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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. |
4. HCF (306, 657) = 9í LCM (306, 657)hTTU |
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Answer» hcf*lcm = 306*657 lcm = 306*657/9 = 22338 |
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| 2. |
H.CE (306, 657)=9 fant LCME( 306, 657)t |
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Answer» LCM * HCF = 306*657 LCM = 306*657/9 LCM = 22338 |
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| 3. |
Given that HCF (306, 657-9, find LCM (306, 657) |
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| 4. |
Given that HCF (306, 657-9, find LCM (306, 657). |
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| 5. |
Given that HCF (306, 657)-9, find LCM (306, 657) |
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Answer» HCF (306,657) = 9 Product of LCM ×HCF =product of two number = LCM × 9 = 306×657LCM = 306×657 by 9 =201042 =22338 |
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| 6. |
4 Eiven that HCF 606, 657)-9, find LCM (306, 657). |
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Answer» HCF × LCM = product of two number9 × LCM = 306 × 657LCM = (306×657)/9 = 22,338 |
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| 7. |
118 164356 336 |
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| 8. |
(D) The value of π is-The product of LC.M. and H.C.F. of twomumbers is 88288.if one of the numbersis 248, find the other number(A) 356) 365(B) 635(D) 653 |
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| 9. |
Solve any two of the following sub-quFactorize : 3x2-2x-8.i) |
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Answer» 3x*x - 2x - 8 = 3x*x - 6x + 4x - 8 = 3x(x-2)+4(x-2) = (3x+4)(x-2) |
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| 10. |
man can complete a piece of work in 20 days. The amount of work done in 8 dA)th2Td3(Brdth33r, end C in 15 days C and A in 20 days. If A |
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Answer» Let amount of work can be done in 20 days = xThen amount of work done in 1 day = x/20 Amount of work done in 8 days= 8x/20 = 2x/5 Hence in 8 days amount of work done is 2/5th of work done in 20 days (b) is correct option |
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| 11. |
5. Sumeet bought a plot of land for 2,25,000. He sold a rd of the plot at a loss of 8%. At whatpercent profit must he sell the remaining plot of land in order to get an overall profit of 12%?3 |
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| 12. |
४ +tan’ AS cot? A sec’A (i) -1 (i) co A हे. नए धा* & .. उत्तर-_बराबर हैं- (M. Q., Set—1 ; | |
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| 13. |
Which of the lines 2x-y+3=0 and x-4y-7=0 is farther from the origin. |
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| 14. |
Section -Find the acute angle between the lines 2x + y -1 = 0and 3x + y + 4 = 0OR |
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| 15. |
345+876÷133 |
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Answer» 345+876/133=345+6.586 =351.586 it's dmas876÷133 +345= 351.58 |
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| 16. |
If in a 3-digit number, the sum of its all three digits is 9, when the middle digit is 5, if the first digit is replaced by thethird digit and middle remain same then the difference between new number and the old number is 198. Find thenumber. |
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Answer» a+b+c = 9 The number can be 351 351-153 = 198 |
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| 17. |
53.How many three digit numbers are divisible by 13 and having middle digit(a) 55?(d) 13(b) 7(c) 10 |
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Answer» For a number to be divisible by 13, we have a small trick.sayxyzis the number whenxis any digit number whiley&zare single digits then (4*x - yz) should be divisible by 13. For 3 digit numberxis single digit as well. Without any logics we can say that upper bound is 9 for our answer. According to given questionyis fixed as 5.xlies from 1 to 9 both included.zlies from 0 to 9 both included. Now lets go one by one : 15z : to satisfy above trick z can be 6. 25z : No value of z exists. Similarly 351,455,559,650,754,858 satisfies while 95z will not. So the answer is7. |
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| 18. |
The middle digit of a number between 100 and 1000 is zero and the sum of tdigit is 13. If the digits are reversed, the number so formed exceedsnumber by 495. Find the number. |
