RRB Clerk Prelims Memory Based Paper (Held On: 14 August 2022 Shift 2) (Previous Year Paper)

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Forward Pair: TU, DE

Backward Pair: EI

Hence, Three such pairs are there

Given series: 

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B& 1 X ^ Z U * 4

The pattern is as follows: 

Number → Consonants → Number 

Lets mark consonants

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4

such consonants are there in the series which is neither immediately preceded nor immediately followed by a number is 

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4

1. O V B
2. V B &
3. ^ Z U

Hence, three is the correct answer.

Given series:

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4

The element which is seventh to the left of the fifth symbol from the right end

fifth symbol from the right end(Only symbol)

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4

Now seventh left of ! is H

Hence, H is the correct answer.

Given series:

Left End  9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4  Right End

Symbols and Consonants are dropped, the series will be

Left End  9 A 3 7 I 8 6 5 E 3 O 1 U 4  Right End

Now, tenth from the right end is I.

Hence, I is the correct answer.

Given series :

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4

Therefore, the position of “E” with respect to “B” is Sixth to the left.

Hence, Sixth to the left is the correct answer.

Given series:

9 @ A 3 H 7 $ I M 8 K ! 6 L 5 E % J 3 O V B & 1 X ^ Z U * 4

Required Order→Symbol→Number

! 6, & 1, and * 4

Hence, three is the correct answer.

Eight persons: A, B, C, D, E, F, G, and H.

Months: January, March, May, June, July, August, November, December.

1) G is celebrating his birthday in one of the months which has 30 days but before November.

2) Three persons celebrate between H and G.

3) C's birthday celebration is just before H.

4) The number of persons who celebrate after C is one less than the number of persons who celebrate before A.

5) The number of birthdays between G and D is the same as the number of birthdays between G and F.

6)  D celebrates his birthday just before C.

7) B celebrates his birthday before E.

Hence, C celebrates his birthday in November.

Eight persons: A, B, C, D, E, F, G, and H.

Months: January, March, May, June, July, August, November, December.

1) G is celebrating his birthday in one of the months which has 30 days but before November.

2) Three persons celebrate between H and G.

3) C's birthday celebration is just before H.

4) The number of persons who celebrate after C is one less than the number of persons who celebrate before A.

5) The number of birthdays between G and D is the same as the number of birthdays between G and F.

6)  D celebrates his birthday just before C.

7) B celebrates his birthday before E.

Hence, E celebrates his birthday in July.

Eight persons: A, B, C, D, E, F, G, and H.

Months: January, March, May, June, July, August, November, December.

1) G is celebrating his birthday in one of the months which has 30 days but before November.

2) Three persons celebrate between H and G.

3) C's birthday celebration is just before H.

4) The number of persons who celebrate after C is one less than the number of persons who celebrate before A.

5) The number of birthdays between G and D is the same as the number of birthdays between G and F.

6)  D celebrates his birthday just before C.

7) B celebrates his birthday before E.

Hence, Three persons celebrate birthday between F and D.

Eight persons: A, B, C, D, E, F, G, and H.

Months: January, March, May, June, July, August, November, December.

1) G is celebrating his birthday in one of the months which has 30 days but before November.

2) Three persons celebrate between H and G.

3) C's birthday celebration is just before H.

4) The number of persons who celebrate after C is one less than the number of persons who celebrate before A.

5) The number of birthdays between G and D is the same as the number of birthdays between G and F.

6)  D celebrates his birthday just before C.

7) B celebrates his birthday before E.

Hence, G celebrates his birthday just after A.

Eight persons: A, B, C, D, E, F, G, and H.

Months: January, March, May, June, July, August, November, December.

1) G is celebrating his birthday in one of the months which has 30 days but before November.

2) Three persons celebrate between H and G.

3) C's birthday celebration is just before H.

4) The number of persons who celebrate after C is one less than the number of persons who celebrate before A.

5) The number of birthdays between G and D is the same as the number of birthdays between G and F.

6)  D celebrates his birthday just before C.

7) B celebrates his birthday before E.

 

F, H, D, and A celebrated their birthdays in the months having 31 days but C celebrated his birthday in November which has 30 days.

Hence, C is different from this group.

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

Hence, H is north east of O.

