RRB Clerk Prelims Memory Based Paper (Held On: 19 August 2023 Shift 1) (Previous Year Paper)

Eight persons: P, Q, R, S, T, U, V and W.

1) Two players have matches between P and Q.

2) R has match on 26th April and immediate before P.

3) T’s match is immediately before S’s match.

4. U’s match is immediately after V’s match but not in the same month.

5.  S has match in the month, which has 30 days.

Thus, the above is the final arrangement.

Hence, four players have matches after R.

Eight persons: P, Q, R, S, T, U, V and W.

1) Two players have matches between P and Q.

2) R has match on 26th April and immediate before P.

3) T’s match is immediately before S’s match.

4) U’s match is immediately after V’s match but not in the same month.

5)  S has match in the month, which has 30 days.

P is on 15th October. 

Hence, correct answer is option 1.

Eight persons: P, Q, R, S, T, U, V and W.

1) Two players have matches between P and Q.

2) R has match on 26th April and immediate before P.

3) T’s match is immediately before S’s match.

4) U’s match is immediately after V’s match but not in the same month.

5)  S has match in the month, which has 30 days.

Two people have matches between S and V - true.

Hence, correct answer is option 2.

Eight persons: P, Q, R, S, T, U, V and W.

1) Two players have matches between P and Q.

2) R has match on 26th April and immediate before P.

3) T’s match is immediately before S’s match.

4) U’s match is immediately after V’s match but not in the same month.

5)  S has match in the month, which has 30 days.

P and V have match in the same month.

Hence, correct answer is option 2.

Eight persons: P, Q, R, S, T, U, V and W.

1) Two players have matches between P and Q.

2) R has match on 26th April and immediate before P.

3) T’s match is immediately before S’s match.

4) U’s match is immediately after V’s match but not in the same month.

5)  S has match in the month, which has 30 days.

S has match in the date of 15th and others have 26th.

Hence, correct answer is option 5.

Given series:- 525       852      347      293      174=

Smallest number = 174

Largest number = 852

Second digit of smallest number:- 7

Second digit of largest number:- 5

Resultant will be:- 7×5 = 35;

Hence, “35” is the correct answer.

Given series:- 525       852      347      293      174

Subtracting 1 from 1^st and 3^rd digit:- 424       751      246      192      073

Interchanging 1^st and 3^rd digit:- 424   157      642      291      370

Hence, second lowest number is 291 which resulted from “293”

Hence, “293” is correct answer.

Given series:- 525       852      347      293      174

Interchanging 2^nd and 3^rd digit:- 552       825        374        239        147

Here, “239” only one number which is a prime number.

Hence, “One” is the correct answer.

Given series:- 525       852      347      293      174

2^nd digit of third highest number is 4 (347 is the 3^rd highest number among the given series)

3^rd digit of the smallest number is 4 (174 is smallest number among the given series)

Hence, resultant will be:- 4 + 4 = 8;

Hence, “8” is the correct answer.

Given series:- 525       852      347      293      174

Arranging in ascending order:- 255       258        347         239          147

Then 2^nd largest number is 258 which resultant from 852.

Hence, “852” is the correct answer.

Given statements: E > R ≥ A > B; T ≤ X ≤ B

On combining: E > R ≥ A > B ≥ X ≥ T

Conclusions:

I. E ≥ T → False (as E > R ≥ A > B ≥ X ≥ T → E > T)

II. X < R → True (as R ≥ A > B ≥ X → R > X)

Hence, only II is True.

Given statements,

F ≥ E = J = K, R = K ≥ T

On combining, we get:

F ≥ E = J = K = R ≥ T

Conclusions**:**

I. T < F → False (F ≥ E = J = K = R ≥ T → F ≥ T).

II. F = T → False (F ≥ E = J = K = R ≥ T → F ≥ T).

Since F ≥ T is true so either conclusion I or II definitely follows.

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

Given statements,

P < D ≤ E < K,  J = E ≤ Z

On combining: 

P < D ≤ E = J ≤ Z, E < K

Conclusions**:** 

I. P < J → True (as P < D ≤ E = J → P < J).

II. D ≤ Z → True (as D ≤ E ≤ Z → D ≤ Z).

Hence, Both conclusions I and II follow.

