RRB Clerk Prelims Memory Based Paper (Held On: 17 August 2024 Shift 1) (Previous Year Paper)

From the given instructions:

Six people live in a three storey building. The ground floor is numbered 1 and the floor above it numbered 2 and so on. Each floor consists of two flats i.e. Flat X and Flat Y. Flat X of floor 2 is just above Flat X of floor 1 and just below Flat X of floor 3. So, we can draw a structure as:

1. S, who lives in Flat X, lives adjacent to the floor where T lives, but in a different building. Only one person lives between P and S in the same building. So, here we have two cases,

Case 1:

Case 2:

2. T and V live on the same floor.

Case 1:

Case 2:

3. Q lives to the southeast of V.

Case 1:

Case 2:

4. R does not live on the floor where S lives. So, here Case 2 gets eliminated.

So, Final Arrangement,

T lives adjacent to R in the same building.

From the given instructions:

Six people live in a three storey building. The ground floor is numbered 1 and the floor above it numbered 2 and so on. Each floor consists of two flats i.e. Flat X and Flat Y. Flat X of floor 2 is just above Flat X of floor 1 and just below Flat X of floor 3. So, we can draw a structure as:

1. S, who lives in Flat X, lives adjacent to the floor where T lives, but in a different building. Only one person lives between P and S in the same building. So, here we have two cases,

Case 1:

Case 2:

2. T and V live on the same floor.

Case 1:

Case 2:

3. Q lives to the southeast of V.

Case 1:

Case 2:

4. R does not live on the floor where S lives. So, here Case 2 gets eliminated.

So, Final Arrangement,

Four people live below R in both flats.

From the given instructions:

Six people live in a three storey building. The ground floor is numbered 1 and the floor above it numbered 2 and so on. Each floor consists of two flats i.e. Flat X and Flat Y. Flat X of floor 2 is just above Flat X of floor 1 and just below Flat X of floor 3. So, we can draw a structure as:

1. S, who lives in Flat X, lives adjacent to the floor where T lives, but in a different building. Only one person lives between P and S in the same building. So, here we have two cases,

Case 1:

Case 2:

2. T and V live on the same floor.

Case 1:

Case 2:

3. Q lives to the southeast of V.

Case 1:

Case 2:

4. R does not live on the floor where S lives. So, here Case 2 gets eliminated.

So, Final Arrangement,

P lives to the northwest of T.

From the given instructions:

Six people live in a three storey building. The ground floor is numbered 1 and the floor above it numbered 2 and so on. Each floor consists of two flats i.e. Flat X and Flat Y. Flat X of floor 2 is just above Flat X of floor 1 and just below Flat X of floor 3. So, we can draw a structure as:

1. S, who lives in Flat X, lives adjacent to the floor where T lives, but in a different building. Only one person lives between P and S in the same building. So, here we have two cases,

Case 1:

Case 2:

2. T and V live on the same floor.

Case 1:

Case 2:

3. Q lives to the southeast of V.

Case 1:

Case 2:

4. R does not live on the floor where S lives. So, here Case 2 gets eliminated.

So, Final Arrangement,

All other options live on odd number floor whereas, V lives on even numbered floor. So, V is the odd one out.

From the given instructions:

Six people live in a three storey building. The ground floor is numbered 1 and the floor above it numbered 2 and so on. Each floor consists of two flats i.e. Flat X and Flat Y. Flat X of floor 2 is just above Flat X of floor 1 and just below Flat X of floor 3. So, we can draw a structure as:

1. S, who lives in Flat X, lives adjacent to the floor where T lives, but in a different building. Only one person lives between P and S in the same building. So, here we have two cases,

Case 1:

Case 2:

2. T and V live on the same floor.

Case 1:

Case 2:

3. Q lives to the southeast of V.

Case 1:

Case 2:

4. R does not live on the floor where S lives. So, here Case 2 gets eliminated.

So, Final Arrangement,

S lives on floor 1 of Flat X.

According to the given information,

Northeast is the correct answer.

According to the given information,

Southeast is the correct answer.

According to the given information,

Total distance covered by Pavan = 5 + 7 + 10 + 15 + 5 + 10 = 52

52 km is the correct answer.

