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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Question numbers 1 to 6 carryImark each.ky Find the largest number which divides 320 and 457 leaving remainders 5 and 7 respectively |
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| 2. |
4(3x+2)-5(6x-1)=2(x-8)-6(7x-4) |
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Answer» ⭐Solution⭐ 4(3x + 2) - 5(6x - 1) = 2(x - 8) - 6(7x - 4) =>12x+8-30x+5= 2x-16-42x+24 => -18x +13 = -40x+8 => -18x+40x = 8-13 =>22x = -5 =>x = -5/22 ✨Thats answer✨ |
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| 3. |
6)12-7x+1 |
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Answer» 12x² - 7x + 1 12x² - 3x - 4x + 1 3x ( 4x - 1) - (4x - 1) (3x - 1) ( 4x - 1) |
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| 4. |
SECTION ASaive the following brieĂy : 2 marks each16the largest number dividing 230 and 142 and leavingremainders 5 and 7 respectively |
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| 5. |
6(3x+2)-5(6x-1)=3(x-8)-5(7x-6)+9x |
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Answer» thanks |
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| 6. |
ExerciseSimplify: सरलीकर |
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Answer» 11/4 is the right answer |
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| 7. |
EXERCISE 7.6Sale |
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Answer» 2/3+1/7= 14+3/21= 17/21 2/3+1/717/22 answer 3/10+7/159+14/3023/30 answer 5/7+1/322/21 answer 9and679y6728859927783938293894739583864858498589 (a)=17/21(b)=23/30(c)=46/63(d)=22/21(e)=17/30 hope this will help you like my answer 2/3+1/717/22 answer. |
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| 8. |
EXAMPLE 1 Evaluate:6 -36 -3 |
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Answer» Like if you find it useful |
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| 9. |
Exercise 11.1Evaluate |
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Answer» 1/9 is the correct answer of the given question |
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| 10. |
\frac { 6 ^ { \frac { 1 } { 3 } } \times \sqrt [ 3 ] { 6 ^ { 7 } } } { \sqrt [ 3 ] { 6 ^ { 6 } } } |
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| 11. |
:Factorise:(i) 49a² + 70ab + 25b² |
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Answer» nhi pta mujhe yaar to tujhe kaha se batau 49a² + 70 ab + 25b²(7a)² + 2× 7a × 5a + (5b)²(7a + 5b)² |
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| 12. |
(iii) 49a? + 70ab + 256% |
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| 13. |
ample 18: Factorise(i) 49a2 + 70ab 25b(ii) 5 - yi4. |
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Answer» i) 49a² + 70ab + 25b² (7a)² + 2 × 7a × 5b + (5b)² (7a + 5b)² ii) 25x²/4 - y²/9 (5x/2)² - (y/3)² (5x/2 - y/3) ( 5x/2 + y/3) |
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| 14. |
(i) 49a2 + 70ab + 25b2 |
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Answer» 49a² + 70ab + 25b² (7a)² + 2 × (7a) × (5b) + (5b)² (7a + 5b)² Please hit like .. |
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| 15. |
10Check which of the following are solutions of the equation 7x - 5y = -3.É (-1,-2)(-4,-5)OR |
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Answer» Given:7x-5y=-3I)(-1,-2)Thus7(-1) -5(-2)=-7+10=3Therefore (-1,-2)is not a solution ii)(-4,-5)7(-4)-5(-5)=-28+25=-3Therefore (-4,-5) is a solution for the equation |
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| 16. |
18, Find the remainder when x+x-2x2+x+1 is divided byc-1 |
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| 17. |
Factorise:(i) 49a2+ 70ab + 25b2 |
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| 18. |
If (2, 3) is a solution of the equation 7x -Ky5, then find K |
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Answer» Given : (2,3) is solution of 7x - ky = 5 put x = 2 and y = 3 7 × 2 - 3k = 5 14 - 5 = 3k 9 = 3k k = 3 |
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| 19. |
()The ratio ofthe sum and the product of two roots of the equation 7x^2-12x+ 18 = 0 |
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| 20. |
Find the discriminant of the quadratic equation root5*x^2-7x+2root5 |
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Answer» Only 4 line is answer |
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| 21. |
कि. -ot 0P 2दे. 70ab + 250 (ii) = X" = S‘N |
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| 22. |
solve the equation 7(x+6)+7x=9 |
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| 23. |
10.Simplify the expression and find its value when a = 5 and b=-3.2(a + ab) +3- ab |
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| 24. |
EXERCISE 25B(a) Draw the graph of the function y 3x.(b) From the graph, find the value of y, when1.(i) x 3(ii) x 5(ii) x - 6 |
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| 25. |
Weights of two friends Aaryan and Neeru are in the ratio4:1 Aaryan's weight increase by 12% and total weight of Aarand Neeru together becomes 50 kg, with an increase of 25By what percent did the weight of Neeru increase?आर्यन और नीरू दो दोस्तों के वजन का अनुपात 4:1 है। आर्यन के वजन में ।की वृद्धि के साथ दोनों का कुल वजन 50 किग्रा हो जाता है जब कुल वजन मेंकी वृद्धि हो। नीरू के वजन में '% वृद्धि ज्ञात करे।(a) 52 (b) 77 (c) 73 (d) 71 |
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Answer» 77%hi hogajust tukkka |
