Saved Bookmarks
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. |
ul Iaub does the point (-4, b) divide the line segment joining the pointsP (2, -2), Q(-14, 6)? Hence, find the value of b.17. The line segment joining A(2, 3) and B(6, -5) is intersected by x-axis at a point KWrite down the ordinate of the point K. Hence, find the ratio in which K divides AB.Also, find the coordinates of the point K.18. If A(-4, 3) and B(8- |
| Answer» | |
| 2. |
13. In figure, S and T trisect the side OR of a right triangle POR, prove thatQ S TR |
|
Answer» related question |
|
| 3. |
Find the value of a , for which point P(a/5,2) is the mid-point of the line segment joining the point B(-5,4) and R(-1,0). |
|
Answer» Coordinate of mid point (x,y)x = (x1 +x2)/2y =(y1 +y2)/2A/qx =(-5 + (-1))/2 = -3⇒a/3 = -3⇒a =-9y = (4+0)/2 = 2coordinate of mid point = (-3,2)so,a= -9 |
|
| 4. |
6. Given 4 ABCA POR, fthen findAB 1PQ'Ä°, 3ar APQR |
|
Answer» Ar(∆ABC)/Ar(∆PQR)=(AB/PQ)^2hencearea ratios will be (1/3)^2=1/9 |
|
| 5. |
15. If (x + a) be a common zero of x2 + px + q andx2 + Ix + m, then the value of a ism+qm-9m+9m-9-) I+ DDI-P2) 1 + P3) 1 + p4) 1-0 |
|
Answer» (-a)² + p·(-a) + q = 0 --> a² -ap + q = 0(-a)² + l·(-a) + m= 0 --> a² -al + m = 0Since they both equal 0, you can set them equal to each other:a² -ap + q = a² -al + m-ap + q + -al + mal - ap = m-qa(l-p) = m-qa=m-q/l-p |
|
| 6. |
7) Sunil has 368 red marbles and 256yellow marbles. How many marblesdoes Sunil have altogether? |
|
Answer» 624 Marbles sunil have in altogether rad marbels Sunil have=368yellow marbels Sunil have=256the total marbels Sunil have= 624 |
|
| 7. |
OROne says, "Give me a hundred, friend! I shall then become twice as rich as you." The other replies, "Ifunt of their weapective)capital |
| Answer» | |
| 8. |
Given Δ ABC ~ Δ PQR, if poAB 1ar Δ-5, then findPthen find AT AABOat A POR |
|
Answer» ΔABC ~ Δ PQR & AB/PQ = 1/3ar(ΔABC)/ar( Δ PQR )= AB²/ PQ² [The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.]ar(ΔABC)/ar( Δ PQR ) = 1²/3² [Given =AB/PQ = ⅓]ar(ΔABC)/ar( Δ PQR ) = 1/9Hence, the Area of ΔABC/Area of ΔPQR = 1/9 |
|
| 9. |
5. In figure if lines PQ and RS intersect at point T. Such that [2]LP RT-40 , LRPT = 95° and LTSQ-75° , find LSOT .95407570 |
| Answer» | |
| 10. |
. How many numbers are therefrom 300 to 650 which arecompletely divisible by both 5and 72/ 300召650 ส币떼 frm |
|
Answer» First take L. C. M. OF 5 & 7 = 35Now mark 35 as common difference (d)First number which is divisible by 35 in between 300 & 650 is (a) = 315Last number which is divisible by 35 is (an) = 630Therefore, an = a + (n - 1)d630 = 315 + (n - 1)35630 - 315 = 35(n - 1)315 = 35 (n - 1)315 / 35 = n - 19 = n - 19 + 1 = n10 = nTherefore, their are 10 numbers which are divisible by both 5 & 7 which lies between 300 & 650 thank u very very much |
|
| 11. |
6 650 6, evaluate |
|