People: A, B, C, D, E, F, G & H

1. D does not sit at the extreme end of the row.

2. B sits second to the left of D.

3. C sits second to the right of F who sits opposite to A.

4. G sits second to the right of H who sits opposite to the one who is an immediate neighbour of E.

5. Thus, the final arrangement is,

H sits opposite to D.

People: A, B, C, D, E, F, G & H

1. D does not sit to the extreme end of the row.

2. B sits second to the left of D.

3. C sits second to the right of F who sits opposite A.

4. G sits second to the right of H, who sits opposite to the one who is an immediate neighbour of E.

5. Thus, the final arrangement is,

 A sits second to the right of E.

People: A, B, C, D, E, F, G & H

1. D does not sit at the extreme end of the row.

2. B sits second to the left of D.

3. C sits second to the right of F who sits opposite to A.

4. G sits second to the right of H who sits opposite to the one who is an immediate neighbour of E.

5. Thus, the final arrangement is,

Except for E, all others are sitting at corners.

 E does not follow the same pattern.

People: A, B, C, D, E, F, G & H

1. D does not sit at the extreme end of the row.

2. B sits second to the left of D.

3. C sits second to the right of F who sits opposite to A.

4. G sits second to the right of H, who sits opposite to the one who is an immediate neighbour of E.

5. Thus, the final arrangement is,

C sits opposite to E.

People: A, B, C, D, E, F, G & H

1. D does not sit at the extreme end of the row.

2. B sits second to the left of D.

3. C sits second to the right of F who sits opposite to A.

4. G sits second to the right of H, who sits opposite to the one, who is an immediate neighbour of E.

5. Thus, the final arrangement is,

 H is an immediate neighbour of F.

The least possible venn diagram for the given statements is:

Conclusion:

I. Some butterfly are party → False (It is possible but not definite, because no definite relation is given between the elements, therefore we will consider it as false)

II. All muffin are home is a possibility → False (As the relation of “only a few” is given between the elements which implies that “some” and “some not” are true, no other relationship is true except these two)

Neither conclusion I nor II follows.

The least possible venn diagram for the given statements is:

Conclusions:

I. Some lobby are not market→ True (as relation of “only a few” is given between the elements which implies that “some” and “some not” both are true)

II. Some garage being lobby is a possibility → True (as no definite positive or negative relation is given between the elements)

Both conclusions I and II follows.

Venn diagram for the given statements are as follows.

Conclusions:

I. All carrot being potato is a possibility - False (Because given A few carrots are radishes and No radish is potato thus the part of carrot which are radish can never be potato so All carrot being potato is a possibility is not followed)

II. Some potato being tomato is a possibility - True (Because given Each radish is tomato and No radish is potato thus the part of tomato which are not radish can be potato so Some potato being tomato is a possibility true)

 Only II follows.

DOG  MAN  HIS  FOR TOY

Adding M at the end of each word,

DOGM   MANM   HISM  FORM  TOYM

One meaningful word is formed.

DOG  MAN  HIS  FOR TOY

After interchaning first and second letter,

ODG  AMN  IHS  OFR  OTY

 Two words have letters in alphabetical order.

Mistake Points 

ODG  AMN  IHS  OFR  OTY 

A =1, M = 13, N = 14 so this is dictionary or alphabetical order.  Similarly OTY O = 15, T = 20, Y = 25

So both are in alphabetical order.

DOG  MAN  HIS  FOR  TOY

After changing the second letter of each word to its succeeding alphabet,

DPG  MBN  HJS  FPR  TPY

None of the words have any vowel.

DOG  MAN  HIS  FOR  TOY

After changing all the consonants to preceeding alphabet and all vowel to its succeeding alphabet ,

CPF   LBM  GJR  EPQ  SPX

One word will have vowel.

DOG  MAN  HIS  FOR TOY

DOG  MAN  HIS  FOR TOY

Second letter of second word from left - A

Third letter of second word from right  - R

ABCDEFGHIJKLMNOPQR

16 letters are betwwen A and R.

Given Word: PNEUMONIA

After arranging all letters of the given word in alphabetical order: 

A E I M N N O P U

After changing all consonants to the previous letter and all vowels to the next letter as per the English alphabetical series: 

B F J L M M P O V

One letter i.e. M is repeating more than once in the above arrangement.

