1. A
The ratio between black and blue pens is 7 to 28 or 7:28. Bring to the lowest terms by dividing both sides by 7 gives 1:4.
2. A
At 100% efficiency 1 machine produces 1450/10 = 145 m of cloth.
At 95% efficiency, 4 machines produce 4 * 145 * 95/100 = 551 m of cloth.
At 90% efficiency, 6 machines produce 6 * 145 * 90/100 = 783 m of cloth.
Total cloth produced by all 10 machines = 551 + 783 = 1334 m
Since the information provided and the question are based on 8 hours, we did not need to use time to reach the answer.
3. D
The turnout at polling station A is 945 out of 1270 registered voters. So, the percentage turnout at station A is:
(945/1270) × 100% = 74.41%
The turnout at polling station B is 860 out of 1050 registered voters. So, the percentage turnout at station B is:
(860/1050) × 100% = 81.90%
The turnout at polling station C is 1210 out of 1440 registered voters. So, the percentage turnout at station C is:
(1210/1440) × 100% = 84.03%
To find the total turnout from all three polling stations, we need to add up the total number of voters and the total number of registered voters from all three stations:
Total number of voters = 945 + 860 + 1210 = 3015
Total number of registered voters = 1270 + 1050 + 1440 = 3760
The overall percentage turnout is:
(3015/3760) × 100% = 80.12%
Therefore, the total turnout from all three polling stations is 80.12% — rounding to 80%.
4. D
This is a simple direct proportion problem:
If Lynn can type 1 page in p minutes, then she can type x pages in 5 minutes
We do cross multiplication: x * p = 5 * 1
Then,
x = 5/p
5. A
This is an inverse ratio problem.
1/x = 1/a + 1/b where a is the time Sally can paint a house, b is the time John can paint a house, x is the time Sally and John can together paint a house.
So,
1/x = 1/4 + 1/6 … We use the least common multiple in the denominator that is 24:
1/x = 6/24 + 4/24
1/x = 10/24
x = 24/10
x = 2.4 hours.
In other words; 2 hours + 0.4 hours = 2 hours + 0.4•60 minutes
= 2 hours 24 minutes
6. D
The original price of the dishwasher is $450. During a 15% off sale, the price of the dishwasher will be reduced by:
15% of $450 = 0.15 x $450 = $67.50
So the sale price of the dishwasher will be:
$450 – $67.50 = $382.50
As an employee, the person receives an additional 20% off the lowest price, which is $382.50. We can calculate the additional discount as:
20% of $382.50 = 0.20 x $382.50 = $76.50
So the final price that the employee will pay for the dishwasher is:
$382.50 – $76.50 = $306.00
Therefore, the employee will pay $306.00 for the dishwasher.
7. D
Original price = x,
80/100 = 12590/X,
80X = 1259000,
X = 15,737.50.
8. D
We are given that each of the n employees earns s amount of salary weekly. This means that one employee earns s salary weekly. So; Richard has ‘ns’ amount of money to employ n employees for a week.
We are asked to find the number of days n employees can be employed with x amount of money. We can do simple direct proportion:
If Richard can employ n employees for 7 days with ‘ns’ amount of money,
Richard can employ n employees for y days with x amount of money … y is the number of days we need to find.
We can do cross multiplication:
y = (x * 7)/(ns)
y = 7x/ns
9. B
The distribution is done at three different rates and in three different amounts:
$6.4 per 20 kilograms to 15 shops … 20•15 = 300 kilograms distributed
$3.4 per 10 kilograms to 12 shops … 10•12 = 120 kilograms distributed
550 – (300 + 120) = 550 – 420 = 130 kilograms left. This 50
amount is distributed in 5 kilogram portions. So, this means that there are 130/5 = 26 shops.
$1.8 per 130 kilograms.
We need to find the amount he earned overall these distributions.
$6.4 per 20 kilograms : 6.4•15 = $96 for 300 kilograms
$3.4 per 10 kilograms : 3.4 *12 = $40.8 for 120 kilograms
$1.8 per 5 kilograms : 1.8 * 26 = $46.8 for 130 kilograms
So, he earned 96 + 40.8 + 46.8 = $ 183.6
The total distribution cost is given as $10
The profit is found by: Money earned – money spent … It is important to remember that he bought 550 kilograms of potatoes for $165 at the beginning:
Profit = 183.6 – 10 – 165 = $8.6
10. B
We check the fractions taking place in the question. We see that there is a “half” (that is 1/2) and 3/7. So, we multiply the denominators of these fractions to decide how to name the total money. We say that Mr. Johnson has 14x at the beginning; he gives half of this, meaning 7x, to his family. $250 to his landlord. He has 3/7 of his money left. 3/7 of 14x is equal to:
14x * (3/7) = 6x
So,
Spent money is: 7x + 250
Unspent money is: 6×51
Total money is: 14x
Write an equation: total money = spent money + unspent money
14x = 7x + 250 + 6x
14x – 7x – 6x = 250
x = 250
We are asked to find the total money that is 14x:
14x = 14 * 250 = $3500
11. D
First calculate total square feet, which is 15 * 24 = 360 ft2. Next, convert this value to square yards, (1 yards2 = 9 ft2) which is 360/9 = 40 yards2. At $0.50 per square yard, the total cost is 40 * 0.50 = $20.
1). A number consists of two digits. The sum of the digits is 10. On reversing the digits of the number, the number decreases by 36. What is the product of the two digits?
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2). If a number is multiplied by three-fourth of itself, the value thus obtained is 10800. What is that number?
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3). The sum of five consecutive numbers is 190. What is the sum of the largest and the smallest numbers?
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4). The product of two consecutive odd numbers is 19043. Which is the smaller one?
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5). If the three-fourth of a number is subtracted from the number, the value so obtained is 163. What is that number?
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6). The sum of two numbers is 10. Their product is 20. Find the sum of the reciprocals of the two numbers.
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7). Five times of a positive integer is equal to 3 less than twice the square of that number. Find the number.
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8). A man has given one-fourth part of his property to his daughter, half part to his sons and one-fifth part given as charity. How much part of his property he has given?
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| 9). A chocolate has 12 equal pieces. Manju gave ( Large frac{1}{4} )th of it to Anju, ( Large frac{1}{3} )rd of it to Sujata and ( Large frac{1}{6} )th of it to Fiza. The number of pieces of chocolate left with Manju is | ||||
| 10). A number consists of two digits whose sum is 10. If the digits of the number are reversed, then the number decreased by 36. Which of the following is/are correct? I. The number is divisible by a composite number. II. The number is a multiple of a prime number.
