Territorial Army Official Paper I (Conducted on 28 Jul 2019) (Previous Year Paper)

Given series;

2, 3, 5, 7, 11, ___, 17

The given series is of prime numbers.

The prime numbers are those which are only divisible by 1 or by itself.

example; 2, 3, 5, 7 etc.

After 11, 13 is the next prime number.

Hence, 13 is the correct answer.

Given series;

23, 48, 99, 203, 413 ___

The pattern followed here is;

Second term = First term × 2 + 2 (Addition of prime numbers)

the next term is determined by multiplying the previous term by 2 and the addition of series of prime numbers.

Hence, 837 is the correct answer.

Given series;

225, 336, 447, 558, ___ 780

The pattern followed here is;

the next term is determined by adding 111 to the previous number.

Hence, 669 is the correct answer.

Given series;

ABC, CAE, EZG, ___ , IXK

the pattern followed is;

Hence, GYI will be the correct answer.

Given series;

___ ___ G ___ C ___ G K ___ P G __

Using option 1)

KCGP / CCGK / PPGK

It is not following any pattern hence, not the correct answer.

Using option 2)

CPGK / CPGK / CPGK

It is following a pattern hence, option 2 is the correct answer

Using option 3)

PKGC / CPGK / KPGP

It is not following any pattern hence, not the correct answer.

Using option 4)

CPGP / CKGK / CPGP

It is not following any pattern hence, not the correct answer.

The given relationship is about what a first term contains.

Menu has food items in it.

Similarly,

The catalogue has books in it.

Hence, book is the correct answer.

The pattern followed here is;

n × (n + 1), where n = any real number which can form the given number in the given format.

42 = 6 × 7;

56 = 7 × 8

Similarly,

110 = 10 × 11;

132 = 11 × 12​

Hence, 132 is the correct answer.

In the given options, all are used to cut the object except one.

Arrow is not used to cut the object.

Hence, Arrow is the odd one from the following.

In the given option;

all the methods given are of cooking except peel.

Hence, the peel is an odd one from the following.

Here, Except 'TUESDAY', only 2 vowels are present in the rest of the words,

whereas, 3 vowels are present in the word 'TUESDAY'.

MONDAY - A and O are 2 vowels present

TUESDAY - A, E, and U are 3 vowels present.

FRIDAY - A and I are 2 vowels present.

SUNDAY - A and U are 2 vowels present.

Note: Out of 26 letters from the English alphabetical series, we have 5 vowels A, E, I, O, and U and the rest of the 21 letters are consonants.

Hence, the word 'TUESDAY' is the odd one out.

In the given options;

all the organs are in pairs in the human body except the heart.

Hence, the heart is odd from the following.

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;

D = 4

D = 4

A = 1

441 is the square of 21.

Hence, 21 is the correct answer.

Given;

 white is called blue

blue is called red

red is called yellow

yellow is called green

green is called black

black is called violet

violet is called orange,

The color of blood is red, and red is called as yellow.

Hence, yellow is the right answer.

In a certain coded language;

From the diagram, Light is coded as 3 

Hence, 3 is the correct answer.

At 12 noon its minute hand points toward North-east

According to the

Statement:

Similarly,

Hour hand point at 9.30 PM will be

Hence, the hour hand will point at South-West direction.

Given statement;

Total strength of class = 60

Boys : Girls = 1 : 2

This means 20 boys and 40 girls are there in the class.

The rank of Kamal from the top is 17^th. 

which means 16 students are ahead of Kamal and nine are girls. So seven are boys.

Total boys = 20

ahead of Kamal = 7

Behind Kamal = 20 - 7 - 1(kamal) = 12 boys. (Kamal is a boy so he will also be counted

Hence, there are 12 boys behind kamal.

According to the statement;

On the right side of Pallavi, there are 20 girls.

On left, girls between Reena and Pallavi is 9.

Malini is 17^th from the left and fourth right of Reena, this means Reena is 17 - 4 = 13^th from left.

So there are 9 + 13 = 22 girls in the left of Pallavi.

Total girls in row are girls on the left of pallavi + girls on the right of pallavi + pallavi self = 22 + 20 + 1

Hence there are 43 girls in row.

The symbol and its meaning is as follow

  

According to the given information;

From the diagram, it is clear that B is the sister-in-law of E.

The symbol and its meaning is as follow

According to the given information;

From the diagram, it is clear that B is the son of D

The symbol and its meaning is as follow

According to the given information;

From the diagram, it is clear that Kapil is the Nephew of Shilpa.

Given statement;

Calculating the number of male members

A family has a man and his wife = +1

They have four sons =  +4

Each four sons has 3 sons and one daughter, = 4 × 3 = 12

Hence, there are 12 + 4 + 1 = 17 male members in the family.

In a certain code;

and,

Similarly,

Hence, 8251896 will be the coded form.

According to English Alphabet series with positional value;

In a certain code;

A = 1; 1 × 2 + 1 = 3;

C = 3; 3 × 2 + 1 = 7;

N = 14; 14 × 2 + 1 = 29;

E = 5; 5 × 2 + 1 = 11

similarly

B = 2; 2 × 2 + 1 = 5;

O = 15; 15 × 2 + 1 = 31;

I = 9; 9 × 2 + 1 = 19;

L = 12; 12 × 2 + 1 = 25

Hence, the coded form will be 5-31-19-25

Given;

A, B, C, D, and E are five friends.

A is shorter than B but taller than E. C is the tallest

 D is shorter than B and taller than A

Hence, D has two persons taller and two persons shorter than him/ her.

