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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2151. |
\frac{7}{24}-\frac{17}{36} |
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Answer» 7/24 - 17/36 = 7/(12×2) - 17/(12×3) = (7×3-17×2)/(12×2×3) = (21-34)/(12×2×3) = -13/72 |
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| 2152. |
Find \frac{7}{24}-\frac{17}{36} |
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Answer» L C M of 24 and 36 is 727*3/24*3 - 17*2/36*221/72 - 34 /72-11/72 7/24 - 17/36 =(7×3-17×2)/72 =(21-34)/72 =-13/72 |
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| 2153. |
व S p दी ८5 ही - ZZ नि, o A iy ae( SmbrcpS 2 अल s s |
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Answer» Given sinθcosθ = 1/2 (sinθ+ cosθ)² = 1 + 2cosθsinθ= 1 + 2× 1/2 = 1 |
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| 2154. |
(ij)Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1mark for each wrong answer. Had 4 marks been awarded for each correct answerand 2 marks been deducted for each incorrect answer, then Yash would havescored 50 marks. How many questions were there in the test?iv) Places A and B are 100 km apart on a highway. One car starts from A and anotherfrom B at the same time. If the cars travel in the same direction at different speeds,they meet in 5 hours. If they travel towards each other, they meet in 1 hour.Whatare the speeds of the two cars? |
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Answer» 3)Let number of correct answers be x and number of incorrect answers be y. As per given conditions, 3x - y = 40 … (1) 4x - 2y = 50 … (2) (1) * (2) 6x - 2y = 80 … (3) Subtracting (2) from (3), we get, 2x = 30 x = 15 y = 3x - 40 = 45 - 40 = 5 Total number of questions in the test = 15 + 5 = 20 hit like if you find it useful 4) |
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| 2155. |
11. Ravi scored 30% marks and failed by 15 marks. Deepak scored 40% marks and obtained 35 marks more tithose required to pass. Find the minimum per cent pass marks. |
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| 2156. |
In a class test in English 10 students scored75 marks, 12 students scored 60 marks, 8scored 40 marks and 3 scored 30 marks, themode for their score is:(a) 30 (b) 25 (c) 60 (d) 75 |
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| 2157. |
(a) Reema scored 40 marks out of 50 in her Mathematics test. Find thepercentage of marks scored by her |
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Answer» = (40/50)* 100 = 80% |
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| 2158. |
Karim scored 15 marks more than Rahim in an examination. Madan scored 10 marksless than Rahim. Totally, they scored 110 marks. What was Karim's score? (2014)(1) 25 (2) 35 (3) 40 (4) 50 |
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Answer» Let the marks of Rahim be xKarim marks= 15+xMadan Marks= x-1015+x+x+x-10= 1105+3x=1103x=110-5x=105/3=35Karim=15+35=50 |
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| 2159. |
12:18 10 4solved 6 out of 10 problems correctly. What is the ratio of the correct problems tothe incorrect problems? |
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Answer» 6 are correctly solved so 4 are incorrectly solvedso ration=6/4=3/2=3:2 |
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| 2160. |
t the sum of an AP whose first term is a, the second termand(at c)(i) + ( -2a)2(b -a)[NCERT Exemplar |
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Answer» like the answer if it helps you |
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| 2161. |
24. Find the common differene of an AP whosefirst term is 4, last term is 49 and the sum[CBSE 2010]of all its terms is 265. |
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| 2162. |
Find 5th term ofan Đ.Đ. whose nth term is 3n-5. |
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| 2163. |
ch2) Find the first four terms of an AP whose first term is-2and common difference is -2fiura in mhat rati dees P divide AB internally |
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Answer» Given sequence is APFirst term = -2 CD = - 2Second term = - 2 + (-2) = -4Third term = -4 + (-2) = -6Fourth term = -6 + (-2) = - 8 |
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| 2164. |
)Findthefirstterm and 16th term of an AP whose 8th andth3erms are 18 and 28 respectively |
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| 2165. |
20105-ZMATHEMATICSe3 HoursSMaximum MarLong Answer Type Questions)Using properties of determinants prove thatb+ cb c+a |
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Answer» 1 2 |
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| 2166. |
Find the A.P. whose nth term is 7 - 3n. Also find the 20th term. |
