RRB Clerk Prelims Memory Based Paper (Held On: 13 Dec 2025 Shift 1) (Previous Year Paper)

Given: 6 persons, namely, P, Q, R, S, T, and U, are going to an amusement park on six different days, Monday to Saturday, but not necessarily in the same order. 

1. T went to the amusement park on Tuesday.

 

2. Only two people went between T and Q.

 

3. U went to the amusement park either immediately after or immediately before Q.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. S went to the amusement park three days before U.

 

5. R went to the amusement park before P.

 

6. P didn't go to the amusement park immediately after Q.

Now, we can eliminate Case 2.

This is our final arrangement,

 

Hence, the correct answer is U.

Given: 6 persons, namely, P, Q, R, S, T, and U, are going to an amusement park on six different days, Monday to Saturday, but not necessarily in the same order. 

1. T went to the amusement park on Tuesday.

 

2. Only two people went between T and Q.

 

3. U went to the amusement park either immediately after or immediately before Q.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. S went to the amusement park three days before U.

 

5. R went to the amusement park before P.

 

6. P didn't go to the amusement park immediately after Q.

Now, we can eliminate Case 2.

This is our final arrangement,

 

Hence, the correct answer is Three.

Given: 6 persons, namely, P, Q, R, S, T, and U, are going to an amusement park on six different days, Monday to Saturday, but not necessarily in the same order. 

1. T went to the amusement park on Tuesday.

 

2. Only two people went between T and Q.

 

3. U went to the amusement park either immediately after or immediately before Q.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. S went to the amusement park three days before U.

 

5. R went to the amusement park before P.

 

6. P didn't go to the amusement park immediately after Q.

Now, we can eliminate Case 2.

This is our final arrangement,

 

Hence, the correct answer is Three.

Given: 6 persons, namely, P, Q, R, S, T, and U, are going to an amusement park on six different days, Monday to Saturday, but not necessarily in the same order. 

1. T went to the amusement park on Tuesday.

 

2. Only two people went between T and Q.

 

3. U went to the amusement park either immediately after or immediately before Q.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. S went to the amusement park three days before U.

 

5. R went to the amusement park before P.

 

6. P didn't go to the amusement park immediately after Q.

Now, we can eliminate Case 2.

This is our final arrangement,

 

Hence, the correct answer is Thursday - P.

Given: 6 persons, namely, P, Q, R, S, T, and U, are going to an amusement park on six different days, Monday to Saturday, but not necessarily in the same order. 

1. T went to the amusement park on Tuesday.

 

2. Only two people went between T and Q.

 

3. U went to the amusement park either immediately after or immediately before Q.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

4. S went to the amusement park three days before U.

 

5. R went to the amusement park before P.

 

6. P didn't go to the amusement park immediately after Q.

Now, we can eliminate Case 2.

This is our final arrangement,

 

Hence, the correct answer is Q.

Given information can be represented as follows,

Thus ‘cricket’ is coded as ‘mo’.

Given information can be represented as follows,

Thus ‘se’ is code for ‘some' or ‘football’.

Given information can be represented as follows,

Code for ‘most’ is ‘pi’.

Code for ‘children’ is ‘ja’.

Code for ‘are’ is ‘da’.

Code for ‘cricket’ is ‘mo’.

Code for ‘good’ and ‘at’ will be some different code which is not present in present codes.

Hence, the possible coding will be “pi ja da su ta mo”.

Given information can be represented as follows,

<br>

Code ‘pi’ is for ‘most’.

Code ‘sun’ is for ‘love’.

Code ‘ko’ is for ‘some’ or ‘football’.

Code ‘ja’ is for ‘children’.

Code ‘za’ is a new code.

Hence, none of the above sentences can be coded as ‘za pi sun ko ja’.

Given information can be represented as follows,

Code for ‘Some’ is ‘se’ or ‘ko’.

Code for ‘are’ is ‘da’

Code for ‘playing’ is ‘sa’.

So, code for ‘kabbadi’ can be ‘jin’ or ‘te’.

