Maths Questions for All Teaching Exam :28th February 2019(Solutions)_00.1
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Maths Questions for All Teaching Exam :28th February 2019(Solutions)

Maths Questions for All Teaching Exam :28th February 2019(Solutions)_40.1

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 2019, DSSSB ,KVS,STET Exam.
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Q1. If sum of six consecutive odd numbers is 168, then find the largest of these numbers.
(a) 23
(b) 27
(c) 31
(d) 33
Q2. Find the sum of all prime numbers less than 20.
(a) 92
(b) 73
(c) 19
(d) 77

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(a) 3
(b) 7
(c) 9
(d) 1
Q4. When the number 3⁹⁸ is divided by 5, the remainder is
(a) 3
(b) 4
(c) 1
(d) 2

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(a) 4
(b) 10
(c) 2
(d) 3
Q6. The product of two whole numbers is 24. The smallest possible sum of these numbers is
(a) 10
(b) 12
(c) 8
(d) 9
Q7. The value of 0.001 + 1.01 + 0.11 is
(a) 1.013
(b) 1.121
(c) 1.111
(d) 1.101
Q8. If x is an integer, then (x + 1)⁴ – (x – 1)⁴ is always divisible by
(a) 6
(b) 8
(c) 9
(d) 12
Q9. The hundred digit of a three digit number is 7 more than the unit digit. The digits of the number are reversed and the resulting number is subtracted from the original three digit number. The unit digit of the final number, so obtained is
(a) 0
(b) 1
(c) 2
(d) 3
Q10. The values of y for which the 4-digit number 51y3 is divisible by 9 is
(a) 2 or 3
(b) 0 or 9
(c) 0 or 3
(d) 3 or 9
Solutions
S1. Ans.(d)
Sol. Let the odd number be 2x + 1.
Then, 1(2x – 5) + (2x – 3) + (2x – 1) + (2x + 1) + (2x + 3) + (2x + 5) = 168
⇒ 12x = 168 ⇒ x = 14
∴ Largest is 2x + 5 = 28 + 5 = 33
S2. Ans.(d)
Sol. 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 = 77

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S4. Ans.(b)
Sol. Unit digit in 3⁹⁸ = Unit digit in (3⁴)²⁴ × Unit digit in (3²) = 9
∴ Dividing 3⁹⁸ by 5 or dividing 9 by 5, we get the remainder 4.

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S6. Ans.(a)
Sol. ∵ Product of two whole numbers is 24.
Possible pairs = (1, 24), (2, 12), (3, 8), (4, 6)
∴ Sum of these pairs
(1, 24) = 1 + 24 = 25
(2, 12) = 2 + 12 = 14
(3, 8) = 3 + 8 = 11
(4, 6) = 4 + 6 = 10
∴ Smallest possible sum = 10

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S8. Ans.(b)
Sol. (x + 1)⁴ – (x – 1)⁴
= (x⁴ + 4x³ + 6x² + 4x + 1) – (x⁴ – 4x³ + 6x² – 4x + 1)
= 2 (4x³ + 4x) = 8 (x³ + x)
∴ It is divisible by 8.
S9. Ans.(d)
Sol. Let three digit number = 111.
Since, digit at hundred place is 7 more than the digit at unit’s place.
∴ Digit at hundred place = 1 + 7 = 8
∴ Required three-digit number = 811
Now, changing the numbers new number = 118
According to the question,
811 – 118 = 693
Now, unit place digit in 693 = 3
∴ Required answer = 3
S10. Ans.(b)
Sol. Given, four digit number is 51y3.
Sum of digits = 5 + 1 + y + 3 = 9 + y
∴ We know that, the given number 51y3 will be divisible by 9 if its sum of digits is divisible by 9. So, 9 + y is divisible by 9. Which is only possible, when we take y = 0 or 9.

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