Only one such pair is there in the number ‘6358427’, each of which has as many digits between them in the number (both forward and backward direction) as they have between them in the Numeric Series.

1. Only three persons were born before Mahi.

2. Two persons born between Mahi and Ravi.

3. Pihu was born in the month having 30 days.

4. Two persons born between Pihu and Rishi. (This eliminates case 2)

5. Juhi was born immediately after Navi. (This eliminates case 1)

Hence, Navi was born in August.

1. Only three persons were born before Mahi.

2. Two persons born between Mahi and Ravi.

3. Pihu was born in the month having 30 days.

4. Two persons born between Pihu and Rishi. (This eliminates case 2)

5. Juhi was born immediately after Navi. (This eliminates case 1)

Hence, three persons born between Ravi and Rishi.

1. Only three persons were born before Mahi.

2. Two persons born between Mahi and Ravi.

3. Pihu was born in the month having 30 days.

4. Two persons born between Pihu and Rishi. (This eliminates case 2)

5. Juhi was born immediately after Navi. (This eliminates case 1)

Hence, the number of people born before Shivi is the same as that of after Rishi.

1. Only three persons were born before Mahi.

2. Two persons born between Mahi and Ravi.

3. Pihu was born in the month having 30 days.

4. Two persons born between Pihu and Rishi. (This eliminates case 2)

5. Juhi was born immediately after Navi. (This eliminates case 1)

All of them born in the month which has 31 days except Pihu.

Hence, Pihu does not belong to the group.

1. Only three persons were born before Mahi.

2. Two persons born between Mahi and Ravi.

3. Pihu was born in the month having 30 days.

4. Two persons born between Pihu and Rishi. (This eliminates case 2)

5. Juhi was born immediately after Navi. (This eliminates case 1)

Hence, Both 1) and 2) are true.

Given statements:

A ≤ B < C < D ≥ E ≥ F < G

Conclusions:

I. A < D → True (A ≤ B < C < D)

II. F ≤ D → True (F ≤ E ≤ D)

 Both I and II follow is the correct answer.

Given statement: - Y = K ≥ V ≤G = C

Conclusions:-

a. K ≤ G – False (when we find path from K to G we get K ≥ V ≤G , so we can’t find definite relation between K and G, so this conclusion holds false).

b. K > G – False (when we find path from K to G we get K ≥ V ≤G, so we can’t find definite relation between K and G, so this conclusion holds false).

Here, our both Conclusion holds false and all the conditions of “either-or” Or Satisfies, so our answer is Either K ≤ G or K > G.

The correct answer is Either Conclusion (a) or (b) follows.

Given Statement: L = M ≥ Q ≥ R ≤ N ≥ O

Conclusions: 

I. L ≥ R → True (Here L = M ≥ Q ≥ R → L ≥ R, so it is true).

II. M ≥ R → True (Here M ≥ Q ≥ R → M ≥ R, so it is true).

Hence, Both I and II follow.

Given Series:

M ! 2 T 4 E @ Q 1 W R $ 6 O ^ P U * J 7 S % D 8 F ? 9 A V + B > 3 Z N L

After numbers are arranged from the right end in ascending order and the position of the rest elements remain the same then the series is,

M ! T E @ Q W R $ O ^ P U * J S % D F ? A V + B > Z N L 9 8 7 6 4 3 2 1

The element that will be placed 20^th from the right end is  %.

Given Series:

M ! 2 T 4 E @ Q 1 W R $ 6 O ^ P U * J 7 S % D 8 F ? 9 A V + B > 3 Z N L

After numbers and symbols are dropped then the series is, 

M T E Q W R O P U J S D F A V B Z N L

The element which is tenth from the right end is J.

J is the correct answer$.

Given Series:

M ! 2 T 4 E @ Q 1 W R $ 6 O ^ P U * J 7 S % D 8 F ? 9 A V + B > 3 Z N L

Three elements are between first and second element except E$. 

E$ is the correct answer$.

Given Series:

M ! 2 T 4 E @ Q 1 W R $ 6 O ^ P U * J 7 S % D 8 F ? 9 A V + B > 3 Z N L

The consonants are immediately followed by the number in the series is, 

T4, Q1, J7, and D8

Four is the correct answer$.

