PIONEER GUESS PAPER +1 CBSE. MATHEMATICS Solutions

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L.K. Gupta (Mathematics Classes) www.pioneermathematics.com MOBILE: 985577, 677 PIONEER GUESS PAPER + CBSE MATHEMATICS Solutions TIME: 3:00 HOURS MAX.

1 L.K. Gupta (Mathematics Classes) MOBILE: , 677 PIONEER GUESS PAPER + CBSE MATHEMATICS Solutions TIME: 3:00 HOURS MAX. MARKS: 00 General Instructions: (i) All questions are compulsory. (ii) The question paper consists of 9 questions divided into three sections, A, B and C. Section A comprises of 0 questions of one mark each, Section B comprises of questions of four marks each and Section C comprises of 7 questions of six marks each (iii) All questions in Section A are to be answered in one word, one sentence or as per the exact requirement of the question. (iv) There is no overall choice. However, an internal choice has been provided in questions of four marks each and questions of six marks each. You have to attempt only one of the alternatives in all such questions. (v) Use of calculators is not permitted. You may ask for logarithmic tables, if required. NAME OF THE CANDIDATE PHONE NUMBER L.K. Gupta (Mathematics Classes) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

2 L.K. Gupta (Mathematics Classes) MOBILE: , 677 Section A Question numbers to 0 carry mark each.. Find the domain of x 3 x As 3 x 0x 3 x > 0 3 x x < 0 x 3 f(x) = x 3 x D f = [,3). How many elements has P (A), if A = φ? 3. Let A = {,, 3, } and R = { a, b} : a A,b A,adivides b}. Write R explicitly. R = {(,),(,),(,3),(,),(,),(,),(3,3),(,) }. Find the locus of P if (, 3, 7). (x 3) + (y ) + (z 5) + (x + ) + (y 3) PA + PB = K, where A and B are the points (3,, 5) and + (z + 7) = k x + y + z + x + y + z + 9 6x 8y z + + x + 9 6y z = k PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

3 L.K. Gupta (Mathematics Classes) MOBILE: , 677 (x + y + z ) x y + z + 09 k = 0 5. Differentiate n y = x tanx dy d n = ( ) + dx dx ( ) = + n x tanxw.r.t. x. d dx n x tanx x tanx n n nx tanx x sec x dy = + dx ( ) n x ntanx xsec x 6. Evaluate: We have, lim π x tanx lim π x π x tanx π x π tan + tan h π tan h π + tan.tanh = lim = lim h 0 h h 0 h + tanh tanh tanh tanh = lim = lim h 0 h h 0 h tanh ( ) tanh tanh = lim = lim = =. h 0h tanh h 0 h tanh ( ) ( ) 7. Check the validity of the statements given below by the method given against it. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 3

4 L.K. Gupta (Mathematics Classes) MOBILE: , 677 p : The sum of an irrational number and a rational number is irrational (by contradiction method). Let abeirrationalnumberandbbea rational number. () b + p a = where p,qare co prime q p a = b.() L.H.S = a = An irrationalnumber q R.H.S. = p b Arationalnumber q = It is a contradiction. Therefore, the sum of a rational and irrational number is irrational. 8. Write the negative of the Statements. Q : All cats scratch q : All cats do not scratch or we may say that there is at least one cat which does not scratch. 9. There are three colored dice of red, white and black color. These dice are placed in a bag. One die is drawn at random from the bag and rolled its color and the number on its uppermost face is noted. Describe the sample space for this experiment. Sample Space = {(R,),(R,),(R,3),(R,),(R,5),(R,6) (B,),(B,),(B,3),(B,),(B,5),(B,6), (W,),(W,),(W,3),(W,),(W,5),(W,6)} PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

