1. Evaluate 13 + 23 + 33 + 43 + 53 = ?
13+ 23+ 33+ 43+ 53
= 1 + 8 + 27 + 64 + 125
= 9 + 27 + 64 + 125
= 36 + 64 + 125
= 100 + 125
2. Evaluate : sqrt(1 + sqrt( 1 + sqrt(64))) = ?
sqrt(1 + sqrt( 1 + sqrt(64))) = sqrt(1 + sqrt(1+ sqrt(64))) = sqrt(1 + sqrt(1 + 8) ) = sqrt(1 + 3) = 2
3. A group of students decided to collect as many paise from each member of group as is the number of members. If the total collection amounts to Rs. 59.29, the number of the member is the group is:
= (59.29 x 100) paise
= 5929 paise.
Number of members
4. What is the value of sqrt(1.5625) ?
sqrt(1.5625) = sqrt(15625 / 10000) = sqrt(15625) / sqrt(10000) = 125 / 100 = 1.25
5. Find the least four digit number which is a perfect square.
1024 = 32^2.
6. If sqrt(2) = 1.414 then find the value of sqrt(20,000) + sqrt(200) = ?
Sqrt (2) = 1.414 =>
Sqrt(20000) = 141.4
Sqrt(200) = 14.14
Adding these two, we get 155.54
7. 1849 students are sitting in an auditorium in such a manner that there are as many students in a row as there columns in the auditorium. Find the number of rows in the auditorium?
Let k be the number of rows and columns, then k x k = 1849 => k 2 = 1849
k = 43
8. The product of two numbers is 120. The sum of their squares is 289. Find the sum of two numbers
Let the numbers be x and y. Given x2 + y2 = 289 and xy = 120.
We know that (x+y)2 = x2 + y2 + 2xy = 289 + 2(120) = 289 + 240 = 529
(x+y)2 = 529 => x+y=23
9. Given n = 12, then find the difference between n2 and (n+1)2.
Given n=12, then n+1 = 13.
(n+1)2 – n2 = 132 – 122 = 169 – 144 = 25
10. The number 252 is written as a + b where a and b are consecutive natural numbers. Find the maximum of these two values.
We know that (2n + 1)2 = (2n2+2n) + (2n2+2n+1),
625 = (25)2 = (2 x 12 + 1)2 = ( 2(12)2 + 2(12) ) + (2 x 122 + 2(12)+1) = 312 + 313
Maximum(312,313) = 313
11. How many natural number lie between the square of the following numbers 12 and 13.
We know that between n2 and (n+1)2 there are 2n non-perfect square numbers.
So, between 122 and 132, there are 2(12) , that is 24 natural numbers.
12. We can express the number 1681 as the sum of first ___?___ odd natural numbers.
The number 1681 = 412 can always be expressed as sum of first 41 odd natural number
13. Evaluate: 1 + 3+ 5 + 7 + … + 49
We know that 1 + 3 + 5 + …+ (2n - 1) = n2.
1 + 3 + 5 + 7 + … + 49 = 252 = 625
14. What will be the unit's digit in the number N^2, if N = 1 + 2 + .. + 35
Given N = 1 + 2 + … + 35 = (35 x 36)/2 = 35 x 18
N is a number which ends with 0, hence N2 also has to end with 0.
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