CBSE Set Qa6a Maths Class X 1998 Sample Test Papers For Class 10th for students online

Q1) Find the value of such that the quadratic equation
( - 3)x² + 4( - 3)x + 4 = 0 has equal roots. (Marks 2)
Ans1) Here a = - 3, b = 4( - 3)
The quadratic equation will have equal roots 
If b² = 4ac
.^.. [4( - 3)]² = 4( - 3) x 4
= (4)²( - 3)² = (4)²( - 3)
= ( - 3)² = ( - 3)
= ( - 3)² - ( - 3) = 0
= ( - 3)[ - 3 - 1] = 0
= ( - 3)( - 4) = 0
= = 3 or =4
But = 3 is not possible since it reduces the equation to a constant. Hence = 4

Q2) The GCD and LCM of two polynomials P(x) and Q(x) are x(x + a) and 12x²(x - a)(x² - a²) respectively. If P(x) = 4x²(x + a), find Q(x)  (Marks 2)
Ans2) GCD x LCM = P(x) Q(x)
x(x + a) 12x²(x - a)(x² - a²) = 4x²(x + a) Q(x)
= Q(x) = (x(x + a) 12x²(x - a)(x² - a²))/(4x²(x + a))
= Q(x) = (12x³(x + a)(x - a)(x² - a²))/(4x²(x + a))
= Q(x) = 3x(x - a)(x² - a²)

Q5) The sales price of a washing machine, inclusive of sales tax, is Rs. 26,160. If sales tax is charged at the rate of 9% of the list price, find the list price of the washing machine.  (Marks 2)
Ans5)  Let the list price of the washing machine = Rs. x
Sales tax = 9%  .^.. Amount paid = Rs 109/100 . x
As per the question 109/100 . x = Rs 26,160
= x = (26160 x 100)/109 = Rs. 24,000

Q9) Without using trigonometric tables, evaluate 
2(sin 43^o/cos 47^o) - (cot 30^o/tan 60^o) - 2 sin 45^o (Marks 2)
Ans9) 2(sin 43^o/cos 47^o) - (cot 30^o/tan 60^o) - 2 sin 45^o
= 2(sin (90o - 47^o)/cos 47^o) - (cot (90o - 60^o)/tan 60^o) - 2 sin 45^o
= 2(cos 47^o/cos 47^o) - (tan 60^o/tan 60^o) - 2 x 1/2
= 2 - 1 - 1 = 0

Q10) The mean of n observations 
x₁, x₂, x₃, ..., x_n is . If each observation is multiplied by p, prove that the mean of the new observation is p   (Marks 2)
Ans10) = (x₁ + x₂ + ... + x_n)/n
= (x₁p + x₂p + ... + x_np)/n
= p[x₁ + x₂ + ... + x_n]/n = p

Q11) The median of following observations, arranged in ascending order, is 25. Find x.
11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46  (Marks 2)
Ans11) n = 10 (even)   Median = 25
Median lies in (5th item +  6th item)/2
= Median = (x + 2 + x + 4)/2 = (2x + 6)/2 = x + 3
= x + 3 = 25   = x = 25 - 3 = 22

Q16) Factorise - Omitted being out of Syllabus.  (Marks 4)

Q18) In a flight of 3000 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 100 km/hour and time increased by one hour, find the original duration of flight.   (Marks 4)
Ans18) Let the usual speed of aircraft = x km/hr 
Reduced speed of aircraft = (x - 100) km/hr 
Distance = 3,000 km
According to the question 
3000/(x - 100)  - 3000/x = 1 hr   (... Time = Distance/speed)
= (3000x - 3000x + 300000)/x(x-100) = 1/1
= x(x - 100) = 300000
= x² - 100x - 300000 = 0 
= x² - 600x + 500x - 300000 = 0
= x(x - 600) + 500(x - 600) = 0
= (x - 600)(x + 500) = 0
= x = 600 or x = -500
^..^. Speed can never be-ve  .^.. speed = 600 km/hr 
Time = D/S = 3000/600 = 5 hours

Q22) In Fig, 5. ABD = CDB = PQB = 90^o. If AB = x units, CD = y units and PQ = z units, prove that 1/x + 1/y = 1/z  (Marks 4)

Ans22) In rt. ABD, 
AB/BD = tan 90^o   = x/BD = tan 90^o   ...(i)
In rt. PQB
PQ/BQ = tan 90^o   = z/BQ = tan 90^o   ...(ii)
From (i) and (ii) , we get
x/BD = z/BQ   = BD/x = BQ/z   ...(I)
Similarly we can prove that from rt. CDB.
y/BD = tan 90^o   ...(iii)
And from rt. PQB, we have z/BQ = tan 90^o   ...(iv)
From (iii) and (iv) y/BD = z/BQ
= BD/y = BQ/z   ...(II)
Adding I and II, we get
BD/x + BD/y = BQ/z + BQ/z
= BD(1/x + 1/y) = 2BQ(1/z)
= 1/x + 1/y = 1/z    ...[^..^. BD = 2BQ]

Q27) Solve for x :
(x² - 25) - (x - 5) = (x² - 16x + 55)  (Marks 6)
Ans27) (x² - 25) - (x - 5) = (x² - 16x + 55) 
= ((x + 5) x (x - 5)) - (x - 5) = (x² - 5x - 11x + 55)
= ((x + 5) x (x - 5)) - (x - 5) = (x(x - 5) - 11(x - 5))
= ((x + 5) x (x - 5)) - (x - 5) - (x - 5)(x - 11) = 0
= (x - 5)[(x + 5) - (x - 5) - (x - 11)] = 0
Either (x - 5) = 0
= x = 5   ...(i)
or (x + 5) - (x - 5) - (x - 11) = 0
(x + 5) - (x - 5) = (x - 11)   ...(ii)
Squaring both sides of (ii), we get 
x + 5 + (x - 5) - 2(x² - 25) = x - 11 
= x + 5 + x - 5 - 2(x² - 25) = x - 11
= x + 5 + x - 5 - x + 11 = 2(x² - 25)
= x + 11 = 2(x² - 25)
Squaring both sides again (x + 11)² = 4(x² - 25)
= x² + 121 + 22x = 4x² - 100
= 4x² - 100 - x² - 121 - 22x = 0
= 3x² - 22x - 221 = 0
= x = (22 + ((22)² + 4 x 3 x 221))/6
= x = (22 + (484 + 2652))/6
= x = (22 + 3136)/6
= x = (22 + 56)/6
= x = 78/6 or -34/6
= x = 13 or -17/3
On checking, we find that x = -17/3 does not satisfy the equation 
.^.. x = 5, 13

next