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Latest Word Problems MCQ Objective Questions
Word Problems Question 1:
‘n’ is a two-digit number such that the product of its digits when added to their sum is ‘n’. The unit place digit of ‘n’ would be —
- 8
- 9
- 1
- 7
Answer (Detailed Solution Below)
Option 2 : 9
Given:
‘n’ is a two-digit number such that the product of its digits when added to their sum is ‘n’.
Calculations:
Let a, b be tens and unit digits of n
Then, n = 10a + b
product(n) + sum(n) = 10a + b
⇒ ab + a + b = 10a + b
⇒ ab = 10a — a
⇒ ab = 9a
∴ b = 9 as, a = 0
∴ units digit is 9
∴ The answer is 9
Word Problems Question 2:
Comprehension:
Consider the word ‘QUESTION’ :
How many 8-letter words with or without meaning, can be formed so that all consonants are together?
- 5760
- 2880
- 1440
- 720
Answer (Detailed Solution Below)
Option 2 : 2880
Concept:
The number of ways of choosing m objects out of n different objects = nCm
The number of ways of arranging n objects = n!
Explanation:
Given word ‘QUESTION’ which has 8 different letters
{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}
If all consonants are together consider them as one string, which can be rearranged internally in 4! ways
⇒ Number of ways of rearranging 4 consonants = 24
Now for forming a word, we have to arrange 4 vowels and one string of consonants
The number of ways of arranging 5 objects = 5! = 120
So, the total number of words formed = 120 × 24 = 2880
∴ The correct answer is option (2).
Word Problems Question 3:
Comprehension:
Consider the word ‘QUESTION’ :
How many 8-letter words with or without meaning, can be formed such that consonants and vowels occupy alternate positions?
- 288
- 576
- 1152
- 2304
Answer (Detailed Solution Below)
Option 3 : 1152
Concept:
The number of ways of choosing m objects out of n different objects = nCm
The number of ways of arranging n objects = n!
Explanation:
Given word ‘QUESTION’ which has 8 different letters
{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}
If vowels occupy odd positions and consonants occupy even positions,
then 4 consonants are to be arranged in 4 positions and 4 vowels are to be arranged in 4 positions
As the number of ways of arranging 4 objects in 4 positions = 4! = 24
⇒ The number of ways arranging 4 vowels in 4 positions and 4 consonants in 4 positions = 24 × 24 = 576
Similarly, for the case when vowels occupy even positions and consonants occupy odd positions
So, the number of 8-letter words that can be formed such that consonants and vowels occupy alternate positions = 576 × 2 = 1152
∴ The correct answer is option (3).
Word Problems Question 4:
Comprehension:
Consider the word ‘QUESTION’ :
How many 4-letter words each of two vowels and two consonants with or without meaning, can be formed ?
- 36
- 144
- 576
- 864
Answer (Detailed Solution Below)
Option 4 : 864
Concept:
The number of ways of choosing m objects out of n different objects = nCm
The number of ways of arranging n objects = n!
Explanation:
Given word ‘QUESTION’ which has 8 different letters
{4 vowels (U,E,I,O) and 4 consonants(Q,S,T,N)}
The number of ways of choosing two vowels from 4 vowels = 4C2 = 6
And the number of ways of choosing two vowels from 4 vowels = 4C2 = 6
⇒ The number of ways of choosing two vowels and two consonants from 4 vowels and 4 consonants = 4C2 × 4C2 = 36
And the number of ways of arranging these four different letters = 4! = 24
⇒ Number of 4-letter words each of two vowels and two consonants with or without meaning = 36 × 24 = 864
∴ The correct option is (4).
Word Problems Question 5:
Two numbers are chosen from 1, 3, 5, 7,… 147, 149, and 151 and multiplied together in all possible ways. The number of ways that will give the product a multiple of 5 is
- 1020
- 435
- 1380
- 945
Answer (Detailed Solution Below)
Option 1 : 1020
Concept:
Combination:
A combination is a way of selecting items from a collection where the order of selection does not matter.
nCk = (frac{n!}{k!(n-k)!}), when n > k
nCk = 0 , when n < k
Where n = distinct object to choose from, C = Combination, k = spaces to fill
Calculation:
In the given numbers 1, 3, 5, 7,…, 147, 149, 151, the numbers which are multiples of 5 are 5, 15, 25, 35, …, 145, which are arithmetic sequences.
Tn = a + (n − 1) × d
⇒ 145 = 5 + (n – 1) 10
⇒ n = 15
and if total number of terms in the given sequence is m, then
⇒ 151 = 1 + (m – 1) 2
⇒ m = 76
So, the number of ways in which product is a multiple of 5
= (both two numbers from 5, 15, 25, 35, …, 145) or (one number from 5, 15, 25, 35, …, 145 and one from remaining numbers ie ’76 — 15 = 61′)
= 15C2 + 15C1 × 76-15C1
= 105 + 15 × 61
= 105 + 915
∴ The required number of ways is 1020.
Top Word Problems MCQ Objective Questions
How many 4-letter words can be formed out of the letters of ‘EGOIST’ having at least one vowel? (repetition is not allowed)
- 12 × 4!
- 15 × 4!
- 21 × 4!
- 24 × 4!
Answer (Detailed Solution Below)
Option 2 : 15 × 4!
Concept:
- The ways of arranging n different things = n!
- The ways of arranging n things, having r same things and rest all are different = (rm n!over r!)
- The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
- The number of ways for selecting r from a group of n (n > r) = nCr
Calculation:
There are 3 vowels O, I, E
Required number of ways N = 1 vowel and 3 consonants + 2 vowel and 2 consonants + 3 vowel and 1 consonants
⇒ N = [(rmleft(^3C_1×{^3C_3}right) + left({^3C_2}×{^3C_2}right) + left({^3C_3}×{^3C_1}right))] × 4!
⇒ N = (3 + 9 + 3) × 4!
⇒ N = 15 × 4! words
In what number of ways can the letters of the word ‘ABLE’ be arranged so that the vowels occupy even places?
- 2
- 4
- 6
- 8
Answer (Detailed Solution Below)
Option 2 : 4
Concept:
Number of ways to arrange n things in r places is given by, nCr
Calculation:
In word ABLE, there are 2 vowels and 2 consonants.
