# NUMBERS & FUNDAMENTALS OF MATHS (PART-III)

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

#### QUERY 21

The difference of two numbers is 3 and their square difference is 39; find the bigger number.

A) 8
B) 7
C) 15
D) 12

MAHA GUPTA
Let the bigger and smaller numbers are x and y respectively

Therefore x – y = 3 —-(i)
And x² – y² = 39 —-(ii)

From (ii)
(x + y)(x – y) = 39
=> (x + y)3 = 39 —putting value of (x – y)
=> x + y = 13 —-(iii)

Solving (i) and (iii)
x = 8
So, the bigger number = 8 (option ‘A’)

#### QUERY 22

A two digit number is 5 times of its digits. The digits get reversed on adding 9 to the number. Find sum of the digits of that number.

A) 9
B) 8
C) 14
D) 17

MAHA GUPTA
Let the digit at unit place be ‘y’; and ten’s place is x
Then the number = 10x + y

Now according to the question
10x + y = 5(x + y) => 5x – 4y = 0 —-(i)

But the places of the digits gets reversed on adding 9 to the number
So, (10x + y) + 9 = 10y + x
=> 9x – 9y = -9
=> x – y = -1 —-(ii)

Solving (i) & (ii)
x = 4; y = 5

So the addition of the digits = 4+5 = 9 (option ‘A’)

NOTE: Such questions, if solved by the answer options take hardly 10 seconds.

#### QUERY 23

A and B hire a grass-field on monthly rent of Rs 100 for one year. A feeds his 10 cows for 11 months there. How many cows can B get fed in the remaining days if he pays Rs 120 less by A?

A) 110
B) 105
C) 85
D) 90

MAHA GUPTA
Total Expenditure of A and B both together = 12*100 = 1200
As expenditure of B is less by Rs 120 that of A; expenditure of B = (1200 – 120)/2 = 540

Therefore expenditure of A = 1200 – 540 = 660

Time left for B = 1 month
Thus expenditure of A in 1 month = 660/11 = 60

But this expenditure of A is on 10 cows; therefore his expenditure on 1 cow = 60/10 = 6

Hence number of cows that B can feed = Total expenditure of B/Expenditure on 1 cow = 540/6 = 90 (option ‘D’)

#### QUERY 24

How many numbers are divisible by 3 in the set of numbers 300, 301, 302,…..499, 500?

A) 80
B) 85
C) 67
D) 73

MAHA GUPTA
In such questions subtract the smallest from the biggest and divide that difference by the divisor; and round-off the quotient thus got to the nearest natural number.

So, the required number = (500 – 300)/3 = 200/3 = 66.67 = 67 (option ‘C’)

#### QUERY 25

Find the number formed by the last two digits of 1218.

A) 71
B) 61
C) 87
D) 92

MAHA GUPTA
When the number ends in 1; and the number is raised to any power

unit digit = 1
ten’s digit = unit digit of the multiplication of the ten’s digit of the number (here 2) with unit digit of the power (here 8)
Therefore it’s unit digit of 2*8 = 16 i.e. 6

Hence the required number is 61 (option ‘B’)

#### QUERY 26

How many three-digit numbers are divisible by 6 in all?

A) 149
B) 150
C) 151
D) 166

MAHA GUPTA
First of all take the lowest three digit number divisible by 6; obviously it’s 102
Therefore all of them will be 6 are: 102, 108, 114,… , 996
We see that this is an A.P. in which a = 102, d = 6 and l = 996

Now, let the number of terms of the above A.P. be n. Then tn = 996.
Hence, a + (n – 1)d = 996
=> 102 + (n – 1) x 6 = 996
=> 6 x (n – 1) = 894
=> (n – 1) = 149
=> n = 150
So there the number of 3-digit numbers divisible by 6 = 150 (option ‘B’)

#### QUERY 27

The product of two whole numbers is 24, the smallest possible sum of these numbers is?

A) 12
B) 8
C) 9
D) 10

MAHA GUPTA
This is only possible when the difference of the factors is the smallest; obviously they are 4 and 6

So their sum = 4 + 6 = 10 (option ‘D’)

#### QUERY 28

Two numbers are such that the ratio of their sum, difference and product is 11 : 1 : 90. Find the sum of the squares of the two numbers.

A) 180
B) 270
C) 549
D) 239

MAHA GUPTA
Let the numbers be a and b
According to the question
a + b = 11k
a – b = 1k
ab = 90k ; where ‘k’ is constant in all

We know (a + b)² = a² + b² + 2ab = (a – b)² + 4ab

Taking (a + b)² = (a – b)² + 4ab
(11k)² = k² + 4(90k)
=> 12k² – k² = 360k
=> k = 3

Now as per question
a + b = 11k = 11*3 = 33
a – b = k = 3
ab = 90k = 90*3 = 270

Taking (a + b)² = a² + b² + 2ab
33² = a² + b² + 2(270)
a² + b² = 1089 – 540 = 549 (option ‘C’)

TRICK
The ratio of two numbers’ sum, difference and product is 11 : 1 : 90
Letting 11, 1 and 90 be above both the numbers must be consecutive because the difference is 1; means they are 6 and 5. But they do not satisfy the third condition, means their product is not 90

The ratio can be written as 22 : 2 : 180; means both the numbers must have difference of 2; means they are 12 and 10. But they do not satisfy the third condition, means their product is not 180

The ratio can be written as 33 : 3 : 270; means both the numbers must have difference of 3; means they are 18 and 15. Here they also satisfy the third condition, i.e, their product is 270.

Now sum of their squares = 18² + 15² = 549 (option ‘C’)

#### QUERY 29

A and B came back home after their exam and their father asked them about the test. A replied that 1/3rd of his answers were wrong and B replied that 5 of his answers were wrong but together we got 3/4 of answers right. How many questions were there for the exam?

MAHA GUPTA
Suppose the total number of questions in the exam be ‘x’
Thus A’s correct attempts = 1 – 1/3 = 2/3 of x = 2x/3
And B’s correct attempts = x – 5

So according to the question
(2x/3) + (x – 5) = 3/4 of the questions = 3/4 of 2*x (x is multiplied by 2 as the total number of questions will become double of the questions in one paper)

Thus, by solving x = 30 (answer)

#### QUERY 30

A sum of money is sufficient to pay A’s wages for 21 days and B’s wages for 28 days. The same money is sufficient to pay the wages of both for how many days?

A) 12 days
B) 11 days
C) 17 days
D) 15 days

MAHA GUPTA
Let the total sum for wages = Rs 8400 (to make calculations easier better to take it product of 100 and LCM of 21 and 28)

Therefore A’s daily wages = 8400/21 = 400
And B’s daily wages = 8400/28 = 300

Thus their total daily wages = 400+300 = 700

So, the required number of days = 8400/700 = 12 days (option ‘A’)

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