Practice Number system - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) The unit digit in the product $(122)^173$ is
(a)
(b)
(c)
(d)
$2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32$
Unit digit in the product of $(122)^173$
= Unit digit in $(122)^1$ = 2
(1 = remainder when 173 is divided by 4).
Q-2) The digit in the unit’s place of $[(251)^98 + (21)^29 – (106)^100 + (705)^35 – 16^4 + 259]$ is :
(a)
(b)
(c)
(d)
$(251)^98$ = ......1
$(21)^29$ = ....1
$(106)^100$ = ......6
$(705)^35$ = ....5
$(16)^4$ = .....6
259 = ......9
∴ Required answer = 1 + 1 – 6 + 5 – 6 + 9 = 16 – 12 = 4
Q-3) The digit in unit’s place of the number $(1570)^2 + (1571)^2 + (1572)^2 + (1573)^2$ is :
(a)
(b)
(c)
(d)
Unit's digit in $(1570)^2$ = 0
Unit's digit in $(1571)^2$ = 1
Unit's digit in $(1572)^2$ = 4
Unit's digit in $(1573)^2$ = 9
∴ Required unit’s digit = Unit’s digit (0 +1+4 + 9) = 4
Q-4) The unit digit in the sum of $(124)^372 + (124)^373$ is
(a)
(b)
(c)
(d)
$4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024$
Remainder on dividing 372 by 4 = 0
Remainder on dividing 373 by 4 = 1
Required unit digit = Unit digit of the sum of 6 + 4 = 0
Q-5) Find the unit digit in the product $(4387)^245 × (621)^72$ .
(a)
(b)
(c)
(d)
$7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5$ = 16807
i.e. The unit’s digit repeats itself after power 4.
Remainder after we divide 245 by 4 = 1
Unit’s digit in the product of $(4387)^245 × (621)^72$ = Unit’s digit in the product of $(4387)^1 × (621)^72$ = 7 × 1 = 7
Q-6) Out of six consecutive natural numbers, if the sum of first three is 27, what is the sum of the other three ?
(a)
(b)
(c)
(d)
x + x + 1 + x + 2 = 27
3x + 3 = 27
3x = 24
x = 8
Three consecutive no's whose sum is 27 are 8, 9,10.
Hence, next 3 consecutive no's having 36 as sum are 11, 12 and 13
Q-7) The sum of three consecutive odd natural numbers each divisible by 3 is 72. What is the largest among them?
(a)
(b)
(c)
(d)
Let the numbers be 3x, 3x + 3 and 3x + 6
3x + 3x + 3 + 3x + 6 = 72
9x + 9 = 72
9x = 72 – 9 = 63
x =$63/9$ = 7
∴ Largest number = 3x + 6 = 3 × 7 + 6 = 27
Q-8) The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is
(a)
(b)
(c)
(d)
$10^2 + 11^2 + 12^2$ = 100 + 121 + 144 = 365
Required sum =10 + 11 + 12 = 33
Q-9) The sum of the squares of three consecutive natural numbers is 2030. Then, what is the middle number?
(a)
(b)
(c)
(d)
Let the three consecutive natural numbers be x, x + 1 and x + 2.
According to question,
$x^2 + (x + 1)^2 + (x + 2)^2$ = 2030
or $x^2 + x^2 + 2x + 1 + x^2 + 4x + 4$ = 2030
or $3x^2 + 6x + 5 = 2030 $
or $3x^2 + 6x – 2025$ = 0
or $x^2 + 2x – 675$ = 0
or $x^2 + 27x – 25x – 675$ = 0
$x (x + 27) – 25 (x + 27)$ = 0
or $(x – 25) (x + 27)$ = 0
$x = 25 and – 27$
∴ Required number = $x$ + 1 = 25 + 1 = 26
Q-10) The sum of all those prime numbers which are not greater than17 is
(a)
(b)
(c)
(d)
Prime numbers upto 17⇒ 2, 3, 5, 7, 11, 13, 17
Required sum = 2 + 3 + 5 + 7 + 11 + 13 + 17 = 58
Q-11) A number when divided by 91 gives a remainder 17. When the same number is divided by 13, the remainder will be ?
(a)
(b)
(c)
(d)
Here, the first divisor (91) is a multiple of the second divisor (13).
∴ Required remainder = Remainder obtained on dividing 17 by 13
⇒ 17 = ( 13 × 1 ) + 4
Hence Required remainder = 4
Q-12) Each member of a picnic party contributed twice as many rupees as the total number of members and the total collection was 3042. The number of members present in the party was ?
(a)
(b)
(c)
(d)
Let the total number of members be X.
Then, each member's contribution =Rs. 2X
Given, X × 2X = 3042
⇒ 2X2 = 3042
⇒ X2 = 1521
⇒ X = 39
Q-13) Find the sum of all positive multiples of 3 less than 50
(a)
(b)
(c)
(d)
Sum of all multiples of 3 upto 50
= 3 + 6 + ..... + 48
= 3 (1 + 2 + 3 + .... + 16)
= 3×${16(16+1)}/2$= 3 × ${272}/2$ = 408
[Since 1+2+3+…+n = ${n(n+1)}/2$]
Q-14) Two positive whole numbers are such that the sum of the first number and twice the second number is 8 and their difference is 2. The numbers are :
(a)
(b)
(c)
(d)
Let the numbers be x and y.
According to the question,
x + 2y = 8 .... (i)
x – y = 2 ....... (ii)
By equation (i) – (ii),
2y + y = 8 – 2
3y = 6; y = 2
From equation (ii),
x – 2 = 2 ; x = 4
Q-15) Sum of three fractions is 2$11/24$ . On dividing the largest fraction by the smallest fraction, $7/6$ is obtained which is $1/3$ greater than the middle fraction. The smallest fraction is
(a)
(b)
(c)
(d)
Let the three fractions be p, q and r,
where p < q < r.
According to the question,
$r/p$ = $7/6$ ⇒ r = $7/6$ p
Again, middle fraction
= q = $7/6$ - $1/3$ = ${7-2}/6$ = $5/6$
∴ p+q+r = 2$11/24$
⇒ p + $5/6$ + $7/6$p = $59/24$
⇒ p+${7p}/6$ = $59/24$ - $5/6$
⇒ ${6p+7p}/6$ = ${59-20}/24$ = $93/24$
⇒ 13p = $39/24$×6 = $39/4$
⇒ p = $39/{4×13}$ = $3/4$
Q-16) Which of the following fraction is the smallest ? $9/13, 17/26, 28/29, 33/52$
(a)
(b)
(c)
(d)
$9/13 = {9×4}/{13×4} = 36/52$
$ 17/26={17×2}/{26×2} = 34/52 $
$33/52 = 33/52 $
Among these $33/52$ is the smallest
Again, $28/29 = 56/58 > 33/52$
Q-17) smallest possible three place decimal number is :
(a)
(b)
(c)
(d)
The smallest possible three place decimal number = 0.001
Q-18) The greatest fraction among $2/3, 5/6, 11/15\text" and "7/8$is
(a)
(b)
(c)
(d)
$2/3 = 0.67 ; 5/6 = 0.83$
$11/15 = 0.73; 7/8 = 0. 875$
Q-19) Which of the following number is NOT divisible by 18 ?
(a)
(b)
(c)
(d)
A number will be exactly divisible by 18 if it is divisible by 2 and 9 both. Clearly 65043 is not divisible by 2.
∴ Required number = 65043
Q-20) $1/0.04$ is equal to :
(a)
(b)
(c)
(d)
$1/0.04$ = $100/4$ = 25