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

Hence, the distance between M and F is 40 so, 40 + 40 = 80 km.

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

Hence, O is west of F.

Given words: NAIL and HEAD.

1. 1^st and 4^th letter letters of the word 'NAIL': N, L

2. 3^rd and 4^th letter of the word 'HEAD': A, D.

The word can be formed with the letters N, L, A and D - LAND.

There is only one word formed.

Hence, the answer is N.

If the second and the third digits are interchanged within each number, we get:

527 → 572

716 → 761

714 → 741

417 → 471

536 → 563

Now, the numbers thus formed are arranged in descending order from right to left:

471 < 563 < 572 < 741 < 761

Hence, 572 is the third number from the right end in the new arrangement.

If the first and the third digits are interchanged within the numbers, we get:

527 → 725

716 → 617

714 → 417

417 → 714

536 → 635

So, arranging the numbers thus formed → 417 < 617 < 635 < 714 < 725

Thus, the second highest number is 714.

Product of the first and second digit of 714 → 7 * 1 = 7

Hence, 7 is the product of the first and second digit of the second-highest number.

If the digits are arranged in descending order from left to right within the numbers, we get:

527 → 752

716 → 761

714 → 741

417 → 741

536 → 653

Thus, 752 is the only number which is divisible by2

Hence, only one number is divisible by 2.

If the digits are arranged in ascending order form left to right within the numbers, we get:

527 → 257

716 → 167

714 → 147

417 → 147

536 → 356

147 = 147 < 167 < 257 < 356

Hence, both 714 and 417 will be the lowest numbers.

If the first digit is interchanged with the second digit within the numbers, we get:

527 → 257

716 → 176

714 → 174

417 → 147

536 → 356

Now, the numbers thus formed are arranged in descending order from left to right:

356 > 257 > 176 > 174 > 147

Hence, 257 is third to the left of first number from the right end in the new arrangement.

1)Three persons are sitting between X and T, who are sitting adjacent to Q.

2. H is sitting 2nd to the right of Q. X is in left of Q.

3. Two persons are sitting between H and S who is sitting at one of the ends.

4. S is not sitting adjacent to Q, who is sitting 2nd to the right of L.

5. L is sitting exactly in the middle of the row.

6. J is sitting 6th from the left end.

7. X is not sitting to the right of J.Hence, Case-1 will be elliminated.

4. Two persons are sitting between J and R.

5. Two persons are sitting between X and K.

Therefore the final arrangement is, 

Hence, the persons are sitting between X and S are nine.

1)Three persons are sitting between X and T, who are sitting adjacent to Q.

2. H is sitting 2nd to the right of Q. X is on the left of Q.

3. Two persons are sitting between H and S who is sitting at one of the ends.

4. S is not sitting adjacent to Q, who is sitting 2nd to the right of L.

5. L is sitting exactly in the middle of the row.

6. J is sitting 6th from the left end.

7. X is not sitting to the right of J.Hence, Case-1 will be eliminated.

 

4. Two persons are sitting between J and R.

5. Two persons are sitting between X and K.

Therefore the final arrangement is, 

Hence, persons are sitting in the row are 15.

1)Three persons are sitting between X and T, who are sitting adjacent to Q.

2. H is sitting 2nd to the right of Q. X is on the left of Q.

3. Two persons are sitting between H and S who is sitting at one of the ends.

4. S is not sitting adjacent to Q, who is sitting 2nd to the right of L.

5. L is sitting exactly in the middle of the row.

6. J is sitting 6th from the left end.

7. X is not sitting to the right of J.Hence, Case-1 will be eliminated.

4. Two persons are sitting between J and R.

5. Two persons are sitting between X and K.

Therefore the final arrangement is, 

Hence**, the person is sitting to the left of Q is nine.**

1)Three persons are sitting between X and T, who are sitting adjacent to Q.

2. H is sitting 2nd to the right of Q. X is on the left of Q.

3. Two persons are sitting between H and S who is sitting at one of the ends.

4. S is not sitting adjacent to Q, who is sitting 2nd to the right of L.

5. L is sitting exactly in the middle of the row.

6. J is sitting 6th from the left end.

7. X is not sitting to the right of J.Hence, Case-1 will be eliminated.

4. Two persons are sitting between J and R.

5. Two persons are sitting between X and K.

Therefore the final arrangement is, 

Hence, the position of H with respect to T is 3rd to the right.

