CLASS X : MATH SOLUTIONS Q1. It is given that x = - 1 is the solution of the quadratic equation 3x +kx-3 = 0. 3 1 1 + k (- ) - 3 =0 3 4 - k - 3=0 k = 3 3 = 9 4 4 Hence, the value of k is 9 4 Q. Let AB and CD be the two towers of heights x and y, respectively. Suppose E is the centre of the line joining the feet of the two towers i.e. BD. Now, in ABE, AAAA BBBB = tan30 xx BBBB = 3 BE = 3x 1 Also,
In CDE, CCCC DDDD = tan60 DE = yy 3. () NOW BE = DE (3) (E is mid point of BD) So, from (1), () and (3) we get 3x = yy 3 xx yy = 1 3 Hence, the ratio of x and y is 1: 3. Q3. There are 6 letters in English alphabets. Total number of outcomes = 6 We know that there are 5 vowels and 1 consonants in English alphabets. Total number of favourable outcomes = 1 Probability that the chosen letter is a consonant = TTTTTTTTTT nnnnnnnnnnnn oooo ffffffffffffffffffff oooooo cccccccccc TTTTTTTTTT nnnnnnnnnnnn oooo oooooooooooooooo = 1 6 Q4. PA and PB are tangents drawn from an external point P to the circle. PA = PB (Length of tangents drawn from an external point to the circle are equal) In PAB, PA = PB PBA = PAB (1) (Angles opposite to equal sides are equal.) Now, APB + PBA + PAB = 180 50 + PAB + PAB = 180 [Using (1)] PAB = 130 PAB = 130 = 65
We know that radius is perpendicular to the tangent at the point of contact OAP = 90 (OA PA) PAB + OAB = 90 65 + OAB = 90 OAB = 90-65 = 5 Hence the measure of OAB is 5 SECTION B Q5. We know that the angle subtended by an arc of a circle at the centre is twice the angle subtended by it at any point on the remaining part of the circle. AOQ = ABQ ABQ = 1 AOQ ABQ = 1 58 = 9 or ABT = 9 We know that the radius is perpendicular to the tangent at the point of contact OAT = 90 (OA AT) Or BAT = 90 Now, in BAT, BAT + ABT + ATB = 180 90 + 9 + ATB= 180 ATB = 61 ATQ = 61 Q6. We have 4x 4a x + (a 4 b 4 ) =0 (4x -4a x +a 4 ) b 4 =0 [(x) - (x) (a) + (a ) ] b 4 = 0
(x a ) (b ) = 0 (x a +b ) (x a b ) =0 x a +b = 0 or x a b = 0 x = a b or x = a + b x = aa bb or aa +bb Q7. TP and TQ are tangents drawn from an external point T to the circle. O is the centre of the circle. Suppose OT intersect PQ at point R. In OPT AND OQT, OP= OQ (Radii of the circle) TP = TQ (Lengths of tangents drawn from an external point to a circle are equal ) OT = OT (Common sides) OPT OQT (BY SSS congruence rule) So, PTO = QTO (BY CPCT). (1) Now, in PRT and QRT, TP =TQ (Lengths of tangents drawn from an external point to a circle are equal) PTO = QTO [From (1)] RT = RT (Common sides)
PRT QRT (By SAS congruence rule) So, PR = QR. () (By CPCT) And, PRT = QRT (BY CPCT) Now, PRT + QRT = 180 (Linear Pair) PRT = 90 PRT = 90 PRT = QRT = 90 (3) From () and (3) we can conclude that OT is the right bisector of the line segment PQ. Q8. The given anthmetic progression is 6, 13, 0., 16 Let 16 be the n th term of the given AP. So, a = 6 d =7 a n = 16 Now, a n = a + (n - 1)d 16 = 6 + (n - 1) 7 7(n-1) = 10 n - 1 = 30 n = 31, which is odd Middle term of the AP = ( 31+1 th term of the AP )P = 16th term of the AP
a 16 =6+16-1 7=6+15 7=6+105=111 Thus, the middle term of the given AP is 111. Q9. Using distance formula, we have AB = ( 5) + ( ) = 9 + 16 = 5. (1) BC = ( ) + (tt + ) = tt + 4tt + 0 () AC = ( 5) + (tt ) = = tt 4tt + 53. (3) Now, it is given that ABC is right angled at B. Using the Pythagorean theorem, we have AB +BC = AC 5 + t + 4t + 0 = t - 4t + 53 [From (1) () and (3)] 45 + 4t = -4t + 53 8t = 8 t = 1 Hence, the value of t is 1. Q10. Let P ( 3 4, 5 1 ) divide AB in the ratio m 1:m. Using the section formula, we have 3 mm mm 1 +mm ( 3, 5 1 4 1 ) = mm 1+ mm, 5mm 1+ mm 1 +mm 3 = mm 1 1+ mm 4 mm 1 +mm.. (1) 5 = 5mm 3 1+ mm 1 mm 1 +mm () From (1) we have 1 mm 3 = mm 1+ 4 mm 1 +mm
3(mm 1 + mm ) = 4(mm 1 + 1 mm ) 3m 1 +3m =8m 1 + m -5 m 1 =- m m 1: m = 1: 5 SECTION - C Q11. Let the coordinates of points B and C of the ABC be (a 1, b 1 ) and (a, b ), respectively. Q is the midpoint of AB. Using midpoint formula, we have (0, -1)= aa 1+1,bb 1 4 aa 1+1 =0 and bb 1 4 a 1 =-1 and b 1 = =-1 Therefore, the coordinates of B are (-1,).
