Indian Bank PO Pre mock test series 2nd, Indian bank po test series

(743.30)² = 552500

M = 6
S = 1.2
DS = 7.2
US = 4.8
x/7.2 + x/4.8 = 1
x = 2.88
D = 2.88 * 2 = 5.76

X^{a3 / abc} * X^{b3 / abc} * X^{c3 / abc
= X(a3 + b3 + c3) / abc}
As a + b + c = 0, (a³ + b³ + c³) / abc = 3
The given expression equals X³

x = 14 + 10√2
x/2 = 7 + 5√2
2/x = (7 – 5√2) / (7 + 5√2)(7 – 5√2) = 5√2 – 7
x/2 + 2/x = 10√2 = 2√50

x(x + 2) – (x + x + 2) = 98

x* 12 : 2x* 6: 3x* 4
1:1:1
1/3 * 18000 = 6000

x/2 – x/3 = 6 => x =6
3 + 6 = 9

S = 180/6 * 18/5 = 108 kmph

C.P. of 50 kg wheat = (30 * 11.50 + 20 * 14.25) = Rs. 630.
S.P. of 50 kg wheat = 130% of Rs. 630 = 130/100 * 630 = Rs. 819.
S.P. per kg = 819/50 = Rs. 16.38 = 16.30.

(385/1001) = 5/13 First Number is 5

Let the son’s present age be x years.
Then, man’s present age = (x + 24) years.
(x + 24) + 2 = 2(x + 2) x + 26 = 2x + 4 => x = 22.

Suppose he has 100 items. Let C.P. of each item be Re. 1.
Total cost = Rs. 100. Number of items left after theft = 80.
S.P. of each item = Rs. 1.10
Total sale = 1.10 * 80 = Rs. 88
Hence, loss % = 12/100 * 100 = 12%

Given that there are three blue marbles, four red marbles, six green marbles and two yellow marbles.
Probability that all the three marbles picked at random are blue = ³C₃/¹⁵C₃ = (1 * 3 * 2 * 1)/(15 * 14 * 13) = 1/455

A+B+C = 1500
A = 1/3(B+C); B = 2/3(A+C)
A/(B+C) = 1/3
A = 1/4 * 1500 => 375

109 * 34 – 108 * 33
= 3706 – 3564 = 142

Amount = [200(1 + 5/100)³ + 200(1 + 5/100)² + 200(1 + 5/100)]
= [200 * 21/20(21/20 * 21/20 + 21/20 + 1)] = Rs. 662.02

5x              6x           2x
50              25           100
250x + 150x + 200x = 4200
600x = 4200
x = 7 => 6x = 42

343 = 7³, 121 = 11², 1331 = 11³, 2197 = 13³ and 125 = 5³.
343, 1331, 2197 and 125 are perfect cubes, but not 121.

1: 2/5 = 5: 2
2/7 * 70 = 20

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

2(4*5 + 5*6 + 4*6) = 148

4x           5x           6x
100        50           25
400x + 350x + 150x = 16000
x = 20
6x = 120

20 = 4² + 4, 42 = 6² + 6, 58 = 7² + 9, 72 = 8² + 8 and 90 = 9² + 9.
20, 42, 72 and 90 can be expressed in n² + n form but not 58.

Clearly, the retailer gets 1 dozen out of 6 dozens free.
Equivalent discount = 1/6 * 100 = 16 2/3%.

A’s marks = 80 ; B’s marks = 60.
Let x% of A = B => x/100 * 80 = 60
=> x = (60 * 100)/80 = 75
B’s marks is 75% of A’s marks.