1)Three persons are sitting between X and T, who are sitting adjacent to Q.

2. H is sitting 2nd to the right of Q. X is on the left of Q.

3. Two persons are sitting between H and S who is sitting at one of the ends.

4. S is not sitting adjacent to Q, who is sitting 2nd to the right of L.

5. L is sitting exactly in the middle of the row.

6. J is sitting 6th from the left end.

7. X is not sitting to the right of J.Hence, Case-1 will be eliminated.

4. Two persons are sitting between J and R.

5. Two persons are sitting between X and K.

Therefore the final arrangement is, 

Hence, X sits exactly in the middle of K and L.

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

Conclusions:

I. All Car is Cycle → False ( Some Car is Train so we cannot say that All Car is Cycle it is possible but not definite so conclusion false )

II.No Bus is Car → False (There is no direct relation between Bus and Car it is possible but not definite so conclusion false )

Hence, None follow.

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

Conclusions:

I. Some Dogs are Zebra ⇒ False ( As it is possible but not definite, as no direct relation between the two is given).

II. All Zebra can be dog ⇒ True ( As there is no relation between Zebra and Dog, so the possibility is true).

Hence, Only conclusion II follows.

The least possible Venn diagram is shown below,

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i) Some Help are Box ⇒ False (This is possible but not definite)

ii) Some Box are Fox ⇒ True (As only a few fox are box hence some box will definitely be fox.)

Hence, only II follows.

The least possible Venn diagram is shown below,

 

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i) Some classroom are music ⇒ True only if case 2 is to be considered otherwise False

ii) No classroom is music ⇒ True only if case 1 is to be considered otherwise False

Hence, either I or II follows.

Given statements: F ≥ L ≥ M < N = P ≥ A ≥ B > C 

Conclusions:

I. F ≥ C → False (As  F ≥ L ≥ M < N = P ≥ A ≥ B > C, there is no direct relation between F and C. So, it is false)

II. F < C → False (As  F ≥ L ≥ M < N = P ≥ A ≥ B > C, there is no direct relation between F and C. So, it is false)

Here, both conclusions I and II forms a complementary pair as both have the same element and all the three symbols (<, =, >) are present between F and C.

Hence, Either I or II follows.

Given Statement: U ≥ T ≥ M = W ≤ E ≥ L = H

Conclusions: 

I. U ≥ E → False (As U ≥ T ≥ M = W ≤ E there is no clear relation between U and E because of the opposite sign between U and E. So, it is false).

II. E > U → False (As U ≥ T ≥ M = W ≤ E there is no clear relation between U and E because of the opposite sign between U and E. So, it is false).

As the elements in the conclusions are the same i.e., U and E, and there are three signs (>, <, and =) between U and E they form a complementary pair.

Hence, Either I or II follows

Given Statement: E ≥ W = A < B ≤ S ≤ T < X

Conclusions: 

I. X > W → True (As W = A < B ≤ S ≤ T < X → X > W. So, it is true).

II. B > X → False (As B ≤ S ≤ T < X → B < X. So, it is false).

Hence, Only I follow.

Given Statement: E ≥ W = A < B ≤ S ≤ T < X

Conclusions: 

I. E > T → False (As E ≥ W = A < B ≤ S ≤ T there is no relation between E and T because of the opposite sign between E and T).

II. A = X → False (As A < B ≤ S ≤ T < X → A < X. So, it is false).

Hence, Neither I nor II follows.

Person: M, N, O, P, Q, R, S, and T

Floor: 1, 2, 3, 4.

1. M lives on a Flat 2 of an odd floor.
2. Only one floor lies between M and Q.
3. Q lives east of N.

 

4. There is one floor between lives between Q and O.

 

5. P lives immediately above N.

 

6. R and T lives in a same floor.

 

7. T lives immediately below Q.

 

Here, N and Q lives on third floor

Hence, the answer is N, Q.

Person: M, N, O, P, Q, R, S, and T

Floor: 1, 2, 3, 4.

