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5th Standard Maths Questions
Maths question helps students to practice the problems without any difficulty and makes more effective towards the learning of the subject. Math questions support with grade text books question for better understanding of the topics.Math Question for Class 5:
Based on the syllabus of class 5, a set of maths question is given for reference of students. The topics covered here are Addition, Subtraction,multiplication,division, fraction, Rounding the number and Factors.
Addition:
Find the missing addend of the group:
1. 2561 = 2000 + ____ + 60 + 1
2. 8565 = _____+ 500 + 60 + 5
3. 5896 = 4000 + 800 + 96 + ____
4. ______= 4000 + 700 + 50 + 1
5. _____ = 900 + 400 + 0 + 6
1. $\frac{5}{2} + \frac{7}{2}$Subtraction:
Find the missing minuend or subtrahend of the following
1. 100 = ___  56
2. 26 = ____ 16
3. 9666 = ____  2564
4. 65434 = ___  42
5. ____  4568 = 8456
6. ____  352 = 4562
Division:
Find the Quotient and remainder by long division method
1. 8 ÷ 56323 2. 4 ÷ 256 3. 5 ÷ 5125 4. 6 ÷ 8564
Find the value of the variables in the multiplication and Division Equation
1. 5x = 25
2. 10 + 2x = 20
3. 4x = 292
4. 48 ÷ y = 24
5. 36 ÷ 6
6. 5d = 125
Add the fractions:
Add the fractions:
2. $\frac{7}{9} + \frac{8}{9}$
3. $\frac{4}{7} + \frac{8}{9}$
4. $\frac{1}{11} + \frac{12}{8}$
5. $\frac{12}{6} + \frac{9}{4}$
Subtract the fractions:
1. $\frac{6}{2}  \frac{7}{2}$
2. $\frac{8}{3}  \frac{4}{3}$
3. $\frac{3}{7}  \frac{2}{9}$
4. $\frac{4}{14}  \frac{10}{4}$
5. $\frac{7}{4}  \frac{3}{2}$
Find the number
 (4 x 2) + (5 x 100) + (0 x 1)
 6 × 1 + 9 × 10 + 2 × 1,000 + 4 × 10,000 + 8 × 100
 6 × 100 + 3× 1 + 7× 10
 5 × 10 + 2 × 1 + 3 × 1,000 + 9 × 100
 9 × 1,000 + 4 × 100 + 5 × 10 + 5 × 1
Rounding the number:
Round the number to nearest Hundred:
1. 1002 2. 256 3. 896 4. 2653 5. 6423
Round the number to nearest Thousand:
Round the number to nearest Thousand:
1. 103015 2. 238442 3. 84602 4. 9632 5. 69642
Factors:
Write the factors of the following number
Factors:
Write the factors of the following number
1. 56 2. 52 3. 68 4. 86 5. 27 6.81 7.45 8.63
Convert the following mixed number into fractions:
1. 4$\tfrac{4}{6}$ 2. 5$\tfrac{2}{6}$ 3. 2$\tfrac{2}{5}$
4. 4$\tfrac{3}{7}$ 5. 7$\tfrac{8}{2}$
Convert the following fractions into mixed fractions:
1. $\frac{10}{6}$ 2. $\frac{30}{10}$ 3. $\frac{26}{12}$
4. $\frac{23}{8}$ 5. $\frac{18}{6}$
Convert the following mixed number into fractions:
1. 4$\tfrac{4}{6}$ 2. 5$\tfrac{2}{6}$ 3. 2$\tfrac{2}{5}$
4. 4$\tfrac{3}{7}$ 5. 7$\tfrac{8}{2}$
Convert the following fractions into mixed fractions:
1. $\frac{10}{6}$ 2. $\frac{30}{10}$ 3. $\frac{26}{12}$
4. $\frac{23}{8}$ 5. $\frac{18}{6}$
Best Results From Yahoo Answers
From Yahoo Answers
Question:easy (but not too easy)
Answers:? ?? ??? ???? ????? where are the q's ??????????
Answers:? ?? ??? ???? ????? where are the q's ??????????
Question:I am checking my daughter's 5th grade math paper can you help me
write in expanded form
20 squared
9 cubed
write in standard form 14 squared
Answers:20*20 9*9*9 14^2
Answers:20*20 9*9*9 14^2
Question:You randomly choose a card from a standard deck of 52 playing cards.
A) find the probability that you choose an ace or a jack.
B) find the probability that you choose a ten or a club.
Answers:well you know two things, there are 52 cards in a deck of cards and there are 4 variations of each card (club, heart, spade, diamond) So the probability of picking out a ace is 4 out of 52, and a jack 4 out of 52 as there are 4 variations of each card, so the overall probability that you'll pick out a jack or an ace is 8/52, or 2/13 in its simplest form What is the probability that you will find a club? well there are 4 variations of every card as you know, and one of those 4 variations is a club, so the chance that you will draw a club is 1/4 the chance that you will draw a club or a ten?: 1/4 of 52 = 13/52 (the chance you will draw a club) and 4/52 is the chance that you will draw a ten, so overall odds are 17/52 of drawing a club or a ten Hope i helped
Answers:well you know two things, there are 52 cards in a deck of cards and there are 4 variations of each card (club, heart, spade, diamond) So the probability of picking out a ace is 4 out of 52, and a jack 4 out of 52 as there are 4 variations of each card, so the overall probability that you'll pick out a jack or an ace is 8/52, or 2/13 in its simplest form What is the probability that you will find a club? well there are 4 variations of every card as you know, and one of those 4 variations is a club, so the chance that you will draw a club is 1/4 the chance that you will draw a club or a ten?: 1/4 of 52 = 13/52 (the chance you will draw a club) and 4/52 is the chance that you will draw a ten, so overall odds are 17/52 of drawing a club or a ten Hope i helped
Question:Tok, so me and mum have been trying to answer this... and cant. show me what the answer is and how ya did... and ya get 10points
his is a 5th grade math problem.
If you can open the spreadsheet, you'll see the list of people who have gotten the correct number.
This is not a trick question.
This is a real math problem so don't say that a bus has no legs.
Good luck with this and have fun!
There are 7 girls in a bus
Each girl has 7 backpacks
In each backpack, there are 7 big cats
For every big cat there are 7 little cats
Question: How many legs are there in the bus?
The number of legs is the PASSWORD to unlock the Excel sheet. If you open it, add your name and send it on to see who else can unlock it.
Answers:7 girls x 2 legs each = 14 legs 49 backpacks x 7 big cats= 343 big cats x 4 legs each = 1372 legs 343 big cats x 7 little cats x 4 legs each = 9604 legs Add the legs together to get a total: 10,990 legs This doesn't include the bus driver of course... TO THE ANSWER BELOW MINE (ZIXAQ) Careful, you seemed to have made a math error 32*7*49 =10,976 not 10,978. Add 14 for the girls legs and you should get 10,990
Answers:7 girls x 2 legs each = 14 legs 49 backpacks x 7 big cats= 343 big cats x 4 legs each = 1372 legs 343 big cats x 7 little cats x 4 legs each = 9604 legs Add the legs together to get a total: 10,990 legs This doesn't include the bus driver of course... TO THE ANSWER BELOW MINE (ZIXAQ) Careful, you seemed to have made a math error 32*7*49 =10,976 not 10,978. Add 14 for the girls legs and you should get 10,990