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**1. Solve this:**

**(a) ****(b) ****(c) **

**(d) Ans:** (c)

Putting the values:

**(a) (b) (c) **

Ans.

ΔABC is equilateral triangle of side 2 cm.

Area of an equilateral triangle

where a is the side of that triangle.

Area of 3 sector of angle 60° each and radius 1 cm each. = 3 x π (radius)^{2} × (θ/360°)

= 3 x π (1)^{2} × (60°/360°) = π/2

Area of shaded portion = Area of ΔABC – Area of 3 sectors**3. Length of two trains is 150 m each. When they are moving in opposite direction. They cross each other in 20 s and when they are moving in same direction. Then the faster train passed slower train in 40 s. Then find the speed of faster train.(a) 10.50 m/s(b) 11.25 m/s**

(d) 25 m/s

Ans.

Let speed of faster train = x m/s

Let speed of slower train = y m/s

Distance = Speed × Time

Relative speed (in opposite direction) = (x + y) m/s

Relative speed (in same direction) = (x – y) m/s

In opposite direction

(x + y) = (150 + 150)/20 = 15 m/s

In same direction,

(x - y) = (150 + 150)/40 = 7.5 m/s

Using these two equations, we get x = 11.25 m/s

Ans.

Let distance from A to B LCM of 24, 10, 40 = 120 km

Time from A to B by bike = 120/40 = 3 h

Time from B to C by cycle = 120/10 = 12 h

All of us know that speed = Distance/time

Time from C to A by autorickshaw = 240/24 = 10 h

Average Speed = Total distance travelled/Total time taken

= (4 x 120)/(3 + 12 + 10) = 480/25 = 19.2 km/h

(a) 70 km from point A

(b) 74 km from point A

(c) 70 km from point B

(d)

Speed of Rohan = 8 km/h Speed of Rahul in start = 4 km/h and Rahul’s speed increases 1 km/h in every hour.

Relative speed in opposite direction AP series is formed = … 12, 13, 14 = Covered distance per hour

Let they meet after n hour.

According to the question, Distance between Rohan and Rahul = Covered distance in n hours

Here, we will use the formula of sum of n terms of an A.P., which is:

Where n = number of terms

a = First term of the A.P.

d = Common difference

⇨ 288 = n(24 + n – 1)

Solving this equation,

n

Splitting this quadratic equation,

(n – 9) (n + 32) = 0

Here, we have two values of n, which are -32 and 9, but negative value of n is not possible in this question.

So, after 9 h Rahul and Rohan meets, then distance covered by Rohan = 9 x 8 = 72 km.

It means, they meet exactly at midway between A and B.

(a) 2

(b) 5

(c) 7

(d) 9

Ans.

5! = 5 x 4 x 3 x 2 x 1 = 120

As we know, unit digit for (5!, 6!, 7!, 8! … ∞) is zero.

Also, (1!)

Unit digit for (1): 1 + 4 + 6 + 6 = 17

∴ Required unit digit = 7.

(b) 170

(d) 175

Ans.

Required ways = 11c

Total possibility = 16

The number of way on applicant may fail in the exam = (16 – 1) = 15

(a) 36

(b) 15

(c) 42

(d) 21

Ans.

SOTICA has three vowel. Then, the required ways = 3! x 3! = 6 x 6 = 36

(a) Square

(b) Circle

(c) Hexagon

(d) Octagon

Ans.

If the perimeter of figures is same then, area of figure which has maximum number of sides is maximum.

Then, circle has maximum area.

(b) 10

(c) 46

(d) 35

Ans.

Number of ways of the gift pack

Number of ways of the gift pack without Kitkat

Then, number of ways of the gift pack with Kitkat = 56 – 10 = 46

Solution.

Using this formula: 114 = P(6/100)

⇨ P = (114 x 100 x 100)/(6 x 6) = 31666.66

(a)

**(b)(c)(d)**

Using the properties of Log, and simplifying this:

Solution.

