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1 Set III. A total of 8 handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the other, the number of people present was : (a) 4 (b) 8 (c) 56 (d) 8 (d) Let No. of people is = x and each one has to hand shake with (x - ) persons. x(x ) x x = 56 x x 56 = 0 = 8 x 8x + 7x 56 = 0 (x 8) (x + 7) = 0 x = 8, x = - 7 No. of person = 8. In the set of equations z x = y x, z =. 4 x, x + y + z = 6, the integral roots in the order x, y, z are : (a) 3, 4, 9 (b) 9, -5, (c), -5, 9 (d) 4, 3, 9 z x = y x, z =.4 x, x +y + = 6. z = +x z = + x Only option D will satisfy. This conditions. Page

2 Options A 9 = + (3), 9 = 7 Options B 9 = + ()(), N/equal Options C = + (9) = 9 option D 9 = + (4), 9 = 9 3. Let D represent a repeating decimal. If P denotes the r figures of D which do not repeat themselves, and Q denotes the s figures which do repeat themselves, then the incorrect expression is : (a) D =. PQQQ (b) 0 r D = P.QQQ (c) 0 r+s D = PQ.QQQ (d) 0 r (0 s ) D = Q(P ) (d) 0 r (0 s ) D = Q(P ) 4. A and B together can do a job in days; B and C can do it four days; and A and C in 5 days. The number of days required for A to do the job alone is : (a) (b) 3 (c) 6 (d) No Answer 5. Two candles of the same height are lighted at the same time. The first is consumed in 4 hours and the second in 3 hours. Assuming that each candle burns at a constant rate, in how many hours after being lighted was the first candle twice the height of the second? (a) 3 4 hr. (b) hr. (c) hr. (d) 5 hr. Page

3 d Let the height in each case be. 4 t = ( - 3 t); t = The points of intersection of xy = and x + y = 5 are joined in succession. The resulting figure is : (a) a straight line (c) a parallelogram Page 3 (b) an equilateral triangle (d) a rectangle x + y = 5 in a circle with centre at the origin. xy = is a hyperbola, symmetric with respect to the line x = y. point of intersection are symmetric with respect to x = y. There are either no intersections, two intersections (if the hyperbola is tangent to the circle) or four intersections. Simultaneous solution of the equations reveals that there are four points of intersection, and that they determine a rectangle 7. A regular octagon is to be formed by cutting equal isosceles right triangles from the corners of a square. If the square has sides of one unit, the leg of each of the triangles has length: (a) 3 (b) (c) (d) 3 x + x = ( x) x = + 4x 4x

4 0 = x 4x + x x 4 4 x, 8. If a and b are two unequal positive numbers; then : (a) ab a ab b (b) a b ab ab a b a b (c) ab a b ab (d) a b ab ab a b a b The Arithmetic Mean is (a + b)/, the Geometric Mean is ab, and the Harmonic Mean is ab/(a + b). The proper order for decreasing magnitude is (e); or Since (a - b) > 0, we have a + b > ab; a + ab + b > 4ab, a + b > ab and (a + b)/ > ab. Since a + ab + b > 4ab, we have > 4ab/(a + b) ; ab > 4a b / (a + b), and ab > ab/ (a + b) 9. In the given Figure I is the Incentre of ABC. AI when produced meets the circumcircle of BID respectively: ABC in D. If BAC = 66 0 and ACB = 80 0, then DBC, IBC & (a) 7 0, 33 0 & 50 0 (b) 33 0, 50 0 & 7 0 (c) 33 0, 7 0 & 50 0 (d) 50 0, 33 0 & 7 0 Page 4

