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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. |
9.The dimension of a rectangular box is in the ratio 2: 3: 4 and the differencebetween the cost of covering it with a sheet of paper Rs. 4/ m' and Rs. 4.50 /m' is Rs. 416. Find the dimensions of the box?ay 8x12x16 b) 6*912c) 10x1520 d ) of these |
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| 2. |
x)-3x2-4x+1 |
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| 3. |
Evaluate: lim 3x2 4x +5. |
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| 4. |
Ä°ftwo positive integers A and B canexpressed as A- ab, B- ab whose a.bare distinct prime numbers then find the LCMof A and B? |
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| 5. |
Q15. AD and BC are perpendiculars to a line segment AB and AD-BC show that AO-0B.At |
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Answer» In triangle OAD & OBCAD=BC (GIVEN)<OAD=<OBC (EACH=90*)<AOD=<BOC (VERTICLE OPP ANGLE)THEREFORE,TRIANGLE ODA=OCB (AAS CRITERIA) OA=OB (CPCT) So, CD bisects AB.please like the solution 👍 ✔️👍✔️ in about writing in Kannada |
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| 6. |
2.If two positive integers m and n are expressible in the form m= pq? and n=pwhere p, q are prime numbers, then HCF (m, n) is |
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om8. Due to some error in the weighing machine. Rahul's weight was shown as 72 kg, as against the actual valueof 80 kg. Find the percentage of error.(একটি পর্ণ একটি দড়িপাল্লায় হওয়ার বাহিলের ১০ কেজি ওজন কে ভুল ভাবে 2 কেজি দেখায়। ভুলের শতাংশ কত ?)। |
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| 8. |
>-1. Seethe folowing fractions and put the suitable sign in the box (<3442(iii)2(iv) 13 |
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| 9. |
hat1+sin θ1-sin2-= (sec θ + tan θ) |
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| 10. |
Multiply and verify the result for given values:(i) (2x - 5)(7 + 4x), X = 2(i) (x + y) (7x - y). x = 1, y = 0fiii) (a? + b)(b? + a), a=-1, b= -2 (iv) (p? - ?)(p-q), p = 2, q=0 |
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| 11. |
\begin{array} { l } { \text { If } p ( x ) = x ^ { 3 } + 2 x ^ { 2 } - 5 x - 6 , \text { find } p ( 2 ) , p ( - 1 ) , p ( - 3 ) \text { and } p ( 0 ) . \text { What } } \\ { \text { do you conclude about the zeros of } p ( x ) ? \text { Is } 0 \text { a zero of } p ( x ) ? } \end{array} |
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| 12. |
hat the sum of pAABC, AD 1L BC. Prove that 3AB -4Aed |
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| 13. |
ImtagnatoSin-l &inde) |
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| 14. |
3 If pand q are positive prime numbers then prove that p+vq is irrational |
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| 15. |
| L2 bbblie [bBE [ b L7LB e 3 i L 00T Lothicly b Malie & kbl htflméfi (&)ol R |
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| 16. |
\operatorname { log } _ { 3 } x ^ { 2 } = 2 \operatorname { log } _ { 3 } 4 - 4 \operatorname { log } _ { 3 } 5 |
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| 17. |
om hett chook Then she tutns to hehhishT anda |
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Answer» it will be 5 km |
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| 18. |
prove hat duk inde bendent one |
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| 19. |
Prove hat SinI nt COS |
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Answer» sin²π/6 +cos²π/3 -tan²π/4= (1/2)² +(1/2)² -1²= 1/4+1/4 -1= 2/4-1= (2-4)/4= -2/4 = -1/2 |
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| 20. |
(a + b)2: (ab), (aIf (p + 3q): (3p - 2q) 43 : 2, find pq |
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Answer» Please hit the like button |
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| 21. |
6) Find PQ ifP 2 1 0 |
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| 22. |
Loglo ta Log 3- log a write expression asLOGN |
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Answer» Log10 + 2log 3 - log2=Log10 +log{(3)2} - log2=log10*9/2=log45Therefore value of n = 45 |
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| 23. |
1. Explain why 13233343563715 is a composite number?QR |
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Answer» Composite number is number which has at least 1 factor other than 1 which can divide it. now in the given number unit digit is 5 which means it will definitely divide by 5 and hence 5 is one of it's factor other than 1.so number is definitely composite. |
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| 24. |
Heh tnen prove hat5 |A|3| 1,2CO-04 |
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| 25. |
5 I o POR, D is the midpoin ofIPM is toPois.0 M o R |
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Answer» medianaltitudeis your correct answer |
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| 26. |
1. Cartesian equation of a line POis 1 Write the drection ratios of a line2x-1 4-y z+iparallel to PQ2 Find2 |
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| 27. |
a square _ b square |
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Answer» a^2-b^2=a^2+ab-ab-b^2=a(a+b)-b(a+b)=(a-b)(a+b) |
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| 28. |
a square + b square is equal to what |
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Answer» (a+b)^2=a^2+2ab+b^2so a^2+b^2=(a+b)^2-2ab |
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| 29. |
1. A book is published in three volumes, the pages being numnumbers are continued from the first volume to the second volume to the third. Theof pages in the second volume is 50 more than that in the first volume, and the numtin the third volume is one and a half times that in the second. The sum of the page raon the first pages of the three volumes is 1709. If n is the last page number, what is the agnprime factor of n? 0numbered Srom 1 owards, Thee |
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Answer» thanks |