5 L.K. Gupta (Mathematics Classes) MOBILE: , The standard deviation of 6 observations is. If each observation is multiplied by 3, find the new standard deviation of the resulting observations. standard deviation = n=6 ( ) = ( ) ( ) Var X S.D. Var X = 6 ( ) ( ) NewVar X = 3 6 = ( ) New S.D. = NewVar X = Section B Questions numbers to carry marks each.. tan Sol. o o o o θtan(θ + 60 ) + tanθtan(θ 60 ) + tan(θ + 60 )tan(θ 60 ) = tanθ + tan60 tanθ tan60 tanθ+ tan60 tanθ tan60 LHS:- tanθ + tanθ 0 0 tanθtan60 + tanθtan tanθtan60 + tanθtan60 tanθ+ 3 tanθ 3 (tanθ + 3)(tanθ 3) = tanθ + tanθ 3tanθ + 3tanθ + ( tan 3)( + 3tanθ) = 3 (tan θ+ 3tanθ)( + 3tanθ) + (tan θ 3tanθ)( 3tanθ) + tan θ 3 3tan θ 3 3 tan θ+ 3tan θ+ 3tanθ+ 3tan θ+ tan θ 3tan θ 3tanθ + 3tan θ + tan θ 3 3tan θ 9tan θ 3 3( 3tan θ) 3tan θ 3tan θ =RHS Hence, Proved. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 5

L.K. Gupta (Mathematics Classes) www.pioneermathematics.com MOBILE: 985577, 677 π. Sketch the graph of y = cosx and y = cos x on the same scale.

6 L.K. Gupta (Mathematics Classes) MOBILE: , 677 π. Sketch the graph of y = cosx and y = cos x on the same scale. OR Solve sin x + 3cosx + = 0 sin x + 3cosx + = 0 ( cos x) + 3cosx + = 0 cos x + 3cosx + = 0 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 6

7 L.K. Gupta (Mathematics Classes) MOBILE: , 677 cos x 3cosx 3 = ± cosx = = 3 ( 3) ()( 3) ± 7 cosx = 3 ± 3 3 cosx = cosx =, ;cosx = 5π 5π x = nπ ± ; cosx = cos,n Z Prove by the principle of mathematical Induction: 3 n n n. = (n ) + 3 n n+ P(n) = n. = (n ) + P() = = ( ) + = Let P(k) be true then k k k. (k ) = + (i) We wish to show that P(k+) is true. For this we have to show that 3 k k+ k k() + (k + ) = k + From - LHS = + (k ) + (k + ) k+ k+ k+ k+ = + (k + k + ) = k + = RHS So P(k) is true P(k + ) is true Hence by the principle of mathematical induction, the given result is true n N. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 7

8 L.K. Gupta (Mathematics Classes) MOBILE: , 677. Find the coefficient of Sol. 0 x in the expansion of Suppose x 0 x in the expansion of x th (r + ) term contains 0 r = x 0 Now Tr + Cr ( x ) = c () (x) ( ) (x) 0 0 r 0 r r r r = (x) ( ) c () () 0 3r r 0 0 r r This term will contain 0 3r = = 3r 30 = 3r r = 0 0 x, if So ( 0 + ) th i.e. th term contain Putting r = 0 in (i), we get ( ) ( ) 0 3(0) x C () x c () r 0 x x 0 x in the binomial expansion of 0 x. 0 x x 0 Coefficient of x 0 is 0 0 c 0() 5. The sum of three numbers in G.P. is 56. If we subtract, 7, from these numbers in that order, we obtain an A.P. Find the numbers. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 8

9 L.K. Gupta (Mathematics Classes) MOBILE: , 677 A. T. Q. a + ar + ar = () (a ), (ar 7), (ar ) (ar 7) = a + ar ar = a + ar ar = a + ar 8 From, we have ar = 56 ar 8 3 ar = 8 ar = 6 a = 6 r Put this value in r + = r r r r r = 56 r 6 + 6r + 6r = 56 r 6 + 6r + 6r = 56r 6r 0r + 6 = 0 r 5r + = 0 r r r + = 0 r(r ) (r ) = 0 (r )(r ) = 0 (r ) = 0,r = 0 r =. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 9

10 L.K. Gupta (Mathematics Classes) MOBILE: , 677 Hence a = 8. no. are 8, 6, 3. OR Sum the series to n terms: S = T S n n n = T 0 = T T T = T T n n = T n n 0 = T T [ ] ( n )( n ) T = T ( )( ) ( )( ) n n n n n Tn = + 6 T = + 3 n n n T = + 3 n n n = + 3n 3n = + Sn 3 n 3 n ( + )( + ) ( + ) n n n n n = n 6 ( + )( + ) n n n 3 = n ( n + ) + n n + 3 = n( n + ) + n ( )[ ] = n n + n + n ( ) 3 ( ) = n n + n = n + n = n n + n n PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 0

