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. |
Sove the following system of equations graphically or a andy3r 212: 54Also. find the coordinates of the point where both the lines meet each other.2 |
|
Answer» point of intersection is x = 2 and y = 3 |
|
| 2. |
15. Solve graphically the following system of linear equations. Also find thecoordinates of the points where the graph lines meet the x-axis.(i) 3x +y 12 0; x -3y +6 0ICBSE SP 20061 |
| Answer» | |
| 3. |
Um, through R, draw a line parallel to PŲ. Let this line meet / at S. What shape5. Make a triangle ABC. From each vertex draw lines parallel to opposite sides. |
|
Answer» Steps of construction: 1. Draw 3 non-collinear points, say, A, B and C such that none of them lies on the same line. 2. Form a triangle ABC by joining AB, BC and CA. 3. Construct a Parallel line to AB Choose B as a centre, draw an arc cutting BC and BA at W and V, respectively. Choose C as a centre and draw an arc on the opposite side of BC to cut BC at P. (Note: use the same radius as in the previous step) With centre P and radius equal to WV, draw an arc cutting the arc drawn in the previous step at Q. Join CQ and produce in both directions to obtain the line parallel to AB. 4. Construct a Parallel line to AC With A as centre, draw an arc cutting AC and AB at T and U, respectively. With centre B and draw an arc on the opposite side of AB to cut AB at X. (Note: use the same radius as in the previous step) With centre X and radius equal to TU, draw an arc cutting the arc drawn in the previous step at Y. Join BY and produce in both directions to obtain the line parallel to AC. 5. Construct a Parallel line to BC With B as a centre, draw an arc cutting BC and BA at W and V, respectively (already drawn). With centre A and draw an arc on the opposite side of AB to cut AB at R. (Note: use the same radius as in the previous step) With centre R and radius equal to WV, draw an arc cutting the arc drawn in the previous step at S. Join AS and produce in both directions to obtain the line parallel to BC. 6. Your construction is ready. |
|
| 4. |
7 The mid-point of the line segment joining A(2, p) and B(g, 4) is (3, 5). Calculate tvalues of p and q |
|
Answer» the mid point in x axis is given by (2+q)/2 = 3 => 2+q = 6 => q = 6-2 = 4 and midpoint in y axis is given by is given by (p+4)/2 = 5 => p+4 = 10 => p = 10-4= 6. |
|
| 5. |
the vahe ontf sum of theperpendicular distances of a variable point P (x, y)() and 3x-2). +7x 0 is always l 0. Show that P must 11ation of the line which is equidistant from parallel lines ghmintl reflects on the r-axis atind |
| Answer» | |
| 6. |
16. In the given figure, ABCD is aparallelogram, E is the mid-point of ABand F is the mid-point of CD. GH is anyline that intersects AD, EF and BC inG, P and H respectively. Prove thatP/GP = PH.17. ABCD is a square. A is joined to a pointPon tRC |
| Answer» | |
| 7. |
