RRB Clerk Prelims Memory Based Paper (Held On: 06 Dec 2025 Shift 1) (Previous Year Paper)

Given: Six persons, namely, P, Z, K, S, T, and M, are going on vacation on two different dates, i.e., 21st and 26th of three different months - March, April, and May, of the same year, but not necessarily in the same order. 

1. P goes on an even-numbered date of the odd-numbered days in the month before May. 

 

2. Three people go between Z and P. 

 

3. K goes in a month that has an odd number of days, and after P. 

 

4. Three people go between S and K. 

 

5. More than one person goes between S and T. 

 

This is our final arrangement,

 

Hence, the correct answer is M.

Given: Six persons, namely, P, Z, K, S, T, and M, are going on vacation on two different dates, i.e., 21st and 26th of three different months - March, April, and May, of the same year, but not necessarily in the same order. 

1. P goes on an even-numbered date of the odd-numbered days in the month before May. 

 

2. Three people go between Z and P. 

 

3. K goes in a month that has an odd number of days, and after P. 

 

4. Three people go between S and K. 

 

5. More than one person goes between S and T. 

 

This is our final arrangement,

 

Hence, the correct answer is Four.

Given: Six persons, namely, P, Z, K, S, T, and M, are going on vacation on two different dates, i.e., 21st and 26th of three different months - March, April, and May, of the same year, but not necessarily in the same order. 

1. P goes on an even-numbered date of the odd-numbered days in the month before May. 

 

2. Three people go between Z and P. 

 

3. K goes in a month that has an odd number of days, and after P. 

 

4. Three people go between S and K. 

 

5. More than one person goes between S and T. 

 

This is our final arrangement,

 

Hence, the correct answer is T.

Given: Six persons, namely, P, Z, K, S, T, and M, are going on vacation on two different dates, i.e., 21st and 26th of three different months - March, April, and May, of the same year, but not necessarily in the same order. 

1. P goes on an even-numbered date of the odd-numbered days in the month before May. 

 

2. Three people go between Z and P. 

 

3. K goes in a month that has an odd number of days, and after P. 

 

4. Three people go between S and K. 

 

5. More than one person goes between S and T. 

 

This is our final arrangement,

 

Hence, the correct answer is More than four.

Given: Six persons, namely, P, Z, K, S, T, and M, are going on vacation on two different dates, i.e., 21st and 26th of three different months - March, April, and May, of the same year, but not necessarily in the same order. 

1. P goes on an even-numbered date of the odd-numbered days in the month before May. 

 

2. Three people go between Z and P. 

 

3. K goes in a month that has an odd number of days, and after P. 

 

4. Three people go between S and K. 

 

5. More than one person goes between S and T. 

 

This is our final arrangement,

 

Hence, the correct answer is 26 April - T.

Thus ‘kit’ is decoded as ‘dish’.

Thus ‘window car’ is coded as ‘tit rit’.

Thus ‘sand’ is coded as ‘fit’.

Thus ‘tit’ is decoded as ‘window’.

Thus ‘lit’ is decoded as ‘bad’.

Statements:

S ≥ M > Y < Z = F > T

Conclusions:

I. F > M → True

II. M > T → False as there is no clear relation between M and T can be established.

Hence, only conclusion I follows.

Statements:

P ≤ Q ≥ R > S < T

Conclusions:

I. P < T → False as there is no clear relation between P and T can be established.

II. T > Q → False as Q > S & T > S. So, there is no clear relation between T and Q can be established.

Hence, neither of the conclusions I nor II follow.

Statements:

L ≤ M > N > O ≥ P

Conclusions:

I. P < M → True

II. L ≤ N → False as L < M & N < M. So, there is no clear relation between L and N can be established.

Hence, only conclusion I follows.

Given statement: D > H > A > I < C < X = L

Conclusion:

I. A ≥ L → Not true as A > I < C < X = L therefore there is no definite relation between them.

II. D > C → Not true as D > H > A > I < C therefore there is no definite relation between them.

Hence neither conclusion I nor conclusion II is true.

