MAHESH TUTORIALS SUBJECT : Maths(012) First Preliminary Exam Model Answer Paper

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1 SET - GSE atch : 0th Std. Eg. Medium MHESH TUTORILS SUJET : Maths(0) First Prelimiar Exam Model swer Paper PRT -. ulike i two distict poits (d) - ad - (d) 0 x does ot exist - (d) x - x - 0 (a) (d) (d) (d) (d) (a) 7 (0, ) rectagle cot (d) (d) 0. l % V rh 00 (d) : πrθ (a) a b 0. 0 Date: Marks : 00 Time: Hrs.

2 PRT - SETION - Solve the followig sums : [ marks]. 6 Now, x + x. x +, x ad Let Here, + ad - + b a b a k, k 0 b k, a -k let Now, c c c p(x). c - a (-) a (-) (-k) k ax + bx + c (-k)x + (k)x + k -kx + kx + k k(-x + x + ) k(x - x - ), k 0 x (i) x (ii) From equatio (i) 7-x Substitutig 7 - x i equatio (ii), we get, x - x - (7 - x) x x x - 7 x + 7 x x x Substitutig x + {(x, )}. i equatio (i), we get 7 7- {(, )} is the solutio of give pair of equatios. Here a, d 6 T 00 a + ( - )d + ( ) + 00 S,? S [a ( )d] S00 OR S00 a l x 0,0

3 00 [ 99 ] 0 [6 + 97] 0 S00 S00 S00. OR Tm T We kow, d m Here T0 T6 + 0 So T0 T6 0 T0 T6 0 6 d. 0 T a + d a + d a+ a 0 a The.P. is,,,,... I, m Now, m + m + m m + m m m I, m + (7) Perimeter of (x, ) m + m Perimeter of is () (, 7) ad (x, ) m (, ) are give poits. P Let P(x, ) be the poit o such that. P 7 m P(x, ) divides from i the ratio. x mx + x m+, m + m+ x + +, x, 6 + x 0, 0

4 x, The co-ordiates of the poit which divides i the ratio : from are (, ). 7. I let m 90. cos k k, sa k, k Now, b pthagoras theorem + (k) (k) + k 6k + k - 6k 9k 9k k k, k, k. Now si L.H.S. (cosec - cot ). cos si si k k k k ta k 7. OR ( cos ) si ( cos ) cos ( cos ) ( cos )( cos ) ( cos ) ( cos ) R.H.S. Here, maximum class frequec is. Modal class 0 Lower limit l 0 class size Frequec of modal class f Frequec of the class precedig modal class f0 Frequec of the class succeedig modal class f Modal Z f f 0 l + f f f c 0 7

5 Hece, mode of the give data is SETION - Solve the followig sums : [ marks] Let the price of sugar be Rs x per kg. The quatit of sugar purchased for Rs. 0 0 Kg x If the price of sugar is purchased b Rs, ew price of sugar Rs ( x - ) per kg The the quatit of sugar received for Rs 0 0 Kg. x Now, if the price of sugar decreases, kg more sugar is received x x Multiplig the equatio b x (x - ), 0 x 0 (x - ) + x (x - ) 0 x 0 x x - x x - x x + x - 0x x ( x + ) - 0 (x + ) 0 (x + ) (x - 0) 0 x + 0 or x x - or x 0 The price of sugar caot be egative x - x 0 Price of sugar is Rs.0 per kg 0. Here, is the height of the buildig. 60m D is the height of the lamp post ostruct DE. The agle of depressio from the top of the buildig to the top of the lamp post is 0. m XD m DE 0. The agle of depressio of the bottom of the lamp post from the top of the buildig is 60. m X m 60. ED is a rectagle DE ad D E I, m 90, m 60. ta ta 60 60

6 60 60 Multiplig b i the umerator ad deomiator (i) 0.7.6m I ED, m E 90, m DE 0. ta D ta 0 E E DE E DE ( DE ) E (from (i) 0 0 ) 0 E 0m Height of lamp post D E - E 60-0 Height of lamp post D 0m Differece betwee height of buildig ad lamp post. - D m (i) Horizotal distace betwee buildig ad lamp post is.6m (ii) height of lamp post is 0m (iii) Differece betwee heights of buildig ad lamp post is 0m.. (i) coi is tossed three times. So outcomes are:hhh, HHT, HTH, THH, HTT, THT, TTH, TTT Total outcomes Let be the evet that atleast two heads are obtaied. outcomes are: HHH, HHT, HTH, THH Number of outcomes Let be the evet that at most oe head is obtaied. (iii) Let be the evet that exactl two heads are obtaied. Outcomes are :- HHT, HTH, THH. Number of outcomes (ii) P() P()

Outcomes are:- HTT, THT, TTH, TTT Number of outcomes (iv) Let D be the evet that more heads tha tails are obtaied Outcomes are : HHH, HHT, HTH, THH Number of outcomes

umulative frequec (cf) lass Frequec (fi) 0-00 6 6 00-00 6 6 + 6 6 00-00 6 + 0 00-00 7 0 + 7 00-00 66 + 66 + 00 00-600 Here 00 00 Nearest value to 00 lies i the class

