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# Maths Questions for CTET,KVS Exam : 12th 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. Factorise x² + 2x + 1
(a) (x+1)²
(b) (x–1)²
(c) (x+2)²
(d) (x–2)²

Q2. Simplify (7x–11y) (49x²+77xy+121y²).
(a) 343x³ –1331 y³
(b) 343x² + 1331y³
(c) (7x+11y)³
(d) (7x–11y)³

Q4. Factorise x²+3x + 2.
(a) (x+2) (x+1)
(b) (x–2) (x–1)
(c) (x+2) (x–1)
(d) (x–2) (x+1)

Q6. Factorise 3x² –5x –2.
(a) (3x+1) (x+2)
(b) (3x–1) (x–2)
(c) (3x+1)(x–2)
(d) (3x+2) (x–1)

Q7. Factorise x² –20x +91.
(a) (x–13)(x+7)
(b) (x+13)(x–7)
(c) (x–13)(x–7)
(d) (x–13)(x–17)

Q8. Find the zero of the polynomial p(x)=2x + 3.
(a) 1.5
(b) 1.25
(c) –1.5
(d) 2

Q9. Solve for x: 3x – 2 = 4
(a) 2
(b) –2
(c) 3
(d) –3

Q10. Solve for a: 9a-23=166
(a) 22
(b) –22
(c) 23
(d) 21

Solutions

S1. Ans.(a)

Sol.
x² + 2x+1
= (x)² + (x) × (1) + (1)²
= (x+1)

S2. Ans.(a)
Sol.
(7x–11y) (49x²77xy+121y)
= (7x–11y)
{(7x)²+(7x)(11y)+(11y)²}
= (7x)³– (11y) ³
= 343x³ –1331 y³

S3. Ans.(b)
Sol.
(3a+5b) (9a²–15ab+25b²)
= (3a+5b) {(3a)²–(3a×5b)+(5b)²}
= 27a3 + 125b3

S4. Ans.(a)
Sol.
x² + 3x + 2
= x² = 2x + x + 2
= x (x+2) +1(x+2)
= (x+2) (x+1)

S6. Ans.(c)
Sol.
After splitting the middle term, we have
3x² – 5x – 2
= 3x – 6x + × –2
= 3x(x–2)+1(x–2)
= (3x+1) (x–2)

S7. Ans.(c)
Sol.
x² – 20x + 91
= x² –13 –7x + 91
= x(x–13) –7(x–13)
= (x–13) (x–7)

S8. Ans.(c)
Sol.
Putting p(x) = 0, we have
2x + 3 = 0
⇒ 2x = –3
⇒ x = – 1.5

S9. Ans.(a)
Sol.
3x – 2 = 4
⇒ 3x = 4 + 2
⇒ 3x = 4
⇒ x = 2

S10. Ans.(d)
Sol.
9a – 23 = 166
⇒ 9a = 166 + 23
⇒ 9a = 189
⇒ a = 21