Territorial Army Official Paper I (Conducted on 30 Jul 2017) (Previous Year Paper)

So, there are a total of 16 triangles in the figure.

Hence, ‘16’ is the correct answer.

The number of rectangles in the given figure is :

 

 

 

So, there is a total of 21 rectangles in the figure.

Hence, ‘21’ is the correct answer.

Key Points

• A rectangle is a quadrilateral with four right angles. Thus, all the angles in a rectangle are equal (360°/ 4 = 90°). Moreover, the opposite sides of a rectangle are parallel and equal, and diagonals bisect each other.
• A square is a special type of rectangle because it possesses all the properties of a rectangle. Similar to a rectangle, a square has interior angles which measure 90∘ each. opposite sides that are parallel and equal.

Important Points

• All the rectangles are quadrilateral.
• All the squares are rectangles.

Given,

 Kamla is 9^th from the left and Veena is 16^th from the right.

If they interchange their positions.Kamla becomes 25^th from the left.

So the total number of  girls in a  row is :

25 + 16 - 1 = 40

Hence,'40' is the correct answer.

1. Malini is behind Anjana, but in front of Gayatri.

Anjana

Malini

Gayatri

2. Meena is in front of Sheena, but behind Gayatri.

Anjana

Malini

Gayatri

Meena

Sheena

Hence, 'Meena' is at second last position.

Sita is elder than Swapna.

Sita > Swapna

Lavanya is elder thanSwapna but younger than Sita.

Sita > Lavanya > Swapna

Suvarna is younger than both Hari and Swapna.

Hari/Swapna > Suvarna

Swapna is elder than Hari

Swapna > Hari > Suvarna

Sita > Lavanya > Swapna >Hari > Suvarna

Suvarna is youngest among all.

Hence, 'Suvarna' is the correct answer.

Preparing the family tree using the following symbols,

C is A’s father’s nephew means C is the cousin of A 

D is A’s Cousin but not the brother of C means D is Sister of C.

 The required family tree is :

Hence, 'Sister' is the correct answer.

Preparing the family tree using the following symbols,

The two brothers of the daughter of my father’s wife mean my sister's brother.

Hence, 'Brother' is the correct answer.

Nitrogen is a part of air while ice is different from the two.

So the possible Venn diagram is:

Hence, the figure in option 4 is the correct answer.

Bus and car are different types of the vehicle but both come under the category of vehicle.

Hence, the figure in option 4 is the correct answer.

The pattern followed here is :

Similarly,

Hence, 'APERTN' is the correct answer.

According to the position of the letter in English alphabet series.

The pattern followed here is: 

 

Similarly,

 

Hence, '14122914' is the correct answer.

The pattern followed here is :

 

Similarly,

 

Hence,'1983' is the correct answer.

The code for 'FADE' is "

As the code for A and E is 8 and 4 respectively.

In the given word 'GAGE' the code for G is common.

Code for A is fixed and from option 3 it is clear that that code for GAGE is '2824' because only in option (3) '2' is repeated with 8 and 4.

Hence, '2824' is the correct answer.

The primary source of all three Wood, Cork and Paper is tree but primary source of stone is not tree.

Hence, 'Stone' is the correct answer.

All except Brigadier are ranks in the Navy.

While brigadier is rank only in the Army.

Hence, 'Brigadier' is the correct answer.

In all other pairs, first is the alloy used to make the second.

While Iron is not an alloy but a metal.

Hence, 'Iron : Rails' is the correct answer.

In Option 2, 3 and 4 the body part and their disabilities are given but Taste is the ability or function of the tongue.

So, Tongue: Taste is odd among all.

Hence,' Tongue: Taste ' is the correct answer.

The pattern followed here is:

1. 25631→ 2 + 5 + 6 + 3 + 1 = 17

2. 33442 → 3 + 3 + 4 + 4 + 2 = 16 (Not follow the pattern)

3. 34424 → 3 + 4 + 4 + 2 + 4 = 17

4. 52163 → 5 + 2 + 1 + 6 + 3 = 17

Hence, '33442' is the correct answer.

All other numbers contain all four consecutive even numbers but not in proper order.

Additional Information 

This logic is given by the conducting body of the very examination (PIB)

The pattern followed here is:

1. 3 : 12

3 → 3 × 4 = 12

2. 4 : 20

4 → 4 × 5 = 20

3. 6 : 42

6 → 6 × 7 = 42

4) 7 : 63

7 → 7 × 9 = 63 (does not follow the pattern)

Hence,'7 : 63' is the correct answer.

The pattern followed here is :

Hence, '468' is the correct answer.

The pattern followed here is:

Each term is Multiply by 3/2.

Hence, option 4 is the correct answer.​

The pattern followed here is :

8 + 21 = 29

29 + (21 × 4) = 113

113 + (84 × 4) = 449

449 + (336 × 4) = 1793

Hence, '1793' is the correct answer.

The pattern followed here is :

Hence, '7/13' is the correct answer.

The positions of letters according to the English alphabet series:

Logic:

In the given series first  letter is decreased by three and the second letter is also decreased by two to form next term

In the number series number added to the given number increases by one to form the next number.

Hence, 'H10R' is the correct answer.

The pattern followed here is;

Hence, L11T is the correct answer.

The pattern followed here is;

Hence, L25O is the correct answer.

