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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. |
4Twenty-four is divided into two parts such that 7 times the first part added to 5 times thesecond part makes 146. Find each part. |
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| 2. |
Dictance R(5,1) S(5,0) |
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Answer» distance between R and S can be calculated as:distance = √(5-5)² + (1-0)² = √0² + 1² = √1 = 1 1 Is right answersgdkssvska Thank you |
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| 3. |
the giventhens bind the lont d |
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Answer» 1 2 |
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| 4. |
the aaaoinirg yure showss ă˛cm bind the |
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| 5. |
bind the principal solution of equations:(11)tan x =-(i- |
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Answer» part 1 |
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| 6. |
2)bind Solution4x + by zarb and bec- aya atb |
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| 7. |
Factories : x^3 + 235x + 1385. if possible |
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Answer» 1st 2nd 3rd |
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| 8. |
For an event E, P (E ) =3÷7, then find P (E ). |
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| 9. |
びtan θ=cot(30° + e), find the value of e. |
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| 10. |
llathematics for Class-8yolume of a sphere(a) 1386 cm 2re is 4851 cm 3, then its surface area is485(b) 1380cm2(c) 1385 cm2(d) none of thesewellowing question consists f tin |
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| 11. |
नव चीन. Find SR e e |
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Answer» 180-125=5555 is the interior angle Sum of 2 interior anlges = exterior angle55+55=xx=110 |
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| 12. |
(e) Find k for which the folowing equation has equal roots. +a44kr + (e-k + 2), |
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| 13. |
con ' lareĺ Ľpansions are terminating.What can be the maximum number of digits in the repeating block ofdigits in the decimal expansion of |
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Answer» The maximum no. of digits in the repeating block of digits can never be more than n, taking n to be the divisor or denominator.So max. no. of digits in repeating block of 5/7 is 65/7 = 0.714285 bar over 714285 |
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| 14. |
EXERCISE 251.Form 3-digit numbers using all the three given digitswithout repeating any digit and ring the largest number:I. 3, 5, 7: _II. 4, 6, 8:III. 9, 6, 3:IV. 5, 7, 0:V. 2, 3, 1:VI. 8, 5, 3:VII. 6, 0, 3:VIII. 4, 5, 6:Form 3-digit numbers as in the above case and ring!omallest number: |
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Answer» 9,11,1310,12,140,-3,-6 9,11,13 10,12,14o,-3,-6 |
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| 15. |
1. Form the greatest and the smallest four-digit numbers from the given digits without repeatingany digit.a. 3,7,8,9b.9,4,3,2c. 5, 6, 8,0ob liit olyance. |
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Answer» 1.(a) greatest number = 9873smallest number = 3789(b) greatest number = 9432Smallest number = 2349(C) greatest number = 8650smallest number = 5068 |
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| 16. |
them in descenumy UluWrite the smallest and the greatest 2-digit numbers (without repeating a diofrom the following digits:SmallestGreatest00(a) 3, 6,516120720...... |
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Answer» 35 smallest number |
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| 17. |
find three digit number which can be formed by using digit 0,3,5 Without repeating any digit |
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Answer» 350,530,503,305 ..... 503, 530,305,350,053,035 are the digits........ 035,530,305,503 is the right answer 530,305,350,305 is the right answer 530,350,305,503,053,035 |
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| 18. |
In each set form the smallest and the greatest 6-digit numbers, withoutrepeating the following digits.a) 4.3, 0, 2, 7, 1 b) 6,7,1,2,5,8 c) 3, 9, 8, 7, 2,1d) 8.0.1, 2, 5,7 e) 7,5,6,9, 2,99,6, 0,5, 1,3om the smallested the best inumbers by repeating any one digit. |
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Answer» a)greatest 6-digit number=743201 a)smallest 6-digit num.102347 |
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| 19. |
Fill the missing numerals suchthat the sum of each side ofthe square is 18. Rememberthat you have to use numeralsfrom 2 to 9 without repeatingany of them |
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| 20. |
(A) HCF> LCM(C) LCM > HCF(B)HCF = LCM |
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| 21. |
