Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

Hence, Nia will go on the trip on 16^th September.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2. Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

 

Hence, Dia goes on the trip exactly before last.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

 

Hence, Except Mia - Pia all are born in the same month.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

 

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

Hence, Only Dia and Ria goes on the trip after Gia.

Eight girls: Dia, Gia, Mia, Nia, Pia, Tia, Sia, and Ria.

Months: January, February, September, and October.

Dates: 16^th and the 27^th

1) Gia goes on the trip in the months which have neither a maximum number of days or a minimum number of days. 

Gia will neither go on the trip in February as there are only 28days nor will go on the trip in January or October as there are 31days. So, She will go on the trip in September.

2) Only two girls are going on the trip between Gia and Mia.

3) Dia goes on the trip after Gia but not on an odd date.

4) Pia goes on the trip before Tia but neither of them will go on the trip on an even number of dates nor in October.

Pia and Tia will go on the trip on the 27^th of any month but both of them will not go on the trip in October.

5) Ria goes on the trip after Tia. 

6) Only three girls are going on the trip between Nia and Sia, who goes on the trip before Tia.

Case II.I and Case II.II eliminate as in both the cases there will be only one person between Nia and Sia.

The final arrangement:

 

Hence, Tia goes on the 27^th february.

The least possible diagram for the given statements is as follows

Conclusions:

I. All balls are tablets → True (All balls are dolls and all dolls are mobile and all mobiles are tablets so all balls are also tablets)

II. Some mobiles are dolls → True (all dolls are mobile so some mobiles are dolls)

Conclusion I and II both follow.

The least possible diagram for the given statements is as follows

Conclusions:

I. No potato is orange → True (Because all potatoes are lime and no lime is orange so no potato is orange)

II. Some apples are limes → False (Because no orange are limes and all apples are orange so some apples are limes it is not possible)

Only conclusion I follow.

Solution:

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

Conclusions:

I. No bucky is captain. →  False (it is possible but not definite, because no direct relation between bucky and captain is given)

II. Some america is bucky. →  False (because only captain can be america)

Hence, Neither I nor II follows is the correct answer.

Additional Information:

Only captain is america means all america is captain and no other element can be america.

Only a few captain are hydra means some captains are hydra and some are not.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

Now, consonants which are immediately preceded by a symbol and immediately followed by a number: Symbol → Consonant → Number

Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £

Hence, there are two consonants N and Z which are immediately preceded by a symbol and immediately followed by a number → @ N 2 and ¥ Z 8.

Note: A, E, I, O, U are the vowels in English alphabet and all else are called consonants.

Mistake Points 

Please note that U is not a consonant.

According to the question, the pattern should be

Symbol → Consonant → Number

Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £

Hence, there are two consonants N and Z which are immediately preceded by a symbol and immediately followed by a number → @ N 2 and ¥ Z 8.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

Here, each element of each term moves 6 steps forward to give the corresponding element of the next term.

@ + 6 = B; B + 6 = 3; 3 + 6 = I

2 + 6 = 6; 6 + 6 = A; A + 6 = Z

U + 6 = S; S + 6 = 7; 7 + 6 = £

Hence, according to this the missing term will be 3 A 7.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

When all the vowels are dropped, the new sequence is;

Y 5 @ N 2 # 9 B 6 % S © 3 Q & 7 W ¥ Z 8 £

Now, the 8^th term from the right in the new sequence:

(Y 5 @ N 2 # 9 B 6 % S © 3 Q & 7 W ¥ Z 8 £) → Q.

Hence, “Q” is the correct answer. 

Note: A, E, I, O, U are the vowels in English alphabet and all else are called consonants.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

The new sequence after reversing the last 12 elements;

Y 5 @ N 2 # U 9 B E 6 % S £ 8 Z ¥ I W 7 & A Q 3 ©

As, Left – Left = Left

21^st from the left – 6^th from the left = 15^th from the left

Hence, 15^th from the left is 8.

Given series:

(Left side) Y 5 @ N 2 # U 9 B E 6 % S © 3 Q A & 7 W I ¥ Z 8 £ (Right side)

Here, the group is formed in which the first and second elements move 2 steps forward and 4 steps backward respectively to give the second and third elements.

