ANSWER KEY & SOLUTIONS


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1 PREHALFYEARLY ASSESSMENT [PHA MATHS SYLLABUS] ANSWER KEY & SOLUTIONS General Instructions:. The question paper comprises of four sections, A, B, C & D.. All questions are compulsory. 3. Section A Q to 6 are mark each 4. Section B Q 7 to are marks each. 5. Section C Q 3 to are 3 marks each. 6. Section D Q 3 to 30 are 4 marks each. 7. Do proper numbering of Answer in copy & Draw Neat Diagram Ans. Non terminating. (the denominator has factor 7 also, except factors and 5 i.e., q m 5 m ) or the prime factorization of q is not in the form m 5 n. Ans. Given, tan 5 p tan b 5 AB and BC 5 According to Pythagoras Theorem AC AB + BC () + ( 5 ) AC 6 cosec 6, sec cosec sec cosec sec Ans 3. Here, ABC ~ LMN Perimeter of ABC AB Perimeter of LMN LM 60 AB 48 8 AB 0 cm Ans 4. Smallest composite number 4 Smallest prime number Hence, HCF of 4 and is. 60Χ8 AB 40 Ans 5. Since (x + a) is a factor of P(x) x + ax + 5x + 0 P (a) 0 a + a(a) + 5(a) a a 5a a a 0 a
2 Ans 6. Yes, Here, a a a b, 4a b b c and b c a a a b c The given system of equations is consistent. a b c Ans.7 DE.5 cm AD BD AD AB 3 DE BC DE BC cm 3 [By Thal s theorem] B D A E C Ans.8 cosec 30 o + 3 sin 60 o 3 4 tan 30 o () Ans. 9 A o sec4a cosec (A 0 o ) Sec4A sec {90 o (A 0 o )} sec (0 o A) 4A 0 o A 5A 0 o A o Ans 0. x y (i) x + y 4 (ii) On adding (i) and (ii), we get x 6 or x 3 From (i), 3 y y a 3, b. Ans. Ans Let the ten s digit be x and unit s digit y Number 0x + y 0x + y 4(x + y) 6x 3y x y Again 0x + y 3xy 0x + x 3x (x) x 6x x (rejecting x 0) x + y y 4 The required number is 4 Ans 3. As two zeros are and (x ) (x + ) x is a factor of the given polynomial. x 3 x  3 x 3x x 6 x 3  x 3x 6 3x x 3 + 3x x 6 (x + ) (x  ) (x 3) For zeros, x + 0, x  0, x +3 0 x , x, x 3 The zeros of given polynomial are, and 3.
3 Ans 4. We have, cot θ + (cosec θ ) + cosec θ + cosec θ 8. [sin θ sin 3 ] 3 Ans 5. Given integers are 56, 96 and 404. First we find the HCF of 56 and ( Ans Applying Euclid s division algorithm, we get x + 40 Since the remainder 40 0, so we apply the division lemma to 56 and ( x Since the remainder 6 0, so we apply the division lemma to 40 and ( 40 6 x + 8 Since 8 0, so we apply the division lemma to 6 and x ( Clearly, HCF of 56 and 96 is 8. Let us find the HCF of 8 and the third number 404 by Euclid s algorithm ( Applying Euclid s division, we get x ( Since the remainder is 4 0. So we apply the division lemma to 8 and x + 0 We observe that the remainder at this stage is zero. Therefore, the divisor of this stage i.e., 4 is the HCF of 56, 96 and 404. As we know, Dividend Quotient x Divisor + Remainder So, we have, x 3 3x + x + (x ) x g(x) + (x +4) x 3 3x + x + + x 4 (x ) g(x) x 3 3x + 3x (x ) g(x) 3 x 3x gx ( ) ( x ) Now we divide x 3 3x +3x by x. We have, x x x x3 3x 3x x x x + 3x  x + x +  x x Hence, g(x) x x
4 Ans 7. Given pair of linear equations is x + 3y 6 x 3y From equation (i), we have table x 0 6 y 0 From equation (ii), we have table x 0 6 y 4 0 According to table we have graph as given (i) (ii) Lines (i) and (ii) intersect yaxis at (0, ) and (0, 4) Ans 8. Ans 9. XP XQ 3 XP XQ 3 We have, PY QZ XY XZ 4 Now, in XPQ and XYZ XP XQ [By BPT] XY XZ X X [Common] XPQ ~ XYZ [SAS similarity] In two similar triangles ar( XPQ) XP XP ar( XYZ XY XY ar( XPQ) 3 XP 3 3 XY 4 ar XPQ 9 x 3 8 cm 6 Area of quadrilateral PYZQ Area of XYZ Area of XPQ 3 cm 8 cm 4 cm Let us assume that 5 3is a rational number. So, 5 3may be written as p 5 3, where p and q are integers, having no common factor except and q 0. q p 5 p p q p Since, 5 p p is a rational number as P and q are integers. p 3 is also a rational number. Which contradicts our assumption. This, our supposition is wrong. Hence, 5 3 is an irrational number.
