Saved Bookmarks
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.
| 72001. |
3.Find the length of the are of a cirele of diameter 12 cm subtending an angle of 60° atthe centre. |
|
Answer» angle=arc/radiusradius=12/2=6cmangle=π/3π/3=arc/6arc=2πcm=6.28cm |
|
| 72002. |
angle of 60°.5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cnmand taking B as centre, draw another cirele of radius 3 cm. Construct tangents to eachcircle from the centre of the other circle. |
| Answer» | |
| 72003. |
Exercise 1What is the ratio ofă100 and 210 ? Express your answer in the simplest form.Sudha has 5. Money with Radha is 3 times the money with Sudha. How much moneydoes Radha have? |
|
Answer» matter |
|
| 72004. |
Rs. 860 are divided among 45 persons consisting of men, womenchildren. The ratio of men's women's and children's share are as 12:15: 16, but the individual shares of a man, a woman and a child are inthe ratio 6: 5:4. Find what a man, woman and a child get. |
|
Answer» Let share of men's, women's and children's are 12s, 15s, 16s Then 12s + 15s + 16s = 86043s = 860s = 20 Thus share of men's, women's and children's are 240, 300 and 320 respectively Ratio of men, women and children = 12/6:15/5:16/4= 2:3:4 Let no. Of men, women and children are2n, 3n, 4n2n + 3n + 4n = 459n = 45n = 5Number of men, women and children are10, 15, 20 Share of man = 240/10 = 24Share of woman = 300/15 = 20Share of child = 320/20 = 16 |
|
| 72005. |
Reema is sister of Vipin. Sudha is mother of Vipin.Jagdish is father-in-law of Sudha. How is Reemarelated to Jagdish? |
|
Answer» Reema is grand child of Jagdish jagdish will be grandfather of reema |
|
| 72006. |
12. Find the equation of the circle with radius 5 whose centre lies on x-axis andasses through the point (2,3). |
| Answer» | |
| 72007. |
21.Acircletouches y -axis at (0, 5) and whose centre les on thecentre lies on theline 2x+y = 13 ; find the equation of the circle. |
|
Answer» Perpendicular distance of Pt. from liner=2(0)+5-13/√5=-8/√5r=8/√5..radius for centre let the X Co ordinate of point =h therefore y Co ordinate =13-2hNow centre=(h, 13-2h) we know radius therefore apply distance formula √(h-0)^2 + (13-2h-5)^2=(8/√5)solve for h and put in point of centre.. now u have centre and radius both.... and we know if (m, n) is centre and r is radius then eqn of circle is (x-m) ^2+(y-n)^2=r^2 |
|
| 72008. |
12310(iii)=lll)2 |
| Answer» | |
| 72009. |
√2+√2=? |
|
Answer» 2√2
|
|
| 72010. |
If a, denotes the general term, then write the next five terms of the sequences given by(5.a, 2 and aa, + 4 for all n 22 |
| Answer» | |
| 72011. |
21. In Mathematics test gycn te 15 students the following marks were obtainede4 38,50, 481 62, 5 43, 096, 54, 198, 42. 45, 54, 62, 6Filrd the mĂŠan and the median |
| Answer» | |
| 72012. |
t cef the cirele pasing throgh eointwhose centre Ilthef the circle with adius 5 whose cenuation of the circle passing through the points (2.3) and (1,1) and1. Find theouwhose centre is on the line 3ytre lies on x-axis an |
|
Answer» thanks |
|
| 72013. |
2) II e USL UT I VITIN 151 10.3) If you spend Rs 15.75 every day as pocket money. Calculate how much will be spenin one month? (1 month = 30 days)1111of 251 |
|
Answer» per one day=15.75per 30 days=30×15.75 =472.5 rupees one day =15.75 30 day =30×15.75 472.50 Per day spend pocket money =Rs 15.75. ° . 1 month =30 days=30×15.75=Rs472.5 one day =15.751 month=30 daysThen=15.75×30=472.5 rupees per one day =15.75per 30 day =30*15.75 =472.5 rs |