Answer» 5cos= 6sintan = sin/cos= 5/6 2) Taking cos commanso12tan -3/12tan +1 = 60/6-3/60/6+1= 7/11answer |
|
| 12. |
In triangle ABC, B = 90°, AB=5 cm and AC=10 cm, then |
| Answer» | |
| 13. |
The heights of 5 boys in a group are152 cm, 170 cm, 156 cm, 164 cm and 158 cm. Find the mean |
|
Answer» Height of 5 boys152cm, 170cm, 156cm, 164cm, 158cm Mean = (152+170+156+164+158)/5 = 800/5 = 160cm |
|
| 14. |
Find the area of rhombus in which AB5 cm and AC8 cm.5 cm |
|
Answer» ABCD is the given rhombus. Side AB = BC = CD = DA = 5cm One diagonal AC = 8cm In right ∆ABD, we have to calculate diagonal BD Using pythagoras theorem, BD^2 = AB^2 + AD^2 = (5)^2 + (5)^2 = 25+25BD = 25√2cm area of rhombus = 1/2* (8)* (25√2) = 4*25√2 = 100√2 cm^2 |
|
| 15. |
In AABC, B =90, AB =5 cm and AC=10 cm, then |
| Answer» | |
| 16. |
21 1 इकाई।(दोनों मुहखुला लाह का बना एक लम्ब वृत्ताकार बेलन की ऊँचाई 42 सेमी है। बेलन यदि | समा० माद।ये बना है तथा उसके बाहरी व्यास की लम्बाई 10 सेमी० होने से बेलन में कितना लोहा लगा है ज्ञातकर लिखेके बाहरी अर्द्धव्यास की लम्बाई = सेमी० = 5 सेमी41 सेमी० मोटा है .. बेलन का भीतरी अर्द्धव्यास = (5-1) सेमी० = 4 सेमी०बेलन में लोहे का आयतन।4) X 42 घन सेमी०1 सेमी०४ (5+4) (54) x 42 घन सेमी० [.a-b = (a+b) (ab)]= x9x 1 x 42 घन सेमी० ।= 22x9 x 6 घन सेमी० ।= ]घन सेमी० ।10 सेमी० |
|
Answer» the question answer is 1,188 |
|
| 17. |
Find the value of k, for which point pć2) is the mid-point of thetne segment jeling thepoints Ql-5,4) and R(-1,0).ten 2dtkr + 1 = 0, find k. |
| Answer» | |
| 18. |
1.The heights of 5 boys in a group are:152 cm, 170 cm, 156 cm, 164 cm and 158 cm. Find the mean height. |
|
Answer» Mean height = sum of all heights/no. of boys = (152 + 170 + 156 + 164 + 158)/5 = 800/5 = 160cm mean = average of 5 heights = (152+170+156+164+158)/5 = 800/5 = 160cm |
|
| 19. |
Find the value of a, for which point P (points Q(-5,4) and R(-1,0)2) is the mid-point of the line segment Joining the |
|
Answer» For midpoint,x=x1+x2/2 a/3 = -5-1/2a=3(-6) a= -18The value of a is -18. |
|
| 20. |
(v) Symmetric and sIC ulII. Show that the relation R in the set A of points in a plane given byR ((P, Q): di tance of the point P from the origin is same as the distance of thepoint Q from th: origin), is an equivalence relation. Further, show that the set ofall points relatel to a point P (0, 0) is the circle passing through P with origin as |
| Answer» | |
| 21. |
(1250 का 0.07%) - (650 का 0.02%) ? |
| Answer» | |
| 22. |
2) is the mid-point of the line segment joining theFind the value of a, for which point Ppoints Q(-5.4) aud R(-1.0) |
| Answer» | |
| 23. |
Find the value of a, for which point Ppoints Q(-5,4) and R(-1,0).2) is the mid-point of the line segment joining the3 |
| Answer» | |
| 24. |
B.37.Findtheslopeof the lines which make an angle of 4S with theline x-2y = 3. |
| Answer» | |
| 25. |
