Subject

Mathematics

Class

CBSE Class 10

Pre Boards

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Sample Papers

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 Multiple Choice QuestionsShort Answer Type

11.

In Fig. , O is the centre of a circle such that diameter AB = 13 cm and AC = 12 cm. BC is joined. Find the area of the shaded region. (Take π = 3.14)

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12.

In Fig., a tent is in the shape of a cylinder surmounted by a conical top of the same diameter. If the height and diameter of cylindrical part are 2.1 m and 3 m respectively and the slant height of conical part is 2.8 m, find the cost of canvas needed to make the tent if the canvas is available at the rate of Rs. 500/sq. metre. (use π = 22/7)

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13.

If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b,a + b). Prove that bx = ay.

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14.

In Fig.  find the area of the shaded region, enclosed between two concentric circles of radii 7 cm and 14 cm where ∠AOC = 40°. (use π =22/7)

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15.

If the ratio of the sum of first n terms of two A.P’s is (7n +1): (4n + 27), find the ratio of their mth terms.


Let a1, a2 be the first terms and d1 , d2 the common differences of the two given A.P's.
Then, we have Sn =straight S subscript straight n space equals straight n over 2 left square bracket 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 right square bracket space and space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 right square bracket
therefore straight S subscript straight n over straight S subscript straight n equals fraction numerator straight n over 2 left square bracket 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 right square bracket over denominator straight n over 2 left square bracket 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 right square bracket end fraction space equals fraction numerator 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 space over denominator 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 end fraction
It space is space given space that space
straight S subscript straight n over straight S subscript straight n space equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction
therefore space fraction numerator 2 straight a subscript 1 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 1 space over denominator 2 straight a subscript 2 plus left parenthesis straight n minus 1 right parenthesis straight d subscript 2 end fraction equals fraction numerator 7 straight n plus 1 over denominator 4 straight n plus 27 end fraction space.. space left parenthesis 1 right parenthesis
To space find space the space ratio space of space the space straight m to the power of th space terms space of space the space tow space given space straight A. straight P apostrophe straight s space
replace space straight n space by space left parenthesis 2 straight m minus 1 right parenthesis space in space equation space 1
therefore space equals fraction numerator 2 straight a subscript 1 plus left parenthesis 2 straight m minus 2 right parenthesis straight d subscript 1 over denominator 2 straight a subscript 2 plus left parenthesis 2 straight m minus 2 right parenthesis straight d subscript 2 end fraction equals fraction numerator 14 straight m minus 7 plus 1 over denominator 8 straight m minus 4 plus 27 end fraction
therefore space fraction numerator straight a subscript 1 plus left parenthesis straight m minus 1 right parenthesis straight d subscript 1 over denominator straight a subscript 2 left parenthesis straight m minus 1 right parenthesis straight d subscript 2 end fraction space equals space fraction numerator 14 straight m minus 6 over denominator 8 straight m plus 23 end fraction
Hence comma space the space ratio space of space the space straight m to the power of th space terms space of space the space two space straight A. straight P apostrophe straight s space is space 14 straight m minus 6 colon 8 straight m plus 23

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16.

Solve for x:
fraction numerator 1 over denominator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x minus 2 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis straight x minus 2 right parenthesis left parenthesis straight x minus 3 right parenthesis end fraction space equals 2 over 3 comma space straight x not equal to 1 comma 2 comma 3

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17.

A conical vessel, with base radius 5 cm and height 24 cm, is full of water. This water is emptied into a cylindrical vessel of base radius 10 cm. Find the height to which the water will rise in the cylindrical vessel. (π = 22/7)

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18.

A sphere of diameter 12 cm, is dropped in a right circular cylindrical vessel, partly filled with water. If the sphere is completely submerged in water, the water level in the cylindrical vessel rises by 3 5 over 9 cm. Find the diameter of the cylindrical vessel.

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19.

A man standing on the deck of a ship, which is 10 m above water level, observes the angle of elevation of the top of a hill as 60° and the angle of depression of the base of a hill as 30°. Find the distance of the hill from the ship and the height of the hill.

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20.

Three different coins are tossed together. Find the probability of getting
(i) exactly two heads
(ii) at least two heads
(iii) at least two tails.

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