A to Z Exams

72

The sequence is 1 * 2, 2 * 3, 3 * 4, 4 * 5, 5 * 6, 6 * 7, 7 * 8.
.'. Missing number = 8 * 9 = 72.

both 3 and 17

100

Each even term of the series is obtained by multiplying the previous term by 2.

2^nd term = (1^st term) x 2 = 6 x 2 = 12

4^th term = (3^rd term) x 2 = 48 x 2 = 96.

6^th term = (5^th term) x 2 = 384 x 2 =768.

 4^th term should be 96 instead of 100.

49.97592

4083 X 102 X 12 = 4997592. Sum of decimal places = 5.

∴ 40.83 X 1.02 X 1.2 = 49.97592.

79

The HCF will be 79 for both as follows
2923 ) 3239 ( 1
       -  2923 
            316 ) 2923 ( 9
                 -  2844
                        79 ) 316 ( 4
                            - 316
                                X

470

(4.7 x 13.26 + 4.7 x 9.43 + 4.7 x 77.31)
= 4.7 (13.26 + 9.43 + 77.31)
= 4.7 x 100
= 470

1.9832

Sum of decimal places = (2 + 2) = 4
2.68 * 0.74 = 1.9832.

24

Any factor of this number should be of the form 2a×3b×5c.

For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4. b can take values 0,2, 4, 6 and c can take values 0,2.

Total number of perfect squares =3×4×2= 24

2^2 x 3^5 x 7^2

3240 = 2^3 x 3^4 x 5; 3600 = 2^4 x 3^2 x 5^2; H.C.F. = 36 = 2^2 x 3^2 Since H.C.F. is the product of lowest powers of common factors, so the third number must have (2^2 x 3^2) as its factors Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 3^5 and 7^2 as its factors. Third number = 2^2 x 3^5 x 7^2

6 and 12 both

(6n² + 6n) = 6n(n + 1), which is always divisible by 6 and 12 both, since n(n + 1) is always even.

10.25%

Let the sum be Rs. 100. Then,

S.I. for first 6 months = Rs.		100 x 10 x 1		= Rs. 5
		100 x 2

S.I. for last 6 months = Rs.		105 x 10 x 1		= Rs. 5.25
		100 x 2

So, amount at the end of 1 year = Rs. (100 + 5 + 5.25) = Rs. 110.25

i.e Effective rate = (110.25 - 100) = 10.25%