Territorial Army Official Paper I (Conducted on 31 Jul 2016) (Previous Year Paper)

According to the English alphabet series with positional value:

The pattern followed is;

Hence, the correct answer is FE.

According to the English alphabet series:

The pattern followed here is;

There are two alternative series going on:

ADG, XVT, BEH, WUS, ?

A + 1 = B        X - 1 = W

D + 1 = E        V - 1 = U

G + 1 = H        T - 1 = S

Similarly,

B + 1 = C

E + 1 = F

H + 1 + I

Hence CFI is the correct answer.

According to the English alphabet series:

The pattern followed here is;

Hence, MSYE is the correct answer.

The pattern followed here is,

11 is added in odd place values. 

3 + 11 = 14

14 + 11 = 25 

25 + 11 = 36

hence, 36 is the correct answer.

The pattern followed here is;

Hence, 60 is the correct answer.

The relationship here mentioned is;

Socket is something into which an electric plug is inserted.

Similarly,

Tooth is inserted or fitted into the gums.

Hence, Tooth is the correct answer.

The relationship here mentioned is:

During the day time, stars are visible.

Similarly,

Sun is only visible during the day time.

Hence, Day is the correct answer.

According to the given information;

 

We sleep at night, night is coded as sunshine.

Hence, "Sunshine" is the correct answer.

In the given pair:

Male : Female

Horse : Mare

Similarly,

Bull : Cow fits in the relationship

Hence, Bull : Cowis the correct answer.

The relationship given here is:

Object : commodity to be kept in object.

Money usually kept in the bank

Similarly,

Goods are kept in transport.

Hence, Goodst  fits besfor transport in given options.

All the given options are related to the journey for exploration or determined effort or fight, except Cruize as it is an aimless journey

Hence, Cruize is an odd one in the option

All the given options are related to the progress or successful which is same, except one which is in bold part.

Hence, Renovate is an odd one in the option

All the given options are related to the droplets of water, except Hailstone.

Hence, Hailstone is an odd one in the option

The above relationship given is;

Object : main component

Circle : arc; many arcs together make a circle

Chair : leg ; chair is nothing without the legs.

 Flower : petal; without petal it is not considered as flower.

Cover : page ; Page is important part of books and not cover.

Hence Cover : page is odd one in options.

In a certain code language:

And,

Similarly,

Hence, 24153 is the coded form of PEARL.

In a certain code language:

And,

Similarly,

Hence, 214673 is the coded form of SEARCH.

In a certain code:

and,

Similarly,

Hence, VMRJACJ is the correct answer.

According to the symbol and their meaning:

According to given information:

Hence, the man is the cousin of the woman.

The given relationship given is:

Person : whom he takes care

Here,

Doctor takes care of the patient.

Similarly,

Politicians takes care of the Public and society.

Hence, Public is the correct answer.

From the given relationship:

Ignorance can be avoided by means of Education.

Similarly,

Medicine can be used to avoid the Diseases.

Hence, medicine is the correct answer.

The relationship given here is as follows:

We usually guilt for our past sins or mistakes.

Similarly,

We hope for good things in future.

Hence, future is the correct answer.

From the given information and diagram:

Hence there are 4 people who are strong army officers but not tall.

From the given information:

C = Criminal

T = Thief

J = Judge

Doing theft is a crime and hence thief is a criminal. The circle of Thief comes under Criminal.

Judge is neither thief nor criminal. So the circle of the judge will be separated from them.

Hence, Option b is satisfying the answer.

According to the given information:

There are:-

One husband and wife;

Two sons and two daughters.

Total 3 men and 3 women.

All the women are invited to dinner.

The husband didn't return from the office and the two sons went to play.

Hence, No one was at home.

According to the given statement:

Artists are generally whimsical.

Some of them are frustrated.

Frustrated people are prone to be drug addicts

The Venn diagram is as follows:

From the following conclusion Some artists be rug addicts is true.

The common representation of blood relation diagram element :

The possible family tree for the given relations are :

Hence,' P is the maternal uncle of A' is the correct statement.

Given,

There are five books A, B, C, D and E.

1) A is placed below E

2)  C is placed above D.

3) B is placed below A and D is placed above E. From this statement, we can make a final arrangement.

So, book 'B' toches the table.

Hence, 'B' is the correct answer.

Checking the given options : 

1) Y is married to C who is an engineer.

It is mention in the question that Y is not married to an engineer. so we can eliminate this option.

2) Z is married to C is a doctor.

It is mention in the question that Z is married to a doctor, C is not a doctor. so we can eliminate this option.

3) X is married to a doctor.

It is mention in the question that X is married to a doctor and A is a lawyer. So we can also eliminate this option.

Hence, 'None of these' is the correct answer.

In figure 1, 2 and 3 each dot lies on a different plane. But in figure 4 both dot lies on the same plane,

Hence, option 4 is odd among all.

In figure 1,2 and 4 the outer figure have one side less then the inner one.

But in figure '3' the outer figure have one side more than the inner figure.

Hence, option 3 is odd among all.

In figure 1, 2 and 4 the angle between two arrows is an acute angle.

But figure 3 the angle between two arrows is an obtuse angle.

An acute angle is less than 90^o and obtuse angle are greater than 90^o.

Hence, option 3 is odd among all.

