RRB Clerk Prelims Memory Based Paper (Held On: 13 August 2023 Shift 1) (Previous Year Paper)

Persons: M, N, O, P, Q, R, S, and T.

1. Only one person sits between S and Q, who sits in the row facing north.

2. Q did not sit any extreme end of the row. S faces P, who sits second to the right of N.

3. P did not sit immediate neighbor of T, who faces O.

4) Q is an immediate neighbor of O.

5) R is an immediate neighbor of P. 

Hence, Q sits between M and O.

Persons: M, N, O, P, Q, R, S, and T.

1. Only one person sits between S and Q, who sits in the row facing north.

2. Q did not sit any extreme end of the row. S faces P, who sits second to the right of N.

3. R is an immediate neighbor of P.

4. P did not sit immediate neighbour of T, who faces O.

5) Q is an immediate neighbor of O. 

Hence, R sits immediate right of N.

Persons: M, N, O, P, Q, R, S, and T.

1. Only one person sits between S and Q, who sits in the row facing north.

2. Q did not sit any extreme end of the row. S faces P, who sits second to the right of N.

3. P did not sit immediate neighbor of T, who faces O.

4) Q is an immediate neighbor of O.

5) R is an immediate neighbor of P. 

M is opposite to R, M is immediate right of S and M is immediate left of Q.

Hence, M is opposite to R, immediate right of S, and immediate left of Q.

Persons: M, N, O, P, Q, R, S, and T.

1. Only one person sits between S and Q, who sits in the row facing north.

2. Q did not sit any extreme end of the row. S faces P, who sits second to the right of N.

3. Q is an immediate neighbor of O.

4. P did not sit immediate neighbor of T, who faces O.

5) R is an immediate neighbor of P.

R lies only at the extreme position, the position of S with respect to R will be three places to the left. The arrangement will be as follows.

Hence, If R lies only at the extreme position, S will be three places to the left of R.

Persons: M, N, O, P, Q, R, S, and T.

1. Only one person sits between S and Q, who sits in the row facing north.

2. Q did not sit any extreme end of the row. S faces P, who sits second to the right of N.

3. P did not sit immediate neighbor of T, who faces O.

4) Q is an immediate neighbor of O.

5) R is an immediate neighbor of P. 

Hence, S and T sits extreme end of the row.

Given Series:- 324       825      583      978      159

The largest Number is=978> 825> 583>324> 158

The largest number is= 978

 Second smallest is= 324

Then we have to look for the digit mentioned in the question-  2nd digit of the largest number is added with 1st digit of 2nd smallest number which is:

Second digit of largest number:- 7 (978 is largest number)

First digit of 2^nd smallest number:- 3 (324 is 2^nd smallest number)

Resultant will be:- 7 + 3 = 10;

Hence, “10” is the correct answer.

Given Series:- 324          825            583              978            159

Adding 1 to second digit:-334          835          593            988            169

Subtraction 1 to third digit:- 333           834           592            987     168

Hence, 592 is third largest number which resulted from 583.

Hence, “583” is correct answer.

Given Series:- 324    825    583    978    159

Numbers arranged in ascending order:- 234             258            358              789            159

Hence, 2^nd largest number is 358 which resulted from 583.

Hence, “583” is correct answer.

Given Series:- 324    825    583    978    159

First and third digits are interchanged:- 423      528       385      879        951

Hence, from the above series none of the numbers is a prime number.

Hence, “Zero” is the correct answer.

Given Series:- 324          825            583              978            159

3^rd digit of 2^nd highest number:- 5(825 is 2^nd highest number)

2^nd digit of smallest number:-  5(159 is smallest number)

Hence, resultant will be:- 5×5 = 25;

Hence, “25” is the correct answer.

Below table represents symbols used to draw a family tree,

We can draw the following Family Tree from the information given above:

So, Arman is husband of Kriti.

Hence, Husband is the correct answer.

According to the given information:

1) M sits third from one of the ends.

2. Three persons sit between M and P.

3. O sits six places away from P but not at the end.

As the position of M cannot me determined, two cases will be formed.

Case 1:

Case 2:

4) Only one persons sit between M and R who has only one neighbour.

Case 1:

Case 2:

5) N neither sits to the right of O nor at any of the ends.

6) T sits to the left of O but not immediate left.

7) Not more than two persons sit to the left of T.

8) T sits to the immediate right of N.

If T sits to the left of O, then point 7 will not be applicable in case 1. So it will be cancelled.

9) Three persons sit between T and Q who sits immediate left of L.

10) Four persons sit between L and S.

11) Not more than one person lives between M and S who sits to the left of M.

12) Not more than 7 persons sit between O and S.

Hence, there are 17 persons in the row.

According to the given information:

1) M sits third from one of the ends.

2. Three persons sit between M and P.

3. O sits six places away from P but not at the end.

As the position of M cannot me determined, two cases will be formed.

