Susan invited 13 of her friends for her birthday party and created return gift hampers comprising one each of $3, $4, and $5 gift certificates. One of her friends did not turn up and Susan decided to rework her gift hampers such that each of the 12 friends who turned up got $13 worth gift certificates. How many gift hampers did not contain $5 gift certificates in the new configuration?
(Solve the following problem)

Created: 7 months ago | Updated: 7 months ago

Updated: 7 months ago

$T o t a l v a l u e o f i n i t i a l c o n f i g e r a t i o n = 13 h a m p e r \times $ 12 = $ 156$

$A g a i n t o t a l v a l u e o f n e w c o n f i g e r a t i o n = 12 h a m p e r \times $ 13 = $ 156$

$∴$ The value of gift certificates will not be changed.

$N o w S u s a n s t a r t e d w i t h 13 h a m p e r s . S u s a n h a s 13 p i e c e o f $ 3 w o r t h g i f t$

$∴ S u s a n h a s 13 p i e c e o f $ 4 w o r t h g i f t ∴ S u s a n h a s 13 p i e c e o f $ 5 w o r t h g i f t ∴ P o s s i b l e s e t {5, 5, 3} {4, 4, 5}, (3, 3, 3, 4) i n w h i c h $ 5 c o n t a i n s i n f i r s t t w o o f t h e m a n d t h e t o t a l$

number is 13.

Now, let Susan creates

x hampers of {3, 3, 3, 4}

y hampers of {4,4,5}

And z hampers of {5, 5, 3}

Number of $3 gift will be, 3x + z =13.....(i)

$∴ N u m b e r o f $ 4 g i f t w i l l b e, x + 2 y = 13 . . . . . (i i)$

$∴ N u m b e r o f $ 5 g i f t w i l l b e, y + 2 z = 13 N o w, m u l t i p l y i n g e q u a t i o n (i i i) b y 2 a n d s u b t r a c t i n g (i i) f r o m (i i i) w e g e t,$

$\frac{2 y + 4 z = 26 x + 2 y = 13 - - -}{4 z - x = 13 . . . . . . . . . . . . . . . (i v)}$

$A g a i n, m u l t i p l y i n g e q u a t i o n (i) b y 4 a n d s u b t r a c t i n g e q u a t i o n (i v) f r o m (i) w e g e t \frac{12 x + 4 z = 52 4 z - x = 13}{13 x = 39} ∴ x = 3 H e r e, x i s t h e n u m b e r o f g i f t h a m p e r s w i t h o u t $ 5 c e r t i f i c a t e s . ∴ 3 g i f t h a m p e r s d i d n o t c o n t a i n $ 5 g i f t c e r t i f i c a t e s .$