**Q1. Two chords of length a unit and b unit of a circle make angles 60°and 90° at the centre of a circle respectively, then the correct relation between a and b is:**

(a) b = √2 a

(b) b = 2a

(c) b = √3 a

(d) b = 3/2a

**Q2. Internal bisectors of angle Q and angle R of triangle PQR intersect at O. If angle ROQ = 96° then the value of angle RPQ is:**

(a) 12°

(b) 24°

(c) 36°

(d) 6°

**Q3. In a given circle, the chord PQ is of length 18 cm. AB is the perpendicular bisector of PQ at M. If MB = 3. Find the length of AB**

(a) 25 cm

(b) 30 cm

(c) 28 cm

(d) 27 cm

**Q4. In triangle ABC, angle BAC = 90° and AD⊥BC. If BD = 3 cm and CD = 4 cm, then length of AD is**

(a) 2√3 cm

(b) 3.5 cm

(c) 6 cm

(d) 5 cm

**Q5. The hypotenuse of right-angles triangles is 39 cm and the difference of other two sides is 21 cm. Then, the area of the triangles is**

(a) 180 sq. cm

(b) 270 sq. cm

(c) 450 sq. cm

(d) 540 sq. cm

**Q6. A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is:**

(a) 45

(b) 40

(c) 50

(d) 55

**Q7. Total weekly emoluments of the workers of a factory is Rs. 1534. Average weekly emolument of a worker is Rs. 118. The number of workers in the factory is:**

(a) 16

(b) 14

(c) 13

(d) 12

**Q8. What is the smallest number which leaves remainder 3 when divided by any of the numbers 5, 6 or 8 but leaves no remainder when it is divided by 9?**

(a) 123

(b) 603

(c) 723

(d) 243

**Q9. LCM of 2/3,4/9,5/6 is**

(a) 8/27

(b) 20/3

(c) 10/3

(d) 20/27

**Q10. A, B and C hired a pasture. A grazed 12 cows 2 hours every day for 4 months , B grazed 16 cows, 4 hours every day for 6 months and C grazed 6 cows 9 hours everyday for 2 months. If B has paid Rs. 1152 as a share of fare. Find the amount of total Rent.**

(a) Rs. 1413

(b) Rs. 1214

(c) Rs. 1764

(d) Rs. 1102

**Solutions**

S6. Ans.(c)

Sol. Let the average scored till ‘10’ innings = x

According to the question

(10x+100)/11=x+5

10x + 100 = 11x + 55

x = 45

Average runs are = 45 + 5 = 50

S7. Ans.(c)

Sol. According to the question

Total weekly emoluments of the workers = Rs. 1534

Number of workers = 1534/118 = 13

S8. Ans.(d)

Sol. LCM (5, 6, 8) = 5 × 6 × 4 = 120

⇒ Required number gives remainder 3 when divided by (5, 6, 8) and zero remainder when divided by 9

∴ (120K + 3)/9=(3K + 3)/9

at K = 2, (3K + 3)/9⇒ Remainder = 0

We get 120K + 3 = 120 × 2 + 3 = 243 which is the required number.

S9. Ans. (b)

Sol. LCM of any fractions is

⇒(LCM of numerator)/(HCF of denominator)

⇒LCM(2/3,4/9,5/6)

⇒LCM(2,4,5)/HCF(3,9,6) =20/3

⇒20/3

S10. Ans. (c)