5 (c) AD is Angle Bisector. DBC DAC (Angle in the save segment) o DBC 33 0 A BC 80 0 o 0 66 B B IBC B 7 o IBC BID x 80, x You are given a sequence of 58 terms; each term has the form P + n where P stands for the product of all prime numbers (a prime number is a number divisible only and itself) less than or equal to 6 and n takes successively the value, 3, 4,., 59. Let N be the number of primes appearing in this sequence. Then N is : (a) 0 (b) 6 (c) 7 (d) 57 a All those numbers P + n of the sequence P +, P + 3,., P + 59 for which n is prime are divisible by n because n already occurs as a factor in P. In the remaining members, n is composite and hence can be factored into primes that are smaller than 59. Hence all terms are divisible by n. all members of the sequence are composite. The sides of a regular polygon of n sides, n > 4, are extended to form a star. The number of degrees at each point of the star is : Page 5

6 (a) 360 n n 80 (c) n n 4 80 (b) n (d) n Qe = x 80i e n n.360 (n 4)80 x 80 n n. Two equal parallel chords are drawn 8 inches apart in a circle of radius 8 inches. The area of that part of the circle that lies between the chords is: (a) (c) (b) b 3 34 (d) 3 By symmetry, the required area is 4(T + S). T = , S = ; A Page 6

7 3. In the figure, CD,AE and BFare one-third of their respective sides, It follows that AN :NN :ND 3:3:, and similarly for lines BE and CF. Then the area of triangle NNN3 is : (a) ABC 0 (b) ABC 9 (c) ABC 7 (d) ABC 6 By subtracting from ABC the sum of CBF, BAE and ACD and restoring CDN + BFN3 + AEN, we have NNN3 CBF = BAE = ACD = ABC. 3 From the assertion made in the statement of the problem, it allows that CDN = BFN3 = AEN =. ABC ABC. 7 3 NN N3 = ABC 3. ABC 3. ABC ABC A circular piece of metal of maximum size is cut out of a square piece and then a square piece of maximum size is cut out of the circular piece. The total amount of metal wasted is : (a) 4 the area of the original square (b) the area of the original square (c) the area of the circular piece (d) the area of the circular piece 4 b Let r be radius of one circle AO = OB = r Page 7

8 AB = (AO) (OB) = r r = r = r radius of innermost circle = AB ( r) r Area = r = r 5. Given triangle PQR with RS bisecting angle R, PQ extended to D and angle n a right angle, then : (a) m p q (b) m p q (c) d q p (d) d m mp d, dq m (there are two vertical angles each m ). mp qm; m ( p q). 6. In the diagram if points A, B, C are points of tangency, then x equals : (a) 3'' 6 (c) '' 6 (b) '' 8 (d) 3'' 3 The distance from the centre of the circle to the intersection point of the tangents Page 8

is 3/8 3 3 9 CD x ; x (cm). 8 6 6 6 7. In the figure PA is tangent to semicircle SAR; PB is tangent to semicircle RBT; SRT is a straight line; the arcs are indicated in the figure.

9 is 3/ CD x ; x (cm) In the figure PA is tangent to semicircle SAR; PB is tangent to semicircle RBT; SRT is a straight line; the arcs are indicated in the figure. Angle APB is measured by : (a) (a b) (b) (a + b) (c) (c a) (d) a b b First, draw the line connecting P and R and denote its other intersections with the circles by M and N; see accompanying figure. The arcs MR and NR contain the same number of degrees; so we may denote each arc by x. To verify this, note that we have two isosceles triangles with a base angle of one equal to a base angle of the other. NOR MO'R APR {(c a c x) a} {c x} BPR {b d d (b x)} {d x} and the sum of angles APR and BPR is BPA = c + d. The desired angle is 360 o BPA = 360 o (c + d) = (80 o - c) + (80 o - d) = a + b 8. In a general triangle ADE (as shown) lines EB and EC are drawn. Which of the following angle relation is true? (a) x + z = a + b (b) y + z = a + b (c) m + x = w + n (d) x + y + n = a + b +m Page 9