7 Outcomes are:- HTT, THT, TTH, TTT Number of outcomes (iv) Let D be the evet that more heads tha tails are obtaied Outcomes are : HHH, HHT, HTH, THH Number of outcomes P() P(D). umulative frequec (cf) lass Frequec (fi) Here Nearest value to 00 lies i the class l 00, cf 6, f, c 00 M cf l+ f c lass Hece media of give data is.0. OR. Frequec umulative frequec x 6 + x x x x x + It is give that 6. So, 9 + x + 6, i.e. x + 70 lso, the media is. which lies i the class -. So, the media class is -. 6., l, cf 6 + x, f, c 9

8 cf M l+ f c. 6 x x x x x 0 but x + 70, So, 0 the value of x ad are respectivel 0 ad 0. SETION - Solve the followig sums : [ marks] Data : circle touches the sides,, of at poits respectivel. D x, E, F z. To Prove : Proof : rea of the D, E, F xz(x + + z) circle touches the sides,, of at poits D, E, F respectivel. For the tagets D ad F draw from, D ad F are the poits of cotact. D F x Similarl, for the tagets E, ad D draw from ad the tagets E, ad F draw from, E D, F E z ca be obtaied The side of c F + F z+x a D + D x + b E + E +z I, s + + (z + x) + (x + ) + ( + z) (x + + z) s x++z Now, the area of, s(s a)(s b)(s... ()... ()... () z)(x z x z ( z) x z (z x) (x... () F c) (x E )) D ( Substitutig the values from (), (), () ad ()) (x xz(x z)(z)(x)() z)

9 . Legth of square Radius of sector l r rea of square l l cm rea of sector m 00cm. cm. 00 cm r 60 D cm rea of sector 6 cm rea of remaiig portio of the plate rea of square - rea of sectors cm rea of remaiig portio of the plate is 6 cm. Hemispherical bottles R cm volume of clidrical bottles 6..7cm Total height of top is cm Height of coe h Total height - radius of hemisphere h -.7 h.cm Now, volume of top Volume of coe + volume of hemisphere r cm Hece, bottles are made. OR Radius of coe Radius of hemisphere. lidrical ottles d cm r h 6cm? volume of hemispherical bowls R rh r h + r r [h + r].7.7 [. +.7] [. +.]

10 cm Hece, volume of top is.66 cm 6. i. SETION - D swer the followig questios : [ marks] Data : ostruct PQR with m P 60, m Q ad PQ 6 cm. To ostruct : ostruct P similar to PQR such that the ratio of their correspodig sides is :. Steps of costructio : ostruct PQR with m P 60, m Q ad PQ 6 cm. X Y R 60 P 6 cm Q P P P P P Z ii. Draw PZ makig a acute agle with such PQ such that Z ad X are i differet half plaes of PQ. iii. Select a radius ad draw a arc with cetre which itersects PZ i P. iv. Similarl with PZ cetre P ad the same radius, draw a arc itersectig X at P such that P P P. Similarl cotiue the procedure uptil poit P. v. vi. Draw P Q.

11 vii. viii. ix. 7. Draw P parallel to P Q itersectig PQ i. Draw parallel to QR itersectig PR i. P is the required triagle of desired measures. If a lie parallel to oe of the sides of a triagle itersects the other two sides i distict poits, the the segmets of the other two sides i oe halfplae are proportioal to the segmets i other halfplae. N M Q P P l N Q l M Give : I the plae of, a lie l ad l itersects ad at poits P ad Q repectivel. To prove : Q P Q P Proof : Let QM, ad PN. ostruct Q ad P. rea of a triagle base altitude rea of PQ P QM rea of PQ P QM rea of PQ rea of PQ P QM P QM P P lso rea of PQ Q PN rea of PQ Q PN rea of PQ rea of PQ Q PN Q Q Q PN (i) (ii) PQ ad PQ are havig commo base PQ ad the are lig betwee two parallel lies PQ ad. rea of PQ rea of PQ Q P From (i), (ii) ad (iii). Q P (iii)

12 7. OR Give : Diagoals of covex D iteresect each other at M at right agles. To prove : + D D + Proof : I MD, m M 90. D M + MD [ Pthagoras Theorem] (i) I M, m M 90. M + M ( Pthagoras Theorem) (ii) I M, m M 90. M + M ( Pthagoras Theorem) (iii) D I MD, m M 90. D M + MD ( Pthagoras Theorem) (iv) Now, + D M + M + M + MD M Rearragig the terms M + MD + M + M D + + D D + **** est of Luck ****

MAHESH TUTORIALS SUBJECT : Maths(012) First Preliminary Exam Model Answer Paper

MAHESH TUTORIALS SUBJECT : Maths(012) First Preliminary Exam Model Answer Paper SET - GSE tch : 0th Std. Eg. Medium MHESH TUTILS SUJET : Mths(0) First Prelimiry Exm Model swer Pper PRT -.. () like does ot exist s biomil surd. () 4.. 6. 7. 8. 9. 0... 4 (c) touches () - d () -4 7 (c)