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.
| 73851. |
s. The pair of equations x 5 and y 3, graphically represents lines which ar(a) parallel(c) coincident(b) intersecting at (3, 5)(d) intersecting at (5, 3) |
| Answer» | |
| 73852. |
DononunatorENERCISE 7.2w number lines and locate the points on themAR1 2 3 7e3335rss the following as mited fractions: |
| Answer» | |
| 73853. |
वि किक को Y दि ८“3£ (0,500+8 NE&N? =§502QUISZ + 6,507 + 6८५15 + 050QUISZ - 9,502 + 6८५1५£(8UIsD +9500D) + (95030 — QuUISD)M =U2 ko bip < (०5-28 0-0 (थ) 122 B f 8 262 (0) 20१६ (प) ०2 (थेe है हे ४४ 88 < (guiso- '9500)e (65020 ‘guysp) Eg न 1099191) Be विके.कक एकहै वि S उनकेहै R Bl (o4 me”mo की 180 न £ 2 miex «oyVfiflrfift )8 * (4 Xy Supy bip «D) B e, ¥4 पक (1 )y |
| Answer» | |
| 73854. |
MALUE BASED OUESTİONNeha went to the market with her mother. There was a lot of traffic on the road. Nehasaw a few traffic signs on the road side.STOP(iv)(vi)(a) Which of these signs have lines of symmetry? Wite the number of lines of symmetry foreach(b)Find out the meaning of each of these signs. Why is it important to obey traffic rules? |
|
Answer» a) ii - one line of symmetry iii) - three line of symmetry iv) - four line of symmetry v) - one line of symmetry vi) - four line of symmetry. b) i) no horn ii) stop ahead iii) give way iv) stop v) pedestrian crossing vi) crossroad. we should obey safety rules in order to avoid accidents and drive safely, and also not to get stuck in jam. |
|
| 73855. |
saw a few traftic signs on the road side.STOP(iv)(vi)(a) Which of these signs have lines of symmetry? Write the number of lines of symmetry foreach(b) Find out the meaning of each of these signs. Why is it important to obey traffic rules? |
|
Answer» a) ii - one line of symmetry iii) - three line of symmetry iv) - four line of symmetry v) - one line of symmetry vi) - four line of symmetry. b) i) no horn ii) stop ahead iii) give way iv) stop v) pedestrian crossing vi) crossroad. we should obey safety rules in order to avoid accidents and drive safely, and also not to get stuck in jam. |
|
| 73856. |
Find the condition for the lines ar+hy+g=0, hx + by + f = 0, gx + fy + c = 0 to be concurrent. |
|
Answer» There is a very popular formula for solving such kind of questions. See below picture. |
|
| 73857. |
o TR (AR P AL [N AL- TR e we :L X +\/_‘x - \/—x_zlV2 2 »\/_—x—lz X+\[—x o A FS? |
|
Answer» Thank you so much |
|
| 73858. |
हु =) AT 2 \wfl o TroTe Lz 4 लिS & e\l o «रा “पड 3_l_, \,fl'):»/ % TS L/ 9 f"/ i O./f"‘v/f’ : |
| Answer» | |
| 73859. |
겨lz |
| Answer» | |
| 73860. |
If a, b, c are in AP, then ax + by + crepresents(a) a straight line(b) a pair of straight lines(c) a set of three concurrent lines(d)/a family of concurrent lines0 |
|
Answer» Hence Option D is the right answer (Not B) |
|
| 73861. |
14. Find the equation of the straight lines passing through (1,2) and making an angle of 60° withV3x+y+2 0.4PAIR OF STRAIGHT LINES |
| Answer» | |
| 73862. |
5. If a straight line falling on twostraight lines makes the interiorangles on the same side of it suchLthat their sum is less than tworight angles, then the two straightlines, if produced indefinitely.meet on that side on which thesum of angles is less than tworight angles. In the given fig.,the line ST falls on the lines LM and PQ such that the sum of theinterior angles 1 and 2 is less than 180° (on the left side of ST)Therefore, the lines LMand PQ will intersect on the left side of STNS |
| Answer» | |
| 73863. |
25. In the given figure, AM L BC and AN is the bisector of ZA. Find themeasure of ZMAN.65°30° |
|
Answer» thanks |
|
| 73864. |
24. The interior angles of a polygulis 80. Find the number of sides of the polygon25.In triangle ABC, D is the mid-point of BC and AE t-BC. If AC > AB, then show that:AB2 = AD2BC2 |
|