|
| 72014. |
the igrit house 543. The horizontal distance between two poles is 15 m. The angle of depression of tetop of the first pole as seen from the top of the second pole is 309. If the height of thesecond pole is 24 m, find the height of the first pole. (3 1.732) ICBSE 2013 |
| Answer» | |
| 72015. |
Sudha ranks 8th from the top and 37th rank fromthe bottom. How many students are there inher class?(1)47 (2) 46 (3)45 (4) 44 |
|
Answer» 37+8=45, but the sudha's rank has been counted twice. so, 1 will be subtracted from 45 ie, 45-1=44 44 is the right answer |
|
| 72016. |
5 6 7 e isThe sequence(a) H.P(c) A.P.ง่า(b) G.P.(d) of these, |
|
Answer» The sequence is in A.P |
|
| 72017. |
10.The line segment joining the points A(4, -5) andB(4,5) is divided bythe point P such thatFind the co-ordinates, of P.28 5APー |
| Answer» | |
| 72018. |
In an A.P the first term is 2 and the sum of the first five terms is one fourth of the sum of the next fiveterms. Show that the 20th term is-112. |
| Answer» | |
| 72019. |
the sum of the next five terms, find the sum of first 30 terms.(v)If the sum of first 7 terms of an A.P. is 10 and that of next 7 terms is 17, find its 28thterm. |
| Answer» | |
| 72020. |
lll) i |
|
Answer» we know that. i⁴ = 1 and i² = -1=> i⁴ⁿ = 1 now i^(318) = i^(316+2) = i^(316).i² = i^(4*79).i²= i⁴ⁿ.i² = 1.i² = -1 hence i^(318) = i² = -1 |
|
| 72021. |
Find the equation of the circle whose radius is 5and whose centre lies on y-axis and circlepasses through the point (3,2)12. |
| Answer» | |
| 72022. |
The area of sector of a circle whose radius is 12 metro and whose angle at the center is 42° is? |
| Answer» | |
| 72023. |
20 labour can dig a pond in 12 days. How many days will it take 16 labour to dig the same pond? |
|
Answer» If 20 Labour dig pond in 12 days1 labour dig pond in 12*20 days Then 16 Labour dig pond in= 12*20/16 = 3*5 = 15 days thx |
|
| 72024. |
17. The sum of n terms of an AP is 3n2 +5m. Find the AP Hence, find its 15te5 teril.l1 and R nre2-2) and (2-4) respectively, find the co-ordin(2, |
|
Answer» The sum of n terms of an A.P. = 3n^2+5n then sum of n-1 terms= 3(n-1)^2+5(n-1) So nth term of A.P will be Tn=3n^2+5n- 3(n-1)^2-5(n-1)= 3n^2+5n-3n2-3+6n-5n+5=6n+2 in this way by putinng values of n= 1,2,3,4,5,6,7,_______n-1, nHence n=15hence6*15+2=90+2=92 |
|
| 72025. |
wellwiceIls Dieduul. Fluuleleiguldlu ut vitaulut mera10. In the centre of a rectangular lawn of dimensions 50 m x 40 m, a rectangular pond has tobe constructed so that the area of the grass surrounding the pond would be 1184 m2Find the length and breadth of the pond.[NCERT EXEMPLAR) |
|
Answer» Solution:-Dimensions = 50 m and 40 mArea of the rectangular lawn = 50 × 40 = 2000 m²Area of the grass surrounding the pond = 1184 m²So, the area of the pond = area of the lawn - area of the grass= 2000 - 1184Area of the pond = 816 m²Let the width of around the pond be 'x' m Then, the length of the pond = (50 - 2x) mand the breadth of the pond = (40 - 2x) mArea of the pond = 816 m²⇒ (50 - 2x) × (40 - 2x) = 8162000 - 80x - 100x + 4x² = 8164x² - 180x + 2000 - 816 = 04x² - 180x + 1184 = 0 dividing it by 4 we getx² - 45x + 296 = 0 x² - 37x - 8x + 296 =0x(x - 37) - 8(x - 37) = 0(x - 37) (x - 8) = 0x = 37 or x = 8suppose if x = 37, thenthe length of the pond will be 50 - 2*37 50 - 74 = - 24m, which is not possible because length cannot be negative.Therefore the length of the pond = 50 - 2*850 - 16 = 34 mand Breadth of the pond = 40 - 2*840 - 16 = 24 mLength = 34 m and breadth = 24 m |
