Abhilasha Classses. Class X (IX to X Moving) Date: MM 150 Mob no (Set-AAA) Sol: Sol: Sol: Sol:

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Class X (IX to X Moving) Date: 0-6 MM 0 Mob no.- 97967 Student Name... School.. Roll No... Contact No.

1 Class X (IX to X Moving) Date: 0-6 MM 0 Mob no Student Name... School.. Roll No... Contact No If = y = 8 z and + + =, then the y z value of is (a) 7 6 (c) 7 8 [A] (b) 7 3 (d) none of these Let = y = 8 z = k. Then, = k /, = k /y and 8 = k /z. Now, 8 = k /z.k /y = k / y z + + = + = 7 6 =. + =. If ( 3) (9) = 3 α 3 3, then α equals (a) (b) 3 (c) (d) d y ( 3) (9) = 3 α 3 3, (3) / (3) = 3 α 3 3/ (3) 3/ = 3 α+3 z y z =. The solution of : (i) = ( ) b b c c + + = (ii) ) = 9and (iii) = 0 are respectively (a) 9, & (c), & 9 (b) 9, & (d), 9 & () ( ) [A] (i) = ( 3/ ) ( ) = 3/ = 3 / = 3 = 9 (ii) = = = = 0 =. (iii) = = 36 b c a a 3/ a ( + ) = 36 = = c 3 b a 3/ a b c a b c α = (after comparing 3. The epression : (b a) + (c a) + (a b) + (c b) + (b c) + (a c) is equal to (a) a b c (b) (c) 0 (d) none of these [B] Given Ep= =. If = 0, then is equal to (a) or 0 (b) or 3 (c) or 9 (d) or P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

2 [A] = 0 y 0y + 9 = 0, where y = 3 (y 9) (y ) = = 0 y = 9 or y = 3 = 3 or 3 = = 3 0 = or = If t t + = 0, then the value of (t 3 + t3) is- (a) (b) 8 (c) (d) 6 [c] t t + = 0 t + t = (t 3 + t 3) = (t + t )3 3( t + t ) = 6 = 7. If /3 + y /3 + z /3 = 0, then (a) + y + z = 0 (b) ( + y + z) 3 = 7 yz (c) + y + z = 3yz (d) 3 + y 3 + z 3 = 0 [B] Using a + b+ c = 0 a 3 + b 3 + c 3 = 3abc, we get : /3 + y /3 + z /3 = 0 + y + z = 3 /3 y /3 z /3 9. If a + b + c = 0, then a + b + c is (a) (ab + bc + ca) (b) (ab + bc + ca) (c) 0 [B] (a + b + c) = 0 (d) a bc a + b + c + (ab + bc + ca) = 0 a + b + c = (ab + bc + ca) 0. The value of n for which the epression n 8 + becomes a perfect square, is (a) (b) 6 (c) 8 (d) (b) 9 + n 8 + This factors into; =(3 +m+)(3 +m+) =9 +6m 3 +(m +) +m+ comparing coefficients, m= -8, 6m= -. m= - (m +)=n n=( +) 8. ( ) is divisible by (a) both ( ) & ( + ) (b) ( ) but not by ( + ) (c) ( + ) but not by ( ) (d) neither ( ) nor ( + ) [b] () put = and =- we get -+- = not satisfy () obviously nd will true. P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web: =6.. If y ( y) + yz (y z) + z (z ) = k ( y) (y z) (z ), then which of these are not true : (a) k = (b) k = 0 (c) k = (d) k = [c]y ( y) + yz(y z) + z (z ) = k ( y) (y z) (z ) y y + y z yz + z z = ( ky + kz) + y (k kz) + z ( k + ky) (y z) + y (z ) + z ( y) = ( ky + kz) + y (k kz) + z ( k + ky) compare the coefficients of, y and z

