CTET/ UPTET Exam Practice Mathematics Questions
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.With proper system, Study Notes, Quizzes, Vocabulary one can quiet his/her nerves and exceed expectations in the blink of an eye. So, for this, we are providing you the daily quiz for all teaching exams i.e CTET Exam 2019, DSSSB ,KVS,STET Exam.
Q1. What least value must be given to n so that the number 6135n2 becomes divisible by 9?
Q2. Find the multiple of 11 in the following numbers.
Q3. The number 89715938* is divisible by 4. The unknown non-zero digit marked as * will be
Q4. A number is divisible by 11 if the difference between the sums of the digits in odd and even places respectively is
(a) a multiple of 3
(b) a multiple of 5
(c) zero or a multiple of 7
(d) zero or a multiple of 11
Q5. If m and n are integers divisible by 5, which of the following is not necessarily true?
(a) m + n is divisible by 10
(b) m – n is divisible by 5
(c) m² – n² is divisible by 25
(d) None of these
Q6. 325325 is a six-digit number. It is divisible by
(a) 7 only
(b) 11 only
(c) 13 only
(d) all 7, 11 and 13
Q7. If a and b are two odd positive integers, by which of the following integers is (a⁴ – b⁴) always divisible?
Q8. A number is multiplied by 11 and 11 is added to the product. If the resulting number is divisible by 13, the smallest original number is
Q9. If the square of an odd natural number is divided by 8, then the remainder will be
Q10. The largest number that exactly divides each number of the sequence 1⁵ – 1, 2⁵ – 2, 3⁵ – 3, …, n⁵ – n, … is
Sol. 6135n2 is divisible by 9 if (6 + 1 + 3 + 5 + n + 2) = (17 + n) is divisible by 9.
This happens when the least value of n is 1.
Sol. In 978626, we have (6 + 6 + 7) – (2 + 8 + 9) = 0.
Hence, 978626 is completely divisible by 11.
Sol. Let the missing digit be x. Then, (80 + x) must be divisible by 4.
Hence, x = 4.
Sol. Clearly, (d) is true.
Sol. Take m = 15 and n = 20. Then, each one of m and n is divisible by 5. But, (m + n) is not divisible by 10.
Hence, (m + n) is divisible by 10 is not true.
Sol. Let a = 2k + 1 & b = 2m + 1. Then,
a⁴ – b⁴ = (a² – b²) (a² + b²) = (a + b) (a – b) (a² + b²)
= (2k + 2m + 2) (2k – 2m) (4k² + 4m² + 2 + 4k + 4m)
= 8(k + m + 1) (k – m) (2k² + 2m² + 1 + 2k + 2m), which is divisible by 8.
Sol. Let the odd natural number be (2n + 1).
n = 1 gives (2n + 1)² = (2 + 1)² = 9.
This when divided by 8 gives 1 as remainder.
n = 2 gives (2n + 1)² = 25.
This when divided by 8 gives 1 as remainder and so on.
Sol. Required number = (2⁵ – 2) = (32 – 2) = 30.
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