1) A man walks down the backside of his house straight 25 metres, then

turns to the right and walks 50 metres again; then he turns towards left and again walks 25 metres. If his house faces to the East, what is his direction from the starting point? (a) South-East (b) South-West (c) North-East (d) North-West

Solution:

His direction from the starting point: Ans: (d) North-West

2) Two statements are given followed by two Conclusions: Statements: All numbers are divisible by 2 All numbers are divisible by 3 Conclusion-I: All numbers are divisible by 6 Conclusion-II: All numbers are divisible by 4 Which of the above Conclusions, logically follows/follow from the two given Statements?

(a) Only Conclusion - I (b) Only Conclusion - II (c) Neither Conclusion – I nor Conclusion II (d) Both Conclusion – I and Conclusion II Solution:

Since numbers are divisible by 2 and 3. 2 and 3 are coprime. No two common factors for 2 and 3. So it's always divisible by 6.

Ans: (a) Only Conclusion - I

3) Two Statements are given followed by two Conclusions: Statements:

All cats are dogs All cats are black

Conclusion-I: All dogs are black

Conclusion-II: Some dogs are not black

Which of the above Conclusions, logically follows/follow from the two given Statements, disregarding commonly known facts?

(a) Only Conclusion - I (b) Only Conclusion – II (c) Neither Conclusion – I nor Conclusion II (d) Both Conclusion – I and Conclusion II

Solution:

Also based on the distribution property of syllogism {“Any term distributed in the conclusion must be distributed in either premise”} dogs and black are not distributed in the premises which are distributed in conclusion I and II. So both the conclusions are invalid. Ans: (c) Neither Conclusion – I nor Conclusion II

4) Consider the following sequence of numbers:

5 1 4 7 3 9 8 5 7 2 6 3 1 5 8 6 3 8 5 2 2 4 3 4 9 6 How many odd numbers are followed by the odd number in the above sequence?

(a) 5 (b) 6 (c) 7 (d) 8

Solution: Following are odd numbers followed by odd numbers:

5 1 7 3 3 9 5 7 3 1 1 5

Ans: (b) 6

5) A is 16th from the left end in a row of boys and V is 18th from the right end. G is 11th from A towards the right and 3rd from V towards the right end. How many boys are there in the row? (a) 40 (b) 41 (c) 42 (d) Cannot be determined due to insufficient data Solution: 15 + A + 6 + V + 2 + G + 15 +1 +1 +1 No of Boys = 15 + 1 + 6 + 1 + 2 + 1 + 15 = 41 Ans: (b) 41

6) Three Statements S1, S2 and S3 are given below followed by a Question :

S1 : C is younger than D, but older than A and B S2 : D is the oldest S3 : A is older than B Question :

Who among A, B, C and D is the youngest? Which one of the following is correct in respect of the above statements and the Question?

(a) S1 alone is sufficient to answer the Question. (b) S1 and S2 together are sufficient to answer the Question (c) S2 and S3 together are sufficient to answer the Question (d) S1 and S3 together are sufficient to answer the Question

Solution:

Age

D

C

A

B

The above conclusion can be drawn based on S1 and S3 alone. S2 is not needed.

Ans: (d) S1 and S3 together are sufficient to answer the Question

7) How many integers are there between 1 and 100 which have 4 as a digit but are not divisible by 4?

(a) 5 (b) 11 (c) 12 (d) 13

Solution:

The numbers which has 4 as a digit are

41, 42, 43, 44, 45, 46, 47, 48, 49, 14, 24, 34, 54, 64, 74, 84 & 94

In these what are all the numbers that are not divisible by 4 are

41, 42, 43, 45, 46, 47, 49, 14, 34, 54, 74 & 94

Ans: (c) 12

8) Let x, y be the volumes; m, n be the masses of two metallic cubes P and Q respectively, Each side of Q is two times that of P and mass of Q is two times that of P. Let u=m/x and v=n/y, which one of the following is correct?

