**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