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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. |
\quad f(x)=\frac{\sqrt{2}-\sqrt{1+\sin x}}{\cos ^{2} x}, \text { for } x \neq \frac{\pi}{2} \text { is continuous at } x=\frac{\pi}{2}, \text { find } f\left(\frac{\pi}{2}\right) \approx |
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| 2. |
\frac{180}{\pi} \quad \text { (B) } \frac{\pi}{270} |
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| 3. |
\begin{array} { l } { \sin \left( \frac { \pi } { 3 } - \sin ^ { - 1 } \left( - \frac { 1 } { 2 } \right) \right) \text { is equal to } } \\ { ( A ) \frac { 1 } { 2 } \quad \text { (B) } \frac { 1 } { 3 } } \end{array} |
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Answer» 1 2 |
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| 4. |
\text { evaluate }\quad \int _ { 0 } ^ { \pi / 2 } \frac { 3 \sin \theta + 4 \cos \theta } { \sin \theta + \cos \theta } |
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| 5. |
(b)CCalculate the median and mode for the following distributionWeight (in kg)35471 525660.lNo. of students131 |
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| 6. |
13. Chalk contains 10% calcium, 3% carbon, 12% oxygen and the remaining sand.Find the anocunt of caron and calcium (in gramy) in 2 kg of chalk. Also find |
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Answer» 10% 1kg means 10*1000/100=100gramscalcium=100grams3% carbon means 3*1000/100=30gramcarbon=30grams12% oxygen means 12*1000/100=120gramsoxygen=120grams |
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| 7. |
classmALeDofeSiaoblifs and express the19ĺpress the TEsult mith psiiveă |
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Answer» [ (x^4)^1/3y * (xy^4)^-1/2]^-4= {(x)^[4/3 - 1/2]*(y)^[1 - 2]}^-4={(x)^[5/6] * (y)^-1}^-4= x^[-10/3]*(y)^4= y^4 / x^10/3 |
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| 8. |
- 1 \text { and } 0 \quad \text { (ii) } - 2 \text { and } - 1 \quad \text { (iii) } \frac { - 4 } { 5 } \text { and } \frac { - 2 } { 3 } |
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Answer» 1)-1and 0-0.1,-0.2,-0.3,-0.4,-0.52)-2and-1-1.1,-1.2,-1.3,-1.4,-1.5 |
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| 9. |
. The vertices of a triangle are A(1, 2), B(5, 7) and C(ii, 13). Find the length of themedian passing through the vertex A. |
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| 10. |
\quad \text { (i) } 9 ^ { \frac { 3 } { 2 } } \quad \text { (ii) } 32 ^ { \frac { 2 } { 5 } } |
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Answer» thanks hey thank you for the ans |
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| 11. |
\quad \text (i) 64 ^ \frac 1 2 \quad \text (ii) 32 ^ \frac 1 5 |
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Answer» 1)(64)^1/2 = (8^2)^1/2=82)(32)^1/5 = (2^5)^1/5=23)(125)^1/3=(5^3)^1/3=5 |
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| 12. |
6 = 2.45(a) 2.54(c) 2.04(b) 421(d) 2.45 |
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Answer» 2.04 is the right answer Given √6=2.45then√2/√3+√3/√2=5/√6taking LCM5/√6= 5/2.45 2.04OptionC |
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| 13. |
हक ज> कमागत प्राकृत संख्याओं के वर्गों का योग 421 है । संख्याओं को ज्ञात करें । [(:85 |
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| 14. |
e o रा e 4 ;f” - p+Xg दरo y 421 gif SE= iy, b |
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| 15. |
(cos(((4*pi)/9)*(quad*(text*(i*(i*i)))))*tan((5*pi)/9))*sin(((7*pi)/18)*(quad*(text*(i*i)))) |
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Answer» here ans is sin 7π/18 (i) : +next both negative |
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| 16. |
\begin{array}{l}{\text { 0) } \frac{x}{2}+\frac{x}{3}+\frac{x}{4}=13} \\ {\text { g) } \frac{m}{2}-\frac{4}{5}=\frac{1}{5}-\frac{m}{5}-\frac{3 m}{10}} \\ {\text { h) } 2\left(\frac{p-5}{3}\right)-\left(\frac{p-2}{4}\right)=\frac{9}{2}} \\ {\text { i) }\left(\frac{1}{m-3}\right)=\left(\frac{3}{m-1}\right)-\frac{9}{2}} \\ {\text { j) }(p-2)(p-3)=(p-4)(p-6)}\end{array} |