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| 26. |
Q.7 Find the sureatest number which divides 2011 & 2623 leaving remainders 9 and 5 respectively. |
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| 27. |
10. Find the greatest number whichrespectivelyreatest number which divide 2011 and 2623 leaving remainders 9 and 5 |
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| 28. |
Hint. Two numbers are co-primes if their HCF is 1.5. Find the greatest number which divides 615 and 963, leaving the remainder 6 in each casewhich divides 2011 and 2623, leaving remainders 9 ana |
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| 29. |
The greatest number which divides2011 & 2623leaving remainder 9&5 respectively each |
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| 30. |
5.Find the greatest number of 4-digit which when divided by 20, 24 and 45 leaves a remainder 18 ineach case. |
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| 31. |
. Finol the angle 아 tnte ( section |
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| 32. |
Find the cubes of the following numbers.a. -23b. 17 491 3 |
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Answer» a. cube of -23³ is (-23) × (-23) × (-23) = - 12,167 b. 17³ = 17×17× 17 = 4,913 |
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| 33. |
b.11. Inthe given figure, plg, then find a and4a 1870°2а +12 |
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| 34. |
ence of diy491QUESTION SET 18 : BINOMIAL DISTRIBUTION- 2. The probability that a certain type of component will survive a check testis 0.5. Find the probability that exactly two of the next four components willsurvive.(Oct. '14) |
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| 35. |
या है-। ऊ4. अनुक्रम 56,58, 62, 70, 84, 118, 18 |
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Answer» 84 is the digit thats answer |
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| 36. |
i)Solve the equation 7x-9 = 16ngle of |
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| 37. |
In Fig.Eio, 10.38,ABC69°, 4 ACB 31°, find4.ngleL BDC |
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| 38. |
21. (a+4b)-5 and ab--3 find the value of (a2 +16b2) |
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Answer» Given : ab = -3 ( a + 4b) = 5 sqauring both sides (a + 4b)² = 5² = 25 a² + 16b² + 8ab = 25 a² + 16b² - 24 = 25 a² + 16b² = 49 |
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| 39. |
5. In the given figure, 4B<LA and C<LD.Show that AD < BC. |
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Answer» GIVEN angle b < angle a angle c < angle d TO PROVE AD < BCPROOF angle b < angle a SO, OA < OB ( SIDE OPPOSITE TO SMALLER ANGLE IS SMALL ITS THEOREM GIVEN IN NCERT) LET THE ABOVE ONE BE 1 NOW, angle c < angle d SO, OD < OC ( SIDE OPPOSITE TO SMALLER ANGLE IS SMALL ITS THEOREM GIVEN IN NCERT) LET THE ABOVE ONE BE 2 NOW, ADDING 1 AND 2 OA + OD < OB + OCADDING WE GET, AD < BC HENCE PROVED |
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| 40. |
18 Factorise:(i) 49 a^{2}+70 a b+25 b^{2} |
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| 41. |
x square-m square-9× n square+6mn |
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Answer» x²-m²-9n²+6mn = x²-(m²+9n²-6mn)= x²-(m²+(3n)²-2(m)(3n))= x²-(m+3n)²= (x-m-3n)(x+m+3n) . |
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| 42. |
28. In one day, a rickshaw puller earned? 137능-Out of this money, he spent? 563 on food.24How much money is left with him?3.5 |
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| 43. |
TheperimielrIn an isosceles triangle, the vertex angle is twice of either base anges |
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Answer» please post the complete question. |
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| 44. |
low many different shapes can you make by changingngle between the matchsticks in each of these? Try. |
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| 45. |
the adjacent figure, CD andare altitudes of an isoscelesngle ABC with AC = AB. ProvetAE = AD |
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| 46. |
A number when divided by 221 leaves18 as remainder. If the same numberis divided by 13, the remainder is? |
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| 47. |
Fmd a quadratic polyiomial, the sum andproduetofr whose zeroes are b and N respectively-FDivide 6x13r+x-2 by 2x +1, and find the quotient and remainder13. Divide -2s by-x +1. find quotient and remainder |
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| 48. |
(a) 17061. Find that smallest number which when divided by 5,6,8,9 and 12 leaves1 as remainder in each case but leaves no remainder when it is dividedby 13. |
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Answer» lcm(5,6,8,912)=360.Hence 361 is divided by 5,6,8,9, 12 we get the remainder as 1.Find a positive integer 360*n+1 such that 13 divides that number. such a n is 10.Hence 3601 is the required number. |
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| 49. |
The remainder when 3815 is divided by 13 is |
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Answer» this not answer |
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| 50. |
13.I+ 1) is divided by (x + 1), find the remainder. |
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Answer» 2 will be remainder of this question |
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