AIC 1st Term Maths Exam for SS 1
- 1. Convert 37 in base ten to base five
A) 122
B) 1225
C) 1023
D) 1001011
- 2. Convert 1264 in base eight to base ten
A) 171
B) 692
C) 629
D) 117
- 3. Convert 211 in base three to base eight
A) 11
B) 26
C) 7
D) 62
- 4. Convert 1.101 to decimal
A) 0.25
B) 0.125
C) 1.625
D) 0.625
- 5. In what base is the addition 465 + 24 + 225 = 1050?
A) 6
B) 9
C) 5
D) 7
- 6. Evaluate (203)2 in base four
A) 11512
B) 103021
C) 10152
D) 113021
A) 1100
B) 110
C) 101
D) 1011
- 8. If (34) in base five = 23x, find x.
A) x = -8
B) x = -16
C) x = 16
D) x = 8
- 9. Round off 163.864 to the nearest whole number
A) 163
B) 200
C) 164
D) 163.9
- 10. Express the sum of 6.03 x 106 and 2.17 x105 in standard form
A) 6.247 x 106
B) 6.247 x 10-6
C) 6.247 x 10-11
D) 6.247 x 1011
- 12. If x is a positive integer, determine the least value of x for which 13 + 2x = 3(mod 8)
A) 5
B) 1
C) 16
D) 3
- 13. Evaluate 3√350 using logarithm table
A) 70.47
B) 7.047
C) 704700
D) 704.7
- 14. Evaluate (3.69 x 105) ÷ (1.64 x 10-3)
A) 2.25 x 10-8
B) 2.25x 102
C) 2.25 x 108
D) 2.25 x 10-2
- 15. Calculate the following using mathematical table. 3√(38.32 x 2.964/8.637 x 6.285)2
A) 4.1106
B) 1.636
C) 0.2137
D) 7.716
- 16. Express 0.0006131 in standard form
A) 6131 x 10-4
B) 6.131 x 104
C) 6.131 x 10-4
D) 6131 x 104
- 17. Simplify (1/16)-¾ + 5 (90)
A) 2/3
B) 13
C) 7/25
D) 3½
- 18. Evaluate 2 ÷ (64/125)-⅔
A) 5 ⅚
B) 3 ½
C) 1 ⁷/25
D) 2 ⅛
- 19. Simplify 125⅓ x 49½ x 10-1
A) 6 ⅞
B) 2 ⅓
C) 3 ½
D) 35/5
- 20. Given that 102 = 100, write the expression in logarithm form
A) 100 = log10 2
B) 2 = log10 100
C) 3 = log10 100
D) 1 = log10 102
|
Students who took this test also took :