1. + 2 = 9; 9 - 4 = N

2) 6 + 2 = S; S + 2 = 3

3. 7 + 2 = I; I - 4 = A

4. ¥ + 2 = 8; 8 - 4 = W

5. @ + 2 = 2; 2 - 4 = Y

Hence, 6 S 3 does not belong to that group.

Given statements : T = V < U ≥ X; V ≤ S; H > U

On Combining : T = V < U ≥ X; V ≤ S; H > U

Conclusions:

I. S ≥ T → True  (As V ≤ S and T = V → S ≥ T)

II. X < H → True (As H > U and U  ≥  X → X < H )

Therefore Both I and II are True

Given Statements : L < O < M ; Q> N ≥ P > M 

On Combining : Q > N ≥ P > M > O > L

Conclusions :

I. P > L → True ( As P > M > O > L → thus clear relation P and L can be determined )

II. M > N → False ( As N ≥ P > M → thus we conclude that M is less than N not greater )

Therefore Only I is true.

Given statements:  X ≤ U < V > T; U < S; H > V

On combining:X ≤ U < V > T; U < S; H > V

Conclusions:

I. X < H → True (H > V and  V > U ≥ X → X < H)

II. T ≤ S → False (U <  S and U < V > T  → thus clear relation between T and  S cannot be determined)

Therfore Only conclusion I is True.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Clearly, two boxes are placed above box U.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Clearly, box V is placed second from the bottom.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Clearly, five boxes are placed between box T and box S.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

Here, box S is placed exactly above box P, and box R is placed exactly above box Q, in the same way, box V is placed exactly above box T.

Clearly, box V is related to box T.

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Seven boxes P, Q, R, S, T, U, and V are stacked one above the other.

1. At most two boxes are placed below box Q. (Here the word ‘at most’ means Maximum.)
2. Box R is placed exactly above box Q.

Following are the possibilities we get,

3. Two boxes are placed between box R and box S, where box S is above box R.
4. At least five boxes are placed below box S.

Here, this statement is violated in case 1, so it is eliminated.

5. Number of boxes above box P is the same as the number of boxes below box V.
6. Box P is placed at one of the places below box S.

Here, this statement is not possible in case 2, so it is eliminated and the following are the possibilities we get in case 3,

7. Box U is placed in one of the places above box T, but not exactly above.
8. Box T is placed near Box V.

Here, this statement is not possible in case 3b, so it is eliminated. So, the final arrangement is,

If we rearrange all the boxes in alphabetical order from bottom to top, then

Clearly, the position of None (None of the boxes) remains unchanged.

Additional Information

The word 'at least' means the minimum number, while 'at most' means the maximum number.

Eight pairs are there in the word ‘MISUNDERSTANDING’, each of which has as many letters between them in the word (both forward and backward direction) as they have between them in the English Alphabet.

Hence, the correct answer is more than four.

Note: Ignore the count from S to A, U to D and U to G as they start the repetition cycle again after Z.

The least possible diagram based on the given information,

Clearly, the distance between points J and P is 3 km.

The least possible diagram based on the given information,

Except U, all others lie in the same line.

Hence, U does not belong to the group.

The least possible diagram based on the given information,

Clearly, point S is to the north-west of point U.

Given word and the new word after the given rearrangement:

 

Hence, no vowel is there in the new word.

Persons–A, B, C, D, E, F and G

1. A got appointment letter on Wednesday.

2.Only one person got appointment letter between A and D.(So, D got appointment letter either on Monday or on Friday)

Case I

Case II

3.Two persons got appointment letter between B and E.

4.B got appointment letter before E.

5.C got appointment letter before D and after B.(So,case I will be eliminated and C got appointment letter on Tuesday, B got appointment letter on Monday and E got appointment letter on Thursday)

6. F got appointment letter in one of the day after D.

7.G does not get appointment letter Sunday.(So, G got appointment letter on Saturday and F got appointment letter on Sunday)

Hence, only two persons got appointment letter after D.

Persons–A, B, C, D, E, F and G

1. A got appointment letter on Wednesday.

2.Only one person got appointment letter between A and D.(So, D got appointment letter either on Monday or on Friday)

Case I

Case II

3.Two persons got appointment letter between B and E.

4.B got appointment letter before E.