Given;

So after correcting the sign it will be written as;

15 × 2 + 900 ÷ 90 – 100

Using BODMAS rule;

15 × 2 + 10 - 100;

30 + 10 - 100

40 - 100

-60

Hence, -60 is the correct answer

Given,

Given equation is :

18 c 14 a 6 b 16 d 4 =?

According to the coding of symbols:

18 × 14 + 6 - 16 ÷ 4 = ?

18 × 14 + 6 - 4 = ?

252 + 6 - 4 = ?

252 + 2 = ?

? = 254

Hence, '254' is the correct answer

The least possible Venn diagram is as follows:

Explanation:

(I) All goats are animals → True (All goats are cows and All cows are animals)

(II) All animals are goats → False (All goats are cows and All cows are animals) 

Hence, only conclusion I follow.

The least possible Venn diagram is as follows:

Explanation:

(I) Some dogs are cats → True (Some cats are dogs)

(II) Some toys are cats → False (it is possible but not definite)

(III) Some cats are not toys → True (Some cats are dogs and No dog is a toy)

(IV) All toys are cats → False (it is possible but not definite)

Hence, only conclusion I and III follow

According to the given information :

1.12.91 is the first Sunday of December 91.

Sunday + 2 days = Tuesday

Thus, 1.12.91 + 2 days = 3.12.91

Thus, the first Tuesday of December 91 is 3.12.91

The day of a week repeats after every 7 days.

Fourth Tuesday will be after 3 weeks (3 × 7 = 21 days)

3.12.91 + 21 days = 24.12.91

Hence, the correct answer is "24.12.91".

Alternate Method

First Tuesday = 3.12.91

Second Tuesday = 3.12.91 + 7 days = 10.12.91 

​Third Tuesday = 10.12.91 + 7 days = 17.12.91 

​Fourth Tuesday = 17.12.91 + 7 days = 24.12.91

The pattern followed here is :

Figure in option 1, 2 and 3 the spikes of outer shape and spikes of the inner circle are in the same direction.

But in option 4 the spikes of outer shape and spikes of the inner circle are in the opposite direction.

  

 

Hence, the figure in option (d) is the correct answer.

The figure present in option 1, 2, and 3 requires one line to make a close figure.

But the figure in option 4 requires two lines to make a close figure.

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

The pattern followed here is :

Figures in options 2, 3, and 4 are the same figure but the orientation is different which is represented from various angles.

Figure 1 is not as same as the other figures. 

Hence, the figure in option 1 is the correct answer.

The pattern followed here is:

The image shift 180^o and the shaded dote moves alternately one end to other end.

Hence, the figure in option 2 is the correct answer.

The pattern followed here is:

The arrow moves 45^o clockwise while pin moves 45^o anticlockwise.

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

The pattern followed here is :

 → Moves anticlockwise from corners.

 → Moves clockwise from corners.

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

By checking the following alternatives:

1. 22

22 + 6 + 16 + 8  = 52

52 ÷ 4 = 13

 2) 25

25 + 6 + 16 + 8 = 55

55 ÷ 4 = 13.75

3) 20

20 + 6 + 16 + 8 = 50

50 ÷ 4 = 12.5

4. 18

18 + 6 + 16 + 8 = 48

48 ÷ 4 = 12

Hence, '22' is the correct answer.

Given,

10 cats caught 10 rats in 10 seconds.

Cats are required to catch 100 rats in 100 seconds.

⟹ 1 cat in 10 secs caught= 1 rat

⟹ 1 cat in 100 secs caught= 10 rats

So, 10 cats can caught 100 rats in 100 seconds.

Hence, '10' is the correct answer.

Glass is used to hold the Juice, cup is used to hold the tea similarly  Spoon is used to hold the water but knife is used to cut the fruits

Hence, 'Knife, Fruit' is odd among all.

Army, Navy and Air force all are defence services but they all are related to each other.

So the relation between Army, Navy and Air force is :

Hence, the figure in option 3 is the correct answer.

Majuli is a river island located in Brahmaputra river in Assam while Godavari river is located in Maharashtra. so the Venn diagram to show the relation between this three is follow:

Hence, figure in option 4 is the correct answer.

According to the positions of alphabets in English alphabet series:

Insert alphabet table:

The pattern followed here is:

Similarly,

Hence, 'OCQIXCJG' is the correct answer.

There are 14 squares namely,

ABEF, BCFG, CGDH, EFIJ, FGJK, GHKL, IJMN, JKNO, KLOP, AICK, BDJL, EGMO, FHNP and ADMP.

Hence, '14' is the correct answer.

The total number of the rectangle in the given figure is :

ABRQ, RQCD, ABCD, AECF, BDFE, DFGH, GHIJ, IJPM, BEGH, BEIJ, DFIJ, DFPM, GHPM, BEPM, IKPL, JNKO, NOML, and JMKL

So there is a total of 18 rectangles in the given figure.

Hence, '18' is the correct answer.

Key Points

All square are rectangles but no rectangle is square.

The paper when unfolded looks like.

Hence, the figure in option (1) is the correct answer.

The pattern followed here is :

Column 1) : 11² - 1² = 120

Column 2) : 72 - 22 = 45

Similarly,

Column 3) : 5² - 3² = 16

Hence, '16' is the correct answer.

The pattern followed here is :

Figure I)

5 × 3 + 4 = 19

Similarly,

Figure II)

5 × 7 + 6 = 41

Figure III)

6 × 4 + 5 = 29

Hence, '41' is the correct answer.

The pattern followed here is:

1. Row I : 12 ÷ 6 = 2; 2+ 1 = 3

2. Row II : 16 ÷ 8 = 2 ; 2 + 2 = 4

Similarly,

3) 128 ÷ 64 = 2; 2 + 3 = 5 

Hence, '64' is the correct answer.