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| 17. |
A dishonest shopkeeper professes to sell pulses at cost price, but he uses a false weight of 940one kilogram. Find his gain per cent.gm i |
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| 18. |
\frac { p } { 2 } - \frac { 1 } { 4 } = \frac { p } { 3 } + \frac { 1 } { 2 } |
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| 19. |
p = 2 - a , \text { value of } a ^ { 3 } + 6 a p + p ^ { 3 } - 8 |
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| 20. |
P(E | F)=\frac{P(E \cap F)}{P(F)}=\frac{\frac{1}{6}}{\frac{3}{4}}= |
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Answer» P(E|F) = P( E INTERSECTION F)/P(F) = (1/6)/(3/4) = 4/(3×6) = 2/(3×3) = 2/9 thnq |
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| 21. |
( 6 ^ { 2 } \times 3 ^ { - 2 } ) ( 6 ^ { 2 } \div 3 ^ { - 2 } ) |
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| 22. |
☺find & of a kilogram. |
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Answer» 875grams is the answer or 0.875KG is the answer 800grams or 0.800 is the answer |
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| 23. |
3. Change into kilograms and grams(a) 7900 g(1) 10 |
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Answer» 1kg = 1000g1g. = 1/1000 kg7900g = 7900/1000 = 7.9 kg 1kg =1000kg 1g =1/10007900=7900/1000=7.9kg 1kg=1000g1g=1/10007900g=7900g/1000g Wight=7.9kg 7.9kg bcoz 1kg=1000gram so, 7000g= 7kg nd 900gram is 0.9kg so, 7900g = 7.9kg 1kg =1000g 1g = 1/ 1000kg 7900g=7900/1000 =7.9kg 7kg 900g i also have checked 7 kg. 900 gramanswer. 7.9 kg because in 1kg 1000gmso that 7900 is divided by 1000 and hence result be 7.9 kg 7900/1000=7.900 kg,= 7 kg 900 g 1kg=1000g1g=1/10007900g=7900/1000Wight=7kg.9g 1kg=1000gramsso7900/1000=7.9kg 1kg=1000g1g.=1/1000kg7900g=7900/1000 =7.9kg 7900gm=7 kilogram and 900 gram right answer is 7 kilo 900 grams 1 kg -10007900/1000 -7979/10=7.9 1kg=1000gr=7x1000=7000=7kilograms 900gr. 7000+900=7900 7•9 kg is correct answer as if we divide 7900 by 1000 then we will get 7•9 as answer 😊 7 kelo 900 gram7900=7.9kg 7.9 kg ans 1kg=1000g7900g=7900/1000 =7.9kg 7 kilo 900 graamin likha jaiga 7900 To convert gram to kilogram we have to divide the number by 1000if we divide 7900/1000 answer will be 7kg 900g 7•9 kg Shi Answer he 7.9kg is the correct answer 7 kilo 900 grams hai 7 kilogram and 900 gram the answer is 7 kg 900g. 1 g=1/1000Kg7900g=7900/1000Kg7.9Kg 1000 g = 1kgSo, 7900g = 7900/1000 kg = 7.9 kg 7.9kg it's correct and questions 7.9kg is best and correct answer please like 👍👍👍👍🥺🥺 1kg=1000g1g=1/1000kg7900g=7900/1000 7.9,kg 7900g=7.9kg is the snswer 1kg=1000g7900g=7.9 kg 1kg=1000kg1g=1/1000kg7900g=7900/1000 =7.9kg 7.9 kg 7.9kg is the correct answers NB as p 7900/1000=7.9Kg That is to change gram to kilogram we have to divide 7900by 1000 The answer is 7 kg 900 g or 7.9 kg The answer is =7900 divided 1000 =7.9kg 7 kg 900 gram this is a answer 7.9 kilogram and 7900 gram 1kg=1000gm1g=1/1000kg7900gm=7900/1000kg or,7900gm=7.9kg 1000g = 1kg7900g = 7900 / 10007900g = 7 kg and 900 g |
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| 24. |
dishonest shopkeeper professes to sell pulses at cost price, but he uses a false weight of 940one kilogram. Find his gain per cent |
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Answer» 186 47 |
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| 25. |
A dishonest shopkeeper professes to sell pulses at cost price, but he uses a false weightone kilogram. Find his gain per cent.of 940 gm |
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| 26. |
\begin { equation } \begin{array}{l}{\tan \theta=\frac{a}{b} \text { then } \sin \theta=} \\ {\text { A) } \frac{a}{\sqrt{a^{2}+b^{2}}} \quad \text { (B) } \frac{b}{\sqrt{a^{2}+b^{2}}} \quad \text { (C) } \frac{b}{a} \quad \text { (D) } \sqrt{a^{2}+b^{2}}}\end{array} \end { equation } |
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Answer» The option (A) is the answer |
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| 27. |
