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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 151. |
The sum of three consecutive natural numbers is 90. Find the greatest number. |
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Answer» Correct Answer - 31 Let the three consecutive natural numbers be (x-1), (x) and (x+1) Given that `(x-1)+(x)+(x+1)=90` `x-1+x+x+1=90` `x=(90)/(3)` x=30 ltbrlt `:.` The three numbers are 30-1= 29, 30 and 30+1=31 `:.` The greatest number is 31 |
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| 152. |
The cost of pizza is `2(1)/(2)` times the cost of a burger. If the sum of cost of one burger a and one pizza is Rs. 105, then find the cost of the burger. |
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Answer» Correct Answer - ₹ 30 Let the cost of the burger be Rs.x The cost of the pizza `=(2(1)/(2))(x)=(5)/(2)x` Given that : `(5x)/(2)+x=105rArr(5x+2x)/(2)=105` `(7x)/(2)=105rArrx=(105xx2)/(7)` `x=15xx2xrArrx=30` `:.` The cost of the burger = Rs. 30 |
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| 153. |
3p2 – 5pq + 2q2 + 6pq – q2 + pq is a(i) Monomial(ii) Binomial(iii) Trinomial(iv) Quadrinomial |
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Answer» (iii) Trinomial |
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| 154. |
Subtract -2mn from 6mn. |
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Answer» 6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn |
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| 155. |
Match the following :(a) x/2 = 10(i) x = 4(b) 20 = 6x - 4(ii) x = 1(c) 2x - 5 = 3 - x(iii) x = 20(d) 7x - 4 - 8x = 20(iv) x = 8/3(e) 4/11 - x = -7/11(v) x = -24(A) (i), (ii), (iv), (iii), (v)(B) (iii), (iv), (i), (ii), (v)(C) (iii), (i), (iv), (v), (ii)(D) (iii), (i), (v), (iv), (ii) |
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Answer» (C) (iii),(i), (iv), (v), (ii) a. x/2 = 10, multiplying by 2 on both sides, we get x/2 x 2 = 10 x 2 ⇒ x = 20 b. 20 = 6x – 4 by transposition ⇒ 20 + 4 = 6x 6x = 24 dividing by 6 on both sides, 6x/6 = 24/6 ⇒ x = 4 c. 2x – 5 = 3 – x By transposing the variable ‘x’, we get 2x – 5 + x = 3 by transposing – 5 to other side, 2x + x = 3 + 5 ∴ 3x = 8 3x/3 = 8/3 ∴ x = 8/3 d. 7x – 4 – 8x = 20 by transposing – 4 to other side, 7x – 8x = 20 + 4 – x = 24 ∴ x = – 24 4/11 – x = −7/11 Transposing 4/11 to other side, – x = −7/11− 4/11 = (−7−4)/11 = −11/11 = – 1 ∴ – x = – 1 ⇒ x = 1 |
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| 156. |
Find the area of the square whose side is (x – 2) |
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Answer» Side of a square = x – 2 ∴ Area = Side × Side = (x – 2) (x – 2) = x(x – 2) – 2(x – 2) = x(x) + (x)(-2) + (-2)(x) + (-2)(-2) = x – 2x – 2x + 4x2 – 4x + 4 |
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| 157. |
“An equation is multiplied or divided by two different numbers on either side”. What will happen to the equation? |
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Answer» When an equation is multiplied or divided by 2 different numbers on either side, there will be a change in the equation & accordingly, solution will also change. |
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| 158. |
A quanity which has no fixed value is called a/an _____ |
| Answer» Correct Answer - variable | |
| 159. |
A quantity which has a fixed numerical value is called a/an_____ |
| Answer» Correct Answer - constant | |
| 160. |
State whether the statement are true or false.The distance between New Delhi and Bhopal is not a variable. |
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Answer» True The distance between New Delhi and Bhopal is not a variable. |
