CBSE Set Qa4 Maths Class X 2000 Sample Test Papers For Class 10th for students online

Q22) In fig. AB is a diameter of circle with centre O, OA = 7 cm. Find the area of the shaded region (use = 22/7). (Marks 4)

Ans22)  Area of circle with OB as diameter = 77/2 cm² = 38.5 cm²
Area of semi-circle with OA as radius = 77 cm²
Area of in the semicircle = 1/2 x 14 x 7 = 49 cm²
Area of the shaded portion in the semi-circle = 28 cm²
.^.. Total area of the shaded region = (38.5 + 28) = 66.5 cm²

Q23) In fig. AB and CD are two parallel tangents to a circle with centre O. ST is tangent segment between two parallel tangents touching the circle at Q. Show that SOT = 90^o. (Marks 4)

Ans23) Given AB and CD are two parallel tangents at P and Q resp. to the circle with centre O. ST is a tgt. at Q, meets AB at S and CD at T.
To Prove. SOT = 90^o
Const. Join OR, OP and OQ.
Proof. OPS = OQS = 90^o  (angle between the tangent and radii drawn through the point of contact.)
OP = OQ = r (radii of the same circle)
OS = OS  (common)
= OPS ~ OQS  (RHS)
= 1 = 2    (CPCT)
similarly 3 = 4
1 + 2 + 3 + 4 = 180^o   (angles on a straight line)
= 22 + 23 = 180^o   ( 1 = 2, 3 = 4)
= 2 + 3 = 90^o
= SOT = 90^o

Q24) Construct a quad. ABCD in which AB = 2.5 cm BC = 3.5 cm, AC = 4.2 cm, CD = 3.5 cm and AD = 2.5 cm. Construct another quad. AB'C'D' with diagonal AC' = 6.3 cm such that it is similar to quad ABCD.  (Marks 4)
Ans24) 

Steps :-
1. Construct quad ABCD with the given measurements.
2. Extend AC to AC' s.t AC' = 6.3 cm
3. C'D' || CD, C' B' || CB
4. AB'C'D' is the required quadrilateral.

Q25) Find the cost of living index number for the year 1995 assuming 1990 as the base year.  (Marks 4)

Ans25) 

Commodity	Quantity qo	Rate per kg (in rupees)		Total expenditure
		in 1990 po	in 1995 p1	1990 poqo	1995 p1qo
A	10	7	10	70	100
B	15	12	20	180	300
C	8	25	25	200	200
D	25	12	20	300	500
E	5	50	60	250	300
				poqo = 1000	p1qo =1400

Cost of living index number = p₁q_o/ p_oq_o x 100
= 1400/1000 x 100
= 140

Section - C

Q26) Solve for x : 9^{x + 2} - 6 x 3^{x + 1} + 1 = 0  (Marks 6)
Ans26) 3^2x . 3⁴ - 6 . 3^x . 3 + 1 = 0
= 81 x (3^x)² - 18 x 3^x + 1 = 0
Put 3^x = y
= 81y² - 18 y + 1 = 0
= (9y - 1)² = 0
= y = 1/9
 at y = 1/9
3^x = y
= 3^x = 1/9
= 3^x = 1/3²
= 3^x = 3^-2
= x = -2

Q27) ) If a radius of the circular and of a conical bucket, which is 45 cm high are 28 cm, 7 cm, find the capacity of the bucket.  (Marks 6)
Ans27)  

OAB ~ OCA By A.A. (^..^. AOB = COD common, ABO = CDO as AB || CD)
= OA/OC = AB/CD  ( corresponding sides prop)
= h/(h + 45) = 7/28
= h = 15 cm
Volume of small cone = 1/3 x 22/7 x 7 x 7 x 15 = 770 cm³
Volume of big cone = 1/3 x 22/7 x 28 x 28 x 60 = 49280 cm³
Volume of the bucket = 49280 - 770 = 48510 cm³

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