If we take z3 out of this, it becomes: z

Hence, (-7x y

(a) 13

(b) 19

(c) 15

(d) 18

We can see that in above figure that there are 19 intersection points.

Here, we are using “Cosine rule”, which is as follows:

Cos 60° = (b

Putting the value of Cos 60°, which is 1/2 and doing the cross multiplication. a

Volume of wood = External volume of box with wood – Internal volume of box without wood = (LBH)external – (LBH)internal

= 20 x 14 x 10 = 19 x 13 x 9

= 577 cm

Total weight of wood = 3.462 kg

Weight of 577 cm

Then, weight of 1 cm

(a) 120

(b) 60

(c) 240

(d) 480

(5!)

5! = 5 x 4 x 3 x 2 x 1 = 120

Now, 120

(a) 30 days

(c) 60 days

(d) 15 days

Ans.

Let efficiency of Gitesh = 2x, and efficiency of Jitesh = x

Jitesh is half efficient as Gitesh, so he will take double time to finish the task.

And, difference between their time is 30 days.

So, Jitesh’s time = 2 x 30 = 60 days

And, Gitesh’s time = x = 30 days

Let us assume the total work to be 60 units.

Jitesh will do 1 unit per day.

Gitesh will do 2 units per day.

Together they will do 3 units per day.

Time taken by both, when working together = 60/3 = 20 days.

Ans.

Water which was added =

Putting the values:

⇨ 6 = 2K/5

⇨ K = 15 litres

(a) 10 min

(b) 20 min

(c) 15 min

(d) 12 min

Ans.

Halt time = [(Faster train’s speed – Slower train’s speed)/Faster train’s speed] x 60

[(60 – 50)/60] x 60 = 10 mins

They had to answer four sections Maths, DI, LR and English.

The number of students who qualified in Maths = 55.

The number of students who qualified in LR= 38.

The number of students who qualified in (Maths + English) = 30.

The number of students who qualified in (LR + English) = 15.

The number of students who qualified in (Maths + LR) = 20.

The number of students who qualified in (Maths + LR + English) = 5.

The number of students who qualified in DI = 22.

The number of students who qualified in (DI + LR) = 5.

The number of students who qualified in (DI + Maths) = 5.

The number of students who qualified in (DI + Maths + LR) = 5.

The number of students who qualified in English = 50.

Those who qualified in English, could not qualify in DI section.

(a) 60

(b) 55

(c) 50

(d) 20

Using the Venn Diagram concepts, we can draw a diagram like this:

Qualified at least two section = 10 + 5 + 5 + 25 + 10 = 55 aspirants

(a) 10

(b) 20

(c) 25

(d) 5

Ans.

Using the Venn Diagram concepts, we can draw a diagram like this:

Qualified in both Maths and LR = 10 aspirants

(a) 20

(b) 10

(d) 5

Ans.

Using the Venn Diagram concepts, we can draw a diagram like this:

None of the student failed i.e. 0.

(a) 10

(b) 25

(c) 20

(d) 17

Ans.

Using the Venn Diagram concepts, we can draw a diagram like this:

Qualified only in DI = 17

(b) 160479 tonne

(d) 120659 tonne

Production of rice in year 1950-51= 127890 tonne Production of rice in year 1950-51 is 09.09% less than that of in year 1949-50

As we know that 09.09% is 1/11. Using it below:

Then, production of rice in 1949-50

= (11/10) x Production of rice in year 1950-51

= (11/10) x 127890 = 140679 tonne

(a) 149830 tonnes

(b) 125600 tonnes

(c) 143980 tonnes

(d) 162500 tonnes

Required difference = (Production of rice in 1979-80) − (Production of rice in 1969-70)

Using the fraction values of %:

= 213465 x (4/3) – 112325 x (6/5)

= 284620 – 134790 = 149830 tonne

Production of rice in 1960-61= 201924 tonne

The production in 1959-60:

5/6 of 201924 = 168270 tonne

(b) 867265029

(c) 913952088

(d) 752643214

119 + 126 + …….. + 113113

This is the series of AP.