The given information can be analysed as follows,

Hence, ‘sns’ means ‘time’ in the coded language.

The given information can be analysed as follows,

Hence, the code of ‘shut down’ is ‘lmo cef’.

The given information can be analysed as follows,

Hence, ‘hidden’ is coded as ‘drs’.

The given information can be analysed as follows,

Hence, ‘land’ is coded as ‘xyn’.

The given information can be analysed as follows,

Hence, ‘cancel’ is coded as ‘axt’

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

Distance between point K and point S is ?

KS = KJ + JS

According to the above figure:

KJ = 4 feet, JS = 4 feet

KS = 4 + 4

KS = 8 feet

Hence, ‘8 feet’ is the correct answer.

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

Distance between point K and point V is ?

KV = KJ + JS + SV

According to the above figure:

KJ = 4 feet, JS = 4 feet, SV = 4 feet

KV = 4 + 4 + 4

KV = 12 feet

Hence, the distance between K and V is 12 feet.

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

From the above diagram, S is South -West of point U.

Given,

Number - 78546932

The digits are arranged in descending order. We get, 

Here, 8, 3 and 2 remain in the same position.

Therefore, three is the correct option.

People: T, U, V, W, X, Y and Z

Days: Monday to Sunday

1. Three persons attended the yoga sessions between T and Y.

2. Y attended the sessions before W.

3. T attended the session on one of the days before Thursday.

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4. No two persons attend the session immediately before or immediately after according to English alphabet.

5. X attended the session one of the days after V.

6. One person attended the session between U and X.

This eliminates case 1 and case 2.

 

Hence, two persons attend Yoga session between Z and W.

People: T, U, V, W, X, Y and Z

Days: Monday to Sunday

1. Three persons attended the yoga sessions between T and Y.

2. Y attended the sessions before W.

3. T attended the session on one of the days before Thursday.

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4. No two persons attend the session immediately before or immediately after according to English alphabet.

5. X attended the session one of the days after V.

6. One person attended the session between U and X.

This eliminates case 1 and case 2.

 

Hence, U attends Yoga session on Friday.

People: T, U, V, W, X, Y and Z

Days: Monday to Sunday

1. Three persons attended the yoga sessions between T and Y.

2. Y attended the sessions before W.

3. T attended the session on one of the days before Thursday.

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4. No two persons attend the session immediately before or immediately after according to English alphabet.

5. X attended the session one of the days after V.

6. One person attended the session between U and X.

This eliminates case 1 and case 2.

 

Hence, V attends Yoga session on Monday.

People: T, U, V, W, X, Y and Z

Days: Monday to Sunday

1. Three persons attended the yoga sessions between T and Y.

2. Y attended the sessions before W.

3. T attended the session on one of the days before Thursday.

<br>

4. No two persons attend the session immediately before or immediately after according to English alphabet.

5. X attended the session one of the days after V.

6. One person attended the session between U and X.

This eliminates case 1 and case 2.

 

Hence, two persons attend session after U.

People: T, U, V, W, X, Y and Z

Days: Monday to Sunday

1. Three persons attended the yoga sessions between T and Y.

2. Y attended the sessions before W.

3. T attended the session on one of the days before Thursday.

<br>

4. No two persons attend the session immediately before or immediately after according to English alphabet.

5. X attended the session one of the days after V.

6. One person attended the session between U and X.

This eliminates case 1 and case 2.

 

i) X attends the Yoga session on Wednesday. → True

ii) Two persons attend Yoga session between T and X. → False

iii) W attends the Yoga session on Sunday. → True

Hence, statement I and III are correct.

Given- SHORTAGE

2nd, 5th, 6th, and 8th letters are- H, T, A, and E

The meaningful words are - HEAT and HATE

Hence,  More than one word can be formed.  the correct answer will be X

There are eight persons P, Q, R, S, T, U, V and W sitting on a rectangular table but not necessarily in the same order. Four of them are sitting on the middle side of the table facing inside and rest of them are sitting on the corners of the table facing outside.