Case 1:

Case 2:

4) Only one persons sit between M and R who has only one neighbour.

Case 1:

Case 2:

5) N neither sits to the right of O nor at any of the ends.

6) T sits to the left of O but not immediate left.

7) Not more than two persons sit to the left of T.

8) T sits to the immediate right of N.

If T sits to the left of O, then point 7 will not be applicable in case 1. So it will be cancelled.

9) Three persons sit between T and Q who sits immediate left of L.

10) Four sit between L and S.

11) Not more than one person lives between M and S who sits to the left of M.

12) Not more than 7 persons sit between O and S.

Hence, two persons are sitting between L and P.

According to the given information:

1) M sits third from one of the ends.

2. Three persons sit between M and P.

3. O sits six places away from P but not at the end.

As the position of M cannot me determined, two cases will be formed.

Case 1:

Case 2:

4) Only one persons sit between M and R who has only one neighbour.

Case 1:

Case 2:

5) N neither sits to the right of O nor at any of the ends.

6) T sits to the left of O but not immediate left.

7) Not more than two persons sit to the left of T.

8) T sits to the immediate right of N.

If T sits to the left of O, then point 7 will not be applicable in case 1. So it will be cancelled.

9) Three persons sit between T and Q who sits immediate left of L.

10) Four persons sit between L and S.

11) Not more than one person lives between M and S who sits to the left of M.

12) Not more than 7 persons sit between O and S.

Hence, S sits fourth to the left of R.

Given statements,

M = F, R ≤ J, M > J, I ≥ F

On combining, we get: 

R ≤ J < M = F ≤ I

Conclusions:

I. R < F → True (R ≤ J < M = F → R < F).

II. I > J → True (J < M = F ≤ I → J < I).

Hence, the correct answer is both conclusions I and II follow.

Given statements,

T ≥ E = W, V < R < Q = W, T ≤ P

On combining, we get:

V < R < Q = W = E ≤ T ≤ P 

Conclusions:

I. Q < P → False (Q = W = E ≤ T ≤ P → Q ≤ P).

II. P = Q → False (Q = W = E ≤ T ≤ P → Q ≤ P).

Since Q ≤ P is true so either conclusion I or II definitely follows.

Hence, the correct answer is either conclusion I or II follows.

Given statements: W > Q > Z ≤ L; N < C ≤ Z

On combining: W > Q > Z ≥ C > N; L ≥ Z ≥ C > N

Conclusions:

I. W > N → True (as W > Q > Z ≥ C > N → W > N)

II. N ≤ L → False (as L ≥ Z ≥ C > N → L > N)

Hence, only I is True.

Given statements: F > K ≥ D; S ≥ D ≥ U > V

On combining: F > K ≥ D ≥ U > V; F > K ≥ D ≤ S

Conclusions:

I. V < K → True (as K ≥ D ≥ U > V → K > V)

II. F > S → False (as F > K ≥ D ≤ S thus clear relation between F and S cannot be determined)

Hence, only I is True.

Six members: A, B, C, D, E and F

1) Only one person purchase article before A. E purchase before B and D but not before A. 

 ____  > A >_____ >_____ >_____ >_____

2) Only one person purchased article after D. 

3) F purchased the article in the last.

_____ > A >_____ >_____ > D > F

4) E purchase before B and D but not before A. 

_____ > A > E > B > D > F

Only C is remaining, so C will be the first person to purchase the article

So, the final arrangement will be as follows:

C > A > E > B > D > F

Hence, C is the first person to purchase the article.

Six members: A, B, C, D, E and F

1) Only one person purchase article before A. E purchase before B and D but not before A. 

 ____  > A >_____ >_____ >_____ >_____

2) Only one person purchased article after D. 

3) F purchased the article in the last.

_____ > A >_____ >_____ > D > F

4) E purchase before B and D but not before A. 

_____ > A > E > B > D > F

Only C is remaining, so C will be the first person to purchase the article

So, the final arrangement will be as follows:

C > A > E > B > D > F

Six members: A, B, C, D, E and F

1) Only one person purchase article before A. E purchase before B and D but not before A. 

 ____  > A >_____ >_____ >_____ >_____

2) Only one person purchased article after D. 

3) F purchased the article in the last.

_____ > A >_____ >_____ > D > F

4) E purchase before B and D but not before A. 

_____ > A > E > B > D > F

Only C is remaining, so C will be the first person to purchase the article

So, the final arrangement will be as follows:

C > A > E > B > D > F

Number of person purchase the article between C and B is same as that of between E and F

Six members: A, B, C, D, E and F

1) Only one person purchase article before A. E purchase before B and D but not before A. 

 ____  > A >_____ >_____ >_____ >_____

2) Only one person purchased article after D. 

3) F purchased the article in the last.

_____ > A >_____ >_____ > D > F

4) E purchase before B and D but not before A. 

_____ > A > E > B > D > F

Only C is remaining, so C will be the first person to purchase the article

So, the final arrangement will be as follows:

C > A > E > B > D > F

Except BC number of person between all the mention pair is two.