English alphabet series is;

In the given series, the letters is increasing by one and after a comma, one letter is skipped.

For example;

AB = 2 Letter (C skipped)

DEF = 3 Letter (G skipped)

HIJK = 4 Letter (L skipped)

MNOPQ = 5 Letter (R skipped)

STUVWX = 6 Letter

Hence, MNOPQ is the correct answer.

The pattern followed here is;

Hence, MQNRO is the correct answer.

According to the given statement;

Hence, North-West is the correct answer.

Given;

Eight people A, B, C, D, E, F, G and H are sitting and all are facing in the outward direction.

All of them move anticlockwise to three place then this is how arrangement looks

Hence, B is facing West direction.

There is a code for determining the day on dates.

Let's find first what will be the day on 1^st March 2013.

Formula:

100 years = 5 odd days.

200 years = 3 odd days

300 years = 1 odd days.

400 years = 0 odd days

500 years = 5 odd days....and the series will continue like above.

This means multiple of 4 will have 0 odd days.

Code for months (Ordinary year:)

Jan = 1

Feb = 4

Mar = 4

April = 0

May = 2

June = 5

July = 0

August = 3

September = 6

October = 1

November = 4

December = 6

Sunday = 0

Monday = 1

Tuesday = 2

Wednesday = 3

Thursday = 4

Friday = 5

Saturday = 6 

Now,

Formula: DD + month code + YE00 + AR + AR/4 (only dividend will be taken) + 5

2013 = YEAR

YE00 = 2000 and AR = 13

1 March 2013 = 1 + 4 + 0 + 13 + 3 + 5

= 26(after 7 days the weeks repeat so 7 + 7 + 7 + 5)

All the 7 digits will be ignored except the last one

5 = Friday

1 March 2013 = Friday

So, 6 March = Wednesday

Hence, 6, 13, 20, 27 is the correct answer.

Total hrs. from 9 A.M. on Sunday to 9 P.M. on next Friday = 132 hr.

132 hr. = 132 × 60 × 60 = 4,75,200 sec

The total gain in this time period = 8 min 15 sec → 495 sec

It gains one second in;

\(475200/495\) = 960 sec → 16 min

To show the correct time it has to gain only 4 min or 240 sec.

It gains 240 sec. in 64 hrs → 2 days and 16 hr.

From 9 A.M. on Sunday, 2 days and 16 hr = 1 A.M. Wednesday.

Hence, 1 A.M. Wednesday is the correct answer.

Formula:

(720/11 - M)(24 × 60/M) mins.

here, M is the intervals of n minutes of the correct time.

Given n = 64

Substituting in the formula:

(720/11 - 64)(24 × 60)64

= 360/11

=.\(32\dfrac{8}{11}\)​ minute gain

Hence, option 2 is the correct answer

Given alphabet series;

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

Fourth to the left of I = E

sixteenth to the right of E = U

Hence, U is the correct answer.

Given alphabet series;

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

In the given alphabet L is in exactly middle of G and Q.

Hence, L is the correct answer

Given alphabet series;

A B C D E F G H I J K L M N O P Q R S T U V W X Y Z

In the given alphabet

The eighteenth letter from the left end = R

Tenth letter from the right end = Q

Hence, no letter is in between Q and R

Given:

Statements:

(I) All teachers are experienced.

(II) Some teachers are spinsters.

According to given statement:

Conclusions:

(I) Some experienced are spinsters. - True (as some spinsters are teachers and all teachers are experienced.)

(II) Some spinsters are experienced. True (as some spinsters are teachers and all teachers are experienced.)

Hence, "both conclusions I and II follow".

Given:

Statements:

(a) Some cats are dogs.

(b) No dog is a toy.

According to the given statement;

Conclusions:

(I) Some dogs are Cats. (True, as some cats are dogs)

(II) Some toys are Cats. (False, it can be possible, but not definite.)

(III) Some cats are not toys. (True, given some cat are dog and No dog is Toys so the part of the Cat which is Dog is also not Toys .)

(IV) All toys are Cats. (False, as no dog is a toy.)

Hence, Only conclusion I and III follow.

The pattern followed is;

Taking the first row 

2 + 9 = 11;

9 - 2 = 7

The same pattern followed in all rows.

In the third row;

Using pattern;

7 + ? = 10; ? = 10 - 7; ? = 3

? -  7 = -4; 3 - 7 = -4 (satisfies)

In fourth row;

Using pattern;

6 + 4 = 10;

4 - 6 = -2

Hence, 3 and (-2) is the correct answer.

The pattern followed here is;

Outer circle number = (Inner circle number × 2) + next number in inner circle clockwise.

3 × 2 = 6; 6 + 4 = 10

4 × 2 = 8; 8 + 6 = 14;

6 × 2 = 12; 12 + 8 = 20

8 × 2 = 16; 16 + 3 = 19

Hence, 19 is the correct answer.

The pattern followed here is:

91 ÷ (6 + 7);

91 ÷ 13 = 7;

7² = 49

Similarly,

90 ÷ (7 + 8);

90 ÷ 15 = 6;

6² = 36

Similarly,

81 ÷ 9 = 9 (In option 72 = 7 + 2 = 9)

9² = 81

Hence, 72 is the correct ansewer

In the given matrix:

The pattern followed here is;

In first column:

7 + 2² + 5 = 16

In second column:

9 + 4² + 7 = 32

Similarly,

8 + 3² + 6 = 23

Hence, 23 is the correct answer.