1. V is sitting at the corner.
2. S sits second to the right of V.

3. U sits fifth to the right of S.
4. Three persons are sitting between U and P.

5. Two people are sitting between P and T.
6. Q and R sit opposite to each other where Q is not sitting near to V.

The final arrangement is shown here,

Hence, Q is the immediate left of T.

There are eight persons P, Q, R, S, T, U, V and W sitting on a rectangular table but not necessarily in the same order. Four of them are sitting on the middle side of the table facing inside and rest of them are sitting on the corners of the table facing outside.

1. V is sitting at the corner.
2. S sits second to the right of V.

3. U sits fifth to the right of S.
4. Three persons are sitting between U and P.

5. Two people are sitting between P and T.
6. Q and R sit opposite to each other where Q is not sitting near to V.

The final arrangement is shown here,

Hence, WTVS are sitting at the corners.

There are eight persons P, Q, R, S, T, U, V and W sitting on a rectangular table but not necessarily in the same order. Four of them are sitting on the middle side of the table facing inside and rest of them are sitting on the corners of the table facing outside.

1. V is sitting at the corner.
2. S sits second to the right of V.

3. U sits fifth to the right of S.
4. Three persons are sitting between U and P.

5. Two people are sitting between P and T.
6. Q and R sit opposite to each other where Q is not sitting near to V.

The final arrangement is shown here,

Hence, From the left of P, four persons are sitting between P and V.

There are eight persons P, Q, R, S, T, U, V and W sitting on a rectangular table but not necessarily in the same order. Four of them are sitting on the middle side of the table facing inside and rest of them are sitting on the corners of the table facing outside.

1. V is sitting at the corner.
2. S sits second to the right of V.

3. U sits fifth to the right of S.
4. Three persons are sitting between U and P.

5. Two people are sitting between P and T.
6. Q and R sit opposite to each other where Q is not sitting near to V.

The final arrangement is shown here,

Hence, T and S are sitting at the corners.

There are eight persons P, Q, R, S, T, U, V and W sitting on a rectangular table but not necessarily in the same order. Four of them are sitting on the middle side of the table facing inside and rest of them are sitting on the corners of the table facing outside.

1. V is sitting at the corner.
2. S sits second to the right of V.

3. U sits fifth to the right of S.
4. Three persons are sitting between U and P.

5. Two people are sitting between P and T.
6. Q and R sit opposite to each other where Q is not sitting near to V.

The final arrangement is shown here,

Hence, W and V sits opposite to each other.

Note: Some + Some not is not considered under the Either-or case.

Either - or is the case when only one of the conclusions can be true at a time, but in the case of 'Some + Some not', both the conditions can be true at a time. Hence, 'Some + Some not' is considered as 'either - or' pair.

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

Conclusions:

I. Some lake is ocean → False (as there is no relation between lake and ocean. So, it is possible but it is not definite. Hence, some lake is ocean is false).

II. Some lake is not ocean → False (as there is no relation between lake and ocean. So, it is possible but it is not definite. Hence, some lake is ocean is false).

Hence, Neither conclusion I nor II follows.

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

Conclusions:

I. Some Online are not Video → False (As At least some Video are Online, we do not know about the other part of online is video or not, so it is false)

II. All Online are Youtube → False (As Some Online is not Youtube then All Online are Youtube is not possible, so it is false).

Hence, Neither I nor II follows.

Additional Information 

1. Some X are Y means some part of the X circle is within the Y circle. 2.  No X is Y means circle X and Y are not intersecting at all.
2. All X is Y means the whole circle X lies within the Y circle.
3. If there is no exact conclusion for the given statements then the possibility case arises.
4. If Only X is Y means all Y is X.
5. Only a few:- Some + Some not.
6. At least, Many, Few, Most, Seldom, Rarely means Some.

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

Conclusions:

I. All Headset are Earphones → False (As only a few Earphones are Headset, some Headset are already Earphones but another portion of Headset is Earphones or not is possible but not definite, So it is false).

II. No Backpack is Headset → False (As some Headset are Backpack, some portion of Headset is already Backpack so No Backpack is Headset will be false).