10 d From triangle AEC, x + n +y + w = 80 o From triangle BED, m + a + w + b = 80 o. x + y + n = a + b + m 9. ABCD is a rectangle (see the diagram) with P any point on AB. PS BD,PR AC, AF BD and PQ AF. Then PR + PS is equal to : (a) PQ (b) AE (c) PT AT (d) AF PTR ATQ; PR / AQ PT/ AT PT AT ( PAT PBS APT); PR AQ, PS QF; PR P S AQ QF AF; or SBP TPA TAP, A, Q, R, P are concylic, arc PR arc AQ, PR AQ, PS QF; PR RS AF. 0. The length of a triangle is of length b, and the altitude is of length h, A rectangle of height x is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is: bx hx (a) ( h x ) (b) ( b x ) h b bx (c) ( h x ) (d) x (b - x) h Page 0

11 Designate the base of the rectangle by y. Then, because of similar triangles, h x h, y b (h x) y b h area = xy = bx (h - x) h. Compute (a),999,000 (b),888,000 (c),999,999 (d),999,999 a s = = ( ) + ( ) + ( ) + ( ) +. + (3 + ) + (3 - ) + = = =, 999, 000. What remainder are obtained when the number consisting of 00 sevens is divided by the number 00? (a) 777 (b) 707 (c) 700 (d) 770 The number 777, 777 is exactly divisible by 00, yielding the quotient 777. Hence the number , yields, upon division by 00, a quotient of the grouping 777, 000 repeated 666 times Moreover, the number 77, 777 yields a quotient of 777 and a remainder of 7000 upon division by 00; and so the quotient obtained by dividing A by 00 has the form Page

777000 777 000... 7777 000 77 the grouping 777, 000 repeated 66 times and there is a remainder of 700. 3. Compute the unique positive integer n such that. + 3. 3 + 4. 4 + 5. 5 +. + n.

12 the grouping 777, 000 repeated 66 times and there is a remainder of Compute the unique positive integer n such that n. n = (n+0) (a) 403 (b) 53 (c) 43 (d) 503 b (. ) n. n = (n + 0) + (). The left side can be summed as: n = n n = n n = n+ 3. =. + n = n n n n n( ) ( ) = n + (n - ) +. since this equals n + 0 +, n = n0, so n = 9 + = 53 n 4. Two equilateral triangle measures cm on each side. They are positioned to form a regular six-pointed star. What is the area of the overlapping figure? (a) 48 3 cm (b) 4 3 cm (c) 36 3 cm (d) 3 cm Page

13 Required Area = 3 3 () 3 (4) (6) c 5. A digital clock displays the correct time on January 0. If the clock loses 5 minutes per day, what will be the next date when the clock displays the correct time? (a) 7 th april (b) 7 th feb (c) 6 th april (d) 8 th feb The clock loses hour in 4 days so it loses hours in 48 days. From January to February is 3 days. Add 7 days to get to 8 February, assuming that only a - hour clock was used. But the problem describes a digital clock specifically. Since virtually all digital clocks show 4-hours or A.M through P.M., this solution is the only one possible. If we assume that a 4 hour clock was used, then solution is 96 days. That is, the 97th day of the year = = 3 = 90, , which is 7 April on 6 April in a leap year. Consider a clock that also shows the month and the year 6. Sachin Verma said: The day before yesterday I was 0, but I will turn (?).yrs in the next year {maximum possible answer is} (a) (b) 3 (c) (d)4 We need to have as much time as possible between the uttering of this sentence and Peter s next birthday. We can manage this if he made this statement on Page 3

14 January, and he was born on December 3. He will turn 3 at the end of the next calendar year 7. The remainder when 7 84 is divided by 34 is (a) 0 (b) (c) 49 (d) 34 b , when divided by 34, leaves a remainder of 7 4 = 40, when divided by 34, leaves a remainder of = 6807, when divided by 34, leaves a remainder of = 7649, when divided by 34, leaves a remainder of. And so on. 84 7, when divided by 34, will leave a remainder of is Equal to (a) (b) (c) 6 (d) 6 () ( ) ( 3) () ( ) () ( 3 ) ( )( 3) ( 3) ( ). 3 3 ( 3) ( 3 ) ( 3) ( ) Page 4