tan6 cot®+= t 0.l1—-cotb 1-tan® thmheeProve that |
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| 22. |
The relation between LCM and HCF of twopositive integers will be:(A) HCF LCM(C) LCM > HCFHCF= LCM(B)(D) of these |
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Answer» LCM X HCF = FIRST NUMBER XSECOND NUMBER. LCM of two numbers a and b is the least common multiple and HCF of two numbers is the highest common factor.The relation between LCM and HCF of two numbers and b |
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| 23. |
What number multiplied by 9 gives 648? |
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Answer» 9 |
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| 24. |
ICa lxlo-1)176 |
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Answer» a² is the correct answer the following question answer is a2 (1/a×1/10)^-2 × 1/100= (1/10a)^-2× 1/100= (10a/1)^2 ×1/100= 100a^2/100= a^2 answer a² is the correct answer of the given question (a-1*10-1)-2+1/100 Final result : 100a - 1299 ——————————— 100 Step by step solution : Step1: 1 Simplify ——— 100 Equation at the end of step1: 1 (a - 13) + ——— 100 Step2: Rewriting the whole as an Equivalent Fraction : 2.1Adding a fraction to a whole Rewrite the whole as a fraction using100as the denominator : a - 13 (a - 13) • 100 a - 13 = —————— = —————————————— 1 100 Equivalent fraction :The fraction thus generated looks different but has the same value as the whole Common denominator :The equivalent fraction and the other fraction involved in the calculation share the same denominator Adding fractions that have a common denominator : 2.2 Adding up the two equivalent fractionsAdd the two equivalent fractions which now have a common denominator Combine the numerators together, put the sum or difference over the common denominator then reduce to lowest terms if possible: (a-13) • 100 + 1 100a - 1299 ———————————————— = ——————————— 100 100 Final result : 100a - 1299 ——————————— 100 a^2 is correct answer. |
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| 25. |
SA ne 1L1o |
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Answer» 1 2 3 |
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| 26. |
what vector added to (1方以) and (2n) willgive a unit vector along negative y axis? |
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Answer» The resultant unit vector should be -1j so , let that vector be a now a + (2i-2j+k) + ( 2i-k) = -1j => a = -4i+j |
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| 27. |
Show that the vectors 2i-3j+ 4k and 4i+6j-Sik are collinear.OK are collinear. |
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Answer» these two vectors are collinear because cross product of these 2 vectors is 0|. I. j. k ||2 -3 4 |=== 0|-4 6 -8 | |
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| 28. |
In an AP of 50terms, the sum of 1st 10 terms is 210 and the sum of last15 terms is 2565.finr AP |
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Answer» AP of an=a+ ( n-1)d, sum of sn=n/2(2a+( n-1)d, sum first 10 terms is 210. 10/2(2a+9d)=210, 5(2a+9d)=210, ==2a+9d=42__________(1), 15 terms the last=(50-15+1)th=35d; 15/2(2a36+(15-1)d)=2565; 15/2(2a+35d)+14d)=2565; 15(a+35d+7d)=2565; a + 42d=171_______(2), from (1) and (2)= d/4; a=3, A.P. are 3, 7, 11, 15 _____,,,,199 S₅₀ - S₃₅ = 2565⇒ 50/2 {2a + (50 - 1)d} - 35/2 {2a + (35 - 1)d} = 256525 (2a + 49d) - 35/2 (2a + 34d) = 2565⇒ 5 (2a + 49d) - 7/2 (2a + 34d) = 513⇒ 10a + 245d - 7a + 119d = 513⇒ 3a + 126d = 513⇒ a + 42d = 171 ........(2)Multiply the equation (2) with 2, we get2a + 84d = 342 .........(3)Subtracting (1) from (3) 2a + 84d = 342 2a + 9d = 42- - -_______________ 75d = 300_______________d= 4Now, substituting the value of d in equation (1)2a + 9d = 422a + 9*4 = 422a = 42 - 362a = 6a = 3So, the required AP is 3, 7, 11, 15, 19, 23, 27, 31, 35, 39 ........ |
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| 29. |
If Ā= 31+ 2ſ and B = 2î +3j - k, then find a unit vector along (Ā-B). |
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| 30. |
21. 111×XIVArrange the following in1· V, IV, IX, VII, Xlll, XXV |
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Answer» Ascending order = IV, V, VII, IX, XIII, XXV |
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| 31. |
1S -The ratio between the LCM and HCF of 5, 15 and 20 is |
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Answer» LCM of 5, 15, 20 5 = 5 15 = 3 × 5 20 = 2² × 5 LCM = 5 × 3 × 2² = 60 HCF of 5, 15, 20 5 = 5 15 = 3 × 5 20 = 2² × 5 HCF = 5 Ratio of LCM to HCF 60 : 5 12 : 1 |
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| 32. |
9046 \div 8 |
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Answer» 9046 ÷ 8 = 1130.75 is the answer full solution pls solution pls |
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| 33. |
3 \div\left[(8-5) \div\left\{(4-2) \div\left(2+\frac{8}{13}\right)\right\}\right] |
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Answer» 3÷[(8-5)÷{(4-2)÷(2+8/13)}] 3÷[(3)÷(2)÷28/13] 3÷[3/2 × 13/28]3÷[39÷56] 3÷[0.696]4.31 |