In figure 2, 3 and 4 one side of shape is made by an arc.

But in figure 1 all sides of shape formed by lines.

Hence, option one is odd among all.

The pattern followed here is:

In the figure, the line adjacent on the end of the middle line get deleted one by one and get added at the opposite side one by one.

So the answer figure is the figure (c).

Hence, option c is the correct answer.

The pattern followed here is:

In the given figure First circle is divided by a horizontal line.

In the next figure, the two parts get separated.

Similarly,

In the next figure, those two parts are divided vertically and separated into four parts in the next figure.

Hence, option 1 is the correct answer.

The pattern followed here is:

The V-shaped image moves anticlockwise 90^o and the shaded end of the image facing each other alternately.

Hence, option 4 is the correct answer.

The original direction of map is:

The directions of map after placing the map in diffrent manner.

So according to the figure South becomes North east .

Hence, 'North - East' is the correct answer.

Total distance covered  = 50 km=50k=50k

Total time taken = 8 hours

Let the distance travelled by him by foot be x km and the distance travelled by him on a bicycle is 50 - x km.

Then, 

x/4 + (50 - x)/ 10 = 8

(5x + 100 - 2x)/ 20 = 8

3x + 100 =20 × 8 

3x = 160 - 100

3x = 60

x = 60 ÷ 3

x = 20  

So, the distance travelled by him on foot is  20km.

Hence, '20km' is the correct answer

Total number of triangle in the given figure is:

So, the given figure contain 28 triangles.

Hence, '28' is the correct answer.

Given,

Rs. 8,000 can maintain a family of 4 persons for 40 days.

Rs. 10,500 maintain a family of 6 persons. 

 x is the number of days to maintain Rs. 10500 for the family of 6 person.

8000 / 40 × 4 = 10500 / 6 × x

8000 / 160 = 10500/ 6 × x

50 × (6 × x) = 10500

300 × x = 10500

x = 10500 ÷ 300

x = 35

So ,Rs. 10500 can maintain a family of 6 persons for 35 days.

Hence, '35' is the correct answer.

Given,

Y can complete a job in 12 days.

X is twice as fast as Y.

So X will complete the work in :

12 ÷ 2 = 6 days 

Both X and Y together can finish the job in 

1/6 + 1/12 = 3 / 12

3/4 =1/4

So, X and Y can complete the work in 4 days

Hence, '4' is the correct naswer.

Given,

Out of 60 students, 5 students play none of the sports

So the number of students playing at least one game is

60–5=55.

The formula for A union B is:

n(A∪B) =n(A) +n(B) - n(A∩B)

where,

n(A) is the number of students playing cricket =45

n(B) is the number of students playing football =30

n(A∪B) is the number of students playing at least cricket or football = 55

n(A∩B) is the number of students playing both cricket and football which we should find out.

By using the above formula,

55=45+30-n(A∩B),

n(A∩B) = 45+30–55 =20.

So the number of students playing both cricket and football is 20

Hence, '20' is the correct answer.

The diagram formed according to given conditions is:

According to the diagram:

'A' is the starting point and 'E' is the ending point.

AB = 7 km

DE = 3 km

CB = DF = 6 km

DC = BF = 3 km

AF = AB - BF

AF = 7 km - 3 km

AF = 4 km

EF = DF - DE

EF = 6 km - 3 km

EF = 3 km

The shortest distance between the starting point and ending point is:

EA² = EF² + AF²

EA ² = 3² + 4²

EA² = 9 + 16

EA² = 25

EA = 5

The shortest distance between from his house to ending point is 5km.

Hence, '5km' is the correct answer.

Probability of getting a queen of clubs = (1 / 52).

Probability of getting a king of hearts = (1 / 52).

The probability of getting either a queen of clubs or a king of hearts

= (1 / 52) + (1 / 52) = (2 / 52) = (1 / 26).

 

Hence'1/26' is the correct answer.

The diagram formed according to given data is :

A is the starting point and D is the ending point of Gangaram.

AB= 15 km, BC = 15 km and DC = 15 km

AD = BC = 15 km

So, Gangaram is 15 m east from the starting point.

Hence, '15 m, East' is the correct answer.

According to the given conditions.

So, I am in the south direction from the starting point.

Hence, 'south' is the correct answer

Given,

Length of train is  260m

Length of platform is 290m

So the total distance train has to be coverd is :

Length of platform + length of train = 260 + 290

Distance = 550m

The speed of train is 55kmph

So we have to make both the quantity in same units

Time = 2 × 18

Time = 36 sec

Hence, '36 seconds' is the correct answer.

Given,

My father's birthday is on Tuesday.

My uncle visit me after 64 days of my father's birthday.

So, my uncle visit on :

64 ÷ 7

remainder become 1 

The number of odd day is 1 

so Tuesday + 1 odd day = Wednesday

Hence, 'Wednesday' is the correct answer.

On arranging the given options in ascending order:

Subedar Major <Captain < Major < Lieutenant Colonel

Subedar Major : This rank is senior rank of junior commissioned officer.

Captain : This rank is commissioned officer rank.

Major :This rank is commissioned officer rank.

Lieutenant Colonel :This rank is commissioned officer rank.

Hence 'Lieutenant Colonel' is the correct answer.

