IPPB Mains Online Test Series 5, IPPB Test Series, IPPB Mock Test

S:E = 1:3
E:M = 3:5
————
S:E:M = 3:9:15
3/27 * 162 = 18

Let E₁ be the event of drawing a red card.
Let E₂ be the event of drawing a king .
P(E₁ ∩ E₂) = P(E₁) . P(E₂)
(As E₁ and E₂ are independent)
= 1/2 * 1/13 = 1/26

a² = 3136 => a = 56
56 * 4 * 3 = 672 – 6 = 666 * 1.5 = 999

L.C.M of 252, 308 and 198 = 2772
So, A, B and C will again meet at the starting point in 2772 sec, i.e., 46 min 12 sec.

760 —- 22
760 —- 20
x —– 19
x*19 = 760*20
x = 800
760
——-
40

A:B = 2:3, B:C = 4:5 = (4 * 3/4):(5 * 3/4) = 3:15/4
and C:D = 6:7 = (6 * 15/24):(7 * 15/24) = 15/4:35/8 A:B:C:D = 2:3:15/4:35/8 = 16:24:30:35

1/30 + 1/40 = 7/120
120/7 = 17 1/7 days

Calculated average height of 35 boys = 180 cm.
Wrong total height of 35 boys = 180 * 35 cm. This was as a result of an actual height of 106 cm being wrongly written as 166 cm. Correct total height of 35 boys = 180 * 35 cm – 166 cm + 106 cm
= 180 * 35 cm – 166 cm + 106 cm/35 = 180 cm – 60 /35 cm
= 180 cm – 1.71 cm = 178.29 cm.

100 —- 100 — 10
100 — 10
——————–
300 —- 20 years

A = 8000(21/20)³ = 9261

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

Given ratio = 1/2:2/3:3/4 = 6:8:9
1st part = 782 * 6/23 = Rs. 204.

A:B:C = (40000 * 36) : (80000 * 12 + 40000 * 24) : (120000 * 24 + 40000 * 12)
= 144:192:336 = 3:4:7

22004 – 29303 + 101299 – 59532
123303 – 888.35 = 34468 = 34500

1 to 11 = 11 * 10.9 = 119.9
1 to 6 = 6 * 10.5 = 63
6 to 11 = 6 * 11.4 = 68.4
63 + 68.4 = 131.4 – 119.9 = 11.5
6^th number = 11.5

Originally, let the number of boys and girls in the college be 7x and 8x respectively.
Their increased number is (120% of 7x) and (110% of 8x).
i.e., (120/100 * 7x) and (110/100 * 8x)
i.e., 42x/5 and 44x/5
Required ratio = 42x/5 : 44x/5 = 21:22

LCM = 10080
10080) 99999 (8
90720
———
9279
99999 – 9279 = 90720

Let the number be x.
30% of x = 30/100 * x = 3x/10
20% of x = 20/100 * x = x/5 = 2x/10.
Given that 3x/10 – 2x/10 = 18 => x = 180.
Required number is 180.

Let the two numbers be 5x and 6x.
Let the number added to both so that their ratio becomes 7:8 be k.
(5x + k)/(6x + k) = 7/8
42x = 7k => k = 2x.
6x – 5x = 10 => x = 10
k = 2x = 20.

(25/100) * (30/100) * (45/100)
1/4 * 3/10 * 9/20
27/800 = 0.03375

D = 25 * 60 + 1500 = 3000 m
T = 3000/60 * 18/5 = 180 sec = 3 mins

C’s 1 day’s work = 1/3 – (1/6 + 1/8) = 1/3 – 7/24 = 1/24
A’s wages : B’s wages : C’s wages
1/6 : 1/8 : 1/24 = 4:3:1
C’s share = 1/8 * 3200 = Rs. 400

CI = 14800{ [ 1 + 13.5/100]² – 1 }
= 14800 { [1 + 27/200]² – 1
= 14800 { 2 + 27/200}{27/200}
= (74)[2 + 27/200](27) =
1998[2 + 27/200] = 3996 + 269.73 = Rs. 4266

1 man’s 1 day work = 1/100
(10 men + 15 women)’s 1 day work = 1/6
15 women’s 1 day work = (1/6 – 10/100) = 1/15
1 woman’s 1 day work = 1/225
1 woman alone can complete the work in 225 days.

l * 30 = 1200 → l = 40

40 + 30 + 50 = 120
120 * 10 = 1200

Ratio of times taken by A and B = 100:130 = 10:13
Suppose B takes x days to do the work.
x = (23 * 13)/10 = 299/10
A’s 1 day work = 1/23; B’s 1 day work = 10/299
(A + B)’s 1 day work = (1/23 + 10/299) = 1/13
A and B together can complete the job in 13 days.

Tiruvanantapuram, Hyderabad, Bangalore and Bhubaneswar are state capitals, while Calicut is not a state capital.

41, 43, 47 and 53 are prime numbers, but not 57.

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

4/18 + (x -7)/27 + x/36 = 1
x = 16 days

If the cost price is Rs.100, then to get a profit of 20%, the selling price should be Rs.120.
If 120kg are to be sold, and the dealer gives only 100kg, to get a profit of 20%.
How many grams he has to give instead of one kilogram(1000 gm).
120 gm —— 100 gm
1000 gm —— ?
(1000 * 100)/120 = 2500/3 = 833 1/3 grams.

1/8 – 1/x = 1/10
x = 40

Relative speed of the thief and policeman = 11 – 10 = 1 km/hr.
Distance covered in 6 minutes = 1/60 * 6 = 1/10 km = 100 m.
Distance between the thief and policeman = 200 – 100 = 100 m.

Let the pipe B be closed after x minutes.
30/16 – x/24 = 1 => x/24 = 30/16 – 1 = 14/16
=> x = 14/16 * 24 = 21.

Number of male employees = 6400 * 65/100 = 4160
Required number of male employees who are less than 50 years old = 4160 * (100 – 25)%
= 4160 * 75/100 = 3120.

[(3.572)³ + (2.428)³] / [(3.572)² – (3.572) * (2.428) + (2.428)²]
= {(3.572)² – (3.572) * (2.428) + (2.428)²}{3.752 + 2.428} / [(3.572)² – (3.572) * (2.428) + (2.428)²]
= 3.752 + 2.428 = 6

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

(60/100) * X – 40 = 50
6X = 900
X = 150

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

110% of S.P. = 616
S.P. = (616 * 100)/110 = Rs. 560
C.P = (110 * 560)/112 = Rs. 500

50/100 * x = 30 * 20 => x = 1200

Let the school ages of Neelam and Shaan be 5x and 6x years respectively.
Then, (1/3 * 5x)/(1/2 * 6x) = 5/9
15 = 15
Thus, Shaan’s age cannot be determined.

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

P(at least one graduate) = 1 – P(no graduates) =
1 – ⁶C₃/¹⁰C₃ = 1 – (6 * 5 * 4)/(10 * 9 * 8) = 5/6

100 * 100 = 10000
125 * 80 = 10000
No change

45 + 31 = 76/15 = 1 (Remainder)

x = √8 – √7 = 1/(√8 + √7) and y = √6 – √5 = 1/(√6 + √5) and z = √10 – 3 = 1/(√10 + 3)
As (√8 + √7) > (√6 + √5), (√8 – √7) (√8 + √7) > (√6 + √5)
z

1/20 + 1/30 – 1/40 = 7/120
120/7 = 17 1/7