Given Series

M ! 2 T 4 E @ Q 1 W R $ 6 O ^ P U * J 7 S % D 8 F ? 9 A V + B > 3 Z N L

The element which is seventh to the right of 20^th from the right end is

20^th from the right is U.

seventh from the right of the 20^th element is 8.

8 is the correct answer$.

Word: SIGNATURE

The second, fourth, seventh, and eighth letters are I, N, U, and R respectively.

There is only one word that can be formed using these letters i.e. RUIN.

Thus, ‘I’ is the second letter from the right end.

All are facing North.

1- Two people are sitting between X and T. 

2- Only one person is sitting between T and M, who is not sitting to the left of T. 

Case-1

Case-2

3- M is sitting third to the right of S. 

4- Two people are sitting between M and Q, who is sitting third to the left of Y. 

Case-1

Case-2

5- P is sitting second to the left of S. So case-1 gets eliminated.

So, Final arrangement is;

Twelve people are sitting in the row.

All are facing North.

1- Two people are sitting between X and T. 

2- Only one person is sitting between T and M, who is not sitting to the left of T. 

Case-1

Case-2

3- M is sitting third to the right of S. 

4- Two people are sitting between M and Q, who is sitting third to the left of Y. 

Case-1

Case-2

5- P is sitting second to the left of S. So case-1 gets eliminated.

So, Final arrangement is;

  X is second to the left of Q.

All are facing North.

1- Two people are sitting between X and T. 

2- Only one person is sitting between T and M, who is not sitting to the left of T. 

Case-1

Case-2

3- M is sitting third to the right of S. 

4- Two people are sitting between M and Q, who is sitting third to the left of Y. 

Case-1

Case-2

5- P is sitting second to the left of S. So case-1 gets eliminated.

So, Final arrangement is;

If A is sitting in between T and M then Q is sitting Fourth to the right of A.

Calculation:

Monday participants = 123, Thursday participants = 127

Percentage change = \(\frac{127 - 123}{123} \times 100 = \frac{4}{123} \times 100 = 3.25%\)

Answer: 3.25% increase

Calculation:

Total participants on Tuesday = 139 + 143 + 155 = 437

Increased by 20% = 437 × 1.20 = 524.4 (rounded to 524)

Difference = 524 - 437 = 87

Answer: 87 more participants

Calculation:

Zumba (Wednesday + Thursday) = 168 + 159 = 327

Cycling (Monday + Tuesday) = 146 + 155 = 301

Ratio = 327 : 301

Answer: 327:301

Calculation:

Monday: |123 - 146| = 23

Tuesday: |139 - 155| = 16

Wednesday: |114 - 132| = 18

Thursday: |127 - 149| = 22

Answer: Monday, Difference = 23

Calculation:

Zumba participants = 157 + 143 + 168 + 159 = 627

10% of Zumba participants on each day = 10% of 627 = 62.7

Unique Zumba participants = 627 - 62.7 = 564.3 (rounded to 564)

Answer: 564 unique participants

Paresh does 40% work in 10 days

Paresh can do the 100% work in

\( = \frac{{10}}{{1 \times \left( {\frac{{40}}{{100}}} \right)}}\)

=10 × 100/40 = 25 days

Rest work =1 – (4/10) = 6 / 10 = 3 / 5

Paresh & Bhave finished the remaining work in 6 days

The whole work will be done by Paresh & Bhave in 5/3 × 6 = 10 days

Paresh’s 1 day’s work = 1/25

And (Paresh + Bhave)’s 1 day’s work = 1/10

∴ Bhave’s 1 day’s work = (1/10) – (1/25)

= 15 / 250

∴ Bhave will do the whole work in 16 2/3 days

Given:

Saving of Rahul = Rs. 9600

Expenditure on Shopping = 15%

Expenditure on School fees = 20% of the remaining = 20% of 85% = 17%

Expenditure on other items = 20%

Concept:

Income = Expenditure + Saving

Calculation:

Let total income be 100%

Savings = 100% – (15% + 17% + 20%)

⇒ 100% – 52%

⇒ 48%

Now, 48% = Rs. 9600

⇒ 100% = 20000

∴ The required result will be Rs. 20,000

Let the length of train be x

Then, the length of platform is also x

Speed of train = 54 km/hr = 54 × (5/18) m/s = 15 m/s

∴ According to question

(x + x)/15 = 36

⇒ 2x = 540

⇒ x = 270 m

The length of the platform is 270 m

Now, a train of length 230 m crosses the platform in 25 sec

Let the speed of train be v m/s

⇒ (270 + 230)/v = 25

⇒ v = 20m/s

Speed of train = 20 m/s = 20 × (18/5) = 72 km/hr

∴ The speed of the train is 72 km/hr

Given:

8, 27, 64, 343, 1331.