Given,

Rakesh rank is 9^th from the top

Rakesh rank 38^th from the bottom.

So the total number of student in the class is :

Total student = 9 + 38 - 1

Total student = 47 - 1

Total student = 46

Hence, '46' is the correct answer.

The 'day after tomorrow' means next day after tomorrow .

The day after tomorrow is 'Sunday' is given.

So, Today is Friday.

The day before yesterday means previous day of yesterday.

So day before yesterday is 'Wednesday'

Hence, 'Wednesday' is the correct answer.

Given - 

0.99 × 14 ÷ 11 ÷ 0.7

Concept used - 

Solution - 

0.99 × 14 ÷ 11 ÷ 0.7

⇒ 0.99 × 14 × (1/11) × (1/0.7)

⇒ 0.09 × 20

⇒ 1.8

∴ Ans = 1.8

Given - 

112 - 62 ÷ 6 × \(\frac{5}{2}\)+ 2 of 10

Concept used - 

Solution - 

112 - 62 ÷ 6 × \(\frac{5}{2}\)+ 2 of 10

⇒ 121 - 6² × (1/6) × (5/2) + 20

⇒ 121 - 15 + 20

⇒ 126

∴ Ans = 126.

Given - 

\(\frac{(24 \times 13) + (28 \div 7)}{(24 + 13) - \frac{14}{3} \ of \ \frac{5}{8}} \)

Concept used - 

Solution - 

\(\frac{(24 \times 13) + (28 \div 7)}{(24 + 13) - \frac{14}{3} \ of \ \frac{5}{8}} \)

⇒ \(\frac{312+4}{37-\frac{14}{3}\times \frac{5}{8}}\)

⇒ \(\frac{316\times 12}{37\times 12-35}\)

⇒ \(\frac{3792}{409}\)

Given - 

LCM of 6, 9, 12, 18

Solution - 

By prime factorization method - 

⇒ 6 = 2 × 3

⇒ 9 = 3 × 3

⇒ 12 = 2 × 2 × 3

⇒ 18 = 2 × 3 × 3

⇒ LCM (6, 9, 12, 18) = 2 × 2 × 3 × 3 = 36

∴ LCM (6, 9, 12, 18) = 36.

Solution:

Concept used:

Let odd no. be ‘o’ even no. be ‘e’

Calculation:

X and Y both are odd numbers taken as ‘o’.

Now check by options,

1. X + Y

⇒ o + o

⇒ e

Possible.

2. X × Y

⇒ o × o

⇒ o

Not possible.

3. XY + 2

⇒ o × o ⇒ o

⇒ o + 2

⇒ o + e ⇒ o

Not possible.

4. 2X + Y

⇒ 2X ⇒ e × o ⇒ e

⇒ e + Y ⇒ e + o

⇒ o

Not Possible

∴ X + Y will give even number.

Alternate method:

Let values of X and Y be 1 and 1. (Taking small values will reduce calculation. As distinction between X and Y is not given assume both numbers be equal)

Now verify the options,

1. X + Y

⇒ 1 + 1 = 2

⇒ 2 ⇒ even

2. X × Y

⇒ 1 × 1 =1

⇒ 1 ⇒ odd

3. XY + 2

⇒ 1 × 1 + 2

⇒ 1 + 2 = 3

⇒ 3 ⇒ odd

4. 2X + Y

⇒ 2 × 1 + 1

⇒ 2 + 1 = 3

⇒ 3 ⇒ odd

∴ X + Y will always result in even number.

Concept used:

Follow BODMAS rule to solve this question, as per the order given below

Calculations:

12.1212 + 17.0005 - 9.1102 = ?

⇒ 29.1217 – 9.1102 = ?

⇒ 20.0115 = ? 

∴ The value of ? is 20.0115

1 + 0.1 + 0.01 + 0.001

⇒ 1.1 + 0.01 + 0.001

⇒ 1.11 + 0.001

⇒ 1.111

∴ 1 + 0.1 + 0.01 + 0.001 = 1.111

GIVEN:

Amount of water added to 12 litres of milk = 3 litres

Amount of water added to 10 litres of milk = 4 litres

CALCULATION: 

Strength of the milk in the first solution = 12/15 = 4/5

Strength of the milk in second solution = 10/14 = 5/7

⇒  The ratio of the strengths of the milk = (4/5) : (5/7)

On cross multiplying

The ratio of the strengths of the milk = 28 : 25

GIVEN:

weight of Gold = 19 × weight of water

weight of Copper = 9 × weight of water

CALCULATION:

Let the ratio of gold and copper be a : l

(19a + 9l)/(a + l) ​= 15/1​

⇒ 19a + 9l = 15a + 15l

⇒ 19a − 15a = 15l − 9l

⇒ 4a = 6l

⇒ 2a = 3l

⇒ l : a​ = 2 : 3​

∴ Gold and copper are taken in the ratio 3:2

GIVEN:

Ratio between Kavita and Reema = 4 : 3

Reema's share = Rs. 2400

CALCULATION:

Let their shares be Rs. 4x and Rs. 3x.

Then 3x = 2400

⇒ x = 800

∴ Total amount = 7x = Rs. 5600

Let the average of runs for 16 innings be A.

