# QUESTIONS ON REASONING (PART-1)

#### QUESTIONS ON REASONING (PART-1)

Most of these questions are taken from the previous examinations conducted by the Staff Selection Commission (SSC) of the General Intelligence and Reasoning section of the following exams as well as other exams. They are all solved and supported by detailed explanation.

1. Combined Graduate Level (CGL) Exam Tier-I

2. Combined Higher Secondary (10+2) Exam (CHSL) Tier-I

3. SI in Delhi Police and CPO Exam Tier-I

4. Stenographers Exam

5. Grade-II DASS Exam conducted by Delhi Staff Subordinate Services (DSSSB)

#### QUERY 1

**If 25/5 = 15, 30/6 = 20; then 35/7 = ?**

**nanda kishore**

25/5 is equal to 5 ……………..multiply 5 with 3

30/6 is equal to 5………………multiply 5 with 4

35/7 is equal to 5……………….multiply 5 with 5……………….so the answer is 25

#### QUERY 2

**If 33+45 = 30, 90+26 = 40; Then 30+45 = ?**

**nanda kishore**

Total of digits in the first case 3+3+4+5 is 15 and in 15 sum of digits is 6. Now multiplying it by 5 = 30

Total of digits in the second case 9+0+2+6 is 17 and in 17 sum of digits is 8. Now multiplying it with 5 = 40

Total of digits in the find expression 3+0+4+5 is 12 and in 12 sum of digits is 3; so multiply it by 5 to get the required answer, which is 3*5 = 15

#### QUERY 3

**FIND THE ODD NUMBER PAIR FROM THE GIVEN ALTERNATIVES
**

120-560

91-299

78-169

104-429

**nanda kishore**

Option 1: Sum of digits = 1+2+0 = 3 & 5+6+0 = 11; again the sum = 1+1 = 2. The difference = 3 — 2 = 1

Option 2: Sum of digits = 9+1 = 10; again the sum = 1+0 = 2; & 2+9+9 = 20; again the sum = 2+0 = 2. The difference = 1-2 = -1

Option 3: Sum of digits = 7+8 = 15; again the sum = 1+5 = 6 & 1+6+9 = 16; again the sum = 1+6 = 7. The difference = 6-7 = -1

Option 4: Sum of the digits = 1+0+4 = 5 & 4+2+9 = 15; again the sum = 1+5 = 6. The difference = 5-6 = -1

SO FROM ABOVE WE CAN SAY 120-560 (option ‘1’) IS THE ANSWER

#### QUERY 4

**PICK THE ODD NUMBER FROM THE GIVEN ALTERNATIVES
**

187

289

143

165

**MAHA GUPTA**

Every number except 289 from above is divisible by 11; so 289 (option ‘2’) is the answer.

#### QUERY 5

**FIND OUT THE SET AMONG THE FOUR SETS WHICH IS LIKE THE GIVEN SET. THE GIVEN SET IS (48, 64, 78).**

13, 28, 44

17, 25, 42

16, 43, 58

74, 48, 64.

**nanda kishore**

The trend of the digits of each term in the given set is such that the digit of the first and last term are in increasing order while in the middle term they are decreasing; this trend is visible only in the terms of option ‘3’ ; hence being the answer.

#### QUERY 6

**PICK THE ODD NUMBER FROM THE GIVEN ALTERNATIVES.**

81

93

66

72

**Maverick Baloo Praful Nigam**

81 is the only perfect square here; so option ‘1’ is the answer.

#### QUERY 7

**If LUXOR is coded as 30; then GUILDS will be coded as**

36

38

24

40

**Dipangshu Talapatra**

L = 12, U = 21, X = 24, O = 15, R = 18. Add these numbers to obtain 90. 90/3 = 30

Similarly, G = 7, U = 21, I = 9, L = 12, D = 4, S = 19. Addition = 72. 72/3 = 24.

So the answer is 24 (option ‘3’)

#### QUERY 8

**5 : 30 : : 8 : ?**

**RONNIE BANSAL**

In such questions one has to establish some logical relation between the first two terms.