Total number of letters = 4
Total number of even place = 2
There are 2 vowels to be filled in 2 places.
Hence, the number of ways = 2C2 = 1
The vowels can arrange among themselves in 2! = 2 ways.
Now, the 2 consonants can fill the remaining 2 places in 2! = 2 ways.
Therefore, total number of ways = 1 × 2 × 2 = 4 ways.
Hence, option (2) is correct.
How many ways the word ‘EQUATION’ can be arranged so that all the vowels will always come together?
- 5!
- 4!
- 4 × 5!
- 4! × 5!
Answer (Detailed Solution Below)
Option 4 : 4! × 5!
Concept:
n objects can be arranged in n! ways
When two or more objects are taken together they are considered as one object.
Calculation:
The total number of vowels in EQUATION is 5 (E, U, A, I, O)
So, we can consider them as single unity and they can arrange themselves in 5! ways.
Rest 3 consonants along with the vowels together can arrange in 4! ways.
So, the total number of ways the word can be arranged = 4! × 5!
In how many ways can the letter of the word ‘SCALE’ be arranged, so that the vowels do not come together?
- 72
- 36
- 48
- 60
Answer (Detailed Solution Below)
Option 1 : 72
The formula to find a number of ways word can be arranged, so that the vowels do not come together is:
Total word(factorial) — (Total word — 1)factorial × (number of vowels)factorial
Given:
Total words = 5
Total word — 1 = 4
Number of vowels = 2
Substituting in the formula:
5! — (4! × 2!) = 120 — 48
= 72
Hence, 72 ways are there to arrange SCALE so that vowels do not come together.
Find the number of different permutations of the letters of the word INDIA
- 20
- 40
- 50
- 60
Answer (Detailed Solution Below)
Option 4 : 60
Concept:
Let there be n objects, of which m objects are alike of one kind, and the remaining (n – m) objects are distinct. Then, the total number of permutations that can be formed from these objects is (rm frac{n!}{m!})
Calculation:
The given word ‘INDIA‘ contains 5 letters, out of which two are alike of one kind (2 I‘s), and the other three letters are distinct.
∴Total number of their permutations = (rm frac{5!}{2!}=frac{5times4times3times2times1}{2times1}=60)
Hence, option (4) is correct.
How many words can be made from the letters of the word BHARAT in which B and H never come together
- 360
- 300
- 240
- 120
Answer (Detailed Solution Below)
Option 3 : 240
Concept:
The number of words from n letters where a letter is repeated m times
= n!/m!
where n! = n(n — 1)(n — 2)….3.2.1
Calculation:
Number of words when B and H did not come together
⇒ Total number of words — Number of words when B and H come together
Now, Total number of words from the letters of the word BHARAT = 6!/2! = 360
{letter A has repeated two times and the total letters are 6}
And, when B and H come together,
let us consider B and H as a single letter (BH),
So, the number of words when B and H come together
= the number of words from the letters of the word (BH)ARAT = (frac{5 !}{2!}times 2=120)
⇒ Number of words when B and H did not come together = 360 — 120 = 240
∴ The correct answer is option (3).
How many words can be formed using all the letters of the word ‘NATION’ so that all the three vowels should never come together?
- 354
- 348
- 288
- None of the above
Answer (Detailed Solution Below)
Option 3 : 288
Concept:
- Suppose a set of n objects has n1 of one kind of object, n2 of a second kind, n3 of a third kind, and so on with n = n1 + n2 + n3 +…+ nk then the number of distinguishable permutations of the n objects is = (frac{{{rm{n}}!}}{{{{rm{n}}_1}! times {{rm{n}}_2}! times {{rm{n}}_3}! ldots ldots ldots {{rm{n}}_{rm{k}}}!}})
Calculation:
The given word is ‘NATION’
Number of letters repeating in the word =2N
Vowels in the word are {A, I, O}
Total number of words that can be formed from given word ‘NATION’ = (frac{{6!}}{{2!}} = 360)
Consider A, I, O as one group
Then the no. of words formed by this group and remaining letters is 4!
The three vowels can be arranged among themselves in 3! Ways
∴ Number of words where all the vowels come together = (rm frac {(4!) (3!)}{2}) =72
Number of words formed so that all the three vowels are never together = Total number of words formed using all the letters in the word NATION − Number of words where all the vowels come together
Thus, the number of words formed so that all the three vowels are never together = 360 – 72 = 288
In how many ways can the word ‘CORONAVIRUS’ be arranged such that all the vowels come together?
- 30 × 5!
- 30 × 7!
- 20 × 5!
- 20 × 7!
Answer (Detailed Solution Below)
Option 2 : 30 × 7!
Concept:
- The ways of arranging n different things = n!
- The ways of arranging n things, having r same things and rest all are different = (rm n!over r!)
- The no. of ways of arranging the n arranged thing and m arranged things together = n! × m!
- The number of ways for selecting r from a group of n (n > r) = nCr
Calculation:
Given: The word CORONAVIRUS
There are 5 vowels in the word (A, I, O, O, U)
Taking vowels a single group (A, I, O, U) and O repeated two times
They can be arranged within themselves in (rm frac {5!}{2!}) ways
Now to arrange letters we have (AIOOU), C, R, N, V, R, S
They can be arranged in = (rm 7!over2!) ways (‘R’ appear 2 times)
Hence the required arrangement = (rm {7!over2!} times {5!over2!}) = 30 × 7! ways
The number of permutations that can be made out of the letters of the word “DEFENCE» which start with a consonant and end with a consonant is:
- 140 ways
- 220 ways
- 120 ways
- 240 ways
Answer (Detailed Solution Below)
Option 4 : 240 ways
Concept:
Suppose a set of n objects has n1 of one kind of object, n2 of a second kind, n3 of a third kind, and so on with n = n1 + n2 + n3 +…+ nk then the number of distinguishable permutations of the n objects is = (frac{{{rm{n}}!}}{{{{rm{n}}_1}! times {{rm{n}}_2}! times {{rm{n}}_3}! ldots ldots ldots {{rm{n}}_{rm{k}}}!}})
Calculation:
The word ‘DEFENCE’ has 3 vowels “E, E, E» and 4 consonants” D, F, N, C».
The consonant for the first place can be selected in 4 ways.
First placed occupied by consonant now remaining consonant are three for the last place.