Hence, Neither I nor II follows​.

Given Word: SANITY

We can use the positional values of letters/alphabets in the English alphabets series so as to map the letters in an easier manner.

According to the English alphabetical series, the positional values of letters are given as follows:

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The letters are mapped as shown:

Hence, no such pairs of letters in the word "SANITY" which have as many letters between them as English alphabetical series (Both in the backward and forward direction)

From the given information,

S is married to A who is mother of G.

R is sister- in- law of A. K is grandson of P and only son of R. P has one granddaughter. 

As gender of P is not defined we cannot find its relation with anyone. 

Hence, P is grandfather of G is incorrect.

From the given information,

S is married to A who is mother of G.

R is sister- in- law of A. K is grandson of P and only son of R. P has one granddaughter. 

As gender of P cannot be define hence number of male member in the organisation cannot be determined.

From the given information,

S is married to A who is mother of G.

R is sister- in- law of A. K is grandson of P and only son of R. P has one granddaughter. 

S is son of P.

Solution:

Number of bikes purchased by company E = 120

Number of bikes purchased by company F = (120 - 40) = 80

Number of cars purchased by the company E = 160

Number of cars purchased by company F = (160 + 40) = 200

Required ratio = (120 + 80): (160 + 200) = 200: 360 = 5: 9

Solution:

Average number of bikes purchased by all the companies = (100 + 80 + 100 + 200 + 120)/5 = 120

Average number of cars purchased by all the companies = (120 + 140 + 60 + 180 + 160)/ 5 = 132

Required difference = (132 - 120) = 12

Solution:

Initial number of bikes and cars purchased in the company D = (200 + 180) = 380

After increasing x% in bike purchase number of bikes and cars purchased in company D = (210 ×  2) = 420

Number of bikes purchase increased = 40

To find the value of x

100 × (240 - 200)/200 = 100 × 40/200 = 20%

Solution:

Number of bikes purchased by the company C = 100

1/5 th of bikes purchased by company C = 100/5 = 20

Number of cars purchased by company C = 60

1/3 rd of cars purchased by company C = 60 × 1/3 = 20

Required difference = (20 - 20) = 0

Solution:

Number of yamaha bikes purchased by company A = 100 × 20/100 = 20

Number of yamaha bikes purchased by company B = 80 × (30/100) = 24

Number of Tata cars purchased by company A = 120 × (25/100) = 30

Number of Tata cars purchased by company B  = 140 × (20/100) = 28

Required total = (20 + 24 + 30 + 28) = 102

Applying BODMAS rule to solve the problem,

Step - 1: Parts of the equation enclosed in ‘Brackets’ must be solved first.

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

Step - 3: Next the part of the equation which contain ‘Division’ and ‘Multiplication’ must be solved.

Step - 4: Last but not the least, the parts of the equation that contains ‘Addition’ and ‘Subtraction’ must be solved.

Now the given expression,

$\begin{array}{l} \Rightarrow \frac{1}{8} + \frac{{16}}{3} \times \frac{9}{2} - \frac{{13}}{4} \times \frac{{23}}{6} = ? \times \frac{5}{3}\ \Rightarrow \frac{1}{8} + 24 - \frac{{299}}{{24}} = ? \times \frac{5}{3}\ \Rightarrow \frac{{3 + 576 - 299}}{{24}} = ? \times \frac{5}{3}\ \Rightarrow \frac{{280}}{{24}} = ? \times \frac{5}{3}\ \Rightarrow \frac{{280 \times 3}}{{24 \times 5}} = ?\ \Rightarrow ? = \frac{{840}}{{120}} \end{array}$

⇒ ? = 7

Given:

43 × 45 + 43 × 55 - 43 × 30 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

43 × 45 + 43 × 55 - 43 × 30 = ?

⇒ 43(45 + 55 - 30) = ?

⇒ 43(100 - 30) = ?

⇒ 43 × 70 = ?

⇒ 3010

∴ The value of ? is 3010.

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

Calculation:

(175 - 25) + √25 - 10 × 6 = ?

⇒ 150 + 5 – 60 = ? 