3 y z = k (y z) k = similarly for y and z k = (A), (B) and (D) are the the correct answers. a = 7+ 3 b = If y coordinate of a point is zero, then this point always lies (a) in I quadrant (c) on - ais [c]obviously lies in ais (b) in II quadrant (D) on y ais 3. If a = + 3 and b = 3, then - a b is equal to (a) (b) (c) 8 3.[d] a=+ 3 = a = (- 3) 3 ( a ) = 7-3 ( b ) = 7+ 3 a - b = -8 3 = If a = and b = (d) 8 3 (a) (b) (c) 8 (d) 8 [a] a= and then a + b is equal to a + b =. In PQR, PQ = PR, QPR is equal to (a) 0º (b) 30º (c) 0º (d) 0º [a] Q= R=(PQ=PR) Now, R=80-00 R=80= Q P+ Q+ R=80 (A.S.P) P =80 P=80-60 P=0 6. If the sides of a triangle are in the ratio : : 3, then the respective altitudes on them will be in the ratio (a) 3 : : (b) : : 3 (c) 0 : : 6 (d) : : 0 -[d] Let the sides are =,, 3 and altitude be h,h, h 3. Area at = h () or area of = h () or area of = 3 h 3 (3) from equation (), (), (3), h : h :h 3 : = : : 3 3 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

4 Multiply by 60 = : : 0 diagonal s of ll gm also divide each other in two equal parts it mean median AO and OC lies on diagonal AC as we know centroid divides the median : 7. If ABCD is a rectangle, E, F are the mid points of BC and AD respectively and G is any point on EF, then GAB equals : (a) (ABCD) (b) 3 (ABCD) then, AE = OA 3 = AC 3 () (c) (ABCD) -[c] Ar GAB = [ ABEF] (d) 6 (ABCD) OE = OA = AC 3 6 Similarly CF = AC 3 () and OF = AC 6 Ar ABEF = [ABCD] Ar GAB = [ABCD] EF = AC 3 from () and (3) EF = AE (3) 8. If a rectangle and a parallelogram are equal in area and have the same base and are situated on the same side, then the quotient: Perimeter of rectangle Perimeter of gm is : (a) equal to (b) greater than (c) less than (d) indeterminate [c] Let ABCD is a Rectangle and ABPQ is a ll gm Perimeter of Rectangle = [AB+BC] Perimeter of ll gm =[AB+BP] But BP>BC (Hypotenus is the longest side) Perimeter of Rectangle < Perimeter of ll gm Perimeter of Rectangle Perimeter of ll gm = Less than 0. If one of the angles of a triangle is 30, then the angle between the bisectors of the other two angles can be (a) 0 (b) 6 (c) (d) A+B+C In OBC B + C B+C = 0 + BOC =80 + BOC =80 BOC =80 - BOC = 9. If ABCD is a parallelogram and E, F are the centroids of s ABD and BCD respectively, then EF equals (a) AE (b) BE (c) CE (d) DE. Angles of a triangle are in the ratio : : 3. The smallest angle of the triangle is (a) 60 (b) 0 (c) 80 (d) 0 [a] As we know medians divide the corresponding sides in two equal part and [b] Let angles are =,, 3, ++3+ = 80 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

5 9 = 80 = 0 Smallest angle = 0 (a) Changes (b) Remains the same (c) Changes in case of multiplication only (d) Changes in case of division only (b) = 0. The graph of the linear equation + 3y = 6 cuts the y-ais at the point (a) (, 0) (b) (0, 3) (c) (3, 0) (d) (0, ) [d] +3y = y = 3. Any point on the line y = is of the form (a) (a, a) (b) (0, a) (c) (a, 0) (d) (a, a) (a) Obviously line passes through (a,a). The value of is - (a) (b) 6 (c) 8 (d) 0 6. If bisectors of A and B of a quadrilateral ABCD intersect each other at P, of B and C at Q, of C and D at R and of D and A at S, then PQRS is a (a) Rectangle (b) rhombus (c) parallelogram (d) Quadrilateral whose opposite angles are supplementary -[d] In Qaud ABCD In APR A+ B A+ B = 360 ( C+ D) A+ B + [APB] = 80 = 80 - ( C+ D) APB = C+ D Similarly DRC = A+ B PSR = C+ D (a) = = PQR = A+ B APB + DRC = A+ B = A+ B+ C+ D = 360 = 80 + C+ D = = = 6 =.. If we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation : Similarly, PSR + PQR = 80 It is clear that it is quad whose opposite s sum is 80 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