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| 161. |
Let, `P=({:(cos""pi/4, -sin""pi/4),(sin""pi/4, cos""pi/4):}) " and " x=({:(1/sqrt(2)),(1/sqrt(2)):})`. Then `P^(3)X` is equal to-A. `({:(0),(1):})`B. `({:(-1/sqrt(2)),(1/sqrt(2)):})`C. `({:(-1),(0):})`D. `({:(-1/sqrt(2)),(-1/sqrt(2)):})` |
| Answer» Correct Answer - C | |
| 162. |
If `f(x)=|(1,x,x+1),(2x,x(x-1),(x+1)x),(3x(x-1),x(x-1)(x-2),(x+1)x(x-1))|,` then f(100) is equal to -(i)`0`(ii)`1`(iii)`100`(iv)`-100`A. 0 (zero)B. 1C. 100D. 10 |
| Answer» Correct Answer - A | |
| 163. |
The value of `lambda` for which the system of equations `2x-y-z=12,x-2y +z = -4, x+y+z=4` has no solution isA. 3B. 1C. 0 (zero)D. -3 |
| Answer» Correct Answer - D | |
| 164. |
Let `a,lambda,mu in R,` Consider the system of linear equations `ax+2y=lambda 3x-2y=mu` Which of the flollowing statement (s) is (are) correct?A. If a=-3, then the system has infinitely many solutions for all values of `lambda` and `mu`.B. If `a ne -3`, then the system has a unique solution for all values of `lambda` and `mu`.C. If `lambda+mu=0`, then the system has infintiely many solutions for a=-3.D. If `lambda+mu ne 0`, then the system has no solution for a=-3. |
| Answer» Correct Answer - B::C::D | |
| 165. |
Find all the zeroes of f(x) = x2 - 2x. |
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Answer» f(x) = x2 - 2x = x (x - 2) f(x) = 0 x = 0 or x = 2 Hence, zeroes are 0 and 2. |
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| 166. |
If x2 + \(\frac{1}{x^2}\) = 23 find x + \(\frac{1}{x}\). |
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Answer» We have (a + b)2 = a2 + 2ab + b2 So (x + \(\frac{1}{x}\))2 = x2 + 2 x (x) x \(\frac{1}{x}\) + \(\frac{1}{x^2}\) = x2 + 2 + \(\frac{1}{x^2}\) = x2 + \(\frac{1}{x^2}\) + 2 = 23 + 2 ∵ x2 + \(\frac{1}{x^2}\) = 23 (x + \(\frac{1}{x}\))2 = 25 (x + \(\frac{1}{x}\))2 = 52 x + \(\frac{1}{x}\) = 5 |
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| 167. |
What should be added to `2xy-3x^(2)+6` so that the sum is `4x^(2)-7xy+15`? |
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Answer» Correct Answer - `7x^(2) - 9xy +9` Let f(x) should be added to `2xy-3x^(2)+6`, so tha rthe sum is `4x^(2)-7xy+15` `f(x) =4x^(2)-7xy+15-(2xy-3x^(2)+6)` `=4x^(2)-7xy+15-2xy+3x^(2)-6` `(4x^(2)+3x^(2))+(-7xt-2xy)+(15-6)` `=7x^(2)-9xy+9` |
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| 168. |
Find the product of (2x+3) and (3x-2) |
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Answer» Correct Answer - `6x^(2) + 5x - 6` `(2x+3)(3x-2)` `=(2x)(3x)-(2x)(2)+3(3x)-(3)(2)` `=6x^(2)-4x+9x-6` `6x^(2)+5x-6` |
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| 169. |
If 4x2 + y2 = 40 and xy = b find the value of 2x + y. |
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Answer» We have (a + b)2 = a2 + 2ab + b2 (2x + y)2 = (2x)2 + (2 (x) 2x (x) y) + y2 = (4x2 + y2) + 4xy = 40 + 4 x 6 = 40 + 24 (2x + y)2 = 64 (2x + y)2 = 82 2x + y = 8 |
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| 170. |
`A=2x+3xy+7x^(2),B=3x-2y-5x^(2) and C=2xy-6x^(2)-3x`. Find the value of A+B+C. |
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Answer» Correct Answer - `8x-xy +8x^(2)` `A+B-c=(2x+3xy+7x^(2))+(3x-2xy-5x^(2))` `-(2xy-6x^(2)-3x)` `=(2x+3xy+7x^(2)+3x-2xy-5x^(2)-2xy+6x^(2)+3x)` `=(2x+3x+3x)+(3xy-2xy-2xy)+7x^(2)-5x^(2)+6x^(2)` `=8x+(3xy-4xy)+(13x^(2)-5x^(2))` `8x-1xy+8x^(2)` |
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| 171. |
Identify the like terms from the following:(i) 2x2y, 2xy2, 3xy2, 14x2y, 7yx(ii) 3x3y2, y3x, y3x2, – y3x, 3y3x(iii) 11pq, -pq, 11pqr, -11pq, pq |