Six members: A, B, C, D, E and F

1) Only one person purchase article before A. E purchase before B and D but not before A. 

 ____  > A >_____ >_____ >_____ >_____

2) Only one person purchased article after D. 

3) F purchased the article in the last.

_____ > A >_____ >_____ > D > F

4) E purchase before B and D but not before A. 

_____ > A > E > B > D > F

Only C is remaining, so C will be the first person to purchase the article

So, the final arrangement will be as follows:

C > A > E > B > D > F

B will be given 13 starting number 10 from C.

From the given instructions:

Seven people A, B, C, D, E, F and G are living in a seven story building, where the bottommost floor is numbered as 1, topmost is numbered as 7. So, we can draw a structure as:

 

1. E lives on one of the floor below third floor. So, here we have two cases;

 

2. Three people are living between F and E.

 

3. Two people are living between C and F.

 

4. G lives immediately below D, who lives on an odd numbered floor.

 

5. A lives above B, who does not live adjacent to C. So, here Case 2 gets eliminated.

 

So, Final Arrangement:

 

Four people live between the floor numbered 6 and B.

From the given instructions:

Seven people A, B, C, D, E, F and G are living in a seven story building, where the bottommost floor is numbered as 1, topmost is numbered as 7. So, we can draw a structure as:

 

1. E lives on one of the floor below third floor. So, here we have two cases;

 

2. Three people are living between F and E.

 

3. Two people are living between C and F.

 

4. G lives immediately below D, who lives on an odd numbered floor.

 

5. A lives above B, who does not live adjacent to C. So, here Case 2 gets eliminated.

 

So, Final Arrangement:

 

G lives immediately above C.

From the given instructions:

Seven people A, B, C, D, E, F and G are living in a seven story building, where the bottommost floor is numbered as 1, topmost is numbered as 7. So, we can draw a structure as:

 

1. E lives on one of the floor below third floor. So, here we have two cases;

 

2. Three people are living between F and E.

 

3. Two people are living between C and F.

 

4. G lives immediately below D, who lives on an odd numbered floor.

 

5. A lives above B, who does not live adjacent to C. So, here Case 2 gets eliminated.

 

So, Final Arrangement:

 

Five people live above E.

From the given instructions:

Seven people A, B, C, D, E, F and G are living in a seven story building, where the bottommost floor is numbered as 1, topmost is numbered as 7. So, we can draw a structure as:

 

1. E lives on one of the floor below third floor. So, here we have two cases;

 

2. Three people are living between F and E.

 

3. Two people are living between C and F.

 

4. G lives immediately below D, who lives on an odd numbered floor.

 

5. A lives above B, who does not live adjacent to C. So, here Case 2 gets eliminated.

 

So, Final Arrangement:

 

All options are true.

From the given instructions:

Seven people A, B, C, D, E, F and G are living in a seven story building, where the bottommost floor is numbered as 1, topmost is numbered as 7. So, we can draw a structure as:

 

1. E lives on one of the floor below third floor. So, here we have two cases;

 

2. Three people are living between F and E.

 

3. Two people are living between C and F.

 

4. G lives immediately below D, who lives on an odd numbered floor.

 

5. A lives above B, who does not live adjacent to C. So, here Case 2 gets eliminated.

 

So, Final Arrangement:

 

Both F and G lives adjacent to D.

The logic follows here is:

The given number can be represented as shown below:

In the forward direction → "89", "14", 46", "35".

In the backward direction → "25", "26", "29", "56", "59", "69", "34".

So, Eleven pairs are there in the number ‘8943651424’, each of which has as many digits between them as in the number series (both forward and backward direction).

Hence, the correct answer is "More than 7".

Given series:

Left Side   R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &   Right Side

Symbol immediately followed by a digit and immediately preceded by an alphabet: Alphabet > Symbol > Digit

R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &

As there is only one symbol immediately followed by a digit and immediately preceded by an alphabet i.e. €

Hence, the element which is seventeenth to the right of € is Z.

Given Series:

Left Side   R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &   Right Side

Alphabets followed immediately by a symbol and preceded immediately by a Digit: Digit > Alphabets > Symbol

R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &

Hence, there are three such alphabets which are followed by a symbol and preceded by a digit: 8 C $, 5 U £, 6 A @

Given Series:

Left side   R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &   Right side

Symbols immediately followed and immediately preceded by a digit: Digit > symbol > Digit.

R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &

So, there are two such symbols which are immediately followed and immediately preceded by a digit i.e. ® and %.

Hence, There are 2 Symbols  ® and %.

Given Series:

Left side   R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &   Right side

If all the symbols and alphabets are removed then the new series will be:

3 8 2 7 5 1 6 4

So, the third digit from the left end = 2

and, the seventh digit from the left end = 6

Now, the average between two numbers is defined as half of the sum of the given two numbers.