According to English Alphabet series with positional value;

The pattern followed is;

A = 1

F = 6

I = 9 

169; is the square of 169 is 13,

Similarly;

A = 1

D = 4

D = 4

144 is the square of 12.

Hence, 12 is the correct answer.

According to the question:

Similarly,

Then the correct mathematical equation will be;

Hence, 0 is the correct answer

According to the given statement;

Option 1) 

7 - 43 ÷ 6 × 1 = 4

Using BODMAS rule:

7 - 7.166 × 1 = 4

Not possible, because there is no decimal number on the right side of equal sign.

Option 2) 

3 × 6 ÷ 2 + 3 - 6 = 6

Using BODMAS rule:

3 × 3 + 3 - 6;

9 + 3 - 6;

12 - 6;

6

It is equal to the R.H.S

Option 3) 

5 + 7 - 3 ÷ 2 = 4

Using BODMAS rule:

5 + 7 - 1.5;

Not possible, because there is no decimal number on the right side of equal sign.

Option 4) 

2 × 5 - 6 + 2 = 4

Using BODMAS rule:

10 - 6 + 2;

4 + 2;

6

It is not equal to R.H.S.

Hence Option 2 is correct.

According to the English Alphabet series;

Given series;

a c b _ c e _ f _

The pattern followed here is;

First alphabet + 2;

Second alphabet - 1

Then, +2 and -1 alternatively continues.

a c b d c e d f e

Hence, the correct answer is dde

According to English Alphabet series;

Given series;

r t x_ s x_z_txy_ _yz

Using option 1

r t x y s x y z / r t x y x s y z

Here no pattern followed. 

Using option 2

r t x y s x y z / s t x y x r y z

Here no pattern followed.

Using option 3

r t x y s x y z / r t x y s x y z

It forms a pattern.

Using option 4

r t x y s x y z / x t x y r s y z

Here no pattern followed.

Hence Option 1 is the correct answer.

Given Unscrambled series;

Using option 1

CBUSERNLMA is not a meaningful word.

Using option 2

EBSUCMRLNS is not a meaningful word.

Using option 3

UBMLASCENR is not a meaningful word.

Using option 4

UNSCRAMBLE is a meaningful word.

Unscramble means to make it in a readable format.

Hence, option 4 is the correct answer.

Concept used:

Calculations:

⇒ \(\dfrac{\cos^2 60^\circ + 4 \sec^2 30^\circ - \tan^2 45^\circ}{\sin^2 30^\circ + \cos^2 30^\circ}\)

⇒ {(1/2)² + 4 × (2/√3)² - 1}/{(1/2)² + (√3/2)²}

⇒ {(1/4) + (4/3) × 4 - 1}/{(1/4) + (3/4)

⇒ (1/4) + (16/3) - 1

⇒ (3 + 64)/12 - 1

⇒ (67/12) - 1

⇒ 55/12

∴ The value is 55/12

Given:

\(\dfrac{\tan 57^\circ + \cot 37^\circ}{\tan 33^\circ + \cot 53^\circ}\)

Concept used:

tan(90° - θ) = cotθ

tanθ = 1/cotθ

Calculations:

⇒ \(\dfrac{\tan 57^\circ + \cot 37^\circ}{\tan 33^\circ + \cot 53^\circ}\)

⇒ {tan(90° - 33) + cot 37°}/{tan 33° + cot(90° - 37°)}

⇒ (cot 33° + cot 37°)/(tan 33° + tan 37°)

⇒ {(1/tan 33°) + cot 37°}/{tan 33° + (1/cot 37°)}

⇒ {(1 + tan 33° cot 37°)/(1 + tan 33° cot 37°)} × (cot 37°)/(tan 33°)

⇒ cot 37° × cot 33° 

⇒ tan 57° cot 37°

∴ The value is tan 57° cot 37°

Given:

\(\dfrac{\sin θ + \cos θ }{\sin θ - \cos θ}=3\)

Calculations:

⇒ \(\dfrac{\sin θ + \cos θ }{\sin θ - \cos θ}=3\)

⇒ sinθ + cosθ = 3 × (sinθ - cosθ)

⇒ sinθ + cosθ = 3sinθ - 3cosθ

⇒ 4cosθ = 2sinθ

⇒ tanθ = 2/1

⇒ sinθ = 2/√5

⇒ sin4 θ = (4/5)²

⇒ sin4 θ = 16/25

∴ The value of sin4 θ is 16/25

Given:

\(\sinθ - \cosθ= \dfrac{7}{13}\)

Formula used:

sin²θ + cos²θ = 1

(sin θ + cos θ)² = sin2θ + cos2θ + 2 × sin θ × cos θ

Calculations:

\(\sinθ - \cosθ= \dfrac{7}{13}\)           ------(1)

Let, sin θ + cos θ = x          ------(2)

⇒ By squaring and adding equaltion (1) and (2)

⇒ sin2θ + cos2θ + 2 × sin θ × cos θ + sin2θ + cos2θ - 2 × sin θ × cos θ = (49/169) + x²

