SEBI Online Mock Test 4, SEBI Online Test in English Series 4

Work done by P alone in one day = 1/6th of the total work done by Q alone in one day = 1/3(of that done by P in one day) = 1/3(1/6 of the total) = 1/18 of the total.
Work done by P and Q, working together in one day = 1/6 + 1/18 = 4/18 = 2/9 of the total
They would take 9/2 days = 4 (1/2) days to complete the work working together.

Let the five numbers be P, Q, R, S and T.
=> P + Q + R + S + T = 655.
(P + Q)/2 = 85 and R = 125
P + Q = 170 and R = 125
P + Q + R = 295
S + T = 655 – (P + Q + R) = 360
Average of the last two numbers = (S + T)/2 = 180.

Except in 862, in all other numbers sum of first two digits is same as the last digit.

x + y = 80
x – 4y = 5
x = 65 y = 15

P = (P*30*R)/100
R = 3 1/3%

15:25 => 3:5
9600*10/100 = 960
9600 – 960 = 8640
8640*3/8 = 3240 + 960
= 4200

RS = 5 ¾ – 4 ½ = 1 ¼
T = 3 ½ h.
D = 5/4 * 7/2 = 35/8 = 4 3/8 km

2/3 x = 28 => x = 42
42 * 1/2 = 21

Speed = 60*(5/18) m/sec = 50/3 m/sec
Length of Train (Distance) = Speed * Time
(50/3) * 9 = 150 meter

Let the amount lent at 7% be Rs. x
Amount lent at 10% is Rs. (6000 – x)
Total interest for one year on the two sums lent
= 7/100 x + 10/100 (6000 – x) = 600 – 3x/100
=> 600 – 3/100 x = 450 => x = 5000
Amount lent at 10% = 1000
Required ratio = 5000 : 1000 = 5 : 1
Learn more problems on : Averages

1/3 * 1/3 * 1/7 * 1/9 * 2140 = ?
? = 3.77

Let sum = x. Then, S.I. = x.
Rate = (100 * S.I.) / (P * T) = (100 * x) / (x * 12)
= 25/3 = 8 1/3 %

Ratio of their shares = 22500 : 35000 = 9:14
Deepak’s share = 13800 * 14/23 = Rs. 8400.

S.I. = (1000 * 10 * 4)/100 = Rs. 400
C.I. = [1000 * (1 + 10/100)⁴ – 1000] = Rs. 464.10 Difference = (464.10 – 400) = Rs. 64.10

Area of the square = s * s = 5(125 * 64)
=> s = 25 * 8 = 200 cm
Perimeter of the square = 4 * 200 = 800 cm.

S.I. for 1 year = (854 – 815) = Rs. 39
S.I. for 3 years = 39 * 3 = Rs. 117
Principal = (815 – 117) = Rs. 698.

Let Rajan’s present age be x years.
Then, his age at the time of marriage = (x – 8) years.
x = 6/5 (x – 8)
5x = 6x – 48 => x = 48
Rajan’s sister’s age at the time of his marriage = (x – 8) – 10 = 30 years.
Rajan’s sister’s present age = (30 + 8) = 38 years.

12 men take 32 days to complete the work.
One man will take (12 * 32) days to complete it.
Similarly one woman will take (16 * 36) days to complete it and one boy will take (48 * 16) days to complete it.
One man, one woman and one boy working together will complete = 1/(12 * 32) + 1/(16 * 36) + 1/(48 * 16)
= 1/(4 * 3 * 16 * 2) + 1/(16 * 4 * 9) + 1/(16 * 3 * 4 * 4)
= 1/64(1/6 + 1/9 + 1/12) = 1/64 * 13/36 of the work in a day.
They will take (64 * 36)/13 days to complete the work working together.

x = -√7
1/x = 1/(3 – √7) = 3 + √7/(3 – √7) (3 + √7)
= (3 + √7)/2 => x + 1/x = (9 – √7)/2

4/12 + x/15 = 1
x = 10

Let x percent of 64 be 80.
64 * x/100 = 80 => x = (80 * 100)/64 => x = 125
80 is 125% of 64.

a/d = (3/4)*(7/9)*(5/7) => 5/12

2x * x = 128 => x= 8

A + B + C = 48
B + C + D = 45
A + D = 13
A – D = 3
A + D = 13

2D = 10
D = 5

Let C’s age be x years.
Then, B’s age = 2x years.
A’s age = (2x + 2) years.
(2x + 2) + 2x + x = 27
5x = 25 => x = 5
Hence, B’s age = 2x = 10 years.