  Then, average of runs for 17 innings = A + 3 

Total runs for 16 innings = 16 A 

⇒ Total runs for 17 innings = 16A + 85 

Given,

(16A + 85)/17 = A + 3

⇒ 16A + 85 = 17A + 51

⇒ A = 34

∴ Average of 17 innings = 34 + 3 = 37

FORMULAE USED:

A^x × A^y = A^{x + y}

CALCULATION:

a⁵ × a⁷ = a^{(5 + 7)} = a¹²

Given:

100% = ?

Concept used:

Using the concept of percentage.

Calculation:

100% = ?

⇒ 100 × 1/100 = ?

⇒ ? = 1

∴ 100% is equivalent to 1.

Given:

6a³b³c² ÷ 2ab²c

Concept used:

Using the law of indices.

X^a/X^b = X^{(a – b)}

Calculation:

6a³b³c² ÷ 2ab²c

⇒ (6a³b³c²)/(2ab²c)

⇒ 6/2 × a^{(3 – 1)} × b^{(3 – 2)} × c(^{2 – 1)}

⇒ 3 × a² × b × c

∴ The value of 6a³b³c² ÷ 2ab²c is 3a²bc.

Given:

\(\frac{{{{\rm{x}}^{ - 2}}{\rm{;}} \times {\rm{;}}{{\rm{y}}^{ - 4}}}}{{{{\rm{x}}^{ - 3}}{\rm{;}} \times {\rm{;}}{{\rm{y}}^{ - 1}}{\rm{;}}}} \div \frac{{{{\rm{y}}^{ - 2}}}}{{{{\rm{x}}^{ - 1}}}}\)

Concept used:

Using the law of Indices.

X^a × X^b = X^{(a + b)}

X^a/X^b = X^{(a – b)}

Calculation:

\(\frac{{{{\rm{x}}^{ - 2}}{\rm{;}} \times {\rm{;}}{{\rm{y}}^{ - 4}}}}{{{{\rm{x}}^{ - 3}}{\rm{;}} \times {\rm{;}}{{\rm{y}}^{ - 1}}{\rm{;}}}} \div \frac{{{{\rm{y}}^{ - 2}}}}{{{{\rm{x}}^{ - 1}}}}\)

\( \Rightarrow ;\frac{{{{\rm{x}}^{ - 2}}{\rm{;}} \times {\rm{;}}{{\rm{y}}^{ - 4}}}}{{{{\rm{x}}^{ - 3}}{\rm{;}} \times {\rm{;}}{{\rm{y}}^{ - 1}}{\rm{;}}}} \times {\rm{;}}\frac{{{{\rm{x}}^{ - 1}}}}{{{{\rm{y}}^{ - 2}}}}\)

⇒ [x^{(-2) – (-3)} × y^{(-4) – (-1)}] × x^-1 × y²                               

⇒ x¹ × y^(-3) × x^(-1) × y²

⇒ x^{1 – 1} × y^{2 – 3}

⇒ x⁰ × y^(-1)

⇒ 1/y

∴ The value of \(\frac{{{{\bf{x}}^{ - 2}}; \times ;{{\bf{y}}^{ - 4}}}}{{{{\bf{x}}^{ - 3}}; \times ;{{\bf{y}}^{ - 1}};}} \div \frac{{{{\bf{y}}^{ - 2}}}}{{{{\bf{x}}^{ - 1}}}};{\bf{is}};\frac{1}{{\bf{y}}};.\)

Given:

170% × 1140 = ?

Concept used:

Using the concept of percentage.

Calculation:

170% of 1140 = ?

⇒ 170% × 1140 = ?

⇒ 170/100 × 1140 = ?

⇒ 17 × 114 = ?

⇒ ? = 1938

∴ 170% of 1140 is 1938.

Given:

14% of 280 + 18% of 350 = ?

Concept used:

Calculation:

14% of 280 + 18% of 350 = ?

⇒ 14% × 280 + 18% × 350 = ?

⇒ 14/100 × 280 + 18/100 × 350 = ?

⇒ 14/10 × 28 + 18/10 × 35 = ?

⇒ 392/10 + 630/10 = ?

⇒ 1022/10 = ?

⇒ ? = 102.2

∴ The value of ? is 102.2.

Given:

Speed from house to office = X₁ km/h

Returning speed = X₂ km/h

Concept used:

Average speed = (Total distance traveled)/(Total time is taken)

Calculation:

Let the Distance from the house to the office be D Km.

Time is taken from house to office, T₁ = D/X₁ hrs.

Time is taken from office to house, T₂ = D/X₂ hrs.

Average speed = (Total distance traveled)/(Total time is taken)

⇒ Average speed = (D + D)/(T₁ + T₂)

⇒ Average speed = 2D/( D/X₁ + D/X₂)

⇒ Average speed = 2D/D(1/X₁ + 1/X₂)

⇒ Average speed = (2X₁X₂)/(X₁ + X₂)

.∴ His average speed is \(\frac{{2{{\bf{X}}_1}{{\bf{X}}_2}}}{{{{\bf{X}}_1} + ;{{\bf{X}}_2}}}\)

Shortcut Trick

Total height of tree = h + x, where h is the breaking point height and x is the hypotenuse.

For 30° angle, h : x = 1 : 2. Total ratio parts = 1 + 2 = 3.

Height of broken point (h) = 1/3 × Total height = 20/3 m.