15 = 9. If f(x) + f( x) =. Then f f... f? (a) 999 (b) 998 (c) 99 (d) f f f f f f f f = (998 times) (f(x) + f( x) = by question) = 996 = 998 = when divided by 7 leaves remainder (a) (b) 3 (c) 4 (d) 5 (c) Award winning Question. Give the solution of the question and get your reward from pioneermathematics.com. Page 5

3. In triangle ABC, the incircle touches the sides, BC, CA and AB at D, E F respectively. If radius of incircle is 4 units and BD, CE, AF be consecutive integers, find the sides of triangle ABC.

16 3. In triangle ABC, the incircle touches the sides, BC, CA and AB at D, E F respectively. If radius of incircle is 4 units and BD, CE, AF be consecutive integers, find the sides of triangle ABC. (a) 9, 0, (b) 3, 4, 5 (c) 4, 5, 6 Area of ABC By Hero s form a a a S 3a 3 3 a (d) None of these D a a a a a a a 3 3a 3 a 3a 3 a a 3a 3 a 4 3 a 3 a a a a a 3a a.() 4 Now, Area of ABC 4a 4 a 4 a Now, 4 a a a. = (3a + 3) = 6 (a + )..() from () and () a a 3 a 36 a 6 (a + 3) (a ) = 6 Page 6

17 a = 3 3. The interior angle of a regular polygon exceeds the exterior angel by 3 0. The number of sides in the polygon is. (a) 7 (b) 8 (c) (d) 5 n i 80 e i e 3 n n 0 n m n 0 80n n n 80n n n n 80n 70 3n 0 48n n = Given the sides of a triangle (a, b and c). then the median ma drawn to the side a can be GIVEN by the formula (a) ma = (c) ma = b c a (b) ma = b c a (d) ma = c a b b c a Page 7

18 . In ABC, the mid-points of the sides BC, CA and AB are D, E and F respectively. The lines, AD, BE and CF are called medians of the triangle ABC, the points of concurrency of three medians is called centroid. Generally it is represented by G. By analytical geometry : AG AD 3 BG BE 3 and CG CF 3 Length of medians and the angles that the median makes with sides In above figure, AD = AC + CD AC.CD. cos C a AD b ab cos C 4 a b a c AD b ab. 4 ab b c a AD 4 or AD b c a AD b c bc cos A 34. Pedal triangle is a triangle formed by joining the foot of the altitudes in a triangle,then the orthocentre of a triangle is the.of the pedal triangle. (a) circumcentre (b) Incentre (c) Centroid (d) orthocenter (b) Incentric Page 8

19 35. Find the highest power of 3 contained in 000! (a) 499 (b) 498 (c) 496 (d) 497 (b) P = 3, n = 000 n [4] 4 5 p 3 n 4 6 p 3 3 n 0 7 p 3 n p 3 3 n 333 [] p 3 n [37] 37 3 p 3 n 37 4 p 3 3 Highest power of 3 contained in 000! n n n n n n n p p p p p p p = = Find the remainder when is divided by 7 (a) (b) 5 (c) 6 (d) 3 (b) 00 = mod 7 Page 9

20 3 00 = 4 mod 7, mod m 4 00 = 4 mod 7, 5 00 = mod 7, a b mod m = mod 7 c d modm a c b d mod m But 5 mod mod 7 Remainder is If,, be the roots of x 3 + x 3x = 0.Find the value of (a) -40 (b) -4 (c) -4 (d) None of these (c) Roots of the equation x 3 + x 3x = 0 are,,. Let us first form an equation whose roots are 3, 3, 3. If y is root of the transformed equation, then y = x 3 To eliminate x between Eqs. (i) and (ii). Eq. (i) can be written as x 3 = - (x 3x) On cubing both sides of above equations x 9 3x 6 + 3x 3 = - [8x 6 7x 3 8x 3 (x 3x)] x 9 3x 6 + 3x 3 = - [8x 6 7x 3 8x 3 ( x 3 )] x 9 3x 6 + 3x 3 = - 8x 6 + 7x 3 + 8x 3 8x 6 x 9 + 3x 6-4x 3 = 0 Putting x 3 = y in this equation, we get y 3 + 3y 4y = 0 Page 0