The pattern followed here is:

Column I) 

Alphabets: A + 6 → G ; G + 6 → M

Number:  

2 + 3 = 5

Column II) 

Alphabets: C + 6 → I ; I + 6 → O

Number: 

5 + 4 = 9

Similarly,

Column II) 

Alphabets: E + 6 → K ; K + 6 → Q

Number: 

6+ ? = 14

? = 14 - 6

? = 8

Hence, 'K8' is the missing number.

The pattern followed here is:

In figure 1) 584 

5 = 5 × 1

8 = 2 × 4

4 = 2 × 2

In figure 2) 694

6 = 3 × 2

9 = 3 × 3

4 = 2 × 2

Similarly, in figure 3) 678

6 = 6 × 1

7 = 7 × 1

8 = 4 × 2 

Hence, '678' si the correct answer.

By solving the given alternatives.

1. 5 4 3 6 7 8 2 1 → K E R B A Y O D

2. 5 4 8 2 6 7 3 1 → K E Y O B A R D

3) 5 4 8 6 2 7 3 1 → K E Y B O A R D

4. 5 4 8 6 2 7 1 3 → K E Y B O A D R

So, code in option (c) has a meaningful combination of words.

Hence, '5 4 8 6 2 7 3 1' is the correct answer.

Given :

Two rational numbers 3/5 and 2/3

Calculations :

According to the question 

We need to insert two rational numbers between 3/5 and 2/3 

Firstly we will find the values of these two numbers 

3/5 = 0.6 and 2/3 = 0.66 

Now let us check the values of the options given

8/13 = 0.61     7/11 = 0.63

15/20 = 0.75   11/12 = 0.91

19/10 = 1.9     10/15 = 0.66

41/20 = 2.05   16/25 = 0.64 

After checking all these options we can find out only option (1) can be inserted between these two numbers.

∴ Option (1) is the correct answer

Given :

Two irrational numbers √2 and √3 

Calculations :

We need to insert two values between √2 and √3 

Firstly find the decimal values of the numbers 

√2 = 1.42     √3 = 1.73 

So the number inserted must be between 1.42 and 1.73 

Now check the values of the given options 

49/28 = 1.75

56/35 = 1.6

63/45 = 1.4

85/66 = 1.28 

We can see that only 56/35 can be put between √2 and √3 

∴ Option (2) is the right choice.

Given :

[(5.49)³ - (1.49)³]/[(5.49)² + 5.49 × 1.49 + (1.49)²] = ?

Formula used :

(A)³ - (B)³ = (A - B).(A² + B² + AB) 

Calculations :

(5.49)³ - (1.49)³ 

⇒ (5.49 - 1.49)/[(5.49)² + 5.49 × 1.49 + (1.49)²]     ----(1)

\(\frac{5.49\times 5.49 \times 5.49 - 1.49 \times 1.49 \times 1.49} {5.49\times 5.49+5.49\times1.49+1.49\times1.49}\)

⇒ [(5.49)3 - (1.49)3]/[(5.49)2 + 5.49 × 1.49 + (1.49)2]  

From eq.(1)

⇒ (5.49 - 1.49)/[(5.49)2 + 5.49 × 1.49 + (1.49)2]/[(5.49)2 + 5.49 × 1.49 + (1.49)2]  

⇒ 5.49 - 1.49 

⇒  4 

∴ option 2 will be the value of given algebraic equation

Given :

 x, x + 2 and x + 4  are all prime numbers 

Concept used :

A number divisible by 1 and itself only is known as a prime number 

calculations :

We will go through the options 

Put x = 3, then the numbers are 3, 5, 7 

Put x = 2 then the numbers are 2, 4, 6 

Put x = 11 then the numbers are 11, 13, 15 

Put x =  17 then the numbers are 17, 19, 21 

After analyzing the options we can see that at x = 3 all numbers are coming prime

∴ The value of x = 3

Concept used :

Prime factors of a number with even powers gives us a factor that is a perfect square

Example : 36 = 2² × 3²  here 2² , 3² and (2×3)² are perfect squares 

calculations :

Perfect squares of 2⁵ × 3⁶ = 1, 2², 2⁴, 3², 3⁴, 3⁶, (2 × 3)², (2 × 3)⁴ (2² × 3⁴), (2⁴ × 3²), (2² × 3⁶), (2⁴ × 3⁶)

1 is also a perfect square factor 

∴ There are total 12 factors which are perfect squares

GIVEN:

\(\sqrt[3][{{\frac{x}{{27}}] = \frac{5}{3}}}\)

CONCEPT:

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

CALCULATION :

\(\sqrt[3][{{\frac{x}{{27}}] = \frac{5}{3}}}\)

⇒(x/27)^{(1/3) × 3} =  (5/3)³

⇒(x/27) =  (5/3)3

⇒(x/27) =  125/27

⇒x = 125

∴ The value of x is 125.

Given :

Sum of 5 consecutive integers is S

Calculations :

Let the smallest of these 5 numbers be x 

5 consecutive numbers will be x, x +1, x + 2, x + 3, x + 4

According to the question 

x + x + 1 + x + 2 + x + 3 + x + 4 = S 

5x + 10 = S 

5x = S - 10 

x = (S - 10)/5 

Largest of these 5 number = x + 4 = [(S - 10)/5] + 4 

⇒ (S + 10)/5 

∴ Option 4 will be the right answer

Given:

A number is as much greater than 36 as it is less than 86

Calculation:

Let x be the number

⇒ x – 36 = 86 – x

⇒ 2x = 122

∴ x = 61

Shortcut TrickAs the number is in midway of 36, 86.