(c) In thegiven figure, C is a point on the minor arc AB of the circle withCentre O. Given <ACB-p", LAOB=g" express "q, in terms of p.Calculate p, if OACB is a parallelogram.14stion 2. |
|
Answer» 1 2 |
|
| 8. |
If from any point P on thex2 + y2 +y2gx+ 2fy + c sin?a + (g?+f?) cosa0, tangents are drawn to the circleclex2 + y2 + 2gx 2fy + c0 then the angle between the tangents is(A) Îą(B) 2 Îą(D) 32 |
| Answer» | |
| 9. |
The line segment joining A(-1 3) and B(G, 5) is divided in the ratio 1 :3 at P,thepoint where the line segment AB intersects y-axis. Calculate(7) the value of a.(ii) the coordinates of P |
|
Answer» If point p is at y axis so coordinate of p is (0,y).so by section formula,(0,y)=(a-3/4,5+5/3*3)(0,y)=(a-3/4,5+5/4)so,a-3/4=0so, a=3and y=5+5/4=5/2so coordinate of p=(0,5/2) |
|
| 10. |
7. A chord of a circle of radius 20 cn subtends an angle of 90° at the centre. Find the aof the orresponding major segment of the circle. (Use π3.14) |
| Answer» | |
| 11. |
$.A chord PQ of a circle with radius 15 cm subtends an angle of 60° with thecentre of the circlc. Find the area of the minor as well as the major segment. |
| Answer» | |
| 12. |
. In Fig. 9.17, PORS and ABRS are parallelograms P AQ Band X is any point on side BR. Show that(G) ar (PQRS)- ar (ABRS)(ii) ar (AX S) = ar (PQRS)R.2Fig. 9.17 |
| Answer» | |
| 13. |
a chord of a circle of radius 20cm subtends an angle of 90 at the centre. find the area of the corresponding major segment of the circle. |
| Answer» | |
| 14. |
6. Find the value of x for each of the fig., given below.14x+1090539 |
|
Answer» 4x+1+143=1804x=36x=9and 3x-2=943x=96x=32 |
|
| 15. |
C.B.S.E. 20145. If the ratio of the first n term of two A.P's is (7n + 1): (4n + 27), find the ratio of their mth ter |
|
Answer» Given ratio of sum of n terms of two AP’s Sn:S'n = (7n+1):(4n+27)..... (1) Let’s consider the ratio these two AP’s mth terms as am: a’m. We know the nth term of AP an= a + (n – 1)dHence, am: a’m= a + (m – 1)d : a’ + (m – 1)d’ On multiplying by 2, we getam: a’m= [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1: S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23] Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23 not able to understand |
|
| 16. |
ăŹz-f9 kg of sugar costs? 238.50, how much sugar can be bought for371? |
| Answer» | |
| 17. |
t ter curved surlace areaow cylindrical pipe has inner circumference 44 dm and outer 45 dmn both sides (inside as well as outside) at 2 per m? if length is 3.5 m9, A hollow |
|
Answer» Inner circumference = 44dm = 4.4mouter circumference = 45dm = 4.5 mlength = 3.5minner CSA = 4.4×3.5 = 15.4 m²outer CSA = 4.5×3.5 = 15.75 m²total = 15.4 + 15.75 = 31.15 m²cost = Rs. 2 per m²total cost = 31.15×2 = Rs.62.30 |
|
| 18. |
1. Why were big European powers met in Berlin in 1885? |
|
Answer» Big European nations met in Berlin to hold a conference regarding the split up of the federated states of Africa. They wanted to split Africa into separate states in order to gain control over the continent’s abundant natural resources for the development of their industrial sector. The conference lasted for three months and resulted in the split up of Africa into fifty states. |