Therefore, average of third and seventh digit from the left end i.e. 2 and 6 = (2 + 6)/2 = 8/2 = 4.

Hence, 4 is the correct answer.

Given Series:

Left side   R € 3 W 8 C $ D © K 2 ® 7 T F 5 U £ Z 1 % 6 A @ X 4 Q H &   Right side

The pattern followed here is:

RW3 → R + 3 = W and W - 1 = 3.

K7®  → K + 3 = 7 and 7 - 1 = ®.

@Q4 → @ + 3 = Q and Q - 1 = 4.

5Z£  → 5 + 3 = Z and Z - 1 = £.

Z6A  → Z + 3 = 6 and 6 - 1 = %. (But A given in option as Z6A)

Hence Z 6 A doesn't belong to that group.

Concept:

1. Only a few is a conditional statement which means only a certain part. For eg. if the statement is only a few A is B means only a specific part of the A is B and the remaining part of the A is not B.

a) Only a few - means only "some" not "all".

b) "few" is same as "some".

c) Only "A is B" is same as "all B are A".

2. In this case following conclusions:

• All A being B is a possibility will be false.
• All B being A is a possibility will be true.
• Some A are not B is true.
• Some B are not A is false.
• Some A are B is true.

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We can say that it is a conditional statement which is a combination of some and some not both.

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

Conclusions:

I. Some Pencil are Ball → False (as No Pencil is Ball, no part of Pencil will be Ball. So, it is false)

II. Some Pencil are not Ball → True (as No Pencil is Ball, some part of Pencil will also not be Ball. So, it is true)

Hence, only conclusion II follows.

The least possible Venn diagram,

Conclusion:

I. Some chairs are not laptops → False (It is possible but not definite as there is not specific relation between chairs and laptops, Hence some chain are not laptops is false)

 II. All chairs can never be lights → True ( as only a few chairs are lights, hence all chairs can never be lights is true)

Hence, only conclusion II follows.

The best possible Venn diagram is, 

Conclusions:

I.Some drop is winter: False( As no relation is given between drop and winter, it is possible but not definite)

II. No drop is winter: False( As no relation is given between drop and winter, it is possible but not definite)

Conclusion I and II form a complementary pair.

Hence, Either I or II follows is the correct answer.

1) Point A is 6 meters to the West of B.

2. B is 4 meters to the North of C.

3. D is 6 meters to the West of C.

4. E is 8 meters to the south of D.

5. F is 16 meters to the to the East of E.
   

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6. G is 7 meters to the West of H and 9 meters to the North of F.

7. H is 2 meters North of I.
   

             

The distance between C and E is :

Applying Pythagoras Theorem

CE = $√\left(6^2+8^2\right)$= $√100$ = 10 meters. 

Hence, the shortest distance between C and E is 10 m.

1) Point A is 6 meters to the West of B.

2. B is 4 meters to the North of C.

3. D is 6 meters to the West of C.

4. E is 8 meters to the south of D.

5. F is 16 meters to the to the East of E.

6. G is 7 meters to the West of H and 9 meters to the North of F.

7. H is 2 meters North of I.

             

Hence,  I is in South East direction with respect to D.

1) Point A is 6 meters to the West of B.

2. B is 4 meters to the North of C.

3. D is 6 meters to the West of C.

4. E is 8 meters to the south of D.

5. F is 16 meters to the to the East of E.

6. G is 7 meters to the West of H and 9 meters to the North of F.

7. H is 2 meters North of I.

             

From the picture we see the distance between D and I is equal to the total distance of EF and GH. 

Hence the distance is = (16 + 7) = 23 meters.

Solution:

Number of males working in place K = 850

Number of males working in place  L = 770

Number of males working in place M = 820

Number of males working in place N = 640

Required average = (850 + 770 + 820 + 640)/4 = 770

Solution:

Number of males working in place K = 850

Number of males working in place M = 820

Number of females working in place L = 720 

Number of females working in place N = 700

Required total = (850 + 820 + 720 + 700) = 3090

Solution:

Number of males working in place N = 640

Number of females working in place K = 750

25% of the number of males working in place N = 640 * (25/100) = 160

1/5 th of number of females working in place K = 750/5 = 150

Required percentage = 160/150 * 100 = 106.667%

Solution:

Number of males working in place L = 770

1/11 th of number of males = 770/11 = 70

Number of females working in place M = 900

1/10 th of number of females working in place M = 900/10 = 90

(90 - 70) = 20

Solution:

Number of males working in place K = 850

Number of females working in place L = 720

Number married males = 850 * (20/100) = 170

Number of married females = (720) * (25/100) = 180

Number of unmarried males = (850 - 170) = 680

Number of unmarried males = (720 - 180) = 540

Required sum = (680 + 540) = 1220

Given:

(4/11) of 143 + (7/15) of 120 = 50% of (?)

Concept Used:

Calculation:

(4/11) of 143 + (7/15) of 120 = 50% of (?)

