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S. Ans.(d) Given, x + x = 5 3x x + 5x = 3x x [(x + x ) 5] 3 (x + ) 5 = 3 0 5 = 3 5 x S. Ans.(c) (a + a ) = 3 a + a = 3 a 3 + = 3 3 3 3 a3 a 3 + a 3 = 0 a 3 + + 3 3 = 0 + 3 3 = 3 3 a3 S3. Ans.(a) a 3 + b 3 + c 3 3abc = (a + b + c)(a + b + c ab bc ca) = (a + b + c)[(a b) + (b c) + (c a) ] = (58 + 60 + 6)[( ) + ( ) + 4 ] = 780 4 = 9360 S4. Ans.(b) Given, pq + qr + rp = 0 qr = pq + rp p p qr + q q rp + r r pq p = p + rp + pq + q q + pq + qr + r = p p + q + r + = p + q + r p + q + r = q p + q + r + r p + q + r r + qr + rp Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
S5. Ans.(c) Given, u 3 + ( v) 3 + ( 3w) 3 = 3 ( )( 3)uvw u + ( v) + ( 3w) = 0 u v 3w = 0 u v = 3w S6. Ans.(c) x + y + z = x y z 3 x + y + z x + y + z + + + = 0 (x x + ) + (y + y + ) + (z + z + ) = 0 (x ) + (y + ) + (z + ) = 0 x =, y = and z = 5() 4( ) + ( ) 5 + 4, 9 = 7 S7. Ans.(b) x + 3 + 8 and x = 3 + 8 3 8 3 8 = 3 8 9 8 = 3 8 x + = 3 + 8 + 3 8 x x + x = 6 x + x = 6 = 36 = 34 S8. Ans.(b) x 4 7x 3 + 7x 7x + 7 = 6x 3 6x 3 x 3 + 6x + x 6x x + 7 = 6x 3 6x 3 6x + 6x + 6x 6x 6 + 7 = 7 6 = S9. Ans.(c) ( y z x y (z + x) ( ) ) 3 + ( z x y ) 3 + ( x y z 3 + ( z (x + y ) 3 ) 3 x (y + z) + ( ) 3 3 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
( y ( y) 3 ) + ( z ( z) 3 ) + ( x ( x) 3 ) = ( y 3 ) + ( z 3 ) + ( x 3 ) = y 3 + z 3 + x 3 (If a + b + c = 0 then a 3 + b 3 + c 3 = 3abc) x 3 + y 3 + z 3 = 3xyz S0. Ans.(a) pqr = r = p q and r = pq Eliminating r from given expression, = + p + q + q = q + pq + + = + q + r + + r + p + q + pq + + p q + p q + q + pq + + q + pq + pq + q + pq = + q + pq + q + pq = S. Ans.(a) (r cos θ 3) + (r sin θ ) = 0 r cos θ 3 = 0 and r sin θ = 0 r cos θ = 3 and r sin θ = r cos θ + r sin θ = 3 + r (sin θ + cos θ) = 4 r = 4 r =, - tanθ = r sinθ r cosθ = 3 And r cosθ = 3 cosθ = 3 r secθ = r 3 r tanθ+secθ r = r secθ+tanθ r = r( 3 ) r + 3 3 + r 3 3 + 3 taking positive value of r = r r + = 4 + = 4 5 4 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
S. Ans.(c) x = a (sin θ + cosθ) and y = b (sinθ cosθ) x = sinθ + cosθ and y = sinθ cosθ a b x a + y b = (sinθ + cosθ) + (sinθ cosθ) = sin θ + cos θ + sinθ. cosθ + sin θ + cos θ sinθ. cosθ = (sin θ + cos θ) = S3. Ans.(a) sin = x y Cos = sin = x = y x y y sec = y x sec sin 69 = sec sin (90 ) = sec cos = y y y x y x y = y (y x ) y y x = x y y x S4. Ans.(b) a cos θ + b sin θ = p a sin θ b cos θ = q On squaring and adding, a cos θ + b sin θ + a b sin θ. cos θ + a sin θ + b cos θ a b sin θ. cosθ = p + q a cos θ + a sin θ + b sin θ + b cos θ = p + q a (cos θ + sin θ) + b (sin θ + cos θ) = p + q a + b = P + q S5. Ans.(a) 5 sec θ + + cot θ + 3 sin θ = 5 cos θ + cosec θ + 3 sin θ = 5 cos θ + sin θ + 3 sin θ 5 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
= 5 (cos θ + sin θ) = 5 S6. Ans.(c) x = a secα. cosβ x α = secα. cosβ Similarly. y b = secα. sinβ, z c = tanα x + y z a b c = sec α. cos β + sec α. sin β tan α = sec α(cos β + sin β) tan α = sec α tan α = S7. Ans.(d) tan θ = e secθ + tan 3 θ. cosecθ = secθ + tan θ. tanθ. cosecθ = secθ + tan θ. sin θ. cosθ sin θ = sec θ. ( + tan θ) = ( + tan θ). ( + tan θ) = ( + tan θ) 3 = ( + e ) 3 = ( e ) 3 S8. Ans.(b) 5 + 8 = 3 8 5 7 + 6 = 7 6 7 3 + 0 = 0 3 9 + = 7 + 6 is the smallest number. S9. Ans.(a) Given: Angle subtended by the pole at the foot of tower = 30 ; Height of tower = H; Distance between tower and pole = d and angle of depression at the foot of the pole. = 60. 6 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