The number will be (36 + 86)/2 = 61

Given:

LCM = 90 × HCF

LCM + HCF = 1456

One of the Numbers = 160

Formula Used:

Product of two numbers = Product of their LCM and HCF

Calculations:

Let HCF be x and the other number be N

⇒ LCM = 90x

The sum of LCM and HCF = 91x

⇒ 91x = 1456

⇒ x = 16

LCM × HCF = 160 × N

⇒ 90 × 16 × 16 = 160 × N

⇒ N = 144

∴ The other Number is 144

Calculation:

\(\sqrt\frac{{0.324 \times 0.64 \times 129.6}}{{0.729 \times 1.024 \times 36}} = \sqrt\frac{{(1.8^2\times0.1) \times 0.8^2 \times (3.6^2 \times 10)}}{{(2.7^2\times0.1) \times (0.32^2 \times10)\times 6^2}}\)

\(⇒ \frac{{1.8 \times 0.8 \times 3.6}}{{2.7 \times 0.32\times 6}} = \frac{{2 \times 1 \times 0.6}}{{3 \times 0.4\times 1}}\)

⇒ 1

∴ The Square root is 1

Given:

2√2 ∶ 3√5

Formula Used:

Duplicate ratio of a ∶ b is a² ∶ b²

Calculation:

⇒ Duplicate ratio of 2√2 ∶ 3√5 is (2√2)² ∶ (3√5)²

∴ The answer is 8 ∶ 45

Given:

log_x 4 + log_x 16 + log_x 64 = 12

Formula Used:

If log_x y = a, then x^a = y

log_x a^b = b log_x a

Calculation:

log_x 4 + log_x 16 + log_x 64 = 12

⇒ log_x 2² + log_x 2⁴ + log_x 2⁶ = 12

⇒ 2 log_x 2 + 4 log_x 2 + 6 log_x 2 = 12

⇒ 12 log_x 2 = 12

⇒ log_x 2 = 1

⇒ 2 = x¹

∴ x = 2

Given:

(a – b) ∶ (a + b) = 1 ∶ 5

Formula Used:

(a + b) × (a – b) = a² – b²

(a + b)² + (a – b)² = 2(a² + b²)

Calculation:

Let a – b = x

⇒ a + b = 5x

⇒ a² – b² = 5x²

(a – b)² + (a + b)² = x² + (5x)²

⇒ 2(a² + b²) = 26x²

⇒ a² + b² = 13x²

∴ (a² – b²) ∶ (a² + b²) = 5 ∶ 13

Given:

\(\frac{{3x - y + 1}}{3} = \frac{{2x + y + 2}}{5} = \frac{{3x + 2y + 1}}{6}\)

Calculation:

Let \(\frac{{3x - y + 1}}{3} = \frac{{2x + y + 2}}{5} = \frac{{3x + 2y + 1}}{6} = k\)

⇒ 3x – y + 1 = 3k    → (1)

2x + y + 2 = 5k         → (2)

3x + 2y + 1 = 6k       → (3)

1. – (1)  →  3y = 3k ⇒ y = k

Put this in eq (1) & (2)

⇒ 3x + 1 = 4k, 2x + 2 = 4k

⇒ 3x + 1 = 2x + 2

⇒ x = 1

⇒ 3×1 + 1 = 4k

⇒ k = 1

∴ Values of  x, y are 1,1

Given:

Lights change after 36, 42, 72 seconds

Concept Used:

For any events happening at some interval of time, to happen all the events at the same time it takes the LCM of individual time intervals.

Calculation:

LCM of 36, 42, 72 is 504

504 seconds = 8 min 24 sec

∴ The answer is 8 min 24 sec.

Given:

x = 6 × y

(x + 4) = 4 × (y + 4)

Calculation:

x = 6 × y      ---- (1)

(x + 4) = 4 × (y + 4)      ---- (2)

Put the value of eq.

(1) in eq. (2) ⇒ 6y + 4 = 4y + 16

⇒ 6y – 4y = 16 – 4

⇒ 2y = 12

⇒ y = 6

∴ The present age of y is 6 years.

Given:

Rohit ∶ Sahil (1 year ago) = 6 ∶ 7

Rohit ∶ Sahil (4 year after) = 7 ∶ 8

Calculation:

Let Rohit’s age 1 year ago be 6x, and Sahil’s age 1 year ago be 7x

Time difference = 4 + 1

⇒ 5 years

(6x + 5)/(7x + 5) = 7/8

⇒ 8 × (6x + 5) = 7 × (7x + 5)

⇒ 48x + 40 = 49x + 35

⇒ 49x – 48x = 40 – 35

⇒ x = 5

Sahil’s age 1 year ago = 7x

⇒ 7 × 5

⇒ 35 years

Sahil’s present age = 35 + 1 = 36 years

∴ The present age of Sahil is 36 years.

Given:

A fraction becomes 2/3 if 1 is added to both its numerator and denominator. Same fraction becomes 1/2 if 1 subtracted both from numerator and denominator.

Calculation:

Let the numerator of fraction be N, and denominator of fraction be D.

The fraction = N/D

When 1 is added to its numerator and denominator both

⇒ (N + 1)/(D + 1) = 2/3

⇒ 3 × (N + 1) = 2 × (D + 1)

⇒ 3N + 3 = 2D + 2

⇒ 2D – 3N = 1      ---- (1)

When 1 is subtracted from its numerator and denominator both

⇒ (N – 1)/(D – 1) = 1/2

⇒ 2 × (N – 1) = D – 1

⇒ 2N – 2 = D – 1

⇒ D = 2N – 1      ---- (2)

Put the value of D in eq.