The least possible diagram for the given statements is as follows

Conclusions:

I. Some candle being light is a possibility.→ False (Any part of wax can never be light)

II. Some wax is not light. → True (It is definitely true)          

Hence, only II follows.

The least possible diagram for the given statements is as follows

Conclusions:

I. Few fruits are green → True (It is definitely true)

II. No fruit is green → False (It is definitely false)

Hence, only I follows.

The least possible diagram for the given statements is as follows

Conclusions:

I. Some cooler is air conditioner. → False (It is possible but not definite)

II. No cooler is air conditioner. → False (It is possible but not definite)

Both are complementary conclusions.

Hence, either I or II follows.

Given: Six persons, namely, J, K, L, M, N, and P, have different weights.

1. J is heavier than exactly two persons. 

_ > _ > _ > J > _ > _

2. P is heavier than exactly one person. 

_ > _ > _ > J > P > _

3. N is neither the heaviest nor the lightest person. 

Case 1: _ > _ > N > J > P > _

Case 2: _ > N > _ > J > P > _

4. K is heavier than L but lighter than M.

Case 1: M > K > N > J > P > L

Case 2: M > N > K > J > P > L

5. N is not heavier than K.

Now, we can eliminate Case 2.

This is our final arrangement,

Case 1: M > K > N > J > P > L

Hence, the correct answer is M.

Given: Six persons, namely, J, K, L, M, N, and P, have different weights.

1. J is heavier than exactly two persons. 

_ > _ > _ > J > _ > _

2. P is heavier than exactly one person. 

_ > _ > _ > J > P > _

3. N is neither the heaviest nor the lightest person. 

Case 1: _ > _ > N > J > P > _

Case 2: _ > N > _ > J > P > _

4. K is heavier than L but lighter than M.

Case 1: M > K > N > J > P > L

Case 2: M > N > K > J > P > L

5. N is not heavier than K.

Now, we can eliminate Case 2.

This is our final arrangement,

Case 1: M > K > N > J > P > L

Hence, the correct answer is Two.

Given: Six persons, namely, J, K, L, M, N, and P, have different weights.

1. J is heavier than exactly two persons. 

_ > _ > _ > J > _ > _

2. P is heavier than exactly one person. 

_ > _ > _ > J > P > _

3. N is neither the heaviest nor the lightest person. 

Case 1: _ > _ > N > J > P > _

Case 2: _ > N > _ > J > P > _

4. K is heavier than L but lighter than M.

Case 1: M > K > N > J > P > L

Case 2: M > N > K > J > P > L

5. N is not heavier than K.

Now, we can eliminate Case 2.

This is our final arrangement,

Case 1: M > K > N > J > P > L

Hence, the correct answer is L.

Given word: ADVENTUREFUL

First, fourth, sixth, and tenth letters of the word = A, E, T, F

Meaningful word - FATE (your future; something that happens to you) and FEAT (something you do that shows great strength, skill or courage)

Hence, the correct answer is Y.

Given: 7 boxes, namely, J, K, L, M, N, O, and P, are placed one above the other in a stack, but not necessarily in the same order.

1. Box N is placed third from the bottom.

 

2. Only one box is placed between N and P.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

3. Three boxes are placed between P and L.

 

4. Box K is placed immediately below L.

Now, we can eliminate Case 2.

 

5. Box M is placed two boxes above Box K.

 

6. Box J is placed above Box O.

This is our final arrangement,

 

Hence, the correct answer is Box P.

Given: 7 boxes, namely, J, K, L, M, N, O, and P, are placed one above the other in a stack, but not necessarily in the same order.

1. Box N is placed third from the bottom.

 

2. Only one box is placed between N and P.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

3. Three boxes are placed between P and L.

 

4. Box K is placed immediately below L.

Now, we can eliminate Case 2.

 

5. Box M is placed two boxes above Box K.

 

6. Box J is placed above Box O.

This is our final arrangement,

 

Hence, the correct answer is Four.

Given: 7 boxes, namely, J, K, L, M, N, O, and P, are placed one above the other in a stack, but not necessarily in the same order.