⇒ ? = 95

∴ The required value is 95

Given:

(16900 ÷ 26) + (53 of 10 ÷ 25) = ? × 70

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(16900 ÷ 26) + (5³ of 10 ÷ 25) = ? × 70

⇒ 650 + (125 of 10 ÷ 25) = ? × 70

⇒ 650 + (1250 ÷ 25) = ? × 70

⇒ 650 + 50 = ? × 70

⇒ 700 = ? × 70

⇒ ? = 10

∴ The value of ? is 10.

Given:

427 - 11² + (32)^2/5 + (9)^3/2 – 35 × 4 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

427 - 11² + (32)^2/5 + (9)^3/2 – 35 × 4 = ?

⇒ 427 – 121 + 4 + 27 – 140 = ?

⇒ 427 + 31 – 261 = ?

⇒ 458 – 261 = ?

⇒ ? = 197

∴ The value of ‘?’ is 197.

Additional Information

(32)^2/5 = (2 × 2 × 2 × 2 × 2)^2/5

⇒ (2⁵)^2/5

⇒ (2)^{5 × (2/5)}

⇒ (2)²

⇒ 4

(9)^3/2 = (3 × 3) ^3/2

⇒ (3²)^3/2

⇒ (3)^{2 × (3/2)}

⇒ (3)³

⇒ 27

Given:

150/25 –  √625 + 183/3 + 1.2 × 50 = ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

150/25 –  √625 + 183/3 + 1.2 × 50 = ?

⇒ 150/25 –  25  + 183/3 + 1.2 × 50 = ?

⇒ 6 –  25 + 61 + 1.2 × 50 = ?

⇒ 6 – 25 + 61 + 60 = ?

⇒ ? = 102

∴ The value of ‘?’ is 102.

Given:

(√0.1024) + (√0.2401) + (√0.1225) – (√0.6400) = (?)2

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

(√0.1024) + (√0.2401) + (√0.1225) – (√0.6400) = (?)2

⇒ (√1024/10000) + (√2401/10000) + (√1225/10000) – (√6400/10000) = (?)²

⇒ (32/100) + (49/100) + (35/100) – (80/100) = (?)²

⇒ (116 – 80)/100 = (?)²

⇒ 36/100 = (?)²

⇒ ? = (√36/100)

⇒ ? = 6/10

⇒ ? = 0.6

∴ The value of ‘?’ is 0.6.

 

(√0.1024) + (√0.2401) + (√0.1225) – (√0.6400) = (?)2

⇒ 0.32 + 0.49 + 0.35 – 0.8 = (?)²

⇒ 1.16 - 0.8 = (?)²

⇒ 0.36 = (?)²

⇒ ? = √0.36

⇒ ? = 0.6

∴ The value of ‘?’ is 0.6.

Given:

(256)3/4 ÷ [1/(8)1/3]  ×  640.5 = 2?

Concept Used:

(a^m)ⁿ = a^{m × n}

a^m × aⁿ = a^{m + n}

If a^m = aⁿ then m = n

Calculation:

(256)3/4 ÷ [1/(8)1/3]  ×  640.5 = 2?

⇒ (4)3 × 2 × (8²)1/2 = 2 ?

⇒ (2²)3 × 2 × 2³ = 2 ?

⇒ 26 × 2 × 23 = 2 ?

⇒ 26 + 1 + 3 = 2 ?

⇒ 6 + 1 + 3 = ?

⇒ ? = 10

∴ The value of (?) is 10

Given:

125% of 120 + 55% of 460 - 21 ÷ 7 = 250% of ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

125% of 120 + 55% of 460 - 21 ÷ 7 = 250% of ?

⇒ (125/100) × 120 + (55/100) × 460 - 3 = (250/100) × ?

⇒ (5/4) × 120 + (11/20) × 460 - 3 = (5/2) × ?

⇒ 150 + 253 - 3 = (5/2) × ?

⇒ 400 = (5/2) × ?

⇒ ? = 160

∴ 160 should come in place of the question mark (?).