P is the midpoint of AC. Now, (, -1) = aa +1, bb 4 aa +1 = and bb 4 a = 3 and b = = -1 Therefore, the coordinates of C are (3,). Thus, the vertices of ABC are A(1, -4), B (-1,) and C(3,). Now, Area of the triangle having vertices (x 1, y 1 ),(x, y ),(x 3,y 3 ) = 1x1y-y3+y3-y1+x3y1-y Area of ABC = 1 [x 1(y - y 3 ) + x (y 3 - y 1) + 3 (y 1 - y )] = 1 [1 ( - ) + -1( + 4) + 3(-4 - )] =-1 Since area is a measure that cannot be negative, we will take the numerical value of 1, that is 1. Thus, the area of ABC is 1 square units. Q1 We have kx + 1 - (k - 1) x + x = 0 This equation can be rearranged as (k + 1) x - (k - 1)x + 1 = 0 Here, a = k+1, b = -(k-1) and c =1 D = b 4ac = [-(k-1) ] 4 (k + 1) 1 =4[(k-1) (k+1)] = 4[k 3k]
= 4[k (k -3)] The given equations will have equal roots, if D = 0. 4[k (k-3)] = 0 k = 0 or k -3 = 0 k= 3 Putting k= 3 in the given equation, we get 4x - 4x + 1 =0 (x-1) = 0 x -1 =0 x =1/ Hence, the roots of the given equation are 1/ and 1/. Q13 Let AB the building and CD be the tower. CD/BD=tan 45 30/BD =1 BD =30m
In ABD, AAAA BBBB = tan 30 AB = BD 1 3 AB= 10 3 m Therefore, the height of the building is 10 3 m. Q14. When two dice are thrown simultaneously, the possible outcomes are Total number of possible outcomes = 36 We know TTTTTTTTTT nnnnnnnnnnnn oooo ffffffffffffffffffff oooooooooooooooo Probability of an event = TTTTTTTTTT nnnnnnnnnnnn oooo oooooooooooooooo (1) The outcomes favorable of the event the sum of the numbers on two dice to be 5 denoted by E are (1, 4), (, 3), (4, 1) and (3, ). Total number of favorable outcomes = 4 PE =4/36 =1/9 (ii) The outcomes favourable to the event even numbers on both dice denoted by F are
(, ), (,4),(,6), (4,),(4,4),(4,6), (6,), (6,4) and (6,6). Total number of favouable outcomes = 9 PF =9/36 = 1/4 Q15 Let a and d be the first term and the common difference of AP, respectively, We know Sn = nn [a+ (n-1) d] S 8 = 8 [a+ (8-1) d and S4 = 8 [a+ (4-1) d] S 8 = 4(a + 7d) and S 4 = (a + 3d) S 8 = 8a + 8d and S 4 =4a + 6d Now, 3(S 8 S 4 ) =3(8a+8d-4a-6d) = 3(4a+d) =6(a+11d) = 1 [a+ (1-1) d] =S 1 S 1 =3(S 8 -S 4 ) 16. Let the radius of the semi-circle APB be r. The radius of the semi- circle AQO = rr Now, Perimeter of the given figure = Length of arc AQO + Length of arc APB + OB
= r ( 3 7 + 1) = r 80 14 = r( 40 7 ) cm r(40 7 ) = 40 r = 7cm Area of the shaded region = Area of semi circle AQO + Area of semi- circle APB = ππ(7 ) + ππ7 = 49 ππ ( 1 8 + 1 ) = 96.5 cm 17. The remaining solid is a frustum of the given cone. Total surface area of the frustum Where h = Height of the frustum = 1-4 = 8 cm r 1 = Larger radius of the frustum = 6 m
r = Smaller radius of the frustum l = Slant height of the frustum In the given figure, ABC ~ ADE by AA similarity criterion. BC/DE = AB/AD rr = 4 6 1 r = cm We know l = 4 5 cm. Total surface area of the frustum = ππππ(r 1 +r ) +πr 1 + πr = 350.59 cm Hence, the total surface area of the remaining solid is 350.59 cm.
