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