⇒ (4/11) × 143 + (7/15) × 120 = (50/100) × (?)

⇒ 52 + 56 = ?/2

⇒ ? = 108 × 2 = 216

∴ The value of (?) is 216

Given:

11.11% of 99.11 + 22.22% of 98.97 – 9.99% of 100 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

Since we need to find out the approximate value, we can write these values to their nearest integers.

11.11% of 99.11 + 22.22% of 98.97 - 9.99% of 100 = ?

⇒ 11.11% of 99 + 22.22% of 99 - 10% of 100 = ?

⇒ (1/9) × 99 + (2/9) × 99 - (1/10) × 100 = ?

⇒ 11 + 22  - 10 = ?

⇒ 33 - 10 = ?

⇒ ? = 23

∴ The value of ‘?’ Is 23.

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Alternate Method

11.11% of 99 + 22.22% of 99 - 10% of 100 = ?

⇒ (1/9) × 99 + (2/9) × 99 - (1/10) × 100 = ?

⇒ {(1/9) + (2/9)} × 99 - 10 = ?

⇒ (1/3) × 99 - 10 = ?

⇒ 33 - 10 = ?

⇒ ? = 23

∴ The value of ‘?’ Is 23.

Given:

(5/6) + (7/5) + (2/3) + (3/10) = ? + (47/15)

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

(5/6) + (7/5) + (2/3) + (3/10) = ? + (47/15)

⇒ (5/6) + (7/5) + (2/3) + (3/10) - (47/15) = ?

⇒ (25 + 42 + 20 + 9 - 94)/30 = ?

⇒ ? = 2/30

⇒ ? = 1/15

∴ The value of ‘?’ Is 1/15.

Given:

$2\frac{3}{5} × 1\frac{{19}}{{26}} × 6\frac{2}{3} = ?$

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

$2\frac{3}{5} × 1\frac{{19}}{{26}} × 6\frac{2}{3} = ?$

⇒ (13/5) × (45/26) × (20/3) = ?

⇒ (9/2) × (20/3) = ?

⇒ 3 × 10 = ?

⇒ ? = 30

∴ 30 will come in place of the question mark ('?').

Given : 

  (16)²×(4)² ÷ (64)² = (4)^?

 Calculations : 

 (16)² × (4)⁴ ÷ (64)² = (4)^?

  ⇒ (4²)² × (4)⁴ ÷ (4³)² = (4)^?

 ⇒ (4)⁴ × (4)⁴ ÷ (4)⁶ = (4)^?

 ⇒ (4)⁸ ÷ (4)⁶ = (4)^?

 ⇒ (4)^(8-6) = (4)^?

 ⇒ (4)² = (4)^?

 ⇒ ? = 2

 Hence 2 will come in the place of '?'.

Given:

230 + 40% of 45 - 55 ÷ 5 = ? - 216 ÷ 6

Concept used:

Follow the 'BODMAS' Rule

B ⇒ Bracket, O ⇒ of, D ⇒ Division, M ⇒ Multiplication, A ⇒ Addition, S ⇒ Subtraction

Calculation:

⇒ 230 + 40% × 45 - 55 ÷ 5 = ? - 216 ÷ 6

⇒ 230 + 18 - 11 = ? - 36

⇒ ? = 273

∴ The value of ? is 273.

Solution :

Concept used :

Follow the BODMAS rule to solve this question, as per the question 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 part of the equation that contains 'Division' and 'multiplication' are calculated,

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

⇒ 16 × 15 + 20% of 350 - √225 × 3 = 53% of ?

⇒ 240 + 70 - (15 × 3) = 53/100 × ?

⇒ 240 + 70 - 45 = 53/100 × ?

⇒ 310 - 45 = 53/100 × ?

⇒ 265 × 100/53 = ?

⇒ 500 = ?

Hence, the correct answer is option 1.

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

78 × 16 + 285 + 648 - 17 × 25 = ?

⇒ 1248 + 285 + 648 - 425 = ?

⇒ 2181 - 425 = ?

⇒ ? = 1756

Hence, 1756 will come in the place of (?).

Concept:

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 the 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculation:

Given expression is 

(953 + 487 - 220) ÷ 5 = 482 - 62 + ?

⇒ 1220 ÷ 5 = 482 - 62 + ?

⇒ 244 = 482 - 62 + ?

⇒ 244 + 62 - 482 = ?

⇒ - 176

Hence, 176 will come in the place of (?).

Concept:

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 the 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculation:

Given expression is 

5.6 ÷ (0.8)² × 6 - 225 ÷ 15 = ?

⇒ 5.6 ÷ 0.64 × 6 - 225 ÷ 15

⇒ 8.75 × 6 - 15

⇒ 52.5 - 15

⇒ 37.5

Hence, 37.5 will come in the place of (?).