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Answer» (i) 2x2y, 2xy2, 3xy2, 14x2y, 7yx (a) 2x2y and 14x2y are like terms. (b) 2xy2 and 3xy2 are like terms. (ii) 3x3y2, y3x, y3x2, – y3x, 3y3x (a) y3x, – y3x and 3y3x are like terms. (iii) 11 pq, -pq, 11pqr, -11 pq, pq (a) 11 pq, -pq, -pq and pq are like terms. |
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| 172. |
Find the product of (12x) and `(5x-(3)/(2))` |
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Answer» Correct Answer - `60x^(2)- 18x` `(12x) (5x-(3)/(2))=(12x)(5x)-(12x)((3)/(2))` `60x^(2)-6x xx3=60x^(2)-18x` |
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| 173. |
If x=2 and y=3, then find the values of `(i) (2x+3y)/(4x-3y) " "(ii) (2x^(2)-7x+2y)/(2y^(2)-7y+2x)` |
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Answer» Correct Answer - (i) `-13` (ii) `0` x=2, y=3 `(i) (2x+3y)/(4x-3y)=(2xx2+3xx3)/(4xx2-3xx3)` `=(4+9)/(8-9)=(13)/(-1)=-13` `(ii) (2x^(2)-7x+2y)/(2y^(2)-7y+2x)=(2(2)^(2)-7(2)+2(3))/(2(3)^(2)-7(3)+2(2))` `=(2xx4-14+6)/(2zxx9-21+4)=(8-8)/(18-17)=(0)/(1)=0` |
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| 174. |
If p = -2, q = 1 and r = 3, find the value of 3p2q2r. |
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Answer» Given p = -2; q = 1; r = 3 ∴ 3p2q2r = 3 x (-2)2 x (1)2 x (3) = 3 x (-2 x 1)2 x (3) [Since am x bm = (a x b)m] = 3 x (-2)2 x (3) = 3 x (-1)2 x 22 x 3 = 31+1 x 1 x 4 [Since am x an = am+n] = 32 x 4 = 9 x 4 ∴ 3p2q2r = 36 |
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| 175. |
Find 27a3 + 64b3, if 3a + 4b = 10 and ab = 2. |
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Answer» 3a + 4b = 10, ab = 2 (3a + 4b)3 = (3a)3 + 3(3a)2 (4b) + 3 (3a) (4b)2 + (4b)3 (27a3 + 64b3) = (3a + 4b)3 – 3 (3a) (4b) (3a + 4b) ∵ x3 + y3 = (x + y)3 – 3xy – (x + y) = 103 – 36 ab (10)= 1000 – 36 x 2 x 10 = 1000 – 720 = 280 |
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| 176. |
Fill in the blanks to make the statements true:The variable used in the equation 2p + 8 = 18 is __________. |
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Answer» The variable used in the equation 2p + 8 = 18 is p. The word ‘variable’ means something that can vary, i.e., change. The value of a variable is not fixed. We use a variable to represent a number and denote it by any letter such as l, m, n, p, x, y, z etc |
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| 177. |
Fill in the blanks to make the statements true:Annual salary at r rupees per month along with a festival bonus of Rs 2000 is __________. |
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Answer» Annual salary at r rupees per month along with a festival bonus of Rs 2000 is ₹12r + 2000. From the question it is given that, Salary per month is r rupees a festival bonus of ₹ 2000 Therefore, Annual salary at r rupees per month along with a festival bonus of Rs 2000 is ₹ 12r + 2000 |
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| 178. |
Let y denote age in year. Write down the value of y for your 10 friends. |
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Answer» Let age of one friend = y years. then age of 10 friends are as follows: 1. (y – 2) years 2. (y + 3) years 3. (y + 1) years 4. (y – 1) years 5. y years 6. (2 y – 10) years 7. (y + 2) years 8. (y – 3) years 9. (y + 3) years 10.(y – 2.5) years |
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| 179. |
The present age of Vimal is p years. then (i) How old was he 10 years ago? (ii) How old will Vimal be 5 years from now? (iii) Vimal’s aunt is thrice as old as vimal. How old is vimal’s aunt? (iv) Age of Vimal’s mother is 5 years less than twice of vimal’s age. How old is his mother? |