Then, the consonant for the last place can be selected in 3 ways.
The other 5 places in between can be occupied by the remaining 5 letters in (5!/3!) ways = 20 ways. (Out of 5 letters, 3 letters are repeated)
Hence, the total number of ways = 4 × 3 × 20 = 240 ways
Find the number of ways in which word ERASER can be arranged if vowels occupy only even places.
- 144
- 72
- 9
- 18
Answer (Detailed Solution Below)
Option 3 : 9
Concept:
- The ways of arranging n different things = n!
- The ways of arranging n things, having r same things and rest all are different = (rm n!over r!)
Calculation:
ERASER has 3 vowels and 3 consonants
1_2_3_4_5_6
Vowels occupy only even places: 3 vowels can be arranged in 3! Ways
Similarly, consonants can be arranged in 3! ways
Total possible arrangement = (3! × 3!)/(2! × 2!)
= (3 × 2 × 1 × 3 × 2 × 1)/(2 × 1 × 2 × 1)
= 9
Question: 1
In each of the following sentences there is a blank space, followed by four choices of words marked (a), (b), (c) and (d).
A. Himalaya is the ___ of the Ganges.
B. The reporter had a valid ___ for the story.
C. He spent hours looking for the ___ of that question.
D. The strategy is to ___ supplies from smaller companies.
(A) source
(B) start
(C) look
(D) mother
Question: 2
In each of the following sentences there is a blank space, followed by four choices of words marked (a), (b), (c) and (d).
A. He has an ___ in ethnic music.
B. They said nothing of great ___.
C. Primary colours can add ___ to a room.
D. How much ____ did you pay for the new loan?
(A) influence
(B) interest
(C) income
(D) interface
Question: 3
In each of the following sentences there is a blank space, followed by four choices of words marked (a), (b), (c) and (d).
A. I cannot ___ the dogma of this church.
B. Please ___ my present.
C. People did not ___ atonal music at that time.
D. I shall have to ___ these unpleasant working conditions.
(A) accept
(B) but
(C) repeat
(D) forget
Question: 4
In each of the following sentences there is a blank space, followed by four choices of words marked (a), (b), (c) and (d).
A. The theatre was her first ___
B. Their __ left them indifferent to their surroundings.
C. It was 40 ___
D. He has a very complicated ___ life.
(A) done
(B) personal
(C) experience
(D) love
Question: 5
In each of the following sentences there is a blank space, followed by four choices of words marked (a), (b), (c) and (d).
A. We ___ the room with an electric heater.
B. The hostess ____ lunch for all the guests.
C. The will ___ that each child should receive half of the money.
D. He ___ for his large family by working their jobs.
(A) added
(B) provided
(C) accepted
(D) notified
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If you looking for different questions on Sets you have come the right way. We have curated Word Problems on Sets with Step by Step Solutions for your understanding. Make use of them while practicing Sets Problems and increase your conceptual knowledge. In this, you will understand how to Solve Sets Word Problems using Venn Diagrams easily. If you need help on different concepts of Sets refer to Set Theory and learn the representation of a set, types of sets, etc. Check out the Solved Examples provided and learn how to solve related problems during your work.
1. Let A and B be two finite sets such that n(A) = 30, n(B) = 18 and n(A ∪ B) = 26, find n(A ∩ B)?
Solution:
Formula for n(A ∪ B) = n(A) + n(B) – n(A ∩ B).
Rearranging it we get the n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
=30+18 – 26
= 22
Therefore, n(A ∩ B) = 22.
2. If If n(A – B) = 12, n(A ∪ B) = 45 and n(A ∩ B) = 15, then find n(B)?
Solution:
n(A∪B) = n(A – B) + n(A ∩ B) + n(B – A)
45 = 12+15 +n(B-A)
n(B-A) = 45-12-15
= 45-27
= 18
n(B) = n(A ∩ B) + n(B – A)
= 15+18
= 33
3. In a group of 80 people, 37 like cold drinks and 52 like hot drinks and each person likes at least one of the two drinks. Find How many people like both coffee and tea?
Solution:
Let A = Set of people who like cold drinks.
B = Set of people who like hot drinks.
Given
(A ∪ B) = 80 n(A) = 37 n(B) = 22 then;
n(A ∩ B) = n(A) + n(B) – n(A ∪ B)
= 37+52-80
= 89 – 80
= 9
Therefore, 9 people like both tea and coffee.
4. Let S={4, 5, 6}. Write all the possible partitions of S?
Solution:
Remember that partition of S is a collection of nonempty sets that are disjoint and their union is S.
Possible Partitions of S are
{4},{5},{6}
{4,5},{6}
{4,6},{5}
{5,6},{4}
{4,5,6}.
5. In a school, there are 30 teachers who teach Mathematics or Physics. Of these, 18 teach Mathematics and 6 teach both Physics and Mathematics. How many teach Physics only?
Solution:
Total Number of Teachers who teach Mathematics or Physics = 30
Number of Teachers who teach Mathematics n(M) = 18
Number of Teachers who teach both Mathematics and Physics n( M ∩ P) = 6
n(M) = 18 n( M ∩ P) = 6 and n (M ∪ P )= 30
n (M ∪ P ) = n(M) + n(P) – n(M ∩ P)
30 = 18 + n(P) – 6
30 = 12 + n(P)
⇒ n(P) = 30 – 12
⇒ n(P) = 18
Number of teachers teach Physics only = n( P – M)
n( P – M) = n(P) – n( M ∩ P )
= 18 – 6
n( P – M) = 12
Number of teachers who teach physics only is 12.
6. In a survey of 80 people, it was found that 35 people read newspaper H, 20 read newspaper T, 15 read the newspaper I, 5 read both H and I, 10 read both H and T, 7 read both T and I, 4 read all three newspapers. Find the number of people who read at least one of the newspapers?
Solution:
n(H) is the Number of People who read the newspaper H i.e. 35
n(I) is the Number of People who read the newspaper I i.e. 15
n(T) is the Number of People who read the newspaper T i.e. 20
n(H ∩ I) is the Number of People who read H and I i.e. 5
n(H ∩ T) is the Number of People who read H and T i.e 10
n(T ∩ I) is the Number of People who read T and I i.e. 7
n(H ∩ T ∩ I) is the Number of People who read all three Newspapers H, T, I i.e. 4
n(H ∪T ∪ I ) is the Number of people who read at least one of the newspapers
n(H ∪T ∪ I ) = n(H) + n(T) + n(I) – n(H ∩ T) – n(T ∩ I) – n(H ∩ I) + n(H ∩ T ∩ I)
n(H ∪T ∪ I )= 35+20+15-10-7-5+4
= 74-22
= 52
Therefore, the number of people who read at least one of the newspapers is 52.
