Indian Bank PO Pre Online Test Series 4th, Indian Bank Online Test

r = 100   r = 101
r² = 10000    r² = 10201
10000 —- 201
100 —- ? => 2.01%

800 — 200
100 — ? => 25%

1347
7431
————
6084

Let the radii of the larger and the smaller circles be l cm and s cm respectively. Let the side of the square be a cm.
a² = 784 = (4)(196) = (2²).(14²)
a = (2)(14) = 28
a = 2l, l = a/2 = 14
l = (7/3)s
Therefore s = (3/7)(l) = 6 Circumference of the smaller circle = 2∏s = 12∏ cm.

(27²/4^-3)^-5/6 = [(3³)²/(2²)^-3]^-5/6
= (3⁶ * 2⁶)^{-5/6
= (66})^-5/6 = 1/6⁵ = 1/ 7776.

504/M = 384/800
(504 * 800) / 384 = M
M = 1050

Average speed = (2 * 3 * 2) / (3 + 2) = 12/5 km/hr.
Distance traveled = 12/5 * 5 = 12 km.
Distance between house and school = 12/2 = 6 km.

The simple interest for the first year is 10200/2 is Rs.5100 and compound interest for first year also is Rs.5100. The compound interest for second year on Rs.5100 for one year
So rate of the interest = (100 * 1530)/ (5100 * 1) = 30% p.a.
So P = (100 * 10200)/ (30 * 2) = Rs.17000

a³ = 1331 => a = 11
b² = 121 => b = ± 11
a ≥ b

Let the required numbers be 33a and 33b.
Then, 33a + 33b = 528 => a + b = 16.
Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9).
Required numbers are (33 * 1, 33 * 15), (33 * 3, 33 * 13), (33 * 5, 33 * 11), (33 * 7, 33 * 9).
The number of such pairs is 4.

Let C.P. = Rs. 4x. Then, S.P. = Rs. 5x
Gain = (5x – 4x) = Rs. x
Gain % = (x * 100)/ 4x = 25%.

Correct sum = (36 * 50 + 48 – 23) = 1825.
Correct mean = 1825/50 = 36.5

Number of students passed = 60% of 140 = 60/100 * 140 = 84
Number of students failed = 140 – 84 = 56.
Required percentage = 28/56 * 100 = 50%.

Let the side of the square be a cm. Let the length and the breadth of the rectangle be l cm and b cm respectively.
4a = 2(l + b)
2a = l + b
l . b = 480
We cannot find ( l + b) only with the help of l . b. Therefore a cannot be found .
Area of the square cannot be found.

g% = 20 + 20 + (20*20)/100
= 44%

221 + 43 = 264/17 = 9 (Remainder)

Let ages of Ramu and Shyam be x and y respectively.
x/y = 3/4 => x = 3/4 y
(x- 7)/(y – 7) = 2/3
=> 3x – 21 = 2y – 14 => 3x = 2y + 7
But x = 3/4 y
3 * 3/4 y = 2y + 7
9y = 8y + 28 => y = 28years
=> x = 21 years
Ratio of their ages five years hence, (21 + 5)/(28 + 5) = 26/33.

The given digits are 1, 2, 3, 5, 7, 9
A number is even when its units digit is even. Of the given digits, two is the only even digit.
Units place is filled with only ‘2’ and the remaining three places can be filled in ⁵P₃ ways.
Number of even numbers = ⁵P₃ = 60.

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

3^4m+1 = 3^7m-5 equating powers of 3 on both sides, 4m+1 = 7m – 5 => 3m = 6 => m = 2.

Let the article be worth Rs. x.
C.P. 90% of Rs. x = Rs. 9x/10
S.P. = 110% of Rs. x = Rs. 11x/10
Gain = (11x/10 – 9x/10) = Rs. x/5
Gain % = x/5 * 10/9x * 100 = 22 2/9 % > 20%

?² = 3/7 * 4/9 * 2/5 * 7560
=> ?² = 576 = 24²
? = 24

96% —- 12
144% —- ?
96/144 * 12 = 8

3/12 + x/15 = 1
x= 11 1/4

(10*10)/100 = 1%loss
100 — 99
? — 1980 => Rs.2000

3240 = 2³ * 3⁴ * 5; 3600 = 2⁴ * 3² * 5^{2
H.C.F = 36 = 22} * 3²
Since H.C.F is the product of lowest powers of common factors, so the third number must have (2² * 3²) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 3⁵ and 7² as its factors.
Third number = 2² * 3⁵ * 7²

5 > 4 and 1/2 > 1/4
5^1/2 > 4^1/4
4^1/4 cannot be the greatest. Hence 3^1/4 also cannot be the greatest.
3^7/10 = (3⁷)^1/10 = (2187)^1/10
5^1/2 = 5^5/10 = (5⁵)^1/10 = (3125)^1/10
6^1/5 = 6^2/10 = (6²)^1/10 = (36)^1/10
As (3125)^1/10 > (2187)^1/10 > (36)^1/10 , 5^1/2 is the greatest.

