# NUMBERS & FUNDAMENTALS OF MATHS (PART-V)

## NUMBERS & FUNDAMENTALS OF MATHS (PART-V)

#### QUERY 41

Find Value of √[1 + {√1 + (√1 …)}]

A) equal to 1
B) between 0 and 1
C) between 1 and 2
D) greater than 2

ABHIK CHAKRABORTY
Let y be the expression given in the bracket {  }

Then the expression can also be written as y = √(1 + y)

Squaring both sides: y² = 1 + y

=> y² – y – 1 = 0

=> y = 1.618 or -1.618

The negative value here is rejected as the given expression is root of a number; and root of a number cannot be negative ever.

Therefore option ‘C’ is correct.

#### QUERY 42

The greatest number among 260, 348, 436 and 524 is?

A) 260
B) 348
C) 436
D) 524

MAHA GUPTA
To compare such questions make all the powers same by taking HCF of the powers.

So, HCF of 60, 48, 36, 24 = 60

Now the numbers = 260, 348, 436, 524

The numbers after conversion of powers = (25)12, (34)12, (43)12, (52)12

= 3212, 8112, 6412, 2512

Therefore 348 is the biggest among all (option ‘B’)

#### QUERY 43

In how many ways word ‘glamour’ can be arranged such that vowels are always together?

A) 600
B) 800
C) 920
D) 720

MAHA GUPTA
The word GLAMOUR has three vowels and 4 constants, where vowels have to stay together
Therefore we can say there are 4+1 = 5 units in all (4 of consonants and 1 of vowels)

Number of ways in which vowels can appear = 3!
Number of ways in which all 5 units can appear = 5!

So, number of ways in which the word GLAMOUR can appear if vowels have to be together = 3!*5! = (3*2*1)*(5*4*3*2*1) = 6*120 = 720 (option ‘D’)

#### QUERY 44

Weight of an empty bottle is 20% of weight when filled with some liquid. Some liquid removed, then weight of bottle along with liquid is half of the original weight. Find fraction of the liquid removed.

A) 3/7
B) 7/20
C) 5/8
D) 7/9

MAHA GUPTA
Let the original weight of bottle along with liquid = 100 gm
Then the weight of bottle = 20% of 100 = 20 gm
And the weight of the liquid it contained = 100 – 20 = 80 gm

Now some of the liquid is removed in such a way that weight of the bottle alongwith the liquid is half of the original weight, therefore weight now = 100/2 = 50 gm

Obviously the weight of bottle still is 20 gm in it; so weight of liquid now = 50 – 20 = 30 gm

So liquid removed = 80 – 30 = 50 gm

Hence the fraction of liquid removed = 50/80 = 5/8 (option ‘C’)

#### QUERY 45

In an examination, a student answered 15 of the first 20 questions correctly. Of the remaining questions, he answered one-third correctly. If each question carries equal marks and he scored 50% marks, how many total questions were there in the paper?

A) 40
B) 45
C) 50
D) 60

MAHA GUPTA
Part of first 20 question answered correctly = 1520 34
Marks scored = 50% = 12
FIRST 20 QUESTIONS                     REMAINING QUESTIONS

 3⁄4 1⁄3 1⁄2 1⁄2 – 1⁄3 = 1⁄6 1⁄2 – 3⁄4= 1⁄4

Hence, ratio of first 20 questions and the remaining question = 16 : 1= 2 : 3

Number of first 20 questions = 2 ratios = 20 questions
So, 1 ratio = 202 = 10
Thus, number of the remaining questions = 3 ratios = 3*10 = 30

Therefore, number of total questions = 20 + 30 = 50 (option ‘C’)

ANOTHER METHOD
Let the total number of questions = x
So, the number of remaining questions = x – 20
And marks scored in the remaining questions = (x-20)3
Marks scored = 50% of x = x2

Now, according to above
15 + (x-20)x2

=> x = 50

TRICK
40. Remaining questions = 40 – 20 = 20 (incorrect as not divisible by 3)

45. Remaining questions = 45 – 20 = 25 (incorrect as not divisible by 3)

50. Remaining questions = 50 – 20 = 30. So, total questions = 20 + 30 = 50. Marks scored = 15 + 30= 25.
We see 25 is 50% of 50, hence correct.

#### QUERY 46

A student gets 4 marks for each correct answer and gets deduction of 1 marks for each wrong attempt. He attempted all 75 questions in the exam and got 125 marks in all. Find the number of his correct attempts.

