IPPB Mains Mock Test Series 2nd, IPPB Mains Online Test Series 2

I. x² + 3x + 2x + 6 = 0
=> (x + 3)(x + 2) = 0 => x = -3 or -2
II. y² + 7y + 2y + 14 = 0
=> (y + 7)(y + 2) = 0 => y = -7 or -2
No relationship can be established between x and y.

Let sum = x. Then, S.I. = 2x/5, Time = 10 years.
Rate = (100 * 2x) / (x * 5 * 10) = 4%

Relative speed = 120 + 80 = 200 km/hr.
= 200 * 5/18 = 500/9 m/sec.
Let the length of the other train be x m.
Then, (x + 270)/9 = 500/9 => x = 230.

r₁ = x  r₂ = 3x
Π * x² * h = 40
Π9 x²h = 40 * 9 = 360

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

Assume X soldiers join the fort. 1200 soldiers have provision for 1200(days for which provisions last them)(rate of consumption of each soldier)
= 1200(30)(3) kg
Also provisions available for (1200 + x) soldiers is (1200 + x)(25)(2.5) kg
As the same provisions are available
=> 1200(30)(3) = (1200 + x)(25)(2.5)
x = [1200(30)(3)]/[(25)(2.5)] – 1200
x = 528

Let the total worth of the stock be Rs. x.
The SP of 20% of the stock = 1/5 * x * 1.1 = 11x/50
The SP of 80% of the stock = 4/5 * x * 0.95 = 19x/25 = 38x/50
Total SP = 11x/50 + 38x/50 = 49x/50
Overall loss = x – 49x/50 = x/50
x/50 = 400 => x = 20000

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.

Here, five seniors out of 12 seniors can be selected in ¹²C₅ ways. Also, five juniors out of ten juniors can be selected ¹⁰C₅ ways. Hence the total number of different ways of selection = ¹²C₅ * ¹⁰C₅ = ¹²C₇ * ¹⁰C₅
= ¹²C₅ = ¹²C₇

There are three ladies and five gentlemen and a committee of 5 members to be formed.
Number of ways such that two ladies are always included in the committee = ⁶C₃ = (6 * 5 * 4)/6 = 20.

Part of the filled by all the three pipes in one minute
= 1/18 + 1/15 – 1/45 = (5 + 6 – 2)/90 = 9/90 = 1/10
So, the tank becomes full in 10 minutes.

1 day work of the three persons = (1/15 + 1/20 + 1/25) = 47/300
So, all three together will complete the work in 300/47 = 6.4 days.-

37.5 % of 976
= 37.5/100 * 976 = 375/1000 * 976 = 3/8 * 976
= 3 * 122 = 366.

X = (Y*NR)/100   Y = (Z*NR)/100
X/Y = NR/100    Y/Z = NR/100
X/Y = Y/Z
Y² = XZ

P(11/10)⁴ – P(6/5)² = 482
P = 2000

1/12 + 1/15 – 1/x = 1/15
x = 12
12 * 45 = 540

Total time taken = (160/64 + 160/8) = 9/2 hrs.
Average speed = 320 * 2/9 = 71.11 km/hr.

B’s 1 day work = (1/12 – 1/20) = 1/30
Now, (A + B)’s 1 day work = (1/20 + 1/60) = 1/15
So, A and B together will complete the work in 15 days.

22/7 r² = 616 => r = 14
2 * 22/7 * 14 = 88

Number = 269 * 68 + 0 = 18292
67) 18292 (273
18291
——–
1
Required number = 1.

Required probability = (⁵C₃ . ⁴C₁)/⁹C₄ = (10 * 4)/126 = 20/63

23 + 29 + 31 + 37 + 41 = 161/5 = 32.2

Let the greater and the smaller number be g and s respectively.
gs = 2560
g + s exceeds g – s by 64 i.e., g + s – (g – s) = 64
i.e., 2s = 64 => s = 32.
g = 2560/s = 80.

1/4 : 3/4 = (1:3)5 = 5:15
1/5 : 4/5 = (1:4)4 = 4:16
——
9:31

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.

Speed = 45 Km/hr = 45*(5/18) m/sec = 25/2 m/sec
Total distance = 360+140 = 500 meter
Time = Distance/speed
= 500 * (2/25) = 40 seconds

(x*3*1)/100 + [(4000 – x)*5*1]/100 = 144
3x/100 + 200 – 5x/100 = 144
2x/100 = 56 → x = 2800

(A + B)’s 1 day work = 1/12;
(B + C)’s 1 day work = 1/15;
(A + C)’s 1 day work = 1/20
Adding, we get: 2(A + B + C)’s 1 day work = (1/12 + 1/15 + 1/20) = 1/5
(A + B + C)’s 1 day work = 1/10
So, A, B and C together can complete the work in 10 days.

3M = 6W —- 20 days
12M + 8W —–?
24W + 8 W = 32W —?
6W —- 20                    32 —–?
6 * 20 = 32 * x => x = 15/4 days

130
14
——-
361—— 144
100%——? => 400

(5/8 * 12/5 – 5/4)/(5/4) (3/2 -5/4)/(5/4) = 1/5

Let the initial quantity of milk in vessel be T litres.
Let us say y litres of the mixture is taken out and replaced by water for n times, alternatively.
Quantity of milk finally in the vessel is then given by [(T – y)/T]ⁿ * T
For the given problem, T = 90, y = 9 and n = 2.
Hence, quantity of  milk finally in the vessel
= [(90 – 9)/90]² (90) =  72.9 litres.

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

Let the number be x. Then,
3(2x + 9) = 75
2x = 16 => x = 8

Let the radii of the smaller and the larger circles be s m and l m respectively.
2∏s = 264 and 2∏l = 352
s = 264/2∏ and l = 352/2∏
Difference between the areas = ∏l² – ∏s²
= ∏{176²/∏² – 132²/∏²}
= 176²/∏ – 132²/∏
= (176 – 132)(176 + 132)/∏
= (44)(308)/(22/7) = (2)(308)(7) = 4312 sq m

1/2:1/3:1/4 = 6:4:3
Ram = 6/13 * 3250 = 1500
Shyam = 4/13 * 3250 = 1000
Mohan = 3/13 * 3250 = 750

(300 – 100)/11 = 200/11 = 18 2/11
18 Numbers

All are blood relation except Daughter-in-law.

2) 65 (32                               3) 32 (10               7) 10 (1
64                                         30                           7
——–                                  ——–                      – —–
1                                           2                            3
=> 65

? = (24 * 5 * 7 * 9) / 216 = 35

Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12
a + 3a = 4a = 12 => a = 3
Product of the roots = 3a² = 3(3)² = 27.

[50/x + 50/(x/2)] – 100/x = 2
x = 25

Given Exp. = 35 + 15 * 3/2
= 35 + 45/2 = 35 + 22.5 = 57.5

Let the two consecutive positive integers be x and x + 1
x² + (x + 1)² – x(x + 1) = 91
x² + x – 90 = 0
(x + 10)(x – 9) = 0 => x = -10 or 9.
As x is positive x = 9
Hence the two consecutive positive integers are 9 and 10.

D = 72 * 5/18 = 25 = 500 – 150 = 350

Total number of balls = 9 + 3 + 4
Two balls can be drawn from 16 balls in ¹⁶C₂ ways.

CP₁ = 100             SP₁ = 90
CP₂ = 80                SP₂ = 80 * (140/100) = 112
22 —– 100
55 —– ? => Rs.250