5.C got appointment letter before D and after B.(So,case I will be eliminated and C got appointment letter on Tuesday, B got appointment letter on Monday and E got appointment letter on Thursday)

6. F got appointment letter in one of the day after D.

7.G does not get appointment letter Sunday.(So, G got appointment letter on Saturday and F got appointment letter on Sunday)

Hence, B got appointment letter on Monday.

Persons–A, B, C, D, E, F and G

1. A got appointment letter on Wednesday.

2.Only one person got appointment letter between A and D.(So, D got appointment letter either on Monday or on Friday)

Case I

Case II

3.Two persons got appointment letter between B and E.

4.B got appointment letter before E.

5.C got appointment letter before D and after B.(So,case I will be eliminated and C got appointment letter on Tuesday, B got appointment letter on Monday and E got appointment letter on Thursday)

6. F got appointment letter in one of the day after D.

7.G does not get appointment letter Sunday.(So, G got appointment letter on Saturday and F got appointment letter on Sunday)

Hence, D got appointment letter on Friday.

Persons–A, B, C, D, E, F and G

1. A got appointment letter on Wednesday.

2.Only one person got appointment letter between A and D.(So, D got appointment letter either on Monday or on Friday)

Case I

Case II

3.Two persons got appointment letter between B and E.

4.B got appointment letter before E.

5.C got appointment letter before D and after B.(So,case I will be eliminated and C got appointment letter on Tuesday, B got appointment letter on Monday and E got appointment letter on Thursday)

6. F got appointment letter in one of the day after D.

7.G does not get appointment letter Sunday.(So, G got appointment letter on Saturday and F got appointment letter on Sunday)

Hence, more than three persons got appointment letter between B and F.

Persons–A, B, C, D, E, F and G

1. A got appointment letter on Wednesday.

2.Only one person got appointment letter between A and D.(So, D got appointment letter either on Monday or on Friday)

Case I

Case II

3.Two persons got appointment letter between B and E.

4.B got appointment letter before E.

5.C got appointment letter before D and after B.(So,case I will be eliminated and C got appointment letter on Tuesday, B got appointment letter on Monday and E got appointment letter on Thursday)

6. F got appointment letter in one of the day after D.

7.G does not get appointment letter Sunday.(So, G got appointment letter on Saturday and F got appointment letter on Sunday)

Hence, E got appointment letter on Thursday.

The positions of the letters according to the English alphabet series :

C (-1) → B (+2) → D

G (-1) → F (+2) → H

M (+1) → N (+1) → O

R (-1) → Q (+2) → S

V (-1) → U (+2) → W

Thus, MNO has a different pattern.

From the given information,

1. W is the son of T, who is the only daughter of P.

2. P is the brother-in-law of V and does not has any siblings.

3) X is the niece of Q, who is the sibling of T.

4) R is the brother-in-law of Q.

5) V is not a female. So, S is the wife of P. 

6. P has only one daughter. So Q will the son of P.

7. Less than four female members are there in the family, so Q will be male member in the family.

Therefore the final arrangement is:

Therefore Q is the son of S.

From the given information,

1. W is the son of T, who is the only daughter of P.

2. P is the brother-in-law of V and does not has any siblings.

3) X is the niece of Q, who is the sibling of T.

4) R is the brother-in-law of Q.

5) V is not a female. So, S is the wife of P. 

6. P has only one daughter. So Q will the son of P.

7. Less than four female members are there in the family, so Q will be male member in the family.

Therefore the final arrangement is:

Therefore, Q is the Maternal uncle of W.

From the given information,

1. W is the son of T, who is the only daughter of P.

2. P is the brother-in-law of V and does not has any siblings.

3) X is the niece of Q, who is the sibling of T.

4) R is the brother-in-law of Q.

5) V is not a female. So, S is the wife of P. 

6. P has only one daughter. So Q will the son of P.

7. Less than four female members are there in the family, so Q will be male member in the family.

Therefore the final arrangement is:

Therefore, there are three females.

