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# Maths Questions for CTET,KVS Exam : 11th 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. Which of the following is equivalent to 25/36÷5/18 ?
(a) 125/648
(b) 12/15
(c) 5/2
(d) 450/540

Q2. The difference between the greatest and the smallest fractions amongst 6/7,8/9,9/10,7/8 is
(a) 2/63
(b) 3/70
(c) 1/56
(d) 1/72 Q4. Which of the following is the simplest form of 35/45?
(a) 7/9
(b) 5/9
(c) 7/5
(d) 35/45

Q5. If a³ = 1 + 7, 3³ = 1 + 7 + b, 4³ = 1 + 7 + c, then the value of a + b + c is
(a) 58
(b) 75
(c) 77
(d) 110

Q6. How many times will I be writing 2 if I wrote down numbers from 11 to 199?
(a) 36
(b) 37
(c) 38
(d) 39

Q7. What should be subtracted from (-3)/7 to get (-1)/3 ?
(a) 2/21
(b) (-2)/21
(c) 1/21
(d) (-1)/21

Q8. The value of (41/3) ÷ 3 is
(a) 9/13
(b) 13
(c) 4
(d) 13/9  Solutions

S1. Ans.(c)
Sol.
25/36÷5/18=25/36×18/5=5/2

S2. Ans.(b)
Sol. Fractions are
6/7,8/9,9/10,7/8
LCM of 7, 9, 10 and 8 = 2520
Now,
6/7=  (6×360)/(7×360)=2160/2520
8/9=(8×280)/(9×280)=2240/2520
9/10=(9×252)/(10×252)=2268/2520
7/8=(7×315)/(8×315)=2205/2520
Here, 2160 is the least and 2268 is the greatest number among 2160, 2240, 2268 and 1890.
Therefore, the smallest fraction = 6/7 and the greatest fraction = 9/10.
Note: When two and more than two positive fractions have equal difference between their numerators and denominators, then the fraction having greater while the fraction having lesser numerator is the lesser fraction.
Difference
=9/10-6/7=(63-60)/70=3/70

S3. Ans.(b)
Sol. 829030000 can be written in the standard form as
8.2903 × 10⁸
Therefore, k = 8.2903 and n = 8
Hence, k + n = 16.2903

S4. Ans.(a)
Sol. HCF of 35 and 45 = 5
∴35/45=(35÷5)/(45÷5)=7/9

S5. Ans.(c)
Sol. a³ = 1 + 7 = 8 ⇒ a = 2
3³ = 1 + 7 + b
⇒ 27 = 8 + b ⇒ b = 27 – 8 = 19
4³ = 1 + 7 + c
⇒ 64 = 8 + c ⇒ c = 64 – 8 = 56
a + b + c = 2 + 19 + 56 = 77

S6. Ans.(d)
Sol. Any digit from 2 to 9 is used 20 times to write from 1 to 100. Only 1 is used 21 times.
∴ Number of times 2 is written from 11 to 100 = 20 – 1 = 19
2 is used 21 times to write from 101 to 200, and all digits from 1 to 9 except 2 are used 20 times.
∴ Number of times 2 is written from 101 to 199 = 20
Thus, 2 is written for 19 + 20 = 39 times

S7. Ans.(b)
Sol. Required number = (-3)/7-((-1)/3)
=(-3)/7+1/3
=(-9+7)/21
=(-2)/21

S8. Ans.(d)
Sol.
4 1/3÷3=13/3×1/3=13/9 