11 L.K. Gupta (Mathematics Classes) MOBILE: , A solution of 8% boric acid is to be diluted by adding a % boric acid solution to it. The resulting mixture is to be more than % but less than 6% boric acid. if there are 60 litres of the 8 % solution, how many litres of % solution will have to be added? Sol. Let x litre of 80% acid solution be added to 60 litre of % solution of acid Total quantity of mixture = (60 + x) Total acid content in 60 + x litre of mixture 8 = x It is given that acid content in the resulting mixture must be between % to 6% 8 % of (60 + x) < x < 6% of ( 60 + x) 8x < 00 + < ( 60 x) ( 60 x) ( + ) < + < ( + ) 60 x 8x x 60 + x 60 < 8x < x 60 Taking st inequality x + 60 < x 8 x + 60 < x x + 30 < x x > 30...() Taking nd inequality ( ) 8x < x x x < 8 3x x < + 30 orx < 80...(ii) From(i)&(ii) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

12 L.K. Gupta (Mathematics Classes) MOBILE: , 677 More than 30 litres but less than 80 litres of % solution will be added. 7. Solve (+ i)x (5 i)x + ( i) = 0 (+ i)x (5 i)x + ( i) = 0 x = 5 i ± (5 i) 8( i)(+ i) + i 5 i ± 5 0i 8( i + i i ) x = + i 5 i ± 0i 6 + 8i 8 x = + i 5 i ± i = + i a + ib = i a b + iab = i a b = 0 ab =, ab = (a + b ) = (a b ) + a b = 0+ = a + b = ± a = ;b = 5 i ± i + i 5 i + i 5 i + i x =,x = + i + i 6 i x =,x = + i + i PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

13 L.K. Gupta (Mathematics Classes) MOBILE: , 677 (6 i)( i) ( i) x =,x = i 8i 6 8i x =,x = 0 0 x = i,x = i 5 5 x = i, i Express sinα + i cosα in the form r(cosθ+ isinθ): sinα + icosα ( ) ( ) z = + sinα + cos α = cosα + sinα cosα + cosαsinα tanθ = = sin sin sin α + α α tanθ = + tanα cosα tanθ = secα + tanα OR ( ) { ( )} θ tan sec tan = α + α ( ) r = cos tan secα + tanα + isin{tan (secα + tan α} Find the modulus and argument of i and hence express each of them in the polar form: i Let z = i.then z = + ( ) = PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 3

14 L.K. Gupta (Mathematics Classes) MOBILE: , 677 let tan α = Im(z).then Re(z) tanα = = α = π Since the point (, ) lies in the fourth quadrant, therefore the argument of z is given by θ = α = π So, in the polar form of z = i is r (cos θ + i sinθ) = π π cos + isin = π π cos isin 9. Show that the product of perpendiculars on the line x cosθ + y sinθ = from the points a b ( ± a b,0) is b. Line given x y cosθ sinθ a + b = xbcosθ + yasinθ = ab from the point ( ± a ) b,0 a b bcosθ ab a b bcosθ ab b cos θ a sin θ + b cos θ + a sin θ ( ) ( ) a b b cos θ ab a b cosθ + ab a b cosθ + a b b cos θ + a sin θ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

15 L.K. Gupta (Mathematics Classes) MOBILE: , 677 b a b cos θ + b cos θ + a b b cos θ + a sin θ ( ) cos θb a + b + a + b b cos θ a sin θ + ( ( + ) + a ) + ( ) ( ( + ) + a ) b cos θ a b b cos θ a cos θ b cos θ a b b cos θ + a a cos θ ( ( + ) + ) cos θ ( b a ) + a b cos θ a b a ( ) Hence Proved. 0. If y = x a +, prove that xy dy = x a a x dx a x x a y = + a x dy =. a. dx a x x a = 3 ax x A.T.Q x a =.x. +. a x ax 3 a 3 x PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 5

16 L.K. Gupta (Mathematics Classes) MOBILE: , 677 x a x x a = 3 + a x ax x x a x a = + a x a x x a = + = a x Hence Proved OR [byusing(a b)(a b) a b ] Find the coordinates of the point of intersection of the axis and the directrix of the parabola whose focus is (3,3) and directrix is 3x y =. Find also its length of the latus rectum. Directrix 3x y = Slope = 3 = 3/ ( ) x 0 /3 y / 0 Axis is to Directrix and passes through focus n eq of axis : PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 6

