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Maths Questions for CTET,KVS Exam : 7th september 2018

Maths Questions for CTET,KVS Exam : 7th september 2018_30.1

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. Sara deposited Rs. 2000 for 3 years and 9 months at the rate of 5% p.a. simple interest. Find the amount in Rs. at maturity.
(a) Rs. 2735
(b) Rs. 2375
(c) Rs. 2357
(d) Rs. 375

Q2. One of the factors of 4x² + y² + 14x – 7y – 4xy + 12 is
(a) 2x + y + 4
(b) 2x + y – 4
(c) 2x – y + 3
(d) 2x – y – 3

Q3. Marshall bought 15 refills and sold them at Rs. 4 each. If it cost him Rs. 55 for the refills, what was his gain or loss percent?
(a) Gain = 9.90%
(b) Loss = 9.09%
(c) Gain = 9.09%
(d) Loss = 9.90%

Q4. A fraction is equivalent to 5/8. Its denominator and numerator add up to 91. What is the difference between the denominator and numerator of this fraction?
(a) 3
(b) 13
(c) 19
(d) 21

Q5. The salaries of Rakesh and Suresh are in the ratio 2 : 3. If the salary of each is increased by Rs. 4000, the new ratio becomes 40 : 57. What is the present salary of Suresh in Rs.?
(a) Rs. 32,000
(b) Rs. 34,000
(c) Rs. 38,000
(d) Rs. 40,000

Q6. 40% of (100 – 20% of 300) is equal to
(a) 16
(b) 20
(c) 64
(d) 140

Q7. Factorise 4x² + 9y² + 16z² – 12xy – 24yz + 16xz.
(a) (2x + 3y – 4z)²
(b) (2x – 3y + 4z)²
(c) (2x – 3y – 4z)²
(d) (2x + 3y + 4z)²

Q8. Express 0.33 ÷ 0.11 as a percentage.
(a) 1/300
(b) 1/3
(c) 30
(d) 300

Q9. Find the LCM of x² + 10x + 25 and x² + 3x – 10.
(a) (x + 5)² (x – 2)
(b) (x – 5)² (x – 2)
(c) (x – 2)
(d) (x + 5)²

Q10. An athlete runs 400 m race in 50 seconds. Find her speed in km/h.
(a) 25 km/h
(b) 30 km/h
(c) 28.8 km/h
(d) 36 km/h

Solutions

S1. Ans.(b)
Sol. P = Rs. 2000, R = 5%
And T = 3 years and 9 months
= 3 years + 9/12 years = 15/4 years
S.I.=  PRT/100
=(2000×5×15)/(100×4)
= Rs. 375
A = P + S.I. = 2000 + 375 = Rs. 2375

S2. Ans.(c)
Sol. Rearranging the terms of the given expression, we get 4x² + y² + 14x – 7y – 4xy + 12
= 4x² + y² – 4xy + 14x – 7y + 12
= (2x – y)² + 12x – 6y + 9 + 2x – y + 3
= (2x – y)² + 6 (2x – y) + 3² + 2x – y + 3
= (2x – y)² + 2 × 3 × (2x – y) + 3² + 2x – y + 3
= (2x – y + 3)² + (2x – y + 3)
= (2x – y + 3) {(2x – y + 3) + 1}
= (2x – y + 3) (2x – y + 4)
Hence, 2x – y + 3 is a factor of the given expression.

S3. Ans.(c)
Sol. Cost price = Rs. 55
Selling price = 4 × 15 = Rs. 60
Since S.P. > C.P., there is again.
Gain = S.P. – C.P. = 60 – 55 = Rs. 5
Gain % = 5/55 × 100 = 9.09%

S4. Ans.(d)
Sol. Let the numerator be x.
∴ The denominator will be 91 – x
According to the question,
x/(91-x)=5/8
p8x = 455 – 5x
p13x = 455
px = 35
∴ Numerator = 35 and denominator = 91 – 35 = 56
∴ Difference = 56 – 35 = 21

S5. Ans.(b)
Sol. Let the present salaries of Rakesh and Suresh be 2x and 3x respectively.
∴(2x+4000)/(3x+4000)=40/57
⇒ 114x + 228000 = 120x + 160000
⇒ 120x – 114x = 228000 – 160000
⇒ 6x = 68000 ⇒ 3x = 34000
∴ Present salary of Suresh = Rs. 34000

S6. Ans.(a)
Sol. 40% of (100 – 20% of 300)
=40% of (100-20/100×300)
= 40% of (100 – 60)
= 40% of 40
=40/100×40
= 16

S7. Ans.(b)
Sol. 4x² + 9y² + 16z² – 12xy – 24yz + 16xz
= (2x)² + (–3y)² + (4z)³ + 2(2x)
(–3y) + 2(–3y)(4z) + 2 (4z)(2x)
∵ {(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca}
= (2x)² + (–3y)² + (4z)³ + 2(2x)
(–3y) + 2 (–3y) (4z) + 2 (4z) (2x)
= (2x – 3y + 4z)²

S8. Ans.(d)
Sol. The given fraction is
In order to convert a fraction into percentage, the fraction is multiplied by 100.
In terms of percentage, the given fraction = 33/11 × 100 = 300%

S9. Ans.(a)
Sol. We have
x² + 10x + 25
= (x)² + 2 (x) (5) + (5)² = (x + 5)²
And x² + 3x – 10 = x² + 5x – 2x – 10
= x (x + 5) – 2 (x + 5) = (x + 5) (x – 2)
∴ LCM = (x + 5)² (x – 2)

S10. Ans.(c)
Sol. Speed of athlete = 400/50
= 8 m/s
Speed in km/h = 8 × 18/5
= 28.8 km/h