18. Let h be the height of the cone and r be the radius of the base of cone. The volume of the wooden toy = ππrr h 3 + 3 πr3 According to the question, 77 6 (h + 7) = 1001 6 Now, h= 6 cm The height of the wooden toy = 6 cm + 3.5 cm = 9.5 cm Curved surface area of the hemispherical part = X(/7)X3.5X3.5 = 7 (3.5) =77 cm Hence, the cost painting the hemispherical part of the toy = 77 X 10 = Rs 770 Q19. Surface area of the remaining block = Surface area of the cuboid + Curved surface area of cylinder - Area of base of cylinder = ( 15 10 + 10 5 +15 5) + ( 7 5 ) ( 7 7 ) 7 7 = 583 cm Thus, the surface area of the remaining block is 583 cm. Q0. From the symmetry of the given figure, we have
Area of the shaded region = Area of the square with side 14 cm (Area of square with side 4 cm + 4 Area of each semicircle with radius cm ) = 14 14 [ 4 4 + 4 1 3.14 () ] = 196 ( 16 + 5.1) = 196 41.1 = 154.88 cm Section - D 1. Let the numerator and the denominator of the fraction be x + 3, respectively. Original fraction = xx xx+3 Now, is added to both the numerator and the denominator. New fraction = (xx+) (xx+3) According to the question, xx xx + 3 + xx + xx + 5 = 9 0 xx(xx + 5) + (xx + )(xx + 3) (xx + 3)(xx + 5) = 9 0 Solving we get
xx = 7 10 Orxx = 45 11 Now, xx ( 45 )as it is a fraction. 11 So, the original fraction becomes (7/10).. Since Ramkali increased her weekly savings uniformly every week by a fixed number, her savings will form an AP. Let S n be the sum of savings in all 1 weeks. Sn= nn (aa + (nn 1)dd) (Here a is the money saved in the first week and d is the fixed increase in the weekly savings.) Sn= 1 ( 100 + (1 1)0)=Rs,50 Ramkali required Rs,500 after 1 weeks, but she saved Rs,50. So, she will be able to send her daughter to school after 1 weeks. It shows that Ramkali is aware of the importance of girl child education. 3. + 3 = 3 xx+1 (xx ) 5xx 4(xx ) + 3(xx + 1) (xx + 1)(xx ) = 3 5xx 5xx 4(xx ) + 3(xx + 1) = 46(xx + 1)(xx ) 11xx 1xx 9 = 0 11xx 44xx + 3xx 9 = 0 11xx(xx 4) 3(xx 4) = 0
Solving we get x =(-3/11) or 4 4. Given: A circle C (o,r) and a tangent l at point A To prove: Construction: OA / Take a point B, other than A, on the tangent l. Join OB. Suppose OB meets the circle in C. Proof: We know that among all line segments joining the point O to a point on /, the perpendicular is shortest to /. OA = OC (Radius of the same circle) Now, OB=OC+BC OB > OC OB > OA OA < OB B is an arbitrary point on the tangent l. Thus, OA is shorter than any other line segment joining O to any point on l. Here, OA l
Hence, the tangent at any point of a circle is perpendicular to the radius through the point of contact. 5. We know that tangents from an external point are equal in length. PQ = PR In PQR, PQ = PR PQR = PRQ (Angles opposite to equal sides are equal). Now in PQR, PQR + PRQ + RQP =180 0 RQP = 180 0-30 0 RQP = 75 0 Also, radius is perpendicular to the tangent at the point of contact. OQP = ORP Now, in PQOR, 0 = 90 ROQ + OQP + QPR + PRO = 360 0 90 0 + 90 0 +30 0 + ROQ = 360 0 ROQ = 150 0
Since, angle subtended by an arc at any point on the circle is half the angle subtended at the centre by the same arc. angle QSR = 75 0 Also, QSR = SQT (Alternate interior angles) SQT =75 0 Now, SQT + PQR + SQR = 180 0 75 0 + 75 0 + SQR = 180 0 SQR =30 0 6. Given: A ABC with BC = 7 cm, Steps of Construction: 0 B = 60 = and AB = 6 cm 1. Draw a line segment AB = 6 cm. 0. With B as centre, B = 60. 3. With B as centre and radius BC=7 cm, draw an arc. 4. Join AC to obtain ABC. 5. Below AB, make an acute angle BAX. 6. Along AX, mark off four points A 1, A, A 3 and A 4 such that AA 1 =A 1 A =A A 3 =A 3 A 4. 7. Join A 4 B. 8. From A 3, draw A 3 B //A 4 B 9. From B draw B C //BC Thus, AB C is the required triangle with each of its side (3/4) times the size of the corresponding sides of ABC.