Given series: YTE   IML   FOP   USK   CTA   JOI

On arranging the words as per the order in which they would appear in the dictionary, we get:

CTA   FOP   IML   JOI   USK   YTE

Hence, JOI is the third word from the right end.

Given series: YTE   IML   FOP   USK   CTA   JOI

The fifth word from the left end is CTA, 

C → D (The letter is increased by one)

T → V (The letter is increased by two)

A → D (The letter is increased by three)

Similarly, the fifth word from the right end is IML,

I → J

M → O

L → O

Hence, the replacement for IML will be JOO.

Given series: YTE   IML   FOP   USK   CTA   JOI

On replacing the first letter by its successor and the next two letters by their predecessors, we get:

ZSD   JLK   GNO   VRJ   DSZ   KNH

Hence, five words don’t have a vowel in them.

Given series: YTE   IML   FOP   USK   CTA   JOI

On interchanging the first and the last letter of each word.

ETY   LMI   POF   KSU   ATC   IOJ

Hence, four words end with consonants.

Given series: YTE   IML   FOP   USK   CTA   JOI

On arranging the letters of each word in alphabetical order, we get:

ETY   ILM   FOP   KSU   ACT   IJO

Hence, two words end with a vowel.

Here, persons are P, Q, R, S, T, U, V and W.

1. Q sits third to the right of P.

2. S and Q sit immediate right to each other.

3. Three persons sit between V and R.

4. T sits second to the left of R.

5. U neither faces outwards nor sits adjacent to Q. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Hence, W sits second to the left of U.

Here, persons are P, Q, R, S, T, U, V and W.

1. Q sits third to the right of P.

2. S and Q sit immediate right to each other.

3. Three persons sit between V and R.

4. T sits second to the left of R.

5. U neither faces outwards nor sits adjacent to Q. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, each pair represents persons sitting opposite to each other whereas V and P sits adjacent to each other.

Hence, the odd one is VP.

Here, persons are P, Q, R, S, T, U, V and W.

1. Q sits third to the right of P.

2. S and Q sit immediate right to each other.

3. Three persons sit between V and R.

4. T sits second to the left of R.

5. U neither faces outwards nor sits adjacent to Q. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, W sits immediate left to Q not immediate right. Therefore, statement I does not follow.

Only two persons sit between W and T when counted from the right of W. Therefore, statement II follows.

W sits opposite to P not T. Therefore, statement III does not follow.

Hence, the correct answer is only II follows.

Here, persons are P, Q, R, S, T, U, V and W.

1. Q sits third to the right of P.

2. S and Q sit immediate right to each other.

3. Three persons sit between V and R.

4. T sits second to the left of R.

5. U neither faces outwards nor sits adjacent to Q. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, P sits second to the right  of U. Therefore, statement I is correct.

Q sits second to the right of R. Therefore, statement II is correct.

S sits opposite to U. Therefore, statement III is correct.

Hence, the incorrect statement is None of these.

Here, persons are P, Q, R, S, T, U, V and W.

1. Q sits third to the right of P.

2. S and Q sit immediate right to each other.

3. Three persons sit between V and R.

4. T sits second to the left of R.

5. U neither faces outwards nor sits adjacent to Q. Therefore, case 2 is eliminated. Hence, the final arrangement is as follows

Since, Q sits opposite to T and P sits third to the right of Q.

Hence, the correct answer is P.

Preparing the family tree using the following symbols:

Hence, G is the daughter of F.

Preparing the family tree using the following symbols:

Hence, A is the grandfather of E.

Preparing the family tree using the following symbols:

Hence, B is the mother-in-law of D.

The least possible Venn diagram for the given statements is as follows

Conclusions:

I. Only a few chemicals are elements → True (As this can be deduced.)

II.  All element are an chemical is a possibility. → True (All reagents are chemical and only a few reagents are elements.)