6 OCD = 0 7. D and E are the mid-points of the sides AB and AC respectively of ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is (a) DAE = EFC (b) AE = EF (c) DE = EF (d) ADE = ECF. [c] AE = EC = We need according to given option only DE = EF 8. Diagonals of a parallelogram ABCD intersect at O. If BOC = 90º and BDC = 0º, then OAB is (a) 90º (b) 0º (c) 0º (d) 0º [c] In DOC CDO + DOC + OCD = OCD = 80 OCD = 0 OCD = [Alt +Int s] = 0 OCD = [Alt +Int s] = If the length of a chord of a circle is 6 cm and is at a distance of cm from the centre of the circle, then the radius of the circle (in cm) is : (a) (b) 6 (c) 7 (d) 3 [c] [By Euclid fifth postulate] 3. If a straight line falling on two straight lines makes the interior angles on the same side of it, whose sum is 0, then the two straight lines, if produced indefinitely, meet on the side on which the sum of angles is (a) less than 0 (b) greater than 0 (c) is equal to 0 (d) greater than 80 [d] Let original v = πr h r = 3 r then h = h New volume v = π( r 3 ) h v v = πr h π r 9 h =9 h h h = 9h 9. In the adjoining figure, if QPR = 67º and SPR = 7º and RP is a diameter of the circle, then QRS is equal to (a) º (b)3º (c) 67º (d)8º [c] In DOC Q R CDO + DOC + OCD = OCD = 80 P S 3. 7 Cards numbered,, are put in a bo and mied thoroughly. One person draws a card from the bo. then the probability that the number on the card is Divisible by 3 and both is (a) 3/ 7 (b) /7 (c) / 7 (d) 0/ 7 (b) If a number is divisible by both 3 and, then it is a multiple of 6. In cards bearing number,, 3..., 7 there are only cards which bear a number divisible by 3 and both i.e. by 6. These cards bear numbers 6 and Favourable number of elementary events = 6 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

7 Hence, P (Getting a card bearing a number divisible by 3 and ) =./ A card is drawn at random from a well-shuffled pack of cards. Find the probability that the card drawn is neither a red card nor a queen. (a) 6/ 3 (b) 7/3 (c) / 3 (d) / (a) here are 6 red cards (including red queens) and more queens are there. Thus, we have to set aside 8 cards. And, we have to draw card out of the remaining ( 8) = cards. Required probability = 6/3 3. The mean of first three terms is and mean of net two terms is 8. The mean of all the five terms is (a). (b).0 (c). (d).6 (d) mean is then sum of 3 numbers = 3= &the sum of numbers= 8 = 36 mean of numbers = +36 = 78 =.6 3. The radius of a wire is decreased to one-third. If volume remains the same, the length will become: (a) time (b) 3 times (c) 6 times (d) 9 times [d] 70- = XAC XAC = In a PQR, PS is bisector of P and Q = 70 0, R = 30 0, then (a) QS > PQ > PR (c) PQ > QS > SR [b] in PQR QPR = 80 0 ( A.S.P) in QPS QPS = 0 0 & QSP = 70 0 then Q = PSQ= 70 0 (b) QS < PQ < PR (d) PQ < QS < SR PQ = PS (sides opposite to equal angle ) and QS < PQ (greater angle formed greater side ) now in PSR PSR = 0 0 & PRS = 30 0 then PR> PS Or PR> PQ it is clear that QS < PQ < PR 38. The median of 0,,, 9, 8,, 6 is (a) 0 (b) (c) (d) [a] 6, 8, 9, 0,,, median 0 new radius = r 3 A.T.Q π( r 3 ) a = π(r) h a 9 = h a = 9 h and new height be a 36. X lines in the interior of BAC. If BAC = 70 0 and BAX = 0 then XAC = (a) 8 0 (b) 9 0 (c) 7 0 (d) 30 0 [a] BAC = BAX+ XAC 39. In ia right angle triangle ABC is right angled at B. Given that AB = 9 cm, AC = cm and D, E are the mid-points of the sides AB and AC respectively, then the area of ADE= (a) 67. cm (b) 3. cm (c) 7 cm - [b] Area of ADE = Ar [ ABC] = [ 9] (d) data insufficient 7 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web: = 7