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| 19. |
15. A three digit number is equal to 17times the sum of its digits. If the digits.are reversed, the new number is 198more than the old number; also thesum of extreme digits is less than themiddle digit by unity. Find theoriginal number. |
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Answer» Let ones digit = x, tens digit = y, hundreds digit = zSo, number becomes = 100z+10y+xAccording to question-100z+10y+x = 17(x+y+z)100z+10y+x = 17x + 17y + 17z100z-17z +10y-17y + x-17x = 083z -7y -16x = 0 ----------------(1) Also,100z+10y+x + 198 = 100x + 10y + z100z- z +10y-10y +x - 100x +198 = 099z - 99x +198 = 099x - 99z = 198 x - z = 2 x = z + 2--------------------(2) And, z + x = y-1By putting (2)y = z+z+2+1 y = 2z+3 ----------------------(3) Putting value of y from eq(3) and x from eq(2) in (1) we get 83z -7(2z+3) -16(z+2) = 0 83z -14z -21 -16z -32 = 053z = 53z = 1put z = 1 in eq(3)y = 2(1) +3y = 5 Put z = 1 in eq(2) x = 1 + 3 = 3So, number = 100(1) + 10(5) + 3 = 153 Therefore, number = 153 |
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| 20. |
345/25 |
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Answer» Given division is 345 ÷ 25 = 13.8. 345÷25=13.8 |
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| 21. |
356+345 |
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Answer» 1234 is answer......... sorry 701 is the answer |
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| 22. |
345*6788 |
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Answer» the product of the above two numbers is 2,341,860 |
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| 23. |
If both roots of x2 -32x + c 0 are prime numbers then possible values of c are [16(A) 60(B) 87(C) 247(D) 231 |
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Answer» There are only two such combinations. 3+29=32 & 13+19=32 Now, c is the product of both roots. Then c = 3*29 = 87 & c = 13*19 = 247 |
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| 24. |
389)247( |
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| 25. |
(5) A three digit number is equal to 17 times the sum of its digits; If the digits arereversed, the new number is 198 more than the old number; also the sum ofextreme digits is less than the middle digit by unity. Find the original number |
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Answer» Let ones digit = x , tens digit = y , hundreds digit= zSo, number becomes = 100z+10y+xAccording to question- 100z+10y+x = 17(x+y+z) 100z+10y+x = 17x + 17y + 17z 100z-17z +10y-17y + x-17x = 0 83z -7y -16x = 0 -----------------------(1) Also,100z+10y+x + 198 = 100x + 10y + z100z- z +10y-10y +x - 100x +198 = 099z - 99x +198 = 099x - 99z = 198 x - z = 2x = z + 2 ----------------------------(2) And, z + x = y-1By putting (2)y = z+z+2+1 y = 2z+3 ----------------------------(3) Putting (3) and (2) in (1) we get83z -7(2z+3) -16(z+2) = 0 83z -14z -21 -16z -32 = 053z = 53z = 1put z = 1 in (3)y = 2(1) +3y = 5 Put z = 1 in (2)x = 1+3= 3So. number = 100(1) + 10(5) + 3 = 153So, number = 153 |
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| 26. |
= x2 + 7x + 10, 3HTIX"(M. Q., Set-II: 2015) |
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| 27. |
x+y=5x-5y=6 |
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Answer» ooo I just check this App I want to do friendship with you my contact number is 7007169848 |
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| 28. |
Practice set 1.1Write the following sets in roster form.Set of even numbers(ii) Set of even prime numbers from 1 to 50(iii) Set of negative integers (iv)Seven basic sounds of a sargam (sur) |
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Answer» 1) 2,4,6,8....2) 23) -1,-2,-3,....4) Sa,Re,Ga,Ma,Pa,Dha,Ni |
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| 29. |
2the value of the polynomial p(yy-5y +6 at y2 |
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Answer» Value at y= -2 will be(-2)^2-5(-2)+64+10+620Answer |
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| 30. |
Find the point of intersection of the lines 2x-3y + 8-0 and 4x + 5y6] |
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| 31. |
Ifx k2 and y k is a solution of equation x-5y +6 0, find value of k? |
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| 32. |
Describe the following sets in set-builder form.(i) The set of all letters in the wordPROBABILITY.(ii) The set of all even natural numbers.(iii) (5, 25, 125, 625) |
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| 33. |
