# “VIVA Session”- on Quant- Problems on Numbers Solutions Updated

“VIVA Session”- on Quant- Problems on Numbers Solutions Updated:
Dear Friends, This session is an initiative of our team which will be conducted daily that mainly concentrates on evaluating skill and knowledge of our viewers. In this session we are going to ask set of questions without giving any options which are important in examination point of view, you need to think on those questions and give answers by commenting below and you can also ask your questions from today’s particular subject. At the end of the day we will give answers.

1. What is the least number must be added to 3000 to obtain a number exactly divisible by 19?
Ans: On dividing 3000 by 19, we get 17 as remainder.
Therefore Number to be added = (19-17) = 2.

2. Find the smallest 6 digit number which is exactly divisible by 11?
Ans: Smallest number of 6 digits is 100000.
On dividing 100000 by 111, we get 100 as remainder.
Therefore number to be added = (111-100) – 11.
Hence required number = 100011.

3. Find the sum of all odd digits upto 100.
Ans: Odd numbers upto 100 = 1,3,5,7,9,………..,99.
The series is in AP with a = 1 and d = 2.
Let it contain n terms. Then 1 + (n-1) x 2 = 99è n = 50.
Therefore required sum = (n/2) x (first term + last term) = (50/2) x (1+99) = 2500.

4. 3200 + 550 – 768 – ? = 842 – 720 + 530
Ans:2300

5. in a division sum, the divisor is 10 times the quotient and 5 times the remainder. If the remainder is 46, what is the dividend?
Ans:Divisor = (5×46) = 230 ===> 10 x Quotient = 230 ===> Quotient = 23
Dividend = (Divisor x Quotient) + Remainder = (230 x 23) + 46 = 5290 +46 = 5336.

6. How many digits will be there to the right of the decimal point in the product of 95.75 and 0.02554
Ans: Seven

7. A number when divided by 296 leaves 75 as remainder. When the same number is divided by 37, the remainder will be:
Ans:Let x = 296q + 75   = (37 x 8q + 37 x 2) + 1   = 37 (8q + 2) + 1
Thus, when the number is divided by 37, the remainder is 1.

8. A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When it is successively divided by 5 and 4, then the respective remainders will be
Ans:
4 |x           y = (5 x 1 + 4) = 9
——–
5 |y -1        x = (4 xy + 1) = (4 x 9 + 1) = 37
——–
| 1 -4
Now, 37 when divided successively by 5 and 4, we get
5 | 37
———
4 | 7 – 2
———
| 1 – 3
Respective remainders are 2 and 3.

9. If 60% of (3 /5) of a number is 36, then the number is:
Ans:Let the number be x. then 60% of (3/5) of x = 36 ===> x = { 36 x (25/9) } = 100
Therefore required number = 100

10. A number was divided successively in order by 4, 5 and 6. The remainders were respectively 2, 3 and 4. The number is
Ans:
4 |x           z = 6 x 1 + 4  = 10
-----------
5 |y -2        y = 5 xz + 3  = 5 x 10 + 3  = 53
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6 |z - 3       x = 4 xy + 2  = 4 x 53 + 2  = 214
-----------
| 1 - 4

Hence, required number = 214.

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