Concept used:

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 second, 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

125 ÷ 25% of 500 × 2 + 800 ÷ 20 × 5 = ? - 325

⇒ 125 ÷ 125 × 2 + 40 × 5 = ? - 325

⇒ 1 × 2 + 200 = ? - 325

⇒ 202 = ? - 325

⇒ ? = 527

Hence, 527 will come in the place of (?).

Concept used:

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 second, 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

20% of 800 + 16 × 32 – 1/6 of 1200 = 80 + ?

⇒ 160 + 16 × 32 – 200 = 80 + ?

⇒ 160 + 512 – 200 = 80 + ?

⇒ 672 – 200 = 80 + ?

⇒ 672 – 200 - 80 = ?

⇒ ? = 392

Hence, 392 will come in the place of (?).

45 × 3 + 7 + ? + 36 ÷ 6 × 9 = 100

⇒ 45 × 3 + 7 + ? + 6 × 9 = 100

⇒ 135 + 7 + ? + 54 = 100

⇒ 196 + ? = 100

⇒ ? = 100 - 196

⇒ ? = -96

Concept used:

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 second, 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

2400 ÷ 12 – {26 × 6 + (32 – 5 × 6)} = ?

⇒ 2400 ÷ 12 – {26 × 6 + (32 – 30)} = ?

⇒ 2400 ÷ 12 – {26 × 6 + 2} = ?

⇒ 2400 ÷ 12 – 158 = ?

⇒ 200 - 158 = ?

⇒ 42

Hence, 42 will come in the place of (?).

Concept used:

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 second, 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 least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Calculations:

(1700 + 1800 – 1300 + 1800 ) ÷ ? = 200 + (5)2 + 25

⇒ 4000 ÷ ? = 200 + 25 + 25

⇒ 4000 ÷ ? = 250

⇒  ? = 4000/250

⇒ ? = 16

Hence, 16 will come in the place of (?).

Shortcut Trick

Combine the ratio of A : B (1 : 3) and A : C (2 : 3) by making A common: A : B : C = 2 : 6 : 3.

Difference between A and C (3 − 2 = 1 unit) is given as 2 years.

Present total units = 2 + 6 + 3 = 11 units ⇒ 11 × 2 = 22 years.

Sum after 14 years = 22 + (3 × 14) = 22 + 42 = 64 years.

∴ The correct answer is 64 years.

Alternate Method

Given: Ratio A : B = 1 : 3 and Ratio A : C = 2 : 3.

Formula: Sum of ages = (A + B + C) + (3 × years hence).

⇒ To unify ratios, multiply A : B by 2: A : B = 2 : 6.

⇒ Unified ratio A : B : C = 2 : 6 : 3.

⇒ Difference (C − A) = 3x − 2x = x.

⇒ Given x = 2 years.

⇒ Present age of A = 2 × 2 = 4, B = 6 × 2 = 12, C = 3 × 2 = 6.

⇒ Present sum = 4 + 12 + 6 = 22 years.

⇒ Sum after 14 years = 22 + (14 + 14 + 14) = 22 + 42 = 64 years.

∴ The correct answer is 64 years.

Additional Information

Ratio Equalization

To combine two ratios (A:B and A:C), make the common term (A) equal by multiplying both ratios with appropriate constants.

Age Sum Property

When calculating the sum of ages of n people after t years, the total sum increases by n × t years.

Age Difference Property

The difference between the ages of two individuals remains constant at any point in time.

Given: A invested for 6 months and B invested for 12 months. Ratio of their investment is 8:9.

Concept Used:

(A's investment) × (Time A invested for)/(B's investment) × (Time B invested for) = (A's share)/(B's share)

Calculation:

Let us assume that A invested 8x for 6 months and B invested 9x for a year, which is 12 months.

(A’s share)/(B’s share) = (8x × 6)/(9x × 12)

⇒ 4 : 9

Total profit is 5200.

The share of A will be = (5200) × 4/13

⇒ 1600

∴, the share of A will be rupees 1600.

Shortcut Trick

Remaining Mixture = 48 − 10 = 38 litres.

Milk = (3/4) × 38 = 28.5 litres | Water = (1/4) × 38 = 9.5 litres.

New Water = 9.5 + 20 = 29.5 litres.

Difference = |28.5 − 29.5| = 1 litre.

∴ The correct answer is 1 litre.

Alternate Method

Initial Ratio of Milk : Water = 3 : 1

Total Parts = 3 + 1 = 4 units

⇒ 4 units = 48 litres → 1 unit = 12 litres

⇒ Initial Milk = 36 litres, Initial Water = 12 litres

⇒ In 10 litres mixture removed: Milk = (3/4) × 10 = 7.5 litres, Water = (1/4) × 10 = 2.5 litres

⇒ Remaining Milk = 36 − 7.5 = 28.5 litres

⇒ Remaining Water = 12 − 2.5 = 9.5 litres

⇒ Final Water = 9.5 + 20 = 29.5 litres

⇒ Final Difference = |28.5 − 29.5| = 1 litre

∴ The correct answer is 1 litre.