∴ The correct answer is 20/3 m.

Alternate Method

Given: Total height of tree = 20 m. Angle of elevation (θ) = 30°.

Formula Used: sin θ = Perpendicular / Hypotenuse

⇒ Let the height of the breaking point from ground be h.

⇒ Let the length of the broken part (hypotenuse) be x.

⇒ Total height = h + x = 20 m ⇒ x = 20 − h

⇒ In ΔABC, sin 30° = h / x

⇒ 1 / 2 = h / (20 − h)

⇒ 20 − h = 2h

⇒ 3h = 20

⇒ h = 20/3 m

∴ The correct answer is 20/3 m.

Additional Information

Trigonometric Ratios in Right Triangles

For a 30°-60°-90° triangle, the sides opposite to these angles are in the ratio 1 : √3 : 2.

Height and Distance Basics

Angle of elevation is the angle between the horizontal line and the line of sight when looking up at an object.

Breaking Tree Logic

In problems involving broken trees, the length of the broken part forms the hypotenuse of the right-angled triangle.

Given:

Height of tree = 20 m

Angle made by top of the tree to ground = 30°

Concept used:

Using the concept of trigonometric ratios,

Sinθ = Perpendicular/Hypotenuse

Calculation:

Let the height of the point where the tree is broken be x m.

AB = x

Height of tree = 20 m

AC = 20 – x

In ΔABC,

Sin30° = AB/AC

⇒ 1/2 = x/(20 – x)

⇒ 20 – x = 2x

⇒ 3x = 20

⇒ x = 20/3 m

∴ The height of the point where the tree is broken is 20/3 m.

Given :

Principal, P = Rs. 8000

Rate of interest, R = 6%

Time, T = 5 years

Returned amount after 3 years = Rs. 7000

Formula used :

Simple Interest (S.I.) = (P × R × T)/100

Amount, A = P + S.I.

Calculation :

S.I. for 3 years

⇒ S.I.₁ = (8000 × 6 × 3)/100

⇒ S.I.₁ = 1440

Amount after 3 years

⇒ A₁ = 8000 + 1440

⇒ A₁ = Rs. 9440

Returned amount = Rs. 7000

Left sum = 9440 - 7000

⇒ Left sum = Rs. 2440

Now, the interest will be applied only on the left sum for the rest of the 2 years.

⇒ S.I.₂ = (2440 × 2 × 6)/100

⇒ S.I.₂ = 292.8

Amount on left sum

⇒ A₂ = 2440 + 292.8

⇒ A₂ = 2732.8

∴ The amount he should return at the end to settle the loan is Rs. 2732.8

Given :

Profit %, P% = 20%

Discount %, D% = 16%

Formula used :

Selling Price (S.P.) = (C.P.)× (100 + P%)/100

Selling Price (S.P.) = (M.P.)× (100 - D%)/100

Calculation :

Let the cost price(C.P.) of book be x

⇒ The selling price of book = x × (120/100)

⇒ The selling price of book = (6x)/5      -----(1)

Also, there is 16% of discount on marked price(M.P.)

Selling price of book = (M.P) × (84/100)

⇒ M.P. = (6x/5) × (100/84)     (from 1)

⇒ M.P. = (10x)/7      -----(2)

Ratio of cost price and printed price

⇒ (C.P./M.P.) = x/(10x/7)

⇒ (C.P./M.P.) = 7/10

∴ C.P. : M.P. is 7 : 10

Given :

The ratio of Physic : Chemistry : Mathematics = 3 : 4 : 6

Increased percent of seats in Physics = 10%

Increased percent of seats in Chemistry = 20%

Increased percent of seats in Mathematics = 50%

Formula used :

Increased value = original value × (1 + increased % /100)

Calculation :

Let the seats in Physics be 3x

Let the seats in Chemistry be 4x

Let the seats in Mathematics be 6x

⇒ Number of seats in Physics after increment = 3x × (110/100)

⇒ Number of seats in Physics after increment = 33x/10      -----(1)

⇒ Number of seats in Chemistry after increment = 4x × (120/100)

⇒ Number of seats in Chemistry after increment = 48x/10      -----(2)

⇒ Number of seats in Mathematics after increment = 6x × (150/100)

⇒ Number of seats in Mathematics after increment = 90x/10      -----(3)

From (1), (2), and (3)

Increased ratio

Physics : Chemistry : Mathematics = (33x/10) : (48x/10) : (90x/10)

⇒ Physics : Chemistry : Mathematics = 33 : 48 : 90

∴ Physics : Chemistry : Mathematics = 11 : 16 : 30

Given :

The diameter of well, d = 14m

Depth of well, h = 15m

Width of well = 21m

Formula used :

The volume of cylinder = π r² h unit cube

The volume of ring = π h (R² - r²) unit cube

Concept used :

The volume of earth dugout = Volume of embankment

Calculation :

Let the height of the embankment be H

The radius of well, r = 14/2 = 7m

The outer radius of embankment, R = 21 + 7 = 28m

The inner radius of the embankment, r = 7m

The volume of earth dugout = Volume of the embankment

⇒ π H(R² - r²) = π r² h

⇒ H × (28² - 7²) = 7² × 15

⇒ H × (28 + 7)(28 - 7) = 7 × 7 × 15

⇒ H × 35 × 21 = 7 × 7 × 15

⇒ H = (7 × 7 × 15)/(35 × 21)

⇒ H = 1m

∴ the height of the embankment is 1m.