7. In a school, all pupils play either Hockey or Football or both. 400 play Football, 150 play Hockey, and 130 play both the games. Find
(i) The number of pupils who play Football only,
(ii) The number of pupils who play Hockey only,
(iii) The total number of pupils in the school.
H = Hockey and F = Football
n (H ) = 150 n (F)= 400
n ( H ∩ F) = 130
(i) The number of pupils who only play Football = n (F – H )
n (F – H ) = n(F) – n( F ∩ H )
= 400 – 130
= 270
(ii) The number of pupils who only play Hockey = n (H – F )
n (H– F ) = n(H) – n( F ∩ H )
= 150 – 130
= 20
(iii) The total number of pupils in school
= n(H) + n(F) – n (F ∩ H)
= 150 + 400 – 130
= 420
Welcome to AMBiPi (Amans Maths Blogs). In this article, you will get SAT 2022 Math Practice Solving Linear Equations Word Problems Questions with Answer Keys.
SAT 2022 Math Multiple Choice Practice Questions
Solving Linear Equations Word Problems Multiple Choice Questions with Answer Keys
SAT Math Practice Test Question No 1:
If 1/2 + 4x = x +2, what is the value of x?
Option A : x = 3/10
Option B : x = 1/2
Option C : x = 5/6
Option D : x = 15/2
Show/Hide Answer Key
Option A : x = 3/10
SAT Math Practice Test Question No 2:
8x + 5 = bx−7
In the equation shown above, b is a constant. For what value of b does the equation have no solutions?
Option A : -8
Option B : -7
Option C : 5
Option D : 8
Show/Hide Answer Key
Option D : 8
SAT Math Practice Test Question No 3:
If -8 – 8y = 6 – 2y, what is the value of y?
Option A : y = -3/7
Option B : y = 3/7
Option C : y = 7/3
Option D : y = -7/3
Show/Hide Answer Key
Option D : y = -7/3
SAT Math Practice Test Question No 4:
2| – v/4 | + 26 < 12
Which of the following best describes the solutions to the inequality shown above
Option A : -7 < 7
Option B : -28 < 28
Option C : No solution
Option D : All real numbers
Show/Hide Answer Key
Option C : No solution
SAT Math Practice Test Question No 5:
2/3 – 5x = bx+ 1/3
In the equation shown above, b is a constant. For what value of b does the equation no solutions?
Option A : 5
Option B : 0
Option C : -5
Option D : 2/3
Show/Hide Answer Key
Option C : -5
SAT Math Practice Test Question No 6:
If Dimitri is helping to plan the school talent show. Each performer for the talent show has 6 minutes for his or her performance, which includes transition time between performances. If the introduction for the talent show is 24 minutes long and the show will last 150 minutes, how many different performances can the talent show accommodate?
Option A : 21
Option B : 24
Option C : 25
Option D : 29
Show/Hide Answer Key
Option A : 21
SAT Math Practice Test Question No 7:
The property taxes in a town decrease as the distance from the local elementary school increases. The greatest property taxes are 4.5,and for every 10 miles from the school, property taxes decrease by 0.5 percentage points. If a house is directly east or west of the school and its property taxes are 3 what is the distance of that house from the school?
Option A : 10 miles
Option B : 20 miles
Option C : 30 miles
Option D : 40 miles
Show/Hide Answer Key
Option C : 30 miles
SAT Math Practice Test Question No 8:
If Dalia is installing a tile floor in a rectangular room. Dalia has 152 tiles available to tile the room. If each row requires 9 1/2 tiles, and 19 tiles break while Dalia is laying the floor, how many full rows of tile can she install before running out of tiles?
Option A : 12
Option B : 14
Option C : 16
Option D : 18
Show/Hide Answer Key
Option B : 14
SAT Math Practice Test Question No 9:
Phytoremediation is the use of plant growth to purify pollutants from soil, water, or air. Suppose that a crop of brake ferns can remove 15 milligrams (mg) per square meter of a particular pollutant from the soil in 20 weeks. After 20 weeks, the ferns are harvested and a new crop is planted. If cc represents the number of crops of brake ferns needed to Phytoremediation soil contaminated with 170mg per square meter of the pollutant down to healthy levels of 5mg per square meter, which equation best models the situation?
Option A : 170 − 20c = 5
Option B : 170 + 20c = 5
Option C : 170 − 15c = 5
Option D : 170 + 20c = 5
Show/Hide Answer Key
Option C : 170 − 15c = 5
SAT Math Practice Test Question No 10:
One of the rules in a public speaking contest requires contestants to speak for as close to 5 minutes (300 seconds) as possible. Contestants lose 3 points for each second they speak either over or under 5 minutes. Which expression below can be used to determine the number of points a contestant loses if she speaks for x seconds?
Option A : 3|x−300|
Option B : 5|x−300|
Option C : 3|x+300|
Option D : 3/5|x+300|
Show/Hide Answer Key
Option A: 3|x−300|
SAT Math Practice Test Question No 11:
Cara is hanging a poster that is 91 centimeters (cm) wide in her room. The center of the wall is 180cm from the right end of the wall. If Cara hangs the poster so that the center of the poster is located at the center of the wall, how far will the left and right edges of the poster be from the right end of the wall?
Option A : 225.5cm and 89cm, respectively
Option B : 271.5cm and 134.5cm, respectively
Option C : 225.5cm and 134.5cm, respectively
Option D : 271cm and 89cm, respectively
Show/Hide Answer Key
Option C : 225.5cm and 134.5cm, respectively
SAT Math Practice Test Question No 12:
Felipe is saving money for a class trip. He already has saved $250 that he will put toward the trip. To save more money for the trip, Felipe gets a job where each month he can add $350 to his savings for the trip. Let m be the number of months that Felipe has worked at his new job. If Felipe needs to save $2700 to go on the trip, which equation best models the situation?