Given:

(? ÷ 7 ÷ 5) × 25 + 125 = 750

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

(? ÷ 7 ÷ 5) × 25 + 125 = 750

⇒ (? ÷ 7 ÷ 5) × 25 = 625

⇒ (? ÷ 7 ÷ 5) = 25

⇒ ? × (1/7) × (1/5) = 25

⇒ ? = 25 × 35

⇒ ? = 875

∴ The value of ? is 875.

<br>

Don't calculate (? ÷ 7/5) = 25

⇒ (? × 5/7) = 25

⇒ ? = 35

This is the wrong way to calculate the value.

Given:​

$17{1\over3} +{21{1\over2}} = 33{1\over3}$% of ?

Concept Used:

​

Calculation:

$17{1\over3} +{21{1\over2}} = 33{1\over3}$% of ?

⇒ (52/3) + (43/2) = (1/3) of ?

⇒ (1/3) of (?) = (104 + 129)/6

⇒ ? = 233/2

⇒ ? = 116.5

∴ The value of (?) is 116.5

Given:

(8)- 3 × (2)5 × (4)5 ÷ {(2)4 × (4)4 } = (8)?/2

Concept Used:

​Formula used:

x^m × xⁿ = x^{m + n}

xm ÷ xn = xm - n

Calculation:

(8)- 3 × (2)5 × (4)5 ÷ {(2)4 × (4)4 } = (8)?/2

⇒ (8)- 3 × (8)5 ÷ (8)4 = (8)?/2

⇒ 8^{(-3 + 5 - 4)} = (8)?/2

⇒ 8^-2 = (8)?/2

Comparing powers from both sides

⇒ ?/2 = - 2

⇒ ? = - 4

∴ The value of (?) is - 4.

 

When degrees are the same and base are different.

Then, you can multiply both bases and the degree remains the same.

for example,  (2)5 × (4)⁵ = (8)⁵.

Given:

√(113 - 31 + 122) ÷ √(62 × 10 + 52 - 23 × 3) = ?

Concept Used:

Calculation:

√(113 - 31 + 122) ÷ √(62 × 10 + 52 - 23 × 3) = ?

⇒ √1444 ÷ √361 = ?

⇒ 38 ÷ 19 = ?

⇒ ? = 2

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∴ The value of ? is 2.

Given:

(256)1/2 × (4)3 × (216)1/3 ÷ 48 = 2?

Concept Used:

Follow the BODMAS rule according to the table given below:

 

Calculation:

(256)1/2 × (4)3 × (216)1/3 ÷ 48 = 2?

⇒ 16 × 64 × 6 ÷ 48 = 2^?

⇒ 16 × 64 × (1/8) = 2^?

⇒ 16 × 8 = 2^?

⇒ 2⁷ = 2^?

⇒ ? = 7

∴ 7 will come in place of the question mark ('?').

Important Points

If (x)^a = (x)^b then a = b.

Given:

∛2197 + ∛1728 + ∛3375 = 8 × ?

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

∛2197 + ∛1728 + ∛3375 = 8 × ?

⇒ 13 + 12 + 15 = 8 × ?

⇒ 40 = 8 × ?

⇒ ? = 40/8 = 5

⇒ ? = 5

∴ The value of ‘?’ is 5.

Solution:

Cost of each magazine in 2010 = x

Cost of each magazine in 2012 = 150/100 × x = 1.5x

Total revenue generated by selling magazine by person A = (20 × x + 1.5x × 12) = 380

(20x + 18x) = 380

38x = 380

x = 10

Solution:

Number of magazine sold by person B in 2010 = 16

1/4 th of the magazine sold by person B in 2010 = 16/4 = 4

Number of magazine sold by person C in 2012 = 21

1/7 th of the number of magazine sold by person C = 21/7 = 3

Number of magazines sold by person C in all the three years = (30 + 14 + 21) = 65

Required percentage = (4 + 3)/65 × 100 = 10.76%

Solution:

Number of magazines sold by the person C in 2011 = 14

1/7 th of the number of magazines sold by the person C in 2011 = 14/7 = 2

Number of magazine sold by the person D in 2010 = 25

1/5 th of the number of magazines sold by person D in 2010 = 25/5 = 5

Required sum = (2 + 5) = 7

Solution:

Number of magazines sold by the person C in 2010 = 30

Number of magazines sold by the person E in 2010 = 30 × (120/100) = 36

If the number of magazines sold by person E in 2011 = 36 × (125/100) = 45

Number of magazines sold by person E in 2012 = 45 × (120/100) = 54

Required sum = (36 + 45 + 54) = 135

Solution:

Total number of magazines sold by person A initially in all the years = (20 + 18 + 12) = 50

Number of increased magazines =(55- 50) = 5

To find the value of x

New increased number of sold magazine = (20 + 5) = 25

Value of x

(25 - 20)/20 ×  100 = 100/4 = 25%

The given series is cubes of prime numbers added with 1

2³ + 1 = 9

3³ + 1 = 28

5³ + 1 = 126

7³ + 1 = 344

11³ + 1 = 1332 = ?

The given series is the squares of all the prime numbers

2² = 4

3² = 9

5² = 25

7² = 49

11² = 121

13² = 169

14 × 2 – 2 = 26

26 × 2 – 2 = 50

50 × 2 – 2 = 98

98 × 2 – 2 = 194

194 × 2 – 2 = 386

The first element in the series is 18.

And,

18 × 1 = 18

18 × 2 = 36

36 × 3 = 108

108 × 4 = 432

432 × 5 = 2160

Given that, mixture of 90 litres of the spirit and water contains 20% water.

In the first mixture

Water = (20/100) × 90 = 18 litres

∴ Spirit = 90 – 18 = 72 litres

After adding some water to the mixture, water became 50% of the new mixture

∴ 72 litres of water in new mixture = 50% of new mixture

∴ Total litres of new mixture = $\left( {\frac{{100}}{{50}} \times 72} \right) = 144{\rm{;litres}}$

So, water to be added = 144 – 90 = 54 litres

Given: average age of 28 students in class = 15

Total age of 28 students in class = 28 × 15 = 420

Let the age of teacher be x years.

If the teacher’s age is included, average age of class = 16

Thus, (420 + x)/29 = 16

⇒ 420 + x = 16 × 29

⇒ 420 + x = 464

∴ x = 44

∴ The age of teacher = 44 years

Given:

Amount = Rs. 15,950

Time = 2 years

New Amount = Rs. 16,820

Assumption:

Let the rate of interest = x%

Formula used:

Amount = Principal + Interest

Calculation:

Case 1:

P + I = 15950 (where, I = PRN/100)

⇒ P + [P × (x/100) × 2] = 15950

⇒ P + (2Px/100) = 15950

⇒ 2Px/100 = 15950 - P      ----(1)

Case 2:

New interest rate = x% + (60% of x) = x + 0.6x = 1.6x

⇒ P + I = 16820

⇒ P + [P × (160x/100)/100 × 2] = 16820

⇒ P + [2Px/100 × (160/100)] = 16820

⇒ P + [160/100(15950 - P)] = 16820

⇒ 100P + 2552000 - 160P = 1682000

⇒ 60P = 870000

⇒ P = Rs. 14500

Put the value in equation 1,

⇒ 14500 × (x/100) × 2 = 15950 - 14500

⇒ 29000 × (x/100) = 1450

⇒ 290x = 1450

⇒ Rate of Interest = 5%

∴ Required value = 5%

 

Ratio (Vessle1 : Vessel2)

⇒ (5/8 − 3/7) : (3/7 − 3/8)

⇒ 11 : 3

Detailed solution:

Given

Milk and water ration in Vessels₁ = 3 : 5

Milk and water ration in Vessels2 = 5 : 3

Calculation

In the first vessel milk = 3/8 of the mixture

In the second vessel milk = 5/8 of the mixture

It is required to have a mixture in which milk is 3/7 of the new mixture. 

Let 1st mixture : 2nd mixture be 1 : x

So the vol. of resulting mixture = 1 + x

(milk in the 1st mix.) + (milk in the 2nd mix.) = (milk in the resulting mix.)

