Q2. If the coordinate of the points A, B, C be (4, 4) (3-2) and (3-16) respectively, then the area of the triangle ABC is?
(a) 12 square unit
(b) 10 square unit
(c) 14 square unit
(d) 7 square unit
Q4. When two circle touch externally, the number of common tangent are:
(a) 4
(b) 3
(c) 2
(d) 1
Q6. M and N are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases state whether MN is parallel to QR
(a) PM = 4, QM = 4.5, PN = 4, NR = 4.5
(b) PQ = 1.28, PR = 2.56, PM = 0.16, PN = 0.32
(c) None of these
(d) Both of them
Q7. Without stoppage a train travels at an average speed of 75 km per hour and with stoppages it covers the same distance at an average speed of 60 km/hr. How many minutes per hour does the train stop?
(a) 10 minutes
(b) 12 minutes
(c) 13 minutes
(d) 14 minutes
Q8. Two trains running in the same direction at 40 kmph and 22 kmph completely pass one another in 1 minute. If the length of first train is 125 metres, the length of the second train is
(a) 125 m
(b) 150 m
(c) 175 m
(d) Can’t be determined
Q9. The distance between two stations, Delhi and Amritsar, is 450 km. A train starts at 4 p.m. from Delhi and moves towards Amritsar at an average speed of 60 km/hr. Another train starts from Amritsar at 3.20 p.m. and moves towards Delhi at an average speed of 80 km/hr. At what time will the both trains meet?
(a) 5.30 p.m.
(b) 5.50 a.m.
(c) 6.50 p.m.
(d) 6.30 a.m.
Q10. A, B and C can walk at the rates of 3, 4 and 5 km per hour respectively. They starts from Pune at 1, 2, 3 o’clock respectively. When B catches A, B sends him back with a message to C. When will C get the message?
(a) 4.15 o’clock
(b) 5.15 o’clock
(c) 6.25 o’clock
(d) Can’t be determined
Solutions
S1. Ans.(a)
Sol. In ∆ADE & ∆ABC
∆ADE ≅ ∆ABC
Therefore,
(AD/DB)=(AE/EC)
(3/5)=(AE)/(AC – AE)
3(5.6 – AE) = 5AE
3 × 5.6 = 8AE
AE=(3 × 5.6)/8
AE = 2.1 cm
S2. Ans.(d)
Sol. Area of a triangle =1/2 [x1(y2-y3)+x2(y3-y1)+x3 (y1-y2)]
=(1/2)[4(-2+16)+3(-16-4)+3(4+2)]
=1/2 [56-60+18]=7
S3. Ans.(a)
Sol.
AO=√(OQ^2-AQ^2 )=√(5^2-4^2 )=√9=3
Now, from similar ∆s QAO & QPR
PR = 2OA = 2 × 3 = 6
S5. Ans.(c)
Sol. We know that,
Sum of exterior angle in any polygon = 360°
So,
x + 90 + 115 + 75 = 360°
x = 80°
S6. Ans.(a)
Sol. The triangle PQR is isosceles
⇒ MR/QR by converse of proportionally theorem.
S7. Ans.(b)
Sol. Time of rest per hour
=(Difference in average speed )/(Speed without stoppage )
=(75 – 60)/75=(1/5) hr. = 12 minutes.
S8. Ans.(c)
Sol. Length of the second train = Relative speed × Time taken to cross each other – Length of first train
={(40-22)x(5/18)}×60-125=175 m.
S9. Ans.(c)
Sol. Let first train also starts at 3 : 20 and add 40 km. to the total distance
Total distance = 490 km
Total speed = 140 km/hr
Total time =D/S=490/140 = 3.5 hr.
∴ They will meet at 3 : 20 p.m. + 3.5 hr.
= 6:50 p.m.
S10. Ans.(b)
Sol. Distance already covered by A when B starts moving at 2o’clock = 3km/hr × (2 – 1) hr. = 3 kms
Now, time taken by B to catch A
=(3 km )/(4-3) km/hr= 3 hrs
B will catch A at (2’o clock + 3 hrs) i.e. at 5 o’clock
Distance where B will catch A = 4 km/hr × (5 – 2) hr. = 12 km.
Also at 5’o clock, C will be at 5 km/hr × (5 – 3) hrs = 10 kms
i.e. at 5’o clock distance between (A or B) and C = (12 – 10)km = 2km
Now, time to meet A & C
(2 km )/(5+3) km/hr
=1/(4 hr)=15 min
Time of meeting of of A & C
= 5 : 15 o’clock
