Question

A has 5 more pencils than B and B has 5 more pencils than
C. What is the least number of pencils which must change
hands in order that B and C now have equal pencils?
(2) 5
(3) 6
(2) 4
(4) 2​

Answer 1

Answer:

this is your answer .

$xrctc < 4 \frac{ \frac{ \frac{6774 \frac{ \frac{ { \gamma \tan(?) }^{2} }{?} }{?} \times \frac{?}{?} }{?} }{?} }{?} \times \ \tan(?) {?}{?} \times \frac{?}{?} \cot( \tan( \ \alpha \cos(?) ( \tan( \cot(? \gamma \cot( \beta \beta \pi\pi) ) ) ) ) ) log_{ \tan( \sec(e \gamma \beta \cot(?) ) ) }(?)$

sdyikfrdssryiifrraerethiiue. sm chl dp di di di di gp vl

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