Given:

[(27 × 10) ÷ 30 + 10500 ÷ 500] = ? × 15

Concept Used:

Calculation:

Suppose x will come in the place of the question mark;

[(27 × 10) ÷ 30 + 10500 ÷ 500] = x × 15

Applying BODMAS Rule;

⇒ [(270) ÷ 30 + 21] = 15x

⇒ 9 + 21 = 15x

⇒ 30 = 15x

∴ x = 2

Follow BODMAS (Brackets, Of, Division, Multiplication, Addition, Subtraction) rule to solve this question, as per the order given below,

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

Step-2-  Then any mathematical 'Of' or 'Exponent' must be solved,

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,

⇒ 999 ÷ 3 – 165 = 75 + 111 - ?

⇒ 333 – 165 = 186 - ?

⇒ 168 = 186 - ?

⇒ ? = 186 – 168

⇒ ? = 18

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

⇒ (71 × 8) + 90 = 132 × 4 + ?

⇒ 568 + 90 = 528 + ?

⇒ 658 = 528 + ?

⇒ ? = 658 – 528

⇒ ? = 130

∴ The value of ? is 130

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.

Now, given expression:

$? = \frac{{26}}{{24}} \times 408 + \frac{{25}}{{4800}} \times 43200$

⇒ ? = 442 + 225

⇒ ? = 667

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

(8 + 6 × 7) ÷ 10 × 3

⇒ (8 + 42) ÷ 10 × 3

⇒ 50 ÷ 10 × 3

⇒ 5 × 3 = 15

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 

5 × √25 + 300 - 100 = ?²

⇒ 5 × 5 + 200 = ?2

⇒ 225 = ?2

∴ 15 = ?

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 

18 - ? + 19 + (11)2 = 120

⇒ 18 - ? + 19 + 121 = 120

⇒ ? = 121 - 120 + 19 + 18

∴ ? = 38

Concept used:

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

Calculations:

2.5% of 40 + 4.5% of 80 = ?% of 10

⇒ (2.5/100) × 40 + (4.5/100) × 80 = (?/100) × 10

⇒ 1 + 3.6 = ?/10

⇒ 4.6 = ?/10

⇒ ? = 46

∴ The value of ? is 46.

450% of ? ÷ 90 + 350% of 200 ÷ 70 = 20

⇒ (450/100 × ?/90) + (350/100 × 200/70) = 20

⇒ ?/20 + 10 = 20

⇒ ?/20 = 10

⇒ ? = 200

935.5 + 555.4 + 64.5 + 1444.6 = 600% of ?

⇒ (935.5 + 64.5) + (555.5 + 1444.6) = 600/100 × ?

⇒ 1000 + 2000 = 6 × ?

⇒ 3000 = 6 × ?

⇒ ? = 3000/6= 500

256 × 2^-5 × ? = (2/25) of 50 × 75

⇒ 2⁸ × 2^-5 × ? = (2/25) of 50 × 75

⇒ 2⁸ × 2^-5 × ? = (2/25) ×  50 × 75

⇒ 2⁸ × 2^-5 × ? = 2 ×  2 × 75

⇒ 2⁸/2⁵ × ? = 4 × 75

⇒ 8 × ? = 4 × 75

⇒ ? = 75/2

⇒ ? = 37.5

232 - 413 ÷ 59 = ? × 18

⇒ 529 - 7 = ? × 18

⇒ 522/18 = ?

⇒ ? = 29

Shortcut Trick

Combine the ratio of A : B : C : D using the common terms.

A : B = 3 : 4 | B : C = 1 : 2 (or 4 : 8) | C : D = 4 : 3 (or 8 : 6).

Combined ratio A : B : C : D = 3 : 4 : 8 : 6.

Given D = 6 units = 72 years → 1 unit = 12 years.

Oldest is C = 8 units = 8 × 12 = 96 years.

∴ The correct answer is 96 yrs.

Alternate Method

Given:

Ratio A : B = 3 : 4

Ratio B : C = 1 : 2

Ratio C : D = 4 : 3

Age of D = 72 years

Calculations:

⇒ To combine, make B same: A : B = 3 : 4, B : C = 4 : 8

⇒ Now make C same: C : D is 4 : 3, but in combined it is 8. So, C : D = 8 : 6

⇒ Combined Ratio A : B : C : D = 3 : 4 : 8 : 6

⇒ Value of D = 6x = 72

⇒ x = 72 ÷ 6 = 12

⇒ Oldest is C (highest ratio part)

⇒ Age of C = 8 × 12 = 96 years

∴ The correct answer is 96 yrs.

