CBSE Set Qa4 Maths Class X 1999 Sample Test Papers For Class 10th for students online

Q22) Through the mid-point M of the side CD of a parallelogram ABCD, the line BM is drawn intersecting AC in L and AD produced in E. Prove that EL = 2BL. (Marks 4)
Ans22)  

Given || gm ABCD. M is the mid-point of CD Line BM is drawn intersecting AC in L and AD produced in E To Prove. EL = 2 BC Proof. In BMC and EMD 1 = 2    (vertically opposite angles) 3 = 4    (interior alternate angles) CM = MD (given) = BMC ~ EMD  (A.A.S.) = BC = DE   (C.P.C.T.) But BC = AD   (opposite sides of ||gm = 2BC = AE   - (i)  (AD + DE = AE)

Consider ALE and CLB
3 = 4   (interior alternate angle)
5 = 6   (vertically opposite angles)
= ALE ~ CLB  (A.A)
= AE/BC = EL/BL   corresponding sides of similar s are proportional
= 2BC/BC = EL/BL   (using (i))
= EL = 2BL

Q23) Flow Chart. Deleted from the syllabus. (Marks 4)

Q24) Algorithm. Deleted from the syllabus  (Marks 4)

Q25) Find the cost of living index number for the year 1998 assuming 1996 as the base year from the following data:  (Marks 4)

Ans25) 

Commodity	Quantity (in units) qo	Price per unit (in rupees)		poqo	p1qo
		1996 po	1998 p1
A	15	13	14	195	210
B	10	12	18	120	180
C	25	9	10	225	250
D	12	25	30	300	360
E	8	70	75	560	600
				poqo = 1400	p1qo =1600

Cost of living index number = p₁q_o/ p_oq_o x 100
= 1600/1400 x 100
= 800/7
= 114.29

Section - C

Q26) Solve for x :
2(x² + 1/x²) - 3(x - 1/x) - 4 = 0  (Marks 6)
Ans26) 2(x² + 1/x²) - 3(x - 1/x) - 4 = 0
Let x - 1/x = y
squaring 
(x - 1/x)² = y² = x² - 1/x² - 2 = ^y2 = x² + 1/x² = y² + 2
Substituting there values in the given equation, we get
2(y² + 2) - 3(y) - 4 = 0
= 2y² + 4 - 3y - 4 = 0
= 2y² - 3y = 0
= y(2y - 3) = 0
= y = 0 or y = 3/2

.^.. x = -1, -1/2, 2, 1

Q27) ) A pole 5 m high is fixed on the top of a tower. The angle of elevation of the top of the pole observed from a point A on the ground is 60^o and the angle of depression of the point A from the top of the tower is 45^o. Find the height of the tower.  (Marks 6)
Ans27)  

CD = height of the pole = 5 cm BC = height of the tower = x m AB = y m

ABC
tan 45^o = BC/AB
= 1 = x/y
= x = y   -(i)
ABD
tan 60^o = BD/AB
= 3 = (5 + x)/y
= 3 = 5 + x / x   using (i)
= x (3 - 1) = 5
= x = 5/(3 - 1) x (3 + 1)/(3 + 1)
= x = 5(1.732 + 1)/2
= x = 5 x 1.366
= x = 6.83
= Height of tower = 6.83 m 

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