(a) u = 4v (b) u = 2v (c) v = u (d) v = 4u

Solution:

Given: Each side of Q is two times that of P

Let's take the side of cube P as ‘a’. Then Volume of cube P is

x = a3

Then the side of cube Q as ‘2a’. Then Volume of cube Q is

y = (2a)3 = 8a3

Given: Mass of Q is two times that of P

Mass of P: m

Mass of Q: n = 2m

u = m/x = m/ (a3)

v = n/y = 2m/ (8a3) = m/ (4a3) = {m / (a3)} / 4 = u/4

v = u/4

=> Ans: u = 4v (a)

9) The average age of a teacher and three students is 20 years. If all the three students are of same age and the difference between the age of the teacher and each student is 20 years, then what is the age of the teacher?

(a) 25 years (b) 30 years (c) 35 years (d) 45 years

Solution:

Given:

T + S1 + S2 + S3 = 4(20) = 80

S1 = S2 = S3 = X

T + X + X + X = T + 3X = 80 …. (1)

T - X = 20 …. (2)

(1) - (2) => 3X + X = 80 - 20

4X = 60

X = 15 =>

Ans: T = 25

10) A person bought a car and sold it for Rs.3,00,000. If he incurred a loss of 20% then how much did he spend to buy the car?

(a) Rs.3,60,000 (b) Rs.3,65,000 (c) Rs.3,70,000 (d) Rs.3,75,000

Solution:

SP = 3,00,000

L = 20%

CP = 100 * SP / (100 - L%)

= 100 * 300000 / (100% - 20%)

= 100*300000/80

= 5*300000/4

= 5*75000

Ans: CP = 3,75,000 (d)

18) Two statements S1 and S2 are given below with regard to four numbers P, Q, R and S followed by a Question:

S1: R is greater than P as well as Q. S2: S is not the largest one. Question:

Among four numbers P, Q, R and S, which one is the largest? Which one of the following is correct in respect of the above Statements and the Question?

(a) S1 alone is sufficient to answer the Question (b) S2 alone is sufficient to answer the Question (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question (d) S1 and S2 together are not sufficient to answer the Question

Solution: S is not the largest one means S may be less than or equal to R.

Height

S

R S

PQ S

Ans: (d) S1 and S2 together are not sufficient to answer the Question 19) Two Statements S1 and S2 are given below followed by a Question

S1: η is a prime number. S2: η leaves a remainder of 1 when divided by 4. Question:

If η is a unique natural number between 10 and 20, then what is η? Which one of the following is correct in respect of the above Statements and the Questions?

(a) S1 alone is sufficient to answer the Question. (b) S2 alone is sufficient to answer the Question. (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question. (d) S1 and S2 together are not sufficient to answer the Question. Solution:

Based on S1 & S2 Possible values for η are 13 & 17. It's not unique.

Ans: (d) S1 and S2 together are not sufficient to answer the Question. 20)

1. Two Statements S1 and S2 are given below with regard to two numbers followed by a Question:

S1: Their product is 21. S2: Their sum is 10. Question:

What are the two numbers? Which one of the following is correct in respect of the above Statements and the Questions?

(a) S1 alone is sufficient to answer the Question. (b) S2 alone is sufficient to answer the Question. (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

(d) S1 and S2 together are not sufficient to answer the Question. Solution: S1: x*y = 21 S2: x + y = 10 Both statements need to find x and y. Ans: (c) S1 and S2 together are sufficient to answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question.

21. In the sum

⊗ +1 ⊗ + 5 ⊗ + ⊗ ⊗ + ⊗ 1=1 ⊗ ⊗ For which digit does the symbol stand?

(a) 2 (b) 3 (c) 4 (d) 5 Solution: Replace the symbol with all 4 options. 2 + 12 + 52 + 22 + 21 ≠ 122 3 + 13 + 53 + 33 + 31 = 133

Ans: (b) 3 22) If you have two straight sticks of length 7.5 feet and 3.25 feet,

what is the minimum length can you measure?

(a) 0.05 foot (b) 0.25 foot (c) 1 foot (d) 3.25 feet

Solution: From 7.5 feet length stick, if we subtract two times 3.25 stick then 7.5 - 2*3.25 = 7.5 - 6.5 = 1 foot Ans (c) 1 foot

23) A simple mathematical operation in each number of the sequence 14, 18, 20, 24, 30, 32, …. Results in a sequence with respect to prime numbers. Which one of the following is the next number in the sequence?