Option A : 250m − 350 = 2700
Option B : 250m + 350 = 2700
Option C : 350m − 250 = 2700
Option D : 350m + 250 = 2700
Show/Hide Answer Key
Option D : 350m + 250 = 2700
SAT Math Practice Test Question No 13:
Camille and Hiroki have decided to start walking for exercise. Camille is going to walk 7 miles the first day and 3 miles each day after that Hiroki is too busy to walk on the first 2 days, so he decides to walk 5 miles each day until he has walked the same number of miles as Camille. If Camille and Hiroki will have walked the same number of miles, how many days will Camille have walked?
Option A : 2
Option B : 3
Option C : 5
Option D : 7
Show/Hide Answer Key
Option D : 7
SAT Math Practice Test Question No 14:
An art gallery displays a large painting in the center of a wall that is 24 feet (ft) wide. The painting is 10ft wide. Which of the following equations can be used to find the distances, x in feet, from the left end of the wall to the edges of the painting?
Option A : |x − 10| = 12
Option B : 2|x − 12| = 10
Option C : 2|x − 10| = 12
Option D : |x − 12| = 10
Show/Hide Answer Key
Option B : 2|x−12|=10
SAT Math Practice Test Question No 15:
At the county fair, the operator of a game guesses a contestant’s weight. For each pound the operator’s guess differs from the contestant’s weight, the contestant will receive $3. If a contestant received $15 when the operator guessed 120 pounds, what are the possible values for the weight of the contestant?
Option A : 105 and 115
Option B : 105 and 125
Option C : 115 and 125
Option D : 115 and 135
Show/Hide Answer Key
Option C : 115 and 125
SAT 2022 Math Grid-in Practice Questions
Solving Linear Equations and Inequalities Grid-in Choice Questions with Answer Keys
SAT Math Practice Test Question No 16:
A park shaped like a pentagon has four equal length sides and one unequal side whose length is 35 feet (ft). The perimeter of the park is 195 ft. What is the length in feet of one of the equal sides?
Show/Hide Answer Key
Correct Answer : 40
SAT Math Practice Test Question No 17:
Sasha has $2.65 in change in her pocket. The $2.65 is made up of one quarter plus an equal number of nickels and dimes. How many nickels does Sasha have in her pocket?
Show/Hide Answer Key
Correct Answer : 16
SAT Math Practice Test Question No 18:
The number of subscribers to a certain newspaper decreases by about 2,000 each year. In 2010, there were 19,000 subscribers. In what year should the newspaper expect to have approximately 7,000 subscribers?
Show/Hide Answer Key
Correct Answer : 2016
SAT Math Practice Test Question No 19:
An airplane begins its descent to land from a height of 35,000 feet (ft) above sea level. The airplane’s height changes by about −4000 ft every 3 minutes. Rounded to the nearest minute, in approximately how many minutes will the plane land?
Assume that the airport runway is at sea level.
Show/Hide Answer Key
Correct Answer : 26
SAT Math Practice Test Question No 20:
A car driving on a straight path travels about 260 feet (ft) in 3 seconds. Rounded to the nearest second, approximately how many seconds will it take for the car to travel 1 mile (mi) at the same rate? 1 mi=5,280 ft.
Show/Hide Answer Key
Correct Answer : 61
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This article is for parents who think about how to help with math and support their children. The math word problems below provide a gentle introduction to common math operations for schoolers of different grades.
What are math word problems?
During long-time education, kids face various hurdles that turn into real challenges. Parents shouldn’t leave their youngsters with their problems. They need an adult’s possible help, but what if the parents themselves aren’t good at mathematics? All’s not lost. You can provide your kid with different types of support. Not let a kid burn the midnight oil! Help him/ her to get over the challenges thanks to these captivating math word examples.
Math word problems are short math questions formulated into one or several sentences. They help schoolers to apply their knowledge to real-life scenarios. Besides, this kind of task helps kids to understand this subject better.
Addition for the first and second grades
These math examples are perfect for kids that just stepped into primary school. Here you find six easy math problems with answers:
1. Peter has eight apples. Dennis gives Peter three more. How many apples does Peter have in all?
Show answer
Answer: 8 apples + 3 apples = 11 apples.
2. Ann has seven candies. Lack gives her seven candies more. How many candies does Ann have in all?
Show answer
Answer: 7 candies + 7 candies = 14 candies.
3. Walter has two books. Matt has nine books. If Matt gives all his books to Walter, how many books will Walter have?
Show answer
Answer: 2 books + 9 books = 11 books.
4. There are three crayons on the table. Albert puts five more crayons on the table. How many crayons are on the table?
Show answer
Answer: 3 crayons + 5 crayons = 8 crayons.
5. Bill has nine oranges. His friend has one orange. If his friend gives his orange to Bill, how many oranges will Bill have?
Show answer
Answer: 9 oranges + 1 orange = 10 oranges.
6. Jassie has four leaves. Ben has two leaves. Ben gives her all his leaves. How many leaves does Jessie have in all?
Show answer
Answer: 4 leaves + 2 leaves = 6 leaves.
Subtraction for the first and second grades
1. There were three books in total at the book shop. A customer bought one book. How many books are left?
Show answer
Answer: 3 books – 1 book = 2 books.
2. There are five pizzas in total at the pizza shop. Andy bought one pizza. How many pizzas are left?
Show answer
Answer: 5 pizzas – 1 pizza = 4 pizzas.
3. Liza had eleven stickers. She gave one of her stickers to Sarah. How many stickers does Liza have?
Show answer
Answer: 11 stickers – 1 sticker = 10 stickers.
4. Adrianna had ten stones. But then she left two stones. How many stones does Adrianna have?
Show answer
Answer: 10 stones – 2 stones = 8 stones.
5. Mary bought a big bag of candy to share with her friends. There were 20 candies in the bag. Mary gave three candies to Marissa. She also gave three candies to Kayla. How many candies were left?
Show answer
Answer: 20 candies – 3 candies – 3 candies = 14 candies.
6. Betty had a pack of 25 pencil crayons. She gave five to her friend Theresa. She gave three to her friend Mary. How many pencil crayons does Betty have left?
Show answer
Answer: 25 crayons – 5 crayons – 3 crayons = 17 crayons.
Multiplication for the 2nd grade and 3rd grade
See the simple multiplication word problems. Make sure that the kid has a concrete understanding of the meaning of multiplication before.
Bill is having his friends over for the game night. He decided to prepare snacks and games.
1. He makes mini sandwiches. If he has five friends coming over and he made three sandwiches for each of them, how many sandwiches did he make?
Show answer
Answer: 5 x 3 = 15 sandwiches.