|
| 19. |
A room is 10.2 m long, 8.5 m wide and 4.6 high. It has a door measuring 1.5 m x 2.6 and a window measuring 2 m * 1.7 m, find the cost of tiling the walls at the rate of ₹13.50 per sq. m. |
| Answer» | |
| 20. |
5. A room is 10.2 m long, 8.5 m wide and 4.6mhigh. It has a door measuring 1.5 m Ă 2.6 mand a window measuring 2 m x 1.7 m. Findthe cost of tiling the walls at the rate of13.50 per sq m |
|
Answer» Area of the 4 walls = 2(l+b)×h sq. units= 2(10.2+8.5)×4.6 m sq.= 2×18.7×4.6 m sq.= 172.04 m sq. Area of a door = l×b sq. units= 1.5×2.6 m sq.= 3.9 m sq. Area of a window = 3.4 m sq. [ l×b ] Total tilling area = Area of 4 walls- (area of a door + area of a window) = 172.04 - (3.9 + 3.4) m sq. = 164.74 m sq. Total cost = Rs. (13.50×164.74)= Rs. 2223.99= Rs. 2224 |
|
| 21. |
z = x + i y \text { and } w = \frac { 1 - i z } { z - i } , \text { show that } | w | = 1 \Rightarrow |
| Answer» | |
| 22. |
5. A room is 10.2 m long, 8.5 m wide and 4.6rmhigh. It has a door measuring 1.5 m x 2.6 mand a window measuring 2 mx 1.7 m. Findthe cost of tiling the walls at the rate of13.50 per sq. m. |
|
Answer» Area of the 4 walls = 2(l+b)×h sq. units= 2(10.2+8.5)×4.6 m sq.= 2×18.7×4.6 m sq.= 172.04 m sq. Area of a door = l×b sq. units= 1.5×2.6 m sq.= 3.9 m sq. Area of a window = 3.4 m sq. [ l×b ] Total tilling area = Area of 4 walls- (area of adoor + area of a window) = 172.04 - (3.9 + 3.4) m sq. = 164.74 m sq. Total cost = Rs. (13.50×164.74)= Rs. 2223.99= Rs. 2224 |
|
| 23. |
If (1x2 + mx + n)2=9x4-36x3+9x2-144x+144 findl, m, n. (3,-6, )o)四1-3, 6,:/e)5 |
| Answer» | |
| 24. |
Mensuration61(a1. The dimensions of a room are 16x 14x 10 metresThere are four windows of 1.3 mx 1.4 m and 2doorsof2 mx I m. What will be the cost of white washingthe walls and painting the doors and window, if therate of white washing is Rs 5 per m and the rate ofpainting is Rs 8 per m(ob) Rs. 313384 |
| Answer» | |
| 25. |
l subtends an angle ofmeasure 450 at the major segmentIfthe chord y : mx + 1 of the circle x2 + y2of the circle then value of m is(A) 2+v2(B) -2+v2(C)-1+N2(D) 1+ V2 |
| Answer» | |
| 26. |
f x + 1 andx-1 are factors of mx^3 + x^2-2x +n, find the value of m and n. |
| Answer» | |
| 27. |
A cuboidal tank of dimensions 8 mx 5 mx 3 m completely filled with water. An outlet pipe canremove water from the tank at the rate of 20 litre per minute. Find the time taken in hours) bythe pipe to completely empty the tank |
| Answer» | |
| 28. |
12) 34(3) 12/8(4) 1210mension of a cuboid is 10 mx 8 mx 6 m. Then the diagonal of the cu$ 717 7 fant 10 m * 8 m 6 m , at Ho fa chouf -(1) 41.44 m(2) 14.14 m(3) 41.14 m(4) 14.41 inter the |
| Answer» | |
| 29. |
7kg 8g = |
|
Answer» Given,7kg 8gw.k.t1kg=1000g7kg=7000g7000g+8g=7008gTherefore 7kg8g=7008g 1kg =1000g7kg=7000g =7000+8g =7008g |
|
| 30. |
13. If 7kg sugar costs Rs.230, what is the cost of 15 kg sugar? |
|