[15000 * (1 + R/100)² – 15000] – (15000 * R * 2)/100 = 96
15000[(1 + R/100)² – 1 – 2R/100] = 96
15000[(100 + R)² – 10000 – 200R]/10000 = 96
R² = (96 * 2)/3 = 64 => R = 8
Rate = 8%

Principal = (P.W. of Rs. 882 due 1 year hence) + (P.W. of Rs. 882 due 2 years hence)
= [882/(1 + 5/100) + 882/(1 + 5/100)²]
= (882 * 20)/21 + (882 * 400)/441 = Rs. 1640.

Simple interest = (18000 * 4 * 15)/100 = Rs. 10800
Amount = P + I = 18000 + 10800 = Rs. 28800

Let the numbers be x and (22 – x). Then,
5x = 6(22 – x)
11x = 132 => x = 12
So, the numbers are 12 and 10.

Let the speed of the man in still water be a kmph and let the speed of the stream be b kmph.
Now 30/(a + b) + 20/(a – b) = 4 and 45/(a + b) + 40/(a – b) = 7
Solving the equation, the speed of man in still water is 12.5 kmph.

2 * 22/7 * 14 * x = 1056 => x = 12

C = 1/5 – 1/6 = 1/30 => 30 days

C.I. = [4000 * (1 + 10/100)² – 4000]
= (4000 * 11/10 * 11/10 – 4000) = Rs. 840.
Sum = (420 * 100)/(3 * 8) = Rs. 1750

Let the price of the article be Rs. 100.
20% of 100 = 20.
New price = 100 + 20 = Rs. 120
Required percentage = (120 – 100)/120 * 100
= 20/120 * 100 = 50/3 = 16 2/3%.

4/3 π * 3 * 3 * 3 = 4/3 π * 1 * 1 * 1 * x
x = 27

1500 —- 270
100 —- ? => 18%

HCF of 323, 437 = 19
323 * 437 = 19 * 19 * x
x = 391

Let the time taken to cover from A to B in car and cycle be x hours and y hours respectively.
x + y = 7 — (1) ; 2x = 4 — (2)
solving both the equations, we get y = 5
So, time taken to cover both ways by cycle = 2y hours = 10 hours.

Simran : Nanda = (50000 * 36) : (80000 * 30) = 3:4
Simran’s share = 24500 * 3/7 = Rs. 10500

LCM = 1440
1440 – 1 = 1439

Relative speed = (42 + 30) * 5/18 = 4 * 5 = 20 mps.
Distance covered in passing each other = 120 + 280 = 400 m.
The time required = d/s = 400/20 = 20 sec.

Except Austriala : Pound, all others are correction combination of country and its currency.

16 years ago, let T = x years and G = 8x years. After 8 years from now, T = (x + 16 + 8) years and G = (8x + 16 + 8) years.
8x + 24 = 3(x + 24) => 5x = 48
8 years ago, T/G = (x + 8)/(8x + 8) = (48/5 + 8) / (8 * 48/5 + 8) = 88/424 = 11/53

Work done by the tank in 1 hour = (1/2 – 1/3) = 1/14 Leak will empty the tank in 14 hrs.

Let the length of each train be x m.
Then, distance covered = 2x m.
Relative speed = 46 – 36 = 10 km/hr.
= 10 * 5/18 = 25/9 m/sec.
2x/36 = 25/9 => x = 50.

4 πr² = 64 => r = 4

Kim’s wages : David’s wages = Kim’s 1 day work : David’s 1 day work = 1/3 : 1/2 = 2:3
Kim’s share = 2/5 * 150 = Rs. 60

(2000 + 3)(2000 + 4) – (2000 + 1)(2000 + 2) = ?
Since (2000 * 2000) – (2000 * 2000) is equal to zero. ?
= (8000 + 6000 + 12) – (4000 + 2000 + 2)
=> ? = 14012 – 6002 = 8010

0.003 * ? * 0.0003 = 0.00000027
3/1000 * ? * 3/10000 = 3/1000 * 3/1000 * 3/100
? = 3/10 = 0.3