(a) 34 (b) 36 (c) 38 (d) 40

Solution: Actually the sequence is formed based on Prime numbers. They are 13, 17, 19, 23, 29, 31…. Based on that 14, 18, 20, 24, 30, 32… So next number would be 38 (Prime number 37) Ans: (c) 38

24) One page is torn from a booklet whose pages are numbered in the usual manner starting from the first page as 1. The sum of the numbers on the remaining pages is 195. The torn page contains which of the following numbers?

(a) 5, 6 (b) 7, 8 (c) 9, 10 (d) 11, 12

Solution: Sum of first n natural numbers

1 + 2 + 3 + ….. N = N(N+1)/2

Since the total number of remaining pages is only 195, we have to guess with some minimum number of pages values. Say for ex: If 10 pages means Sum of the pages = 10*11/2 = 5*11 = 55 Not Possible

If 20 pages means Sum of the pages = 20*21/2 = 10*21 = 210 Possible Difference 210 - 195 = 15 = 7 + 8

Ans (b) 7, 8

25) Consider the following arrangement that has some missing letters:

abab_b_bcb_dcdcded_d The missing letters which complete the arrangement are

(a) a, b, c, d (b) a, b, d, e (c) a, c, c, e (d) b, c, d, e Solution:

Option (a) abababbbcbcdcdcdeddd Option (b) abababbbcbddcdcdeded Option (c) abababcbcbcdcdcdeded Ans: (c) a, c, c, e

26) Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3. If the sum of the numbers is 15902, then what is the difference between the values of A and D?

(a) 1 (b) 2 (c) 3 (d) 4

Solution:

Each letter represents a different digit greater than 3 => 4, 5, 6, 7, 8 & 9

A 3 B C

D E 2 F

1 5 9 0 2 Unit digit: C + F = 12 C, F may be 4 & 8 or 5 & 7 ----(1) Ten digits: 1 (carry) + B + 2 = 0

B = 7 ----(2) => (1) => C & F will be 4 & 8

Hundred digits: 1 (carry) + 3 + E = 4 + E = 9

E = 5 ----(3) Thousand digits: A + D = 15

A & D will be 6 & 9

Difference between A and D = 9 - 6 = 3 Ans: (c) 3

27) Two Statements S1 and S2 are given below followed by a Question :

S1 : There are not more than two figures on any page of a 51- page book.

S2 : There is at least one figure on every page. Question :

Are there more than 100 figures in that book?

Which one of the following is correct in respect of the above Statements and the Question?

(a) Both S1 and S2 are sufficient to Answer the Question, but neither S1 alone nor S2 alone is sufficient to answer the Question. (b) S1 alone is sufficient to answer the Question. (c) S1 and S2 together are not sufficient to answer the Question. (d) S2 alone is sufficient to answer the Question. Solution: If all pages have one figure means total figures count will be 51. If all pages have 2 figures means total figures count will be 102 Since we don't know the no of one figure or two figure pages count, Ans: (c) S1 and S2 together are not sufficient to answer the Question.

28) Consider the following data :

Average marks in English

Average marks in Hindi

Girls 9 8

Boys 8 7

Overall average marks 8.8 x

(a) 7.8 (b) 7.6 (c) 7.4 (d) 7.2

Solution:

(x - 7)/(8 - x) = 4/1 x - 7 = 4(8 - x) 32 - 4x = x - 7 4x + x = 32 + 7 5x = 39 x = 7.8

Ans (a) 7.8

29) A family of two generations consisting of six members P, Q, R, S, T and U has three males and three females. There are two married couples and two unmarried siblings. U is P’s daughter and Q is R’s mother-in-law. T is an unmarried male and S is a male. Which one of the following is correct?