(1) ⇒ 2 × (2N – 1) – 3N = 1

⇒ 4N – 2 – 3N = 1

⇒ N = 3

Put the value of N in eq. (2) ⇒ D = 2 × 3 – 1

⇒ D = 6 – 1

⇒ D = 5

∴ The fraction is 3/5.

Given:

The selling price of both the houses = Rs. 29,700 each

On one = 10% loss

On other = 10% profit

Formula Used:

Net effect = x + y + (x × y)/100

Where x → Profit or Loss on one

y → Profit or Loss on other

Calculation:

Net effect = x + y + (x × y)/100

⇒ -10 + 10 + [(-10) × 10]/100

⇒ -1%

∴ Man will have loss in the whole transaction.

Given:

The ratio of income of A and B = 4 ∶ 3

The ratio of annual expenses of A and B = 3 ∶ 2

Savings = Rs. 60000 each

Concept:

Income = Expenditure + Saving

Calculation:

Let the income of A and B be 4x and 3x respectively.

Their expenditure = (4x – 60000)/(3x – 60000) = 3/2

⇒ 2 × (4x – 60000) = 3 × (3x – 60000)

⇒ 8x – 120000 = 9x – 180000

⇒ 9x – 8x = 180000 – 120000

⇒ x = 60000

Income of A = 4x

⇒ 4 × 60000

⇒ 240000

∴ The income of A is Rs. 240000.

Given:

Area of the field = 25 hectare

Speed of the boy = 10km/hr

Formula Used:

Area of the square = a²

Perimeter of the square = 4 × a

Where a → side of the area

Distance = Speed × Time

1 hectare = 10000 m²

Calculation:

Area of the field = 25 hectare

⇒ a² = 25 hectare

⇒ a² = 25 × 10000 m²

⇒ a = √250000 m²

⇒ a = 500 m

Perimeter of the field = 4 × a

⇒ 4 × 500

⇒ 2000 m

⇒ 2 km

Time = Distance/Speed

⇒ 2/10

⇒ (1/5) hr

⇒ (1/5) × 60 min

⇒ 12 min

∴ A boy will take 12 min to run around the field.

Given:

The price of the cooking gas rises by 15%

Formula Used:

New cost price = Old cost price × (100 + increase %)/100

Consumption reduce = [(New cost price – Old cost price)/New cost price] × 100

Calculation:

Let the cost price of the cooking gas be 100.

New cost price of the cooking gas = 100 × (100 + 15)/100

⇒ 115

Consumption reduce = [(New cost price – Old cost price)/New cost price] × 100

⇒ [(115 – 100)/115] × 100

⇒ 1500/115

⇒ 300/23

\( \Rightarrow 13\frac{1}{{23}}% \)

\(\therefore {\bf{Consumption}};{\bf{reduced}};{\bf{by}};{\bf{family}};{\bf{is}};13\frac{1}{23}% .\)

Given:

Population of a city in 2004 = 10,00,000

Increase in 2005 = 15%

Decrease in 2006 = 35%

Increase in 2007 = 45%

Formula Used:

Final Population = Initial Population × [(100 + increase %)/100] × [(100 – loss %)/100]

Calculation:

Final Population = Initial Population × [(100 + increase %)/100] × [(100 – loss %)/100]

⇒ 10,00,000 × [(100 + 15)/100] × [(100 – 35)/100] × [(100 + 45)/100]

⇒ 10,00,000 × (115/100) × (65/100) × (145/100)

⇒ 115 × 65 × 145

⇒ 10,83,875

∴ The population of the city at the end of year 2007 is 10,83875.

Given:

A sum of money becomes 3 times in 5 years at simple interest.

Concept used:

3 times ⇒ 100% to 300%

6 times ⇒ 100% to 600%

Calculations:

3 times in 5 years

⇒ (300 - 100)% in 5 years

⇒ 40% in 1 year

⇒ Rate of interest per annum = 40%.

Now, 

Time taken to become 6 times

⇒ (600 - 100)%/40%

⇒ 12.5 

∴ The sum of money becomes 6 times in 12.5 years.

Given:

A sum of money on compound interest amount to Rs. 9680 in 2 years and Rs.10648 in 3 years.

Concept used:

In case of compound interest,

Rate of interest per annum = {Amount of (n + 1)^th year/Amount of (n)th year} × 100 - 100

Calculations:

Rate of interest per annum = {Amount of (n + 1)th year/Amount of (n)th year} × 100 - 100

⇒ (10648/9680) × 100 - 100

⇒ (11/10) × 100 - 100

⇒ 10

∴ The rate of interest per annum is 10%.

Given:

Two numbers are less than third number by 30% and 37% respectively.

Calculations:

Let the third number(C) be '100'.

⇒ The first number(A) = (100 - 30)% of 100

⇒ A = 70

⇒ The second number(B) = (100 - 37)% of 100

⇒ B = 63

Now,

Percent by which B is less than by A is

⇒ {(70 - 63)/70} × 100

⇒ 10%

∴ The percentage by which second number is less than the first is 10%.

Given:

Total poplulation of a town = 3600

Population of men = (8/9) of total population

Married men = 10% of population of men.

Concept used:

Number of married men is equal to Number of married women.

Calculations:

Population of men = (8/9) of total population

Population of men = (8/9) × 3600

= 3200

Population of women = (1/9) of total population

= (1/9) × 3600

= 400

Now,

Married men = 10% of population of men.

= (10/100) × 3200

= 320

Number of married men is equal to Number of married women.

Number of married women = 320

Number of unmarried women = (400 - 320)

= 80

% of unmarried women = (Number of unmarried women/ Total number of women) × 100

= (80/400) × 100

= 20

∴ The percentage of unmarried women in the town is 20%.

Given:

Boating at 6/7^th of regular speed in a lake, the tourist got late by 30 minutes.