1. M lives on a Flat 2 of an odd floor.
2. Only one floor lies between M and Q.
3. Q lives east of N.

 

4. There is one floor between lives between Q and O.

 

5. P lives immediately above N.

 

6. R and T lives in a same floor.

 

7. T lives immediately below Q.

 

Here four persons lives above R but Exactly only 2 persons are living above R. (exactly above R means Living vertically above R which same numbered flat in different floors)

Hence, the answer is Two.

Person: M, N, O, P, Q, R, S, and T

Floor: 1, 2, 3, 4.

1. M lives on a Flat 2 of an odd floor.
2. Only one floor lies between M and Q.
3. Q lives east of N.

 

4. There is one floor between lives between Q and O.

 

5. P lives immediately above N.

 

6. R and T lives in a same floor.

 

7. T lives immediately below Q.

 

Here M lives in Ground floor and In ground floor M lives in Second flat

Hence, the answer is Ground floor second flat.

Person: M, N, O, P, Q, R, S, and T

Floor: 1, 2, 3, 4.

1. M lives on a Flat 2 of an odd floor.
2. Only one floor lies between M and Q.
3. Q lives east of N.

 

4. There is one floor between lives between Q and O.

 

5. P lives immediately above N.

 

6. R and T lives in a same floor.

 

7. T lives immediately below Q.

 

Here all the pairs are neighbours which means they are living in a same floor but different flats except O, T (they live in different floors)

Hence, the answer is O, T.

Person: M, N, O, P, Q, R, S, and T

Floor: 1, 2, 3, 4.

1. M lives on a Flat 2 of an odd floor.
2. Only one floor lies between M and Q.
3. Q lives east of N.

 

4. There is one floor between lives between Q and O.

 

5. P lives immediately above N.

 

6. R and T lives in a same floor.

 

7. T lives immediately below Q.

 

Here neighbour of S is P.

Hence, the answer is P.

The pattern followed here is:

1. BQCR

2. DSEU

3. FVGW

 

4)HXIY

Hence, 'DSEU' is odd among all.

Let’s draw the diagram by using the information given below.

1. Biology teacher is sitting in between Economics and Chemistry teacher.

2. Economics teacher who is closest to Hindi teacher is sitting in adjacent right of Biology teacher. 

3. History and Civics teachers are sitting together.

4. Chemistry teacher is sitting opposite to History teacher.

5. Physics and Hindi teachers are sitting farthest from each other.

6. The only left choice is English teacher as asked in the question.

Hence the correct answer is Economics teacher.

Concept used: 

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

Calculation:

672 ÷ 4 + 73 × 3 = ?

⇒ 168 + 219 = ?

⇒ 387 = ?

∴ The value of ? is 387

Given:

$56\frac{1}{2}$

Concept used:

We have to follow the BODMAS rule

Calculation:

113/200 of 1400 – ? = 230

⇒ 791 – ? = 230

⇒ ? = 791 – 230 = 561

∴ The value of ? is 561.

Concept used:

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

Calculations:

900 + 90 × 50 ÷ (144 ÷ 8) = ?

⇒ 900 + 90 × 50 ÷ 18 = ?

⇒ 900 + 90 × 50 ÷ 18 = ?

⇒ 900 + 250 = ?

⇒ 1150 = ?

∴ The value of ? is 1150.

Concept used:

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

Calculation:

785 – 215 + 615 – 385 = ?

⇒ 785 + 615 – 215 – 385 = ?

⇒ 1400 – 600 = ?

⇒ 800 = ?

∴ The value of ? is 800.

Concept used:

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

Calculation:

(560 ÷ 32) × (720 ÷ 48) = ?

⇒ 17.5 × 15

⇒ 262.5

∴ The value of ? is 262.5.

Given:

14.28% of 490 – 71.43% of 63 = ?

Concept used:

To solve this type of question, we can take use of conversion of percentage into fraction. 

Also, follow the BODMAS rule according to the table given below:

Calculation:

Converting given percentages into fraction,

⇒ 14.28% = 1/7

⇒ 71.43% = 5/7

According to the given equation,

⇒ 1/7 of 490 – 5/7 of 63 = ?

⇒ 1/7 × 490 – 5/7 × 63 = ?

⇒ 70 – 45 = ?

⇒ 25 = ?

∴ The value of ? is 25.

Given:

42% of 250 + 115% of 480 = ?

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

42% of 250 + 115% of 480 = ?