Given :

Length of small box, l = 2m

Breadth of small box, b = 4m

Height of small box, h = 5m

Length of larger box, L = 20m

Breadth of larger box, B = 10m

Height of larger box, H = 5m

Formula used :

Volume of cuboid = (length) × (breadth) × (height)

Calculation :

Let the number of smaller box placed in larger box be x

Volume of x small box = volume of large box

⇒ x × (l × b × h) = L × B × H

⇒ x × (2 × 4 × 5) = 20 × 10 × 5

⇒ x = (20 × 10 × 5)/(2 × 4 × 5)

⇒ x = 25

∴ the number of small box can be placed in larger box is 25.

Given :

Marked Price of fan = Rs. 2400

Discount offered = 10%

Formula used :

Selling price = Marked price × (1 - discount%/100)

Calculation :

Selling price of fan = 2400 × (90/100)

⇒ Selling price of fan = 2160

Let the further discount be x%

New selling price = 2160 × (100 - x)/100

⇒ 2160 × (100 - x)/100 = 2000

⇒ 100 - x = (2000 × 10)/216

⇒ 100 - x = 92.59

⇒ x = 100 - 92.59

⇒ x = 7.41

∴ additional discount must be offered to the customer is approx 8%

Given :

a + b + c = 10

ab + bc + ca = 31

To find :

a² + b² + c²

Formula used :

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Calculation :

a2 + b2 + c2 + 2(ab + bc + ca) = (10)²

⇒ a2 + b2 + c2 + 2 × 31 = 100

⇒ a2 + b2 + c2 + 62 = 100

⇒ a2 + b2 + c2 = 100 - 62

⇒ a2 + b2 + c2  = 38

∴ the value of a2 + b2 + c2 is 38

Given :

a cos θ - b sin θ = c

Formula used :

(a + b)² = a² + 2ab + b²

Solution : 

a cos θ - b sin θ = c

Squaring both sides

⇒ (a cos θ - b sin θ)² = c²

⇒ a² cos² θ + b² sin² θ - 2ab sin θ cos θ = c² 

⇒ a² (1 - sin² θ) + b² (1 - cos² θ) - 2ab sin θ cos θ = c² 

⇒ a² - a² sin² θ + b² - b² cos² θ - 2ab sin θ cos θ = c² 

⇒ a² + b² - c² = a² sin² θ + b² cos² θ + 2ab sin θ cos θ 

⇒ ( √(a² + b² - c²) )² = ( a sin θ + b cos θ )² 

∴ a sin θ + b cos θ = ± √(a² + b² - c²)

Given:

Radius = 3 cm, height = 4 cm

Formula used:

The surface area of hemisphere = 2πr²

The surface area of cone = πrl

l² = r² + h²

Where,

l = slant height of the cone

h = height of the cone

r = radius of the sphere

Calculation:

l² = r² + h²

⇒ l² = 3² + 4²

⇒ l = 5 cm

The total surface area of the toy

⇒ The surface area of the hemisphere + surface area of the cone

⇒ 2πr² + πrl

⇒ πr(2r + l)

⇒ π × 3 (2 × 3 + 5)

⇒ 33π cm²

∴ The total surface area of the toy is 33π cm².

Given: 

Calculation:

Corresponding angles

⇒ ∠2 = ∠6, ∠1 = ∠5, ∠3 = ∠7, ∠4 = ∠8

Alternate angles

⇒ ∠4 = ∠6, ∠3 = ∠5

Exterior alternate angles

⇒ ∠2 = ∠8, ∠1 = ∠7

Here all three statements satisfy this condition

∴ I, II, and III statements are true.

Given:

x = (√2 + 1)/(√2 – 1) and y = (√2 – 1)/(√2 + 1)

Calculation:

x = (√2 + 1)/(√2 – 1)

By using rationalization

⇒ x = (√2 + 1)/(√2 – 1) × (√2 + 1)/(√2 + 1)

⇒ x = (√2 + 1)²/(√2)² – 1²

⇒ x = (2 + 1 + 2√2)/(2 – 1)

⇒ x = 3 + 2√2

y = (√2 – 1)/(√2 + 1)

By using rationalization

⇒ y = (√2 - 1)/(√2 + 1) × (√2 - 1)/(√2 - 1)

⇒ y = (√2 - 1)²/(√2)² – 1²

⇒ y = (2 + 1 – 2√2)/(2 – 1)

⇒ y = 3 – 2√2

x² + y²

⇒ (3 + 2√2)² + (3 – 2√2)²

⇒ 9 + 8 + 12√2 + 9 + 8 – 12√2

⇒ 17 + 17

⇒ 34

∴ The value of x² + y² is 34.

Given:

Price of scooter = Rs. 24000

Calculation:

Depreciation price in 1^st year

⇒ 24000 × (5/100)

⇒ 1200

Price of the scooter after depreciation of 1^st year

⇒ 24000 – 1200

⇒ 22800

Depreciation price in 2^nd year

⇒ 22800 × (5/100)

⇒ 1140

Price of the scooter after depreciation of 2^nd year

⇒ 22800 – 1140

⇒ 21660

Depreciation price in 3^rd year

⇒ 21660 × (5/100)

⇒ 1083

Price of the scooter after depreciation of 3^rd year

⇒ 21660 – 1083

⇒ 20577

∴ The Scooter value after three years is Rs. 20577.