Additional Information

Combined Ratio Rule

To combine A:B and B:C, multiply the ratios so that the common term (B) has the same value in both sets.

Ratio & Proportion Basics

If a : b = c : d, then the product of extremes equals the product of means (a × d = b × c).

Age Related Problems

The difference between the ages of two persons remains constant throughout their lives regardless of the time elapsed.

Shortcut Trick

Speed of train = 45 × (5/18) = 12.5 m/s.

Relative Speed with man = 12.5 − 5 = 7.5 m/s.

Length of Train (L_T) = 7.5 × 15 = 112.5 m.

Total Length (L_T + L_P) = 12.5 × 10 = 125 m.

Length of Platform (L_P) = 125 − 112.5 = 12.5 m.

Difference = 112.5 − 12.5 = 100 m.

∴ The correct answer is 100 m.

Alternate Method

Given:

Speed of train = 45 km/h = 45 × (5/18) m/s = 12.5 m/s.

Speed of man = 5 m/s.

Formula Used:

Distance = Speed × Time

Relative Speed (Same direction) = S₁ − S₂

Calculations:

⇒ Case 1 (Crossing man): Train length (L_T) = (Speed_Train − Speed_Man) × Time

⇒ L_T = (12.5 − 5) × 15 = 7.5 × 15 = 112.5 m.

⇒ Case 2 (Crossing platform): L_T + L_P = Speed_Train × Time

⇒ 112.5 + L_P = 12.5 × 10 = 125 m.

⇒ L_P = 125 − 112.5 = 12.5 m.

⇒ Difference = L_T − L_P = 112.5 − 12.5 = 100 m.

∴ The correct answer is 100 m.

Additional Information

Unit Conversion

To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5.

Relative Speed

When two objects move in the same direction, relative speed is the difference of their speeds. In opposite directions, it is the sum.

Train Crossing Lengths

When a train crosses a point object (man/pole), distance = Length of train. When it crosses a platform/bridge, distance = Length of train + Length of platform.

Shortcut Trick

Perimeter of square = 4 × 30 = 120 cm.

Perimeter of small circle = 110% of 120 = 132 cm.

Radius of small circle (r₁) = 132 ÷ (2 × 22/7) = 21 cm.

Ratio of radii 3 : 5 → 3 units = 21 cm, so 5 units (r₂) = 35 cm.

Area of larger circle = (22/7) × 35 × 35 = 110 × 35 = 3850 cm².

∴ The correct answer is 3850 cm².

Alternate Method

Given: Ratio r₁ : r₂ = 3 : 5. Side of square (a) = 30 cm. Perimeter of small circle = 1.10 × Perimeter of square.

Formula: Perimeter of square = 4a; Circumference of circle = 2πr; Area of circle = πr².

⇒ Perimeter of square = 4 × 30 = 120 cm.

⇒ Circumference of smaller circle (C₁) = 1.10 × 120 = 132 cm.

⇒ 2 × (22/7) × r₁ = 132

⇒ r₁ = (132 × 7) ÷ 44 = 3 × 7 = 21 cm.

⇒ According to ratio 3 : 5, r₂ = (21 ÷ 3) × 5 = 35 cm.

⇒ Area of larger circle = (22/7) × 35² = (22/7) × 1225 = 22 × 175 = 3850 cm².

∴ The correct answer is 3850 cm².

Additional Information

Circle Mensuration

The ratio of the circumferences of two circles is the same as the ratio of their radii (r₁:r₂), while the ratio of their areas is the square of the ratio of their radii (r₁²:r₂²).

Square Properties

For a square with side 'a', the perimeter is 4a and the area is a². The diagonal of a square is a√2.

Percentage Increase

To find a value 10% more than X, multiply X by 1.10 or 11/10. For 10% less, multiply by 0.90.