7. Let RQ be the tower and SR be the flag staff. In PQR Tan30 0 = (RQ/PQ) 1 3 = h xx ---------------------------(i)
x = h 3 In PQS, Tan60 0 = SSSS PPPP 3 = h+5 ----------------(ii) xx From (i) and (ii), we get 3h = h+5 h = 5 H =.5 m Hence, the height of the tower is.5metres. 8. Total number of cards = 0 (i) Numbers divisible by or 3 are,3,4,6,8,9,10,1,14,15,16,18and 0. Total favorable number of cards =13 Probability that the number on the drawn card is divisible by or 3=(Total favorable number of cards /Total number of cards) (ii) Prime numbers are, 3, 5, 7, 11, 13,17and19 Total favorable number of cards =8 = (13/0) Probability that the number on the drawn card is a prime number= (Total favorable number of cards/ Total number of cards)=(8/0)=(/5) 9.
Let the vertices of the quadrilateral be A(-4,8), B(-3,-4), C(0,-5)and D(5,6). Join AC to form two triangles, namely, ABC and ACD. Area of quadrilateral ABCD = Area of ABC+ Area of ACD We know Area of triangle having vertices (x 1,y 1 ),(x,y ),(x 3,y 3 ) = (1/)(x 1 (y -y 3 )+x (y 3 -y 1 )+x 3 (y 1 -y )) Now, Area of ABC = (1/)((-4)(-4+5) + (-3)(-5-8) + 0 (8+4)) = (1/)(-4+39) = (35/) Area of ACD =(1/)((-4)(-6-5) +5(-5-8) +0 (8-6)) = (1/)(-44-65) = (109/) Area of quadrilateral ABCD=Area of ABC + Area of ACD = (35/)+(109/)=7 square units Thus, the area of quadrilateral ABCD is 7 square units. 30.
Depth (h 1 ) of the well = 14 m Radius (r 1 ) of the circular end of the well = (4/) m = m Height (h ) of embankment = 40 cm = 0.4 m Let the width of embankment be x. From the figure, it can be observed that the embankment will be cylindrical in shape having outer radius (r ) as ( +x) m and inner radius (r 1 ) as m. Volume of earth dug from the well = Volume of earth used to form embankment Πr 1 h 1 = π(r r 1 )h Π() 14 = π((+x) ) X +4x-40 =0 (x-10)(x+14) =0 X =10 m Therefore, the width of the embankment will be 10 m. 31. Increase in the level of water in half an hour, h=3.15 m=315 cm Radius of the water tank, r=40 cm Volume of water that falls in the tank in half an hour = πr h = π(40) (315)
= 5,04,000π cm 3 Rate of flow of water =.5 km/h Length of water column in half an hour =.5 = 1.6km = 1,6,000 cm Let the internal diameter of the cylindrical pipe be d. Volume of the water that flows through the pipe in half an hour =π (d/) (16000) cm 3 We know Volume of the water that flows through the pipe in half an hour = Volume of water that falls in the tank in half an hour Π(d/) (16000) =5,04,000π (d/) = 4 d =16 d =4 Thus, the internal diameter of the pipe is 4 cm.