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# Maths Questions for CTET,KVS Exam : 21th 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. The sum of two numbers is 85. If one number exceeds the other by 7, find the smallest number.
(a) 39
(b) 32
(c) 46
(d) 53

Q2. A car covers a distance of 178.2 km in 4.4 hours. The distance (in km) covered by the car in 1.2 hours is
(a) 40.5
(b) 40.7
(c) 48.6
(d) 50.2

Q3. In a class, the number of boys is 4/5 of the number of girls. Find the number of girls in the class if the total number of students is 180.
(a) 8
(b) 100
(c) 120
(d) 60

Q4. n² – (n + 1)² – (n + 2)² + (n + 3)² is equal to
(a) 0
(b) 2
(c) 4
(d) 6

Q5. If (3x – 2)/3+(2x + 3)/2=x+7/6, then the value of (5x – 2)/4 is:
(a) (-1)/12
(b) 1/12
(c) (-1)/3
(d) 1/4

Q6. If 4x² + 12xy – 8x + 9y² – 12y = (ax + by) (ax + by – 4), then the value of a² + b² is
(a) 25
(b) 5
(c) 10
(d) 13

Q7. Which of the following is a trinomial?
(a) 6x²y – 4xy² – 7x
(b) 6x²c – 2x²y
(c) 6x²yc – 4x²y² + 4xyz + 2y
(d) 6x²

Q8. What should be added to 3a²b² + 2ac² + 4b²c to get 9a²b² + 7ac² + 2b²c?
(a) 6a²b² – 5ac² – 2b²c
(b) 6a²b² + 5ac² + 2b²c
(c) 6a²b² + 5ac² – 3b²c
(d) 6a²b² + 5ac² – 2b²c

Q9. What should be added to 3x²y – 2xyz² + 4x²y²z² to get 5x²y – 9xyz² – 6x²y²z²?
(a) 2x²y – 7xyz² – 10x²y²z²
(b) 2x²y + 12xyz² + 10x²y²z²
(c) 2x²y + 12xyz² – 10x²y²z²
(d) 2x²y – 12xyz² + 10x²y²z²

Q10. What should be added to 2a² + 3b² + 4c² – 4bc to get 19a² – 9b² – 7c² – 2ab?
(a) 17a² – 12b² – 11c² – 2ab + 4bc
(b) 17a² – 12b² – 11c² – 2ab – 4bc
(c) 17a² – 12b² – 11c² + 2bc
(d) 17a² – 12b² – 11c² + 2ab

Solutions

S1. Ans.(a)
Sol. Let the first number be x. Then,
The second number = x + 7
x + x + 7 = 85
2x = 78
x = 39
∴ The smallest number = 39

S2. Ans.(c)
Sol. Distance covered by the car in 4.4 hours = 178.2 km
Distance covered by the car in 1 hour = 178.2/4.4 km
∴ Distance covered by the car in 1.2 hours = (178.2 × 1.2)/4.4 km = 48.6 km

S3. Ans.(b)
Sol. Let the number of girls be x.
Then, the number of boys = 4/5x
According to the question,
x+4/5 x=180
⇒(5x+4x)/5=180
⇒9x/5=180
⇒x=5/9×180=100
Number of girls = x = 100

S4. Ans.(c)
Sol. The given expression is equal to (n + 3)² – (n + 2)² + n² – (n + 1)²
= (n + 3 + n + 2) (n + 3 – n – 2) + (n + n + 1) (n – n – 1)
= (n + 5) (1) + (2n + 1) (–1)
= 2n + 5 – 2n – 1 = 4

S5. Ans.(a)
Sol.
(3x-2)/3+(2x+3)/2=x+7/6
⇒(6x-4+6x+9)/6=x+7/6
⇒(12x+5)/6=x+7/6
⇒2x+5/6=x+7/6
⇒2x-x=7/6-5/6
⇒x=2/6=1/3
∴(5x-2)/4=(5×1/3-2)/4=(5/3-2)/4
=((5-6)/3)/4=(-1)/12

S6. Ans.(d)
Sol. 4x² + 12xy – 8x + 9y² – 12y
= (ax + by) (ax + by – 4)
⇒ 4x² + 12xy – 8x + 9y² – 12y
= a²x² + abxy – 4ax + abxy + b²y² – 4by
⇒ 4x² + 12xy – 8x + 9y² – 12y
= a²x² + 2abxy – 4ax + b²y² – 4by
Comparing the coefficients of x² and y² on both sides, we get a² = 4 and b² = 9
∴ a² + b² = 4 + 9 = 13

S7. Ans.(a)
Sol. A trinomial has three terms.
∴ 6x²y – 4xy² – 7x is a trinomial.

S8. Ans.(d)
Sol. Let A be added, then
A + 3a²b² + 2ac² + 4b²c
= 9a²b² + 7ac² + 2b²c
⇒ A = 9a²b² + 7ac² + 2b²c – 3a²b² – 2ac² – 4b²c
= 6a²b²+ 5ac² – 2b²c

S9. Ans.(a)
Sol. Let A be added, then
A + 3x²y – 2xyz² + 4x²y²z²
= 5x²y – 9xyz² – 6x²y²z² – 3x²y + 2xyz² – 4x²y²z²
= 2x²y – 7xyz² – 10x²y²z²

S10. Ans.(a)
Sol. Let A be added, then
A + 2a² + 3b² + 4c² – 4bc
= 19a² – 9b² – 7c² – 2ab
⇒ A = 19a² – 9b² – 7c² – 2ab – 2a² – 3b² – 4c² + 4bc
= 17a² – 12b² – 11c² – 2ab + 4bc