Answer» PROOF: [for easy convenience, let me denote all squares by * and multiplication by 'x'] 1) Since AE is perpendicular to BC, ABE is a right triangle with angle E = 90 deg. 2) Hence applying Pythagoras theorem we have, AB* = AE* + BE* 3) In similar way from the right triangle, ADE, AD* = AE* + DE*, => AE* = AD* - DE* 4) Substituting for AE* from step 3 in step 2, we have, AB* = AD* - DE* + BE* 5) From the figure (Kindly make according to the description given in data), BE = BD - DE 6) Substituting this in (4), AB* = AD* - DE* + (BD-DE)* 7) Expanding, AB* = AD* - DE* + BD* + DE* - 2 x BD x DE 8) Cancelling -DE* and +DE* and substituting BD = (1/2)BC {Since, D is mid point of BC} we get, AB* = AD* + (1/4)BC* - BC x DE Thus it is proved that "AB^2=AD^2-BC*DE+(1/4)BC^2" |
|
| 73865. |
%l tanB= 4 €14, A\ :e : 5 3 sin0 - cosO4 Heu Akl |
|
Answer» tanα=4 (5sinα+cosα)/(3sinα-cosα)=(5tan(α)+1)/(3tan(α)-1)=(5*4+1)/(3*4-1)=(21/11) |
|
| 73866. |
हिना (330 - $in bo Y1~ (60 -~ SCN3O |
| Answer» | |
| 73867. |
LEVELEGUESTIONS(MCs)alternative in each of the following:1 to three times its supplement. The measure of the a(b) 135(c) 90AB and CD intersect one another at the point O. 1(d) 120f 19ht linesstraight lines2740, then LAOD =BveBOD(b) 90(c) 94(d) 137 |
|
Answer» 90 Degree is the answer. x+3x = 180 4x = 180 X = 180/4 X = 45 degrees for the smaller angle. 180 – 45 = 135 degrees for the larger angle. Proof: 45+3*45 = 180 45+135 = 180 180/2 = 90 |
|
| 73868. |
ABC is a triangle, right-angled at C. If AB = 25 cm and AC= 7 cm, find BC. |
|
Answer» By phythagorean triplet. AB^2=AC^2+BC^2 (25*25)^2=7*7+BC^2 625=49+BC^2 BC^2=625-49 BC^2=576 BC=root of 576 BC=24. |
|
| 73869. |
The length of a rectangular hall is 5 metres more than its breadth. If the perimeter of thehall is 74 metres, find its length and breadth. |
| Answer» | |
| 73870. |
HOLIDAYSActivity BookWorksheetClass v SectionA. How many line segments are there in each of the following figuresEcoleme TriangleHolidays Activity Book-VB. Construct the line segments with lengths1.5 cm2.9 cm3.7 cm4.3 cm5.4 cm6.8 cmTeacher's SignGraneGoodBestExcellent |
| Answer» | |
| 73871. |
15The sum of the two opposite angles of a parallelogram is 1500. Find themeasure of any one of the other two angles. |
|
Answer» sum of the one opposite angle is eqal to other so each angles are 75° each bro I don't know this answer how to solve me |
|
| 73872. |
quadrlateral2- Two opposite angles of a paralleloOgram are (3x-2)' and (63-2x). Find all the angles of a parallelogram. |
|
Answer» opposite angles are equal in parallelogram3x-2=63-2x5x=65X=13henceangles will be3x-2=3*13-2=39-2=3763-2x=63-2*13=63-26=37now sum of angles of a parallelogram is 360hence 37+37+2y=360y=143 |
|
| 73873. |
byenlouch itElectrical appliances are connected in parallel. Whatadvantages of this arrangement?are(1 mark)the |
|
Answer» (i) Will have equal potential difference across each appliance. (ii) Each appliance has separate switch to the flow of current through it, so that each circuit can work independently. |
|
| 73874. |
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced ifnecessary) at E. Prove that AE = AD |
| Answer» | |
| 73875. |
12. How long will a train 330 m long travelling at 40 km/h take to cross a platform 110 m long?How long will it take to pass a man sitting on the platform? |
|
Answer» Given,Length of train = 330 mSpeed of train = 40 km/hrLength of platform = 110 m Distance = Speed*Time Time taken to pass a man sitting on platform= Distance / Speed= (330+110)/40= 440/40 = 11 hours |
|
| 73876. |
in Δ ABC, right-angled at B, AB-24 crn, BC-71 em. Determine:sin A, cos Asin C, cos C |
| Answer» | |
| 73877. |
lo Fig. 643, the line segmentXYÄŤsparallel to side AC of Î ABC and it0rites the triangle into two parts of equalAXFad the raio ABFig. 643 |
|
Answer» Like my answer if you find it useful! |
|
