1. I started counting from 11 to 28, 11, 12, 13 …27, 28. The first number I counted was 11 and the last number was 28. How many numbers did I count in total?

Ans: If you started from 1 you would have counted 28 numbers but you did not count the first 10, so you only counted 28 – 10 = 18.

How many numbers are there from 43 to 87? There are 87 numbers from 1 to 83 but 1 to 42 are excluded, so there are only 87 – 42 of 45 numbers from 43 to 87.

The class was assigned to read chapter IV of the book which covers pp 38 to 55. How many pages were the class assigned to read? 55 – 37 = 18 pages.

2. Which of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 should we remove so that the sum of the remaining 8 numbers is 40.

Ans: Using “completing the whole”, the following number pairs add up to 10: 1 and 9, 2 and 8, 3 and 7 and 4 and 6. Those four pairs add up to 40. Therefore, remove the 5.

3. If we add 19 and 33, what is the sum?

Ans: Use “completing the whole” to simplify the computation. 19 is only 1 less than 20, so we can add 20 to 33 to get 53 and then subtract 1 to have a total of 52.

We can also use “dagdag-bawas”. Add 1 to 19 and at the same time subtract 1 from 33 to transform the problem into 20 + 32 = 52.

4. Solve the following expression: 11 – 1 + 12 – 2 + 13 – 3 + 14 – 4 + 15 – 5 + 16 – 7

Ans: Again use “completing the whole” and group the numbers to form tens:

(11 – 1) + (12 – 2) + (13 – 3) + (14 – 4) + (15 – 5) + (16 – 7) = 10 + 10 + 10 + 10 + 10 + 9 = 59

5. Write the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 on your scratch paper. Which two numbers should you erase so that the remaining numbers will add up to 37?

Ans:The numbers 1 to 10 has 4 pairs adding up to 10, a 5 and a 10.  If you will erase two numbers, only 8 will remain. These 8 can form 4 pairs: 3 pairs adding up to 10 and 1 pair adding up to 7. The 5 needs a 2 to be 7 so we must break up the 2 and 8 pair. We will remove the 10 and the 8.

1. It is 9:00 o’clock at the moment. What time would it be in 26 hours?

Ans: 26 hours is 2 hours + a whole day (24 hours). 9:00 + 2 hours = 11:00 o’clock

2. Write the numbers 1, 2, 3, 4, 5, 6, 7, 8 and 9 in your scratch paper. Which number must be removed so that the sum of the remaining numbers is 41.

Ans: there are 4 pairs of numbers adding up to 10 and one 5 in the numbers from 1 to 9. If we remove the 5 we will have only 40. So 5 must remain and we must remove 1 less than 5 or 4 to increase the total to 41.

3. How many two digit numbers are there in which the tens digit is greater than the units digit?

Ans: The two digits number run from 10 to 99. If the ten’s digit is 1 we have only 1 number 10 meeting this criterion. If the ten’s digit is 2 we have 2 – 20 and 21. If the tens digit is three we have 3 – 30, 31 and 32 and so on. So the total of such numbers is the sum of the numbers from 1 to 9 or 45.

4. New Year’s day in 2020 falls on a Wednesday. What days occurs 5 times in January 2020?

Ans:  January 29 will also be a Wed, Jan 30 a Thursday and Jan 31, Friday. So There will be 5 Wednesdays, 5 Thursday and 5 Fridays in January 2020.

5. There are 15 tables in a snack bar. Some tables can seat 6 people while some can seat only 4 people. If a maximum of 78 people can be seated in the snack bar, how many 6-people tables are there?

Ans: By using the Vedic Math Sutra “By Alternate Elimination and Retention”, we first eliminate the 6-people tables and assume that all 15 tables can seat only 4 each. The snack bar can accommodate only 60 people. The extra 18 people can be  seated in 18 ÷ (6 – 4) = 9 six – people tables.

1. Isabel had -1 peso and 5-peso coins in her bag totaling P22. If she had a total of 10 coins, how many of these are P5 coins?

Ans: Use the Vedic Math Sutra “ By Alternate Elimination and retention”. Assume that all 10 coins are 1 peso coins. That would only be 10 pesos. The difference of P22 – 10 or P12 must have come from the difference of P 5 – P1 or P4 between the denominations. Therefore there must be  P12/ P4 = 3. So there a 3 5 peso coins.

2. What is one half of three-fifths of 60?

Ans: one half of three-fifths is three tenths and thee tenths of 60 is 18.

3. What is the sum of the first 50 integers?

Ans: the 50 integers form 25 pairs of digits adding up to 51: 1 + 50; 2 + 49, etc so the total is 51 x 25 = 1275

4. What is the largest two digit number that gives a remainder of 7 when divided by 8?

Ans: The biggest two digit- number divisible by 8 is 96. So 95 will give a remainder of 7.

5. In a farm there are 36 ducks. There are six times as many chickens as there are ducks. How many more chickens are there than ducks.

Ans: There are 5 times more chickens than ducks so there are 36 x 5 = 180 more.