NUMBERS & FUNDAMENTALS OF MATHS (PART-IV)

QUERY 31

The price of a necklace varies directly as the number of pearls in it. Also it varies directly as the square root of radius of a pearl. The price of necklace was Rs 150 when it had 75 pearls each of radius 1cm in it. Find the radius of the pearl of necklace having 100 pearls whose cost is Rs 600.

A) 7
B) 9
C) 15
D) 12

MAHA GUPTA
Be the number of pearls at a given situation = n
and the radius of the pearl at a given situation =√r

Therefore the price at that situation = kn√r; where k is constant

Now when the price of the necklace with 75 pearls each of radius 1 cm in it is Rs 150
Hence 150 = k75√1
=> k = 2

When the price of the necklace with 100 pearls in it Rs 600
Therefore 600 = 2*100*√r
=> √r = 3
=> r = 9 (option ‘B’)

QUERY 32

Five lines are intersecting each other. What is the maximum number of intersecting points?

A) 11
B) 10
C) 12
D) 25

MAHA GUPTA
When every number is meeting all the others once we find the sum of all numbers except the biggest.

So here the required number = 4+3+2+1 = 10 (option ‘B’)

QUERY 33

Three friends had dinner at a restaurant. When the bill was received, Amita paid 2/3 as much as Veena paid and Veena paid 1/2 as much as Tanya paid. What fraction of the bill did Veena pay?

A) 1/3
B) 3/11
C) 12/13
D) 5/8

MAHA GUPTA
This sum can be done in mind easily
Let Tanya paid = 60
Therefore Veena paid = 1/2 of 60 = 30
and Amita paid = 2/3 of 30 = 20

So total amount of the bill = 60+30+20 = 110

So Veena’s fraction of the bill = 30/110 = 3/11 (option ‘B’)

QUERY 34

The sum of ages of 5 children born at the intervals of 3 years each is 50 years. What is the age of the youngest child?

A) 4 years
B) 8 years
C) 10 years
D) None of these

MAHA GUPTA
Deduct the time gap from the average; see how
Average = 50/5 = 10 years (means the age of the third child)
The time gap between the third and the first child = 3+3 = 6 years

Therefore the youngest’s age = 10 – 6 = 4 years (option ‘A’)

QUERY 35

An intelligent trader travels from one place to another carrying 3 sacks having 30 coconuts each. No sack can hold more than 30 coconuts. On the way he passes through 30 checkpoints and on each checkpoint he has to give 1 coconut for each sack he is carrying. How many coconuts are left in the end?

A) 90
B) 35
C) 25
D) 30

MAHA GUPTA
Total of coconuts = 3*30 = 90
To save coconuts he’ll start giving coconuts from one sack first
So, the first sack will run for = 30/3 = 10 checkpoints

Now he has left with only 2 sacks; therefore now he’ll have to give 2 coconuts at each checkpoint
So, the second sack will run for = 30/2 = 15 checkpoints

Finally he has one sack only; therefore he’ll need to give only 1 coconut at each checkpoint

Now the checkpoints left = 30 – (10+15) = 5
So the number of coconuts given = 5

Hence the number of coconuts left with him in the end = 90 – (Coconuts of two sacks + 5) = 90 – (30+30+5) = 25 (option ‘C’)

QUERY 36

By reversing the digit of the number 13, the number increased by 18. How many other two digit number increase by 18 when their digits are reserved?

A) 5
B) 6
C) 7
D) 8

SHIV KISHOR
Let the digit at unit place of those numbers be ‘a’ and at tens place be ‘b’; then according to the question
(10a+b) – (10b+a) = 18
=> a – b = 2

Means all such numbers that have their unit digits greater than the tens digit by 2

Except 13, obviously they are 24, 35, 46, 57, 68, 79
So they are 6 in number (option ‘B’)

QUERY 37

The right most non zero digit of the number 30²⁷²⁰ is?
A) 1
B) 3
C) 7
D) 9

MAHA GUPTA
We know that (ab)ⁿ = aⁿbⁿ
Thus, 30²⁷²⁰ = (3*10)²⁷²⁰ = (3²⁷²⁰)(10²⁷²⁰)

10 raised to power of any number will give all zeroes next to 1 like 10, 100, 1000, 10000, etc; so we have just to find the unit digit of 3²⁷²⁰

Unit digit of 3 raised to any natural number repeats after every 4 intervals; so divide 2720 by 4 to see what is left; it is ‘0’ off course; means we have just to find the unit digit of 3⁴

Now 3⁴ = 81
So the required digit = 1 (option ‘A’)

QUERY 38

On children’s day, sweets were to be equally distributed among 175 children in a school. Actually on the children’s day 35 children were absent and therefore each child got 4 sweets extra. How many sweets in total were available for distribution?

A) 3000
B) 2800
C) 2600
D) 2500

MAHA GUPTA
Children that were present = 175 – 35 = 140
Therefore extra sweets in total = 140*4 = 560
Means they were to be distributed among 35 children

Thus sweets for one child = 560/35 = 16

Hence total sweets available for distribution = 175*16 = 2800 (option ‘B’)

QUERY 39

A man ate 100 bananas in 5 days, each day eating 6 more than the previous day. How many bananas did he eat on the first day?

A) 5
B) 6
C) 2
D) 8

MAHA GUPTA
It means the man eats those additional bananas in 4 days
Therefore number of additional bananas ate = 6+12+18+24 = 60
Hence the remaining bananas = 100 – 60 = 40

Means he eats the remaining bananas in 5 days equally
So number of bananas ate on the first day = 40/5 = 8 (option ‘D’)

NOTE: This way longer questions of this type can be solved in quick time.

QUERY 40

When a number is reversed, difference is always a multiple of?

A) 5
B) 6
C) 9
D) 7

MAHA GUPTA
Let the number is a 2 digit number
Therefore the number = 10a + b
Number after reversal of digits = 10b + a

Therefore the difference = (10a + b) – (10b + a)
= 10a – a + b – 10b
= 9a – 9b
= 9(a – b)

We see that 9 is a factor of the difference, so we can say the asked difference is always a multiple of 9 (option ‘C’)

NOTE: Try it on a 2 digit number only as with other number the result could be more than 9 also. But 9 will be among all.