We know that in CBD, tan 30 = h d Similarly, in ABD, tan 60 = H d Dividing equation (i) by (ii), we get tan 30 = h tan 60 H or 3 = h H or 3 3 = h H or h = H 3 (i) (ii) S0. Ans.(c) Given: Height of building = h and angles of elevation = p and q. Let BP be the hill, AD be the building, PC = x and AB = y. Therefore height of hill (BP) = (h + x) PDC = p and PAB = q. We know that in PAB, tan q = h+x y Similarly, in PDC, (i) tan p = x or y = x cot p y Substituting this value of y in equation (i), we get tan q = h + x x cot p or x cot p = (h + x) cot q or x cot p = hcot q + x cot q or x (cot p cot q) = h cot q or x = Therefore height of hill (h + x) = h + h cot q cot p cot q h cot q cot p cot q 7 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
= h cot p h cot q + h cot q cot p cot q = h cot p cot p cot q. S. Ans.(d) Given: Height of aeroplane from the ground AD = km; Initial angle of elevation = 60 and angle of elevation after 0 second = 30. Let A be the initial position of the aeroplane and E be the position of observer. And B be the position of the aeroplane after 0 sec. Therefore AED = 60, BEC = 30 and AB = CD. We know that in AED, or AD DE DE = 3 or DE = 3 Similarly, in BEC, or DE + CD = 3 or = tan 60 = 3 BC DE + CD = tan 30 DE + CD = 3 or CD = 3 DE = 3 3 = 3. Therefore speed of the aeroplane per hour = = 60 60 = 40 3 km/h. 3 0 S. Ans.(b) Length of median of triangle= 3 8 = 4 3 radius of the in circle = 3 4 3cm = 4 3 cm Area of the in circle = π ( 4 3 ) cm = 6 3 π cm radius of circumcircle = 3 4 3 = 8 3 cm Area of the circum circle = π ( 8 3 ) Distance AB = 3 0 Time taken to travel 60 60 = 64 3 π cm 8 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
Area of the required region = ( 64 3 π 6 3 π) cm = 48π = 6π cm 3 6 = = 35 7 7 = 50 7 cm S3. Ans.(b) Circumference of circle= π r = π 3 =6 π cm Area of circle = πr = π 3 3 = 9π cm Required ratio= 6 π: 9 π= :3 S4. Ans.(d) Let the length of the rectangle be x units and breadth be y units. Perimeter of rectangle = (x + y)cm According to the question, x x + y = 5 6 x x + y = 5 8 x + y x = 8 5 x x + y x = 8 5 y x = 8 5 y x = 3 5 x y = 5 3 S5. Ans.(b) Length of semicircular sheet (ACB) = πr = 4 = 44 cm 7 Slant height of cone = l =4 cm Circumference of the base of the cone = πr = 44 7 r 44 = 44 7 r r = 7 cm h = l r = 4 7 9 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
= 7 3 cm = 7.73 = cm S6. Ans.(a) Volume of bucket = 3 πh(r + r + r r ) = 3 7 45(8 + 7 + 8 7) = 3 7 = 3 7 45(784 + 49 + 96) 45 09 = 4850 cm3 S7. Ans.(d) APQ ~ ABC AP = PQ PB BC = PQ BC BC = PQ BC = (PR + RQ) BC = 6 BC = cm S8. Ans.(b) D is mid point of BC and E is mid point of AD. Draw a line parallel to BF from D to G. G is a point on AC. 0 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
DG BF DGC ~ BFC CD : DB = CG : GF CG = GF Now AEF ~ ADG AE : ED = AF : FG AF = FG AF = AF = AF FC (FG+CG) AF AF : FC = : S9. Ans.(d) AG = BC GD = BD = DC Let BGD = x GBD = x BDG + CDG = 80 80 x + 80 y = 80 x + y = 90 S30. Ans.(a) OP = 5 3 OP = 4 OQ = 5 4 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
