Number System - Study Mode

[#511] The number which can be written in the form of n(n + 1)(n + 2), where n is a natural number, is:
Correct Answer

(D) 6

Explanation

Solution: Let the natural number (n) be 1. ∴ n(n + 1)(n + 2) = 1(1 + 1)(1 + 2) = 6

[#512] What is the value of 9991 × 10009?
Correct Answer

(C) 99999919

Explanation

Solution: 9991 × 10009 = 9991(10000 + 9) = 99910000 + 89919 = 99999919 Alternate solution 9991 × 10009 = (10000 - 9)(10000 + 9) = (10000) 2 - (9) 2 = 100000000 - 81 = 99999919

[#513] Twenty one times of a positive number is less than its square by 100. The value of the positive number is:
Correct Answer

(A) 25

Explanation

Solution: Let the number be = x Then according to question 21 × x + 100 = x 2 x 2 - 21x - 100 = 0 x 2 - 25x + 4x - 100 = 0 x(x - 25) + 4(x - 25) = 0 (x - 25)(x + 4) = 0 x = -4 is not possible x = 25

[#514] Let x = (633) 24 - (277) 38 + (266) 54 . What is the units digits of x?
Correct Answer

(A) 8

Explanation

Solution: x = (633) 24 - (277) 38 + (266) 54 x = 1 - 9 + 6 x = 7 - 9 x = . . . . . 7 - 9 x = 17 - 9 x = 8

[#515] What is the sum of all the common terms between the given series S 1 and S 2 ? S 1 = 2, 9, 16, . . . . . ., 632 S 2 = 7, 11, 15, . . . . . ., 743
Correct Answer

(C) 6974

Explanation

Solution: S 1 = 2, 9, 16, . . . . . ., 632 S 2 = 7, 11, 15, . . . . . ., 743 S 1 = 2, 9, 16, 23, 30, 37, 44, 51 . . . . . . S 2 = 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, . . . . . . Number of terms in $${{ ext{S}}_1} = frac{{632 - 2}}{7} + 1 = 91$$ ∵ In S 1 there is no further terms beyond 632 therefore there is no common terms. ∴ Number of terms in series S 1 = 91 If this series start from the 23 then number of terms = 91 - 3 = 88 Now, number of common terms in both series S 1 and S 2 = $$frac{{88}}{7}$$ = 22 = n Now, common terms 23, 51, . . . . . . 22 $$eqalign{
& { ext{Sum}} = frac{n}{2}left[ {2a + left( {n - 1}
ight)d}
ight] cr
& = 11left[ {46 + 21 imes 28}
ight] cr
& = 11left[ {46 + 588}
ight] cr
& = 11 imes 634 cr
& = 6974 cr} $$