Calculation:

The logic follows here is:

(2)³ = 8

(3)³ = 27

(5)³ = 125

(7)³ = 343

(11)³ = 1331

By observing the above pattern, we can see that

Except 64, all numbers are cube of prime numbers

∴ 64 is the wrong term.

Given:

8, 4, 4, 8, 33, 256

Calculation:

The series follows the following pattern,

8 × (1/2) = 4

4 × 1 = 4

4 × 2 = 8

8 × 4 = 32 ≠ 33

32 × 8 = 256

Since 32 will come in place of 33

∴ The required wrong number will be 33.

Given:

25, 27, 33, 40, 51, 65

Calculation:

The pattern of the given series can be expressed as,

Since 32 will come in place of 33

∴ The required wrong number will be 33.

Given:

50, 51, 59, 86, 140

Calculation:

The series follows the following pattern,

50 + 1³ = 51

51 + 2³ = 59

59 + 3³ = 86

86 + 4³ = 150 ≠ 140

Since 150 will come in place of 140

∴ The required wrong number will be 140.

Given:

154, 145, 161, 136, 174, 123, 187

Calculation:

The series follows the following pattern

⇒ 154 - 3² = 145

⇒ 145 + 4² = 161

⇒ 161 - 5² = 136

⇒ 136 + 6² = 172 ≠ 174

⇒ 172 - 7² = 123

⇒ 123 + 8² = 187

∴ 174 is wrong and should be replaced by 172.

Alternate Method

∴ 174 is wrong and should be replaced by 172.

Calculation:

Number of laptops sold in 2010 = 150

Number of laptops sold in 2014 = 350

Percentage increase = \(\frac{350 - 150}{150} \times 100 = \frac{200}{150} \times 100 = 133.33%\)

Answer: 133.33%

Calculation:

Total laptops sold over 5 years = 150 + 250 + 200 + 300 + 350 = 1250

Average = \( \frac{1250}{5} = 250\)

Answer: 250 laptops

Calculation:

Number of laptops sold in 2013 = 300

Revenue per laptop without discount = ₹15,000

Revenue lost per laptop due to 10% discount = 15,000 × 10% = ₹1,500

Total revenue lost = 300 × ₹1,500 = ₹450,000

Answer: ₹450,000

Calculation:

Highest sales = 350 (in 2014)

Lowest sales = 150 (in 2010)

Difference = 350 - 150 = 200

Answer: 200 laptops

Calculation:

Total laptops needed for an average of 320 over 6 years = 320 × 6 = 1920

Total laptops sold from 2010 to 2014 = 150 + 250 + 200 + 300 + 350 = 1250

Laptops needed in 2015 = 1920 - 1250 = 670

Answer: 670 laptops

Given :

Quantity of mixture = 20 L

Ratio of milk and water = 2 : 3

Calculations :

Quantity of milk = (2/5) × 20 = 8 L

Quantity of water in mixture = 20 - 8 = 12 L

Let x L of water be mixed

⇒ 8/(12 + x) = 1/3

⇒ x = 12

∴ 12 L of water should be mixed.

Given:

A is 4 years older than B.

The sum of present age of A and C is 48 years.

The ratio of present age of B and C is 7 : 4.

Calculation:

Let the age of A be x.

The age of C = 48 - x

A is 4 years older than B.

The age of B = x - 4

The ratio of present age of B and C is 7 : 4.

⇒(x - 4)/(48 - x) = 7/4

⇒ 4x - 16 = 336 - 7x

⇒ 11x = 352

⇒ x = 32

The age of A two years ago = 32 - 2 = 30 years

∴ The age of A two years ago is 30 years.