13 CP = 11 SP
11 — 2 CP
100 — ? =>18 2/11%

Let the present ages of Anand and Bala be ‘a’ and ‘b’ respectively.
a – 10 = 1/3 (b – 10) — (1)
b = a + 12
Substituting b = a + 12 in first equation,
a – 10 = 1/3 (a + 2) => 3a – 30 = a + 2
=> 2a = 32 => a = 16.

a/b = 4/1 => a = 4b
(a – 3b)/(2a – b) = (4b – 3b)/(8b – b)
= b/7b => 1/7

Length = Speed * time
Speed = L/T
S = 400/20
S = 20 M/Sec
Speed= 20*18/5 (To convert M/Sec in to Kmph multiply by 18/5)
Speed = 72 Kmph

A = P(1 + TR/100)
=> 3600 = 3000[1 + (4 * R)/100] => R = 5%
Now R = 6%
=> A = 3000[1 + (4 * 6)/100] = Rs. 3720.

Average = (11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99)/9
= [(11 + 99) + (22 + 88) + (33 + 77) + (44 + 66) + 55]/9
= [(4 * 110) + 55]/9 = 495/9 = 55.

A = 20000(11/10)³
= 26620
= 20000
———-
6620

(830 + ? * 3 * 14 / 100) – (830 * 14 * 3 / 100) = 20.16
=> 830 + ? * 42 – 830 * 42 = 2018
=> ? * 42 = 2018 + 830 * 42 – 830
=> ? * 42 = 42 * 48 + 830 * 42 – 830
=> ? = 48 + 830 – (830 / 42)
=> ? = 48 + 830 – 19.76
=> ? = 858.24

Let the rate of interest be R% p.a.
4400{[1 + R/100]² – 1} = 11193.60
[1 + R/100]² = (44000 + 11193.60)/44000
[1 + R/100]² = 1 + 2544/1000 = 1 + 159/625
[1 + R/100]² = 784/625 = (28/25)²
1 + R/100 = 28/25
R/100 = 3/25
Therefore R = 12 SI on Rs.44000 at 12% p.a. for two years = 44000(2)(12)/100
=Rs.10560

Let the ten’s and unit’s digit be x and 8/x respectively.
Then,
(10x + 8/x) + 18 = 10 * 8/x + x
9x² + 18x – 72 = 0
x² + 2x – 8 = 0
(x + 4)(x – 2) = 0
x = 2
So, ten’s digit = 2 and unit’s digit = 4.
Hence, required number = 24.

N (N + 1)

2x² + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = -3 or x = 3/2.

The numbers less than 696 and are squares of certain numbers are 676, 625.
The least number that should be subtracted from 696 to make it perfect square = 696 – 676 = 20.

Area of the path = Area of the outer circle – Area of the inner circle = ∏{4/2 + 25/100}² – ∏[4/2]²
= ∏[2.25² – 2²] = ∏(0.25)(4.25)  { (a² – b² = (a – b)(a + b) }
= (3.14)(1/4)(17/4) = 53.38/16 = 3.34 sq m

Time taken to meet for the first time anywhere on the track
= length of the track / relative speed
= 300 / (15 + 25)5/18 = 300* 18 / 40 * 5 = 27 seconds.

SI = 40 + 40
CI = 40 + 40 + 1.6 = 81.6

2000 —- 50
2000 —- 40
x —– 50
x*50 = 2000*40
x=1600
2000
——-
400

Let the cistern be filled by pipe A alone in x hours.
Then, pipe B will fill it in (x + 6) hours.
1/x + 1/(x + 6) = 1/4
x² – 2x – 24 = 0
(x – 6)(x + 4) = 0 => x = 6.

Speed downstream = 60 + 20 = 80 kmph
Speed upstream = 60 – 20 = 40 kmph

B = 1/16 – 1/24 = 1/48 => 48 days

Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n – 48n = 69 => n = 23
Therefore there were 23 members in the group initially.

W = 62%    L = 38%
62% – 38% = 24%
24% ——– 288
62% ——– ? => 744

The group contain six men and seven women
Three men can be selected from six men in ⁶C₃ ways.
Four women can be selected from seven women in ⁷C₄ ways.
Total number of ways = (⁷C₄)(⁶C₃).