Find the value of a, for which point Ppoints Q(-5,4) and R(-1,0).2) is the mid-point of the line segment joining the1 |
| Answer» | |
| 26. |
0) 42- 4-1-(2s-1 |
|
Answer» RHS = (2s - 1)² = (2s)² + (1)² - 2*1*2x = 4s² - 4s + 1 = LHS |
|
| 27. |
\int _ { 0 } ^ { 1 } x ( 1 - x ) ^ { 5 } d x = \frac { 1 } { 42 } |
| Answer» | |
| 28. |
H.C.F. of 650 and 1170 |
|
Answer» 650 = 2×5×5×13 1170 = 2×3×3×5×13 HCF = 2×5×13 = 130 |
|
| 29. |
. Find the value of :(1) 20% of 650 |
|
Answer» 20/100*650=130 is the answer 130 is the correct and best answer 10%=650/10=65 . 20%=10%× 2=65×2=130 20/100=20/650=2*65=130 answer |
|
| 30. |
In a circle of radius 5 cm, AB and AC are two chords such that AB AC6 cm. Find the length of the chord BC. |
|
Answer» See the diagram.we need to find 2y. In the right angled triangle AMC6² = x² + y²⇒ y² = 36-x² In the right angled triangle OMC5² = y² + (5-x)²⇒ 25 = (36-x²) + 25 + x² - 10x⇒ 25 = 36 - x²+ 25 + x² - 10x⇒ 0 = 36 - 10x⇒ x = 36/10 = 3.6 y =√(36-3.6²) =√23.04 = 4.8 cmBC = 2y = 4.8×2 = 9.6 cm |
|
| 31. |
In a circle of radius 5 cm, AB and AC are two chords suchthat AB - AC 6 cm, as shown in the figure. Find thelength of the chord BC |
| Answer» | |
| 32. |
AB 5 cm, BC-4 cm and AC-6 cm. State, with reason, vpossible to draw AABC or not. |
|
Answer» is this line segment |
|
| 33. |
11. In the given fig AB and AC are oppositerays. If(a-3b)=20". find La and 268 |
|
Answer» thank you |
|
| 34. |
हकन, ८ 20 Tatb-cतो प्रत्येक अनुपात का मान होगा।) 1 (b 3()R 2 e 5€ D& |
|
Answer» add all the three .you will get a+b+c/a+b+c hence answer will be 1 |
|
| 35. |
19. In the given figure, in AABC it is given thatLB-40° and 2C = 50°. DE 11 BC and EF | | AB.Find: (6) ZADE+ MEN (ii) ZBDE and(iii) BFE.40°50 |
| Answer» | |
| 36. |
3)37#, find the value of tan θ, (θ is an acute angle)35If sec θ |
| Answer» | |
| 37. |
37Find the equation of the circle whose radius is 4 andcentre is (0, 1). |
|
Answer» Equationwill be(x-0)^2+(y-1)^2= 4^2 |
|
| 38. |
find the value of k for which point P is the mid point of the the segment joining points Q(-5,4) and R(-1,0) |
|
Answer» let a = k |
|
| 39. |
37. Find the area of the region bounded by the curve yline y x and positiveaxis. |
| Answer» | |
| 40. |
hrs?The cost of 15 benches is Rs. 37), find the cost of each bench?the5 m toward |
|
Answer» cost of 15 benches=37.5cost of one bench =37.5/15=2.5rs |
|
| 41. |
Find the value of k, for which point Ppoints Ql-5,4) and R(-1,0).2) is the mid-point of theine segment joining the |
| Answer» | |
| 42. |
EXAMPLE 31 If the coefficientsfthat n2 -n(4r +1)+42 -2-0.r-1 r r+11 in the binomial expansion of(1 + a)" are in A.P, roINCER |
| Answer» | |
| 43. |
What is the heightof the table?170 cm130 cmITIn |
|
