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# Maths Questions for CTET,KVS Exam : 4th 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 reciprocal of (-3)/8×((-7)/13) is
(a) 104/21
(b) (-104)/21
(c) 21/104
(d) (-21)/104 Q3. Which of the numbers –20, -3/4,1/2, 10 is greater than its square?
(a) 1/2
(b) -3/4
(c) 10
(d) –20

Q4. The smallest value of ‘y’ in the number 71y8978 so that it is divisible by 11 is
(a) 3
(b) 1
(c) 0
(d) 2 Q6. The LCM of 121, 125 and 135 is
(a) 3³ × 5³ × 11²
(b) 3³ × 5² × 11²
(c) 3² × 5³ × 11³
(d) 3² × 5² × 11²

Q7. The additive inverse of S, where S = 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + ……………… + 49 – 50, is:
(a) 1
(b) 0
(c) 25
(d) –25 Solutions

S1. Ans.(a)
Sol.
(-3)/8×(-7)/13=21/104
Reciprocal of (-3)/8×(-7)/13
= Reciprocal of 21/104
=104/21 S4. Ans.(a)
Sol. A number is divisible by 11 if the difference of sum of digits at even places and odd places is either 0 or divisible by 11.
Here, the difference of sum of digits at even places and odd places in 71y8978
= (7 + y + 9 + 8) – (1 + 8 + 7)
= 24 + y – 16
= y + 8
The smallest value of y for which y + 8 is divisible by 11 is 3. S6. Ans.(a)
Sol. We have
121 = 11 × 11 = 11²
125 = 5 × 5 × 5 = 5³
135 = 3 × 3 × 3 × 5 = 3³ × 5
Therefore, the LCM of 121, 125 and 135 = 3³ × 5³ × 11²

S7. Ans.(c)
Sol. We have
S = 1 – 2 + 3 – 4 + 5 – 6 + 7 – 8 + ………. + 49 – 50
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + ………. + (49 – 50);
Here, there would be 25 parentheses.
= (1 – 2) + (3 – 4) + (5 – 6) + (7 – 8) + ………. + (49 – 50);
= (–1) + (–1) + (–1) + (–1) + ……… + (–1);
Here, there would be 25 parentheses
= –25
Therefore, the additive inverse of S would be 25. 