1. Box N is placed third from the bottom.

 

2. Only one box is placed between N and P.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

3. Three boxes are placed between P and L.

 

4. Box K is placed immediately below L.

Now, we can eliminate Case 2.

 

5. Box M is placed two boxes above Box K.

 

6. Box J is placed above Box O.

This is our final arrangement,

 

Hence, the correct answer is Box M.

Given: 7 boxes, namely, J, K, L, M, N, O, and P, are placed one above the other in a stack, but not necessarily in the same order.

1. Box N is placed third from the bottom.

 

2. Only one box is placed between N and P.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

3. Three boxes are placed between P and L.

 

4. Box K is placed immediately below L.

Now, we can eliminate Case 2.

 

5. Box M is placed two boxes above Box K.

 

6. Box J is placed above Box O.

This is our final arrangement,

 

Hence, the correct answer is More than four.

Given: 7 boxes, namely, J, K, L, M, N, O, and P, are placed one above the other in a stack, but not necessarily in the same order.

1. Box N is placed third from the bottom.

 

2. Only one box is placed between N and P.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

 

3. Three boxes are placed between P and L.

 

4. Box K is placed immediately below L.

Now, we can eliminate Case 2.

 

5. Box M is placed two boxes above Box K.

 

6. Box J is placed above Box O.

This is our final arrangement,

 

Except for K-P, in all other pairs, only one box is placed between them.

Hence, the correct answer is K - P.

Given statements:

L ≤ T< N > O ≥ P

Conclusions:

I. O < T → According to statement T< N > O → Thus clear relationship between T and O cannot be established, hence false.

II. P ≤ N → According to statement N > O ≥ P → N > P → Hence false.

Hence neither conclusion I nor II follows.

Given statement:

U ≥ Q = R > S > T

Conclusions:

I. U ≥ T → According to statement U ≥ Q = R > S > T → U ≥ R > S > T → U > S > T → U> T → hence false.

II. T < Q → According to statement Q = R > S > T → Q > S > T → Q > T → hence true.

Hence, only conclusion II follows.

Given statement: S ≥ K < M > W ≥ Z

Conclusions:

I. W < K → False (as K < M > W → relationship between W and K is not definite)

II. Z = S → False (as S ≥ K < M > W ≥ Z → relationship between Z and S is not definite)

Hence neither conclusion I nor II follows.

Here, S represents Symbol,

V represents Vowels,

C represents Consonants.

Thus in the given sequence there are only two vowels ( C A S and L O Q ) satisfies the given condition.

Hence the answer is 2.

By following the procedure in the sequence,

Thus, the letter S replaces Thursday, Hence S is the answer.

Thus, the answer is A.

Observe the pattern from the sequence.

Hence the answer is EDZ.

When alternate letter/ number is dropped, the sequence thus formed is,

Hence the answer is V.

Given: Eight persons, namely, A, B, C, D, E, F, G, and H, sit around the square table. Four people sit on the corners, while four people sit on the sides. Persons on each corner face towards the center while persons on each side face away from the center.

1. C sits third to the right of D.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. Only two people sit between C and F.

3. B sits immediately to the left of F.

4. Three people sit between B and H.

5. E sits immediately left of H and doesn't sit at the side of the table.

Now, we can eliminate Case 2.

6. E and G face in the same direction.

This is our final arrangement,

Hence, the correct answer is A.

Given: Eight persons, namely, A, B, C, D, E, F, G, and H, sit around the square table. Four people sit on the corners, while four people sit on the sides. Persons on each corner face towards the center while persons on each side face away from the center.

1. C sits third to the right of D.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. Only two people sit between C and F.

3. B sits immediately to the left of F.

4. Three people sit between B and H.

5. E sits immediately left of H and doesn't sit at the side of the table.

Now, we can eliminate Case 2.

6. E and G face in the same direction.

This is our final arrangement,

Hence, the correct answer is Three.

Given: Eight persons, namely, A, B, C, D, E, F, G, and H, sit around the square table. Four people sit on the corners, while four people sit on the sides. Persons on each corner face towards the center while persons on each side face away from the center.

1. C sits third to the right of D.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. Only two people sit between C and F.