2y+148 If3Ń+8 272ytl, then find the value of12y289 4913216 |
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| 34. |
How many three digits number are divisible by 7 |
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| 35. |
28965°6.24 |
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Answer» since PQ _|_ PS , then angle P = 90° now angle x = QSR .... ( alternate interior angles) and x+28 = 65 ..... ( exterior angle property)=> x = 65-28 = 37° now, in ∆PQS , 90+37+y = 180=> y = 180-90-37 = 53° |
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| 36. |
D.A quadratic function f(x) attains a maximum of 11at x I. The value of the function at x 0 is 8.What is the value of f(x) at x =-9?B. -289D. -294247C. -286E. of these |
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| 37. |
345 + 289 = |
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Answer» 634 is the answer of the following 634 is the answer of given questions 345+289=634 is the correct answer The right answer is 634. 634is your answer of questions 634 is your answer our following The right answer is 634 634 is the correct answer 345+289 = 634 is the answer of given questions 634 is your answer your question 634 is the right answer 634 is the answer of given questions 634 is the answer of given questions |
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| 38. |
4/cot(30^circ)^2 %2B 1/(sin(60^circ)^2) - cos(45^circ)^2 |
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Answer» 19/12 is the correct answer of the given question need help in any other questions 4/ cot^2 30 + 1/sin^2 60-cos^2 45= 4/(V3)^2+ (2/V2)^2-(V2/2)2=4/3+4/3-2/4=16+16-6/12=32-6/12=26/12=13/6 |
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| 39. |
4/cot(30^circ)^2 %2B 1/(sin(30^circ)^2) - cos(45^circ)^2 |
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Answer» 29/6 |
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| 40. |
4: Factorise y- 5y 6 |
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Answer» Factorize:y2- 5y + 6 (To split the middle term we have to take two integers whose sum gives -5y and product gives 6y^2)y2- 3y - 2y + 6 (-3y+(-2y) = -5y and -3y*-2y = 6y2)y(y-3) -2(y-3)(y-3)(y-2) |
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| 41. |
4 Factorise the followinga 4a2-4a1 |
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Answer» 4a² - 4a + 1 4a² - 2a - 2a + 1 2a( 2a - 1) - ( 2a - 1) (2a - 1) ( 2a - 1) (2a - 1)² |
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| 42. |
4. Factorise:12x2-7x + 1(ii) 2x2 + 7x+3 |
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| 43. |
Factorise the following expression :x^4-(y+z)^4 |
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| 44. |
Factorise : x3- 5x2 -2x + 4 |
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Answer» x*x*x - 5x*x - 2x + 4 = x*x*x + x*x - 6x*x - 6x + 4x + 4 = (x+1)(x*x - 6x + 4) If you find this answer helpful then like it. |
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| 45. |
20. The distance between two stations is 300 km. Two motorcyclists start simultaneously fromthese stations and move towards each other. The speed of one of them is 7 km/h more thathat of the other. If the distance between them after 2 hours of their start is 34 km, find thespeed of each motorcyclist. Check your solution. |
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| 46. |
cos 38° cos 52° —sin 38° आए 529& |
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Answer» We know,cosA cosB - sinA sinB = cos(A+B) Here we have A = 38° and B = 52° cos(38°)cos(52°) - sin(38°)sin(52°)= cos(38°+52°)= cos(90°)= 0 |
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| 47. |
From the given alternatives select twosuch signs when there are mutuallyinterchanged the following equationbecomes correct.20.36+12 x 6+9-638 |
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Answer» option d IS RIGHT AS AFTER REPLACING WE CAN GET OUR REQUIRED RESULT |
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| 48. |
3. Select the related word/letters/ number from the givenalternatives.64:8::289:? |
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Answer» 17As the above relation is between square roots. √64 = 8√289 = 7 |
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| 49. |
नगर की जनसंख्या 31 दिसम्बर 1978 को 1 पददर 10% वार्षिक हो, तो 31 दिसम्बर 1981 को उसे2 |
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| 50. |
\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 60^{\circ}}{\tan 120^{\circ}}= |
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