How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?

Question: How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?

[A].

8

[B].

11

[C].

12

[D].

13

Answer: Option D

Explanation:

Required numbers are 24, 30, 36, 42, …, 96

This is an A.P. in which a = 24, d = 6 and l = 96

Let the number of terms in it be n.

Then tn = 96    a + (n – 1)d = 96

24 + (n – 1) x 6 = 96

(n – 1) x 6 = 72

(n – 1) = 12

n = 13

Required number of numbers = 13.

(51+ 52 + 53 + … + 100) = ?

Question:
(51+ 52 + 53 + … + 100) = ?

[A].

2525

[B].

2975

[C].

3225

[D].

3775

Answer: Option D

Explanation:

This is an A.P. in which a = 51, l = 100 and n = 50.

Sum = n (a + l) = 50 x (51 + 100)   = (25 x 151)   = 3775.
2 2

The sum all even natural numbers between 1 and 31 is:

Question:
The sum all even natural numbers between 1 and 31 is:

[A].

16

[B].

128

[C].

240

[D].

512

Answer: Option C

Explanation:

Required sum = (2 + 4 + 6 + … + 30)

This is an A.P. in which a = 2, d = (4 – 2) = 2 and l = 30.

Let the number of terms be n. Then,

tn = 30    a + (n – 1)d = 30

2 + (n – 1) x 2 = 30

n – 1 = 14

n = 15

Sn = n (a + l) = 15 x (2 + 30)   = 240.
2 2

(4300731) – ? = 2535618

Question:
(4300731) – ? = 2535618

[A].

1865113

[B].

1775123

[C].

1765113

[D].

1675123

Answer: Option C

Explanation:

Let 4300731 – x = 2535618

Then x, = 4300731 – 2535618 = 1765113

The sum of first 45 natural numbers is:

Question:
The sum of first 45 natural numbers is:

[A].

1035

[B].

1280

[C].

2070

[D].

2140

Answer: Option A

Explanation:

Let Sn = (1 + 2 + 3 + … + 45)

This is an A.P. in which a = 1, d = 1, n = 45 and l = 45

Sn = n (a + l) = 45 x (1 + 45)   = (45 x 23)   = 1035
2 2

Required sum = 1035.

How many 3-digit numbers are completely divisible 6 ?

Question:
How many 3-digit numbers are completely divisible 6 ?

[A].

149

[B].

150

[C].

151

[D].

166

Answer: Option B

Explanation:

3-digit number divisible by 6 are: 102, 108, 114,… , 996

This is an A.P. in which a = 102, d = 6 and l = 996

Let the number of terms be n. Then tn = 996.

a + (n – 1)d = 996

102 + (n – 1) x 6 = 996

6 x (n – 1) = 894

(n – 1) = 149

n = 150

Number of terms = 150.

A boy multiplied 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong and the other digits are correct, then the correct answer would be:

Question: A boy multiplied 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong and the other digits are correct, then the correct answer would be:
[A].

553681

[B].

555181

[C].

555681

[D].

556581

Answer: Option C

Explanation:

987 = 3 x 7 x 47

So, the required number must be divisible by each one of 3, 7, 47

553681 (Sum of digits = 28, not divisible by 3)

555181 (Sum of digits = 25, not divisible by 3)

555681 is divisible by 3, 7, 47.

When a number is divided by 13, the remainder is 11. When the same number is divided by 17, then remainder is 9. What is the number ?

Question:
When a number is divided by 13, the remainder is 11. When the same number is divided by 17, then remainder is 9. What is the number ?

[A].

339

[B].

349

[C].

369

[D].

Data inadequate

Answer: Option B

Explanation:

x = 13p + 11 and x = 17q + 9

13p + 11 = 17q + 9

17q – 13p = 2

q = 2 + 13p
17
The least value of p for which q = 2 + 13p is a whole number is p = 26
17

x = (13 x 26 + 11)

   = (338 + 11)

   = 349

In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors 585) and got the remainders 4, 8, 12 respectively. If he had divided the number by 585, the remainder would have been

Question:
In dividing a number by 585, a student employed the method of short division. He divided the number successively by 5, 9 and 13 (factors 585) and got the remainders 4, 8, 12 respectively. If he had divided the number by 585, the remainder would have been

[A].

24

[B].

144

[C].

292

[D].

584

Answer: Option D

Explanation:

5 | x z = 13 x 1 + 12 = 25
————–
9 | y – 4 y = 9 x z + 8 = 9 x 25 + 8 = 233
————–
13| z – 8 x = 5 x y + 4 = 5 x 233 + 4 = 1169
————–
| 1 -12

585) 1169 (1
585

584

Therefore, on dividing the number by 585, remainder = 584.