CIE Mathematics Paper-1: Specimen Questions 171 - 172 of 175

Question number: 171

Short Answer Question▾

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(a )8p×83=26 . Find the value of p.

(b) 9q÷95=312 . Find the value of q.

(c) 6x=1216 . Find the value of x.

Explanation

Here,

(a )8p×83=26

Now, we need to convert right hand sided base into 8.

So, 26=23×23=8×8=82

So, 8p×83=828p+3=82 because multiplication of same based term will give addition of their exponent.

p+3=2 because base is same of all exponents.

p=23p=1.

(b) 9q÷95=312

Now, we need to convert right hand sided base into 9.

So, 312=32×32×32×32×32×32=9×9×9×9×9×9=96

So, 9q÷95=969q5=96 because division of same based term will give subtraction of their exponent.

q5=6 because base is same of all exponents.

q=6+5q=11.

(c) 6x=12166x=1636x=63x=3 .

Question number: 172

Short Answer Question▾

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The bearing of a temple T on a hill from the ground G is 1700 . Find the bearing of G from T.

Temple on the hill and have bearing from ground

Temple on the hill and have bearing from ground

Showing a temple on the hill with some bearing with the ground

Explanation

Here,

The bearing of temple from the ground is 1700 , and line TG is straight line which gives angle of 1800 .

So, needed bearing of ground from temple will be: 18001700=100 .

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