21 it roots are 3, 3, 3. Changing y to, Eq. (iii) becomes y y y y y 3 + 4y 3y = 0 Its roots are,, ,, If we Solve the system b( x y) c( z x) a x y cxy x z bxz c( y z) a( x y) b y z ayz x y cxy a( x z) b( y z) c. x z bxz y z ayz = sum of roots of Eq. (iv) = - 4 then the value of x= (a) a. a b c (c) a. a b c x y y z Put ; x y cxy y z ayz (b) a. a b c (d)none of these x z and x z bxz Then, the system takes the form b c a ; c b; b c Page

22 a b or ; ; c b bc a c ac c b a ab a b c abc b b c abc and consequently b c a a c b ; ; bc ac a b c ab x y cxy Further x y or or cx x y x y cxy Finally c x y Analogously, we get b a ; x z y z Where from we find x, y, z 39. Which of the following TRUE (a) (3) > (7) 7 (b) (30) 00 > () 567 (c) 7 9 >8 9 (d) (50) 300 >(0000) 00 (00) 00 (d) () Now, 3 < 3 Raising the power to Page

23 (3) < (3) (3) < ( 5 ) = 60 (i) Now, 60 < 68 = 4 7 (ii) ( 4 ) 7 < (6) 7 (6) 7 < (7) 7 (iii) From (i), (ii) and (iii),we get (3) < 60 < 68 < (7) 7 (3) < (7) 7 (7) 7 > (3) (ii) (30) 00 < (3) 00 So, ( 5 ) 00 = ( 0 ) 50 = (04) 50 Now, (04) 50 < (04) 54 = ((04) ) 7 = ( 0 ) 7 Now, 0 < ( 0 ) 7 < ( ) 7 = 567 (30) 00 < > (30) 00 (iii) If a, b are +ve and n is a natural number, then (a + b) n = a n + na n- b + (term involving powers of a and b). Also, (a + b) n a n + na n- b (equality for n = ) now, (8) 9 = (7 + ) 9 > = 7 90 (7 + 9) = 7 90 (98) Now, 7 90 (98) > = = 7 Hence, (8) 9 > 7 9 (iv) We know (50) 3 = = = Now, > Hence, (50) 3 > Page 3

24 Raising the power to 00 (50) 300 > (0000) 00 (00) If P, Q, R, S are the sides of a quadrilateral. Find the minimum value of p q r s (a)/ (b) /3 (c)/3 (d)3/ We have AB = p BC = q CD = r AD = s We know that (p - q) + (q - r) + (r - p) 0 (p + q + r ) (pq + qr + rp) 3 (p + q + r ) (p + q + r ) + (pq + qr + rp) [on adding p + q + r to both sides] 3(p + q + r ) (p + q + r) [ sum of any three sides of a quadrilateral is greater than fourth one] 3(p + q + r ) (p + q + r) > s p q r s 3 Minimum value of p q r is s 3 4. A student on vacation for D days observed that (i) it rained 7 times morning or afternoon (ii) when it rained in the afternoon it was clear in the morning Page 4

25 (iii) there were 5 clear afternoons, and (iv) there were 6 clear mornings. Find D. (a)9 (b) 7 (c)0 (d)5 Let the set of days it rained in the morning be M. Let Ar be the set of days it rained in afternoon. M be the set of days, when there were clear morning. ' A r be the set of days when there were clear afternoon. By condition (ii), we get Mr Ar = By (iv), we get ' M r = 6 By (iii), we get ' A r = 5 By (i), we get Mr Ar = 5 Mr and Ar are disjoint sets n(mr) = d 6 n(ar) = d 5 Applying the principal of inclusion-exclusion n(mr A r ) n(m r ) n(a r ) n(mr A r ) 7 = (d - 6) + (d - 5) 0 d = 8 d = 9 4. OPQ is a quadrant of a circle and semicircles are drawn on OP and OQ. Then (a)a>b (b) a<b (c)a=b (d)can t be determined (c) Area of quadrant = areas of two semi circles + b a. Page 5