⇒ (42/100) × 250 + (115/100) × 480 = ?

⇒ 105 + 552 = ?

⇒ 657 = ?

∴ The value of ? is 657

Given:

(63 + 7 – 3) × 3 = 142 + ?

Concept used:

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

Calculation:

(63 + 7 – 3) × 3 = 142 + ?

⇒ (70 – 3) × 3 = 142 + ?

⇒ 67 × 3 = 142 + ?

⇒ 201 = 196 + ?

⇒ ? = 201 – 196 

⇒ ? = 5

∴ Required number is 5

Given:

42% of 360 + ? % of 320 = 348

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

Using the BODMAS rule

⇒ 42% of 360 + ? % of 320 = 348

⇒ 42/100 × 360 + ? % of 320 = 348

⇒ 151.2 + ? % of 320 = 348

⇒ ? % of 320 = 348 – 151.2

⇒ ? % of 320 = 196.8

⇒ ?/100 × 320 = 196.8

⇒ ? = 196.8/3.2

⇒ ? = 61.50

∴ 61.50 will come in place of ‘?’

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation :

15% of 600 + 20% of 1000 = ?

⇒ (15/100) × 600 + ( 20/100) × 1000 = ?

⇒ 90 + 200 = ?

⇒ ? = 290

∴ The value of ? is 290

Concept:

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

Calculations:

The given expression is:

(45²) + (21²) = (?)² + 257

⇒ (?)² = (45²) + (21²) - 257

⇒ (?)² = 2025 + 441 - 257 = 2209

⇒ ? = $\sqrt {2209}$

⇒ ? = 47

Hence, 47 is the correct answer.

Concept Used:

Calculation:

⇒ 13 × 14 ÷ 7 + 12 = 20 + ?

⇒ 13 × 2 + 12 = 20 + ?

⇒ 26 + 12 = 20 + ?

⇒ 38 = 20 + ?

⇒ ? = 38 - 20 = 18

∴ ? = 18

Concept used:

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

Calculation:

(324 ÷ 4)^1/2 × 5 + 11 = ?

⇒ (81)^1/2 × 5 + 11 = ?

⇒ 9 × 5 + 11 = ?

⇒ 45 + 11 = ?

⇒ 56 = ?

∴ The value of ? is 56.

Concept used:

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

Calculation:

(1/4) + (22/7) + (32/15) + ? = 6

⇒ 6 – [(1/4) + (22/7) + (32/15)] = ?

⇒ 6 – [(105 + 1320 + 896)/420] = ?

⇒ 6 – (2321/420) = ?

⇒ (2520 – 2321)/420 = ?

⇒ 199/420 = ?

∴ The value of ? is 199/420

Concept Used:

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

Calculation:

25% of 900 ÷ 0.75 + √289 = ?

⇒ 25 × 9 ÷ 0.75 + 17 = ?

⇒ 25 × 9 × (4/3) + 17 = ?

⇒ 300 + 17 = ?

⇒ 317 = ?

∴ The value of ? is 317

Given:

Formula Used:

Required Percentage = (First Quantity/Second Quantity) × 100

Calculation:

On the basis of given bar graph the following table can be formed

 

From the table formed above total number of male in all four years = (70 + 55 + 60 + 45)% = 230%

And total number of female in all four years = (30 + 45 + 40 + 55)% = 170%

∴ Required Percentage = (230/170) × 100 ≈ 136%

Hence, option (3) is correct

Given:

The number of male join the gym in 2016 is 750 more than the number of male who join the gym in 2017.

Formula Used:

Basic concept of arithmetic and percentage

Calculation:

On the basis of given bar graph the following table can be formed

 

It is given that the number of male join the gym in 2016 is 750 more than the number of male who join the gym in 2017. The number of male join the gym in 2016 is 70% and the number of male who join the gym in 2017 is 55%

∴ 70% - 55% = 750

⇒ 15% = 750

⇒ 45% = 750 × 3 = 2250

Now, number of female in 2017 = 45% = 2250

Hence, option (1) is correct

Formula Used:

Required Percentage = (First Quantity/Second Quantity) × 100

Calculation:

On the basis of given bar graph the following table can be formed

 

From the table formed above the number of male joining the gym in 2016 is 70% and the number of female who join the gym in 2017 is 45%

