CBSE Set Qa4 Q22 Sample Test Papers For Class 10th for students online

Q22) In Fig. 4, AB is the chord of a circle with centre O. AB is produced to C such that BC = OB. CO is joined and produced to meet the circle in D. If ACD = y^o and AOD= x^o, prove that x = 3y (Marks 4)

Ans22)  BC = OB  (given) 
OCB = BOC = y^o [Angles opposite to equal sides are equal
OBA = BOC + OCB  [Ext. angles of a is equal to the sum of the opposite interior angles 
= OBA = y + y = 2y
OA = OB   …[Radii of the same circle]
OAB = OBA   ...[Angles opp. equal sides of a ]
= 2y
AOD = OAC + OCA
= 2y + y = 3y
x = 3y^o

Q23) Algorithm - Omitted being out Syllabus.  (Marks 4)

Q24) Flow Chart - Omitted, being out of Syllabus.  (Marks 4)

Q25) The population of two towns A and B are 8,76,000 and 6,90,000 respectively and the respective crude death rates are 14.3 and 16.2 respectively. Find to the nearest whole number, the crude death rate for the two towns taken together.  (Marks 4)
Ans25) Crude death rate 
= No. of deaths/ Population of the year x 100
.^.. 14.3 = (N₁/876000) x 1000
And 16.2 = (N₂/690000) x 1000
N₁ = 14.3 x 876 = 12526.8 
N₂ = 690 x 16.2 = 11178
.^.. Combined crude death rate 
= ((12526.8 + 11178)/(86000 + 690000)) x 1000
= 23704.8/1566 = 15.137

Section - C

Q26) Solve for x :
(x² - 16) - (x - 4) = (x² - 5x + 4)  (Marks 6)
Ans26) (x² - 16) - (x - 4) = (x² - 5x + 4) 
= ((x - 4) x (x + 4)) - ((x - 4) x (x - 4)) = (x² - 4x - x - 4)
= ((x - 4) x (x + 4)) - ((x - 4) x (x - 4)) = (x(x - 4) -1(x - 4))
= ((x - 4) x (x + 4)) - ((x - 4) x (x - 4)) = ((x - 4)(x - 1))
= (x - 4)[(x + 4) - (x - 4) = (x - 1)]
= (x - 4)[(x + 4) - (x - 4) - (x - 1)] = 0
Either (x - 4) = 0
= x = 4   ...(i)
or (x + 4) - (x - 4) - (x - 1) = 0
(x + 4) - (x - 4) = (x - 1)   ...(ii)
Squaring both sides of (ii), we get 
x + 4 + x - 4 - 2((x + 4)(x - 4)) = x - 1 
= x + x - x + 4 - 4 + 1 = 2((x + 4)(x - 4))
= x + 1 = 2(x² - 16)
Squaring both sides again (x + 1)² = 4(x² - 16)
= x² + 1 + 2x = 4x² - 64
= 4x² - x² - 2x - 64 - 1 = 0
= 3x² - 2x - 65 = 0
= 3x² - 15x + 13x - 65 = 0
= 3x(x - 5) + 13(x - 5) = 0
= (3x + 13)(x - 5) = 0
= 3x = -13 or x = 5
= x = -13/3 or x = 5
On checking, we find that x = -13/3 does not satisfy the equation 
.^.. x = 4, 5

Q27) A man on the deck of a ship is 16 m above water level. He observes that the angle of elevation of the top of a cliff is 45^o and the angle of depression of the base is 30^o, Calculate the distance of the cliff from the ship and the height of the cliff.  (Marks 6)
Ans27)  

Height of deck BC = 16 m  In rt OBC, BC/OB = tan 30o = 16/OB = 1/3 = OB = 163 m   ...(i)  In rt OBA BA/OB = tan 45o = BA/163 = 1   = BA = 163 m  ... Height of the cliff = AB + BC = 163 + 16 = 16(3 + 1) = 16(1.732 + 1.00) m  = 16 x 2.732 = 43.712 m

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