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Answer» Since you gave an explicit formula for the AP, we can just plug in values for n, as shown below: 0 -> 7 1 -> 4 2 -> 1 3 -> -2 … So the AP is; 7, 4, 1, -2, -5, -8, … With a common difference of -3. Before I answer your second part, I have to talk about something first. You might have noticed that the first term of the sequence is n = 0, not 1. I personally like this better for reasons beyond the scope of this answer (but I will say that it’s because of arithmetic series), but it is reasonable to have 0 or 1 as your first term. If you choose to have your first term as n = 1, then the sequence would be the same, but without the 7. 4, 1, -2, -5, -8, -11… So the answer is different depending on how you define your sequence. If the first term is n = 0, then to find the 20th term, set n = 20 - 1 = 19 7 -3(19) = -50 If the first term is n = 1, then to find the 20th term, set n = 20 7 -3(20) = -53 nth term = 7-3nn=1 = 7-3[1]=4,n=2 = 7 -3[2] =1,n=3 =7-3[3]=7-9=-2 ... and so onso the common difference is 320th term = 7-3[20]=-53 |
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| 2167. |
6)Write the 7th term of the sequence, whose nth term is an23=n,2n) |
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Answer» Put n=7 in the nth terma7=7^2/2^7=49/2^7 |
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| 2168. |
Example 3. Find the 17th and 24th term of the sequence whose nth term is4n-3. |
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Answer» nth term= 4n - 3so, 17th term= 4*17 - 3 = 65and 24th term = 4*24 - 3 = 93 thankuuu |
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| 2169. |
1. Find the sum of n term of that series whose nth term is :li) 3n2 +2n + 5(ii) 4n3 +7n + 1(iii) nin + 1 (n + 2) |
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Answer» A similar example. |
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| 2170. |
Show that sequence (Tn) whose nth term Tn=2n +7, is an A.P. |
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| 2171. |
Write the first five terms of the sequences whose nth term is a,=12+1 |
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Answer» 1/2,2/3,3/4,4/5and5/6 |
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| 2172. |
ए o S X% _:‘\\_ डर = i R& Y,m\ae/%\' (M\o*%a’: FdbeD |
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Answer» Like if you find it useful |
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| 2173. |
Write the first five terms of the sequences whose nth term is a, = n(n+2) |
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Answer» 3,8,15,24 and 35 is the correct answer |
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| 2174. |
LUmmon difference of an A.P. whose n term is 3n+7.. If the areas of t |
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| 2175. |
,In Fig 10.37POR-10, where PO and R ae Ppints on a circle with centre O. Fiod OPR |
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Answer» 1 2 |
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| 2176. |
Fig. 3.58 RIn figure 3.59, point D and E are onside BC of A ABCsuch that BD CE and AD AE.Show that Δ ABDΔ ACE.Fig. 3.59 |
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| 2177. |
BC. IfIn the figure, DEAE =x + 2 and EC =x-1, find the valueof r.(a)AD =x, DB = x-2. |
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Answer» According to BPTx/x-2= x+2/x-1x(x-1)= x^2-4x^2-x= x^2-4x= 4 |
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| 2178. |
nauus RoP Class 922 D is the midpoint of side BC of ÎABC and E isthe midpoint of BD. If O is the midpoint of AE,prove thatar(ABOE)=8ar(AABC)no23. In a trapezium ARCD R |
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| 2179. |
nown below1. In the figure, o is the centre of the circle. Name the parts shown he(a) Radius of the circle(b) Diameter of the circle(c) Chords of the circle (except diameter)and2. In the above problem, if radius is 4 cm, find thediameter PQ.3. Write the missing data.aрRadius 16 cmRadiusRadiusRadiusDiameterDiameter 26 cmDiameter 10 cmDiameter4. Use your compass to draw the circles of the given radii.(a) 4 cm (b) 6 cm (C) 5.5 cm (d) 7.2 cm (e) 3. |
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Answer» what is your question (a) radius of circle is OP(b) diameter of the circle PQ(c)chord of circle AB and CD a) diameter=32cmb)radius=13cmc)radius=5cm |
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| 2180. |
If a right circular cone has radius 4 cm and slant height 5 cm then what is itsvolume?(a) 16 t cm (b) 14 a cm (c) 12 cm' (d) 18 t cm |
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Answer» h^2=l^2 -r^2 = (5)^2-(4)^2 = 25-16 =9h=√9=3volume of cone= 1/3πr^2h = 1/3π(4)^2×3 = 16π cm^3 |
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| 2181. |
3. If a right circular cone has radius 4 cm and slant height 5 cm then what is itsvolume?(a) 16 t cm (b) 14 r сm (c) 12 7 cm (d) 18 t cm |
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Answer» b option is correct answer (a) 16πc³ is the correct 16πcm3 is the correct answer option a is correct answer 16π cm cube option a is correct answer |
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| 2182. |
The perimeter of a sector of a circle of radius 4 cmand formed by an arc of measure 11 cm is:(i) 15 cnm(ii) 30 cm(ii) 19 cm(iv) 44 cm |
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| 2183. |