3. B sits immediately to the left of F.

4. Three people sit between B and H.

 

5

. E sits immediately left of H and doesn't sit at the side of the table.

Now, we can eliminate Case 2.

6. E and G face in the same direction.

This is our final arrangement,

Hence, the correct answer is G.

Given: Eight persons, namely, A, B, C, D, E, F, G, and H, sit around the square table. Four people sit on the corners, while four people sit on the sides. Persons on each corner face towards the center while persons on each side face away from the center.

1. C sits third to the right of D.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. Only two people sit between C and F.

3. B sits immediately to the left of F.

4. Three people sit between B and H.

5. E sits immediately left of H and doesn't sit at the side of the table.

Now, we can eliminate Case 2.

6. E and G face in the same direction.

This is our final arrangement,

Except for E, all others face outside the table.

Hence, the correct answer is E.

Given: Eight persons, namely, A, B, C, D, E, F, G, and H, sit around the square table. Four people sit on the corners, while four people sit on the sides. Persons on each corner face towards the center while persons on each side face away from the center.

1. C sits third to the right of D.

Here, we have 2 Cases, i.e., Case 1 and Case 2.

2. Only two people sit between C and F.

3. B sits immediately to the left of F.

4. Three people sit between B and H.

5. E sits immediately left of H and doesn't sit at the side of the table.

Now, we can eliminate Case 2.

6. E and G face in the same direction.

This is our final arrangement,

Hence, the correct answer is C.

Given word: CONFIDENTS

 

Clearly, 4 vowels are present.

Hence, the correct answer is Four.

We have drawn the figure according to the information given in the question,

In the above figure, the dashed line represents the distance between G and C.

By Pythagoras theorem,

GC² = 12² + 5² (because CD = 12 km and GD = DE – GE = 5 km)

GC² = 169

GC = 13 km

Hence, the distance between point G and point C is 13 km.

We have drawn the figure according to the information given in the question,

 Clearly, Suresh reaches at point E.

We have drawn the figure according to the information given in the question,

In the above figure, the dashed line represents the distance between Suresh’s final and starting point i.e., the distance between A and F.

By Pythagoras theorem,

AF² = 8² + 8² (because BA = 8 km and BF = 8 km)

AF² = 128

AF = $8\sqrt 2$ km

Hence, Suresh is $8\sqrt 2$ km far from the starting point.

Given Number: 729148536

 

Clearly, only 2 remain unchanged.

Hence, the correct answer is one.

CALCULATION

Total profit of FIAT

= 1500 + 1600 + 1200 + 1400

= 5700

<br>

Total profit of AMCAT

= 1400 + 1800 + 1600 + 1100

= 5900

<br>

Required ratio

= 5700 : 5900

= 57 : 59

CALCULATION

Total profit of AMCAT = 5900

Average profit

= 5900 / 4

= 1475

Answer: 1475

CALCULATION

Profit in April = 1600

Profit in May = 1200

<br>

Decrease = 1600 − 1200 = 400

<br>

Percentage decrease

= (400 / 1600) × 100

= 25%

<br>

Answer: 25%

CALCULATION

Profit in June = 1400

<br>

Increase = 20% of 1400 = 280

<br>

Profit in July

= 1400 + 280

= 1680

<br>

Answer: 1680

CALCULATION

Combined profit in April

= 1600 + 1800

= 3400

<br>

x − 100 = 3400

x = 3500

Given:

Simple Interest (SI) = Rs. 3420

Rate (R) = 20%

Time (T) = 3 years

Formula used:

Simple Interest = (P × R × T) ÷ 100

Calculations:

SI = (P × 20 × 3) ÷ 100

⇒ 3420 = (60P) ÷ 100

⇒ 3420 = 0.6P

⇒ P = 3420 ÷ 0.6 = 5700

New principal = P − 200 = 5700 − 200 = 5500

Time = 2 years, Rate = 20%

SI = (5500 × 20 × 2) ÷ 100

⇒ SI = 2200

∴ The simple interest is Rs. 2200.