(a) R is U’s husband. (b) R is S’s wife. (c) S is unmarried. (d) None of the above

Solution:

Based on given conditions,

Gen 1 P QF

Gen 2 UF R TM

Since there are only 2 generations and 2 couples are there… we have to club all these scenarios

Gen 1 PM QF

Gen 2 RF SM UF TM

PM & QF are couple

RF and SM are couple

SM, UF and TM are siblings

Ans: (b) R is S’s wife

30) If in a particular year 12th January is a Sunday, then which one of the following is correct?

(a) 15th July is a Sunday of the year is a leap year.

(b) 15th July is a Sunday if the year is not a leap year.

(c) 12th July is a Sunday if the year is a leap year.

(d) 12th July is not a Sunday if the year is a leap year.

Solution:

Since in 3 options Leap year is there… will take the year as Leap year first

12th Jan - Sunday

12th July -

ODD Days Calculation

Month Days Odd days

Jan 19 5

Feb 29 1

Mar 31 3

Apr 30 2

May 31 3

June 30 2

July 12 5

Final Total 21

Odd days (21/7) 0

12th July = Day (12th Jan Sunday) + Odd days = 0 + 0 = 0 => Sunday

Ans (c) 12th July is a Sunday if the year is a leap year.

31) How many Zeroes are there at the end of the following product?

1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60 (a) 10 (b) 12 (c) 14 (d) 15 Solution: To calculate zeros we need to find number of 5’s and 2’s in the multiplication

1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

= 1 × 5 × (5 × 2) × (5 × 3) × (5 × 2 × 2) × (5 × 5) × (5 × 2 × 3) × (5 × 7) × (5 × 2 × 2 × 2) × (5 × 9) × (5 × 5 × 2) × (5 × 11) × (5 × 2 × 2 × 3)

= 514 × 210 × …

=> (5 × 2)10 × 54

=> 1010 × 54

10 number of zeros are there at the end of the product

Ans (a) 10

38) Let XYZ be a three-digit number, where (X + Y + Z) is not a multiple of 3. Then (XYZ + YZX + ZXY) is not divisible by

(a) 3 (b) 9 (c) 37 (d) (X + Y + Z)

Solution: (XYZ + YZX + ZXY) = (100X + 10Y + Z) + (100Y + 10Z + X) + (100Z + 10X + Y) (XYZ + YZX + ZXY) = (111X + 111Y + 111Z)

= 111(X + Y + Z) (XYZ + YZX + ZXY) = 3*37*(X + Y + Z) => (XYZ + YZX + ZXY) is divisible by 3, 37 & (X + Y + Z) Ans: (b) 9

39) Let p, q, r and s be natural numbers such that p – 2016 = q + 2017 = r – 2018 = s + 2019 Which one of the following is the largest natural number?

(a) p (b) q (c) r (d) s Solution:

Given: p – 2016 = q + 2017 = r – 2018 = s + 2019 Let p – 2016 = q + 2017 = r – 2018 = s + 2019 = x p = x + 2016 q = x - 2017 = x - 2019 + 2 = s + 2 r = x + 2018 = x + 2016 + 2 = p + 2 s = x - 2019 {s = x - 2019} < {q = x - 2017} < {p = x + 2016} < {r = x + 2018} s < q < p < r Ans: (c) r

40) How many five-digit prime numbers can be obtained by using all the digits 1, 2, 3, 4 and 5 without repetition of digits?

(a) Zero (b) One (c) Nine (d) Ten

Solution: Sum of the digits 1 + 2 + 3 + 4 + 5 = 15 Since 15 is divisible by 3, the 5 digit number formed by these numbers without repetition is also divisible by 3. So its not a PRIME NUMBER

Ans: (a) Zero

41) A person X can complete 20% of work in 8 days and another person Y can complete 25% of the same work in 6 days. If they work together, in how many days will 40% of the work be completed?