QUESTIONS7. CSA of a hemisphere having radius 4 cm will be :(A) 32 1 cm (B) 16 cm(C) 64 a cm (D) of these.[H.B. 2018]Ans. (A) 32 t cm. |
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| 2184. |
)126(a) 153the given figure, PA and PB are two tangentsfrom an external point P to a circle withtre C and radius 4 cm. If PA L PB, then theCBSE 201awngth of each tangent is3 cm) 5 cm(b) 4 cm(d) 6 cm |
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Answer» thankful |
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| 2185. |
Construct a tangent to a circle of radius 4 cm from a pointradius 6 cm and measure its length. Also verify the measura1n |
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| 2186. |
Write the first term and common difference of an A.P. 4, 10, 16, 22, ..... |
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Answer» first term of the given AP is 4 and the common difference is (10-4)=6 |
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| 2187. |
Solve: \frac{2}{3}(4 x-1)-\left(4 x-\frac{1-3 x}{2}\right)=\frac{x-7}{2} |
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| 2188. |
Write the common difference of an A.P. whose n'h term is a -3n+7 |
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| 2189. |
1. Write the common difference of the AP0.3,-0.3,-0.9, |
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Answer» Common difference =a2-a1d=-0.3-0.3=-0.6 |
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| 2190. |
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the areaof the remaining sheet. (Take Tt 3.14) |
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| 2191. |
16. In an equilateral triangle, prove that three times the square of one side is equal to fourtimes the square of one of its altitudes. |
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| 2192. |
Question 16. In an equilateral triangle, prove that three times thesquare of one side is equal to four times the square of one of itsaltitudes.(Say) |
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| 2193. |
the sum of(1) Prove that three times the sum of the squares of the side of a triangle is equal to four timessquares of the medians of that triangle. |
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| 2194. |
वित उत्तरीय प्रश्न31. १कजिए कि :-41 30° + cos 60°)-3{cos 45 |
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Answer» Perhaps this question is incomplete. |
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| 2195. |
Q. 6 Solve the following questions: (Any one)(1) Prove that three times the sum of the squares of the side of atriangle is equal to four times the sum of squares of the medians ofthat triangle.(2) In AABC, seg MN I side BC. A-M-B and A-N-C. Seg MNBMABdivides A ABC into two parts equal in area. Determine |
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Answer» theorem states that the sum of the squares of two sides of a triangle is equal to twice the square of the median on the third side plus half the square of the third side.Hence AB2+ AC2= 2BD2+2AD2=2×(½BC)2+2AD2 = ½ BC2+ 2AD2∴ 2AB2+ 2AC2= BC2+ 4AD2 → (1)Similarly, we get2AB2+ 2BC2= AC2+ 4BE2→ (2)2BC2+ 2AC2= AB2+ 4CF2→ (3)Adding (1) (2) and (3), we get4AB2+ 4BC2+ 4AC2= AB2+ BC2+ AC2+ 4AD2+ 4BE2+ 4CF23(AB2+ BC2+ AC2) = 4(AD2+ BE2+ CF2) Hence, three times the sum of squares of the sides of a triangle is equal to four times the sum of squares of the medians of the triangle. |
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| 2196. |
selpuilaeral triangle ABC, D is a point on side BC such that BDBCIS In anejuilateral. Prove thatta prove that three times the square of one side is eual to four |
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Answer» ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC . Now, ∆ABE and ∆AEC ∠AEB = ∠ACE = 90° AE is common side of both triangles , AB = AC [ all sides of equilateral triangle are equal ]From R - H - S congruence rule , ∆ABE ≡ ∆ACE ∴ BE = EC = BC/2 Now, from Pythagoras theorem , ∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2) From equation (1) and (2) AB² - AD² = BE² - DE² = (BC/2)² - (BE - BD)² = BC²/4 - {(BC/2) - (BC/3)}² = BC²/4 - (BC/6)² = BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9 ∵AB = BC = CA So, AB² = AD² + 2AB²/9 9AB² - 2AB² = 9AD²Hence, 9AD² = 7AB² |
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| 2197. |
41. Which is greater out of 17" and 31". |
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Answer» Divide 31^11 and 17^14 by 17^11 NOW31^11/17^11and17^14/17^11 (1.82)^11and 17^3 { Consider 1.82 approx. = 2 } 2^11 = 1024 × 2 = 2048 17^3 = 4913 2048 < 4913 |
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| 2198. |
. Divide V27 by V625 |
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Answer» (27)^1/3= 3(625)^1/4= 5By dividing= 3/5Will be AnswerPlease like the solution 👍 ✔️ |
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| 2199. |
Find the 10th term of the sequence v3, v12,V27, |
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| 2200. |
The value of v27-75-15 +44% +431, willwillbe |
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