Additional Information

Mixture Removal Property

When a portion of a mixture is removed, the ratio of components in the remaining mixture stays the same as the original ratio.

Alligation Rule

Quantity of Cheaper / Quantity of Dearer = (Rate of Dearer − Mean Rate) / (Mean Rate − Rate of Cheaper). This is used for mixing two varieties.

Replacement Formula

Final Quantity = Initial Quantity × [1 − (Removed Amount / Total Capacity)]ⁿ, where n is the number of replacement operations.

Give

Area of circle = 154 sq.m

The breadth of the rectangle = 1/7 of the circle's radius

Formula used:

Area of Circle = πr²

Area of Rectangle = half of the Area of Circle = 154/2 = 77 sq.m

Perimeter of Rectangle = 2 × (Length + Breadth)

Calculation:

First, we need to determine the radius of the circle from its area.

⇒ Area of circle = πr2

⇒ 154 = πr2

⇒ r2 = 154/π = 49

So, r (radius) = √49 = 7 m (taking the square root of both sides)

Then, we calculate the breadth of the rectangle, which is 1/7th of the radius of the circle:

⇒ Breadth = r/7 = 7/7 = 1 m

To determine the length of the rectangle, we use the fact that the Area of the rectangle is equal to Length × Breadth:

⇒ 77 sq.m (Area of the rectangle) = Length × 1 m

So, Length = 77 / 1 = 77 m

Finally, the perimeter of the rectangle is:

⇒ Perimeter = 2 × (Length + Breadth) = 2  ×  (77 m + 1 m) = 2  ×  78 m = 156 m

So, the perimeter of the rectangle is 156 meters.

Given:

Amount after 3 years = Rs. 1440

Rate of interest = 12%

Amount invested after 2 years at 10% ci annually.

Formula used:

SI = PRT / 100

A = P × [1 + (R / 100)]^T

Where , A = Final Amount

P = Principal

R = Rate of interest

T = Time period

Calculation:

1440 = (P × 12 × 3) / 100

⇒ P = Rs. 4000

Now, CI = 4000(1 + (10/100)]²

⇒ Rs. 4000 × (1.10)²

⇒ 4000 × 1.21

⇒ Rs. 4840

∴ Amount obtained on the same sum is Rs. 4840

Given:

Marked price of the laptop is Rs. 48,000

The profit of the shopkeeper is 20%

Given discount on the laptop is 12%

Formula used:

SP = {(100 + P%)/100} × CP

Where, SP is selling price, CP is cost price and P is profit

SP = {(100 - D%)/100} × MP

Where, D is discount and MP is marked price

Calculations:

The marked price of the laptop is Rs. 48,000

According to question, we have

The selling price of the laptop is

⇒ SP = {(100 - D%)/100} × MP

⇒ SP = {(100 - 12)/100} × 48,000

⇒ SP = (88/100) × MP

⇒ SP = 0.88 × 48,000

⇒ SP = Rs. 42,240

So, the selling price of the laptop is Rs. 42,240

And, the cost price of the laptop is

⇒ SP = {(100 + P%)/100} × CP

⇒ 42,240 = {(100 + 20)/100} × CP

⇒ 42,240 = 1.2 × CP

⇒ CP = 42,240/1.2

⇒ CP = Rs. 35,200

∴ The cost price of the laptop is Rs. 35,200

Alternate Method

Discount = 12% of marked price

SP = MP - Discount = 25 -3 = 22 unit

Gain = 20%

SP = CP + Profit = 5 + 1 = 6 unit

So, CP = 110 units

SP = 132 units

MP = 150 units

According to question, we have

Marked price is Rs. 48,000

⇒ 150 units = Rs. 48,000

⇒ 1 units = Rs. 320

⇒ 110 units = Rs. 320 × 100

⇒ 110 units = Rs. 35,200

So, the CP is Rs. 35,200

∴ The cost price of the laptop is Rs. 35,200

 

Required cost price is

⇒ CP = {48,000 × (100 – 12)}/(100 + 20)

⇒ CP = (48,000 × 88)/120

⇒ CP = Rs. 35,200

∴ The cost price of the laptop is Rs. 35,200

Solution:

Given:

Pipe A can fill tank = 6 hours

Pipe A and leakage pipe both can fill tank = 9 hours

Calculation:

⇒ 1/6 + 1/B = 1/9

⇒ 1/B = 1/9 – 1/6

⇒ 1/B = (2 – 3)/18

⇒ 1/B = - 1/18

∴ Leakage can empty tank in 18 hours.

