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# Maths Questions for CTET,KVS Exam : 2nd october 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. The degree of the polynomial 3a² + 4b³ – 5ab³ + 7 – 5a⁷b⁵ is

(a) 2
(b) 3
(c) 4
(d) 12

Q2. Find the product of (2x² + x + 3) and (x + 2).
(a) 2x³ + 5x² + 5x + 6
(b) x³ + 5x² + 5x + 6
(c) 2x³ + 4x² + 5x + 6
(d) 2x³ + 5x² + 3x + 6

Q3. Find the product of (x² – 3x + 2) and (2x – 1).
(a) 2x³ – 7x² + 7x + 2
(b) 2x³ – 7x² + 7x – 2
(c) 2x³ – 7x² – 7x – 2
(d) 2x³ + 7x² + 7x – 2

Q4. One of the factors of x² – z² + y² – 2xy is
(a) x – y + z
(b) x + y – z
(c) x + y + z
(d) y + z – x

Q5. In the product (x² – 2) (1 – 3x + 2x²), the sum of coefficients of x² and x is

(a) 5
(b) 6
(c) 2
(d) 3

Q6. If x is an integer, then (x + 1)⁴ – (x – 1)⁴ is always divisible by
(a) 6
(b) 8
(c) 9
(d) 12

Q7. The expression x² – y² + x + y – z² + 2yz – z has one factor, which is
(a) y – x + z
(b) x – y + z + 1
(c) x + y – z + 1
(d) x – y – z + 1

Q8. A factor common to x² + 7x + 10 and x² – 3x – 10 is
(a) x – 5
(b) x + 5
(c) x + 2
(d) x – 2

Q9. The sum of three consecutive integers is 99. Find the smallest integer.
(a) 33
(b) 34
(c) 32
(d) 35

Q10. The sum of two positive numbers is 63. If one number x is double the other, then the equation is
(a) x/(x – 63)=2
(b) (63 – x)/x=2
(c) (x – 63)/x=2
(d) x/(63 – x)=2

Solutions

S1. Ans.(d)
Sol. The highest power of the variable is 12(7 + 5 in 5a⁷b⁵).
So, the degree of the polynomial is 12.

S2. Ans.(a)
Sol. (x + 2) (2x² + x + 3)
= x(2x² + x + 3) + 2(2x² + x + 3)
= 2x³ + x² + 3x + 4x² + 2x + 6
= 2x³ + 5x² + 5x + 6

S3. Ans.(b)
Sol. (2x – 1)(x² – 3x + 2)
= 2x(x² – 3x + 2) – 1(x² – 3x + 2)
= 2x³ – 6x² + 4x – x² + 3x – 2
= 2x³ – 7x² + 7x – 2

S4. Ans.(a)
Sol. We have
x² – z² + y² – 2xy
= x² – 2xy + y² – z²
= (x – y)² – z²
= (x – y + z) (x – y – z)
Hence, (x – y + z) is one of the factors

S5. Ans.(d)
Sol. We have
(x² – 2) (1 – 3x + 2x²)
= x² – 3x³ + 2x⁴ – 2 + 6x – 4x²
= 2x⁴ – 3x² – 3x² + 6x – 2
Coefficient of x² = –3
Coefficient of x = 6

S8. Ans.(c)
Sol. x² + 7x + 10 = x² + 5x + 2x + 10
= x (x + 5) + 2 (x + 5)
= (x + 5) (x + 2)
x² – 3x – 10 = x² – 5x + 2x – 10
= x (x – 5) + 2 (x – 5)
= (x – 5) (x + 2)
Hence, the common factor x + 2.

S9. Ans.(c)
Sol. Let the smallest integer be x.
∴ Two other integers are
x + 1 and x + 2
x + x + 1 + x + 2 = 99
⇒ 3x + 3 = 99
⇒ 3x = 96
⇒ x = 32
Three integers are 32, 33 and 34.
Hence, the smallest integer = 32.

S10. Ans.(d)
Sol. One number = x
∴ Another number = x/2
According to the question,
x+x/2 = 63
⇒ x/2 = 63 –x
⇒ x/(63 – x) = 2