Numbers

Question:
-84 x 29 + 365 = ?

[A].

2436

[B].

2801

[C].

-2801

[D].

-2071

Answer: Option D

Explanation:

Given Exp.	= -84 x (30 – 1) + 365
	= -(84 x 30) + 84 + 365
	= -2520 + 449
	= -2071

Question:
3 + 33 + 333 + 3.33 = ?

[A].

362.3

[B].

372.33

[C].

702.33

[D].

702

Answer: Option B

Explanation:

3
+ 33
+ 333
+ 3.33
——
372.33
——

Question: (112 + 122 + 132 + … + 202) = ?
[A].

385

[B].

2485

[C].

2870

[D].

3255

Answer: Option B

Explanation:

(112 + 122 + 132 + … + 202) = (12 + 22 + 32 + … + 202) – (12 + 22 + 32 + … + 102)

	Ref: (12 + 22 + 32 + … + n2) =	1	n(n + 1)(2n + 1)
		6

=		20 x 21 x 41	–	10 x 11 x 21
		6		6

= (2870 – 385)

= 2485.

Question:
(?) – 19657 – 33994 = 9999

[A].

63650

[B].

53760

[C].

59640

[D].

61560

Answer: Option A

Explanation:

19657 Let x – 53651 = 9999
33994 Then, x = 9999 + 53651 = 63650
—–
53651
—–

Question:
(22 + 42 + 62 + … + 202) = ?

[A].

770

[B].

1155

[C].

1540

[D].

385 x 385

Answer: Option C

Explanation:

(22 + 42 + 62 + … + 202) = (1 x 2)2 + (2 x 2)2 + (2 x 3)2 + … + (2 x 10)2

= (22 x 12) + (22 x 22) + (22 x 32) + … + (22 x 102)

= 22 x [12 + 22 + 32 + … + 102]

	Ref: (12 + 22 + 32 + … + n2) =	1	n(n + 1)(2n + 1)
		6

=		4 x	1	x 10 x 11 x 21
			6

= (4 x 5 x 77)

= 1540.

Question:
(12 + 22 + 32 + … + 102) = ?

[A].

330

[B].

345

[C].

365

[D].

385

Answer: Option D

Explanation:

We know that (12 + 22 + 32 + … + n2) =	1	n(n + 1)(2n + 1)
	6

Putting n = 10, required sum =		1	x 10 x 11 x 21		= 385
		6

Question:
2 + 22 + 23 + … + 29 = ?

[A].

2044

[B].

1022

[C].

1056

[D].

None of these

Answer: Option B

Explanation:

This is a G.P. in which a = 2, r =	22	= 2 and n = 9.
	2

Sn =	a(rn – 1)	=	2 x (29 – 1)	= 2 x (512 – 1)   = 2 x 511   = 1022.
	(r – 1)		(2 – 1)

Question:
How many terms are there in the G.P. 3, 6, 12, 24, … , 384 ?

[A].

8

[B].

9

[C].

10

[D].

11

Answer: Option A

Explanation:

Here a = 3 and r =	6	= 2. Let the number of terms be n.
	3

Then, tn = 384     arn-1 = 384

3 x 2n – 1 = 384

2n-1 = 128 = 27

n – 1 = 7

n = 8

Number of terms = 8.

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