CBSE Set Qa7a Maths Class X 1998 Sample Test Papers For Class 10th for students online

Q2) Find the value of such that the quadratic equation ( - 4)x² + 2( - 4)x + 4 = 0  (Marks 2)
Ans2)  Here a = - 4, b = 2( - 4), c = 4
The quadratic equation has equal roots if 
b² - 4ac = 0
= [2( - 4)]² - 4( - 4)4 = 0
= 4( - 4)² - 4( - 4)4 = 0
= ( - 4)² - 4( - 4) = 0
= ( - 4)[ - 4 - 4] = 0
= ( - 4)( - 8) = 0
= = 4 or = 8
The value = 4 will reduce the equation to a constant only.
.^.. = 8

Q3) The GCD of two polynomials P(x) = 4x²(x² - 3x + 2) and Q(x) = 12x(x - 2)(x² - 4) is 4x(x - 2). Find the LCM of the polynomials P(x) and Q(x).  (Marks 2)
Ans3) LCM x GCD
= Product of the two polynomials 
.^.. [4x(x - 2)] x LCM
= 4x²(x² - 3x + 2) 12x(x - 2)(x² - 4)
= 4x²(x² - 2x - x + 2) 12x(x - 2)(x - 2)(x + 2)
= 48x³(x - 2)(x - 1)(x - 2)²(x + 2)
= 48x³(x - 2)³(x - 1)(x + 2)
.^.. LCM = (48x³(x - 2)³(x - 1)(x + 2))/4x (x - 2)
= 12x²(x - 2)²(x - 1)(x + 2) 
= 12x²(x - 2)(x - 1)(x - 2)(x + 2)
= 12x²(x² - 3x + 2)(^x2 - 4)

Q5) The sales price of a motor-cycle, inclusive of sales tax, is Rs. 38,520. If sales tax is charged at the rate of 7 % of the list price, find the list price of the motor-cycle.  (Marks 2)
Ans5) Let the list price of the motor-cycle = Rs x
Sales tax = 7%   .^.. Sales price = Rs. 107/100 . x
As per the question 107/100 . x = Rs 38,520
= x = 38520 x 100/107 = Rs. 36,000

Q8) Without using trigonometric tables, evaluate (cos² 20^o + cos² 70^o)/(sin² 59^o + sin² 31^o)  (Marks 2)
Ans8) cos 20^o = cos (90^o - 70^o) = sin 70^o
sin 59^o = sin (90^o - 31^o) = cos 31^o
.^.. (cos² 20^o + cos² 70^o)/(sin² 59^o + sin² 31^o) = (sin² 70^o + cos² 70^o)/(cos² 31^o + sin² 31^o) = 1

Q10) The median of the following observation, arranged is ascending order is 22. Find x.
8, 11, 13, 15, x + 1, x + 3, 30, 35, 40, 43  (Marks 2)
Ans10)  Median = 22    n = 10 (even)
Median lies in (5th item + 6th item)/2
= Median = ((x + 1) + (x + 3))/2 = 22
= 2x + 4 = 44
= 2x = 44 - 4
= 2x = 40 
= x = 20

Q11) The mean of n observations x₁, x₂, x₃, ..., x_n is . If each observation is divided by p, prove that the mean of the new observation is /p.  (Marks 2)
Ans11) = (x₁ + x₂ + x₃ + ... + x_n)/n
New mean = (x₁/p + x₂/p + x₃/p + ... + x_n/p)/n
= [x₁ + x₂ + x₃ + ... + x_n]/n = /p

Q16) In a flight of 6000 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 400 km/ hour and time increased by 30 minutes. Find the original duration of flight.   (Marks 4)
Ans16) Let the usual speed of aircraft = x km/hr 
Reduced speed of aircraft = (x - 400) km/ hr 
Distance = 6000 km 
According to the question 
6000/(x - 400) - 6000/x = 30 minutes   [^..^. Time = Distance/Speed]
(6000x - 6000x + 2400000)/x(x - 400) = 1/2 hr 
= x² - 400x = 4800000
= x² - 400x - 4800000 = 0
= x² - 2400x + 2000x - 4800000 = 0
= x(x - 2400) + 2000(x - 2400) = 0
= (x - 2400)(x + 2000) = 0
= x = 2400 or x = -2000
^..^. Speed can never be -ve
.^.. Speed = 2400 km/hr 
Time = D/S = 6000/2400 = 5/1 = 2 ½hours

Q21) In ABC, If AD is the median, show that AB² + AC² = 2(AD² + BD²)   (Marks 4)
Ans21) Given ABC, AD is the median.

To prove AB² + AC² = 2(AD² + BD²)
Proof ABC is a triangle and AD is the median. Draw AE  |  BC
AB² = BE² + AE²    …(I)
AC² = EC² + AE²    …(II)
Adding (I) and (II) 
AB² + AC²
= 2AE² + BE² + EC²
= 2AE² + (BD - ED)² + (DC + ED)²
= 2AE² + BD² + ED² + DC² + ED²
= 2AE² + 2ED² + BD² + DC²
= 2(AE² + ED²) + BD² + DC²
= 2AD² + BD² + BD²
= 2[AD² + BD²]

Q26) Solve for x :
(x² - 36) - (x - 6) = 2(x² - 15x + 54)  (Marks 6)
Ans26) (x² - 36) - (x - 6) = 2(x² - 15x + 54) 
= ((x - 6) x (x + 6)) - (x - 6) = 2(x² - 6x - 9x - 54)
= ((x - 6) x (x + 6)) - (x - 6)² = 2(x(x - 6) - 9(x - 6))
= ((x - 6) x (x + 6)) - (x - 6)² = 2((x - 6)(x - 9))
= (x - 6) x (x + 6) - (x - 6) x (x - 6) - 2(x - 6)(x - 9) = 0
= (x - 6)[(x + 6) - (x - 6) - 2(x - 9)] = 0
Either (x - 6) = 0
= x = 6   ...(i)
or (x + 6) - (x - 6) - 2(x - 9) = 0
= (x + 6) - (x - 6) = 2(x - 9)   ...(ii)
Squaring both sides of (ii), we get 
x + 6 + x - 6 - 2((x + 6)(x - 6)) = 4(x - 9) 
= 2x - 2(x² - 6x + 6x - 36) = 4x - 36
= -2x + 36 = 2(x² - 36)
= -x + 18 = (x² - 36)
Squaring both sides again
= x² + 324 + 2(-x)(18) = x² - 36
= 324 - 36x = -36
= 36x = 324 + 36 360
= x = 10
By checking, we find that x = 6, 10