Let total number of terms be n.

Last number of an AP = a + (n – 1)d

Where a = first term

n = number of terms

d = common difference

Putting the values in the formula:

113113 = 119 + (n – 1) x 7

n = 16143

There is another formula, and it is for sum of n terms of an AP when first and last terms are given.

Sum = n/2(a + l)

= 16143/2 x (119 + 113113)

= 913952088

(a) 5

(b) 2

(c) 3

(d) 1

Ans.

ABCD is rectangle. Then, ∠B = 90°

Using the Pythagoras Theorem:

Length of the Robot = 4 m.

Travelled distance by the Robot = 34 – 4 = 30 m

Speed of the Robot = 30/15 = 2 m/s

(a) 0.034

(c) 0.091

(d) 0.030

Let total circuit board manufacture in factory = 100k

Then, the circuit board manufactured by the Robots:

A = 100k (25/100) = 25k

B = 100k (35/100) = 35k

C = 100k (40/100) = 40k

Number of faulty circuit boards by Robots:

A = 25k (5/100) = 1.25k

B = 35k (4/100) = 1.4k

C = 40k (2/100) = 0.8k

Total faulty boards = 3.45k

Probability = 3.45k/100k = 0.034

(a) 47

(b) 43

(c) 31

(d) 37

Ans.

Let the minimum stipend = x

Then, the maximum stipend = x + 45

Let number of students = n

Then, total stipend of a group of students = 50n …(i)

New average = 49

Then, total stipend of a group of students = 49 (n – 2) ….. (ii)

By Equations. (i) and (ii), we get

The stipend of two students

50n - 49 (n – 2) = n + 98

As per the question,

n + 98 = x + x + 45

Minimum earning lie between 42 to 47. Putting x = 45.

n + 98 = 2(45) + 45

n = 37

For x = 45, n is a prime number. Number of students = 37.

(a) 5 h

(b) 3 h

(d) 6 h

Valve A fills a bathtub = 10 h Valve B fills the bathtub = 15 h Let us assume the capacity of the bathtub = 150 litres A’s 1 hour filling work = 15 litres B’s 1 hour filling work = 10 litres A was on for 5 hours, it would have filled 15 x 5 = 75 litres.

Remaining 75 litres were filled by A + B, time taken by them to fill this: 75/(15 + 10) = 3 h B was on for 3 hours.

Percentage of employees in different departments of branch ‘XYZ’ in the year 2014

Total number of employees = 450

(a) 41

(b) 42

(c) 58

(d) 54

Ans.

Total number of employees in department C

450 x 0.26 = 117

Female employees in department C = 117 x (5/13) = 45

Now, total number of employees in department F = 450 x (22/100) = 99

Female employees in department F = 45 – 1 = 41

Male employees in F department = 99 – 41 = 58

(b) 60

(c) 65

(d) 70

As per the pie chart, E has 12%.

A and C together have 26 + 15 = 41%

Required Percentage = [(41 – 12)/41] x 100 = 70% (Approximately)

Full circle pie chart makes the angle of 360. 15 degree is for failed students, and as per the question, there are 5 failed students.

15 degree means 5 students.

360 degree means (5/15) x 360 = 120.

(a) 30

(b) 37

(c) 40

(d) 45

Ans.

135 degree is for passed female candidates.

Calculating the percentage = (135/360) x 100 = 37.5%

(a) 60.8

(b) 56

(c) 71

(d) 58

Ans.

210 degree is for passed male candidates.

Percentage of passed male candidates with respect to total passed candidates is: (210/345) x 100 = 60.8 %

(a) 9: 1

(b) 1: 14

(c) 14: 9

(d) 9: 14

Ans.

Ratio of passed male candidates to the successful female candidates is: 210/135 = 14 : 9

(a) 1

(b) -2

(c) -1

(d) 2

Ans.

Simplifying for n, and using the properties of powers:

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