Concept used:

LCM method

Calculations:

Let regular speed and changed speed be 7x and 6x respectively.

⇒ Difference of time is 60 minutes = 30 minutes

⇒ 1 minute = (1/2) minutes

⇒ Regular time = 6 hours × (1/2)

⇒ 3 hours.

∴ The time taken at usual speed is 3 hours.

 

Topper's Approach:

In 6/7, 6 represents usual time and 7 represents late time.

⇒ Difference = (7 - 6)

⇒ 1 = 30 minutes or (1/2) hour

⇒ 6 = 6 × (1/2) hour

⇒ 3 hours.

∴ The time taken at usual speed is 3 hours.

Given:

Length of the rectangular field = 242 m

Area of the rectangular field = 4840 m²

Cost of fencing = 0.50 Rs./meter

Formula used:

Area of the rectangular field = (length × breadth)

Perimeter of the rectangular field = 2 × (length + breadth)

Total cost of fencing of the rectangular field = (Perimeter × cost of fencing per meter)

Calculations:

Area of the rectangular field = (length × breadth)

⇒ 4840 = (242 × breadth)

⇒ breadth = 20 m

Perimeter of the rectangular field = 2 × (length + breadth)

⇒ 2 × (242 + 20)

⇒ 524 m

Total cost of fencing of the rectangular field = (Perimeter × cost of fencing per meter)

⇒ (524 × 0.50)

⇒ 262

∴ The total cost of fencing of the rectangular field is Rs. 262.

Given:

Area of four walls of a room = 660 m²

l = 2b, where 'l' is length of the room and 'b' is breadth of the room.

Height of the room(h) = 11 m

Formula used:

Area of four walls of a room = 2(l + b) × h, where 'l' is length of the room, 'b' is breadth of the room and 'h' is height of the room.

Area of ceiling of a room = (l × b), where 'l' is length of the room and 'b' is breadth of the room.

Calculations:

Let the length and breadth of the room be 2x and x respectively.

Area of four walls of a room = 2(l + b) × h

⇒ 2 × (2x + x) × 11 = 660

⇒ 2 × (3x) × 11 = 660 

⇒ x = 10

Area of ceiling of a room = (l × b)

⇒ ( 2x × x)

⇒ 200

∴ The area of ceiling of the room is 200 m².

Given:

Ratio of two complementary angles is 1 : 5

Concept used:

Sum of two complementary angles is 90°.

Calculations:

Let the two angles be 1x and 5x.

Sum of two complementary angles is 90°.

⇒ (1x + 5x) = 90°

⇒ x = 15°

⇒ (5x - x) = 60° 

∴ The difference between two complementary angles is 60°.

Additional Information

Two angles are called complementary when their measures add to 90 degrees.

Two angles are called supplementary when their measures add up to 180 degrees.

Given:

Speed of man = 2/5 of his original speed

He is late by 15 minutes

Formula used:

Distance = Speed × Time

Calculation:

Let the original speed of man be 5x and time taken by him be t minutes

Distance = Speed × Time

⇒ Distance = 5x × t = 5xt      ---- (1)

Reduced speed = 2/5 of 5x 

⇒ Reduced speed = 2x

New time taken (t') = t + 15

Distance = Reduced speed × t'

⇒ Distance = 2x × (t + 15)

⇒ Distance = 2xt + 30x      ---- (2)

By equating (1) and (2) we'll get

5xt = 2xt + 30x

⇒ 3xt = 30x

⇒ t = 30x/3x

⇒ t = 10

∴ Time taken with original speed is 10 minutes

Let original time be 2x and new time be 5x

5x – 2x = 15 minutes

⇒ 3x = 15 minutes

⇒ x = 5 minutes

Original time = 2x = 2 × 5

∴ Time taken with original speed is 10 minute

Calculation:

Draw a line segment EF which is parallel to PA and passes through the point making angle x and name the angles y and z.

y = 45° (Alternate interior angles)

z = 30° (Alternate interior angles)

x + y + z = 360° (Complete angle)

⇒ x = 360° – z – y

⇒ x = 360° – 30° – 45° = 360° – 75° 

∴ x is 285°

Shortcut TrickSpeed of boat in still water = (Downstream speed + Upstream speed)/2

Speed of stream = (Downstream speed - Upstream speed)/2

⇒ Speed of boat in still water = (13 + 7)/2 = 10 km/h

⇒ Speed of stream = (13 - 7)/2 = 3 km/h

Basic Mehtod:

Given:

Speed of boat upstream = 7 km/hr

Speed of boat downstream = 13 km/hr

Formula used:

Downstream speed (S_D) = Speed of boat in still water + Speed of stream

Upstream speed (S_U) = Speed of boat in still water – Speed of stream

T_D = D/S_D

T_U = D/S_U

Calculation:

Let the speed of boat in still water be x km/hr and speed of stream = y km/hr

S_D = x + y

⇒ SD = 13 

⇒ x + y = 13      ---- (1)

S_U = x – y

⇒ SU = 7

⇒ x – y = 7

⇒ x = 7 + y      ---- (2)

Solving equation (1) and (2)

7 + y + y = 13

⇒ 2y = 13 – 7

⇒ y = 3 km/hr

⇒ x = 7 + 3 = 10

⇒ x = 10 km/hr

∴ Speed of boat in still water is 10 km/hr and speed of stream is 3 km/hr

Alternate Method

We can directly check options for

Downstream speed (SD) = Speed of boat in still water + Speed of stream

Upstream speed (SU) = Speed of boat in still water – Speed of stream

∴ Speed of boat in still water is 10 km/hr and speed of the stream is 3 km/hr

Given:

Ajay can do the work in 10 days

Harshal can do the work in 12 days

Ajay left the work 2 days before completion of work

Concept used:

If a person does the work in x days then work done by the person in 1 day = 1/x of total work.