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Answer» Vimal’s present age = P years (i) His age 10 years ago = (p – 10) years (ii) Vimal’s age after 5 years from now = (p + 5) years (iii) Vimal’s aunt age = 3 × p = 3p years (iv) Vimal’s mother age = 5 years less than twice of vimal’s age = 2 × p – 5 = (2p – 5) years |
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| 180. |
The height of a triangle is 5 more than twice of its base. What is its height if base is b? |
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Answer» Base of triangle= b unit Height of triangle = 5 more than twice of base = 2 × b + 5 = (2b + 5) unit |
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| 181. |
Derive statement from 3x – 7 = 11. |
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Answer» Subtracting 7 from thrice of a number gives 11. |
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| 182. |
Nathu has Rs. x with him Then, (i) How much money do Bina have if she owns twice as much as Nathu does? (ii) How much money is Nathu left with after buying books worth Rs. 150? (iii) How much money do Seema have if she owns half as much as Nathu has initially? (iv) How much money do Milli have if she owns thrice as much as Nathu does? |
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Answer» (i) Nathu has Rs. = x Bina has money = Twice money as much as Nathu’s money = 2 × x = Rs. 2x (ii) Nathu purchased books of = Rs. 150 money remains = Rs. (x – 150) (iii) Nathu has intial money = Rs. x Seema has money = half money as much as Nathu’s initial money = Rs. (iv) Milli has money = Thrice money as much as Nath’s money = 3 × x = Rs. 3x |
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| 183. |
Adding 15 in twice of a number results 51 Form equations form this statement |
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Answer» Let number x. double of number + 15 = 51 ⇒ 2 × x + 15 = 51 ⇒ 2x + 15 = 51. |
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| 184. |
Identify the operations(addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed.(a) z + 1, z – 1, y + 17, y – 17(b) 17y, y/17, 5z(c) 2y + 17, 2y – 17(d) 7m, -7m + 3, -7m – 3 |
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Answer» (a) z + 1, z – 1, y + 17, y – 17 Addition as 1 is added to z Subtraction as 1 is subtracted from z Addition as 17 is added to y Subtraction as 17 is subtracted from y (b) 17y, y/17, 5z Multiplication as y is multiplied as y is multiplied with 17 Division as y is divided by 17 Multiplication as z is multiplied with 5 Multiplication as z is multiplied with 5 (c) 2y + 17, 2y – 17 Multiplication and addition y is multiplied with 2, and 17 is added to the result (d) 7m, -7m + 3, -7m – 3 Multiplication with 2 and 17 is subtracted from the result Multiplication and subtractions m is multiplied by 7 and 3 is subtracted from the result |
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| 185. |
With the numbers 3, 7 and 4, form arithmetic expressions using (i) Only addition and subtraction operations. (ii) Only multiplication and addition operations. |
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Answer» (i) 3 + 7 + 4, 3 – 7 + 4, 3 + 7 – 4, – 3 + 7 + 4, – 3 + 7 – 4, – 3 – 7 + 4, – 3 – 7 – 4. (ii) 3 × 7 + 4, 3 × 7 × 4, 3 + 7 × 4, 3 + 4 × 7, 3 × 4 + 7. |
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| 186. |
A cube is a three-dimensional figure as shown in Fig 11.11. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. Find the formula for the total length of the edges of a cube. |
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Answer» Total number of edges in a cube `= 12` Length of an edge ` = l` So, total length of all the edges `= 12**l=12l` |
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| 187. |
Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different. |
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Answer» 2y + 7, 2y – 7, 7y + 2…. |
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| 188. |
Take Sarita’s present age to be y years(i) What will be her age 5 years from now?(ii) What was her age 3 years back(iii) Sarita’s grandfather is 6 times her age. what is the age of her grandfather?(iv) Grandmother is 2 years younger than grandfather. What is grandfather’s age?(v) Saritha’s father’s age is 5 years more than 3 times Saritha’s age. What is her father’s age? |
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Answer» (i) Saritha’s present age + 5 = y + 5 (ii) 3 years ago, Saritha’s age = Saritha’s present age – 3 y – 3 (iii) Grand father’s age = 6 × Sarita’s present age = 6y (iv) Grand father’s age = Grand father’s present age – 2 = 6y – 2 (v) Father ’s age = 5 + 3 x saritha’s persent age = 5 + 3y |
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| 189. |
Solve for x:2x-15=3x-20 |
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Answer» Correct Answer - 5 2x-15=3x-20 `-15+20=3x-2x` 5=x `:.x=5` |
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| 190. |
Simplify `8x-2[2x-3(4x+5-bar(2-x))]` |
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Answer» Correct Answer - `34 x + 18` `8x-2[2x-3(4x+5-bar(2-x))]` `=8x-2[2x-3(4x+5-2+x)]` `=8x-2[2x-3(5x+3)]` `=8x-2[2x-15x-9]` `=8x-2[-13x-9]` 6x+26x+18 =34x+18 |
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| 191. |
Solve for `x: (2x)/(3)+4=(8)/(3)(2x-6)` |
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Answer» Correct Answer - `(30)/(7)` `(2x)/(3)+4=(8)/(3)(2x-6)` `(2x)/(3)+4=(16x)/(3)-16rArr 4+16=(16x)/(3)-(2x)/(3)` `20=(16-2x)/(3)rArr20=(14x)/(3)` `20xx3=14xrArr60=14x` `(60)/(14)=xrArr(30)/(7)=x` `:. x=(30)/(7)` |
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| 192. |
Complete the table and find the solution of the equation x/3 = 4 using the table. |
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Answer» For, the table can be constructed as follows
12/3 = 4 |
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| 193. |
Complete the table and by inspection of the table, find the solution to the equation 5t = 35 |
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Answer» For 5t, the table can be constructed as follows
By inspection, we can find that t = 7 is the solution of the above equation as for t = 7, 5t = 5 × 7 = 35 |
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| 194. |
Complete the table and by inspection of the table find the solution to the equation m +10 = 6 |
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Answer» By inspection, we can find that m = 6 is the solution of the above equation as for m = 6, m + 10 = 6 + 10 = 16 |
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| 195. |
(a) Complete the table and by inspection of the table find the solution to the equation `m + 10 = 16`. (i) Complete the table and by inspection of the table, find the solution to the equation 5t = 35. (q) Complete the table and find the solution of the equation `z/3 = 4`using the table. (y) Complete the table and find the solution to the equation `m − 3 = 7`.. |
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Answer» a)6+10=16 b)5*7=35 c)12/3=4 d)10-7=3. |
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| 196. |
Pick out the solution from the values given in the bracket next to each equation. show that the other values do not satisfy the equation.(a) 5m = 60 (10, 5, 12, 15)(b) n + 12 = 20 (12, 8, 20, 0)(c) p – 5 = 5 (0, 10, 5 – 5)(d) q/2 = 7 (7, 2, 10, 14)(e) r – 4 = 0 (4, -4, 8, 0)(f) x + 4 = 2 (-2, 9, 2, 4) |