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| 34. |
( - 40 + 80 a ) \div 8 a |
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Answer» -40+80a×1/8a-40+10a -40/8a + 80a/8a=-5/a+10=-5(1/a-2) |
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| 35. |
2 \frac { 1 } { 4 } \div 8 = |
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| 36. |
8 व्यक्तियों की एक टीम ने तीरन्दाजी प्रतियोगिता में भाग लिया।सबसे अच्छे तीरन्दाज ने 85 अंक प्राप्त किये यदि इसने 92 अंकप्राप्त किये होते तो औसत प्राप्तांक 84 होता। पूरी टीम ने कुल कितनेअंक प्राप्त किये?(a) 672(b) 665folk15( 58845 के बीचसे कम कि(a) 38.5"(c) 45० ।(इ) इनमें ।14. 17 संख्या |
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Answer» Sum of all observations =average ×no of observations =84×8=672.One of the observations is 85 in stead of 92 which is 7 less than the actual observation calculated in the average. Therefore the required total points is 672-7=665 |
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| 37. |
ig.) The sum of 1st three terms of the GP. is to thesum of 1st 6 terms as 152:665. Find the commonratio. |
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| 38. |
1,2づ13 A man buys Diwali cards at 850 per 100 cards and sells then at 136 a ten. Findhis gain per cent. |
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Answer» Thanks !☺️👍 |
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| 39. |
1029=.............. .. . ...2. Write the Hindu-Arabic Numerals for the following Roman Numerals:(a) MCIV(b) CMXLIII(c) DCCL(d) MD(e) MCCLXXV =Match the following:11 |
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Answer» 1143 (a). 1104(b). 943(c). 750(d). 1500(e). 1275 |
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| 40. |
Th radi of two circlcs are 19 am and 9 cm respactivdlynd the radius of the circle which has circumerence cqualsu m of the circumErcnces of the two circles. |
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| 41. |
1. Write the Hindu-Arabic numerals for:(2) LVI (6) LXXIXc) XCII |
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Answer» (a) 58(b) 79(c) 92 a.58b.79c.92is the most correct answer the Hindu Arabic numericals. 58,72,92 |
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| 42. |
\left[ 4 \frac { 1 } { 2 } \times 2 \frac { 1 } { 5 } \times 2 \frac { 2 } { 3 } \right] - \left[ \frac { 3 } { 5 } \times 2 \frac { 2 } { 3 } \times 3 \frac { 3 } { 4 } \right] |
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Answer» 102/5 will be the answer 9/2*11/5*8/3-(3/5*8/3*15/4)so after solving 102/5 will be the answer |
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| 43. |
If H.C.F. of 26 and 91 is 13, then their LCM1s:(a) 182(c) 14(b) 2366(d) of these |
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Answer» Answer:a)182Given:HCF of 26 and 91= 13Formula: LCM= product of two number/HCFSolution:LCM= 26*91/13=2366/13=182 |
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| 44. |
6. Write the answer in Roman Numerals(2) XXX - XXV V ML-1(c) XLIV - XL CIV (d) - |
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Answer» xxx=30and xxiv=2430-24=6vi |
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| 45. |
\left. \begin{array} { l } { \text { ss } 6.4 \times 10 ^ { 5 } \text { in usua } } \\ { \frac { 16 \times 2 ^ { n + 1 } - 8 \times 2 ^ { n } } { 16 \times 2 ^ { n + 2 } - 4 \times 2 ^ { n + 1 } } } \end{array} \right. |
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Answer» 16×2^n+1-8×2^n/16×2n+2-4×2^n+132^n+1-16^n/32^n+2-8^n+132^n-16^n/32^n-8^n+2+116^n/24^n+316/24+316/27 3/7 3/7 |
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| 46. |
( - 2 ) ^ { 2 } \times ( - 2 ) ^ { 3 } \times ( - 2 ) ^ { 2 } \times ( - 2 ) ^ { 3 } |
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Answer» (-2)² × (-2)³ × (-2)² × (-2)³= (-2)^(10) = 2^(10) |
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| 47. |
\frac{2 \times 2}{3 \times 2}=\frac{4}{6} \text { and } \frac{5 \times 3}{2 \times 3}=\frac{15}{6} |
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Answer» Yes, the given answer is already given there. samjh nahi aa raha answer kesse aya |
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| 48. |
2 ^ { 6 } \times 2 ^ { 2 } \times 2 ^ { 5 } \div 8 ^ { 3 } |
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| 49. |
17. Which number nearest to one lakh is exactly divisible by 2, 3, 4, 5, 6 and 7?t: The number ncarest to I lakh divisible by the given numbers may be a little less than one lakh, or alittle greater than one lakh depending on the difference of the number and one lakh.] |
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Answer» LCM(2, 3, 4, 5, 6, 7) = 420Number required = 100000 - 420 = 100580 |
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| 50. |
nd the amount and the compound interest of乏2500 for 2 years on 12% per annum.Fi |
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