Rajat has two congruent sheets of paper in triangular shape. He built a parallelogram with them by three different methods. It is known that the perimeters of the three parallelograms are 32, 41 and 43. Find the perimeter of the triangular piece of paper.
There are 10 consecutive positive integers written on a blackboard. One number is erased. The sum of remaining nine integers is 2011. Which number was erased?
Find the number of subsets S of {1,2,….., 63}the sum of whose elements is 2008.
1 + 2 + _ _ _ + 63 = 2016. So the problem is equivalent to finding the number of subsets of{1,2,….., 63} whose sum of elements is 8. We can count this manually : {8}, {1,7} , {2,6} , {3,5} , {1,2,5} , { 1,3,4} Hence , 6
A 5-member cricket team has to be selected for a new format of cricket named Hong Kong Super Five! It must be composed of 2 batswomen, 2 bowlers and one wicket-keeper.There are three batting contenders viz. Supriya, Sumaya and Sanjana. There are two wicket-keeping contenders viz. Sunita and Sarita. There are three bowling contenders viz. Savita, Sujata and Suprita. If Savita is selected in the team, then Suprita cannot be selected. If Sanjana is selected, then Savita cannot be selected. If Sumaya is selected, then Sanjana also has to be selected.
If Savita and Sujata are selected, then which of the following statements must be true?
I. Sarita is selected.
II. Both Supriya and Sumaya are selected as batswomen.
III. Sunita is selected.
1. Only I
2. Only II and III
3. Only III
4. None of these
For solution check : Puzzles of the Day: 96
Let f(x) be a quadratic polynomial with f(2) = 10 and f(–2) = – 2. Then the coefficient of x in f(x) is
1. 3
2. 2
3. 5
4. Cannot be determined
f(2) = 10, f(–2) = –2 f(x) = ax2 + bx + c 10 = 4a + 2b + c ….. (1) –2 = 4a – 2b + c …… (2) From (1) and (2) b = 3
A 5-member cricket team has to be selected for a new format of cricket named Hong Kong Super Five! It must be composed of 2 batswomen, 2 bowlers and one wicket-keeper.There are three batting contenders viz. Supriya, Sumaya and Sanjana. There are two wicket-keeping contenders viz. Sunita and Sarita. There are three bowling contenders viz. Savita, Sujata and Suprita. If Savita is selected in the team, then Suprita cannot be selected. If Sanjana is selected, then Savita cannot be selected. If Sumaya is selected, then Sanjana also has to be selected.
Which of the following player must be selected in the team?
1. Supriya
2. Sumaya
3. Savita
4. Sujata
For solution check : Puzzles of the Day: 96
How many positive integers less than 1200 have no repeating digits ?
(no repeating digits means no digit occurs more than once)
There are 9 such 1-digit numbers, Similarly the number of such 3-digit numbers is 9×9×8 = 648, since the third digit cannot equal to either of the first two. For 4-digit numbers beginning 10, there are 8×7 and there are none beginning 11.
9×9 = 81 such 2- digit numbers (since the first digit can be 1 to 9 and the second digit any other number from 0 to 9).
Hence , 9 + 81 + 648 + 56 = 794
A 5-member cricket team has to be selected for a new format of cricket named Hong Kong Super Five! It must be composed of 2 batswomen, 2 bowlers and one wicket-keeper.There are three batting contenders viz. Supriya, Sumaya and Sanjana. There are two wicket-keeping contenders viz. Sunita and Sarita. There are three bowling contenders viz. Savita, Sujata and Suprita. If Savita is selected in the team, then Suprita cannot be selected. If Sanjana is selected, then Savita cannot be selected. If Sumaya is selected, then Sanjana also has to be selected.
Which of the following is the list of people who can be selected in a team?
1. Sujata,Sumaya,Sunita,Suprita,Supriya
2. Sanjana,Sarita,Savita,Sujata,Sumaya
3. Sanjana,Sarita,Sujata,Sumaya,Suprita
4. Sarita,Savita,Sumaya,Suprita,Supriya
For solution check : Puzzles of the Day: 96