Given:

Ratio of length and breadth = 8 : 3

Area of rectangle = 384 m²

Formula used:

Area of rectangle = Length × Breadth

Calculations:

Let length = 8x

Let breadth = 3x

⇒ Area = 8x × 3x = 24x²

⇒ 24x² = 384

⇒ x² = 384 ÷ 24 = 16

⇒ x = 4

Length = 8 × 4 = 32 m

Breadth = 3 × 4 = 12 m

∴ Length = 32 m and Breadth = 12 m.

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.

{(8 + 8 ÷ 2 × 3) of 8}/2 – 3 = 2.5 × ?

⇒ {(8 + 4 × 3) × 8)}/2 – 3 = 2.5 × ?

⇒ 160/2 - 3 = 2.5 × ?

⇒ ? = 77/2.5

⇒ ? = 30.8

Concept used:

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

Calculation:

$42{\rm{}} \times {\rm{}}\left[ {\frac{1}{7}; \times ;\left( {91 + 49} \right) + 17\frac{1}{3}} \right]{\rm{}} = {\rm{}}?$

⇒ 42 × [(1/7) × 140 + (52/3)] = ?

⇒ 42 × [20 + (52/3)] = ?

⇒ ? = 42 × (112/3)

⇒ ? = 1568

∴ The value of ? is 1568

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

55 % of 6620 +∛681472 × (1/11) – 32² = ?

⇒ (55/100) × 6620 + ∛681472 × (1/11) - 32²  = ?

⇒ (11/20) × 6620 + ∛681472 × (1/11) – 1024 = ?

⇒ (11 × 331) + 88 × (1/11) – 1024 = ?

⇒ 3641 + 8 – 1024 = ?

⇒ 3649 – 1024 = ?

∴ ? = 2625

Given,

64 ÷ √? + 12 = 20

⇒ 64 ÷ √? = 8

⇒ 64 ÷ 8 = √?

⇒  8 = √?

∴  ? = 8² = 64.

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 

27 – [162 – (273 + 281) ÷ 2]

= 27 – [256 – 554/2]

= 27 – 256 + 277

= 48

555 - 210 + √? = 192 + 27 × 7 ÷ 21

⇒ 345 + √? = 361 + 9

⇒ 345 + √? = 370

⇒ √? = 25

⇒ ? = 625

GIVEN :

50% of 7800 ÷ 26 ÷ 25 = ?

CONCEPT :

CALCULATION :

50% of 7800 ÷ 26 ÷ 25 = ?

⇒ 3900 ÷ 26 ÷ 25 = ?

⇒ ? = 150 ÷ 25

∴ ? = 6

(4 × 3)4 ÷ (9 × 16)3 × (64 × 27)5 = (12)?

⇒ 12⁴ ÷ 12⁶ × 12¹⁵ = 12^?

⇒ 12^{4 - 6 + 15} = 12^?

Hence, ? = 4 - 6 + 15

∴ ? = 13

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 

[28 - {8 + 12 ÷ (16 - 12)}] = ?

⇒ [28 - {8 + 12 ÷ 4}] = ?

⇒ [28 - {8 + 3}] = ?

⇒ [28 - 11] = ?

∴ ? = 17

1280% of 75 ÷ 30 + ? = 80

⇒ 960 ÷ 30 + ? = 80

⇒ 32 + ? = 80

⇒ ? = 80 - 32 = 48

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 

(√361 × 36) ÷ 12 + √441 = ?

⇒ (19 × 36) ÷ 12 + 21 = ?

⇒ 684 ÷ 12 + 21 = ?

⇒ 57 + 21 = ?

∴ ? = 78

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

700 ÷ 14 × 8 – ? = 156 + 244

⇒ 50 × 8 – ? = 400

⇒ 400 – ? = 400

⇒ ? = 400 – 400 = 0

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

<br>

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

<br>

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

<br>

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

<br>

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

<br>

Given expression is

? + 303 = 519 – 90 + 182

⇒ ? + 303 = 519 – 90 + 324

⇒ ? + 303 = 843 – 90 

⇒ ?  = 753 – 303

⇒  ? = 450

∴ ? = 450

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

<br>

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

<br>

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

<br>

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

<br>

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

<br>

Given expression is

39 × 24 – 19 × 14 + (82 × 50%) = ? + 2

⇒ (936) – (266) + (64 × 50%) = ? + 2

⇒ (670) + (32) = ? + 2

⇒ 702 – 2 = ?