∴ Required percentage = (45/70) × 100 = 64.28%

Hence, option (2) is correct

Given:

The difference of male and female who join the gym in 2019 is 450

Formula Used:

Basic concept of arithmetic and percentage

Calculation:

On the basis of given bar graph the following table can be formed

 

From the table formed above 45% male and 55% female join the gym in 2019

∴ 55% - 45% = 450

⇒ 10% = 450

⇒ 1% = 45

Now average of number of male and female who join gym in 2019 = 50% = 50 × 45 = 2250

Hence, option (1) is correct

Given:

The number of male who join the gym in 2018 is three times the number of male join the gym in 2017 and the number of female in 2017 is 900

Formula Used:

Basic concept of arithmetic and percentage

Calculation:

On the basis of given bar graph the following table can be formed

 

Let the total number of male and female in 2017 is x and the number of female in 2017 is 45%

∴ 45% of x = 900

⇒ x = 2000

Now, number of male in 2018 is three time of male in 2017

∴ Number of male in 2018 = 3 × (55% of 2000) = 3300

Number of male in 2018 is 60%and number of female in 40%

∴ 60% = 3300

⇒ 1% = 55

Now, number of female in 2018 = 40% = 40 × 55 = 2200

Hence, option (1) is correct

Concept:

Distance = Speed × Time

To convert km/hr into m/s, multiply by 5/18

Calculation:

Total distance covered by train in 48 seconds with the speed of 120 km/hr.

Distance = 120 km/hr × 48 sec

= 120 × 5/18 × 48

= 1600 metres.

Length of train + Length of platform = 1600 metres

600 + length of platform = 1600

Length of platform = 1000 m

Shortcut Trick

Growth rate 5% → Fraction 1/20. Multiplier for 1 year = 21/20.

For 2 years (2019 to 2021), Multiplier = (21/20) × (21/20) = 441/400.

Given 441 units = 44,100 ⇒ 1 unit = 100.

Initial population (400 units) = 400 × 100 = 40,000.

∴ The correct answer is 40,000.

Alternate Method

Given: Final Population (A) = 44,100, Rate (R) = 5%, Time (n) = 2 years.

Formula Used: A = P × (1 + R/100)ⁿ

⇒ 44,100 = P × (1 + 5/100)²

⇒ 44,100 = P × (105/100) × (105/100)

⇒ 44,100 = P × (21/20) × (21/20)

⇒ 44,100 = P × (441/400)

⇒ P = (44,100 × 400) ÷ 441

⇒ P = 100 × 400

⇒ P = 40,000

∴ The correct answer is 40,000.

Additional Information

Population Growth Formula

Population after 'n' years is given by P(1 + R/100)ⁿ, where P is the initial population and R is the annual growth rate.

Population Decrease Formula

If the population decreases, the formula becomes P(1 − R/100)ⁿ, similar to the depreciation formula in accounting.

Successive Percentage Change

For two years of growth R₁ and R₂, the net effective growth rate is (R₁ + R₂ + R₁R₂/100)%.

Given:

A is 50% more efficient than B.

Time is taken by A to finish the work = time taken by B – 15

Concept used:

Work = time × efficiency

When work done by A and B is the same, t ∝ 1/e; where ‘t’ is time and ‘e’ is efficiency.

Calculation:

The ratio of efficiency of A to B = 3 ∶ 2

The ratio of time of A to B = 2 ∶ 3

Then, time is taken by A to finish the task = time taken by B – 15

⇒ 2x = 3x – 15

⇒ x = 15

Time taken by A to finish work = 2x = 30 hrs

Then, time is taken by A to do double the work = 2 × 30 = 60 hours.

∴ Time is taken by A to finish double the work is 60 hours.

Given:

 A lent 8000 to B for 3 years

 A lent 10000 to C for 4 years

 Interest earned in total = 6400

 Rate is the same in both

Concept:

Since rate is same, simple interest method can be used to find the rate

Calculation:

Let the rate be R

Then, ((8000 × R × 3)/100) + ((10000 × R × 4)/100) = 6400

⇒ 640 × R = 6400

⇒ R = 10

∴ The rate of interest is 10%

Given:

Time taken to travel 72 km upstream = 8 hours

Time taken to travel 72 km downstream = 4 hours

Formula used:

Speed = Distance/Time

Speed of boat in still water = (Upstream speed + Downstream speed)/2

Calculation:

Speed of boat in upstream = 72/8 = 9 kmph

Speed of boat in downstream = 72/4 = 18 kmph

∴ Speed of boat in still water = (9 + 18)/2 = 27/2 = 13.5 kmph

Calculation:

The series follows the following pattern

22 + 1² = 23

23 + 2² = 27

27 + 3² = 36

36 + 4² = 52

52 + 5² = 77

∴ The missing number is 27.