26 i.e., r b a 4 r r 4 4 r r b a b a = 0 a = b 43. Let ABC be equilateral. On side AB produced, we choose a point P such that A lies between P and B. We now denote a as the length of sides of ABC; r as the radius of incircle of PAC; and r as the extradius of PBC with respect to side BC. Determine the sum r + r as a function of a alone. (a) a 3 (b) a 3 (a) (c) a 5 3 (d) 3 3a we see that TOR = 60 o since it is the supplement of TAR = 0 o (as an exterior angle for ABC). Hence, AOR = 30 o. Similarly, we obtain BOS = 30 o. Since, tangents drawn to a circle an external point are equal, we have TT = TA + AB + BT = RA + AB + SB r = r tan 30 o + a + r tan 30 o r = a 3 and T T T C CT ' ' ' ' r r = CR + CS = (a - RA) + (a - SB) = a - 3 Page 6

27 Since, common external tangents to two circles are equal TT = T T ' ' r r r r Hence, a a, 3 3 Whence we find that r + r = a The sum of n term of the series is (a) n 3 (b) (c) n 3 3 (d) n 3 n n 3 n n 3 n n 3 n n 3 n n 3 n n 3 n n 3 n n 3 3 Page 7

28 45. In ABC, the medians AD, BE and CF meet at G, then (a) 4(AD + BE + CF) > 3(AB + BC + AC) (b) 3(AD + BE + CF) > (AB + BC + AC) (c) 3(AD + BE + CF) > 4(AB + BC + AC) (d) (AD + BE + CF) >3 (AB + BC + AC) (a) 4(AD + BE + CF) > 3(AB + BC + AC) 46. The point of concurrency of the perpendicular bisectors of a triangle is called (a) Incentre (b) Orthocentre (c) Circumcentre (d) Centroid (c) Circumcentre 47. Some friends are sitting on a bench. Vijay is sitting next to Anita and Sanjay is next to Geeta.Geeta is not sitting with Ajay. Ajay is on the left end of the bench and Sanjay is in second position from right hand side. Vijay is on the right side of Anita and to the right side of Ajay, Vijay and Sanjay are sitting together. Who is sitting in the centre? (a) Ajay (b) Vijay (c) Geeta (d) Sanjay (b) 48. If x k divides x 3 6x + x 6 = 0, then k can t be equal to (a) (b) (c) 3 (d) 4 Page 8

29 (d) x = k is zero of polynomial Now put k = 3 6() + () 6 = = 0 0 = 0 k = () 3 6() + 6 = = 0 = 0 k = 3 (3) 3 6(3) = = 0 = 0 k = 4 (4) 3 4(4) = = = 0 k = A convex polygon has 44 diagonals. The number of its sides is (a) 0 (b) (c) (d) 3 (b) n n n C n n No. of diagonal. = n n n Page 9

30 44 n n n 88 = n 3n 88 = n(n 3) 88 = 8 = ( 3) n = 50. Roman numeral for the greatest three digit number is (a) IXIXIX (c) CMIXIX (b) CMXCIX (b) CMXCIX (d) CMIIC 5. In the new budget, the price of a petrol rose by 0%, the percent by which one must reduce the consumption so that the expenditure does not increase is : (a) (c) 6 % (b) 6 % % (c) Let price of petrol = Rs x price hike = 0% i.e. 0 x x 00 0 x x New price = x 0 0 earlier consumption = y litra earlier investment = xy. A.T.Q., Present investment = previous investment (d) 0% Page 30