8 =3.cm 0. The sum of the interior angles of polygon is three times the sum of its eterior angles. Then numbers of sides in polygon is (a) 6 (b) 7 (c) 8 (d) 9 [c] Sum of all s = 80 (n-) A.T.Q 80(n-) = (a) 8 (b) 6 (c) (d) 3 -[d] A.M = = 7 () = =00 () From () and () Ecluding no. = 3-00 = 3 = 7 n - = 6 n = 8 sides. In ABC, AD is a median and P is a point is AD such that AP : PD = : then the area of ABP = 3. Any cyclic parallelogram is a. (a) rectangle (b) rhombus (c) trapezium (d) square [a] It is clear that is rectangle +=80(opposite angle off cyclic Quadrilateral) (a) Area of ABC (c) 3 Area of ABC (b) 3 Area of ABC (d) 6 Area of ABC =90 Similarly y = 90 [c] Construct BQ median Ar ABQ = Ar( QBP) = Ar (PBD) = ABD 3 But Ar ABD = [ABC] Ar [ABP] = [ABD] = 3 [ABC] = [ABC] 6. The locus of the centre of all circles of given radius r, in the same planes, passing through a fied point is : (a) a point (b) a circle (c) a straight line (d) two straight lines -[b] obviously a circle.. In a cyclic quadrilateral if A - C = 70 0, then the greater of the angles A and C is equal to : (a) 9 0 (b) 0 0 (c) 0 (d) 0 - [c] A+ C =80 [opposite s cr A+ C =70. The arithmetic mean of numbers is 7. If one of the numbers be ecluded, their mean is. The ecluded number is : A =0 A = 8 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

9 % increase = 30a 6a In the sides of a triangle are doubled, then its area (a) remains the same (b) becomes doubled (c) becomes three times (d) becomes four times -[d] Area become times. 7. Inside a triangular garden there is a flower bed in the form of a similar triangle. Around the flower bed runs a uniform path of such a width that the side of the garden are double of the corresponding sides of the flower bed. The areas of the path and the flower bed are in the ratio : (a) : (b) : (c) :3 (d) 3 : -[d] Area of the path = A -A =3A Ratio = 3A A =% 9. The ratio of three numbers is 3 : : and the sum of their squares is 0. The sum of the numbers is : (a) 30 (b) 0 (c) 60 (d) 90 -[c] (3) + () +() = =0 0 = 0 = 0 0 X= Sides =, 0, Sum of no = +0+ =60 =3: 8. The percentage increase in the surface area of a cube when each side is increased to 3 times the original length is (a) (b) 00 (c) 7 (d) [d] Surface area of cube = 6a after increase side = 6( 3 a) = 9 6a 0. The difference between a number and its twofifth is 0. What is 0% of that number? (a).7 (b) 8 (c) 0 (d) none of these [b] let the number be - = 0 3 = 0 = 70 = 80 its 0% = 0 80= 8 00 Increases part = a -6a = a a = 30a 9 P a g e CONTACT- Study Center- CL-69 D. D. Nagar Gwalior, MOB web:

Class IX Chapter 8 Quadrilaterals Maths

Class IX Chapter 8 Quadrilaterals Maths Exercise 8.1 Question 1: The angles of quadrilateral are in the ratio 3: 5: 9: 13. Find all the angles of the quadrilateral. Answer: Let the common ratio between