11.11 \% \quad \text b ) 11 \% \quad \text c) 11.19 \% \text d ) 11.25 \% |
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Answer» He purchased 10 orange for 251 orange= 2.5total amount = 22.5 for 9now selling price = 25for 9gain percentage= SP- CP/CP= 25-22.5 gain percentage= 25-22.5/22.5= 0.11 or 11.11 percentage |
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| 28. |
Time Allowed: 3 hoursGeneral Instructions: Same as Model Question Paper1ет-SECTION-A26460, HCF of (a, b) 27 and b 540, find a.1. If the LCM of (a, b) |
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Answer» LCM×HCM=product of 2 number26460×27=a×540a=1323 |
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| 29. |
2480 के परिवृत्त पर बिन्दु A से PAQ स्पर्श रेखा इस प्रकार से है कि PAQ || BC, तो सिद्ध कीजिए कि ∆ABCसमद्रिबाहु त्रिभुज होगा। |
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| 30. |
13 पुरूष एवं 4 लड़के मिलकर 7 दिनों में 756 रू कमाते हैं। 11पुरूष एवं 13 लड़के मिलकर 8 दिनों में 3008 रू. कमाते हैं।कितने समय में 7 पुरूष एवं 9 लड़के मिलकर 2480 रू कमापायेंगे?(a) 8 दिन(b) 10 दिन) 15 दिन(d) 20 दिन |
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Answer» 3M + 4B = 756/7 ----- (i) and,11M + 13B = 3008/8 ------ (ii)solve the above equation (i) and (ii)we get M=20 and B=12 put these value in equationLet X be the number of day to earn Rs.2480 by 7 men and 9 boys7M + 9B = (2480/X) put the value of M and B,we get X=10 Days they man and children are taking 10 day |
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| 31. |
\begin{array} { l } { \text { The direction cosines of } \overline { A B } \text { are } } \\ { - 2 , \quad 2 , \quad 1 . \text { If } \quad A = ( 4,1,5 ) \text { and } } \\ { I ( A B ) = 6 \text { units, find the } } \\ { \text { coordinates of B. } } \end{array}. |
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| 32. |
1 kilogram=grams |
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Answer» 1 kg = 1000= 10^3 gm |
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| 33. |
) 1 kilogram=grams |
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Answer» 1 kg = 10^3= 1000grams 1 kilogram=1000grams 1 ku = 10^3= 1000grams 1kg=10^3grams1kg=1000grams 1kilogram =1000 gram 1 kilogram =1000grams |
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| 34. |
The value of N6+ N6+N6+... too |
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Answer» Let us assume : x= root(6+root(6 + root(6 + ……… So, x= root(6+x) x^2=6+x x^2-x-6=0 (x-3)(x+2)=0 So, x=3 or -2. But taking -2 as the answer would mean we are taking root of (-2) . Considering the fact that x belongs to the set of real numbers , we discard the solution x=-2. So the answer is x =3. |
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| 35. |
Express the following in Kilograms190 grams (ii) 3240 grams |
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Answer» a) 190 × 100 = 19000 b) 3240 × 100 = 324000 |
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| 36. |
| उदाहरण 9:2x4 - 3x3 - 3x + 6x-2 के सभी शून्यक ज्ञात कीजिए, यदि आपको इसके_दो शून्यक 2 और .-.-2 ज्ञात हैं। |
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| 37. |
-Ce)Express into kilometres-8888 metres.㤠|
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| 38. |
प्ररणापला 5.21. निम्नलिखित सारणी में, रिक्त स्थानों को भरिए, जहाँ AP का प्रथम पद a, सार्व अंतर d औरवाँ पद है:CIC0088881892.523635 |
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Answer» i) =7+(8-1)×3 =7+21 =28 ans i)=7+(8-1)×3=7+21=28 ans i)=7+(8-1)×3=7+21=28 answer I) 7+(8-1)×3 =7+21 =28 is correct answer. |
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| 39. |
Convert the following into hours, minutes and seconds:(a) 7935 seconds (b) 5138 seconds (C) 8888 seconds (d) 2480 sen |
