L - 120 = 0 ⇒ L(L + 12) - 10 x (L + 12) = 0 ⇒ (L + 12)(L - 10) = 0 ⇒ L + 12 = 0 or L - 10 = 0 ⇒ L = -12 or 10 ⇒ L = 10(since, L cannot be negative) ∴ Original rate per metre of cloth = 120/L = 120/10 = ₹ 12

Question 14

If √0.03 x 0.3 a = 0.03 x 0.3 x √ b then value of a is b

Accepted Answer

Question 15

Area of the triangle formed by the graph of the line 2x – 3y + 6 = 0 along with the coordinate axes is:

Accepted Answer

Question 16

If the distance between two points (0,–5) and (x,0) is 13 unit, then x =

Accepted Answer

According to question , we have √( x - 0 )2 + ( 0 + 5 )2 = 13 ⇒ x2 + 25 = 169 ⇒ x2 = 169 - 25 = 144 ∴ x = √144 = 12

Question 17

If x = 997, y = 998, z = 999, then the value of x2 + y 2 + z 2 – xy – yz – zx will be

Accepted Answer

Question 18

If, a + b + c = 0, the value of ( a 2 + b 2 + c 2 ) is bc ca ab

Accepted Answer

Question 19

The equation of this graph is

Accepted Answer

Question 20

Angry Arjun carried some arrows for fighting with Bheshhm. With half the arrows, he cut down the arrows thrown by Bheeshm on him and with six other arrows he killed the rath driver of Bheeshm. with one arrows each he Knocked down respectively the ra…

Accepted Answer

Let Arjun had x arrows. ∴ x/2 + 6 + 3 + 4√x + 1 = x ⇒ x = 20 + 8√x ⇒ x - 20 =8√x On squaring both sides and by solving the formed quadratics equation, we get x = 100

Elementary Algebra Questions