Number System - Study Mode
[#521] Let x be the least number which when divided by 15,18, 20 and 27, the remainder in each case is 10 and x is a multiple of 31. What least number should be added to x to make it a perfect square?
Correct Answer
(A) 39
Explanation
Solution: L.C.M. of 15,18, 20, 27 = 540 x = 540K + 10 x = 540 × 4 + 10 = 2170 46 2 < 2170 > 47 2 47 2 = 2209 To make perfect square = 2209 - 2170 = 39
[#522] If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
Correct Answer
(C) $$frac{5}{{52}}$$
Explanation
Solution: $$eqalign{
& { ext{Lets first number}} = a cr
& { ext{Second number}} = b cr
& a + b = 65 cr
& sqrt {a imes b} = 26 cr
& ab = 676 cr
& { ext{Sum of reciprocal}} = frac{1}{a} + frac{1}{b} cr
& = frac{{b + a}}{{ab}} cr
& = frac{{65}}{{676}} cr
& = frac{5}{{52}} cr} $$
[#523] The sum of three consecutive natural numbers is always divisible by . . . . . . . .?
Correct Answer
(A) 3
Explanation
Solution: Let the number be x, x + 1, x + 2 Sum = x + x + 1 + x + 2 = 3x + 3 = 3(x + 1) It is always divisible by 3
[#524] Three numbers are in Arithmetic progression (AP) whose sum is 30 and the product is 910. Then the greatest number in the AP is:
Correct Answer
(C) 13
Explanation
Solution: Let the three number is a - d, a, a + d a is first term, d is common difference a + d + a + a - d = 30 xa0 xa0 (Given) 3a = 30 a = 10 (a + d)(a)(a - d) = 910 (10 + d)(10)(10 - d) = 91 × 10 (10 + d)(10 - d) = 91 Put d = 3 $$x08oxed{left( {10 + 3}
ight)left( {10 - 3}
ight) = 91}$$ So, d = 3 So, greater number is = a + b = 10 + 3 = 13
[#525] What is the value of $$99frac{{11}}{{99}} + 99frac{{13}}{{99}} + 99frac{{15}}{{99}} + ........ + 99frac{{67}}{{99}}?$$
Correct Answer
(A) $$frac{{95120}}{{33}}$$
Explanation
Solution: $$eqalign{
& 99frac{{11}}{{99}} + 99frac{{13}}{{99}} + 99frac{{15}}{{99}} + ........ + 99frac{{67}}{{99}} cr
& { ext{Number of term}} = frac{{67 - 11}}{2} + 1 = 29 cr
& = 99 imes 29 + frac{1}{{99}}left( {11 + 13 + 15 + ........ + 67}
ight) cr
& = 2871 + frac{1}{{99}}left[ {frac{{29}}{2}left( {2 imes 11 + left( {29 - 1}
ight)2}
ight)}
ight] cr
& = 2871 + frac{1}{{99}}left[ {frac{{99}}{2}left( {22 + 56}
ight)}
ight] cr
& = 2871 + frac{1}{{99}}left( {29 imes 39}
ight) cr
& = 2871 + frac{{1131}}{{99}} cr
& = 2871 + frac{{377}}{{33}} cr
& = frac{{95120}}{{33}} cr} $$