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Answer» (a) 5m = 60 (10, 5, 12, 15) m = 12 a solution to the given equation because form = 12 5m = 5 × 12 = 60 and hence, the equation is satisfied m = 10 is not a solution to the given equation because for m = 10. 5m = 5 × 10 = 50, and net 60 m = 5 is net a solution to the given equation because for m = 5, 5m = 5 × 5 = 25, and net 60 m = 15 is net a solution to the given equation because for m = 15 5m = 5 × 15 = 75, and net 6. (b) n + 12 = 20 (12, 8, 20, 0) n = 8 is a solution to the given equation because for n = 8 n + 12 = 8 + 12 = 20 and hence, the equation is satisfied n +12 = 12 + 12 = 24, and net 20 n = 20 is net a solution to the given equation because for n = 20 n + 12 = 20 + 12 = 32, and net 20 (c) p – 5 = 5 (0, 10, 5 – 5) p = 10 is a solution to the given equation because for p = 10, p – 5 = 10 – 5 = 5 and hence, the equation is satisfied. p = 0 is net a solution to the given equation because for p = 0 . p – 5 = 5 – 5 = 0, and net 5 p = -5 is net s solution to the given equation because for p = -5 (d) q/2 = 7 (7, 2, 10, 14) q = 14 is a solution to the given equation because for q = 7 (q/2) = 7/2 and not 7 q/2 = 7/2 and net 7 q = 2 is not a solution to the given equation because for q = 7 q/2 = 7/2 and net 7 q = 2 is net a solution to the given equation because for q = 2 q/2 = 2/2 = 1, and net 7 q = 10 is net a solution to the given equation because for qn = 10, q/2 = 10/2 = 5, and net 7 (e) r – 4 = 0 (4, -4, 8, 0) r = 4 is a solution to the given equation because for r = 4 r = 4 = 4 – 4 = 0 and hence, the equation is satisfied r = -4 is net a solution to-the given equation because for r = -4 r – 4 = -4 – 4 = -8 and net 0 r = 8 is net a solution to the given equation because for r = 8 r – 4 = 8 – 4 = 4 and net 0 r = 0 is net a solution to the given equation because for r = 0 r – 4 = 0 – 4 = -4 and net 0 . (f) x + 4 = 2 (-2, 9, 2, 4) x = -2 is a solution to the given equation because for x – 2 x + 4 = -2 + 4 = 2 and hence, the equation is satisfied x = 0 is net a solution to the given equation because for x = 0 x + 4 = 0 + 4 = 4 and net 2 x = 2 is net a solution to the given equation because for x = 2 x + 4 = 2 + 4 = 6 and net 2 x = 4 is net a solution to the given equation because for x = 4 x + 4 = 4 + 4 = 8, and net 2. |
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| 197. |
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable . (a) `17 = x + 7 ` (b) `(t - 7 ) > 5` (c) `4/2 = 2` (d) `(7 xx 3) -19 = 8` (e) `5 xx 4 - 8 = 2 x` (f) `x - = 0 ` (g) `2 m < 30` (h) `2 n + 1 = 11 ` (i) `7 = (11 xx 5) - (12 xx 4)` (j) `7 = (11 xx 2 ) + p` (k) `20 = 5 y ` (l) `(3 q)/ 2 < 5` (m) `z + 12 > 24 ` (n) `20 = (10 - 5) = 3 xx 5 ` (o) `7 - x = 5` |
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Answer» a)Variable is x b)Variable is t c)No variable d)No variable e)Variable is x f)Variable is x g)Variable is m h)Variable is n i)No variable j)Variable is p k)Variable is y l)Variable is q m)Variable is z n)No vaiable o)Variable is x. |
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| 198. |
Complete the entire in the third column of the table.S.No.EquationValue of variableEquation satisfied yes/noa10y = 80y = 10b10y = 80y = 8c10y = 80y = 5d4l = 20l = 20e4l = 20l = 80f4l = 20l = 5gb + 5 = 9b = 5hb +5 = 9b = 9ib + 5 = 9b = 4jh - 8 = 5h = 13kh - 8 = 5h = 8lh - 8 = 5h = 0mp + 3 = 1p = 3np + 3 = 1p = 1op + 3 = 1p = 0pp + 3 = 1p = -1qp + 3 = 1p = - 2 |
