UP Police Constable Mock Test - 9 - UP Police Constable MCQ

Hence, Option B is correct.

Total females = 100% - 52% = 48%
So the answer = (100 – 60)% of 52% + (100 – 70)% of 48% = 35.20%
Hence, Option B is correct.

Between 5 and 78:
First number = 6
Last number = 72

Hence, Option D is correct.

Perimeter of the rectangle
= 72/2 = 36 cm
Let the width of the rectangle = x
Length = x + 2
2 × (x + x + 2) = 36
2x + 2 = 18
x = 8 cm
Hence, Option A is correct.

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM
⇒ 864 × Second number
= 96 × 1296 ⇒ Second number

X : Y = 4% : 6% = 2 : 3
Hence, Option B is correct.

x = 12 and y = 4
∴  (x + y)^x/y = (12 + 4)^12/4 = (16)³
= 16 × 16 × 16 = 4096 

According to the question,
Every day adding 1 rs extra to previous day.
Let us assume after n day, total saving will become perfect square.
1 + 2 + 3 + 4 + 5 + 6 +.......................+ tn.
Apply the algebra A.P formula,
sum of total rupees after n days = n(n+1)/2
Hit and trail method, Put the value of n = 2 , 3 , 4, 5 .... and so on to get the perfect square.
If n = 2
n(n+1)/2 = 2 x 3 / 2 = 3 which is not perfect Square.
If n = 3
n(n+1)/2 = 3 x 4 / 2 = 6 which is not perfect Square.
If n = 4
n(n+1)/2 = 4 x 5 / 2 = 10 which is not perfect Square.
If n = 5
n(n+1)/2 = 5 x 6 / 2 = 15 which is not perfect Square.
If n = 6
n(n+1)/2 = 6 x 7 / 2 = 21 which is not perfect Square.
If n = 7
n(n+1)/2 = 7 x 8 / 2 = 28 which is not perfect Square.
If n = 8
n(n+1)/2 = 8 x 9 / 2 = 36 which is perfect Square.

n(n+1)/2 should be a perfect square . The first value of n when this occurs would be for n = 8. thus , on the 8th of March of the required condition would come true.

I. x² – 14x + 45 = 0
x² – 9x – 5x + 45 =0
x = 9 and 5
II. y² + 2y – 35 = 0
y² + 7y – 5y – 35 = 0
y = – 7 and 5
Answer: x ≥ y
Hence, Option C is correct.

Here 11^th term = 1/13
13^th term = 1/11
Let the first term of the AP be a and common difference be d

Subtracting (1) from (2)
We get,

Now, substituting value of d in equation 1
We get,

Given:

n^th term for the first series = n^th term for the second series.

Concept:

Arithmetic Progression: 

• An arithmetic progression is a list of numbers in which each term is obtained by adding a fixed number to the preceding term except the first term.
• This fixed number is called the common difference of the AP and it can be positive, negative or zero.

Consider the AP, whose first term is 'a' and common difference is 'd' and there are n number of terms.

a, a + d, a + 2d, a + 3d........a + (n - 1)d

Hence, n^th term of AP is given by

T_n = a + (n - 1)d

Calculation:

Consider the first series

3 + 10 + 17 + ... 

a = 3 and d = 10 - 3 = 7

n^th term of AP is given by,

(T_n)₁ = 3 + (n - 1) × 7 

⇒ (Tn)1 = 7n - 4       .......(1)

Consider the second series 

63 + 65 + 67 + ....

a = 63 and d = 2

nth term of AP is given by,

(Tn)₂ = 63 + (n - 1) × 2

⇒ (Tn)₂ = 2n + 61      .......(2)

According to the question

(Tn)1 = (Tn)2

⇒ 7n - 4 = 2n + 61

⇒ 5n = 65

⇒ n = 13

Hence, for the given series, 13^th term will be equal.

terms after square root is

(15/27 - 5/12) / (4/9)

= 5/16

Let milk = 3x and water = 2x
In the final mixture:

Using Statement I :
3x + 8 – (2x + 12) = 16
x = 20
2x + 12 – (3x + 8) = 16
x = – 12
So, statement I alone is sufficient.

Using Statement II :
(3x + 8) : (2x + 12) = 17 : 13
x = 20
So, statement II is alone sufficient.
So, statement I alone or II alone is sufficient.
Hence, option C is correct.

Let the 3 boys be B₁, B₂, B₃ and 4 prizes be P₁, P₂, P₃ and P₄
Now B₁ is eligible to receive any of the 4 available prizes (so 4 ways)
B₂ will receive prize from rest 3 available prizes(so 3 ways)
B₃ will receive his prize from the rest 2 prizes available(so 2 ways)
So total ways would be: 4 × 3 × 2 × 1 = 24 Ways
Hence, the 4 prizes can be distributed in 24 ways

57 after 12 o’clock 12:21……= 1 from 1 to 9 it is 1.01, 1.11, 1.21, 1.31, 1.41, 1.51 = 6 similarly 2.02, 2.12… 6*9 = 54+1 = 55 after 10 o’clock 10:01, 11:11 Ans: 55 + 2 = 57
Option 3 is the correct answer.

Upstream speed of the boat 

Speed of the stream = 15 – 7.50 = 7.50 km/hr
Hence, Option A is correct.

Reqd. time = 60 × (1 + 3) = 240 seconds
Hence, Option D is correct.

Using Rule 7,
Part of the cistern filled by pipe Q in 1 minute

Required time = 60 minutes

When B runs 25 m, A runs 45/2 m.
When B runs 1000 m,

∴ B beats A by 100 m.

Let B joined for x months.
Thus their one month equivalent capitals
A = 85000*12
B = 42500*x = 42500x
Therefore, (85000*12)⁄42500x = 3/1
=> x = 8 months (option ‘D’)

Let CP = 3x
SP = 2x

Hence, Option B is correct.

There should be 230 in place of 232.
Hence, Option D is correct. 

Let AB be the tower and C and D be the two positions of the car.
Then, from figure
AB/AC = tan 60 = √3 => AB = √3AC
AB/AD = tan 45 = 1 => AB = AD
AB = AC + CD
CD = AB - AC = √3AC - AC = AC (√3 - 1)
CD = AC (√3 - 1) => 10 min
 AC => ?
AC/(AC(√3 - 1)) x 10 =? = 10/(√3 - 1) = 13.66 = 13 min 20 sec(approx)

Total possible numbers = 4 × 3 × 2 × 1 = 24
Possibility of 5921= 1

∵ Distance covered in 20 min = 5 km
∴ Distance covered in 1 min = 5/20 km
∴ Distance covered in 100 min = (5/20) x 100
= 5 x 5 = 25 km

CD || AB (Given)
Produce RQ to meet AB in S
∠CRS = ∠PSR (at. int. ∠s)
But ∠CRS = 55°
∴ ∠PSR = 55°
Now in QSP
∠QSP + ∠QPS + ∠PQS = 180°
55° + 38° + ∠SQP = 180°
∴ ∠SQP = 180° – 93° = 87°
But angle a and ∠PQS are linear
∠a = 180° – 87°
∠a = 93°

Hence, Option B is correct.

18% of x = 13.50% of y

50% of x = 37.50% of y
Hence, Option D is correct.

Sum of 100 numbers =100×210=21000
Sum of 120 numbers =120×200=24000
Sum of 20 numbers =24000−21000=3000
∴ Average of 20 new numbers =3000/20=150

Hence, Option D is correct.