2. He also decided to get some juice from fresh oranges. If he used two oranges per glass of juice and made six glasses of juice, how many oranges did he use?
Show answer
Answer: 2 x 6 = 12 oranges.
3. Then Bill prepared the games for his five friends. If each game takes 7 minutes to prepare and he prepared a total of four games, how many minutes did it take for Bill to prepare all the games?
Show answer
Answer: 7 x 4 = 28 minutes.
4. Bill decided to have takeout food as well. If each friend and Bill eat three slices of pizza, how many slices of pizza do they have in total?
Show answer
Answer: 6 (5 friends and Bill) x 3 slices of pizza = 18 slices of pizza.
Mike is having a party at his house to celebrate his birthday. He invited some friends and family.
1. He and his mother prepared cupcakes for dessert. Each box had 8 cupcakes, and they prepared four boxes. How many cupcakes have they prepared in the total?
Show answer
Answer: 8 x 4 = 32 cupcakes.
2. They also baked some cookies. If they baked 6 pans of cookies, and there were 7 cookies per pan, how many cookies did they bake?
Show answer
Answer: 6 x 7 = 42 cookies.
3. Mike planned to serve some cold drinks as well. If they make 7 pitchers of drinks and each pitcher can fill 5 glasses, how many glasses of drinks are they preparing?
Show answer
Answer: 7 x 5 = 35 glasses.
4. At the end of the party, Mike wants to give away some souvenirs to his 6 closest friends. If he gives 2 souvenir items for each friend, how many souvenirs does Mike prepare?
Show answer
Answer: 6 x 2 = 12 souvenirs.
Division: best for 3rd and 4th grades
1. If you have 10 books split evenly into 2 bags, how many books are in each bag?
Show answer
Answer: 10 : 2 = 5 books.
2. You have 40 tickets for the fair. Each ride costs 2 tickets. How many rides can you go on?
3. The school has $20,000 to buy new equipment. If each piece of equipment costs $100, how many pieces can the school buy in total?
Show answer
Answer: $20,000 : $100= 200.
4. Melissa has 2 packs of tennis balls for $10 in total. How much does 1 pack of tennis balls cost?
5. Jack has 25 books. He has a bookshelf with 5 shelves on it. If Jack puts the same number of books on each shelf, how many books will be on each shelf?
6. Matt is having a picnic for his family. He has 36 cookies. There are 6 people in his family. If each person gets the same number of cookies, how many cookies will each person get?
Division with remainders for fourth and fifth grades
1. Sarah sold 35 boxes of cookies. How many cases of ten boxes, plus extra boxes does Sarah need to deliver?
Show answer
Answer: 35 boxes divided by 10 boxes per case = 3 cases and 5 boxes.
2. Candies come in packages of 16. Mat ate 46 candies. How many whole packages of candies did he eat, and how many candies did he leave? 46 candies divided by 16 candies = 2 packages and 2 candies left over.
3. Mary sold 24 boxes of chocolate biscuits. How many cases of ten boxes, plus extra boxes does she need to deliver?
Show answer
Answer: 24 boxes divided by 10 boxes per case = 2 cases and four boxes.
4. Gummy bears come in packages of 25. Suzie and Tom ate 30 gummy bears. How many whole packages did they eat? How many gummy bears did they leave?
Show answer
Answer: 30 divided by 25 = 1 package they have eaten and 20 gummy bears left over.
5. Darel sold 55 ice-creams. How many cases of ten boxes, plus extra boxes does he need to deliver?
Show answer
Answer: 55 boxes divided by 10 boxes per case = 5 cases and 5 boxes.
6. Crackers come in packages of 8. Mat ate 20 crackers. How many whole packages of crackers did he eat, and how many crackers did he leave?
Show answer
Answer: 20 divided by 8 = 2 packages eaten and 4 crackers are left.
Mixed operations for the fifth grade
These math word problems involve four basic operations: addition, multiplication, subtraction, and division. They suit best for the fifth-grade schoolers.
200 planes are taking off from the airport daily. During the Christmas holidays, the airport is busier — 240 planes are taking off every day from the airport.
1. During the Christmas holidays, how many planes take off from the airport in each hour if the airport opens 12 hours daily?
Show answer
Answer: 240÷12=20 planes take off from this airport each hour during the Christmas holidays.
2. Each plane takes 220 passengers. How many passengers depart from the airport every hour during the Christmas holidays? 20 x 220 = 4400.
Show answer
Answer: 4400 passengers depart from the airport every hour.
3. Compared with a normal day, how many more passengers are departing from the airport in a day during the Christmas holidays?
Show answer
Answer: (240-200) x 220 = 8800 more passengers departing from the airport in a day during the Christmas holidays.
4. During normal days on average 650 passengers are late for their plane daily. During the Christmas holidays, 1300 passengers are late for their plane. That’s why 14 planes couldn’t take off and are delayed. How many more passengers are late for their planes during Christmas week?
Show answer
Answer: 1300 – 650 = 650 more passengers are late for their planes each day during the Christmas holidays.
5. According to the administration’s study, an additional 5 minutes of delay in the overall operation of the airport is caused for every 27 passengers that are late for their flights. What is the delay in the overall operation if there are 732 passengers late for their flights?
Show answer
Answer: 732 ÷ 27 × 5 = 136. There will be a delay of 136 minutes in the overall operation of the airport.
Extra info math problems for the fifth grade
1. Ann has 7 pairs of red socks and 8 pairs of pink socks. Her sister has 12 pairs of white socks. How many pairs of socks does Ann have?
2. Kurt spent 17 minutes doing home tasks. He took a 3-minute snack break. Then he studied for 10 more minutes. How long did Kurt study altogether?
Show answer
Answer: 17 + 10 = 27 minutes.
3. There were 15 spelling words on the test. The first schooler spelled 9 words correctly. Miguel spelled 8 words correctly. How many words did Miguel spell incorrectly?
4. In the morning, Jack gave his friend 2 gummies. His friend ate 1 of them. Later Jack gave his friend 7 more gummies. How many gummies did Jack give his friend in all?
5. Peter wants to buy 2 candy bars. They cost 8 cents, and the gum costs 5 cents. How much will Peter pay?
Finding averages for 5th grade
We need to find averages in many situations in everyday life.
1. The dog slept 8 hours on Monday, 10 hours on Tuesday, and 900 minutes on Wednesday. What was the
average number of hours the dog slept per day?