Given:

The number series:

16, 8, 8, ?, 64, 512

Concept Used:

Number series following a fixed pattern.

Calculation;

The given number series follows the pattern:

16 × 2^-1 = 8

8 × 2⁰ = 8

8 × 2¹ = 16

16 × 2² = 64

64 × 2³ = 512

∴ The number coming in place of “?” in the given number series is 16

Calculation:

The series follows following pattern

⇒ 128 × 0.5 = 64

⇒ 64 × 0.5 = 32

⇒ 32 × 0.5 = 16

⇒ 16 × 0.5 = 8

⇒ 8 × 0.5 = 4

∴ The value of ? is 16

Calculation:

The pattern of the number series is-

⇒ 7 × 1 = 7

⇒ 7 × 2 = 14

⇒ 7 × 3 = 21

⇒ 7 × 4 = 28

⇒ 7 × 5 = 35

⇒ 7 × 6 = 42

∴ The value of ? is 28

Given:

The number series:

12, 13, 28, ?, 352, 1765

Concept Used:

Number series following a fixed pattern.

Calculation;

The given number series follows the pattern:

12 × 1 + 1 =  13

13 × 2 + 2 =  28

28 × 3 + 3 =  87

87 × 4 + 4 =  352

352 × 5 + 5 =  1765

∴ The number in place of “?” in the given number series is 87

Given:

Number of science students in college A = (3/5)th of the total number of students in college S

Formula used:

Percentage = (Favourable value/Base value) × 100

Calculation:

Total number of students in college S = 100 + 65 + 35

⇒ 200 students

So, number of science students in college A = (3/5) × 200

⇒ 120 students

Number of science students in college R = 75

Now, required percentage = (75/120) × 100

⇒ 62.5%

∴ The number of students in college R is 62.5% of the number of science students in college A.

Calculation:

Number of arts students in colleges P and T = 60 + 80

⇒ 140 students

Number of commerce students in colleges Q, S and T = 50 + 65 + 95

⇒ 210 students

Now, required ratio = 140 ∶ 210

⇒ 2 ∶ 3

∴ The ratio of arts students in colleges P and T together to commerce students in colleges Q, S and T together is 2 ∶ 3.

Formula used:

Average = (Sum of values)/(Number of values)

Calculation:

Total number of students in college P = 85 + 45 + 60 = 190 students

Total number of students in college Q = 70 + 50 + 55 = 175 students

Total number of students in college R = 75 + 85 + 50 = 210 students

Total number of students in college S = 100 + 65 + 35 = 200 students

and Total number of students in college T = 60 + 95 + 80 = 235 students

So, total number of students in all the colleges = 190 + 175 + 210 + 200 + 235 = 1010 students

Now, required average = 1010/5

⇒ 202 students

∴ The average number of students in all the colleges together is 202.

Formula used:

Percentage = (Favourable value/Base value) × 100

Calculation:

Total number of students in college Q = 70 + 50 + 55

⇒ 175 students

Total number of students in college R = 75 + 85 + 50

⇒ 210 students

Now, required difference = 210 - 175

⇒ 35 students

Then, required percent = (35/175) × 100

⇒ 20%

∴ The total number of students in college R is 20% more than that in college Q.

Calculation:

Total number of students in science stream = 85 + 70 + 75 + 100 + 60

⇒ 390 students

Total number of students in arts stream = 60 + 55 + 50 + 35 + 80

⇒ 280 students

Now, required difference = 390 - 280

⇒ 110 students

∴ The difference between the total number of students in science stream and the total number of students in arts stream is 110.

Shortcut Trick

Initial Alcohol % = 9.09% = 1/11 → Alcohol : Water = 1 : 10.