Total number of students = 1000

Number of students appeared in the NEET exam = 75% of 1000 = 750

Number of girls who appeared in the exam = (2/5) × 750 = 300

Number of girls cleared the exam = 30% of 300 = 90

Number of boys cleared the exam = 2 × 90 = 180

Required percentage = (270/750) × 100 = 36%

∴ Students who cleared the exam is 36% of the students who appeared in the exam

Total number of students = 1000

Number of students appeared in the NEET exam = 75% of 1000 = 750

Number of students attempts Only Physics sections = 30% of 750 = 225

Number of students attempts Chemistry and Biology section = 750 – 225 = 525

Let the number of students attempts Biology section = x

⇒ x + (x + 75) = 525

⇒ 2x + 75 = 525

⇒ 2x = 450

⇒ x = 225

Number of students attempts Chemistry section = 225 + 75 = 300

Number of girls who attempts Chemistry section = 300/3 = 100

Number of Boys who attempts Chemistry section = 200

Required percentage = (200/300) × 100 = 66.66%

∴ The total number of boys who attempts the Chemistry section is 66.66% of total number of students who appeared in Chemistry section

Total number of students = 1000

Number of students appeared in the NEET exam = 75% of 1000 = 750

Number of girls who appeared in the exam = (2/5) × 750 = 300

Number of girls cleared the exam = 30% of 300 = 90

Number of boys cleared the exam = 2 × 90 = 180

Required percentage = (180/750) × 100 = 24%

∴ The total number of boys who cleared the NEET exam is 24% of total number of students who appeared in the exam

Shortcut Trick

Weighted Average = (Price₁ × Ratio₁ + Price₂ × Ratio₂) ÷ Total Ratio

Average Price = (10 × 2 + 15 × 3) ÷ (2 + 3)

Average Price = (20 + 45) ÷ 5 = 65 ÷ 5 = 13

∴ The correct answer is Rs. 13 per litre.

Alternate Method

Given:

Price of 1^st variety (P₁) = Rs. 10/litre

Price of 2^nd variety (P₂) = Rs. 15/litre

Ratio of mixing (n₁ : n₂) = 2 : 3

Formula Used:

Average Price = (Total Cost) ÷ (Total Quantity)

Calculations:

⇒ Let quantities be 2L and 3L respectively.

⇒ Cost of 1^st type = 10 × 2 = Rs. 20

⇒ Cost of 2^nd type = 15 × 3 = Rs. 45

⇒ Total Cost = 20 + 45 = Rs. 65

⇒ Total Quantity = 2 + 3 = 5 litres

⇒ Average Price = 65 ÷ 5 = Rs. 13 per litre

∴ The correct answer is Rs. 13 per litre.

Additional Information

Alligation Rule

It states that (Quantity of Cheaper) / (Quantity of Dearer) = (CP of Dearer − Mean Price) / (Mean Price − CP of Cheaper).

Weighted Average Formula

Average = (n₁x₁ + n₂x₂) ÷ (n₁ + n₂), where n represents the weights/ratios and x represents the values/prices.

Shortcut Trick

Total runs of excluded innings (H + L) = (50 × 60) − (48 × 58)

H + L = 3000 − 2784 = 216 runs

Given: H − L = 182 runs

Lowest score (L) = (Sum − Difference) ÷ 2 = (216 − 182) ÷ 2 = 34 ÷ 2 = 17 runs

∴ The correct answer is 17 runs.

Alternate Method

Given: Total innings = 50, Initial Average = 60, H − L = 182, Average of 48 innings = 58.

⇒ Total runs in 50 innings = 50 × 60 = 3000

⇒ Total runs in 48 innings = 48 × 58 = 2784

⇒ Sum of Highest (H) and Lowest (L) scores = 3000 − 2784 = 216

⇒ H + L = 216

⇒ H − L = 182

⇒ Subtracting equations: (H + L) − (H − L) = 216 − 182

⇒ 2L = 34

⇒ L = 17

∴ The correct answer is 17 runs.

Additional Information

Average Formula

Average = Total Sum of Observations ÷ Number of Observations.

Sum and Difference Rule

If (x + y) = S and (x − y) = D, then the smaller value y = (S − D) ÷ 2 and the larger value x = (S + D) ÷ 2.

Weighted Average

Total Sum = (n₁ × Avg₁) + (n₂ × Avg₂). Useful when combining groups or sets of data.

Shortcut Trick

Given Upstream speed (U) = 6 kmph.

Total time for 30 km Upstream and 30 km Downstream = 8 hours.

