then the value of a b b c c a a b b c c a a b b c c a a b b c c a

Transcription

1 E OF PE (VNE SOLUTION OE & 4) 1. If x = a b,y b c, z = c a a b b c c a 1 x1 y1 z then the value of 1 x1 y1 z is () abc () a b c () 1 () 1 () a b b c c a x, y, z a b b c c a a b b c c a x1 y1 z a b b c c a 1 x1 y1 z a b b c c a a b b c c a a b c a b b c c a 1 b c a a b b c c a. If p = ( a) then a + 6ap + p 8 is () () 1 () () () P a a P a P 8 a P ap a P 8 a P ap 8 a 6aP P 8 x y z. If b c b c a c a c a b a b a b c then x + y + z = () 1 () () 1 () () x y z k b cb c a c ac a b a ba b c x y z k b cb c a c ac a b a ba b c k b c a b c c a bc a a b ca b 4. The angle and are right angles. If 9 cm and 16cm the length of is () 1 cm () 16 cm () 15 cm () 5 cm ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 1

2 E OF PE (VNE SOLUTION OE & 4) () & are 9 9 x 16 Similar x 9 16 x 1cm 9 5. s salary is reduced by 1%. In order to have the salary back to the same original position it must be raised by 1 () 8% () 1% () 11 % () 1.5 % 9 () Let salary of be x. It is reduced by 1%, therefore salary.9x Now to have salary back to the same original position increment should be.1 x % raised.1x %.9x 9 6. In the adjoining figure O, O are angle bisectors of external angles of. Then O is () () 9 () 1 1 P O Q 9 () 18 () P O Q ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE #

3 18 O 9 18 O 9 O E OF PE (VNE SOLUTION OE & 4) 7. In the adjoining figure M and N is the bisector of. If MN is 7 and 5 then () () 17.5 () () M N 7.5 () M N N 75 MN In parallelogram, 1 cm. The altitudes corresponding to the sides and are respectively 9 cm and 11 cm. Find. 1 cm N 9cm 11 cm M () 18 cm 11 () 18 cm 1 () 99 cm 1 () 18 cm 17 ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE #

4 E OF PE (VNE SOLUTION OE & 4) () 1 N 9 x 11 M rea 19 11x x cm Inside a triangular garden there is a flowerbed in the form of a similar triangle. round the flower bed runs a uniform path of such a width that the sides of the garden are double the corresponding sides of the flowerbed. The area of the path and the flowerbed are in the ratio of () 1 : 1 () 1 : () 1 : () : 1 () a a a a a a rea of flowerbed a 4 a 4 4 a 4 1 rea of Path a Ratio : and start walking from the same point. goes North and covers kms, then turns right and covers 4kms. goes West and covers 5kms, then turns right and covers kms. How far apart are they from each other? () 1kms () 9kms () 8kms () 5kms () (-5, ) (4, ) istance 9kms ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 4

5 E OF PE (VNE SOLUTION OE & 4) irections for Q.No.11: There are three circles with for number out side them. While in the first two circles a number has been filled according a rule, in the third circle the mark? Is given. Out of the five alternatives given below the circles one alternative has the number which should occupy? ccording to the rule Identify it and encircle its serial number on the answer sheet against the question ? () 4 () 5 () () 1 () ? If we add the age of three brothers Sunil, Sanjay and Sonu, then it becomes 6 years today. If 6 year ago the Sonu was of half the age of Sanjay and 1/ rd to the age of Sunil, then find out the present age of Sanjay. () () 18 () 14 () () x y z z 6 y 6 x 6 z 6 1 y 6 y 6 x y 6 6 y 6 y 6 y 6 y 6 y 1. If tan and tan are the roots of x px + q =, cot and cot are the roots of x rx + s =, then rs is necessarily () pq () 1/pq () p/q () q/p () tan x px q tan ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 5

6 tan tan P. (1) tan tan q.. () x rx s cot cot cot cot r... () cot cot s.. (4) 1 1 From () cot cot r tan tan tan tan r tan tan p r tan tan p p rs tan tan q E OF PE (VNE SOLUTION OE & 4) log1 x 14. The product of all real roots of the equations x = 1 is () 1 () 1 () 1 () 1 1 () log 1 x x 1 1 x 1, 1 Product of roots Suppose x and y are inversely proportional and positive. If x increases by p%, then y decreases by p () p% () % () 1 % () 1p % 1 p p 1 p () k x y k y x k 1k y' px x x 1 p 1 k 1k k p y y' x x 1 p x 1 p 1 P %derease % 1 P 16. cylindrical box of radius 5cm contains 1 solid spherical balls each of radius 5cm. If the topmost ball touches the upper cover of the box, then the volume of the empty space in the box is () 5 cm () 5 cm p/., () 5 cm () 5 cm () Volume of cylinder r h Volume of balls 1 5 ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 6