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Answer» (a) 10y = 80 y = 10 is net a solution to the given equation because for y = 10 10y = 10 × 10 = 100, and net 80 (b) 10y = 80 y = 8 is a solution to the given equation because for y = 8 10y = 10 × 8 = 80 and hence, the equation is satisfied (c) 10y = 80 y = 5 is net a solution to the given equation because for y = 5 10y = 10 × 5 = 50, and net 80. (d) 4l = 20 l = 20 is net a solution to the given equation because for l = 20 4l = 4 × 20 = 80, and net 20 (e) 4l = 80 l = 80 is net a solution to the given equation because for l = 80 4l = 4 × 80 = 320, and net 20 (f) 4l = 20 4l = 4 × 5 = 20 and hence, the equation because for l = 5 4l = 4 × 5 = 20 and hence, the equation is satisfied (g) b + 5 = 9 b = 5 is net a solution to the given equation because for b = 5 (h) b + 5 = p b = 9 is net a solution to the given equation because for b = 9 b + 5 = 9 + 5 = 14, and net 9 (i) b + 5 = 9 b + 5 = 4 + 5 = 9 and hence, the equation is satisfied (j) h – 8 = 5 h = 13 is a solution to the given equation because for h = B h – 8 = 13 – 8 = 5 and hence, the equation is satisfied (k) h – 8 = 5 h = 8 is net a solution to the given equation because for h = 8 h = 8 = -8 = 0, and net 5 (l) h – 8 = 5 h = 0 is net a solution to the given equation because for h = 0, h – 8 = 0 – 8 = -8, and net 5 (m) p + 3 = 1 p + 3 = 3 + 3 = 6, and net 1 (n) p + 3 = 1 p + 3 = 1 + 3 = 4 , and net 1 (o) p + 3 = 1 p + 3 = 0 + 3 = 3, and net 1 (p) p + 3 = 1 p + 3 = -1 + 3 = 2, and net 1 p + 3 = -2 + 3 = 1 and hence, the equation is satisfied. |
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| 199. |
(a) Given Munnu’s age to be x years, can you guess what `(x + 2)` may show? (Hint : Think of Munnu’s younger brother.) Can you guess what `(x + 4)` may show? What `(3 x +7 )` may show? (b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate? `y + 7, y - 3, y + (4) 1/2, (2) 1/2.` (c) Given n students in the class like football, what may 2 n show ? What may `n/2` show ? (Hint : Think of games others than football )Type here in ASCII with maths in back tick : |
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Answer» a)x-2 represent the age of the person who is 2 years younger than munnu x+4 represents the age of the person than can be his elder brother who is 4 years older than munnu. 3x+7 represents the age of the person whose age is seven years more than 3 time munnus age than can be his father. b)y+7 represents saras age 7 years late y-3 represents saras age 3 years ago y+4 1/2 represents saras age 4 1/2 years later. y-2 1/2 represents saras age 2 1/2 years ago. c)2n may show the number of students who like cricket or football n/2 may show the number of students who like both football and other games like cricket. |
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| 200. |
State which of the following are equations (with a variable). Give reason for your answer. Identify the variable from the equations with a variable.(a) 17 = x + 7(b) (t – 7) > 5(c) 4/2 = 2(d) (7 × 3) – 19 = 8(e) 5 × 4 – 8 = 2x(f) x – 2 = 0(g) 2m < 30(h) 2n + 1 = 11(i) 7 = (11 × 5) – (12 × 4)(j) 7 = (11 × 2 ) + p(k) 20 = 5y(l) 3p/2 < 5 (n) 20 – (10 – 5) = 3 × 5(o) 7 – x = 5 |
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Answer» (a) 17 = x + 7 An equation with variable x (b) (t – 7) > 5 An inequality (c) 4/2 = 2 No, it is a numerical equation (d) (7 × 3) – 19 = 8 No, it is a numerical equation (e) 5 × 4 – 8 = 2x An equation with variable x (f) x – 2 = 0 An equation with variable x (g) 2m < 30 An inequality (h) 2n + 1 = 11 An equation with variable n (i) 7 = (11 × 5) – (12 × 4) No, it is a numerical equation (j) 7 = (11 × 2 ) + p An equation with variable p (k) 20 = 5y An equation with variable y (l) 3p/2 < 5 (An inequality m) z + 12 > 24 An inequality (n) 20 – (10 – 5) = 3 × 5 No, it is a numerical equation (o) 7 – x = 5 An equation with variable x. |
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