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| 30. |
kao 04PoiS |
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| 31. |
Total 04 â |
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Answer» In triangle with the help of Pythagoras theoremh^2= 64+36h= √100= 10cmnow area of Triangle = 1/2* base x height = 1/2*6*8= 24 cm^2now hypotanius is the diameter of circle which is 10 so radius will be 10/2= 5cmnow area will be 3.14*25/2= 39.25total area = 39.25+10= 49.25 cm^2 |
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| 32. |
71000 " 04 o_jâ |
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Answer» align them on top of each other like this72002000 then you start from the ones and go down and add 72002000_______9200 |
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| 33. |
if a-b=4 & a+b=6find a square + b square |
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Answer» thank you |
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| 34. |
50100050000555056015462350080207,0 ,0 68 ,1 7 9 6 7 5 2 4 66311 |
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Answer» 11321 is the correct answer |
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| 35. |
\operatorname { cos } 52 ^ { \circ } + \operatorname { cos } 68 ^ { \circ } + \operatorname { cos } 172 ^ { \circ } = 0 |
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Answer» First of all cos 172 = -cos 8 as 172=180-8So now we have= cos 52 + cos 68 - cos 8 Applying Identity,cos A − cos B = −2 sin ½ (A + B) sin ½ (A − B) let's look at our cos 52 - cos8, here let's A = 52, B=8cos 52 - cos 8 = -2sin[60/2]sin[44/2]cos52 -cos8 = - sin 22socos 52 + cos 68 + cos172= cos52 + cos68 - cos8= cos52 - cos8 + cos68= - sin22 + cos68but by the complementary relationshipsin 22 = cos68 , (since 22+68 = 90)so finally= -cos 68 + cos 68= 0 |
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| 36. |
the tis 36agn |
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Answer» h = √3a/2 = √3×12/2 = 6√3 cm = 10.39cm |
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| 37. |
L. Find theperimeter of1on,of each shape if each side of a small square on squareb.C. |
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Answer» Rectangle=2(l+b) Square=4l |
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| 38. |
Roadif the cost per m2 isMohan wants to buy a trapezium shaped field.Its side along the river is parallel to and twicethe side along the road. If the area of this field is4.8.100 mbetween the two parallel sides is 100 m, find theaaicc22length of the side along the river.River |
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| 39. |
N(बल 2४2 बजा 15 एटा 2o 081 °SLE & Likk % kb Sl |
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Answer» Prime factors of 378 = 2× 3× 3× 3× 7Prime factors of 180 = 2× 2× 3× 3× 5Prime factors of 420 = 2× 2× 3× 5× 7So, HCF of 378, 180 and 420 is 2× 3 = 6And LCM of 378, 180 and 420 is 2× 2× 3× 3× 3× 5× 7 = 3780So, let us check whetherLCM (378, 180 and 420)× HCF (378, 180 and 420) is equal to theproduct of thethree numbers3780× 6 = 378× 180× 42022680≠ 28576800Hence LCM*HCF is not equal to the product of the three numbers. |
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| 40. |
ool=l | 1 |
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Answer» (36+9+20)/2465/24is correct answer LCM OF 4,6,8 IS 24NOW WE HAVE TO MAKE THE DENOMINATOR SAME THEREFORE, (36+9+20) /24=65/24 ANSWER HOPE IT HELPS🖕👌 lcm of 4, 8 , 6=(2×2×6)=24, 6/4+3/8+5/6= =36+9+20/24= =65/24 L.C.M of 4,6and8 is 24 6/4 + 3/8 + 5/6LCM = 2436/24 + 9/24 + 20/2465/24 = 2 17/24 |
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| 41. |
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the lengthhinor arc of the chord.ulac arrs of the same length subtend angles 60° and 75° at |
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| 42. |
8. If the arcs of the same length in two circles subtend angles 75째 and 120째 atthe centre, find the ratio of their radii. |
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Answer» The length of an arc (L) is : x/360 . 2πr given L1 = L2 75/360 . 2πr1 = 120/360 . 2πr2 r1/r2 = 8/5 |
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| 43. |
In two circles, the arcs of same lengths subtendangles 65° and 110° at the centre. The ratio oftheir radii are: |
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| 44. |
Example·If the arcs of the same lengths in two circles subtend angles 65°and 110°at the centre, find the ratio of their radi. |
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| 45. |
2. If two equal chords of a circle intersect within the circle, prove that the segmentsone chord are equal to corresponding segments of the other chord. |
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| 46. |
If two equal chords of a circle intersect within the circle, prove that the segments ofone chord are equal to corresponding segments of the other chord.2. |
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| 47. |
Ny कि मर न २४% रगि dsl =2=-1Sle जा.=Jx |
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| 48. |
e3> Con A —+Coree ) & M.)L ले का नी जलवा=% Sy . |
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Answer» (sinA + cosecA)² + ( cosA + secA)²(sin²A+cosec²A+2)+(cos²A+sec²A + 2)sin²A + cos²A + 4 + 1+ tan²A + 1+ cot²A7 + tan²A + cot²A |
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| 49. |
ilehomo10 3neont |
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| 50. |
deTwo chords AB and CD intersect at P insithe circle. Prove that the sum of the anglessubtended by the arcs AC and BD at thecentre O is equal to twice the angle APC |
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Answer» Given: two chords AB and CD intersect each other at P inside the circle. OA, OB, OC and OD are joined. Proof: Arc AC subtends ∠AOCat the centre and ∠ADCat the remaining Part of the circle. angle AOC= 2<ADC-----(1)similarlyangle BOD = 2 <BAD-----(2) adding bothangle AOC + angle BOD = 2(angle ADC+ angle BAD) But in triangle PAD thnx a lot |
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