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# Maths Questions for CTET,KVS Exam : 27th september 2018

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. What should be subtracted from 3ab + 2ac – 3bc to get 8ab – 9ac – 17bc?
(a) –5ab + 11ac – 14bc
(b) –5ab – 11ac + 14bc
(c) 5ab + 11ac + 14bc
(d) –5ab + 11ac + 14bc

Q3. Find the remainder on dividing p(x) = x³ + 1 by x – 1.
(a) 2
(b) 3
(c) –2
(d) –3

Q4. Subtract 3abc² – 2b²c + 9abc – 7bc² from 7abc² + 6b²c – 3abc + 2bc².
(a) 4abc² + 8b²c – 12abc – 9bc²
(b) 4abc² + 8b²c + 12abc + 9bc²
(c) 4abc² – 8b²c – 12abc + 9bc²
(d) 4abc² + 8b²c – 12abc + 9bc²

Q5. Find the sum of 2a²b² + c² – 3ac² and 2a²b² + c² – 3a²c.
(a) 4a²b² + 2c² – 3ac² – 3a²c
(b) 4a²b² + 2c² – 6ac²
(c) 4a²b² + 2c² – 3ac²
(d) 4a²b² + 2c² – 3a²c

Q6. What should be added to 2ac – 3abc – ab² + 9a²b to get –ac + 9abc + 3ab² – 11a²b?
(a) –3ac + 12abc + 4ab² + 20a²b
(b) –3ac – 12abc + 4ab² – 20a²b
(c) –3ac + 12abc + 4ab² – 20a²b
(d) –3ac + 12abc – 6ab² – 20a²b

Q7. What should be subtracted from –xy + 3xyz + xy² to get 7xy + 3xyz – 9x²y ?
(a) –8xy + xy² + 10x²y
(b) –8x + xy² + 9x²y
(c) –8xy – xy² + 9x²y
(d) 7xy + xy² + 9x²y

Q9. Simplify (27a + 31b) (729a² – 837ab + 961b²).
(a) 19683a³ – 2979b³
(b) 19683a³ + 29791b³
(c) (19683a + 29791b)³
(d) (19683a – 29791b)³

Solutions

S1. Ans.(d)
Sol. Let A be added, then
3ab + 2ac – 3bc – A
= 8ab – 9ac – 17bc
⇒ A = 3ab + 2ac – 3bc – 8ab + 9ac + 17bc – 5ab + 11ac + 14bc

S3. Ans.(a)
Sol. According to the polynomial remainder theorem, when a polynomial p(x) is divided by a linear polynomial x – a, then the remainder is p(a).
∴ The remainder when x³ + 1 is divided by x – 1 is p(1).
p(1) = (1)³ + 1 = 2

S6. Ans.(c)
Sol. Let A be added, then
A + 2ac – 3abc – ab² + 9a²b
= –ac + 9abc + 3ab² – 11a²b
⇒ A = –ac + 9abc + 3ab² – 11a²b – 2ac + 3abc + ab² – 9a²b
= –3ac + 12abc + 4ab² – 20a²b

S7. Ans.(b)
Sol. Let A be subtracted, then
–xy + 3xyz + xy² – A
= 7xy + 3xyz – 9x²y
⇒ A = –xy + 3xyz + xy² – 7xy – 3xyz + 9x²y
= –8xy + xy² + 9x²y

S9. Ans.(b)
Sol. (27a + 31b)(729a² – 837ab + 961b²) = (27a + 31b)
{(27a)² – (27a)(31b) + (31b)²}
= (27a)³ + (31b)³
= 19683a³ + 29791b³

S10. Ans.(d)
Sol. The exponents of x and y in are not whole numbers.
Hence, it is not a polynomial.