17 L.K. Gupta (Mathematics Classes) MOBILE: , 677 y 3 = (x 3) 3/ 3y 9= x + x + 3y = axis and directrix intersect at Q 3x y = () x + 3y = () 3 x 6 y = 8 x + 9 y = 63 5y = 55 8 y = and x = Q, 5 5 Length of latus pectum = (SQ) = Evaluate: Sol. tanx lim sinx π x PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 7

18 L.K. Gupta (Mathematics Classes) MOBILE: , 677 tanx lim π x sinx π Putx = + h π tan + h lim h 0 π sin + h + tanh tanh lim h 0 cosh+ sinh tanh lim h 0 tanh cosh sinh ( )( ) tanh lim h 0 ( ) ( cosh )( + cosh tanh ) sinh ( cosh) + tanh lim h 0 sin h ( tanh) sinh ( cosh ) + sinh lim h 0 sin h ( cosh sinh) sinh ( cosh ) + lim h 0 sinh ( cosh sinh) ( cosh ) ( )( ). What is the probability that a leap year has 53 Sundays and 53 Mondays? PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 8

19 L.K. Gupta (Mathematics Classes) MOBILE: , 677 In a leap year, there are 366 days We have, 366 days = 5 weeks and days Thus, a leap year has always 5 Sundays and 5 Mondays. The remaining two days can be: ) Sunday and Monday ) Monday and Tuesday 3) Tuesday and Wednesday ) Wednesday and Thursday 5) Thursday and Friday 6) Friday and Saturday 7) Saturday and Sunday Let A be the event that a leap year has 53 Sundays and 53 Mondays. A will happen in case () only. Required probability = 7 Section C Question numbers 3 to 9 carry 6 marks each. 3. Prove that 3π 3π (i) cos + x cos x = sinx π π (ii) cos + x + cos x = cos x 3π 3π (i) cos + x cos x = sinx 3π 3π LHS: cos + x cos x = cos (35 o + x) cos (35 o x) o o o o 35 + x + 35 x 35 + x 35 + x = sin sin PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 9

20 L.K. Gupta (Mathematics Classes) MOBILE: , 677 o 70 x = sin sin = sin 35 o sin x 0 0 = sin( )sinx = (sin 90 o cos 5 o + cos 90 o sin 5 o ) sin x = () + 0 sin x = sinx = RHS Hence Proved. π π (ii) cos + x + cos x = cos x π π LHS: cos + x + cos x = cos(5+ x) + cos(5 x) 5+ x + 5 x 5+ x 5 + x = cos cos 0 = cos5 cosx = cosx = cosx = RHS Hence proved.. In a survey of 60 people, it was found that 5 people read newspaper H, 6 read newspaper T, 6 read newspapers I, 9 read both H and I, read both H and T, 8 read both T and I, 3 read all three newspapers. Find: (i) the number of people who read at least one of the newspapers. (ii) the number of people who read exactly one newspaper. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 0

21 L.K. Gupta (Mathematics Classes) MOBILE: , 677 n(h) = 5 n(t) = 6 n(i) = 6 n(h I) = 9 n(h I) = n(t I) = 8 n(h T S) = 3 (i)n(t' H' T) =? [ ] n T Ι H' =? N n(t I H) =? n(t I H) = n(t) + n(i) + n(h) n(t I) n(i H) n(t H) + n(f T H) = = 80 8 = 5 (ii)n(t' H' I) + n(t H' I') + n(t' H I') =? n(t' H' I) = n(t H)' I) = n(i) n(t H) I) = n(i) n((t I) (H I)) = n(i) [n(t I) + n(h I) n(t H I) = 6 [ ] = 6 [] = PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