Solution:

Number of movie tickets sold on Monday = 60

Number of movie tickets sold for movie A = 60 * 1/6 = 10

Number of play tickets sold for Thursday = 60

Number of play tickets sold for game P on Thursday= 60 * 1/4 = 15

Required total = (10 + 15) = 25

Solution:

Number of play tickets sold on Thursday = 60

Number of play tickets sold on Friday = (60 + 30) = 90

Number of movie tickets sold on Wednesday = 70

Number of movie tickets sold on Wednesday = (70 + 40) = 110

Required total = (90 + 110) = 200

Solution:

Number of movie tickets sold on Monday = 60

Number of movie tickets sold on Thursday = 80

Number of play tickets sold on Tuesday = 30

Number of play tickets sold on Wednesday =  80

Required total = (60 + 80 + 30 + 80) = 250

Solution:

Number of movie tickets sold on Monday = 60

1/3 rd of movie tickets sold on Monday = 60/3 = 20

Number of play tickets sold on Wednesday = 80

1/8 of the play tickets sold on Wednesday = 80* 1/8 = 10

Required percentage = 20/10 * 100 = 200%

Solution:

Number of movie tickets sold all the days = (60 + 50 + 70 + 80) = 260

1/2 of the number of movie tickets sold = (260/2) = 130

Number of play tickets sold in all the given days = (40 + 30 +  80 + 60) = 210

1/3rd of the play tickets sold = 210/3 = 70

Required difference = (130 - 70) = 60

Given series:

 12, 25, 50, 150, 600, 3000

Calculation :

 The series follows the following pattern:

 ⇒ 25 x 1 = 25

 ⇒ 25 x 2 = 50

 ⇒ 50 x 3 = 150

⇒ 150 x 4 = 600

⇒ 600 x 5 = 3000

 Hence the wrong term in the series is 12.

Given series:

225, 204, 179, 146, 97, 10

Calculation :

 The series follows the following pattern:

⇒ 225 - 21 = 204

⇒ 204 - 25 = 179

⇒ 179 - 33 = 146

⇒ 146 - 49 = 97

⇒ 97 - 81 = 16

 The wrong term is 10

 Hence the wrong term in the series is 10.

Given series:

25, 14, 14, 21, 42, 105

Calculation :

 The series follows the following pattern:

⇒ 28 x 0.5 = 14

⇒ 14 x 1 = 14

⇒ 14 x 1.5 = 21

⇒ 21 x 2 = 42

⇒ 42 x 2.5 = 105

 The wrong term is 25

 Hence the wrong term in the series is 25.

Given series:

130, 154, 178, 204, 232, 262

Calculation :

 The series follows the following pattern:

⇒ 11 + 11² = 132

⇒ 10 + 12² = 154

⇒ 9 + 13² = 178

⇒ 8 + 14² = 204

⇒ 7 + 15² = 232

⇒ 6 + 16² = 262

 The wrong term is 130

 Hence the wrong term in the series is 130.

Given series:

29, 60, 31, 54, 35, 51

Calculation :

 The series follows the following pattern:

⇒ 29 + 31 = 60

⇒ 60 - 29 = 31

⇒ 31 + 23 = 54

⇒ 54 - 19 = 35

⇒ 35 + 17 = 52

 The wrong term is 51

 Hence the wrong term in the series is 51.

Given:

Total students = 15

After replacement the average of weight reduced by 2 kg.

Formula used:

Average weight = Total weight of all students/Number of students

Calculation:

Let the average weight of 15 students before replacement be y kg.

Sum of weight of all 15 students before replacement = 15y

Now, a student of 52 kg is removed from the group and is replaced by another student.

Let the weight of this new student be x kg.

New sum of weights of students = (15y – 52 + x ) kg

According to question,

The new average weight of the group = y – 2

⇒ y – 2 = (15y - 52 + x)/15

⇒ 15y - 30 = 15y – 52 + x

⇒ x = 22

∴ Weight of the new student is 22 kg.

Formula:

D = S × T

Effective upstream speed = x – y

where "x" and "y" are the speed of the swimmer and river respectively. 

D = Distance in km.

S = Speed in km/hr

T = Time in hours

Calculation:

Effective upstream speed = 12 - 2 = 10 km/hr

Time taken to swim 6 km upstream = $6\over10$ hour

Time taken in minutes = $6\over10$ × 60 = 36 minutes.

The total time taken is 36 minutes.

Given:-

Efficiency ratio = 2 : 3 : 5

Time for C = 20 days.

Concept:-

Efficiency is inversely proportional to time.

Calculation:-

Efficiency ratio of A and C = 2 : 5

Time ratio of A and C = 5 : 2

Time of A = (5 / 2) × 20 = 50 hours.

Efficiency ratio of B and C = 3 : 5

Time ratio of B and C = 5 : 3

Time of B = (5 / 3) × 20 = 100 / 3 hours.

Time for A and B = 1 / 50 + 3 / 100 = 20 hours.

 

Time can directly be found out by adding efficiency of A and B.

Given:-

Efficiency ratio of A, B & C = 2 : 3 : 5

Calculation:-

Efficiency of (A + B) : C = 5 : 5 = 1 : 1

∴ Time required by (A + B) = Time required by C = 20 hours.