(a) 6 (b) 8 (c) 10 (d) 12

Solution: X

20% => 8 days 100% => 40 days Y 25% => 6 days 100% => 24 days X & Y => 1 day work = (1/40) + (1/24) = (3 + 5)/120 = 8/120 X & Y => 1 day work = 1/15 X & Y completes a work (100%) in 15 days 10% in 1.5 days 40% in 6 days

Ans: (a) 6

42) A car travels from a place X to place Y at an average speed of v km/hr, from Y to X at an average speed of 2v km/hr, again from X to Y at an average speed of 3v km/hr and again from Y to X at an average speed 4v km/hr. Then the average speed of the car for the entire journey

(a) is less than v km/hr (b) lies between v and 2v km/hr

(c) lies between 2v and 3v km/hr (d) lies between 3v and 4v km/hr

Solution: Let’s take distance between X and Y as D Time taken for X to Y => T1 = D/v Time taken for Y to X => T2 = D/2v Time taken for X to Y again => T3 = D/3v Time taken for Y to X again => T4 = D/4v Average Speed = Total distance travelled / Total time taken = D + D + D + D

( ) + ( ) + ( ) + ( )vD D

2vD3v

D4v

= 4DD ( ) + ( ) + ( ) + ( )v

1 12v

13v

14v

= 4 ( ) + ( ) + ( ) + ( )v

1 12v

13v

14v

= 4

( )12v12 + 6 + 4 + 3

= 4 ( ) 25

12v

= 4*12v/ 25 = 48v/ 25

= 1.92v

Ans: (b) lies between v and 2v km/hr

43) Consider the following statements :

1. The minimum number of points of intersection of a square and a circle is 2.

2. The maximum number of points of intersection of a square and a circle is 8.

Which of the above statements is/are correct? (a) 1 only (b) 2 only (c) Both 1 and 2 (d) Neither 1 nor 2 Solution: For 1 and 2 possible diagrams are

Hence point 1 wrong Ans: (b) 2 only

44) A man takes half time in rowing a certain distance downstream than upstream. What is the ratio of the speed in still water to the speed of current?

(a) 1 : 2 (b) 2 : 1 (c) 1 : 3 (d) 3 : 1

Solution: Sd => Speed of Downstream Su => Speed of Upstream Sc => Speed of Current Sb => Speed of Boat Time taken ration between Downstream and Upstream => (½) : 1

=> 1 : 2 Speed Ratio between Downstream and Upstream => 2 : 1 Sd = 2k Su = k Sc = (Sd - Su)/2 = (2k - k)/2 = k/2 Sb = (Sd + Su)/2 = (2k + k)/2 = 3k/2

The ratio of the speed in still water to the speed of current

= Sb : Sc

= 3k/2 : k/2

= 3 : 1

Ans (d) 3 : 1

45) How many pairs of natural numbers are there such that the difference of whose squares is 63?

(a) 3 (b) 4 (c) 5 (d) 2

Solution: a2 - b2 = 63 (a + b)(a - b) = 63 Possible two factor options for 63 are 21*3 & 9*7. Try to write those in the format (a + b)(a - b) 21*3 = (12 + 9)(12 - 9) 9*7 = (8 + 1)(8 - 1) 1*63 = (32 + 31)(32 - 31)

Ans: (a) 3

46) Which one of the following will have a minimum change in its value if 5 is added to both numerator and the denominator of the fractions 2/3, 3/4, 4/5 and 5/6?

(a) 2/3 (b) 3/4 (c) 4/5 (d) 5/6

Solution:

If 5 is added to both numerator and the denominator of the fractions

2/3 becomes 7/8

(7/8 - 2/3) / (2/3) = (0.875 - 0.66)/0.66

3/4 becomes 8/9

(8/9 - 3/4) / (3/4) = (0.88 - 0.75)/0.75

4/5 becomes 9/10

(9/10 - 4/5) / (4/5) = (0.9 - 0.8)/0.8

5/6 becomes 10/11

(10/11 - 5/6) / (5/6) = (0.90 - 0.83)/0.83

Numerator: Difference is less for Option d

Denominator: Option d is big

Based on that we can say the change is minimum for Option d

Ans: (d) 5/6

47) A digit n > 3 is divisible by 3 but not divisible by 6. Which one of the following is divisible by 4?