⇒ 2 × (sin2θ + cos2θ) = (49/169) + x²

⇒ 2 = (49/169) + x²

⇒ x² = 2 - (49/169)

⇒ x2 = (289/169)

⇒ x = √289/√169

⇒ x = 17/13

∴ The value of sin θ + cos θ is is 17/13

Given:

\(a^2 \sec^2 x - b^2 \tan^2 x = c^2\)

Concept used:

1 + tan² θ = sec² θ

sec2 θ - tan2 θ = 1

Calcualtions:

⇒ \(a^2 \sec^2 x - b^2 \tan^2 x = c^2\)

⇒ a² (1 + tan2 x) - b² tan² x = c²

⇒ a² + a²  tan2 x - b2 tan2 x = c²

⇒ tan2 x (a² - b²) = c² - a²

⇒  tan2 x = (c2 - a²)/(a2 - b2)

⇒  tan2 x = (a² - c²)/(b² - a²)

⇒ 1 + tan2 θ = sec2 θ

⇒ 1 + (a2 - c2)/(b2 - a2)

⇒ (b2 - a² + a² - c²)/(b2 - a2)

⇒ sec2 θ = (b² - c²)/(b2 - a2)

⇒ \(\sec^2 x + \tan^2 x\)

⇒ {(c2 - a2)/(a2 - b²)} + {(b2 - c2)/(b2 - a2)}

⇒ \(\dfrac{b^2+ a^2 - 2c^2}{b^2 - a^2}\)

∴ The value is \(\dfrac{b^2+ a^2 - 2c^2}{b^2 - a^2}\)

Given:

\(x + \frac{1}{x} = 5\)

Calculation:

\(\frac{{2x}}{{3{x^2}: -: 5x :+: 3}}\)

Divide by x in both dominator and nominator

⇒ \(\frac{2}{{3x: -: 5: + ;\frac{3}{x}}}\)

⇒ \(\frac{2}{{3\left( {x: +: \frac{1}{x}} \right) - :5}}\)

⇒ 2/(15 - 5) = 1/5

Given:

\(\left( {1 - \frac{{2xy}}{{{x^2}; + ;{y^2}}}} \right) ÷ \frac{{\left( {{x^3}; - ;{y^3}} \right)}}{{x;-;y}} - 3xy\)

Concept used:

(a - b)² = a² - 2ab + b²

(a - b)³ = a³ - b³ - 3ab(a - b)

Calculation:

\(\frac{{{x^2}: + ;{y^2}: -: 2xy}}{{x^2: +: y^2}}; ÷ ;\frac{{;{x^3}: - ;{y^3}: -:3xy\left( {x: -: y} \right)}}{{x: -:y}}\)

by using formula,

⇒ (x - y)²/(x² + y²) ÷ (x - y)³/(x - y)

⇒ (x - y)2/(x2 + y2)(x - y)²

⇒ 1/(x² + y²)

∴ The correct answer is 1/(x2 + y2).

Given:

\(\frac{1}{5} + 999\frac{{494}}{{495}} \times 99\)

Concept:

\(;{\rm{a}}\frac{{\rm{b}}}{{\rm{c}}} = {\rm{a}} + \frac{{\rm{b}}}{{\rm{c}}}\)

Calculation:

\(\frac{1}{5} + \left( {999 + \frac{{494}}{{495}} + \frac{1}{{495}} - \frac{1}{{495}}} \right) \times 99\)

\( ⇒ \frac{1}{5} + \left( {999 + 1 - \frac{1}{{495}}} \right) \times 99\)

\(⇒ \frac{1}{5} + \left( {1000\times 99 - \frac{{99}}{{495}};} \right)\)

\(⇒ \frac{1}{5} + 99000 - \frac{1}{5}\)

⇒ 99000

Given:

x = 11

Calculation:

Putting x = 11 in

x⁵ - 12x⁴ + 12x³ - 12x² + 12x - 1

we get,

x⁵ - 11x⁴ - x⁴ + 11x³ + x³ - 11x² - x² + 11x + x - 1

⇒ x⁵ - x⁵ - x⁴ + x⁴ + x³ - x³ - x² + x² + x - 1

⇒ x - 1 = 11 - 1 = 10

Given:

p = 101

Concept used:

(a - b)³ = a³ - b³ - 3a²b + 3ab²

Calculation:

\(\sqrt[3]{{{p^3} - 3{p^2} + 3p - 1}}\)

\( ⇒ \sqrt[3]{{{{\left( {p - 1} \right)}^3}}}\)

⇒ p - 1 = 101 - 1 = 100

∴ The value of this data is 100.

