Dear Students!!! There is most general as well as a scoring section in all the competitive entrance examinations in the teaching field i.e “Mathematics”.Because in this section only one thing is work i.e your accuracy and that could be nourished with the daily practice. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2018, DSSSB ,KVS,STET Exam.
Q1. A man deposited Rs. 9000 in a bank at 6% per annum for 4 yr. For how many years, must another man deposit an amount of Rs. 5400 at 5% per annum in another bank so that both of them get the same interest?
(a) 6 yr
(b) 4 yr
(c) 3 yr
Q2. The compound interest on a sum of money at 5% per annum for 3 yr is Rs. 2522. What would be the simple interest on this sum at the same rate and for the same period?
(a) Rs. 2200
(b) Rs. 2800
(c) Rs. 2500
(d) Rs. 2400
Q3. A number is selected at random from the set {1, 2, 3, …, 50}. The probability that it is a prime, is
(a) 0.1
(b) 0.2
(c) 0.3
(d) 0.7
Q4. A and B are two events such that P (A) = 0.3 and P (A ∪ B) = 0.8. If A and B are independent events, then P (B) is
(a) 2/3
(b) 3/8
(c) 5/7
(d) 1/5
Q5. A cyclist at C is cycling towards B. How far will he have to cycle from C before he is equidistant from both A and B?
(a) 3 km
(b) 6 km
(c) 5 km
(d) 4 km
Q6. In the given figure, EC || AB, ∠ECD = 70°, ∠BDO = 20°, then ∠OBD is
(a) 70°
(b) 60°
(c) 50°
(d) 20°
Q7. In ∆PQR, PS is the bisector of ∠P and PT⊥QR, then ∠TPS is equal to
Q8. In the figure, XAY is a tangent to the circle with centre at O if ∠BAX = 70°, ∠BAQ = 40°, then ∠ABQ is equal to
(a) 20°
(b) 30°
(c) 35°
(d) 40°
Q9. The sides AB and CD of a cyclic quadrilateral ABCD are produced to meet at P. The sides AD and BC are produced to meet at Q. If ∠ADC = 85° and ∠BPC = 40°, then ∠BAD is equal to
(a) 45°
(b) 30°
(c) 95°
(d) 55°
Q10. In the given figure, PQ is a tangent at a point R of the circle with centre O. If ∠TRQ = 30°, then ∠PRS is equal to
(a) 60°
(b) 65°
(c) 50°
(d) 90°
Solutions
S8. Ans.(b)
Sol. ∠QAX = ∠BAX – ∠BAQ
= 70° – 40° = 30°
∠BAY = 180° – ∠BAX = 180° – 70° = 110°
∠EBA = 90° [∵ angle in semi-circle]
∠BAY = ∠AQB = 110°
∠ABQ = 180° – (∠ BAQ + ∠AQB)
= 180° – (40° + 110°) = 30°
S9. Ans.(d)
Sol. In ∆ADP,
∠PAD + ∠ADP + ∠APD = 180°
⇒ ∠BAD + ∠ADP + ∠APD = 180°
⇒ ∠BAD + 85° + 40° = 180°
∴ ∠BAD = 180° – 125° = 55°
S10. Ans.(a)
Sol. ∠SRT = 90° [∵ angle in semi-circle]
∠QRT + ∠SRT + ∠PRS = 180°
∠PRS = 180° – (30° + 90°)
⇒ ∠PRS = 60°
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