Number System - Study Mode
[#311] Find the sum of all positive multiples of 5 less than 100.
Correct Answer
(C) 950
Explanation
Solution: Sum = 5 + 10 + 15 + 20 + .....+ 95 , It is an arithmetic progression Number of terms (n) = $$frac{{{ ext{Last term }} - { ext{ First term}}}}{{{ ext{Differnce}}}}$$ xa0 xa0 + 1 ⇒ n = $$frac{95 - 5}{5}$$ + 1 = 19 So, sum = $$frac{n}{2}$$ [2a + (n - 1)d] Here, a = 5, n = 19, d = 5 Sum = $$frac{19}{2}$$ [10 + (19 - 1)5] = 950
[#312] A number divided by 52 gives remainder 45. If the number divided by 13 , the remainder will be :
Correct Answer
(B) 6
Explanation
Solution: Since 13 is factor of 52. So divide its remainder directly by 13. $$eqalign{
& Rightarrow frac{{{ ext{remainder}}}}{{13}} = frac{{45}}{{13}} cr
& Rightarrow { ext{remainder }} = 6 cr} $$
[#313] The greatest among following numbers is : $${left( 3
ight)^{frac{1}{3}}},$$ $${left( 2
ight)^{frac{1}{2}}}$$ , 1, $${left( 6
ight)^{frac{1}{6}}}$$
Correct Answer
(D) $${left( 3
ight)^{frac{1}{3}}}$$
Explanation
Solution: L.C.M. of powers (3, 2, 1 and 6) = 6 $$eqalign{
& {left( 3
ight)^{frac{1}{3}}} Rightarrow {left( {{3^2}}
ight)^{frac{1}{6}}} = {{x08f{9}}^{frac{{x08f{1}}}{{x08f{6}}}}} cr
& {left( 2
ight)^{frac{1}{2}}} Rightarrow {left( {{2^3}}
ight)^{frac{1}{6}}} = {8^{frac{1}{6}}} cr
& (1) Rightarrow {left( {{1^6}}
ight)^{frac{1}{6}}} = {1^{frac{1}{6}}} cr
& {left( 6
ight)^{frac{1}{6}}} Rightarrow {left( {{6^1}}
ight)^{frac{1}{6}}} = {6^{frac{1}{6}}} cr} $$
[#314] Of the three numbers, the second is twice the first and is also thrice the third. If the average of these three numbers is 44, the largest number is -
Correct Answer
(C) 72
Explanation
Solution: No. → I II III 3x 6x 2x = 11x $$eqalign{
& frac{{11x}}{3} = 44 Rightarrow x = 12 cr
& { ext{Largest number = 6x = 72}} cr} $$
[#315] A positive integer when divided by 425 gives a remainder 45. When the same number is divided by 17, the remainder will be :
Correct Answer
(C) 11
Explanation
Solution: $$eqalign{
& frac{{{ ext{Remainder of number }}}}{{17}} = frac{{45}}{{17}} cr
& Rightarrow { ext{Remainder = 11}} cr} $$