22 L.K. Gupta (Mathematics Classes) MOBILE: , 677 n(t H' I') = n(t (H I)') = n(t) n(h I) t) = n(t) (n(h T) n(i I) = n(t) {n(h T) + n(i T) n(h I I)} = 6 { + 8 3} = 6 6 = 0 n(t' H' I') = n(h (T I)') = n(h) (n(t I) (H)) = n(h) (n(t H) n(i H)) [ ] = n(h) n(t H) + n(i H) n(t H I) = 5 { + 9 3} = 5 {7} = 8 n(t' H' I) + n(t' H I') + n(t H' I') = = 30 Or n(t) + n(h) + n(i) {n(t H) + n(h I) + n(i I)} + 3n [(T H I) ] { } + 3(3) = = If sec Sol. α (θ+ α) + sec(θ α) = secθ,prove that cosθ = ± cos + = cos(θ + α) cos(θ α) cosθ + = cosθcosα sinθsinα cosθcosα + sinθsinα cosθ PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

23 L.K. Gupta (Mathematics Classes) MOBILE: , 677 cosθ cosα = cos θcos α sin θsin α cosθ cosα = cos θcos α sin θsin α cos θ cosα cos θcos α ( cos θ)( cos α) = cos θ cosα cos θcos α cos α cos θ cos θcos α = + cos θ cosα = cos αcos θ cos θ cosαcos θ= cos α + cos θ cosαcos θ cos α + cos θ = 0 cos θ(cosα ) + ( cos α) = 0 cos θ(cosα ) = + cos α cos θ = cos α cos θ = cosα = + cos θ cosα (cosα + )(cosα ) (cosα ) α = + cos θ cos cos θ cos α = α cosθ cos = ± or If the arcs of the same length in two circles subtend angles 65 o and 0 o at the centre, find the ratio of their radii. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 3

24 L.K. Gupta (Mathematics Classes) MOBILE: , 677 Let r andr be the radii of the given circles and let this of same length (l) 0 0 subtends angles of 65 and 0 at their centers π 3π 65 = = π π 0 = = π π r = S; r = S π π r = r 36 8 r r 3 = c s θ = r n n n n 37 n is positive integer for all n N n n n n 37 n is an integer P() = = which is an integer. P() is true, Let P (k) be true. Then k k k k 37 k is an integer Let = k k k k 37 k = λ;λ Z We small now show that P(k + ) is true for which we have show that PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH

25 L.K. Gupta (Mathematics Classes) MOBILE: , (k + ) (k + ) (k + ) (k + ) 37(k + ) is an integer ( k + 7k + k + 35k + 35k + k + 7k + ) 7 k 5 5k 0k 3 0k 5k k 3 3k 3k ( ) + ( ) ( ) k k k k 37k = k 3k 6k 7k 7k k = integer + k + 3k + 6k + 7k + 7k + k + from.(i) Which is an integer. P(m) is true (m +) is true Hence by the principle of mathematical induction P(n) is true for All n N. i.e n n n 6 n is an integer ( + ) 37k 37 + k + + k 0 7. A committee of 7 has to be formed from 9 boys and girls. In how many ways can this be done when the committee consists of (i) Exactly 3 girls? (ii) At least 3 girls (iii) At most 3 girls? Sol. (i) A committee consisting of 3 girls and boys can be formed in C 9 C ways =! C C = = 50 ways. (ii) A committee having at least 3 girls will consists of (a) 3 girls boys (b) girls 3 boys. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 5

26 L.K. Gupta (Mathematics Classes) MOBILE: , 677 This can be done in C 9 C + C 9 Cways 3 3 = ways 3 3 = ways = 588 ways. (iii) A committee having atmost 3 girls will consists of (i) No girls, 7 boys (ii) girl, 6 boys (iii) girls,5 boys (iv) 3 girls, boys. This can be done in 9 C + C 9 C + C 9 C + C 9 C ways = C C C C C C C ways = = ways = (6 + ) ways = ways = A man arranges to pay off a debt of Rs 3600 by 0 annual installment which form an arithmetic series. When 30 of the installments are paid, he dies leaving one third of the debt unpaid, find the value of the first installment. Let the first installment be=rs x Difference in each installment=rs d Total depth =3600 No of installment=0 n Sn = [ a + (n )d ], PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 6

27 L.K. Gupta (Mathematics Classes) MOBILE: , = [ a + (0 )d ], 3600 = 0[ a + 39d ], 80 = a + 39d...[] now according to question Left loan=\3 rd paid loan=\3 rd s30 = 3600 = 00 3 n Sn = [ a + (n )d ], = [ a + (30 )d ], 00 = 5[ a + 9d ], a + 9d = 60...[] solving () and () a + 39d = 80 a + 9d = d = 0... d = Put d= in equation [] a + 39() = 80 a + 78 = 80 a = 0 a = 5 the firstinstalmentisrs5 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 7