Calculation:

Let the total time taken by them to complete the work be x days.

Ajay did the work for (x – 2) days

Work done by Ajay in 1 day = 1/10 of the total work

Work done by Ajay in (x – 2) days = (x – 2)/10 

Work done by Harshal in 1 day = 1/12 of the total work

Work done by Harshal in x days = x/12 

They will complete the whole work in x days

⇒ (x – 2)/10 + x/12 = 1

\( ⇒ ;\frac{{6x; - ;12; + ;5x}}{{60}}; = ;1\)

⇒ 11x = 72

⇒ x = 72/11

∴ The work is completed by them in \(6 \frac{6}{11}\) days

Given:

3 Men can do the work in 43 days

4 women can do the work in 43 days

Concept used:

If n persons can do a work in x days then 1 person will do the same work in nx days

If a person does the work in x days then work done by the person in 1 day = 1/x of total work

Calculation:

3 men can do the work in 43 days

⇒ 1 man can do the work in 43 × 3 = 129 days

⇒ 7 men can do the work in 129/7 days

4 women can do the work in 43 days

⇒ 1 woman can do the work in 43 × 4 = 172 days

⇒ 5 women can do the work in 172/5 days

Work done by 7 men in a day = 7/129 of total work

Work done by 5 women in a day = 5/172 of the total work

⇒ Work done by 7 men and 5 women in a day = 7/129 + 5/172

⇒ Work done by 7 men and 5 women in a day = 1/12

∴ 7 men and 5 women can complete the work in 12 days

Given:

A can do the work in 70 days

B is 40% more efficient than A

Concept used:

If a person does p unit work in a day then, 

The person can complete the work in (Total units)/p days

Calculation;

Let the work done by A in one day be x unit

⇒ Work done by A in 70 days = 70 × x = 70x unit

Total work = 70x unit

As, B is 40% more efficient than A 

⇒ B can do x + 40% of x units in a day

⇒ B can do 1.4x unit in a day

B can complete the work in 70x/1.4x days

∴ B can complete the work in 50 days

Given:

Dimensions of the box are

Length = 10 cm

Breadth = 6 cm

Height = 5 cm

Thickness = 2 cm

Formula used:

Volume of cuboid = Length × Breadth × Height

Volume of material used when thickness is given = Outer volume – Inenr volume

Calculation:

Inner length = 10 – 2 × thickness = 10 – 4 

⇒ Innner length = 6 cm

Inner breadth = 6 – 2 × thickness = 6 – 4

⇒ Inner breadth = 2 cm

Inner height = 5 – 2 × thickness = 5 – 4

⇒ Inner height = 1 cm

Inner volume = 6 × 2 × 1 = 12 cm³

Outer volume = 10 × 6 × 5 = 300 cm³ 

Volume of wood used = Outer volume – Inner volume 

⇒ Volume of wood used = 300 – 12 

∴ Volume of wood used = 288 cm³

Formula used:

tanA = sinA/cosA

cosecA × sinA = 1

sec²A – 1 = tan²A

secA + 1 = tan²A/(secA + 1)

Calculation:

\(\frac{{\tan A - \sin A}}{{{{\sin }^3}A}}\)

\(⇒ ;\frac{{\frac{{sinA}}{{cosA}}; - ;sinA}}{{si{n^3}A}}\)

\(⇒ ;\frac{{\left( {1 - cosA} \right)}}{{si{n^2}A; × ;cosA}}\)

⇒ cosec²A × (secA – 1)

⇒ cosec²A × (tan²A/(secA + 1)

⇒ cosec²A × [sin²A/cos²A × (secA + 1)]

⇒ 1/(cos²A × (secA + 1)

⇒ sec²A/(1/cosA + 1)

⇒ (sec²A × cosA)/(1 + cosA)

∴ secA/(1 + cosA)

We can do this question by putting value of A

let A = 45° 

Questions simplifies to (tan45° – sin45°)/sin³45° 

⇒ (1 – 1/√2)/(1/√2)³ 

⇒ [(√2 – 1)/√2] × 2√2

⇒ 2√2 – 2

Checking option 1

sec45°/(1 – cos45°)

⇒ 2√2 + 1

Checking option 2

sec45°/(1 + cos³45°)

⇒ 4/7 × (2√2 – 1)

Checking option 3

sec45°/(1 + cos45°)

⇒ 2√2 – 2

∴ Option 3 i.e. secA/(1 + cosA) is correct answer

GIVEN:

The length of a pendulum = 60 cm and an arc of length = 16.5 cm 

FORMULA USED:

Length of arc = θ/360° × 2π Radius

CALCULATION:

Length of arc = θ/360° × 2π × Radius

⇒ 16.5 × 360° = 2 × 22/7 × 60 × θ 

⇒ θ = 63°/4

⇒ θ = 15° 45'

∴ The angle through which it swings is 15° 45'

GIVEN:

\(\frac{{\sin θ }}{{\cos \left( {90^\circ + θ } \right)}} + \frac{{\sin θ }}{{\sin \left( {180^\circ + θ } \right)}} + \frac{{\tan \left( {90^\circ + θ } \right)}}{{\cot θ }}\)

FORMULA USED:

cos(90°+ θ) = - sinθ , sin(180°+ θ) = - sinθ and tan(90°+ θ) = - cotθ 

CALCULATION:

\(\frac{{\sin θ }}{{\cos \left( {90^\circ + θ } \right)}} + \frac{{\sin θ }}{{\sin \left( {180^\circ + θ } \right)}} + \frac{{\tan \left( {90^\circ + θ } \right)}}{{\cot θ }}\)