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Answer» Ok wait for my next solution (a) 2h 12min 15sec(b)1h 35min 38sec 2 hour 12 minute 15 sec a-2 hours 12 minutes 15 second sb-1 hour 25 minutes 36 second sc-2 hours 14 minutes 40 second sd-41 minutes 20 seconds is correct answer of this question (a) 2hour 12minute 15sec a- 2 hours 12 minute 15 secb- 1 hour 25 minute 36 sexc- 2 hour 14 minute 40 sexd-41 min 20 sec is correct answer (a)2h 12min 15sec(b)1h 35min 38sec a) 2 hours 12 minute 15 secondb) 1 hour 25 minute 36 secondc) 2 hours 14 minute 40 secondd)41 minute 20 second 2 hours 12 minutes 15 second (a)2 hour 12 minutes 15 second(b)1 hour 25 minutes 38 second(c)2 hour 14 minutes 48 seconds(d)41 minutes 20 second (a)2 hours 12 minutes 15 second (a) 2 h. 12 m 15 s. (b)1 h 15 m 6 s a) 2.2 hrb) 1.42 hrc) 2.46888... hrd) 0.68888... hr 2hours12 minutes 15second 2 h 12min 15 sec is the answer a) 2 hour 12 minute 15 secondb) 1 hour 25 minute 36 secondc) 2 hour 14 minute 40 secondd) 41 minute 20 second 2h 12min 15second, 1h 35min 36 second 2hrs 12min 15sec is answer of first 2hr 12 min 15 sec is correct answer of first a-2 hours 12 minutes 15 secondb-1 hours 25 minutes 36 secondc-2 hours 14 minutes 40 secondd-41 minutes 20 second a)..2 hours 12 minute 15 secondb,). 1 hour 50 minute 36 secondc,). 2 hours 14 minute 40 secondd).. 41 minut 20 second is the right answer of your given question..... a) 2 hours 12 minutes 15 seconds b) 1 hour 35 minute 38 secondc) 2 hours 28 minutes 8 seconds 2 hour 12 minute 15 second is the best answer |
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| 40. |
1. Fill in the blanks.The standard unit of mass isThe smaller unit of mass is,.........grams make 1 kilogram |
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Answer» a) kilogramb)milligramc)1000 |
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| 41. |
We can also makedo it?the design on page 88 like this. How did you |
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Answer» how do this |
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| 42. |
A recipe to make lasagna for 5 people uses 300 grams of minced beemuch minced beef would be needed to serve 9 people? (3 pts)t. How2. |
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Answer» For 5 people 300 grams is needed So for 1 person 300/5 = 60 grams is neededtherefore 9 people 60×9 =540 grams is needed |
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| 43. |
1. Zinc and copper are in the ratioof 5 3 in 200 gm of an alloyHow much grams of copper beadded to make the ratio as 3: 5? |
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| 44. |
23. Without actual division, prove that 2x4 – 5x° + 2x2 – x + 2 is divisible byx - 3x+2. |
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Answer» x^2-3x+2=>x^2-2x-1x+2=>x(x-2)-1(x-2)=>(x-2)(x-1)Therefore,(x-2)(x-1)are the factors.i) (x-2)x-2=0=>x=2So,p(x)=2p(x)=2x^4-5x^3+2x^2-x+2p(2)=2(2)^4-5(2)^3+2(2)^2-2+2=32-40+8= -40+40=0Hence,it proves that (x-2) is a factor .ii) (x-1)=>x-1=0=>x=1So,p(x)=1p(x)=2x^4-5x^3+2x^2-x+2p(1)=2(1)^4-5(1)^3+2(1)^2-1+2=2-5+2-1+2=6-6=0Hence,it proves that (x-1) is a factor. |
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| 45. |
f(x) = 2x4-6x3 + 2x2-x+2,8(x) = x + 2 |
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| 46. |
Find all the zeroes of p(x) = 2x4-3x3-9x2+15x-5, if two of its zeroes are ±VS. |
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| 47. |
(ii) (2x4-3x2-15)-(x + 2). |
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| 48. |
23. Without actual division, prove that 2x4 – 5x + 2x2 - x + 2 is divisible byx? - 3x + 2.-4-3,2 |
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Answer» x^2-3x+2=>x^2-2x-1x+2=>x(x-2)-1(x-2)=>(x-2)(x-1)Therefore,(x-2)(x-1)are the factors.i) (x-2)x-2=0=>x=2So,p(x)=2p(x)=2x^4-5x^3+2x^2-x+2p(2)=2(2)^4-5(2)^3+2(2)^2-2+2=32-40+8= -40+40=0Hence,it proves that (x-2) is a factor .ii) (x-1)=>x-1=0=>x=1So,p(x)=1p(x)=2x^4-5x^3+2x^2-x+2p(1)=2(1)^4-5(1)^3+2(1)^2-1+2=2-5+2-1+2=6-6=0Hence,it proves that (x-1) is a factor. theoremthanks |
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| 49. |
cosec θ cot θ28. If 5 cos θ = 3, evaluate : cosec θ + cot θ |
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| 50. |
27.DividepolynomialP(x)-2x4 3x3 - 2x2-9x-2 byq(x)x-3 & find what should abstract so that it isdivisible by q(x) |
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