CBSE X Mathematics All India 0 Solution (SET ) Section D Q9. The numerator of a fraction is less than its denominator. If is added to the denominator, the fraction is decreased by. Find the fraction. 5 Let the denominator of the fraction be x then the numerator will be (x ). x Original fraction = x It is given that the new fraction is obtained after adding to the denominator. x New fraction = x According to the given condition, x x x x 5 x x x x 5 x x x x x x 5 5 x x x x x x 5(x ) = x + x 5x 45 = x + x x + x 5x + 45 = 0 x 4x + 45 = 0 x 9x 5x + 45 = 0 x(x 9) 5(x 9) = 0 (x 5) (x 9) = 0 x = 5 or x = 9 Denominator = 5 or 9 Numerator = 5 = or 9 = 6 6 Hence, the fraction is or. 5 9 OR In a flight of 800 km, an aircraft was slowed down due to bad weather. Its average speed is reduced by 00 km/h and time increased by 0 minutes. Find the original duration of the flight.
CBSE X Mathematics All India 0 Solution (SET ) Let the original speed of the plane be x km/hr. If the speed of the plane is reduced by 00 km/hr, then Reduced speed of the plane = (x 00) km/hr 800 Time taken by the plane to reach its destination at original speed, t hr x 800 Time taken by the plane to reach its destination at reduced speed, t hr x 00 Given, Time taken by the plane to reach its destination at reduced speed Time taken by the plane to reach its destination at original speed = 0 minutes t t hr 800 800 x00 x 800x800 x00 x x00 800 00 x 00 x x 00 560000 0 x x x x x x x 800 x 700 0 800 700 560000 0 800 700 800 0 x800 0 or x 700 0 x 800 or x 700 x 800 Speed cannot be negative Original speed of flight = 800 km/hr 800 Original duration of flight, t 800 x 7 Hrs Thus, the original duration of the flight is hours 0 minutes. Q0. Find the common difference of an A. P. whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. Let the common difference of the given A. P. be d. First term (a) = 5 (Given) We know that sum of the first n terms of an A.P. is given as:
CBSE X Mathematics All India 0 Solution (SET ) n Sn a n d 4 Sum of first four terms S4 5 4 d 0 d 0 6d And, sum of next four terms = S 8 S 4 8 5 8 (0 6 ) d d 40 8d 0 6d 0 d According to the given condition, S4 S8 S4 0 6d 0 d 0 6d 0 d d 6d 0 0 5d 0 d d 0 5 Hence, the common difference of the given A.P. is. Q. Prove that the length of tangents drawn from an external point to a circle are equal. Here is the link for the solution. http://www.meritnation.com/discuss/question/465554/how-to-prove-that-thelength-of-tangents-drawn-from-an-external-point-to-a-circle-are-equal Q. A hemispherical tank, full of water, is emptied by a pipe at the rate of 5 litres per sec. 7 How much time will it take to empty half the tank if the diameter of the base of the tank is m? It is given that, diameter of base of tank = m
CBSE X Mathematics All India 0 Solution (SET ) Radius, r m Volume of water in the hemispherical tank π r m 7 99 m 4 5 Rate of flow of water out of the pipe litres / sec 7 Let the time taken to empty half the tank be t sec. Rate of flow of water t sec Volumeof waterin thehemispherical tank 5 99 t litre m 7 4 5 99 t m m litre m 7 000 4 000 t 990 Time taken to empty half the tank is (960 + 0) sec = 6 min 0 sec OR A drinking glass is in the shape of the frustum of a cone of height 4 cm. The diameters of its two circular ends are 4 cm and cm. Find the capacity of the glass. Use 7 Here is the link for the solution. http://cbse.meritnation.com/studyonline/solution/math/avhmqqvrvi@qlo9qijjoa!! Q. A military tent of height 8.5 m is in the form of a right circular cylinder of base diameter 0 m and height 5.5 m surmounted by a right circular cone of same base radius. Find the length of the canvas use in making the tent, if the breadth of the canvas is.5 m.
CBSE X Mathematics All India 0 Solution (SET ) 0 It is given that, radius of cylinder (r ) m = 5 m Height of cylinder (h ) = 5.5 m And, height of the tent (H) = 8.5 m So, height of cone (h ) = 8.5 m 5.50 m =.75 m And, radius of cone (r ) = 5 m Let the slant height of the cone be l m. l 5m.75m l (5 7.565) m l.565 m l 5.5 m Curved surface area of the tent = Curved surface area of cylinder + Curved surface area of cone πr h πr l 5 5.5 cm 5 5.5 cm 7 7 (58.57 78.9) cm = 7.5m Now, curved surface area of the tent is equal to the area of rectangular piece of canvas. It is given that, breadth of canvas =.5 m Let l be the length of canvas. l.5 = 7.5
CBSE X Mathematics All India 0 Solution (SET ) 7.5 l 85m.5 Hence, the length of canvas used in making the tent is 85 m. Q4. The angles of elevation and depression of the top and bottom of a light-house from the top of a 60 m high building are 0 and 60 respectively. Find (i) the difference between the heights of the light-house and the building. (ii) the distance between the light-house and the building. Here is the link for the solution. http://www.meritnation.com/discuss/question/865658/the-angles-of-elevation-andthe-depression-of-the-top-and-bottom-of-a-light-house-from-the-top-of-a-60-mhigh-building-are-0-and-60-respectively