Q1. If (x – 2) and (x + 3) are the factors of the equation x2 + k1x + k2 = 0, then what are the values of k1 and k2?
(a) k1 = 6, k2 = – 1
(b) k1 = 1, k2 = – 6
(c) k1 = 1, k2 = 6
(d) k1 = – 6, k2 = 1
Q2. If (x – y) = 7, then what is the value of (x – 15)^3 – (y – 8)^3?
(a) 0
(b) 343
(c) 392
(d) 2863
Q3. If x – y – √18 = –1 and x + y – 3√2 = 1, then what is the value of 12xy(x^2 – y^2)?
(a) 0
(b) 1
(c)512√2
(d) 612√2
Q5. In triangle ABC, a line is drawn from the vertex A to a point D on BC. If BC = 9 cm and DC = 3 cm, then what is the ratio of the areas of triangle ABD and triangle ADC respectively?
(a) 1 : 1
(b) 2 : 1
(c) 3 : 1
(d) 4 : 1
Q6. PQR is a right angled triangle in which ∠R = 90°. If RS ⊥ PQ, PR = 3 cm and RQ = 4 cm, then what is the value of RS (in cm)?
(a) 12/5
(b) 36/5
(c) 5
(d) 2.5
Q7. In triangle PQR, A is the point of intersection of all the altitudes and B is the point of intersection of all the angle bisectors of the triangle. If ∠PBR =105°, then what is the value of ∠PAR (in degrees)?
(a) 60
(b) 100
(c) 150
(d) 115
Q8. If there are four lines in a plane, then what cannot be the number of points of intersection of these lines?
(a) 0
(b) 5
(c) 4
(d) 7
Q9. How many times the digit “3” appears in numbers from 1 to 100
(a) 18
(b) 19
(c) 20
(d) 21
Q10. Sum of the greatest 8-digit number and the smaller 9 digit number is
(a) 19999999
(b) 199999999
(c) 999999999
(d) 10000999
Solutions
S1. Ans.(b)
Sol. If (x– 2) & (x + 3) are factors, the x = 2 and –3 satisfied the eqn.
4 + k1 × 2 + k2 = 0…(i)
9 – 3k1 + k2 = 0…(ii)
From these eqn k1 = 1. k2 = –6
S2. Ans.(a)
Sol. x – y = 7
take
x = 15
& y = 8 then satisfied the eqn.
So, (x – 15)³ – (y – 8)³ = (15 – 15)³ – (8 – 8)³
= 0
S3. Ans.(d)
Sol. x – y -√18= -1, x + y -3√2=1
x-y=√18-1 ….(i)
x+y=1+3√2….(ii)
So,
(x^2-y^2 )= 17
From (i) & (ii)
x=√18
y = 1
So,
12xy (x² – y²) = 12 × √18×1 (17)
=612√2
S4. Ans.(b)
Sol. p/q=r/s=t/4=√5
then, ((3P^2+4r^2+5t^2)/(3q^2+4r^2+〖5u〗^2 ))=(√5)^2 = 5
S5. Ans.(b)
Sol.
BD = 6 cm
DC = 3 cm
Height will be the same of both triangles
So,
Area of ∆ = 1/2 × base × height
(Area of ABD)/(Area of ADC)=BD/DC=6/3=2/1