⇒ sinθ/(- sinθ) + sinθ/(- sinθ) + (- cotθ)/cotθ 

⇒ (-1) + (-1) + (-1) 

⇒ -3

GIVEN:

Diameter AB bisect chord CD at point E such that CE = ED = 8 cm and EB = 4 cm

FORMULA USED:

Pythagoras Theorem, (Hypotenuse)² = (Perpendicular)² + (Base)²

CALCULATION:

Let the radius of the circle be r cm and CE = ED = 8 cm and EB = 4 cm

⇒ OA = OD = OB = r cm

⇒ OE = OB - BE = (r - 4) cm

⇒ In Δ ODE, OD² = OE² + DE² (a/c Pythagoras Theorem)

⇒ r² = (r - 4)² + 8²

⇒ r² = r² + 16 - 8r + 64

⇒ 8r = 80

⇒ r = 10 cm

∴ Radius of the circle is 10 cm

GIVEN:

cos A = cos B × cos C

FORMULAE USED:

tan A = sin A/cos A

CALCULATION:

tan A - tan B - tan C = (sin A/cos A) + (sin B/cos B) + (sin C/cos C)

⇒ (sin A/cos A) - {(sin B cos C + sin C cos B)/(cos B × cos C)}

⇒ (sin A/cos A) - (sin (B + C)/ cos A)      ----(1)

As in a triangle, A + B + C = π

So, B + C = π - A

Equation (1) becomes,

⇒ sin A - sin (π - A)/ cos A

⇒ sin A - sin A/ cos A = 0

∴ The value of tan A- tan B - tan C = 0

GIVEN:

One plane is flying at height of 300 m above the ground and passes another plane 

CALCULATION:

Let one plane is flying at height of x m and passed by another plane flying at a height of 300 m from the ground

⇒ In Δ ACD, tan45° = AD/AC

⇒1 = x/AC

⇒ AC = x cm

⇒ In Δ ABC

⇒ tan60° = AB/AC

⇒ √3 = 300/x

⇒ x = 300/√3 = 100√3

∴ Height of lower plane flying = 100√3 m

GIVEN:

The mean of 20 observations = 15 and two observations was wrongly copied as 3 and 6 instead of 8 and 4

FORMULA USED:

New mean = Old mean + (Sum of right observation - the sum of wrong observation)/Number of observation

CALCULATION: 

New mean = Old mean + (Sum of right observation - the sum of wrong observation)/Number of observation

⇒ New mean = 15 + [(8 + 4) - (6 + 3)]/20

⇒ 15 + (12 - 9)/20

⇒ 15 + 3/20

⇒ 15 + .15 = 15.15

∴ Correct Mean is 15.15

GIVEN:

Distance b/w the two poles is 100 m and angles of elevation of tops of poles are 30° and 60° 

FORMULA USED:

Pythagoras Theorem 

(Hypotenuse)² = (Perpendicular)² + (Base)²

CALCULATION:

Let the distance of one pole from a point on the road be x m and the other be (100 - x) m

⇒ In Δ ABE, tan30° = AB/BE

⇒1/√3 = H/(100 -x)

⇒ √3 H = 100 - x ....(1)

⇒ Similarly, In Δ CED

⇒ tan60° = CD/DE

⇒ √3 = H/x

⇒ √3 x = H.....(2)

⇒ On putting the value of H from equn (2) in (1)

⇒ √3 × √3 x = 100 - x

⇒ 3x + x = 100

⇒ 4x = 100

⇒ x = 25

⇒ Height of the poles = √3 x = √3 × 25

∴ Height of the poles is 25√3 m

GIVEN:

an electric pump can fill a tank in 3 hours and Because of a leak in the tank, it took 3.5 hours to fill the tank

FORMULA USED:

The capacity of the tank = Efficiency of pipe × Time taken

CALCULATION:

The capacity of the tank = L.C.M (3 and 3.5)

⇒ 21litre

⇒ The capacity of the tank = Efficiency of pipe × Time taken

⇒ The efficiency of pipe = The capacity of the tank/Time taken

⇒ 21/3

⇒ 7 l/hr

⇒ The efficiency of Leak = The capacity of the tank/Time taken

⇒ 21/3.5

⇒ 6 l/hr

⇒ The volume of water filled in 1 hr = Volume of water filled by pipe - Volume of water emptied by the leak 

⇒ 6 - 7

⇒ -1 l/hr

⇒ The time required to empty the tank = 21/1 hr

∴ The time required to empty tank by Leak = 21 hr

Given:

A spherical ball of radius R = 3 cm

Ball one radius (r₁) = 1.5 cm

Ball two radius (r₂) = 2 cm

Ball three radius (r₃) = x cm

Formula used:

Volume of sphere = (4/3)πR³

Calculations:

Volume of big spherical ball = Volume of total spherical balls

⇒ (4/3)πR³ = (4/3)π(r1)³ + (4/3)π(r2)³ + (4/3)π(r3)³

⇒ 3³ = 2³ + (1.5)³ + x³

⇒ 27 = 8 + 3.375 + x³

⇒ x³ = 15.625

⇒ x = 2.5 cm

∴ The value of x is 2.5 cm.

Given:

Circumference of the base = 44 m

High conical tent = 9 m

Formula used:

Circumference of circle = 2πr

Volume of cone = (1/3)πr²h

Calculations:

⇒ 2πr = 44

⇒ r = 7 m

⇒ Volume of air contained = (1/3) × (22/7) × 7 × 7 × 9

⇒ The volume of air contained = 462 m³

∴ The volume of air contained is 462 m3