CIE Mathematics Paper-1: Specimen Questions 164 - 166 of 175

Question number: 164

One Liner Question▾

Write in Brief

Solve 383×65×4 .


Here, 383×65×4

  • So, as per BODMAS calculation, expression will be: (383×6)÷(5×4).
  • So, first of all as per BODMAS (Brackets, Order, Division, Multiplication, Addition, Subtraction) brackets will be removed, in that also first of all multiplication will be: (3818)÷(5×4).
  • Then in brackets subtraction will be: 20÷(5×4).
  • Now, in brackets multiplication will be: 20÷20
  • Now, division will be: 1 .

Question number: 165

Short Answer Question▾

Write in Short

(a) Write down a common multiple of 52 and 91

(b) Solve 3563219

Give your answer as a fraction in its lowest terms.



(a) We need to find common multiple of 52 and 91.

So, 52=13×4and91=13×7

So, here common multiple is 13.

(b) Here, 3563219=5×79×77×33×3=5973=5219=169 .

Question number: 166

Essay Question▾

Describe in Detail


(a) a0b2

(b) xaxb ,

(c) (31)3



(a) a0b21b2=1×b2=b2 Because 0 powers always gives value 1 and minus power of denominator will become plus when denominator multiply by its numerator.

(b) xaxbxab Because division of two terms with same base will subtract denominator’s exponent from numerator’s exponent.

(c) (31)331×3=33=13×3×3=127 Because it is an exponent expression that is raised to a power, so we can multiply the exponent and power which is second exponent rule.