CBSE Set Qa1 Section Sample Test Papers For Class 10th for students online

Section - A

Q1) Determine the value of c for which the following system of linear equation has no solution:
cx + 3y = 3; 12x + cy = 6  (Marks 2)
Ans1)  a₁/a₂ = b₁/b₂ c₁/c₂
Here a₁ = c, b₁ = 3, c₁ = 3
        a₂ = 12, b₂ = c, c₂ = 6
a₁/a₂ = b₁/b₂    =c/12 = 3/c
= c² = 36   = c = + 6
if c = 6 then a₁/a₂ = b₁/b₂ = c₁/c₂  .^.. c = -6

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) = 4(x + a)², find Q(x)  (Marks 2)
Ans2) Product of the polynomials 
= LCM x GCD
= P(x) Q(x) = LCM x GCD
= 4x(x + a)² Q(x) = x(x + a) 12x²(x + a)(x² - a²)
= Q(x) = x(x + a) 12x²(x + a)(x + a)(x - a)/4x(x + a)²
= x . 12x²(x + a)²(x + a)(x - a)/4x(x + a)²
= 3x²(x + a)(x - a)
= 3x²(x² - a²)

Q3) If q is the mean proportional between p and r, show that pqr(p + q + r)³ = (pq + qr + pr)³  (Marks 2)
Ans3) q is the mean proportional between 
p and r = q² = pr
L.H.S = pqr(p + q + r)³
= q(pr)(p + q + r)³
= q(q²)(p + q + r)³    ...[^..^. q² = pr
= q³(p + q + r)   ...(i)
R.H.S.= (pq + qr +pr)³ = (pq + qr + q²)³
= [q(p + r + q)]³
= q³(p + r + q)³   ...(ii)
(i) = (ii) proves the result

Q4) Find the value of such that the quadratic equation ( - 12)x² + 2(  - 12)x + 2 = 0 has equal roots.  (Marks 2)
Ans4) The given quadratic equation will have equal roots if 
D = 0    = b² - 4ac = 0
Here a = ( - 12), b = 2( - 12), c = 2
b² - 4ac = 4( -12)² - 4x( - 12) x 2
= 4( - 12)( - 12) - 4( - 12) x 2
= ( - 12)[4( - 12) - 4 x 2]
= ( - 12)[4( - 12) - 8]
= ( - 12) 4[ - 12 - 2]
= 4( - 12)( - 14)
Now b² - 4ac = 0   = 4( - 12)( - 14) = 0
= = 12 or = 14
But 12 because in that case the given equation will imply 2 = 0 which is not true
.^.. = 14 

Q5) In Fig 1, OPQR is a rhombus, three of whose vertices lie on a circle with centre O. If the area of the rhombus is 323 cm², find the radius of the circle.  (Marks 2)

Ans5)  Const. Join OQ.
Proof. Since PQRO is a rhombus.
.^.. PQ = QR = OR = OP ...(i)
(In a rhombus, all sides are equal)
Since the diagonal of a rhombus divides it into two s which are equal in area 
.^.. ar (PQO) = ar(ROQ)
= 1/2 (ar of rhombus PORQ/2)
.^.. (PQO) =  323 / 2   = ar(POQ) = 163
In PQO, PO = OQ   ...[from (i)
.^.. PQO is an equilateral
.^.. ar PQO = 3/4 (side)²
= 163 = 3/4(OP)²
= OP² = 163x 4/3   = PO² = 64
.^.. PO = 64 = 8 cm  ^..^. sides of can't be -ve
Radius of circle = 8 cm

Q6) The sales price of a television, inclusive of sales tax, is Rs. 13,500. If sales tax is charged at the rate of 8% of the list price, find the list price of the television.  (Marks 2)
Ans6) Let the list price of T.V. = Rs. x
Sales tax = 8%
.^.. Sales price = Rs. 108/100 . x
As per the question 
108/100 . x = Rs. 13,500 
= x = 13500 x 100/108 = Rs. 12,500

Q7) Without using trigonometric tables evaluate : 2(cos 67^o/sin 23^o) - (tan 40^o/cot 50^o) - sin90^o  (Marks 2)
Ans7) The given expression
= 2(cos 67^o/sin 23^o) - (tan 40^o/cot 50^o) - sin90^o
= 2(cos (90^o - 23^o)/sin 23^o) - tan (90^o - 50^o)/cot 50^o - sin90^o
= 2(sin 23^o/sin 23^o) - cot 50^o/cot 50^o - sin90^o
= 2 - 1- 1 = 0

Q8) If cos /cos = m and cos /sin = n, show that
(m² + n²)cos² = n² (Marks 2)
Ans8) m/n = cos /cos x sin /cos = tan
By squaring both sides tan² = m²/n²
We know that sec² - tan² = 1
= sec² - m²/n² = 1
= sec² = 1+ m²/n²   = sec² = (m² + n²)/n²
= 1/cos² = (m² + n²)/n²
= (m² + n²) cos² =

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