Target Alcohol : Water = 1 : 2. Equate Water (multiply by 5) → 5 : 10.

Initial mixture (1 + 10) = 11 units = 121 liters ⇒ 1 unit = 11 liters.

Alcohol added = (5 − 1) = 4 units = 4 × 11 = 44 liters.

∴ The correct answer is 44 liters.

Alternate Method

Given: Total weight = 121 liters, Alcohol % = 9.09%.

Formula: Ratio = Quantity 1 ÷ Quantity 2

⇒ Quantity of Alcohol = 9.09% of 121 = (1/11) × 121 = 11 liters.

⇒ Quantity of Water = 121 − 11 = 110 liters.

⇒ Let x liters of Alcohol be added.

⇒ (11 + x) / 110 = 1 / 2

⇒ 2 × (11 + x) = 110

⇒ 22 + 2x = 110 ⇒ 2x = 88

⇒ x = 44 liters.

∴ The correct answer is 44 liters.

Additional Information

Percentage to Fraction Conversion

Recognizing 9.09% as 1/11 and 11.11% as 1/9 is crucial for Banking and SSC exams to speed up calculations.

Constant Quantity Logic

When one component (Alcohol) is added, the other component (Water) remains constant. Use this to equate ratio parts quickly.

Alligation Method

Can be solved by taking Alcohol %: 9.09% (Initial) vs 100% (Added Alcohol) to get 33.33% (Resultant 1/3) in the center.

The correct answer is: 3

Given:

The ratio of incomes = 3:2

the ratio of expenditures = 5:3

Formula used:

Income = expenditure + savings

Calculation:

Let the income ratio be x 

∴ their incomes are 3x and 2x

also let the expenditure ratio be y

∴ their expenditures are 5y and 3y

then, 3x - 5y = 1000             —(1)

and   2x - 3y = 1000             —(2)

∴ from (1) and (2) we get,

   6x - 10y = 2000

   6x - 9y = 3000

on solving,

 y = 1000

and x = 2000

⇒ X's income = 3x = 6000

∴ X's income is Rs 6000

Given:

Age of Leo : Age of David = 4 : 3

Age of Leo after 5 years : Age of David after 5 years = 9 : 7

Calculation:

Let us assume,

Present age of Leo = 4x

Present age of David = 3x

Difference in their ages = 4x – 3x = x

ATQ,

(4x + 5)/(3x + 5) = 9/7

⇒ 28x + 35 = 27x + 45

⇒ x = 10

∴ Difference in their ages is 10 years.

Given∶       

Seema and Abhi enter into a partnership with capitals in the ratio 9 ∶ 7.

Seema withdraws her capital at the end of 7 months.

The ratio of profit received by them is 9 ∶ 7.

Calculation∶

Let Seema invested 9x for 7 months.

And Abhi invested 7x for y months.

Seema’s profit = (share × Time)_S = 9x × 7

Abhi’s profit = (share × Time)_A = 7x × y

Ratio of their profits = 9 ∶ 11

So, $\frac{{9{\rm{x}} \times 7{\rm{}}}}{{7{\rm{x}} \times {\rm{y}}}} = \frac{9}{{11}}$         

⇒ $\frac{9}{{y}} = \frac{9}{{11}}$

⇒ y = 11

⇒ y = 11 months

Given:

Ravi purchased two articles each article cost = Rs. 1,500

When he sold,

Gain on first article = 6% 

Loss on second article = 4%

Calculations:

Selling  price of first article = 1500 + 1500 × 6/100

⇒ 1500× (1 + 6/100) = 1500 × ( 100 + 6) ÷ 100

⇒ 15 × 106 = 1590

Selling  price of second article = 1500 - 1500 × 4/100

⇒ 1500 × (1 - 4/100) = 1500 × ( 100 - 6) ÷ 100

⇒ 15 × 96 = 1440

The total cost price = cost price of first article + cost price of second article

⇒ 1500 + 1500 = 3000

The total selling price = selling price of first article + selling price of second article 

⇒ 1590 + 1440 = 3030

The total cost difference of selling price and cost price whole transaction = total selling price - total cost price 

⇒ 3030 - 3000 = 30

Profit = (30/3000) × 100   (∵ selling price > cost price)

⇒ (1/100) × 100 = 1%

∴ The total transaction gain 1%