# NUMBERS & FUNDAMENTALS OF MATHS (PART-VI)

#### QUERY 51

A 3-digit number 4a3 is added to another 3-digit number 984 to give a 4-digit number 13b7, which is divisible by 11. Then, (a + b) = ?

A) 10
B) 11
C) 12
D) 15

MAHA GUPTA

4     a     3
9     8    4
————
13   b    7
————

You see addition of the unit digits of the first number and the second number is less than 10, so nothing to carry forward from this to the total of tens digits of those numbers; therefore,
a + 8 = b
=> b – a = 8                                                      (i)

Also, 13 b7 is divisible by 11. A number is divisible by 11 if the difference of the sum of the digits at the even places and sum of digits at odd places is either 0 or a multiple of 11. Taking it as
(7 + 3) – (b + 1) = 0

=> 9 – b = 0

=> b = 9

Substituting the value in (i)
9 – a = 8 => a = 1

Therefore, a + b = 1 + 9 = 10 (option ‘A’)

#### QUERY 52

(11² + 12² + 13² + … + 20²) = ?

A) 385
B) 2485
C) 2870
D) 3255

MAHA GUPTA
(11² + 12² + 13² + … + 20²) = (1² + 2² + 3² + … + 20²) – (1² + 2² + 3² + … + 10²)

We know that (1² + 2² + 3² + … + n²) = 16n(n + 1)(2n + 1)

Therefore,
(11² + 12² + 13² + … + 20²) = (20*21*41)/6 – (10*11*21)/6
= 2870 – 385
= 2485 (option ‘B’)

#### QUERY 53

A boy multiplied 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong and the other digits are correct, then the correct answer would be?

A) 553681
B) 555181
C) 555681
D) 556581

MAHA GUPTA
The required number must be divisible by the factors of 987

Now, 987 = 3 x 7 x 47

So, the required number must be divisible by each one of 3, 7, 47

In such a question we should check the answer options with the divisibility of the factors; better to start with the smallest as it’s easier as compared.

We see the first two options reject as neither of them is divisible by 3. Now check the other options with 7.
Here we find that option ‘D’ i.e. 556581 is not divisible by 7. Therefore, the answer is 555681 (option ‘C’)

#### QUERY 54

In a question of division the dividend is 37693, quotient is 52 and the remainder lies between 52 and 104. Find the divisor.

A) 724
B) 723
C) 722
D) 721

MAHA GUPTA
According to the question, 37693 ÷ 52 = 724 as divisor, and 45 as remainder

Means we need to lessen the divisor by 1 so that the earlier quotient can add to the remainder and increase its value.
Thus the new remainder = 45 + 52 = 97 (we see it’s between 52 and 104)

Therefore, the new divisor = 724 – 1 = 723 (option ‘B’)

#### QUERY 55

A bucket full of water weighs 50 kg, while half-filled bucket weighs 30 kg. Find the weight of the empty bucket.

A) 10 kg
B) 15 kg
C) 20 kg
D) 25 kg

MAHA GUPTA
Weight of the completely-filled bucket = 50 kg
Weight of the half-filled bucket = 30 kg

Hence, water in the first half bucket = 50 – 30 = 20 kg
We can say water in the second half of the bucket is also 20 kg

Therefore, weight of the empty bucket = 30 – 20 = 10 kg (option ‘A’)

#### QUERY 56

The weight of a container full of water is 28 kg. When is 1/4 water in it, it weighs 19 kg. Its weight when it is 2/3 full of water will be?

A) 8 kg
B) 20 kg
C) 24 kg
D) 18.67 kg

MAHA GUPTA
Weight of container + total water = 28 kg
Weight of container + 1/4 water = 19 kg

Hence, weight of 3/4 water = 28 – 19 = 9 kg

Thus weight of total water = 9 × 4/3 = 12 kg
And weight of empty container = 28 – 12 = 16 kg

And weight of 2/3 water = 12 × 2/3 = 8 kg
Therefore, weight of the container with 2/3 full of water = 16 + 8 = 24 kg (option ‘C’)

#### QUERY 57

How many digits are required to write numbers from 1 to 400?

A) 1002
B) 1092
C) 1103
D) 1209

MAHA GUPTA
One digit numbers  = 1 to 9 = 9
Two digit numbers = 10 to 99 = 99 – 9 = 90
Three digit numbers = 100 to 400 = 400 – 99 = 301

Therefore total digits required from 1 to 400 = (9*1) + (90*2) + (301*3)  = 1092 (option ‘B’)

#### QUERY 58

How many numbers less than 1000 are multiples of both 10 and 13?

A) 9
B) 8
C) 6
D) 7

MAHA GUPTA
As the numbers are multiples of both 10 and 13, those numbers must be divided by the LCM of 10 and 13.

Now the LCM of 10 and 13 = 130

Now, 1000/130 = 7 as quotient and 90 as remainder

Therefore, the required number = 7 (option ‘D’)

#### QUERY 59

In a three digit number, the digit in the unit’s place is 75% of the digit in ten’s place. The digit in ten’s place is greater than the digit in the hundred’s place by 1. If the sum of the digits in ten’s place and hundred’s place is 15, what is the number?

A) 795
B) 786
C) 687
D) 867

MAHA GUPTA
Unit’s digit is 75% of the ten’s digit, means 3/4 of the ten’s digit
Hence, ten’s digit is either 4 or 8 as it must be divisible by 3/4
It means unit’s digit is either 3 or 6

Now sum of hundred’s digit and ten’s digit is 15, therefore hundred’s digit is either 15 – 4 = 11 or 15 – 8 =7
But a digit can’t be greater than 9, so hundred’s digit is 7

Hence the number = 786 (option ‘B’)

TRICK
Sum of hundred’s digit and ten’s digit is 15; in the answer option’s only 786 (option ‘B’) satisfies the condition.

#### QUERY 60

A three digit number is divisible by 11 and has its digit in the unit’s place equal to 1. The number is 297 more than the number obtained by reversing the digits. What is the number?

A) 121
B) 231
C) 561
D) 451

MAHA GUPTA
Unit’s digit is given already, letting the digit at ten’s place as y and the digit at hundred’s place as x, the number = 100x + 10y + 1

Hence, the number after reversing the digits = 100 + 10y + x

As the number is 297 more than the number obtained by reversing the digits, therefore
(100x + 10y + 1) – (100 + 10y + x) = 297
=> 99x – 99 = 297
=> x = 4

Now the digit at hundred’s place is 4 and at unit’s place is 1, so the number is 451 as the number is divisible by 11. (option ‘D’)

TRICK
Subtract 297 from the numbers given in the answer option and see if the digits are reversed. We see 451 is such a number, hence our answer. (option ‘D’)

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