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Maths Quiz for KVS and CTET Exams 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

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