The second term is the multiplication of the first term and the number next to it i.e. 30 = 5 x 6

so the last term will be 8 x 9 = 72

#### QUERY 9

**12 : 30 : : 20 : ?**

**RONNIE BANSAL**

Here the second term i.e. 30 = Double of First term + Half of First Term i.e. 2*12 + 6

So the Last Term= 2*20 + 10 = 50

#### QUERY 10

**3 : 28 : : 5 : ?**

**RONNIE BANSAL**

The second term i.e. 28 = First Term^3 + 1 i.e. 3^3 + 1=27 + 1 = 28

So the Last Term = 5^3 + 1=125 + 1 = 126 or it could be 5^5 + 1

So, answer options should be checked carefully.

#### QUERY 11

**Identify wrong number in the sequence 0, 7, 28, 63, 124, 215**

28

215

7

63

**RONNIE BANSAL**

The sequence is i^3 -1, 2^3 – 1, 3^3 – 1, 4^3 – 1 and so on

So option ‘1’ i.e. 28 is the wrong number.

#### QUERY 12

**98 – 64 = 14, 86 – 23 = 27, 40 – 11 = ?**

6

7

8

9

**RONNIE BANSAL**

(9+8=17) – (6+4=10) = 7; then 7*2=14

(8+6=14) – (2+3=5) = 9; then 9*3=27

similarly (4+0=4) – (1+1=2) = 2; then 2*4=8

means difference of the sum of digits of each term of every statement is multiplied by 2, 3, 4 and so on.

#### QUERY 13

**Which one set of letters when sequentially placed at the gaps in the given letter series shall complete it?**

** _bcc_ac_aabb_ab_cc**

abaca

bcaca

aabca

bacab

**RONNIE BANSAL**

In questions like this one needs to recognize the sequence being used. If we see above cc, aa, bb are there, Again aa is just preceding bb. I think it can easily be made out that the series is

bbccaa / ccaabb / aabbcc

Hence the option ‘4’ is correct.

You’ll see pairs of letters are moving in cyclic order.

#### QUERY 14

**If 9+3 = 31, 12+15 = 45; then 72+12 = ?**

244

245

246

247

**MAHA GUPTA**

Divide each term of the LHS of each statement by 3 to get the RHS

Now 72/3 = 24 and 12/3 =4

Thus the required number is 244 (option ‘1’)

#### QUERY 15

**In a question paper, there are 12 questions in all out of which only six are to be answered. Six questions have alternative each. Each question has four parts. How many questions including the parts are there in the question paper?**

72

96

24

48

**R@HuL~R@NJ@N**

Total qustn = 12

Number of questions which having alternatives = 6 (means 6 are without alternatives)

Means total questions = 6 (without alternatives) + 6*2 (with alternatives) = 6+12 = 18

Now each has 4 parts, so 18*4 = 72 (option ‘1’)

#### QUERY 16

**3917, 3526, ? , 2857**

3389

2682

3082

3174

**R@HuL~R@NJ@N**

Subtract the number comprising the first three digit from the number itself to get the next number.

As 3917 – 391 = 3526

3526 – 352 = 3174 (option ‘4’)

#### QUERY 17

**There are 19 hockey players in a club. On a particular day 14 were wearing the hockey shirts prescribed. None of them were without either hockey pants or shirts. 11 were wearing the prescribed hockey pants. How many were in complete uniform?**

6

9

7

8

**SUCHI SHARMA**

As 11 were wearing hocking pants, 14 wearing shirts, so

players in complete uniform = 11 + 14 – 19 = 6 (option ‘1’)

#### QUERY 18

**If 63+25 = 16; 12+18 = 12; 23+17 = 13; then 54+22 = ?**

**MAHA GUPTA**

Add all the digits in each equality, such as

6+3+2+5 = 16

1+2+1+8 = 12

2+3+1+7 = 13

So 5+4+2+2 = 13

#### QUERY 19

**14, 28, 20, 40, 32, 64, ?**

52

56

96

128

**MAHA GUPTA**

We have to find the 7th term of the series, which is odd. You see every number at the odd term of the series is less by 8 from the previous one.

So the required number = 64 – 8 = 56 (option ‘2’)

#### QUERY 20

**15 : 50 :: 18 : ?**

**MAHA GUPTA**

We have two reasoning here.

1. first term/3 x 10 = 50

So 18/3 x 10 = 60

2.15 x 3 + 5 = second term i.e. 50

So 18 x 3 + 5 = 59

CONCLUSION: Sometimes the things depend on answer options. If 59 is there and 60 is not, the answer will be 59.

If 60 is there and 59 is not, the answer will be 60

If both the options are available to SSC ke experts ka bhagwaan hi maalik h.