Thus, All follows

The least possible diagram is as follows:

Conclusions:

I. Some Markets being Ground is a possibility - True

II. All Grounds being Markets is a possibility - True

Hence, both conclusions I and II are true

The least possible diagram is as follows:

Conclusions:

I. Some Cups are Plates - False (it is possible but not definite)

II. Some Pens are Plates - False (it is possible but not definite)

Hence, neither of the conclusions is true

Given Word: CHAMBER

The given word can be represented as follows,

We can see that only one Pair: A → C (backward)

Hence, One is the correct answer.

Given: Seven staff members, namely, G, H, I, J, K, L, and M, work in a company, each holding a different position among the following roles arranged from highest to lowest: Director, Deputy Director, Senior Manager, Manager, Assistant Manager, Supervisor, Clerk, but not necessarily in the same order.

1. Exactly two employees hold positions above H. 

 

2. There is exactly one position gap between H and M.

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

 

3. More than two positions separate I and M. 4. I is not the junior-most employee. 

 

5. More than two positions also separate J and I, and J holds a senior position compared to I.

Now, we can eliminate Case 2.

 

6. K does not work as the Manager or the Clerk.

 

7. L holds a position junior to G.

This is our final arrangement,

 

Hence, the correct answer is I.

Given: Seven staff members, namely, G, H, I, J, K, L, and M, work in a company, each holding a different position among the following roles arranged from highest to lowest: Director, Deputy Director, Senior Manager, Manager, Assistant Manager, Supervisor, Clerk, but not necessarily in the same order.

1. Exactly two employees hold positions above H. 

 

2. There is exactly one position gap between H and M.

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

 

3. More than two positions separate I and M. 4. I is not the junior-most employee. 

 

5. More than two positions also separate J and I, and J holds a senior position compared to I.

Now, we can eliminate Case 2.

 

6. K does not work as the Manager or the Clerk.

 

7. L holds a position junior to G.

This is our final arrangement,

 

Hence, the correct answer is One.

Given: Seven staff members, namely, G, H, I, J, K, L, and M, work in a company, each holding a different position among the following roles arranged from highest to lowest: Director, Deputy Director, Senior Manager, Manager, Assistant Manager, Supervisor, Clerk, but not necessarily in the same order.

1. Exactly two employees hold positions above H. 

 

2. There is exactly one position gap between H and M.

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

 

3. More than two positions separate I and M. 4. I is not the junior-most employee. 

 

5. More than two positions also separate J and I, and J holds a senior position compared to I.

Now, we can eliminate Case 2.

 

6. K does not work as the Manager or the Clerk.

 

7. L holds a position junior to G.

This is our final arrangement,

 

Hence, the correct answer is L.

Given: Seven staff members, namely, G, H, I, J, K, L, and M, work in a company, each holding a different position among the following roles arranged from highest to lowest: Director, Deputy Director, Senior Manager, Manager, Assistant Manager, Supervisor, Clerk, but not necessarily in the same order.

1. Exactly two employees hold positions above H. 

 

2. There is exactly one position gap between H and M.

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

 

3. More than two positions separate I and M. 4. I is not the junior-most employee. 

 

5. More than two positions also separate J and I, and J holds a senior position compared to I.

Now, we can eliminate Case 2.

 

6. K does not work as the Manager or the Clerk.

 

7. L holds a position junior to G.

This is our final arrangement,

 

Except for K, L, in all other pairs first person is immediately senior to the second person.

Hence, the correct answer is K, L.

Given: Seven staff members, namely, G, H, I, J, K, L, and M, work in a company, each holding a different position among the following roles arranged from highest to lowest: Director, Deputy Director, Senior Manager, Manager, Assistant Manager, Supervisor, Clerk, but not necessarily in the same order.