|
| 72026. |
2. which fraction comes next in the sequence13 5 7 |
|
Answer» 9/32 is the next sequence |
|
| 72027. |
The seventh term of an Arithmetic progression is four times its secondterm and twelfth term is 2 more than three times of its fourth term. Findthe progression.OR |
|
Answer» A.T.Q. a7 = 4 × a2 a + 6d = 4 × (a + d) a + 6d = 4a + 4d 6d - 4d = 4a - a 2d = 3a -> ( 1 ) a12 = 2 + 3 × (a4) a + 11d = 2 + 3 × (a + 3d) a + 11d = 2 + 3a + 9d 11d - 9d = 2 + 3a - a 2d = 2 + 2a From eq. ( 1 ) 3a = 2 + 2a 3a - 2a = 2 a = 2-> ( 2 ) Putting value of eq. ( 2 ) in eq. ( 1 ) 2d = 3a 2d = 3(2) 2d = 6 d = 6/2 = 3 Hence, the A.P. is a, a+d, a+2d, a+3d..............a+(n-1)d 2, 5, 8, 11..............so on. . |
|
| 72028. |
Ju. It hrst term of an arithmetic progression is 2 and sum of its first five terms is one fourth of newfive terms, then prove that the 20th term of the arithmetic progression is -112. |
| Answer» | |
| 72029. |
Ethemeanofx+2.2x +3,3x+44x + 5 is x +2. findand medlaitbythe3,. If the mean of x +2, 2x +3,3x+ 4, 4x +5 is +2, find x.4 and 12 resetively. then find the |
|
Answer» mean=(x+2)+(2x+3)+(3x+4)+(4x+5)/4x+2=(10x+14)/44x+8=10x+146x=6x=1 hit like if you find it useful |
|
| 72030. |
275 cm can be painted out of this container?lll aiea equal to 9.375 m2. How many bricks ofIf the rainfall recorded on a certain day was 2.5 cm. Find the volume of water that fell on a 5 hectareield. |
| Answer» | |
| 72031. |
a) 154cm36. If the diagonel of a cube is 4v3cm.then its side isSO.b) 3c) 2d) 1a) 4 |
| Answer» | |
| 72032. |
18. The side of an equilateral triangle is equal to the radius of a circle whoseitsarea is 154 cm2. The area of the triangle is |
| Answer» | |
| 72033. |
7.Two candles of the same height are lighted at thesame time. The first is consumed in 4 hours andthe second in 3 hours. Assuming that each candleburns at a constant rate, in how many hours aftietbeing lighted, was the first candle twice the beigihtof the second? |
|
Answer» Let the height of candles = hFirst candle consumed in 4 hoursSecond candle consumed in 3 hours Burning rate of first candle = h/4Burning rate of second candle = h/3 Let after x hours height of second candle is y and height of first candle is 2y Then, y = h - xh/3.......(1)2y = h - xh/4.....(2) 2(h - xh/3) = h - xh/42h - h = 2xh/3 - xh/4h = (8xh - 3xh)/12x = 12/5 = 2.5 Therefore, after 2.5 hours height of first candle is twice the height of second candle |
|
| 72034. |
4. (i) First, second andthirdterms a proportionofare respectively, 4, 6 and 8. Find its fourtterm. |
|
Answer» Let the fourth term be x4: 6=8: x4/6=8/x48/4 = xx= 12 |
|
| 72035. |
Find the angle between the vectors č -314 J+sk,d -21+ j+2k. |
|
Answer» If the two vectors are given asa⃗a→andb⃗b→then its dot product is expressed as a.b. Suppose these two vectors are separated by angleθθ.To know what's the anglemeasurement we solve with the below formula Cos theta = (-6+4+10)/(√9+25+16)(4+4+1)cos theta= 8/√50*3theeta= cos inverse = 8/3√50 |
|
| 72036. |
Acircular pond covers an area of 314 cm2, What is the distance round the pond?(Take T 3.14) |
|
Answer» A=Pi×r^2r^2= A/Pi = 314/3.14=100r=10cmP=2×Pi×r =2×3.14×10 = 6.28×10 =62.cm = 63cmAns)The circumference of the circle is 63cm. |
|
| 72037. |
