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# Maths Questions for CTET,KVS Exam : 26th 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. If 1957 – a9 = 18b8, the sum of the digits a and b is
(a) 15
(b) 14
(c) 13
(d) 12

Q2. Which of the following numbers is a perfect square?
(a) 548543213
(b) 548543215
(c) 548543251
(d) 548543241

3. If a × b = a² + b² and a.b = a² – b², the value of (5 × 2). 25 is
(a) 215
(b) 225
(c) 226
(d) 216

Q4. Estimate the value of 113 × 89.
(a) 10000
(b) 100000
(c) 9000
(d) 1000

Q5. LCM of two prime numbers x and y, (x > y), is 161. The value of 3y – x is
(a) 2
(b) –2
(c) –5
(d) 62

Q7. When we add 1 to a three-digit greatest number, we get
(a) a three-digit greatest number
(b) a four-digit least number
(c) a four-digit greatest number
(d) a three-digit least number

Q8. The smallest number by which 68600 must be multiplied to get a perfect cube is
(a) 5
(b) 10
(c) 8
(d) 12

Q9. The square of 9 is divided by the code root of 125. The remainder is
(a) 1
(b) 2
(c) 3
(d) 4

Solutions

S1. Ans.(b)

Sol.
Here, the next digit to b is 8.
Therefore, a must be greater than 4.
If a = 5; b = 9, a + b = 14
If a = 6; b = 8, a + b = 14
If a = 7; b = 7, a + b = 14
If a = 8; b = 6, a + b = 14
If a = 9; b = 5, a + b = 14
∴ a + b = 14

S2. Ans.(d)
Sol. For a perfect square, the unit’s digit cannot be 2, 3, 7 or 8; therefore, the first option is eliminated.
If the unit’s digit of a perfect square is 5, its ten’s digit is always 2; therefore, the second option is eliminated.
If a perfect square ends in 1, its ten’s digit is always even; thus, option (c) is also wrong.
So, we are left with only option (d), which is the answer.

S3. Ans.(d)
Sol. 5 × 2 = 5² + 2² = 25 + 4 = 29
Then,
(5 × 2) . 25 = 29 . 25 = 29² – 25²
= (29 + 25) (29 – 25)
= 54 × 4 = 216

S4. Ans.(a)
Sol. The estimated value of 113 in hundreds is 100 and that of 89 in hundreds is 100.
Multiplication = 100 × 100 = 10000

S5. Ans.(b)
Sol. We know that the LCM of two prime numbers is the product of those numbers.
161 is the product of prime numbers 23 and 7.
Therefore, 161 is the LCM of 23 and 7.
We get x = 23 and y = 7
Now, 3y – x = 3 × 7 – 23 = 21 – 23 = –2

S7. Ans.(b)
Sol. The three-digit greatest number = 999
999 + 1 = 1000
1000 is the four-digit least number.

S8. Ans.(a)
Sol. 68600 = 2 × 2 × 2 × 5 × 5 × 7 × 7 × 7
= (2 × 7)³ × 5 × 5
Therefore, it should be multiplied by 5 to make it a perfect cube.

S10. Ans.(c)
Sol. Unit’s digit of will be the same as the unit’s digit of .
Now,
Unit’s digit of 3⁴ will be 1.
Unit’s digit of all integral powers of 3⁴ will be 1.
∴ Unit’s digit of (3⁴)⁵⁰⁰ = will be 1.
And unit’s digit of 3³ will be 7.
∴ Unit’s digit of = Unit digit of × Unit digit of 3³ = 1 × 7 = 7
Hence, unit’s digit of will be 7.