A) 35
B) 40
C) 42
D) 46

MAHA GUPTA
Number of correct questions without the effect of negative marking = 125/4

Remaining questions in he scored 0  = 75 – 1254 = 1754. Because of negative marking, obviously, some of them are correct answers and some are incorrect. So we need to know correct attempts out of this also to know the total correct attempts. Let’s do it by allegation method:

CORRECT                           WRONG

 4 -1 0 0 – (-1) = 1 0 – 4 = 4

So, ratio of correct and wrong answers = 1 : 4

Thus, correct attempts out of 1754 = 15 × 1754 = 354

Therefore, total correct attempts = 1254 + 354 = 1604 = 40 (option ‘B’)

ANOTHER METHOD
Suppose all the answers are correct, then his total marks = 75*4 = 300

Marks lost for every incorrect answer = marks for correct answer + negative marks = 4 + 1 = 5
Actual marks short by = 300 – 125 = 175
Thus, wrong attempts = 1755 = 35

Therefore, total of correct attempts = 75 – 35 = 40 (option ‘B’)

RAMAN KUMAR
Let the number of correct questions = x
Therefore incorrect questions = 75 – x

Now according to the question
4x – 1(75 – x) = 125

=> 4x -75 + x = 125

=> 5x = 200

=> x = 40
Hence number of correct attempts = 40 (option ‘B’)

#### QUERY 47

A student gets 3 marks for each correct answer and gets deduction of 1 marks for each wrong attempt. If he scores 0 marks in a paper of 100 questions, how many of his answers were correct assuming he attempts all the questions?

A) 25
B) 50
C) 60
D) 75

MAHA GUPTA
Because his score is 0, so because of negative marking, obviously, some of the attempts are correct  and some are incorrect. Let’s know his correct attempts by the allegation method:

CORRECT                           WRONG

 3 -1 0 0 – (-1) = 1 0 – 3 = 3

So, ratio of correct and wrong answers = 1 : 3

Thus, his correct attempts = 1× 100 = 25 (option ‘A’)

ANOTHER METHOD
Suppose all the answers are correct, then his total marks = 100*3 = 300

Marks lost for every incorrect answer = marks for correct answer + negative marks = 3 + 1 = 4
Thus, wrong attempts = 3004 = 75

Therefore, total of correct attempts = 100 – 75 = 25 (option ‘A’)

ONE MORE METHOD
Let the number of correct questions = x
Therefore incorrect questions = 100 – x

Now according to the question
3x – 1(100 – x) = 0

=> 3x – 100 + x = 0

=> 4x = 100

=> x = 25
Hence number of correct attempts = 25 (option ‘A’)

#### QUERY 48

The weights of equal volume of gold and water are as 37 is to 2. If a liter of water weighs 1 kg, find the weight of a cubic centimeter of gold.

A) 20.5 gm
B) 29.5 gm
C) 16.5 gm
D) 18.5 gm

MAHA GUPTA
Weight of 1 liter of water = 1 kg
Therefore weight of 2 liter of water = 2 kg

But the weights of equal volume of gold and water are in the ratio of 37 : 2, it does mean that the weight of the gold equaling 2 liter of volume is 37 kg

Now you need to know volume of 1 cubic of centimeter has how many liters. Well, 1 cm³ = 1/1000 liter

Weight of 2 liter of gold = 37 kg
Therefore weight of 1/1000 liter of gold = (37/2)*(1/1000) = 37/2000 kg = 18.5 gm (option ‘D’)

#### QUERY 49

A tourist spends daily as many rupees as the number of days of the total tour. If the total expenses were Rs 361, then how many days did his tour last?

A) 21
B) 31
C) 17
D) 19

MAHA GUPTA
Just find √361 here, which is 19 (option ‘D’)

#### QUERY 50

How to do questions on finding the number of teams or participants?

SHIV KISHOR
Note the two words used in such questions: ROUND-ROBIN and KNOCK-OUT. In ROUND-ROBIN basis each team plays a game with all other participating teams .

In ROUND-ROBIN if the number of participants is ‘n’, then the number of games in one round = n(n – 1)/2.

And in KNOCK-OUT basis the team is out of the tournament once it gets defeated. In KNOCK-OUT if the number of games played is ‘n’, then the number of participants is (n + 1), and if the number of participants is ‘n’, then the number of games played is (n -1).

#### EXAMPLE-I

A football tournament is played on the round-robin basis. If the number of participants is 12, how many matches were played in one round?

A) 13
B) 11
C) 66
D) 55

SOLUTION
The number of games played in one round = n(n – 1)/2 = 12(12 – 1)/2 = 66 (option ‘C’)

#### EXAMPLE-II

A Chess World Cup is played on knockout basis in which the number of games played is 96. Find the number of participants.

A) 96
B) 48
C) 97
D) 47

SOLUTION
The number of participants = (n + 1) = 96 + 1 = 97 (option ‘C’)

#### EXAMPLE-III

A football championship is played on a knock-out base. If the 12 team are playing, find how many matches will be played?

A) 12
B) 11
C) 24
D) 66

SOLUTION
The number of matches = (n – 1) = 12 – 1 = 11 (option ‘B’)

#### EXAMPLE-IV

7 people attended a meeting. And each of them shook his hand with each other once. Find the total number of hand-shakes.

A) 21
B) 8
C) 66
D) 14

SOLUTION
This is just like ROUND ROBIN. So number of hand-shakes = n(n -1)/2 = 7(7 -1)/2 = 21 (option ‘A’)

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