What will be the value of (1^{2} + 2^{2} + 3^{2} + ... + 10^{2}) = ?
We know that (1^{2} + 2^{2} + 3^{2} + ... + n^{2})
Putting n=10, required sum
= 385
What will be the sum of (51+ 52 + 53 + ... + 100) = ?
This is an A.P. in which a = 51, l = 100 and n = 50.
What will be the last digit in (7^{95}– 3^{58})?
Unit digit in 7^{95} = Unit digit in [(7^{4})^{23} × 7^{3}]
= Unit digit in [(Unit digit in(2401))^{23} × (343)]
= Unit digit in (1^{23} × 343) = Unit digit in (343) = 3
Unit digit in 358 = Unit digit in [(3^{4})^{14} × 3^{2}]
= Unit digit in [Unit digit in (81)^{14} × 3^{2}]
= Unit digit in [(1)^{14} × 3^{2}]
= Unit digit in (1 × 9)
= Unit digit in (9) = 9
Unit digit in (7^{95} – 3^{58})
= Unit digit in (343 – 9)
= Unit digit in (334) = 4.
Find the value of 1904 × 1904 = ?
904 × 1904 = (1904)^{2}
= (1900 + 4)^{2}
= (1900)^{2} + (4)^{2} + (2 × 1900 × 4)
= 3610000 + 16 + 15200
= 3625216
The least six digit number completely divisible by 111 is:
The least six digit number is 100000.
When 100000 ÷ 111,
Quotient 990 and Remainder = 100
Therefore required number
=100000 + ( 111 – 100)
= 100000 + 11
= 100011
Find the number of terms in the given G.P. series 3, 6, 12, 24... 384.
Here a = 3 and r =6/3 =2.
Let the number of terms be n.
Then, t_{n} = 384
ar^{n1} = 384
3 × 2^{n – 1} = 384
2^{n1} = 128 = 2^{7}
n – 1 = 7
n = 8
Number of terms = 8.
What will be the largest 5 digit number completely divisible by 91?
Largest 5digit number = 99999
When 99999÷91, Quotient = 1098 and
Remainder = 81
Required number = (99999 – 81)
= 99918.
When we multiply a certain number by 7, we obtain product whose each digit is 3. What will be that number?
By hit and trial, we find that
47619 × 7 = 333333.
If 25% of 2/5 of a certain number is 125 then the required number is:
Let the required number be n,
Then 25% × 2/5 × n = 125
By solving the above equation,
∴ n = 1250
Which of the following cannot be the square of counting number?
The square of a natural number never ends in 7.
42437 is not the square of a natural number.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 