Given:

40% if a number is added to 42, the number becomes itself.

Calculation:

Let the number be ‘X’.

⇒ X × (40/100) + 42 = X

⇒ (2/5)X + 42 = X

⇒ 3X/5 = 42

⇒ X = 70

∴ The number is 70.

Given:

Area of circle = Area of square

Formula used:

Area of circle = πr²

Where,

r = radius of the circle

Area of square = side²

Calculation:

Let the side of the square be 'a'.

⇒ πr² = a²

⇒ a²/ r² = π

⇒ (a/r) = √π

∴ The ratio of the side of the square and the radius of the circle is √π : 1.

Given:

The greatest six-digit number is a perfect square

Concept used:

Long division method of the square root.

Calculation:

Largest six digit number = 999999

Do the long division method of square root for the Largest six-digit number

Subtract the remainder from the highest six-digit number

⇒ 999999 – 1998

⇒ 998001

∴ The greatest six-digit number is a perfect square is 998001.

Given:

A Manson can build a tank in 12 h.

Calculation:

Let the Manson be ‘A’ and boy be ‘B’.

A can complete the work in 12 h.

A one hour work

⇒ 1/12

‘A’ work in 6 h

⇒ (1/12) × 6

⇒ 1/2

Remaining work

⇒ 1 – 1/2

⇒ 1/2

The remaining work can be finished in 5 h by (A + B)

Whole Work can be finished by A + B

⇒ 5 × 2

⇒ 10 h

B can complete the whole work

⇒ (1/12) + (1/B) = (1/10)

⇒ 1/B = 1/10 – 1/12

⇒ 1/B = 1/60

⇒ B = 60 h

∴ The time that the boy will take alone to complete the work is 60 h.

Alternate Method:

Let the Manson be ‘A’ and boy be ‘B’.

A can complete the work in 12 h.

In 6 h A can complete 1/2 work.

(A + B) completed 1/2 in 5 h.

(A + B) can complete whole work in 10 h.

B efficiency

⇒ 6 – 5 = 1

B alone can finish the work

⇒ 60/1

⇒ 60 h

∴ The time that the boy will take alone to complete the work is 60 h.

Given

In an Army camp ration is available for 100 soldiers for 10 days

After 2 days 60 soldiers joined

Calculation

After 2 days :

⇒ For 100 soldiers, ration will last (10 - 2) days = 8 days

⇒ For 1 soldiers, the ration will last (100 × 8) days

Now, 

⇒ Number of soldiers = 100 + 60 = 160

⇒ For 160 soldiers, the ration will last for = (100 × 8)/160

⇒ For 160 soldiers, the ration will last for = 5 days.

∴ Number of days will the remaining ratio last is 5 days.

Given

sec A (1 - sin A) (sec A + tan A)

Concept

sec²A - tan²A = 1

sec A = (1/cos A)

(sin A/cos A) = tan A

(a + b) (a - b) = (a² - b²)

Calculation

⇒ (sec A - sec A × sin A) (sec A + tan A)

⇒ (sec A - tan A) (sec A + tan A)

⇒ a = sec A , b = tan A

Now,

⇒ (sec A - tan A) (sec A + tan A) = sec² A - tan² A

⇒ sec2 A - tan2 A = 1

∴ sec A (1 - sin A) (sec A + tan A) = 1

Given

Divide the number 26244 by the smallest number

Concept

Using factorization method 

Calculation

Factor of 

⇒ 26244 = 2 × 2 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3

For perfect cube number must be in 3 pair so unpaired part is

⇒ 2 × 2 × 3 × 3 = 36 

⇒ So the smallest number by which number 26244 must be divided so that the quotient is perfect cube = 36 

∴ The smallest number by which number 26244 must be divided so that the quotient is perfect cube = 36

Given

A cubical vessel can hold 1331 litre of water

Concept

As vessel is cubical so here we can use 

Volume of cube = (length)³

To convert liter into meter we divide liter by 1000.

Calculation

⇒ (1331/1000) = (length)³

⇒ Length = ∛(1331/1000)

⇒ Length = (11/10)

⇒ Length = 1.1 m

∴ The length of side of the vessel in m is 1.1 m

Given

A spherical ball made of iron has diameter = 6 cm

Density of iron =  8 g/cm³

π = 3.142

Concept

Volume of sphere = (4/3) × π × r³

Mass = (Density × Volume)

Diameter = 2 × Radius

Radius = (Diameter/2)

Calculation

⇒ Radius = (6/2) = 3 cm

⇒ Volume of sphere = (4/3) × 3.14 × (3)³

⇒ Volume of sphere = 113.04 cm³

Now, 

⇒ Mass of the ball = 113.04 cm³ × 8 (g/cm³)

⇒ Mass of the ball = 904.32 g

To convert mass from gram to kilogram we must divide the value by 1000

⇒ Mass of the ball = (904.32/1000)