Given:

Speed in downstream = 30 km/h

Speed in upstream = 20 km/h

Formula:

Speed of downstream = Speed of a boat in still water + Speed of water current

Speed of upstream = Speed of a boat in still water - Speed of water current

Speed = Distance/Time

Calculation:

Let the speed of the boat in still water be x km/h and speed of water current be y km/h

Now, according to question

x + y = 30  ----(1)

x – y = 20  ----(2)

Now,

From equation (1) and (2), we get

⇒ 2x = 50

⇒ x = 50/2

⇒ x = 25

∴ The Speed of the boat in still water is 25 km/h

Calculation:

The product is sold at a 20% discount.

Discounted price = MP - 20% of MP = 3750 - (20/100 × 3750)

= 3750 - 750 = ₹3000

So, Selling Price (SP) = ₹3000

Profit = SP - CP

CP = SP - Profit

CP = ₹3000 - ₹250 = ₹2750

Difference = MP - CP

Difference = ₹3750 - ₹2750 = ₹1000

Answer: The difference between the marked price and the cost price is ₹1000.

Concept used:

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

494 - 15 × 8 + 43 - 77 = ?2

⇒ 494 - 120 + 64 - 77 = ?2

⇒ 494 -120 - 13 = ?2

⇒ 494 - 133 = ?2

⇒ 361 = ?2

∴ The value of ? is 19.

Given:

336 ÷ 24 × ∛343 + (7)2 = ?

Concept:

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

Calculation:

336 ÷ 24 × ∛343 + (7)2 = ?

⇒ 14 × 7 + 49 = ?

⇒ 98 + 49 = ?

∴ ? = 147

GIVEN :

\(1{2 \over 13}\)× 338 + ? = 92 × 3 + 237

CONCEPT :

CALCULATION :

\(1{2 \over 13}\) × 338 + ? = 92 × 3 + 237

⇒ 15/13 × 338 + ? = 243 + 237

⇒ 390 + ? = 480

⇒ ? = 90

Given:

60% of 800 + 192 ÷ 32 = 6 × ?

Concept:

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

Calculation:

60% of 800 + 192 ÷ 32 = 6 × ?

⇒ 480 + 192 ÷ 32 = 6 × ?

⇒ 480 + 6 = 6 × ?

⇒ 486 = 6 × ?

⇒ 81 = ?

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, 

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next, 

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, 

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated

25% of (15 × 24 ÷ 18 + 12) = ?

⇒ 25% of (15 × 4/3 + 12) = ?

⇒ 25% of (5 × 4 + 12) = ?

⇒ 25% of (20 + 12) = ?

⇒ 25% of 32 = ?

⇒ ? = 8

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, 

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next, 

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, 

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

383.99 ÷ 11.92 × 4.14 – 119.99 = ∛?

⇒ 384 ÷ 12 × 4 – 120 = ∛?

⇒ 32 × 4 – 120 = ∛?

⇒ 128 – 120 = ∛?

⇒ 8 = ∛?

∴ ? = 512

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, 

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next, 

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, 

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

165 + 156 ÷ 13 – 3² × 19 = ?

⇒ 165 + 156 ÷ 13 – 9 × 19 = ?

⇒ 165 + 12 – 9 × 19 = ?

⇒ 165 + 12 – 9 × 19 = ?

⇒ 177 – 171 = ?

∴ ? = 6

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, 

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next, 

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, 

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

(20 × 45) – (70 × 3) + 360 = ? × 6

⇒ 900 – 210 + 360 = ? × 6

⇒ 1260 – 210 = ? × 6

⇒ 1050 = ? × 6

∴ ? = 175

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

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and in the bracket, 

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next, 

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated, 

Step-4: Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated 

40% of 450 + 85% of 800 = ?

⇒ 180 + 680 = ?

∴ ? = 860

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

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Given expression is

(16/23 × 47/288 × 92/141) + ?  = 1/9

⇒ (16/23 × 47/288 × 92/141) + ?  = 1/9

⇒ (4 × 1/72 × 4/3) + ?  = 1/9

⇒ (4 × 1/18 × 1/3) + ?  = 1/9

⇒ (2/27) + ?  = 1/9 × 3/3

⇒  ? = 3/27 – 2/27

∴ ? = 1/27