5 Ans 0. cot 54 tan (90 54) tan 36 tan 33 cot (90 33) cot 57 cos 45 sec 68 cosec (90 68) cosec sin Putting these values in the given expression we get sec 36 tan 36. sin.. cos ec 57 cot 57 sin Ans. 0 Let a be any positive integer. Then it is of the form 3q, 3q + or 3q +, so we have the following cases. Case (i) When a 3q a 3 (3q) 3 7q 3 9(3q 3 ) 9m where m 3q 3 Case (ii) When a 3q + a 3 (3q +) 3 (3q) 3 + 3(3q). +3 (3q) q 3 + 7q + 9q + 9q (3q + 3q + ) + 9m +, where m q (3q + 3q + ) Case (iii) When a 3q + a 3 (3q + ) 3 (3q) 3 + 3(3q). + 3 (3q) q q q + 8 9q(3q + 6q + 4) + 8 9m + 8, where m q (3q + 6q + 4) Hence, a 3 is either of the form 9m or 9m + or 9m + 8 Ans. Given: ABC and DBC are on the same base BC and AD intersects BC at O. To Prove: ar( ABC) AO ar( DBC) DO Construction: Draw AL BC and DM BC Proof: In ALO and DMO, we have ALO DMO 90 and AOL DOM (Vertically opposite angles) ALO ~ DMO (By AASimilarity) AL AO DM DO (i) BC AL ar( ABC) AL AO ar( DBC BC DM DM DO (Using (i)) Hence, ar( ABC) AO ar( DBC DO
6 Ans.3. Let the number of rows be x. and number of plants in each row y Total number of plants xy. A.T.Q. xy (x ) (y + 3) xy xy + 3x y 3 or 3x y 3 () Again xy (x + ) (y 3) xy xy 3x + y 6 or 3x y 6 () On solving () and () y 9 From eqn () 3x 9 3 3x x 4 Total plants x y Total trees 36 (ii) Word problem relating to linear equations in two variables. (iii) Planting more trees will help to save environment. (iv) Yes, such type of activities will help the society active in large scale and create awareness in them about their environment. Ans4. In ABC, DE AC By Basic Proportionally Theorem, we have A D BD BE DA EC B In ABE, DF AE [given] C F E By Basic Proportionality Theorem, we have BD BF DA FE From (i) and (ii), we have BF BE FE EC Ans. 5 Then given system of equations: x + y 5 3x + ky 5 a a 3, b b k (i) c c For unique solutions a b a b 3 k k 6 For no solution a a b b c c 3 k 3 k 6 and k 6 which is not possible.
7 Ans 6. We have 5 x y 6 3 x y Let u and v x y We have 5u + v (i) 6u 3v (ii) Multiplying (i) by 3 and adding it with (ii), we have 5u + 3v 6 6u 3v u 7 u 7 3 Putting the value of u in equation (i), we have v v Here u 3 x 3 x 3 x 4 and v 3 y 3 y 3 y 5 Hence, solution of the given equations is x 4 and y 5 Ans 7. Since α and β are the zeros of the quadratic polynomial f(x) x 5x + 7. ( 5) 5 α + β and αβ 7 Let S and P denote respectively the sum and product of the zeros of the required polynomial. 5 5 Then, S (α +3β) + (3α +β) 5(α +β) 5 and P (α +3β) (3α +β) P 6α +6β + 3αβ 6α + 6β + αβ + αβ 6(α + β + αβ) + αβ 6 (α + β) + αβ P 6 4 Hence, the required polynomial g(x) is given by g(x) k (x Sx + P) 5 or g(x) k x x4, where k is anyzero real number. Ans 8. Given: Two triangles ABC and PQR such that ABC ~ PQR ar( ABC) AB BC CA To Prove: ar( PQR) PQ QR RP Construction: Draw AM BC and PN QR. Proof: ar (ABC) BC AM and ar (PQR) QR PN
8 So, BC AM ar( ABC) BC AM ar( PQR) QR PN QR PN (i) Now, in ABM and PQN, B Q (As ABC ~ PQR) and AMB PNQ (Each 90 ) So, ABM ~ PQN (AA similarity criterion) Therefore, AM AB PN PQ (ii) Also, ABC ~ PQR (Given) So, AB BC CA PQ QR RP (iii) Therefore, ar( ABC) AB AM ar( PQR) PQ PN [From (i) and (iii)] AB AB AB PQ PQ PQ Now using (iii), we get [From (ii)] ar( ABC) AB BC CA ar( PQR) PQ QR RP Let two triangles ABC and PQR are similar triangles. Here, area of ABC 8 cm and area of PQR 44 cm. Side AB 7 cm and let PQ x cm. ABC ~ PQR area of ( ABC) AB area of ( PQR) PQ x x 9 7 x 9x 7 7 x 9 x 36 cm Largest side of the larger triangle 36 cm. Ans.9 By long division method: Remainder x + ax + b On comparing a, b
9 Ans.30 L.H.S. cot A +cosec AcosecA cot AcosecA +cot A cot A +cosec A  cosec A cot A cot A +cosec A  cot A cosec A + cot A cosec A + cot A cosec A + cosec A +cot A [ coseca cota ] cot A cosec A + cosec A +cot A [cosec A +cot A] cot A cosec A + cosa + cosa cosec A +cot A + R.H.S. sin A sina sina
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