Determine the AP whose 5th term is 19 and the difference of the 8th term from the 13thterm is 20. |
|
Answer» Let a and d are first term and common difference of an A . P. nth term = tn = a + ( n - 1 )d----------( 1 ) i) Given t 5 = 19 a + 4d = 19 -------(2) ii ) differnce of the eighth term from the thirteeth term = 20 t 13 - t 8 = 20 a + 12d - ( a + 7d ) = 20 a + 12d - a - 7d = 20 5d = 20 d = 20/ 5 d = 5 Put d = 5 in ( 2 ) a + 4d = 19 a + 4 × 5 = 19 a + 20 = 19 a = 19 - 20 a = - 1 Therefore,a = -1 , d = 5 Required A .P is a , a+ d , a + 2d , a + 3d , ..... -1 , 4 , 9 , 14, 19 , ... |
|
| 72038. |
50. Determine the AP whose 5th term is 19 and the difference of the 8th term from the 13thNCERT Exemplar)term is 20. |
|
Answer» Let a and d are first term and common difference of an A . P. nth term = tn = a + ( n - 1 )d----------( 1 ) i) Given t 5 = 19 a + 4d = 19 -------(2) ii ) differnce of the eighth term from the thirteeth term = 20 t 13 - t 8 = 20 a + 12d - ( a + 7d ) = 20 a + 12d - a - 7d = 20 5d = 20 d = 20/ 5 d = 5 Put d = 5 in ( 2 ) a + 4d = 19 a + 4 × 5 = 19 a + 20 = 19 a = 19 - 20 a = - 1 Therefore,a = -1 , d = 5 Required A .P is a , a+ d , a + 2d , a + 3d , ..... -1 , 4 , 9 , 14, 19 , ... |
|
| 72039. |
How many terms of the arithmatic sequence 5,7,9 must be added to get140? |
| Answer» | |
| 72040. |
L. Find:(a) 15th term of the progression 2, 5, 8, 11,. . . .. |
|
Answer» a = 2d = 3so a(n) = a + (n-1)d a(15) = 2 + 14×3 = 2+52 = 54 |
|
| 72041. |
2 Find the value of x for which the numbers(x+-in AP |
|
Answer» I not understand |
|
| 72042. |
Question 10.(a) 6 isrs x and y and 48 is thethe mean proportion between two numbers x and y anproportional of x and y. Find the numbers1 00o ield T 3,972 as compound in |
|
Answer» X/6 = 6/y. ........from 1st conditionso,x=36/y.............1)x/y =y/48........from 2nd condition48x=. y ×yfrom 1)- 48×36/y. =y×yy×y×y=48×36so, y= 12substitute in 1)x=3 hit like if you find it useful |
|
| 72043. |
2y-x=0; 10x + 15y= 105 |
| Answer» | |
| 72044. |
6.Two sides of a triangular field are 85 m and 154 m in length and its perimeter is 324 m. Find(i) the area of the field, and(ii) the length of the perpendicular from the opposite vertex on the side measuring 154 |
| Answer» | |
| 72045. |
river 10 meters deep and 100 meters wide is flowing at themany cubic metre of water runs into the sea per second.4. Arate of 4.5 km per hour. Find how |
|
Answer» Area of cross section of river = 10×100 =1000sqm Volume of water flow in 1 hour = Area of cross section ×Speed of water flow V=1000× 4.5 km/h V= 1000 ×4.5 ×5/18 m/sec V=1250 m^3 1m ^3 =1000 liter 1250 m^3 = 1000×1250= 1250000 liter in 1 sec |
|
| 72046. |
123+314 |
|
Answer» 123+314437 answer 123+313437 is the right answer |
|
| 72047. |
Find the common différence of an A.P. in which thedifference of 16th and 12th term is 16 |
| Answer» | |
| 72048. |
(N.C.E.I5. (i) In a G.P, the third term is 24 and the 6this 192. Find the(a) 10th term(b) 8th term.(N.C.E.(H.P.B. 2(ii) Find the 12th term of a G.P., whose 8th ter192 and the common ratio is 2. (NC.E. |
| Answer» | |
| 72049. |
Arithmatic Progressionlustration 5Find out the nth and 12th term of the series:5' 5' 5' 5on |
|
Answer» a1=3/5d=1/5-3/5=-2/5hence an=a+(n-1)d=3/5-(n-1)2/5=3/5-2/5n+2/5=1-2/5nhence 12th term will bea12=3/5+(11)*2/5=3/5+22/5=25/5=5 |
|
| 72050. |
(v) Numbers x and y both squared and added |
|
Answer» The Equation will bex² + y² |
|