It is very easy! suppose $ a_i : $ the value of rice in basket $ i $ $ b_i : $ the number of eggs in basket $ j $ now we prove that there are two baskets $ i,j $ such that $ a_i+a_j > 10 , b_i+b_j>30 $ $ ( $ $ * $ $ ) $ suppose that $ \forall 1 \le i \le 100 : b_m \ge b_i $ if $ b_m < 30 $ then $ b_1+b_2+...+b_100 < 100b_m \Rightarrow 30000<100b_m<30000 $ and it is impossible. so $ b_m \ge 30 $ thus $ \forall 1 \le i \le 100 , i \neq m : b_i+b_m \ge 30+b_i>30 $ if $ a_m \ge 10 $ then $ ( $ $ * $ $ ) $ is true. otherwise suppose $ a_m<10 $ now we prove there is a basket $ j $ such that $ a_m+a_j>10 $ suppose this is false. so $ \forall i \neq m : a_i+a_m \le 10 $ $ a_1+a_m \le 10 $ $ a_2+a_m \le 10 $ $ ... $ $ a_{m-1}+a_m \le 10 $ $ a_{m+1}+a_m \le 10 $ $ ... $ $ a_100+a_m \le 10 $ so $ (a_1+a_2+...+a_{m-1})+(a_{m+1}+a_{m+2}+...+a_100 ) + 99 a_m \le 9900 $ but $ a_m < 10 $ so $ (a_1+a_2+...+a_100)+99a_m < 9900+10 =10000 $ thus $ 10000+99a_m<10000 $ and it is impposible. so there is a basket $ j $ such that $ a_m+a_j > 10 $ but we know $ b_m+b_j >30 $ with this we can simply prove in one step we can make a basket with $ 30 $ eggs and $ 10kg $ rice. continue this way with other baskets. the answer is $ 99 .$ but don't forget you must in an example prove that you need at least $ 99 $ steps! it is easy to find this example ( how?! ) so it's done