OQ = 3 PQ = 7 cm S3. Ans.(b) ADO = ACD = 90 OD = OC (radius of circle) AO is common in both triangle ADO and ACO ADO ACO AOD = AOC = x (say) In same way we can say, BOC = BOE = y (say) And, OD XY OE X Y XY X Y DOE = 80 = (x + y) (x + y) = 0 AOB = (x + y) = 90 S3. Ans.(d) Sides of the trapezium = x and 3x cm (x + 3x) = 480 5x = 480 6 = 80 x = 80 5 = 6 Longer side = 6 3 = 48 cm Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
S33. Ans.(b) COD = 0 BAC = 30 CAD = 0 = 60 (angle made on other part of circle is half of angle made at centre by same are) BAD = 90 BCD= 80-90 = 90 (cyclic quadrilateral) S34. Ans.(d) 360 n 360 n + = 6 60 ( n n + ) = 60( + ) = (n )(n + ) = n + n n + n 8 = 0 (n + 4)(n 3) = 0 n = 4 or n = 3 S35. Ans.(b) Each interior angle of a regular polygon= 80 3 5 = 08 Each exterior angle = 80 08 = 7 So, number of sides = 360 7 = 5 S36. Ans.(d) We have 3x + 3y = 6 Or 3y = 3x + 6 Or, y = 3 x + Comparing the above equation with y = mx + c We get m = 3 and c = Hence slope is ( ) and intercept on the y-axis is. 3 3 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
S37. Ans.(c) We have m = 5 4 and (x, y ) = (, 3) The equation of the line as point slope form is y y = m(x x ) Or y ( 3) = 5 (x ) 4 Or y + 3 = 5 (x ) 4 Or 5x 4y = S38. Ans.(b) (a b) = a ab + b x 4 x + K = (x ) x + K K = () = S39. Ans.(c) Given, x = 6 = 4 6 + P = 3 8 + P = P = 4 P = S40. Ans.(c) cos α cos β = a cos α cos β = a sin α = sin β a sin α = a ( sin β) b sin β = a a sin β a = b sin β a sin β a = (b a ) sin β sin β = a b a = a a b S4. Ans.(c) x = a (sin θ + cosθ) and y = b (sinθ cosθ) x = sinθ + cosθ and y = sinθ cosθ a b x a + y b = (sinθ + cosθ) + (sinθ cosθ) = sin θ + cos θ + sinθ. cosθ + sin θ + cos θ sinθ. cosθ = (sin θ + cos θ) = 4 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
S4. Ans.(c) As we know that median divide area is two equal parts : Area of BDE = 0 cm area of ADB = 40 cm area of ABC = 80 cm S43. Ans.(b) D is mid point of AB and M is mid point of AP DM BP Hence DM = BP E is mid point of BC and N is mid point of PC EN BP EN = BP DM : EN = : S44. Ans.(d) CD = radius = OC = OD COD = 60 CAD = COD 60 = 30.(i) Now ADB = 90 [Angle of semicircle] ADP = 80 90 = 90 Now in ADP, P = 80 ( PAD + ADP) 5 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
80 (30 + 90 ) = 60 S45. Ans.(b) OB + OC = BC OC + OD = CD OD + OA =AD OA + OB = AB (OB + OA + OD + OC ) = AB + BC + CD + DA AB + CD = BC + DA S46. Ans.(c) Sides are of a right angle triangle Orthocentre will be point B And circumcentre will be mid point of AC which is D BD = AD = CD (circumradius) BD = 0 cm S47. Ans.(c) x + + x a = x + x a = x + + x x 4 4 a = x + x 4 + x x 4 4 ax = x + x x 4 = x + x 4 = x + x x 4 6 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
a ax = x + x x 4 a ax = S48. Ans.(b) x + x x 4 x = = 3 and y = + 3 3 = + 3 8xy(x + y ) = 8 ( + 3 ) ( ) (4 + 3 4 3 + 4 + 3 + 4 3) 3 = 8(4) = S49. Ans.(a) cos(40 θ) sin(50 + θ) + cos 40 + cos 50 sin 40 + sin 50 sin[90 (40 θ)] sin(50 + θ) + cos 40 + cos (90 40 ) sin 40 + sin (90 40 ) sin(50 + θ) sin(50 + θ) + cos 40 + sin 40 sin 40 + cos 40 0 + = S50. Ans.(b) cot cot 38 cot 5 cot 60 cot 78 (cot cot 78 )(cot 38 cot 5 ) (cot 60 ) [cot cot (90 )] [cot 38 cot (90 38 )] cot 60 (cot tan ) (cot 38 tan 38 ) cot 60 3 7 Adda47 No. APP for Banking & SSC Preparation Website:store.adda47.com Email:ebooks@adda47.com
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