1. Exactly two employees hold positions above H. 

 

2. There is exactly one position gap between H and M.

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

 

3. More than two positions separate I and M. 4. I is not the junior-most employee. 

 

5. More than two positions also separate J and I, and J holds a senior position compared to I.

Now, we can eliminate Case 2.

 

6. K does not work as the Manager or the Clerk.

 

7. L holds a position junior to G.

This is our final arrangement,

 

Hence, the correct answer is The employee working as Supervisor.

Given word: COMPREHENSION

Operation: Consonants replaced with the next letter and vowels replaced with the previous letter

 

Since, after rearrangement O comes fifth from the right end.

Hence, the correct answer is 'O'.

Here, persons are A, B, C, D, E and F.

1. B got more marks than two persons.

2. C got more marks than B and D.

3. C does not get second highest marks.

Case – 1

              C  >    >     > B  >     >     

Case – 2  

                 >      >  C  >  B  >    >     

4. A got more marks than E and C. Therefore, case 1 is eliminated.

5. E got more marks than B and F.

6. D does not get least marks.

        A  >  E  >  C  >  B  >  D  >  F

7. The second lowest marks is 42 marks. Hence, the final arrangement is as follows

       A  >  E  >  C  >  B  >  D  >  F

                                        42        

Hence, F scores least marks.

Here, persons are A, B, C, D, E and F.

1. B got more marks than two persons.

2. C got more marks than B and D.

3. C does not get second highest marks.

Case – 1

              C  >    >     > B  >     >     

Case – 2  

                 >      >  C  >  B  >    >     

4. A got more marks than E and C. Therefore, case 1 is eliminated.

5. E got more marks than B and F.

6. D does not get least marks.

        A  >  E  >  C  >  B  >  D  >  F

7. The second lowest marks is 42 marks. Hence, the final arrangement is as follows

       A  >  E  >  C  >  B  >  D  >  F

                                        42        

Since, one person scores more marks than E and one person score less marks than D. Therefore, statement I is correct.

A score the highest marks. Therefore, statement II is correct.

C scores less marks than E not more marks. Therefore, statement III is incorrect.

Hence, the correct answer is Only I and II.

Here, persons are A, B, C, D, E and F.

1. B got more marks than two persons.

2. C got more marks than B and D.

3. C does not get second highest marks.

Case – 1

              C  >    >     > B  >     >     

Case – 2  

                 >      >  C  >  B  >    >     

4. A got more marks than E and C. Therefore, case 1 is eliminated.

5. E got more marks than B and F.

6. D does not get least marks.

        A  >  E  >  C  >  B  >  D  >  F

7. The second lowest marks is 42 marks. Hence, the final arrangement is as follows

       A  >  E  >  C  >  B  >  D  >  F

                                        42        

Since, A obtained more marks than E who obtained 68 marks. Therefore, possible marks obtained by A should be more than 68 marks.

Hence, the correct answer is 70 marks.

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,

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.

(12 × 5)² ÷ 60 + 8²

⇒ 60² ÷ 60 + 8²

⇒ 60 + 64

∴ Value is 124

⇒ {2/3 × 5/6 + 8/9 – 9/5}

⇒ {5/9 + 8/9 – 9/5}

⇒ 13/9 – 9/5}

⇒ - 16/45

Solve the given question, using following laws of indices,

Laws of Indices:

1. a^m × aⁿ = a^{{m + n}}
2. a^m ÷ aⁿ = a^{{m - n}}
3. (a^m)ⁿ = a^mn
4. (a)^(1/m) = ^m√a
5. (a)^-m = 1/a^m
6. (a)^m/n = ⁿ√a^m
7. (a)⁰ = 1

3√3 × 9³ ÷ 3⁵ = 3^{? + 2}

⇒ 3 × 3^1/2 × (3²)⁵ ÷ 3³ = 3^{? + 2}

⇒ 3^{1 + 1/2 + 10} ÷ 3³ = 3^{? + 2}

⇒ 3^11.5-3 = 3^{? + 2}

⇒ 11.5 - 3 = ? + 2

⇒ ? = 11.5 - 3 - 2 = 6.5

Given:

5/3 + 2/8 - ? × 8/2 = 4/5

Calculation:

⇒ ? × 8/2 = 5/3 + 2/8 - 4/5

⇒ ? × 4 = (5 × 8)/(3 × 8) + (2 × 3)/(8 × 3) - 4/5

⇒ ? × 4 = (40 + 6)/24 - 4/5

⇒ ? × 4 = (46 × 5)/(24 × 5) - (4 × 24)/(5 × 24)

⇒ ? × 4 = (230 - 96)/120

⇒ ? = 134/120 × ¼

∴ ? = 134/480 = 67/240

Concept used:

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

Calculation:

⇒ (2772 ÷ 42) + 184 = 28 × 34 + 184 - ?

⇒ 66 + 184 = 28 × 34 + 184 - ?

⇒ 66 + 184 = 952 + 184 - ?

⇒ 250 = 1136 - ?

⇒ ? = 1136 - 250

⇒ ? = 886

∴ The value of ? is 886

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.

Given expression is,

914 + 772 - 386 = 2 × ? + 600

⇒ 1686 - 386 = 2 × ? + 600

⇒ 1300 = 2 × ? + 600

⇒ 2 × ? = 700

⇒ ? = 350

Follow these BODMAS rules to solve the question

Step-1∶ The part of the equation containing '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 contains 'Division' and 'Multiplication' is solved

Step-4∶ At last, the part of the equation that contains 'Addition' and 'Subtraction' should be solved.

Given expression is,

60% of 80 ÷ 16 × 30% of 70 = ?

⇒ 48 ÷​ 16 ×​ 21 = ?

⇒ 3 ×​ 21 = ?​

∴ ? = 63

By using the law of indices∶

Laws of Indices∶-

1-∶ am × an = a{m + n}

2-∶ am ÷ an = a{m – n}

3-∶ [(am)n] = amn

4-∶ (a)(1/m) = m√a

5-∶ (a)(-m) = 1/am

6-∶ (a)(m/n) = n√am

7-∶ (a)0 = 1

Given expression is,

√324​ – √225​ + √289 = ? + √100

⇒ 18 – 15 + 17 = ? + 10

⇒ 20 = ? + 10

∴ ? = 10

√6889 ÷ 20.75 × 8 = ? ÷ 5 × 2.5

⇒ 83 ÷ 20.75 × 8 = ? ÷ 5 × 2.5

Applying BODMAS Rule;

⇒ 4 × 8 = ? ÷ 5 × 2.5

⇒ 32 = ? ÷ 5 × 2.5

⇒ 12.8 = ?/5

∴ ? = 64

Given 

28 + 5 × (?) ÷ 3 = 73

⇒ 5 × (?) ÷ 3 = 45

⇒ ? = 45 × 3/5

∴ ? = 27.

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.

⇒ 27⁸ × 3⁶ ÷ 27^3.4 × 27^2.4 = 3^?

⇒ 27⁸ × (3³)² ÷ 27^3.4 × 27^2.4 = 3^?

⇒ 27⁸ × (27)^-1.4 × 27^2.4 = 3^?

⇒ 27⁸ × (27)¹ = 3^?

⇒ 27⁹ = 3^?

⇒ (3³)⁹ = 3^?

⇒ 3²⁷ = 3^?

∴ ? = 27

Calculation:

81.25 ÷ 1.3 ÷ 2$\frac{1}{2}$ × 103 = 2.5 × ?

⇒ 62.5 ÷ $\frac{5}{2}$ × 1000 = 2.5 × ?

⇒ 62.5 × 2/5 × 1000 = 2.5 × ?

⇒ 25000 = 2.5 × ?

⇒ ? = 10000

∴ The correct answer is option (3).