⇒ Mass of the ball = 0.90432 ≈ 0.9 kg

∴ Mass of the ball = 0.9 kg

Given

A hollow sphere of internal and external diameter 4cm and 8 cm 

Melted and recasted into a cone of base diameter 8 cm

D = 8 cm , d = 4 cm

Concept

Volume of hollow sphere = (4/3) × π × ((R)³ - (r)³)

Volume of cone = (1/3) × π × (r)² × h

As sphere is melted and recast into cone so in this case volume never change i.e

Volume of sphere = Volume of cone 

Diameter = 2 × Radius

Radius = (Diameter/2)

Calculation

⇒ R = (8/2) = 4 cm

⇒ r = (4/2) = 2 cm

⇒ Radius of cone = (8/2) = 4 cm

Volume of sphere = Volume of cone 

⇒ (4/3) × π × ((R)3 - (r)3) = (1/3) × π × (r)2 × h

⇒ (4/3) × π × ((4)³ - (2)³) = (1/3) × π × (4)² × h

⇒ 4 × (64 - 8) = 16 × h 

⇒ 4 × 56 = 16 × h

⇒ h = (4 × 56)/16

⇒ h = 14 cm 

∴ Height of cone = 14 cm

Given

Tanθ = (4/3)

Concept

Tanθ = (P/B)

cosθ = (B/H)

Using Pythagoras theorem 

H = √(P² + B²)

Calculation

⇒ H = √(4² + 3²)

⇒ H = √(16 + 9) 

⇒ H = √(25)

⇒ H = 5 

⇒ cosθ = (3/5)

Now,

⇒ (1 - cosθ)/(1 + cosθ)

Put the value of cosθ 

⇒ (1 - (3/5))/(1 + (3/5))

⇒((5 - 3)/5)/(5 + 3)/5

⇒ (2/5)/(8/5)

⇒ (1/4)

∴ (1 - cosθ)/(1 + cosθ) = (1/4)

Given

2xsin 60° - 3ytan 45° = 0

xtan 60° + (√2) ycosec 45° = 5

Concept

Sin60° = (√3/2)

Tan45° = 1 

Tan60° = √3

Cosec45° = √2

Calculation

Put the value of trigonometry identities 

⇒ 2xsin 60° - 3ytan 45° = 0

⇒ (2x × (√3/2)) - 3y = 0

⇒ (√3)x - 3y = 0    ....(1)

⇒ xtan 60° + (√2)y cosec 45° = 5

⇒ (x × √3) + (y × √2 × √2) = 5

⇒ (√3)x + 2y = 5    ....(2)

Subtract equation (1) from (2)

⇒ ((√3)x + 2y) - ((√3)x - 3y) = 5 -0 

⇒ (√3)x + 2y - (√3)x + 3y = 5

⇒ 5y = 5

⇒ y = 1

Put the value of y in equation (1)

⇒ (√3)x - 3 = 0

⇒ (√3)x = 3 

⇒ x = √3

∴ x = √3 and y = 1

Given:

5sinX + 3cosX we have to find the maximum and minimum value of the given expression

Concept Used:

Maximum value of a.sinθ + b.cosθ is √(a² + b²)

Minimum value of a.sinθ + b.cosθ is {- √(a² + b²)}

Calculation:

Here the given expression is 5sinX +3cosX

Comparing with a.sinθ + b.cosθ get, a = 5 and b = 3

Maximum value is √(5² + 3²)

⇒ √(25 + 9)

⇒ √34

Minimum value is {- √(5² – 3²)}

⇒ {- √(25 + 9)}

⇒ - √34

⇒ - √34

∴ Maximum and minimum values are √34 and (- √34)

Given:

Mean of 5 number is 22. If one number is excluded, the mean becomes 19.

Concept Used:

Sum of numbers = Mean × Total number of numbers

Calculation:

Mean of 5 number is 22

Sum of 5 number = 5 × 22

⇒ 110

If one number is excluded, the mean becomes 19

Mean of 4 number = 4 × 19

⇒ 76

The excluded number is (110 – 76)

⇒ 34

∴ The excluded number is 34

Given:

Mean marks of 60 students are 63 and mean marks of the other 40 students are 60

Concept Used:

Total marks = Mean marks × Number of students

Calculation:

Total marks of 60 students = 60 × 63

⇒ 3780

Total marks of 40 students = 40 × 60

⇒ 2400

Total marks of (60 + 40) students = 3780 + 2400

⇒ 6180

Mean marks of all students = 6180/100

⇒ 61.8

The mean marks of all taken together are 61.8

Given:

7, 11, 23, 36, 42, 50, 61, 73, 110 and 120 we have to find the median of these given numbers

Concept Used:

First, we have to arrange the given data as ascending order then we have to see how many numbers of data (n) are there

If n is odd then, Median = {(n/2) + 1} th term

If n is even then, Median = [(n/2) th term + {(n/2) + 1} th term]/2

Calculation:

Arranging the given numbers in ascending order get 7, 11, 23, 36, 42, 50, 61, 73, 110 and 120

Here total number of numbers are 10, which is an even number

Median = [(n/2) th term + {(n/2) + 1} th term]/2

⇒ [(10/2)th term + {(10/2) + 1}th term]/2

⇒ (5^th term + 6^th term)/2

⇒ (42 + 50)/2

⇒ 46

∴ The required median of the given numbers is 46