WC of A: B = 1:2
B: C = 1:2
———————
A: B: C = 1:2:4

HCF of numerators = 1
LCM of denominators = 48
=> 1/48

1/10 + 1/15 = 1/6 * 4 = 2/3
1 – 2/3 = 1/3
1/10 + 1/15 – 1/x = 1/3
x = 8

22/7 * 7 * 7 = 154

1/2 * 15 * 20 = 150

Speed = [45 X 5/18] m/sec = [25/2] m/sec Time = 30 sec Let the length of bridge be x metres. Then, (130 + x)/30 = 25/2 => 2(130 + x) = 750 => x = 245 m.

Let the length and the breadth of the rectangle be l cm and b cm respectively. Let the side of the square be a cm.
a2 = 4096 = 2¹²
a = (2¹²)^1/2 = 2⁶ = 64
L = 2a and b = a – 24
b : l = a – 24 : 2a = 40 : 128 = 5 : 16

Present total age of the members = 26(5) = 130 years.
Present age of the youngest member = 10 years
Present total age of the remaining four members = 130 -10 = 120 years
Their average age at the time of the birth of the youngest member = [120 – (4 * 10)] / 4 = 30 – 10 = 20 years.

SP = CP + g
50 SP = 50 CP + 15 SP
35 SP = 50 CP
35 — 15 CP gain
100 — ? => 42 6/7%

We have M₁ D₁ H₁ / W₁ = M₂ D₂ H₂ / W₂ (Variation rule)
(60 * 30 * 8)/ 200 = (45 * D₂ * 6) / 300
D₂ = (60 * 30 * 8 * 300) / (200 * 45 * 6)  => D₂ = 80.

Sum of the 35 numbers = 35 * 25 = 875
If each number is multiplied by 5, the sum also gets multiplied by 5 and the average also gets multiplied by 5.
Thus, the new average = 25 * 5 = 125.

20 + 24 + 8 * 2 = 60/2 = 30

4 * 0.4 * 0.04 * 0.004 => 4 * (4/10) * (4/100) * (4/1000) = (256/1000000) = 0.000256

Let the number of coins of each kind be x.
=> 5x + 2x + 1x = 1152
=> 8x = 1152 => x = 144

1 – {1 + (a² – 1)^-1}^-1 = 1 – {1 + 1/a² – 1}^-1 = 1 – {a² / a² – 1}^-1 = 1 – (a² – 1)/a² = 1/a².

144) 10000 (69
9936
——–
64
10000 + 144 – 64 = 10080

(30*30)/100 = 9%loss

Let the shares of A, B and C be a, b and c respectively.
a : b : c = 1/2 : 1/3 : 1/4
Let us express each term with a common denominator which is the last number divisible by the denominators of each term i.e., 12.
a : b : c = 6/12 : 4/12 : 3/12 = 6 : 4 : 3.
Share of C = 3/13 * 1560 = Rs. 360.

x Not younger_______ ↑
The last ball can be thrown by any of the remaining 6 players. The first 6 players can throw the ball in ⁶P₆ ways.
The required number of ways = 6(6!) = 4320

Let the sum be Rs. 100. Then,
S.I. for first 6 months = (100 * 10 *1) / (100 * 2) = Rs. 5
S.I. for last 6 months = (105 * 10 * 1) / (100 * 2) = Rs. 5.25
So, amount at the end of 1 year = (100 + 5 + 5.25) = Rs. 110.25
Effective rate = (110.25 – 100) = 10.25%.

Since, 40.1 = 40
52 * 40 = 2080

? = 17^14+16-8 = 17²²

Let the middle digit be x. Then, 2x = 10 or x = 5.
So, the number is either 253 or 352. Since the number increases on reversing the digits, so the hundred’s digit is smaller than the unit’s digit. Hence, required number = 253.