Given:

a^1/3 = 11

Calculation:

a^1/3 = 11

cubing both side we get,

a = 1331

a² - 331a = a(a - 331)

⇒ 1331(1331 - 331)

⇒ 1331 × 1000 = 1331000

Given:

\(11√ n = √ {112} + √ {343} \)

Concept:

(a + b)² = a² + 2ab + b²

Calculation:

\(11√ n = √ {112} + √ {343} \)

Squaring both side we get,

121n = (√112)² + 2 × √(112 × 343) + (√343)²

⇒ 121n = 112 + 343 + 2 × √(2 × 2 × 2 × 2 × 7 × 7 × 7 × 7)

⇒ 121n = 455 + 2 × (2 × 2 × 7 × 7)

⇒ 121n = 455 + 2 × 196

⇒ 121n = 455 + 392 = 847

⇒ n = 847/121 = 7

Given:

x + y = √3

x - y = √2

Concept:

If x + y = a, x - y = b

than x = (a + b)/2 and y = (a - b)/2

a² - b² = (a + b)(a - b)

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

Calculation:

x = (√3 + √2)/2

y = (√3 - √2)/2

xy = (√3 + 2)(√3 - 2)/4 = 1/4

x² + y² = {(√3 + √2)/2}² + {(√3 - √2)/2}²

⇒ (5 + 2√6)/4 + (5 - 2√6)/4

⇒ 5/4 + 2√6/4 + 5/4 - 2√6/4

⇒ 5/4 + 5/4 = 10/4

8xy(x² + y²) = 8 × 1/4 × 10/4 = 5

∴ We get the value of 8xy(x² + y²) is 5.

Given: 

2^x = 3^y = 6^-z

Concept: 

If a^m = b, then b^1/m = a

a^m × aⁿ = a^{m + n}

a^m × b^m = (a × b)^m

Calculation: 

Let 2x = 3y = 6-z = K

Now, we can write 

K^1/x = 2, K^1/y = 3, K^-1/z = 6

Now, 

6 = 2 × 3

Putting the value of 6, 2, and 3, then

⇒ K^-1/z = K^1/x × K^1/y

⇒ K^-1/z = K^{1/x + 1/y}

Now, Comparing the power of both side 

⇒ -1/z = 1/x + 1/y

⇒ 1/x + 1/y + 1/z = 0

∴ The required value is Zero.

Given: 

x = 3 + √8

Concept: 

(a + b)² = a² + b² + 2ab

Calculation:

First, we find the value of 1/x

1/x = 1/(3 + √8)

Rationalize the denominator

⇒ [1/(3 + √8)] × [(3 - √8)/(3 - √8)]

⇒ (3 - √8)/(9 - 8)

⇒ 3 - √8

Now,

x + 1/x = 3 + √8 + 3 - √8

⇒ 6

Now, 

(x + 1/x)² = x² + 1/x² + 2 × x × 1/x

⇒ (6)² = x2 + 1/x2 + 2

⇒ x² + 1/x² = 36 - 2

⇒ x² + 1/x² = 34

∴ The required value is 34.

Given: 

The length of the parallel chords is 8 cm and the diameter of the circle is 10 cm

Concept:  

A line passing through the center perpendicularly bisect the chord into two equal parts.

Equal Chords are equal distance from the center.

Calculation:  

Construct a circle with center 'O' and draw parallel chord AB and CD, EF is the perpendicular bisector of the chords AB and CD.

AB = CD = 8 cm, R = 10/2 = 5cm

We have to find the length of EF.

In ΔBEO, ∠E is 90°, OB = 5 cm, BE = 8/2 = 4 cm,

By Pythagoras theorem

EO = (OB² - BE²)^1/2

⇒ (5² - 4²)^1/2

⇒ (25 - 16)^1/2

⇒ (9)^1/2

⇒ 3 cm

Now, The length of the EF is

⇒ 2 × OE

⇒ 2 × 3

⇒ 6 cm

∴ The required distance between the chords is 6 cm.

Given: 

ABCD is a rhombus.

AB = BC = CD = DA

DP = 1/2(AB)

DP/AB = 1/2

Calculation: 

In ΔAPQ and ΔBCQ

∠PAB = ∠CBQ         ( Due to the corresponding angle)

∠ AQP = ∠BQC

∴ ΔAPQ ∼ ΔBCQ    (Due to AA similarity)

Now,

AQ/BQ = AP/BC

⇒ AQ/BQ = 3/2

⇒ (AB + BQ)/BQ = 3/2

⇒ (AB/BQ) + 1 = 3/2

⇒ AB/BQ = (3/2) - 1

⇒ AB/BQ = 1/2

⇒ BQ/AB = 2/1

∴ The required ratio is 2 : 1.

Given: 

The ratio of the side of the triangle is \(3:1\frac{1}{4}:3\frac{1}{4}\)

Concept: 

If a, b, c is the side of a triangle and c is the greatest side, then

For, Acute tringle

a² + b² > c²

For, Right triangle

a² + b² = c²

For, Obtuse triangle

a² + b² < c²

Calculation: 

The ratio of the side of the triangle is 

⇒ 3 : 5/4 : 13/4

⇒ 12 : 5 : 13

Now, Let the side of the triangle is 5x, 12x and 13x

⇒ (5x)² + (12x)² = (13x)²

⇒ 25x² + 144x² = 169x²

⇒ 169x² = 169x²

Now, We say that the triangle is Right triangle.

∴ Right triangle

Given: 

The side of the equilateral triangle is 6 cm and the triangle is inscribed in the circle

Concept: 

For equilateral triangle

The radius of the circumcircle is 

R = a/√3

The radius of the incircle is 

r = a/2√3

Here, 'a' is the side of an equilateral triangle

Calculation: 

The equilateral triangle is inscribed in a circle that means the circle is the circumcircle and let 'R' be the radius of the circle

Now, 

The radius of the circumcircle is 

⇒ R = 6/√3

⇒ R = 2√3 cm

∴ The required radius of the circle is 2√3 cm.