28 L.K. Gupta (Mathematics Classes) MOBILE: , 677 or There are 5 trees at equal distance of 5 meters in a line with a well, the distance of the well from the nearest tree being 0 meters. A gardener waters all the trees separately starting from the well and he returns to the well after watering each tree to get water for the next. Find the total distance the gardener will cover in order to water all the trees. we have, d = distanceof thewellformnearesttree = 0m d = distanceof thewellformnearesttree = (0 + 5)m d = distanceof thewellformnearesttree = (0 + 5)m d = distanceof thewellformnearesttree = (0 + 5)m 5 Therefore, total distance the garden will cover in order to water All the trees= d + d + d d 5 {( 0) + (0 + 5) + (0 + 5) (0 + 5) } meter = [0 + (0 + 5) + (0 + 5) ( )meter = [( (5times)] + [ ]meter = [50 + 5( ) 5 = [ ( + 5)] 5 = [50 + 6] PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 8

29 L.K. Gupta (Mathematics Classes) MOBILE: , 677 = [ ] = [ ] = 875 = 3750meters 9. From the data given below state which group is more variable GorG? Marks Group G GroupG For Group G Class-interval Frequency f i Mid-values x i ui= xi 5 0 fu fu i i i i N= Here N=50 fu i i = 6 fu = 3 i i PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 9

30 L.K. Gupta (Mathematics Classes) MOBILE: , 677 N X = = 5 Mean = X = A + h fu i i = [ ] = = 7.8 ( ) Var( X) = h fu i i fu i i N N ( ) 6 = 00 ( 3) = For Group G Class-interval Frequency f i Mid-values x i ui= xi 5 0 fu fu i i i i N= PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 30

31 L.K. Gupta (Mathematics Classes) MOBILE: , 677 Here N=50, fu i i = 6, fu i i = 366 N X = = 5 Mean = X = A + h fu i i = [ ] ( ) Var( X) = h fu i i fu i i N N ( ) 6 = 00 ( 366) = = = = 3.8 G is more variable. PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 3

L.K. Gupta (Mathematics Classes) www.pioneermathematics.

32 L.K. Gupta (Mathematics Classes) MOBILE: , 677 PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 3

33 L.K. Gupta (Mathematics Classes) MOBILE: , 677 Online services [ (9 th, 0 th, +, +, CBSE, ICSE, IIT-JEE, AIEEE, SAT, NTSE, OLYMPIAD) Ask a teacher. NCERT solutions. Discussion of R.D. Sharma, R.S. Aggarwal etc. Discussion of T.M.H., A. Das Gupta, S. L. Loney, Hall & knight, Thomas Finney etc. Mathematics Olympiad (RMO, INMO, IMO). Objective chapter wise tests with immediate result & review. Objective practice + smart solutions. Vedic mathematics (Meaningful& applicable). Test series for IIT-JEE and AIEEE. IIT-JEE sure shot Tips & Tricks. AIEEE (Twisters, Tips & Tricks). Boards, IIT-JEE, AIEEE previous year papers and solutions. Latest sample papers for boards, IIT-JEE & AIEEE. Correspondence course (9 th /0 th CBSE/ICSE SA-I, SA-II, +, + IIT-JEE, AIEEE) Self study material (For AIEEE + IIT-JEE). Maths projects. Personality development, IQ sharpeners, jokes etc. And many more L.K.Gupta (M.Sc. Mathematics) Introduces (Individual Approach Guaranteed) PIONEER EDUCATION (THE BEST WAY TO SUCCESS): S.C.O. 30, SECTOR 0 D, CHANDIGARH 33

PIONEER GUESS PAPER MATHEMATICS

PIONEER GUESS PAPER MATHEMATICS PIONEER GUESS PAPER 9 th CBSE (SA-II) MATHEMATICS TIME: 2:0 HOURS MAX. MARKS: 80 GENERAL INSTRUCTIONS & MARKING SCHEME All questions are compulsory. The questions paper consists of 4 questions divided