⇒ ?  = 700

∴ ? = 700

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

<br>

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

<br>

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

<br>

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

<br>

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

<br>

Given expression is

?2 = 27.27% × 7 × (272 – 162) – 3

⇒ ?2 = (3/11) × 7 × (729 – 256) – 3

⇒ ?2 = (3/11) × 7 × (473) – 3

⇒ ?2 = (3) × 7 × (43) – 3

⇒ ?2  = 903 – 3

∴ ? = 30

Given:

Speed of train = 72 km/hr

Time to cross pole = 25 sec

Increase in speed = 18 km/hr

Formula used:

Speed = Distance ÷ Time

Time = Distance ÷ Speed

Calculations:

Convert speed into m/sec

72 km/hr = 72 × 5 ÷ 18 = 20 m/sec

Length of train = Speed × Time

⇒ Length = 20 × 25 = 500 m

New speed = 72 + 18 = 90 km/hr

90 km/hr = 90 × 5 ÷ 18 = 25 m/sec

New time to cross pole = Distance ÷ Speed

⇒ Time = 500 ÷ 25 = 20 sec

Less time taken = 25 − 20 = 5 sec

∴ The train will take 5 seconds less to cross the pole.

Given:

Investment of Rina = Rs. 5600

Investment of Tina = Rs. 4000

Time = 1 year (same for both)

Profit of Rina = Rs. 8400

Formula used:

Profit ratio = Investment ratio (when time is same)

Calculations:

Ratio of investments = 5600 : 4000

⇒ Divide by 800

⇒ Investment ratio = 7 : 5

So, profit ratio of Rina : Tina = 7 : 5

Profit of Rina (7 parts) = 8400

⇒ Value of 1 part = 8400 ÷ 7 = 1200

⇒ Profit of Tina = 5 × 1200 = 6000

∴ The profit amount of Tina is Rs. 6000.

Calculation

Total students of College B = (x + 22)

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Total students in College B = 126 + 131 + 145 = 402

<br>

x + 22 = 402

x = 380

Calculation

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Difference = 145 − 118 = 27

<br>

Percentage = (27 / 118) × 100 = 22.88%

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Answer: 22.88% more

Calculation

Total Science students = 102 + 131 + 105 = 338

Average = 338 / 3 = 112.67 students

Answer: 112.67 students

Calculation

<br>

Total students in College A = 148 + 102 + 127 = 377

Total students in College C = 118 + 105 + 134 = 357

Required difference = 377 - 357 = 20

Calculation

Average Economics = (148 + 126 + 118) / 3 = 130.67

Average Sociology = (127 + 145 + 134) / 3 = 135.33

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Difference = 135.33 − 130.67 = 4.66

Given:

Cost Price (CP) = 100

Increase in CP = 40%

Discount given = 20%

Formula used:

Increased Price = CP × (1 + Increase%)

Selling Price = Marked Price × (1 − Discount%)

Profit % = (Profit ÷ CP) × 100

Calculations:

Let CP = 100

Increased price = 100 × (1 + 40/100)

⇒ Increased price = 100 × 1.4 = 140

Discount = 20% on 140

Selling Price (SP) = 140 × (1 − 20/100)

⇒ SP = 140 × 0.8 = 112

Profit = SP − CP

⇒ Profit = 112 − 100 = 12

⇒ Profit % = (12 ÷ 100) × 100 = 12%

∴ The profit percentage is 12%.

Given:

Weight of item A = 200 kg

Weight of item C is 20% more than item A

Weight of item B is 25% more than item C

Formula used:

Increased weight = Original weight × (1 + Increase%)

Calculations:

Weight of item C = 200 × (1 + 20/100)

⇒ Weight of item C = 200 × 1.2 = 240 kg

Weight of item B = 240 × (1 + 25/100)

⇒ Weight of item B = 240 × 1.25 = 300 kg

Total weight = Weight of A + Weight of B + Weight of C

⇒ Total weight = 200 + 300 + 240

⇒ Total weight = 740 kg

∴ The total weight of all three items together is 740 kg.