(?)² = 805 – 625 – 36

⇒ (?)² = 144

⇒ ? = 12

224/7 + 132 - 30% of 560 = ? × 7

$\Rightarrow 32 + 169 - \frac{{30}}{{100}} \times 560 = 7x$

⇒ 32 + 169 - 168 = 7x

⇒ 201 - 168 = 7x

⇒ 33 = 7x

$\Rightarrow x = \frac{{33}}{7}$

$\therefore x = 4\frac{5}{7}$

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

⇒ (74 × 4 – 66) ÷ 5 = 89 + 37 - ?

⇒ (296 – 66) ÷ 5 = 89 + 37 – ?

⇒ 230 ÷ 5 = 126 – ?

⇒ ? = 126 – 46

⇒ ? = 80

∴ The value of ? is 80

Given:

Sum of three consecutive numbers = 105

Calculations:

If numbers are (n − 1), n, (n + 1)

Sum = 3n

Average of highest and second highest = (n + (n + 1)) ÷ 2

⇒ 3n = 105

⇒ n = 105 ÷ 3

⇒ n = 35

Numbers are: 34, 35, 36

⇒ Average of 35 and 36 = (35 + 36) ÷ 2

⇒ 71 ÷ 2 = 35.5

∴ Required average = 35.5.

Given:

A invests = 8000

B invests = 6000

B withdraws after 8 months

B’s profit share = 4200

Formula used:

Profit share ∝ Capital × Time

Calculations:

B’s ratio = 6000 × 8

A’s ratio = 8000 × 12

⇒ A’s value = 8000 × 12 = 96000

⇒ B’s value = 6000 × 8 = 48000

⇒ Ratio A : B = 96000 : 48000

⇒ = 2 : 1

If B’s share = 4200

⇒ 1 part = 4200

⇒ A’s share = 2 × 4200

⇒ = 8400

∴ A’s share of profit = Rs. 8400.

Given:

Total mixture = 44 litres

Milk is 6 litres more than water

Calculations:

Let water = x

Milk = x + 6

x + (x + 6) = 44

To make milk : water = 1 : 1 → Milk = New water

⇒ 2x + 6 = 44

⇒ 2x = 38

⇒ x = 19

⇒ Water = 19 L

⇒ Milk = 25 L

To make ratio 1 : 1 ⇒ Water must become 25 L

⇒ Extra water needed = 25 − 19

⇒ = 6 L

∴ Amount of water to be added = 6 litres.

Calculation

Male in A = 44.

If ratio = 4k : 5k and 4k = 44 → k = 11.

Females = 5k = 55.

Calculation

(60 + F) / 2 = 39 → 60 + F = 78 → F = 18.

Calculation

Males in E = 24.

Males in F = 24 × 1.25 = 30.

Females in F = 30 + 6 = 36.

Calculation

C = 32, D = 48 → Ratio = 32 : 48 = 2 : 3.

Calculation

A + C = 44 + 32 = 76.

7t + 6 = 76 → 7t = 70 → t = 10.

2 × 1 + 1 = 3

3 × 2 – 2 = 4

4 × 3 + 3 = 15

15 × 4 - 4 = 56

56 × 5 + 5 = 285

GIVEN :

8, 27, 125, 343, ?, 2197

EXPLANATION :

Logic: Given series follows the logic of the cube of prime numbers starting from 2.

8 = 2³

27 = 3³

125 = 5³

343 = 7³

? = 11³ = 1331

2197 = 13³

Hence, the missing term is 1331.

