Practice Unit place - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) The last digit of $(1001)^2008$ + 1002 is
(a)
(b)
(c)
(d)
Last digit of $(1001)^2008$ + 1002 = 1 + 2 = 3
Q-2) The unit digit in the expansion of $(2137)^754$ is
(a)
(b)
(c)
(d)
Expression = $(2137)^754$
Unit’s digit in 2137 = 7
Now, $7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5$ = 16807, ...
Clearly, after index 4, the unit’s digit follow the same order.
Dividing index 754 by 4 we get remainder = 2
∴ Unit’s digit in the expansion of $(2137)^754$ = Unit’s digit in the expansion of $(2137)^2$ = 9
Q-3) One’s digit of the number $(22)^23$ is
(a)
(b)
(c)
(d)
Unit’s digit in the expansion of $(22)^23$
= Unit’s digit in the expansion of $(2)^23$
Now, $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32$
i.e. 2 repeats itself after the index 4.
On dividing 23 by 4, remainder = 3
∴ Unit’s digit in $(2)^23$ = Unit’s digit in $(2)^3$ = 8
Q-4) The digit in the unit’s place of the product $(2464)^1793×(615)^317×(131)^491$ is
(a)
(b)
(c)
(d)
$(4)^{2m}$ gives 6 at unit digit.
$(4)^{2m +1}$ gives 4 at unit digit.
$(5)^n$ gives 5.
The same is the case with 1.
Required digit = Unit’s digit in the product of 4×5 × 1 = 0
Q-5) 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-6) 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-7) 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-8) 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-9) 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-10) Unit digit in $(264)^102 + (264)^103$ is :
(a)
(b)
(c)
(d)
Unit digit in $(264)^4$ i.e. 4 × 4 × 4 × 4 is 6
Unit digit in $(264)^100$ is also 6.
Now, $(264)^102 = (264)^100 × (264)^2$
= (Unit digit 6) × (Unit digit 6)=36
∴ Unit digit is 6
Similarly,
$(264)^103 + (264)^100 × (264)^3$
= (Unit digit 6) × (Unit digit 4)=24
∴ Unit digit is 4
Therefore, the unit digit in $(264)^102 + (264)^103$ is
6 + 4 = 10 i.e. 0.