Given: 

Time is 2 years

Rate is 8% per annum

The difference between simple interest and compound interest is Rs. 768

Concept: 

S.I = (P × R × T)/100

C.I = P[(1 + r/100)n - 1]

Here, n is the number of terms in the year

Calculation: 

Let sum is Rs. P 

Now, 

The simple interest for 2 years at 8% per annum is

⇒ (P × 8 × 2)/100

⇒ 4P/25 

Now, 

The compound interest for 2 years at 8% per annum is

⇒ P[(1 + 8/100)2 - 1]

⇒ P[(27/25)2 - 1]

⇒ P[(729/625) - 1]

⇒ 104P/625 

So, 

The difference between Compound interest and simple interest is 

⇒ 104P/625 - 4P/25 = 768

⇒ 104P - 100P = 768 × 625

⇒ 4P = 768 × 625

⇒ P = 192 × 625

⇒ P = Rs. 1,20,000 

∴ The required sum of money is Rs. 1,20,000 . 

 

Shortcut Trick

The difference between compound interest and simple interest for 2 years is 

⇒ P(r/100)2

According to question 

⇒ P(8/100)2 = 768

⇒ P × 4/625 = 768

⇒ P = (768 × 625)/4

⇒ P = 192 × 625 

⇒ P = Rs. 120,000 

∴ The required sum of money is Rs. 1,20,000 .

Given: 

Time is 2 years

Rate is 5% per annum

The difference between simple interest and compound interest is Rs. 63.

Concept: 

S.I = (P × R × T)/100

C.I = P[(1 + r/100)ⁿ - 1]

Here, n is the number of terms in the year

Calculation: 

Let sum be Rs. P 

Now, 

The simple interest for 2 years at 5% per annum is

(P × 5 × 2)/100

⇒ P/10 

Now, 

The compound interest for 2 years at 5% per annum is

P[(1 + 5/100)² - 1]

⇒ P[(21/20)² - 1]

⇒ P[(441/400) - 1]

⇒ 41P/400 

So, 

The difference between Compound interest and simple interest is 

⇒ 41p/400 - P/10 = 63

⇒ 41P - 40P = 63 × 400

⇒ P = Rs. 25200 

∴ The required sum of money is Rs. 25200.

 

Shortcut Trick

The difference between compound interest and simple interest for 2 years is 

⇒ P(r/100)²

According to question 

⇒ P(5/100)² = 63

⇒ P × 1/400 = 63

⇒ P = 63 × 400 

⇒ P = 25200 

∴ The required sum of money is Rs. 25200

Given:

A sells an article to B making a profit of 1/5 of his outlay 

B sells it to C gaining 20% 

C sells it for Rs. 600 and incurs a loss of 1/6 of his outlay

Concept used:

S.P = C.P × (100 + P%)/100 

S.P = C.P × (100 – L%)/100

Where,

S.P → Selling price 

C.P → Cost price 

P% → Profit% 

L% → Loss%

Calculations:

Let C.P of the product for A be 100x 

Profit gained by A = 100x × 1/5 = 20x 

S.P of the product for A = 100x + 20x = 120x 

C.P of the product for B = S.P of the product for A 

C.P of the product for B = 120x 

S.P of the product for B = 120x × (100 + 20)/100 = 144x 

C.P of the product for C = S.P of the product for B 

C.P of the product for C = 144x 

Loss incurred by C = 144x × 1/6 = 24x 

S.P of product for C = 144x – 24x = 120x 

According to the question 

⇒ 120x = 600 

⇒ x = 5 

C.P of the product for A = 100x = 500

∴ C.P of the product for A be Rs. 500

Given:

Total number of cycles = 10 

C.P of each cycle = Rs. 500 

Money spent on repairing = Rs. 2000

Selling price of first 5 cycle = Rs. 750 

Selling price of second 5 cycle = Rs. 550 

Concept used:

L = C.P – S.P

L% = (L/C.P) × 100 

Where,

L → Loss 

C.P → Cost price 

S.P → Selling price 

L% → Loss%

Calculations:

Cost price of 10 cycles = 10 × 500 = Rs. 5000

Total cost price after repairing = Rs. 5000 + Rs. 2000 = Rs. 7000

Total selling price of 10 cycles = 750 × 5 + 550 × 5 = Rs. 6500

Loss = Rs. 7000 – Rs. 6500 = Rs. 500

Loss% = (500/7000) × 100

⇒ (50/7)% = \(7 \dfrac{1}{7}%\)

∴ Loss% is \(7 \dfrac{1}{7}%\)

Given:

Time taken by A to finish the work = 18 days 

Time taken by B to finish the work = (1/2) of time taken by A 

Concept used:

Total work = LCM

Efficiency = (Total work)/(Total time)

Calculations:

Time taken by B to finish the work = (1/2) × 18 = 9 days 

Total work = LCM (18, 9) = 18 

Total efficiency of A and B together = 1 + 2 = 3 

Work done by A and B together in a day = (3/18) = (1/6)^th part of total work

∴ Work done by A and B together in a day is (1/6)th part of total work

Given:

The rate of working A and B = 2 ∶ 3 

Concept used:

Time taken by is always inversely proportional to efficiency

Calculations:

Let the efficiency of A and B be 2x and 3x respectively 

Time taken by A = 1/2x 

Time taken by B = 1/3x 

The ratio of days taken by A and B = 1/2x ∶ 1/3x 

⇒ 3x ∶ 2x = 3 ∶ 2 

∴ The ratio of days taken by A and B is 3 ∶ 2

Given:

The ratio of the number of boys and girls = 3 ∶ 2 

Percentage of scholarship holder boys = 20% of the boys 

Percentage of scholarship holder girls = 25% of the girls 

Calculations:

Let the number of boys and girls be 300x and 200x respectively 

Number of scholarship holder boys = 300x × 20/100 = 60x 

Number of scholarship holder girls = 200x × 25/100 = 50x 

Total number of students = 300x + 200x = 500x 

Total number of students who are scholarship holder = 60x + 50x = 110x 

Total number of students who are not scholarship holder = 500x – 110x = 390x 

Percentage of students who are not scholarship holder = (390x/500x) × 100 = 78%

∴ Percentage of students who are not scholarship holder is 78%

Given:

Length of first bridge = 800 m 

Length of second bridge = 400 m 

Time taken by train to cross first bridge = 100 sec 

Time taken by train to cross second bridge = 60 sec 

Concept used:

Total distance = Length of train + length of bridge 

Speed = (Distance)/(Time)

Calculations:

Let the length of train be x 

Total distance covered by the train when it passes through first bridge = 800 + x 

Speed of train = (800 + x)/100      ----(1)

Total distance covered by the train when it passes through second bridge = 400 + x 

Speed of train = (400 + x)/60      ----(2)

From equation (1) and (2)

⇒ (800 + x)/100 = (400 + x)/60

⇒ 48000 + 60x = 40000 + 100x 

⇒ 40x = 8000

⇒ x = 200 m 

∴ Lenght of the train is 200 m

Given:

Percentage of students failed in Hindi = 52% 

Percentage of students failed in English = 42%

Percentage of students failed in both the subject = 17% 

Concept used:

(A ∪ B) = (A) + (B) − (A ∩ B)

Where,

(A ∪ B) → Total A and B 

(A) → Total A

(B) → Total B 

(A ∩ B) → Total common part of A and B 

Calculations:

Total failed in Hindi and English = (H ∪ E) = (H) + (E) − (H ∩ E)

⇒ 52% + 42% – 17% = 77% 

Total percentage of student passed in Hindi and English = 100% – 77% = 23%

∴ Total percentage of student passed in Hindi and English is 23%

Given:

Total number of innings = 40 

Average run = 50 runs 

Difference between the highest score and lowest score = 172 runs 

The average of 38 innings = 48 runs

Concept used:

Average = (Total sum)/(Total number)

Calculations:

Total run score by the batsman in 40 innings = 40 × 50 = 2000      ----(1)

Let highest run be x 

So, lowest run = x – 172 

Total run score by the batsman in 38 innings = 38 × 48 = 1824 

Total run score  by the batsman in 40 innings = 1824 + x + x – 172

⇒ 1652 + 2x      ----(2)

From equation (1) and (2)

⇒ 2000 = 1652 + 2x 

⇒ 2x = 348 

⇒ x = 174 

∴ The highest run scores by the batsman is 174

Given:

Three successive or series discounts of 15%, 20% and 30%

Calculation:

Let the Marked price of the article be Rs.100

After 1^st discount of 15%, the price will be 

100 - 100 × (15/100) = 85

After 2^nd discount of 20%, the price will be

85 - 85 × (20/100) = 68

After 3^rd discount of 30%, the price will be

68 - 68 × (30/100) = 47.6

Final price after all three discounts is Rs.47.6

Single discount in this case will be

((100 - 47.6)/100) × 100

⇒ 52.4 %

∴ The single discount for 15%, 20% and 30% is 52.4%

Given:

A dishonest dealer defrauds to the extent of x% in buying as well as selling his goods by using faulty weight. 

Formula Used:

successive change% = (a + b + a × b/100)%

Calculation:

The dishonest dealer defrauds to the extent of x% in buying as well as selling

so his net profit = (x + x + x × x/100)%

⇒ gain percent = (2x + x²/100)%

∴ The gain percent on his outlay is (2x + x2/100)%.

Given:

A and B are in a partnership with investment in the ratio 7 : 9

Calculation:

Let the investment be 7x and 9x for A and B respectively. Then

A → 7x × 3 + (7x/3) × 6      (∵ after 3 months A withdraws its 2/3^rd of initial investment)

B → 9x × 4 + (6x) × 5         (∵ after 4 months B withdraws its \( 33\frac{1}{3}% \) i.e. 1/3^rd of initial investment)

A→ 35x

B→ 66x

As per question 35x + 66x = 10201

⇒ 101x = 10201

⇒ x = 101

∴ Share of A in profit is Rs. 3535 and that of B is Rs. 6666

Given:

Mean marks of 20 students = 15

Wrongly copied marks are 3 and 6

Correct values are 8 and 4

Calculation:

\(Mean = \frac{{Sum;of;total;observations}}{{Total;number;of;observations}}\)

As per question

Sum of total observation = 15 × 20

⇒ 300

Correction required = (8 + 4) - (3 + 6)

⇒ 3

The new sum  will be 300 + 3 = 303

∴ The correct mean is 303/20 or 15.15

Given:

Three circles are bound together by a rubber band

All three circles are touching one another

Concept used:

Calculation:

Length of the rubber band = 10 cm + 10 cm + 10 cm + 2πr  (∵ the band is wrapped over all three circles having 1/3^rd of circumference each)

⇒ 30 + 2π × 5

⇒ 30 + 10π 

∴ The length of the rubber band is 30 + 10π cm

Given:

A river is flowing at the rate of 2 km/hr                                                       

Depth and width of the river are 3 m and 40 m respectively

Concept used:

The amount of water flow will be equal to the volume of the river 

Calculation:

2 km/hr = 2000 m/60 minute

⇒ 100/3 m/min.