Answer» 170000000000000000000000000000 cm 135cm is the right answer........ i think the answer is 40because =170-130=40if you don't mind if it is wrong please give me right answer 120 is the right answer 135 cm is the right answer 125.77cm is the correct answer 135 is the correct one 130 is the right answer 40 is the right answer 170000000000000000000000000cm the correct answer is 135CM 135 is the correct answer 150 is right answer of your question 170 is the correct answer The right answer is 135cm |
|
| 44. |
CBSEClassVALUEQUESTIONSFind Sin3 and 4rationce number betweesin |
| Answer» | |
| 45. |
The perimeter of a rectangle is 130 cm. If the breadth of the rectangleis30 cm, find its length. Also find the area of the rectangle. |
|
Answer» Perimeter = 2(l+b)=> 130 = 2(l+30)=> 65 = l+30=> l = 35 cm. Area = l*b= 30*35= 1050 sq cm |
|
| 46. |
2. The area of a rectangle is 650 cm2 and its breadth is 13 cm. The perimeterof the rectangle is(a) 63 cm(b) 130 cm(c) 100 cm(d) 126 cm |
| Answer» | |
| 47. |
11. Neha has 3.75 Rs. Sheila has twice as much as Neha and 1.20 Rs. more thanHeena. How much money does Heena have? |
|
Answer» 6.3 rupeessheila-2×3.75=7.5Heena-7. 5-1.20=6.3 sheila 2×3.75=7.5heena 7.5-1.20=6.3 answer. 6.3 is correct answer Henna have 6.3 money is the best answer. |
|
| 48. |
etween the length and breadth of a rectangular park is 3:2. If a man cycling along the2 km/h completes one round in 8 minutes, then find theQ16. The ratio bboundary of the park at the speed of 1area of the park. |
| Answer» | |
| 49. |
In a circle of radius 5 cm, AB and AC aretwo equal chords such that AB AC 6 cm. cFind the length of the chord BC. |
|
Answer» AB and AC are two equal chords of a circle, therefore the centre of the circle lies on the bisector of ∠BAC. ⇒ OA is the bisector of ∠BAC. ∴ P divides BC in the ratio = 6 : 6 = 1 : 1. ⇒ P is mid-point of BC. ⇒ OP ⊥ BC. In ΔABP, by pythagoras theorem, AB2= AP2+ BP2 ⇒ BP2= 62- AP2.............(1) In right triangle OBP, we have OB2= OP2+ BP2 ⇒ 52= (5 - AP)2+ BP2 ⇒ BP2= 25 - (5 - AP)2...........(2) Equating (1) and (2), we get 62- AP2= 25 - (5 - AP)2 ⇒ 11 - AP2= -25 - AP2+ 10AP ⇒ 36 = 10AP ⇒ AP = 3.6 cm putting AP in (1), we get BP2= 62- (3.6)2= 23.04 ⇒ BP = 4.8 cm ⇒ BC = 2BP = 2× 4.8 = 9.6 cm |
|
| 50. |
57.TA, TB are tangents to a circle with centre 0. ChordAB intersects TO at C.Given11find AB.04 TA361.8cm2. 12cm.....3.9cm1.4cm |
|
Answer» 1/(OA)² + 1/(TA)² = 1/36 Multiply both sides by (OA)² (TA)²(TA)² + (OA)² = (OA)² (TA)² / 36 Since TA is a tangent of the circle, then TA is perpendicular to OA, so △OAT has a right angle at A. By Pythagorean theorem:(TA)² + (OA)² = (OT)² Therefore:(OT)² = (OA)² (TA)² / 36OT = OA * TA / 6 AB is perpendicular to OT. Therefore, AC is perpendicular to OTSince △OAT is a right triangle, dropping perpendicular from right angle A to side OT (at C) creates 2 right triangles that are both similar to △OAT △OAT ~ △OCA ~ △ACT By similar triangles:AC/OA = TA/OTAC = OA * TA / OTAC = OA * TA / (OA * TA / 6)AC = 6 AB = 2 * ACAB = 12 Option 2) 12 cm is correct. |
|