Answer» Cost of 7 kg sugar = 230Cost of 1 kg sugar = 230/7Cost of 15 kg sugar = (230/7)x15= 32.8x15= 492 ans thanks |
|
| 31. |
Two positive integers a and b can be written as a-ry2 and b.xy3. x,yare prime numbers. Find LCM (a, b). |
|
Answer» Given:a = x³y² , b= xy³ LCM = The highest indices of the unknowns (x and y) LCM =x³y³ |
|
| 32. |
0' If two positive integers a and b are written as a-x4y2 and b-хзуз, where x, y are prime numbers,find the IICF and LCM (a,b) |
|
Answer» Like if you find it useful |
|
| 33. |
hatural number.If two positive integers p & q are written as p ab& q a bwhere a, b are prime numbers then verify |
|
Answer» Like if you find it useful |
|
| 34. |
8. If two positive integers p and q are written as p-a bland q-a'b where a and b are prime numbersthen verify: |
|
Answer» p = a²b³ q = a³b HCF ( p,q ) = a²b [∵Product of the smallest powerof each common prime factors in the numbers ] LCM ( p , q ) = a³b³ [∵ Product of the greatest power of each prime factors , in the numbers ] Now , HCF ( p , q )× LCM ( p , q ) = a²b× a³b³ = a∧5b∧4 --------( 1 ) [∵ a∧m× b∧n = a∧m+n ] pq = a²b³× a³b = a∧5 b∧4 ---------------( 2 ) from ( 1 ) and ( 2 ) , we conclude HCF ( p , q )× LCM ( p ,q ) = pq |
|
| 35. |
If two positive integers a & b are written as a xy & b xy3 where x & y areprime numbers then verify LCM x HCF ax b. |
|
Answer» a = x*x*x*x*xy*y b = xy*y*y HCF = xy*y LCM = x*x*x*x*xy*y*y a × b = x*x*x*x*xy*y*y*y*y HCF × LCM = x*x*x*x*x*xy*y*y*y*y If you find this answer helpful then like it. |
|
| 36. |
1. If two positive integers 'a' and 'b' arewritten as a ry2 and b xy, x, y being!]prime numbers, then find HCF (a, b) |
|
Answer» HCF is termed as highest common factor of given numbers Given,a = x^3*y^2a = x*x*x*y*y b = x*y^3b = x*y*y*y Then HCF of a and b= x*y*y= xy^2 |
|
| 37. |
Ajay bought a TV for?! 5000 and sold it for14100. Find the loss percent |
|
Answer» CP of the T.V. = Rs 15000SP of the T.V. = Rs 14100 Loss= CP-SPtherefore, here loss= Rs(15000-14100) = Rs 900 Now, loss percentage= Loss/CP multiplied by 100 Therefore, here loss percentage = 900/15000 * 100= 6 % hence, the loss percent = 6% |
|
| 38. |
ना कुााघाध 1222५.हे लि नाz (e v (c LG£ 2 ॥थाड 25: 1७७ BED | (G2BB (€ i (2 रा६ ३0० 1215 छु।2 (20 [82 (28० 1का>डे | िटरि (४200६ (६ 02 (2 092 (1¢ 2 pho pR PR 02 (६2o of hE ouh b00100T |
|
Answer» आलेख दिसत नाही . पुन्हा एकदा निट फोटो पाठवा |
|
| 39. |
rabbitit on theLet ustraction questions.ayonorOh yestto theonso addor subtract fora situation iFUN AT HOME CDear Mom/DadAsk,ne at and tell me the answer.Iwildekte answer·ldn...--need To add ar sthe nedfil andan exampledt had 8 steters. She gave 5 stickera to her braherdecorafte a card. Hax many stickera are lit with harkand Keep aCreate mare story auns lke this and keep askinaIwill check your answers and give you painta.en poainta will fetch you a hwg!Love |
| Answer» | |
| 40. |
8. Ravi had 25 marbles he gave 10 marbles to his brother. What percent of the marbles did he give to hisbrother?a. 10%b. 25%c. 405d. 20% |