The volume of water that falls into the sea in a minute

= 3 m × 40 m × (100/3) m

⇒ 4000 m³

∴ The amount of water that falls into the sea in one minute is 4000 m³

Given:

The radius and height of a right circular cylinder is increased by 10%

Calculation:

The volume of a right circular cylinder = πr²h

Let the initial radius and height of cylinder be 'r' and 'h' respectively

V₁ = πr²h

The increased radius will be equal to 1.1r

Increased height will be equal to 1.1h

New volume of cylinder = π(1.1r)²(1.1h)

⇒ V₂ = 1.331πr²h

% Increment in the volume = (((1.331 - 1)πr²h)/πr²h) × 100

⇒ 33.1

∴ The volume of the cylinder is increased by 33.1%

Given:

Pillar is in a cylindrical shape with perimeter of the base as 8.8 meter

Curved surface area = 17.6 m²

Concept used:

Concrete required to build pillar = Volume of the cylindrical pillar

The curved surface area of a cylinder = 2πrh

The base of a cylinder is a circle whose perimeter = 2πr

Calculation:

The perimeter of the base of the pillar = 8.8 m

⇒ 2πr = 8.8

⇒ r = 1.4 m

Also, Curved surface area = 17.6 m

⇒ 2πrh = 17.6

⇒ h = 17.6/2πr

⇒ h = 2 m

Now, Volume of cylindrical pillar = πr²h

⇒ πr²h = (22/7)(1.4)² × 2

⇒ 308/25 = 12.32 m3

∴ Concrete required to build the pillar is 12.32 m³

Given:

Some bricks are arranged in an area measuring 20 m3. If the length, breadth and height of each brick is 25 cm, 12.5 cm and 8 cm respectively, 

Concept used:

Volume of cuboid = Length × Breadth × Height

Calculation:

Some bricks are arranged in an area measuring 20 m3. If the length, breadth and height of each brick is 25 cm, 12.5 cm and 8 cm respectively, 

Volume occupied by bricks = 20 m³

⇒ 20 × 1000000 cm³

Volume of brick = 25 × 12.5 × 8

Volume of brick = 2500 cm³

Number of bricks = \(\frac{{Volume\ of\ pile}}{{Volume\ of\ bricks}}\)

Number of bricks = \(\frac{{20 \ \times\ 1000000}}{{2500}}\)

Number of bricks = 8000

Given:

The length, breadth and height of a room is 5m, 4m and 3m respectively.

Concept used:

Diagonal of cuboid = \(\sqrt {{l^2} + {b^2} + {h^2}} \)

Calculation:

Room is always of cuboidal shape.

To fit the largest side of bamboo in it we need to find the diagonal of the cuboid.

Diagonal = \(\sqrt {{5^2} + {4^2} + {3^2}} \)

Diagonal = 5√2 m

∴ The length of the largest bamboo that can be kept = 5√2 m

Given:

A solid metallic spherical ball of diameter 6 cm is melted and re-casted into a cone with a diameter of the base as 12 cm.

Concept used:

Volume of sphere = (4/3)πr³

Volume of cone = (1/3)πr²h

Calculation:

Radius × 2 = Diameter

Radius of spherical ball = 6/2 = 3 cm

Radius of cone = 12/2 = 6 cm

Volume of spherical ball = Volume of cone

⇒ \(\frac{4}{3}\pi {r^3} = \frac{1}{3}\pi {r^2}h\)

⇒ \(\frac{4}{3} \times \pi \times 27 = \frac{1}{3} \times \pi \times 36 \times h\)

∴ Height = 3 cm

Given:

The ratio of the diameter of two right circular cones of equal height be 3 ∶ 4

Concept used:

Volume of cone = (1/3)πr²h

Calculation:

Height of two right circular cones is equal.

Diameter of first cone is 3x

Radius × 2 = Diameter

Radius of first cone = 3x/2

Radius of the second cone = 4x/2 = 2x

Ratio of volume of cone 

⇒ \(\frac{{\frac{1}{3}\pi {r_1}^2h}}{{\frac{1}{3}\pi {r_2}^2h}}\)

⇒ \(\frac{9}{{4 \times 4}} = \frac{9}{{16}}\)

∴ Ratio of volume of cones = 9 : 16

Given:

log₂ (log₃ 81)

Concept used:

log_a a = 1

log_a bⁿ = n × log_a b

Calculation:

⇒ log₂ (log₃ 3⁴)

⇒ log₂ (4log₃ 3)

⇒ log₂ 4

⇒ log₂ 2²

⇒ 2 log₂ 2  

∴ 2