7 E OF PE (VNE SOLUTION OE & 4) 5 Volume of empty space: 5 cm If the line y = mx + c passes through the points (, ); (4, ) then its y intercept is () 4 () () 4 () () y mx c c (Passes through, ) 18. onsider the figure below: If O : O :1,, Then the value of a is () 1 () 11 () 1 () 14 () We have: O : O :1 O x, O x (assume) O O O 18 9 x x 18 x 9 x O 6, O O O 6 s Now, M O 18 LP a 6 18 a The expression is written in a simplified form as a b 5 c. The value of a b c is. () () 4 () () 7 () To rationalize the denominator of the given expression, we proceed as follows: ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 7

8 E OF PE (VNE SOLUTION OE & 4) Now, we do a second rationalization step: omparing with the expression given in the question, we have: a, b, c a b c What is the value of x in the following figure? () 7 () 5 () 9 () 8 () We note that 4. Using the angle sum property in, we have: x The length of the longest pole which can be put in a room of dimension 1m 1m 5m is () 1 m () 15 m () m () 5 m () onsider the following figure: ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 8

9 E OF PE (VNE SOLUTION OE & 4) The length of the diagonal of this cuboid can be calculated as follows: l b h Thus, the length of the longest pole will be the length of the diagonal, which will be. If m. xy, then the point, x y may lie in () quadrants I and III () quadrant II and IV () quadrants I and II () quadrants III and IV () The point x, y may lie in quadrants I or III, because any point in these two quadrants will have coordinates of the same signs, so that their product will be positive.. The abscissa of a point is positive in the () I and II quadrant () I and IV quadrants () I quadrant only () II quadrant only () The abscissa of a point is positive in the I and IV quadrants 4. In the following figure, E : The length of is () 7.5 cm () 7.5 cm () 7.75 cm () 8.5 cm () Using the PT, we have: E E.5cm.75cm cm E E 4.5cm Thus, E E 8.5cm 5. The area of an equilateral triangle is 16 cm. Its perimeter is () 16 cm () 4 cm () 16 cm () 4 cm () ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 9

10 If l is the length of each of the triangle s sides, we have: 16 8cm 4 l l Thus, the perimeter of the triangle is l, or 4 cm E OF PE (VNE SOLUTION OE & 4) 6. onsider the following figure in which (a diameter of the circle) is of length 5 cm, while = 7cm. The area of the shaded region is closed to () 16 cm () 17cm () 18cm () 19cm () Since is an angle is a semi-circle, We have 9. Thus, 4cm Now, the area of the (upper) semi-circle on is r 45.cm The area of is cm Thus, the area of the shaded region is 1 161cm 7. If the arithmetic mean of 7, 8, x, 11, 14, is x, then x is () 9 () 9.5 () 1 () 1.5 () sum of the quantities rithmetic mean Number of the quantities 8. For cone of radius R, the ratio of its height to radius is : 1. The cone is horizontally halved: The ratio of the volume of the bottom half to that of the top half will be () 5 : 1 () 6 : 1 () 7 : 1 () 8 : 1 () ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 1

11 E OF PE (VNE SOLUTION OE & 4) The height of the cone is H R. Now, the radius and height of the top (conical) half of the cone is R H and R. We have: 1 1 Ventire cone R H R R R 1 R H 1 Vtopconical half R 1 V V V bottom half entirecone top conicalhalf 1 7 R R R 1 1 The required ration is 7 V R bottom half 1 7 V 1 topconicalhalf R 1 9. The surface area of a sphere is 14cm. Its radius is approximately equal to () 5 () 7 () 9 () 11 () 4 r 14 r r 5 r 5 The radius of the sphere is 5 cm.. fair die is rolled. The event E is defined as follows:- E: multiple of shows up. What is the probability of E occurring? () 1 4 () 1 () 1 () 1 5 () No.of favorable out comes 1 P E No.of total out comes 6 1. The surface area of a solid cylinder with diameter 14 cm and height 1 cm is: () 784 cm () 748cm () 874cm () 478cm () The net of the cylinder shows two circles and a rectangle. rea of the two circles rea of the rectangle 71 ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 11