(a) 2n (b) 3n (c) 2n + 4 (d) 3n + 1

Solution: Divisible by 3 but not divisible by 6 means it’s divisible by 9. n = 9x means let’s take 9 2n = 2*9 = 18 => not divisible by 4 3n = 3*9 = 27 => not divisible by 4 2n + 4 = 18 + 4 = 22 => not divisible by 4 3n + 1 = 27 + 1 = 28 => divisible by 4

Ans: (d) 3n + 1

48) If 1 litre of water weighs 1 kg, then how many cubic millimetres of water will weigh 0.1 gm?

(a) 1 (b) 10 (c) 100 (d) 1000

Solution: 1 kg = 1 m 1000 gm = 1000 millimetre 1 gm = 1 millimetre 0.1 gm = 0.1 millimetre Ans: Options wrong

49) A vessel full of water weighs 40 kg. If it is one-third filled, its weight becomes 20 kg. What is the weight of the empty vessel?

(a) 10 kg (b) 15 kg (c) 20 kg (d) 25 kg

Solution: Let’s take weight of empty vessel as V Water weight as W

Vessel Full of water: V + W = 40 … (1) Vessel with ⅓ rd filled water: V + (⅓)W = 20 … (2)

(1) - (2) = (2/3)W = 20

= W = 20*3/2 = 10*3 = 30 => V = 10

Ans: (a) 10 kg

50) A frog tries to come out of a dried well 4.5 m deep with slippery walls. Every time the frog jumps 30 cm, slides down 15 cm. What is the number of jumps required for the frog to come out of the well?

(a) 28 (b) 29 (c) 30 (d) 31

Solution: Jump 1: + 30 - 15 = 15 = 1*15 Jump 2: +15 + 30 - 15 = 30 = 2*15 . . . Jump 28: 28*15 = 420 Jump 29: 420 + 30 => here itself the frog comes out of the well. After that sliding down doesn’t issue. Ans: (b) 29 57) How many different 5-letter words (with or without meaning) can be constructed using all the letters of the word ‘DELHI’ so that each word has to start with D and end with I?

(a) 24 (b) 18 (c) 12 (d) 6

Solution: D _ _ _ I ELH => 3! Ways = 3*2*1 = 6 ways Ans: (d) 6

58) A bottle contains 20 litres of liquid A. 4 litres of liquid A is taken out of it and replaced by same quantity of liquid B. Again 4 litres of the mixture is taken out and replaced by same quantity of liquid B. What is the ratio of quantity of liquid A to that of liquid B in the final mixture?

(a) 4 : 1 (b) 5 : 1 (c) 16 : 9 (d) 17 : 8

Solution: Liquid A Liquid B Total Mixture

20 0 20

Iteration1: 4 litres of Liquid A taken out

-4 0

16 0 16

Iteration1: 4 litres of Liquid B replaced

0 4

16 4 20

Iteration 2: 4 litres of mixture taken out

3.2 0.8

12.8 3.2 16

Iteration 2: 0 4

4 litres of Liquid B replaced

12.8 7.2 20

Final Ratio: 12.8 : 7.2

=> 128 : 72

=> 16 : 9

Ans: (c) 16 : 9

59) The average score of a batsman after his 50th innings was 46.4. After 60th innings, his average score increases by 2.6. What was his average score in the last ten innings?

(a) 122 (b) 91 (c) 62 (d) 49

Solution:

Formula X1 + X2 + …. Xn = n*A

=> X1 + X2 + …. X50 = 50*(46.4) …. (1)

=> X1 + X2 + …. X60 = 60*(46.4 + 2.4)

=> X1 + X2 + …. X60 = 60*49 …. (2)

(2) - (1)

=> (X1 + X2 + …. X60) - (X1 + X2 + …. X50)= 60*49 - 50*46.4

=> (X51 + X52 + …. X60) = (50 + 10)*49 - 50*46.4

= 50*49 + 10*49 - 50*46.4

= 50(49 - 46.4) + 10*49

= 50*2.6 + 490

= 130 + 490

(X51 + X52 + …. X60) = 620

(X51 + X52 + …. X60) / 10 = 62

Ans: (c) 62

60) As a result of 25% hike in the price of rice per kg, a person is able to purchase 6 kg less rice for Rs.1,200. What was the original price of rice per kg?