⇒ (3/8) × 1 + (5/8) × x = (1 + x) × (3/7)

⇒ (3/8) + (5x/8) = (3/7) + (3x/7)

⇒ (5x/8 − 3x/7) = (3/7 − 3/8)

⇒ (35x − 24x)/56 =  (24 − 21)/56

⇒ 11x = 3

∴ x = 3/11

∴ These mixtures are to be mixed in 1 : 3/11 i.e. 11 : 3

Given:

Amount invested by Bani = Rs. 55,000

Share of profit of Bani and Sandeep = 11 ∶ 14

Formula used:

(Capital of Bani × Time period of Bani) ∶ (Capital of Sandeep × Time period of Sandeep) = Profit of Bani : Profit of Sandeep

Calculation:

Let the amount invested by Sandeep be Rs. ‘x’.

(Capital of Bani × Time period of Bani) : (Capital of Sandeep × Time period of Sandeep) = Profit of Bani : Profit of Sandeep

⇒ (55000 × 12) ∶ (x × 7) = 11 ∶ 14

⇒ (55000 × 12)/7x = 11/14

⇒ (5000 × 12)/x = 1/2

⇒ x = 1,20,000

∴ The amount invested by Sandeep is Rs. 1,20,000.

1. If two partners A and B are investing their money for different period of time to run a business then:

(Capital of A × Time period of A) ∶ (Capital of B × Time period of B) = (Profit or Loss of A) ∶ (Profit or Loss of B)

2. If no information has been provided about the time of investment of a partner, it means that he invests for the entire time.

Given:

MRP = 20% more than C.P

Discount = 30%

Concept Used:

Cost price × (100 + Profit %)/100 = Selling price

Discount % = {(MRP - SP)/MRP} × 100

Calculation:

Cost price = CP,  Selling price = SP,  Marked price = MRP

Let CP be 100.

MRP = 20% more than CP

⇒ MRP = 120

Discount = 120 × 30% 

⇒ 36

SP = 120 - 36

⇒ 84 

⇒ Loss = {(100 - 84)/100} × 100

∴ Loss percent is 16%

GIVEN:

Ages are consecutive odd numbers.

Youngest : Oldest (5 years ago) = 4 : 5

EXPLANATION:

Let the ages be x, x + 2, x + 4

A.T.Q

(x - 5)/(x + 4 - 5) = 4/5

⇒ 5x - 25 = 4x - 4

⇒ x = 21 years

Required age = 21 + 2 + 2 = 25 years

∴ The answer is 25 years.

Given:

One side of the square is 4 cm = Breadth of the rectangle

The area of the rectangle is thrice the area of the square

Formula used:

Area of a square = side²

Area of a rectangle = (Length × Breadth)

Calculation:

So, the area of the square = 4² cm² = 16 cm²

According to the question, the area of the rectangle = (16 × 3) cm² = 48 cm²

As per the question, the breadth of the rectangle is 4 cm

So, the length = (48/4) cm = 12 cm

∴ The length of the rectangle is 12 cm.

Given:

Time is taken by boat in still water to cover 48 km = 3 hours

Speed of the current = 4 km/hr

Formula used:

Speed = Distance/Time taken

Upstream speed = Speed of boat in still water - Speed of current

Calculation:

Speed of boat in still water = 48/3 km/hr

⇒ 16 km/hr

Upstream speed = (16 – 4) km/hr

⇒ 12 km/hr

Time is taken to cover the distance upstream = 48/12 hours

⇒ 4 hours

Difference in time = (4 – 3) hour

⇒ 1 hour

∴ The difference in time taken is 1 hour

Let A and B take x and (x + 6) hours respectively. Then,

$\Rightarrow \frac{1}{x} + \frac{1}{{x + 6}} = \frac{1}{4}$

$\Rightarrow \frac{{x + 6 + x}}{{x\left( {x + 6} \right)}} = \frac{1}{4}$

⇒ 8x + 24 = x (x + 6)

⇒ x² – 2x – 24 = 0

⇒ x = 6

Hence, time taken by A = 6 hours.

The number series looks like a difference series by first look.

Series:         67        82     98       118      147     192

Differences:        15      16       20       29        45

Double Diff:               1        4          9       16

Hence, 192 is correct.