Arithmatical - Surds and Indices
| 6. |
Given that 100.48 = x, 100.70 = y and xz = y², then the value of z is close to |
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A. 1.45 |
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B. 1.88 |
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C. 2.9 |
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D. 3.7 |
| Solution |
| xz = y² | <=> (10 0.48 )z |
| <=> (100.70)2 |
| = (10 0.48z) |
| = (10(2×0.70)) |
| = 101.40 |
| 0.48z = 1.40 | z= 140/48 |
| = 35/12 |
| = 2.9(approx.). |
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| 7. |
The value of (256)5/4 is |
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A. 984 |
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B. 512 |
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C. 1014 |
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D. 1024 |
| Solution |
| (256)5/4 | =(44)5/4 |
| =4(4x5/4) |
| = 45
|
| = 1024. |
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| 8. |
If 3(x-y)= 27 and 3(x+y)=243, then x is equal to |
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A. 0 |
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B. 2 |
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C. 4 |
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D. 6 |
| Solution |
| 3(x - y) | = 27 |
| = 33 |
| x - y = 3. ......(1) |
| 3(x+y) | = 243 |
| = 35 |
| x + y = 5. .......(2) |
| On solving equations 1 and 2, we get | = x=4. |
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| 9. |
If ax =b, by= c and cz= a, then the value of xyz is |
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A. 0 |
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B. 1 |
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C. 2 |
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D. abc |
| Solution |
| a1 | = cz |
| =(by)y |
| =(byz) |
| =(ax)yz |
| Therefore, | xyz= 1 |
|
| 10. |
49× 49× 49× 49 = ? |
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A. 4 |
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B. 7 |
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C. 8 |
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D. 16 |
| Solution |
| 49× 49× 49× 49 | = (72 ×72 ×72 ×72 |
| = 7(2+2++2+2) |
| = 78 |
| So, the correct answer is 8. |
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