Time Upstream = 30 ÷ 6 = 5 hours ⇒ Time Downstream = 8 − 5 = 3 hours.

Downstream speed (D) = Distance ÷ Time = 30 ÷ 3 = 10 kmph.

Time for 50 km Downstream = 50 ÷ 10 = 5 hours.

∴ The correct answer is 5 hr.

Alternate Method

Given:

Total distance Upstream (D_u) = 30 km

Total distance Downstream (D_d) = 30 km

Total time taken (T) = 8 hours

Upstream speed (U) = 6 kmph

Formula Used:

Time = Distance ÷ Speed

Calculations:

⇒ Time taken Upstream (T_u) = Distance ÷ Speed = 30 ÷ 6 = 5 hours

⇒ Total Time = T_u + T_d

⇒ 8 = 5 + T_d

⇒ T_d = 3 hours

⇒ Downstream speed (D) = Distance ÷ Time = 30 ÷ 3 = 10 kmph

⇒ Required time for 50 km Downstream = 50 ÷ 10

⇒ Required time = 5 hours

∴ The correct answer is 5 hr.

Additional Information

Speed Formulas

Speed in Still Water (x) = (D + U) ÷ 2 | Speed of Stream (y) = (D − U) ÷ 2

Downstream vs Upstream

Downstream speed (D) = x + y | Upstream speed (U) = x − y

Average Speed

If distance traveled is equal, Average Speed = (2 × D × U) ÷ (D + U)

Shortcut Trick

Total quantity = 26 + 30 = 56 kg.

Total Cost Price (CP) = (26 × 20) + (30 × 36) = 520 + 1080 = Rs. 1600.

Total Selling Price (SP) = 56 × 30 = Rs. 1680.

Profit = 1680 − 1600 = Rs. 80.

Profit% = (80 ÷ 1600) × 100 = 5%.

∴ The correct answer is 5%.

Alternate Method

Given → Quantity 1 = 26 kg at Rs. 20/kg, Quantity 2 = 30 kg at Rs. 36/kg, SP = Rs. 30/kg. Formulas: Total CP = (Q₁ × P₁) + (Q₂ × P₂) | Profit% = ((Total SP − Total CP) ÷ Total CP) × 100.

⇒ Weighted average CP per kg = ((26 × 20) + (30 × 36)) ÷ (26 + 30) ⇒ Average CP = (520 + 1080) ÷ 56 = 1600 ÷ 56 = Rs. 28.57 per kg. ⇒ Profit per kg = 30 − 28.57 = Rs. 1.43 per kg. ⇒ Profit% = (1.43 ÷ 28.57) × 100 = 5%.

∴ The correct answer is 5%.

Additional Information

Weighted Average Method

When two varieties are mixed, the average cost price lies between the prices of the individual varieties. Formula: CP_avg = (n₁c₁ + n₂c₂) / (n₁ + n₂).

Profit Percentage

Profit percentage is always calculated on the Cost Price (CP) unless stated otherwise in the question.

Shortcut Trick

Interest rate hike = 20% − 12% = 8% per year.

Number of years remaining = 5 − 2 = 3 years.

Total extra interest % = 8% × 3 = 24%.

Extra amount = 24% of ₹50,000 = ₹12,000.

∴ The correct answer is ₹12,000.

Alternate Method

Given: Principal (P) = ₹50,000, Original Rate (R₁) = 12%, New Rate (R₂) = 20%, Total Time = 5 years, Hike after 2 years.

Formula Used: Simple Interest (SI) = (P × R × T) ÷ 100

⇒ Remaining time (T) for which rate is increased = 5 − 2 = 3 years.

⇒ Difference in rates (ΔR) = 20% − 12% = 8%.

⇒ Extra Amount = (P × ΔR × T) ÷ 100

⇒ Extra Amount = (50000 × 8 × 3) ÷ 100

⇒ Extra Amount = 500 × 24 = ₹12,000.

∴ The correct answer is ₹12,000.

Additional Information

Simple Interest Concept

Simple Interest is calculated only on the principal amount. It remains constant every year if the rate and principal don't change.

Rate Change Impact

The extra interest is simply the Principal multiplied by the change in rate and the duration for which the change applies.

Total Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI). In this case, the extra amount is the difference in SI.