(a) Rs. 30 (b) Rs. 40 (c) Rs. 50 (d) Rs. 60

Solution:

No of Kg Price per kg Total Amount

N X NX = 1200

N - 6 1.25X (N-6)*1.25X = 1200 1.25NX - 6*1.25X = 1200 1.25*1200 - 7.5X = 1200 1500 - 1200 = 7.5X 7.5X = 300 X = 300/7.5 X = 3000/75 = 40

Ans: (b) Rs. 40

61) What is the greatest length x such that 3½m and 8¾m are integral multiples of x?

(a) m1 21 (b) m1 3

1 (c) m1 41 (d) m1 4

3

Solution: Integral multiples of 3 and 8 is HCF of the numbers2

143

3 = 21

27

8 = 43

435

HCF (Fractions) = HCF (Numerator) / LCM (Denominator) HCF ( , ) = HCF(7,35) / LCM (2,4) = 7/4 = 12

74

3543

Ans: (d) m1 43

62) Consider the following data:

Year Birth Rate Death Rate

1911-1921 48.1 35.5 1921-1931 46.4 36.3 1931-1941 45.2 31.2 1941-1951 39.9 27.4 1951-1961 41.1 22.8 1961-1971 41.1 18.9 1971-1981 37.1 14.8

For which period was the natural growth rate maximum?

(a) 1911-1921 (b) 1941-1951 (c) 1961-1971 (d) 1971-1981 Solution:

Year Birth Rate Death Rate Difference 1911-1921 48.1 35.5 12.6 1921-1931 46.4 36.3 10.1

1931-1941 45.2 31.2 14 1941-1951 39.9 27.4 12.5 1951-1961 41.1 22.8 18.3 1961-1971 41.1 18.9 22.2 1971-1981 37.1 14.8 22.3 Maximum

Ans: (d) 1971-1981

63) The recurring decimal representation 1.272727... is equivalent to (a) 13 /11 (b) 14/11 (c) 127/99 (d) 137/99 Solution: Ans (b) 14/11 = 1.272727… 64) What is the least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 in each case? (a) 1012 (b) 1022 (c) 1122 (d) 1222 Solution: LCM(3, 4, 5, 6) = 60 Least four-digit number when divided by 3, 4, 5 and 6 leaves a remainder 2 = n60 + 2 = 1020 + 2 = 17*60 + 2 Ans: (b) 1022 66. What is the remainder when 51 × 27 × 35 × 62 × 75 is divided by 100? (a) 50 (b) 25 (c) 5 (d) 1 Solution: 51 × 27 × 35 × 62 × 75

51 × 27 × 5 × 7 × 31 × (2 × 5) × 5 × 3 => (One zero is there at the end) Ans: (a) 50 67. A sum of Rs. 2,500 is distributed among X, Y and Z in the ratio : : .What is the difference between the maximum share and the2

143

65

minimum share? (a) Rs. 300 (b) Rs. 350 (c) Rs. 400 (d) Rs. 450 Solution: Simplify the fraction ration ( : : ) to numerator ratio.2

143

65

Multiply by 12 ( : : ) = (12* : 12* : 12* )2

143

65

21

43

65

= 6: 3*3 : 2*5 = 6: 9 : 10

6k + 9k + 10k = 2500 25k = 2500 k = 100 Maximum share - Minimum share = 10k - 6k = 1000 - 600 = 400 Ans: (c) Rs. 400

68. For what value of n, the sum of digits in the number (10n + 1) is 2? (a) For n = 0 only (b) For any whole number n (c) For any positive integer n only (d) For any real number n Solution: Sum of digits in the number

n (10n + 1) Sum of the digits

n = 0 (100 + 1) = 1 + 1 = 2 2

n = 1 (101 + 1) = 10 + 1 = 11 1 + 1 = 2

n = 2 (102 + 1) = 100 + 1 = 101

1 + 0 + 1 = 2

n = 2 (102 + 1) = 1000 + 1 = 1001

1 + 0 + 0 + 1 = 2

n = (½) (101/2 + 1) = 3.162 + 1 = 4.162

4 + 1 + 6 + 2 ≠ 2

Ans: (b) For any whole number n 69. In a class, there are three groups A B and C. If one student from Group A and two students from Group B are shifted to Group C, then what happens to the average weight of the students of the class? (a) It increases. (b) It decreases. (c) It remains the same. (d) No conclusion can be drawn due to insufficient data. Solution:

Group A

Group B

Group C

Weighted Average

No of Students

N1 N2 N3

Average of students

(X1 + X2… Xn)

(Y1 + Y2 + … Yn)

(Z1 + Z2 + … Zn)

{(X1 + X2… Xn) + (Y1 + Y2 + … Yn) + (Z1 + Z2 + … Zn)}/(N1 + N2 + N3)

After shifting

No of Students

N1 - 1 N2 - 2 N3 + 3

Average of students

(X1 + X2… Xn-1)

(Y1 + Y2 + … Yn-2)

(Z1 + Z2 + … Zn + Xn + Yn-1 + Yn)

{(X1 + X2… Xn-1) + (Y1 + Y2 + … Yn-2) + (Z1 + Z2 + … Zn + Xn + Yn-1 + Yn)}/(N1 - 1 + N2 - 2 + N3 + 3) = {(X1 + X2… Xn) + (Y1 + Y2 + … Yn) + (Z1 + Z2 + … Zn)}/(N1 + N2 + N3)

Ans: (c) It remains the same 70. How many different sums can be formed with the denominations Rs. 50, Rs. 100, Rs. 200, Rs. 500 and Rs. 2,000 taking at least three denominations at a time? (a) 16 (b) 15 (c) 14 (d) 10 Solution: Since order is not imp 50 + 100 + 200 = 250 100 + 50 + 200 = 250 200 + 100 + 50 = 250 So it's not permutation. Its combination. nCr = 5C3 = 5C2 = 5*4/(1*2) = 5*2 = 10 Ans: (d) 10

77) A shop owner offers the following discount options on an article to a customer.

1. Successive discounts of 10% and 20%, and then pay a service tax of 10%.

2. Successive discounts of 20% and 10%, and then pay a service tax of 10%.

3. Pay a service tax of 10% first, then successive discounts of 20% and 10%.

Which one of the following is correct? (a) 1 only is the best option for the customer. (b) 2 only is the best option for the customer. (c) 3 only is the best option for the customer. (d) All the options are equally good for the customer. Solution:

Point 1 Point 2 Point 3

Discount 1 100% = 100 10% = 10 Remaining 90

100% = 100 20% = 20 Remaining 80

Service Tax

100% = 100 10% = 10 Remaining 110

Discount 2 100% = 90 20% = 18 Remaining 72

100% = 80 10% = 8 Remaining 72

Discount 1 100% = 110 20% = 22 Remaining 88

Service Tax 100% = 72 10% = 7.2 Final = 79.2

100% = 72 10% = 7.2 Final = 79.2

Discount 2 100% = 88 10% = 8.8 Remaining 79.2

Ans: (d) All the options are equally good for the customer.

78) The letters from A to Z are numbered from 1 to 26 respectively. If GHI = 1578, and DEF = 912, then what is ABC equal to?

(a) 492 (b) 468 (c) 262 (d) 246

Solution: GHI coded as 789 by alphabetical order. And then 789 + 789 = 1578

Similarly DEF coded as 456 by alphabetical order. And then 456 + 456 = 912 Then ABC coded as 123 by alphabetical order. And then 123 + 123 = 246

Ans: (d) 246

79) What is the missing term @ in the following?

ACPQ : BESU :: MNGI : @ (a) NPJL (b) NOJM (c) NPIL (d) NPJM

LHS A C P Q

Missing D QR RST

B E S U

RHS M N G I

Missing O HI JKL

N P J M Ans: (d) NPJM

80) What is the largest number among the following?

(a) ) ( 21 −6

(b) ) ( 41 −3

(c) ) ( 31 −4

(d) ) ( 61 −2

Solution:

= ) (a ) ( a1 −m = −1 −m am

= = 64) ( 21 −6

26

= = 64) ( 41 −3

43

= = 81) ( 31 −4

34

= = 36) ( 61 −2

62

Ans: (c) ) ( 31 −4