4 × 0.5 = 2

2 - 0.5 = 1.5

1.5 × 1 = 1.5

1.5 - 1 = 0.5

0.5 × 1.5 = 0.75

The pattern of the above series is as follows,

64 × 1.5 = 96

96 × 2.5 = 240

240 × 3.5 = 840

840 × 4.5 = 3780

3780 × 5.5 = 20790

GIVEN :

100, 150, 250, 350, 450, 650

CALCULATION :

The pattern of the above series is as follows,

50 is multiplied by prime numbers

50 × 2 = 100

50 × 3 = 150

50 × 5 = 250

50 × 7 = 350

50 × 11 = 550

50 × 13 = 650

Given:

40% of first number = 30% of second number + 68

Sum of numbers = 520

Calculations:

Let first number = x, second number = y

0.4x = 0.3y + 68

x + y = 520

⇒ 0.4x − 0.3y = 68

⇒ 4x − 3y = 680 (multiplying by 10)

x + y = 520

⇒ Multiply second eqn by 3:

⇒ 3x + 3y = 1560

Add with first:

⇒ (4x − 3y) + (3x + 3y) = 680 + 1560

⇒ 7x = 2240

⇒ x = 320

⇒ y = 520 − 320 = 200

Larger number = 320

∴ The larger number is 320.

Given:

Train length = 240 m

Platform length = 160 m

Time to cross platform = 20 sec

Formula used:

Total distance while crossing platform = Train length + Platform length

Speed = Distance ÷ Time

Time to cross pole = Train length ÷ Speed

Calculations:

⇒ Total distance = 240 + 160 = 400 m

⇒ Speed = 400 ÷ 20 = 20 m/s

⇒ Time to cross pole = 240 ÷ 20

⇒ Time = 12 sec

∴ Train will cross the pole in 12 seconds.

Given:

Cost of 6 pens + 9 pencils = 150

Calculations:

Cost of 12 pens + 18 pencils = 2 × (6 pens + 9 pencils)

Required cost = 3/4 × Total cost

⇒ Cost of (12 pens + 18 pencils) = 2 × 150 = 300

⇒ Required cost = 3/4 × 300

⇒ 225

∴ Cost of ¾ of (12 pens + 18 pencils) = 225.

Calculation

Cars (A+B) = 180 + 160 = 340

Scooters (D+E) = 200 + 180 = 380

Ratio = 340 : 380 = 17 : 19

Calculation

Electric scooters in D:

40% of 200 = 80

Calculation

Average number of scooters in all showrooms:

Total scooters = 120 + 250 + 240 + 200 + 180 = 990

Average = 990 ÷ 5 = 198

Calculation

Cars in showroom F:

(220 + F) ÷ 2 = 195

220 + F = 390 → F = 170

Calculation

B total = 160 + 250 = 410

E total = 220 + 180 = 400

Difference = 10

Calculation

Scooters in D = 5x:

200 = 5x → x = 40

Given:

Profit % = 40%

Marked up % = 50%

Discount = 40

Calculations:

Let CP = x

Marked Price = x + 50% of x = 1.5x

Selling Price = CP + 40% of CP = 1.4x

Discount = MP − SP

⇒ 1.5x − 1.4x = 40

⇒ 0.1x = 40

⇒ x = 40 ÷ 0.1

⇒ x = 400

∴ Cost Price = Rs. 400.

Given:

Circumference (C) = 176 m

Formula used:

C = 2πr

Area = πr²

Calculations:

⇒ 176 = 2 × 22/7 × r

⇒ r = 176 × 7 ÷ (44)

⇒ r = 28 m

Area = 22/7 × 28 × 28

⇒ Area = 22/7 × 784

⇒ Area = 22 × 112

⇒ Area = 2464 m²

∴ Area of the circle = 2464 m².

Given:

Present age ratio A : B = 4 : 3

6 years ago ratio A : B = 17 : 12

Calculations

Age relation: (A−6)/(B−6) = 17/12

Present ages: A = 17k + 6, B = 12k + 6

⇒ (17k + 6) : (12k + 6) = 4 : 3

⇒ 3(17k + 6) = 4(12k + 6)

⇒ 51k + 18 = 48k + 24

⇒ 3k = 6

⇒ k = 2

⇒ A (present) = 17×2 + 6 = 40

Age of A after 6 years = 40 + 6 = 46

∴ Age of A after 6 years = 46 years.