Given:

Ratio of milk and water = 9 : 5

Water added = 10 L

New ratio of milk and water = 9 : 7

Formula used:

If ratio is a : b, then quantities = ax and bx

Calculations:

Let initial quantity of milk = 9x

Let initial quantity of water = 5x

After adding water:

Milk = 9x

Water = 5x + 10

⇒ (9x) : (5x + 10) = 9 : 7

⇒ 7 × 9x = 9 × (5x + 10)

⇒ 63x = 45x + 90

⇒ 63x − 45x = 90

⇒ 18x = 90

⇒ x = 5

Initial quantity of milk = 9x = 9 × 5 = 45 L

∴ The initial quantity of milk is 45 L.

Considering the given series

5, 4, 7, 20, 79, 386, 2363

The logic of the given series can be explained as

5 × 1 – 1 = 4

4 × 2 – 1 =7

7 × 3 – 1 = 20

20 × 4 – 1 = 79

79 × 5 – 1 = 394

394 × 6 – 1 = 2363

∴ wrong term of the series is 386

The pattern followed by the series is

123 - 4 = 119 

119 - 8 = 111

111 - 12 = 99

99 - 16 = 83 

83 - 20 = 63

The series follows the following pattern:

12 × 2 = 24

24 ÷ 3 = 8

8 × 4 = 32

32 ÷ 5 = 6.4

6.4 × 6 = 38.4

∴ The wrong term in the series is 6.2.

The series follows the following pattern:

11² - 2 = 119

12² - 3 = 141

13² - 5 = 164

14² - 7 = 189

15² - 11 = 214

16² - 13 = 243

∴ The wrong term in the series is 213.

The series follows the following pattern:

1440 ÷ 6 = 240

240 ÷ 5 = 48

48 ÷ 4 = 12

12 ÷ 3 = 4 

4 ÷ 2 = 2

∴ The wrong term in the series is 16.

Given:

Ratio of present age of Dimpi and Kajal = 3 : 4

Ratio of age of Dimpi before 3 years and Kajal after 6 years = 4 : 7

Formula used:

If present ages are in ratio a : b, then ages = ax and bx

Calculations:

Let present age of Dimpi = 3x

Let present age of Kajal = 4x

Dimpi’s age before 3 years = 3x − 3

Kajal’s age after 6 years = 4x + 6

⇒ (3x − 3) : (4x + 6) = 4 : 7

⇒ 7 × (3x − 3) = 4 × (4x + 6)

⇒ 21x − 21 = 16x + 24

⇒ 21x − 16x = 24 + 21

⇒ 5x = 45

⇒ x = 9

Present age of Dimpi = 3x = 3 × 9 = 27 years

∴ The present age of Dimpi is 27 years.

Given:

Increase of number = 40%

Decrease of number = 80%

Difference between both results = 192

Formula used:

Increased value = Number × (1 + increase%)

Decreased value = Number × (1 − decrease%)

Calculations:

Let the number = x

Increased number = x × (1 + 40/100) = 1.4x

Decreased number = x × (1 − 80/100) = 0.2x

⇒ Difference = 1.4x − 0.2x

⇒ 1.2x = 192

⇒ x = 192 ÷ 1.2 = 160

25% of the number = 25/100 × 160

⇒ 25% = 40

∴ 25% of the number is 40.

Detailed Solution:

Cricketer makes total runs in 26 innings = 28 × 26 = 728

Average after he makes a total 272 runs in his next five innings = (728 + 272)/31 = 1000/31 = 32.258

Increase in his average = 32.258 – 28 = 4.258

Short Trick:

For maintain the average at 28 after he makes a total 272 runs in his next five innings, he needs 140 runs.

∴ Remaining runs = 272 – 140 = 132

∴ These 132 runs will spend in increasing the average of cricketer after 31 innings.

⇒ Increase in his average = 132/31 = 4.258