31 x (present petrol consumption) = xy present petrol consumption = (xy) 0 0 x = 0y Reduction in consumption = y - 0y y /00 = y/ % age = y 00 9 % 5. Like dozen is articles,what is score equals to (a) 0 (b) 30 (c) 4 (d) 36 (a) Three traffic lights at three different road crossing change after 48 seconds, 7 seconds and 00 seconds respectively, If they all change simultaneously at 8 a. m., at what time will they again change simultaneously? (a) 0 a.m. (c) a.m. (b) L.C.M of 48, 7, 00 is = = 3600 sec 48 = 3 = hour 7 = = 5 5 (b) 9 a.m. (d) 0.30 a.m. Page 3

32 54. Tell the number of triangles in the following figures (a)40 (b) 45 (c) 47 (d) 50 (c) A school bus travels from Delhi to Chandigarh. There are 4 children, teacher and driver in the bus. Each child has 4 backpacks with him. There are 4 dogs sitting in each backpack and every dog has 4 puppies. What is the total number of eyes in the bus. (a) 56 (b) 8 (c) 657 (d) 65 (d) No. of teacher = No. of driver= eyes of teacher and driver= (+)X=4 No. of children=4 eyes of children= 4 x =8 No. of dogs in each backpack= 4x4=6x4=64x=8 eyes eyes of puppies= 64x4=56x=5 eyes Total eyes= =65 eyes 56. Ravi is not wearing white and Ajay is not wearing blue. Ravi and sohan wear different colour. Sachin alone wear red. What is sohan colured, if all four them are wearing different colour. (a) red (b) blue (c) white (d) can t say (d) Page 3

33 The fourth colour and some more information are required 57. Who is the father of Geometry? (a) Pythagoras (c) Archimedes (d) Euclid. (b) Thales (d) Euclid. 58. Which of the following correctly shows according to International place value chart? (a), 853, 67, 49 (b) 8, 536, 74, 9 (c) 85, 367, 49 (d) None of these (c) 85, 367, (x% of y + y% of x) = (a) x% of y (c) % of xy x y y x xy xy (b) y% of x (d) x% of xy Page 33

34 60. A conical vessel of radius 6 cm and height 8 cm is completely filled with water.a sphere is lowered into the water and its size is such that when it touches the sides, it is just immersed. What fraction of the water overflows? (a) 5 (b) 3 8 (c) 3 5 (d) 3 4 (b) A vertical section of the conical vessel and the sphere when immersed are shown in the figure. From right angled AMB, AB = AM + MB = = = 0 AB = 0 cm. CB is tangent to the circle at M and AB is tangent to it at P. PB = MB = 6 ( lengths of tangents from an external point to a circle are equal in length) AP = AB - PB = (0 6) cm = 4 cm. Let r cm be the radius of the circle, then OP = OM = r AO = AM OM = (8 r) cm. From right angled OAP, OA = AP + OP (8 r) = 4 + r 64 6r + r = 6 + r Page 34

35 48 = 6r r = 3. Radius of circle i.e. of the sphere = 3 cm. Volume of sphere = cm 36 cm. The volume of water which overflows = volume of the sphere = 36 cm 3. Volume of water in the cone before immersing the sphere = volume of the cone = cm3 = 96 cm 3. The fraction of water which overflows = Volumeof wateroverflows Total volumeof water 96 8 For online solutions visit Page 35

Set II. 1. The value of. (a) 0 (b) 1 (c) 2 (d) 4 Sol : (1) ( 2) Similarly. and so on, we get = 3 1 = 2

Set II. 1. The value of. (a) 0 (b) 1 (c) 2 (d) 4 Sol : (1) ( 2) Similarly. and so on, we get = 3 1 = 2 Set II 1. The value of 1 1 1 1 1 1 1 1 1 3 3 4 4 5 5 6 6 7 7 8 8 9 is (a) 0 (b) 1 (c) (d) 4 1 1 1 1 1 1 1 1 1 1 (1) ( ) 1 1 Similarly 1 3 3 3 1 and so on, we get 1 3 4 3 5 4 6 5 7 6 8 7 9 8 = 3 1 =. If