| 73878. |
13. 15 men can build a 16.25-m-long wall up to a certain height in one day. How many menshould be employed to build a wall of the same height but of length 26 metres in one day? |
|
Answer» 26x15/16.25=24 men should be employed |
|
| 73879. |
2. ABC is a triangle, right-angled at C. If AB 25em and AC 7 cm, find BC. |
| Answer» | |
| 73880. |
The length of a line segment AB is 38 cm. A point X on it divides it in the ratio 9:10. Findthe lengths of the line segments AX and XB..B |
|
Answer» AB = 38 AX:XB = 9:10AX + XB = 389x +10x = 38 19x = 38 x = 38/19 x = 2Then AX = 9x = 9×2 = 18 XB = 10x = 10×2 = 20 20 is the best answer for this question |
|
| 73881. |
1. Find the roots of the following quadratic equations by factorisation: |
| Answer» | |
| 73882. |
Two opposite angles of a parallelogram are (3x-2)" and (50-xy. Find the measure of each angleof the parallelogram. |
| Answer» | |
| 73883. |
The equation of in-circle of an equilateral triangle is2x2 + 2y2+3x-y-50. Find the area of the triangle. |
| Answer» | |
| 73884. |
2. जपि - H या आजाद कि -[11 11137-31-13-172. यदि A = 111.तो सिद्ध कीजिए कि A" = 31-132-13111131-130-130- |
| Answer» | |
| 73885. |
5. In the given figure ABCDE is any pentagon. BP drawnparallel to AC meets DC produced at P and EQ drawrnparallel to AD meets CD produced at Q Prove thatar(ABCDE) ar (AAPQ) |
| Answer» | |
| 73886. |
2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by4 men and 4 boys. How long would it take one man and one boy to do it ?8. |
| Answer» | |
| 73887. |
5. A hostel has enough food to feed 160 students for 90 days. How long would the food last if therewere 20 more students in the hostel?8 months for that he employed 240 men. Due to |
|
Answer» Product of students and days is constant.So, 160*90 = 180*X=> X = 80 days. Please hit the like button if this helped you |
|
| 73888. |
-AAAFind a unit vector in the direction of vector 3 a-2b where a = 2 i +3j+kand b = 31+2)-2 |
| Answer» | |
| 73889. |
Construct a triangle ABC in which BC 7cm, B-75 and AB+AC 13cm.1. |
| Answer» | |
| 73890. |
Q. 1. Construct a triangle ABC in whichBC =7cm, angle B= 75 and AB +AC 13 cm. |
|
Answer» 1 2 |
|
| 73891. |
8. in Fig 934, ABC is a right triangle night angled at A. BCED, ACFG ad Asquares on the sides BC, CA and AB respectively Line segment AX L DEat Y. Show that:D XFig. 9.34(i) MBC# ABD(ii) ar (BYXD) ar (ABMN)(v) ar (CYXE) = 2 ar (FCB)(vi) ar (BCED)ar (ABMN) +ar (ACFG)(ii) ar (BYXD)= 2 ar (MBC)(iv) FCB ACE(vi) ar(CYXE)= ar (ACEC) |
| Answer» | |
| 73892. |
1.find thequadraticnature of the rootsequations. If real rootsof the followingexast, find them.2x - 3x + 5 = |
|
Answer» discriminate D = b^2-4achere a=2, b= -3, c= 5D= (-3)^2-4×2×5 = 9-40 = -31<0so root are immiginary |
|
| 73893. |
x ^ { 2 } + \frac { 1 } { x ^ { 2 } } = 27 , \text { then the positive value of } x - \frac { 1 } { x } |
| Answer» | |
| 73894. |
Find the nature of the roots of the quadratic equation 2x2-6x+3=0 . If real rootsexist, then find the roots. |
| Answer» | |
| 73895. |
EXERCISE 4.41. Find the nature of the roots of the following quadratic equations. If the real roots exisfind them2r-3x 50() 3 -43r+4 0 |
| Answer» | |
| 73896. |
Find the nature of the roots of the quadraticequation\sqrt{3} x^{2}-2 \sqrt{2} x-2 \sqrt{3}=0the real roots exist, find them |
| Answer» | |
| 73897. |
1.Find x in the following figures.125°125° |
|
Answer» Thanks |
|
| 73898. |
12 men can construct a 15 km long road in one day. What length would 8 men construct in one day? |
| Answer» | |
| 73899. |
Q3. In a cyclic quadrilateral ABCD, if 2A = 1050,LB = 70°. Find LCand LD |
| Answer» | |
| 73900. |
3122) 6(i)(125 )x(16 |
|
Answer» {(1/125)² * √(1/16³)}^-1/6 = {(1/(5*5*5*5*5*5))*(1/(4*4*4))}^-1/6 = {(5*5*5*5*5*5) * (2*2*2*2*2*2)}^1/6 = 5*2 = 10 Like my answer if you find it useful! |
|