Aptitude

Question: The sum of how many terms of the series 6 + 12 + 18 + 24 + … is 1800 ?
[A].

16

[B].

24

[C].

20

[D].

18

Answer: Option B

Explanation:

This is an A.P. in which a = 6, d = 6 and Sn = 1800

Then,	n	[2a + (n – 1)d] = 1800
	2

	n	[2 x 6 + (n – 1) x 6] = 1800
	2

3n (n + 1) = 1800

n(n + 1) = 600

n2 + n – 600 = 0

n2 + 25n – 24n – 600 = 0

n(n + 25) – 24(n + 25) = 0

(n + 25)(n – 24) = 0

n = 24

Number of terms = 24.

Question:
The sum of all two digit numbers divisible by 5 is:

[A].

1035

[B].

1245

[C].

1230

[D].

945

Answer: Option D

Explanation:

Required numbers are 10, 15, 20, 25, …, 95

This is an A.P. in which a = 10, d = 5 and l = 95.

tn = 95       a + (n – 1)d = 95

10 + (n – 1) x 5 = 95

(n – 1) x 5 = 85

(n – 1) = 17

n = 18

Requuired Sum =	n	(a + l)	=	18	x (10 + 95)   = (9 x 105)   = 945.
	2			2

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.

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

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

Question:
(4300731) – ? = 2535618

[A].

1865113

[B].

1775123

[C].

1765113

[D].

1675123

Answer: Option C

Explanation:

Let 4300731 – x = 2535618

Then x, = 4300731 – 2535618 = 1765113

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.

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.

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.

Question: A number when divided by 296 leaves 75 as remainder. When the same number is divided by 37, the remainder will be:
[A].

1

[B].

2

[C].

8

[D].

11

Answer: Option A

Explanation:

Let x = 296q + 75

   = (37 x 8q + 37 x 2) + 1

   = 37 (8q + 2) + 1

Thus, when the number is divided by 37, the remainder is 1.