12 7 1 7 Total surface area of the cylinder cm E OF PE (VNE SOLUTION OE & 4). If the number x48 is divisible by 9, what is the value of x is: () 1 () () () 4 () number is divisible by 9 if the sum of its digits is divisible by 9. Sum of digits x 4 8 x 17 x 17 is divisible by 9 for x 1. The value of is () 4 () 5 () 6 () 7 () Then given expression The diameter of the driving wheel of a bus is 14 cm. The number of revolutions per minute must the wheel make in order to keep a speed of 66 km per minute is: () 5 () 5 () 5 () 7 () istance to be covered in 1 minute m 6 ircumference of the wheel.7 m 4.4m 7 11 Therefore the number of revolution per minute ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 1

13 E OF PE (VNE SOLUTION OE & 4) 5. In the diagram given FGH GE, then the value of y is, H 4 F y G 6 8 E () 1 () () () 4 () FGH GE FH FG 4 y E G 8 6 y 6. If x and x 1 are factors of polynomial then, the value of a b is x 5x ax b () () 5 () 8 () 1 () f x x 5x ax b f b f 1 5 a b a b a a b 7. Given circle with centre O and is a chord as shown in fig. find x, if x 6 and x 1 O () 8 () 16 () () 64 () x 6 x 1 x Viewed from the outside inward, the figure below depicts a square circle square circle, each enclosed within other. If the area of square is 4 sq. units. Then the area of shaded region as shown in fig. is ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 1

14 E OF PE (VNE SOLUTION OE & 4) E F H O G () 4 () () 8 () 8 () E F O H G rea =4 sq. units Radius of circle FGH EOF is right angle OE OF EF EF 11 EF 1 Radius of smaller circle is rea of shaded rea of square EFGH rea of smo circle 1 9. shopkeeper doubled his selling price, and the profit becomes thrice to original. The profit percentage is, () 6% () 1% () % () 1% ().P x s.p y y x y x y x Profit x x RS.x Profit % 1% 4. y what number should we divide () () () x to obtain () () None ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 14

15 E OF PE (VNE SOLUTION OE & 4) x school collected Rs. 89 as fees from its students. If each student paid as many ruppes as there were students in the school, the no. of students in the school? () 51 () 47 () 5 () 6 () x x 89 x 89 x 5 4. Given and x y xy 4 Then the value of x () 81 5 () x y xy y is () x y 5 1x 5y ----(1) () Solve, (1), () 1x 5y 5x 5y x 7 x y () 11 5 () None 4. The dimensions of a metallic cuboid are 5cmcm cm. It is melted and recast into a cube, then the surface area of the cube is () 84cm () () bb a a 5 a 51 a 8 48, 4cm () 6a 6 8 8, 4cm Surface area 44. The perimeter of the following shown figure 6, 4cm () 46,4cm 4 () 4 () 1 4 () () 8 ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 15

16 E OF PE (VNE SOLUTION OE & 4) () 4 x x 4 Perimeter x x x 16 x 16 x 8 x Perimeter metallic sheet is of rectangular shape with dimension 6cm 4cm. From each one of the corner, a square of 8 cm is cutoff. n open box is made of the remaining sheet. Then the volume of box () 56 () 18 () 64 () None () 46. The value of P if the mean of, 4,, 5, 6, 4,, 6 and mean of,, 5, 7, 9,, 1, P is same then P? () 1 () () () 4 () Mean Mean P 8 9 P 4 8 P 47. piece of wire 1m long is cut into two pieces one piece is bent into a square and other is bent into a circle of radius m. Then the area of square is, (in m ) () 4 () 8 () 6 () 1 () P Let a be the length of side of square 4a 1 4a 8 a rea of square 4 ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 16

17 48. See the figure and the value of angle E OF PE (VNE SOLUTION OE & 4) x is, x () () 1 () 15 () () None 6 x 9 x E In E 6 x 9 x 18 x x In the shown figure is the angle bisector of and given E 4 area E area then the ratio of E () : 4 () 6 : 9 () 9 : 16 () None ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 17

18 E OF PE (VNE SOLUTION OE & 4) () 9 So, 6 6 E E 1 E rea of E E rea of E 1 le le ut in and E all similar E E 4 E E can do a work in days and in days. If they work on it together for 5 days then the fraction of the work that left is, () () 7 () 7 () () do the work in one day 1 do the work in one day in 5 days Remaining work ENTERS: MUMI / ELHI / KOL / KOLKT / LUKNOW / NSHIK / GO / PUNE # 18