Problems
on Numbers – Solved Questions
1. Find the difference between place value of 7 and face value of 3
in 927384?
a)6997 b)3648 c)6700 d)6450
Answer: A
Explanation:
The face value is the number itself => The face
value of 3 is 3
The place value of a number 7 => 7000
The difference between the place value of 7 and the
face value of 3 is 7000 – 3 = 6997
2. Find the sum of the first 15 natural
numbers?
a)60 b)120 c)180 d)225
Answer : B
Explanation:
The sum of the first n natural
numbers = $\frac{n(n+1)}{2}$
Therefore, the sum of the
first 15 natural numbers = $ \frac{15\times(15+1)}{2}$
= 15 x $\frac{16}{2}$
=120
3. Find the sum of the natural
numbers from 51 to 100
a)3500 b)3685 c)3775 d)5050
Answer : C
Explanation :
We don’t have any direct formula to find the sum of
the natural numbers from M to N.
So, we find the sum of the natural numbers from 1
to 100 and then subtract the sum of the first 50 natural numbers.
Sum of the natural numbers from 51 to 100 = Sum of
natural numbers from 1 to 100 - Sum of natural numbers from 1 to 50
= $\frac{100\times(100+1)}{2}$
- $\frac{50\times(50+1)}{2}$
= (100 x $ \frac{101}{2}$ )- (50 x $\frac{51}{2}$ ) =(5050 -1275) = 3775
= (100 x $ \frac{101}{2}$ )- (50 x $\frac{51}{2}$ ) =(5050 -1275) = 3775
4. Find the sum of squares of
first 12 natural numbers
a) 2250 b)1950 c)2125 d)640
Answer : B
Explanation:
Sum of the square of the first n
natural numbers =$\frac{n\times(n+1)(2n+1)}{6}$
Here n=12
So sum of the squares first 12
natural numbers = $\frac{12\times(12+1)(2 \times 12}{6}$
= 12 x 13 x $\frac{25}{2}$ = 1950
5. Find the sum of the squares of
natural numbers from 21 to 30?
a) 9455 b)2870 c)6400 d)6585
Answer : D
Explanation :
To find the sum of the square of the natural number
from 21 to 30. First, we have to find the sum of squares of the first 30
natural numbers and then subtract the sum of the first 20 natural numbers
212 + 222 +
.. 302 = (12 + 22 + .. +302)
– (12+ 22 + 32 + .. +202)
=
[$\frac{30\times(30+1)((2\times30)+1)}{6} $
] – [ $\frac{20\times(20+1)((2\times20)+1)}{6} $ ]
= [30 x 31 x $\frac{61}{6}$ ] - [20 x 21 x $\frac{41}{6}$ ] =(9455 -2870)= 6585
6. What is the sum of the cubes of the
first 20 natural numbers?
a)88200 b)22050 c)44100 d)12250
Answer : C
Explanation:
Sum of the cubes of the first n
natural numbers = $\left[\frac{n(n+1)}{2}\right]
^{2} $
Hence, the Sum of the cubes of the first 20 natural
numbers = $\left[\frac{20(20+1)}{2}\right]
^{2}$
=
$\left[\frac{20\times21}{2}\right]
^{2}$
=
2102 =44100
7. Find the sum of the even numbers in
the first 120 natural numbers?
a)3660 b)7260 c)14400 d)3600
Answer : A
Explanation:
Sum of the first n even
natural numbers = n(n+1)
In the first 120 natural numbers, we
have $\frac{120}{2}$
= 60, even natural numbers.
Sum
of the even numbers in the first 120 natural numbers = 60(60+1) = 3660
8. Find the sum of the even number
from 101 to 150
a)12450 b)6300 c)3150 d)7200
Answer : C
Explanation :
As we don’t have any direct formula to find the sum
of the even numbers in a range of natural numbers, we find the sum of the even
numbers in first 150 natural numbers, then subtract the sum of the even natural
numbers in the first 100 natural numbers.
Number evens in first 150 natural numbers are $\frac{120}{2}$
= 75 and in first 100 natural numbers is $\frac{100}{2}$
= 50
102+ 104+ 106 +.. +150 =
[75 x (75+1) ]– [ 50 x (50 +1)]
=
(75 x 76 ) - (50 x 51) = 5700 – 2550 =3150
9. Find the sum of the odd numbers in the
first 150 natural numbers?
a)22500 b)5625 c)5700 d)11250
Answer : B
Explanation:
The sum of the first n odd natural
numbers = n2
In the first 150 natural numbers, there are $\frac{150}{2}$
=
75 odd natural numbers.
So the sum of the odd numbers in the first
150 natural numbers = 752 = 5625
10. Find the sum of the squares of
the first 20 even natural numbers?
a)11480 b)5640 c)3600 d)22960
Answer : A
Explanation :
Sum of the first n even natural numbers =
$\frac{2n(n+1)(2n+1)}{3}$
Sum of the squares of first 20 even natural
numbers =$ \frac{2\times20\times(20+1)\times((2\times20)+1)}{3}$ =
$\frac{2\times20\times21\times41}{3} $ = 11480
$\frac{2\times20\times21\times41}{3} $ = 11480
11. Find the sum of the cubes of the
first 30 odd natural numbers?
a)42850 b)35990 c)2680 d)38240
Answer : B
Explanation :
Sum of the first n even odd
natural numbers = $\frac{n(2n-1)(2n+1)}{3}$
Sum of the squares of the first 30 odd natural numbers =
$\frac{30\times((2\times30)-1)((2\times30)+1)}{3}$ = $\frac{30\times59\times61}{3}$ = 35990
12. What is the sum of the cubes of
odd numbers in the first 45 natural numbers?
a)628437 b)384397 c)559153 d)389734
Answer : C
Explanation :
The sum of the cubes of first n even natural numbers =
n2(2n2 - 1)
Number of even numbers in first 45 natural numbers n=
$\frac{45+1}{2}$
= 23
Therefore, the sum of the cubes
of odd natural numbers in 1st 45 natural numbers
=
232 x ((2 x 232)- 1) = 232 x
( 2x 529 )-1)
= 529 x 1057=559153
= 529 x 1057=559153
13. Find the sum of the cubes of even
numbers in the first 40 natural numbers?
a)428430 b)284280 c)523800 d)352800
Answer : D
Explanation:
The sum of the cubes of first n even natural numbers =
$n^{2}(2n^{2}-1)$
Number of even numbers in the first 40
natural numbers n= $\frac{40}{2}$
= 20
Therefore, sum of the cubes
of even natural numbers in 1st 40 natural numbers
= 2 x 202 x (20+1)2 = 2 x 400 x 441 =352800
= 2 x 202 x (20+1)2 = 2 x 400 x 441 =352800
14. Find the sum of the first seven
prime numbers?
a)49 b)58 c)63 d)72
Answer : B
Explanation:
We don’t have any direct
formula to find the sum of the seven prime numbers.
So We have to add all the prime numbers.
Sum of the first 7 prime numbers = (2+3+5+7+11+13+17)=58
15. What is the sum of all prime
numbers between 1 and 60?
a)360 b)421 c)381 d)440
Answer : D
Explanation :
The prime numbers between 1 and 60 are :
1, 2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 ,31
, 37 ,41 , 43 , 47 ,53 and 59
The required sum is 440.