| Answer» | |
| 41. |
Raju and Ravi both are on cigher side of temple of clevation 30' and 6o1'rsmeivetFind the distance between Raju and Ravi? |
| Answer» | |
| 42. |
7 Divide the amount ofAjay1425 among Ajay, Ravi, and Leela such thatgets 5 times as much as Ravi gets and 3 times as Leela. |
|
Answer» Let leena be xTherefore, Ravi = 3x & Ajay = 5(3x) = 15xx+3x+15x=142519x=1425x = 1425/19 = 75Therefore,Leena = x = 75 RsRavi = 3x = 75 • 3 = 225 RsAjay = 15x = 75 • 15 = 1125 Rs Like my answer if you find it useful! |
|
| 43. |
|42. आव्यूह पूर्ण करने के लिए सही विकल्प चुनिए।40 160 ? |80 5 2020 40 10(a) 360 (b) 120 (c) 140 ) 80 |
|
Answer» option D (80) is the correct option answer option d is the correct answer of the given option d is the correct answer. the correct answer is optiona D option d is correct answer |
|
| 44. |
की i _w.. e S0°+ cot 78° 1 0° 3 45° 3 vt =कौजिए। |
|
Answer» Sec 50° +cot 78°= sec(90-40)+ cot(90-12)= cosec 40°+tan 12° |
|
| 45. |
Classes --T10--20T 20.-30T30ä¸40T40-50T50-60-160-70T70-80FrequeneyFind the mean and mode for the following data1210 |
| Answer» | |
| 46. |
\cos 40 \cos 80 \cos 160 = \frac { - 1 } { 8 } |
|
Answer» cos 40.cos 80.cos 160 = (1/2) (2 cos 40.cos 80).cos 160= (1/2) ( cos(40+80) + cos(40 - 80) ).cos 160 { 2cosAcosB = cos(A + B) + cos(A - B) }= (1/2) ( cos(120) + cos(-40) ).cos 160= (1/2) ( -1/2 + cos(40) ).cos 160 { cos(-A) = cos(A) }= (1/2) ( -cos(160)/2 + cos(160)cos(40) )= (1/2) ( -cos(160)/2 + (1/2)2*cos(160)cos(40) )= (1/2) ( -cos(160)/2 + (1/2) [ cos(200) + cos(120) ] )= (1/2) ( -cos(160)/2 + cos(200)/2 + cos(120)/2 )= (1/2) ( -cos(160)/2 + cos(200)/2 - 1/4 )= (1/2) ( 1/2 [cos(200) - cos(160)] - 1/4 )= (1/2) ( 1/2 [-2sin(180)*sin(20))] - 1/4 ) { cos(A)-cos(B)= -2(sin(A+B)/2)*(sin(A-B)/2) }= (1/2) ( 0 - 1/4 ) { sin(180) = 0 }= -1/8 Tq |
|
| 47. |
33.A man on tour travels first 160 km at64 kmph and the next 160 km. at80 kmph. The average speed of the1st 320 km of the tour is: |
| Answer» | |
| 48. |
(7)The side of a cube is 4 cm. Then its suarea will beA 16 cm(B) 96 cm(C) 64 cm (D) 12 cm |
|
Answer» A=6a^2a=4cmA=6×(4)^2=96cm^2 TSA of cube=6a^2=6(16) =96cm^2 |
|
| 49. |
TRS b407. A teak wood log is cut first in the form of cuboid of length 2.3m, width 0.75m and ofthickness. Its volume is 1.104m3. How many rectangular planks of size 2.3mX0.75mx0.04mcut from the cuboid?(16) |
|
Answer» As Volume = l*b*h ⇒1.104 = 2.3*0.7* h ⇒height, h = 1.104/1.61 = 0.6857 Now the cuboid is cut into cubes of size 2.3m*0.75m*0.04m, then Volume(V1) = 0.069m^3 Let the given cuboid is cut into n cuboids then volume of the given cuboid = n*vol of smaller cuboid n = V/V1 = 1.104/0.069 = 16 |
|
| 50. |
2. निम्नलिखित आँकड़े, 225 बिजली उपकरणों के प्रेक्षित जावन कण (LI ।। ।।जीवनकाल ( घंटों में )0-20 20.40 140-60] 60-80 | 80-100| 100-120103552 | 6l 38 | 29बारबारता*बदन जीवनकाल ज्ञात कीजिए।--- |
|
Answer» 10+35+52+61+38+29= 325/ 325/6=65 10+35+52+61+38+29=325/6=65 65 is the correct answer of the given question |
|