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 Exam & STET Exam.

**Q1. What should come in place of ‘a’ so that the number 873a24 is divisible by 4?**

(a) 1

(b) 3

(c) 6

(d) Any of these

**Q3. If 5432*7 is divisible by 9, then the digit in place of * is**

(a) 9

(b) 6

(c) 3

(d) 1

**Q4. If 78*3945 is divisible by 11, where * is a digit, then * is equal to**

(a) 0

(b) 1

(c) 3

(d) 5

**Q5. A number when divided by 280 leaves 115 as remainder. When the same number is divided by 35, the remainder is**

(a) 10

(b) 15

(c) 17

(d) 20

**Q6. The number which can be written in the form of n (n + 1) (n + 2), where n is a natural number, is**

(a) 7

(b) 3

(c) 5

(d) 6

**Q7. The remainder, when 2851 × (2862) ² × (2873) ² is divided by 23 is**

(a) 5

(b) 10

(c) 11

(d) 18

**Q8. Find the unit digit of 1 ×2 × 3 × 4 × 5 ×6 .**

(a) 1

(b) 2

(c) 4

(d) 0

**Q9. Write 94 in roman numeral.**(a) CXIV

(b) XCVI

(c) XCIV

(d) IICX

**Q10. Write 493 in roman numeral.**(a) LDXCIII

(b) CDXCIII

(c) CDCXIII

(d) DCCXIII

**Solutions**

**S1. Ans.(d)**

**Sol.**A number is divisible by 4 if the number made by last two digits is divisible by 4. In the given number, last two digits i.e. 24 is divisible by 4. So, ‘a’ can take any of the values.

**S3. Ans.(b)**

**Sol.**A number is divisible by 9 if the sum of its digits is divisible by 9. Let the number be x.

∴ 5 + 4 + 3 + 2 + x + 7 = 21 + x

If we put x = 6, then it is divisible by 9.

∴ x = 6

Hence, the digit in place of * is 6.

**S4. Ans.(d)**

**Sol.**A number is divisible by 11, if the difference of the sum of its digits at odd places and the sum of its digit of even places, is either 0 or a number divisible by 11.

∴ (5 + 9 + * + 7) – (4 + 3 + 8) = 0

Or multiple of 11

⇒ (21 + *) – 15 = 0 or multiple of 11.

⇒ 6 + * = 0 or multiple of 11.

⇒ * + 6 = a multiple of 11.

⇒ * + 6 = 11

⇒ * = 5

**S5. Ans.(a)**

**Sol**. Here, 280 is a multiple of 35.

∴ Required remainder = Remainder obtained on dividing 115 by 35 = 10

**S6. Ans.(d)**

**Sol.**Given, n(n + 1) (n + 2)

Now, putting n = 1, we have

1 (1 + 1) (1 + 2) = 1 × 2 × 3 = 6

**S8. Ans.(d)**

**Sol**. Unit digit of this expression is 0. Because this expression has the term 2 × 5=0