Show answer
Answer: (8+10+(900:60)) : 3 = 11 hours.
2. Jakarta can get a lot of rain in the rainy season. The rainfall during 6 days was 90 mm, 74 mm, 112 mm, 30 mm, 100 mm, and 44 mm. What was the average daily rainfall during this period?
Show answer
Answer: (90+74+112+30+100+44) : 6 = 75 mm.
3. Mary bought 4 books. The prices of the first 3 books were $30, $15, and $18. The average price she paid for the 4 books was $25 per books. How much did she pay for the 4th books?
Math & logic courses for kids
Times more complex than school, extremely fun, interactive and rewarding to keep 7-13 years old kids engaged. We’re gonna make them love math!
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Ordering and number sense for the 5th grade
1. There are 135 pencils, 200 pens, 167 crayons, and 555 books in the bookshop. How would you write these numbers in ascending order?
Show answer
Answer: 135, 167, 200, 555
2. There are five carrots, one cabbage, eleven eggs, and 15 apples in the fridge. How would you write these numbers in descending order?
3. Peter has completed exercises on pages 279, 256, 264, 259, and 192. How would you write these numbers in ascending order?
Show answer
Answer: 192, 256, 259, 264, 279.
4. Mary picked 32 pants, 15 dresses, 26 pairs of socks, 10 purses. Put all these numbers in order.
5. The family bought 12 cans of tuna, 23 potatoes, 11 onions, and 33 pears. Put all these numbers in order.
Fractions for the 6th-8th grades
1. Jannet cooked 12 lemon biscuits for her daughter, Jill. She ate up 4 biscuits. What fraction of lemon biscuits did Jill eat?
Show answer
Answer: 1/3 of the lemon biscuits.
2. Guinet travels a distance of 7 miles to reach her school. The bus covers only 5 miles. Then she has to walk 2 miles to reach the school. What fraction of the distance does Guinet travel by bus?
Show answer
Answer: 5/7 of the distance
3. Bob has 24 pencils in a box. Eighteen pencils have #2 marked on them, and the 6 are marked #3. What fraction of pencils are marked #3?
Show answer
Answer: 1/4 of the pencils.
4. My mother places 15 tulips in a glass vase. It holds 6 yellow tulips and 9 red tulips. What fraction of tulips are red?
Show answer
Answer: 3/5 of the tulips.
5. Bill owns 14 pairs of socks, of which 7 pairs are white, and the rest are brown. What fraction of pairs of socks are brown?
Show answer
Answer: 1/2 of the pairs of socks.
6. Bred spotted a total of 39 birds in an aviary at the Zoo. He counted 18 macaws and 21 cockatoos. What fraction of macaws did Bred spot at the aviary?
Show answer
Answer: 6/13 of the birds.
Decimals for the 6th grade
Write in words the following decimals:
- 0,004
- 0,07
- 2,1
- 0,725
- 46,36
- 2000,19
Show answer
Answer:
- 0,004 = four thousandths.
- 0,07 = seven hundredths.
- 2,1 = two and one tenth.
- 0,725 = seven hundred twenty five thousandths.
- 46,36 = foury six and thirty six hundredths.
- 2000,19 = two thousand and nineteen hundredths.
Comparing and sequencing for the 6th grade
1. The older brother picked 42 apples at the orchard. The younger brother picked only 22 apples. How many more apples did the older brother pick?
Show answer
Answer: 42 – 22 = 20 apples more.
2. There were 16 oranges in a basket and 66 oranges in a barrel. How many fewer oranges were in the basket than were in the barrel?
Show answer
Answer: 66 – 16 = 50 fewer oranges.
3. There were 40 parrots in the flock. Some of them flew away. Then there were 25 parrots in the flock. How many parrots flew away?
Show answer
Answer: 40 – 25 = 15 parrots flew away.
4. One hundred fifty is how much greater than fifty-three?
5. On Monday, the temperature was 13°C. The next day, the temperature dropped by 8 degrees. What was the temperature on Tuesday?
6. Zoie picked 15 dandelions. Her sister picked 22 ones. How many more dandelions did her sister pick than Zoie?
Show answer
Answer: 22-15 = 7 dandelions more.
Time for the 4th grade
1. The bus was scheduled to arrive at 7:10 p.m. However, it was delayed for 45 minutes. What time was it when the bus arrived?
2. My mother starts her 7-hour work at 9:15 a.m. What time does she get off from work?
3. Jack’s walk started at 6:45 p.m. and ended at 7:25 p.m. How long did his walk last?
4. The school closes at 9:00 p.m. Today, the school’s principal left 15 minutes after the office closed, and his secretary left the office 25 minutes after he left. When did the secretary leave work?
5. Suzie arrives at school at 8:20 a.m. How much time does she need to wait before the school opens? The school opens at 8:35 a.m.
6. The class starts at 9:15 a.m.. The first bell will ring 20 minutes before the class starts. When will the first bell ring?
Money word problems for the fourth grade
1. James had $20. He bought a chocolate bar for $2.30 and a coffee cup for $5.50. How much money did he have left?
Show answer
Answer: $20.00 – $2.30 – $5.50 = $12.20. James had $12.20 left.
2. Coffee mugs cost $1.50 each. How much do 7 coffee mugs cost?
Show answer
Answer: $1.5 x 7 = $10.5.
3. The father gives $32 to his four children to share equally. How much will each of his children get?
4. Each donut costs $1.20. How much do 6 donuts cost?
Show answer
Answer: $1.20 * 6 = $7,2.
5. Bill and Bob went out for takeout food. They bought 4 hamburgers for $10. Fries cost $2 each. How much does one hamburger with fries cost?
Show answer
Answer: $10 ÷ 4 = $2.50. One hamburger costs $2.50. $2.50 + $2.00 = $4.50. One hamburger with fries costs $4.50.
6. A bottle of juice costs $2.80, and a can is $1.50. What would it cost to buy two cans of soft drinks and a bottle of juice?
Show answer
Answer: $1.50 x 2 + $2.80 = $5.80.
Measurement word problems for the 6th grade
The task is to convert the given measures to new units. It best suits the sixth-grade schoolers.
- 55 yd = ____ in.
- 43 ft = ____ yd.
- 31 in = ____ ft.
- 29 ft = ____ in.
- 72 in = ____ ft.
- 13 ft = ____ yd.
- 54 lb = ____ t.
- 26 t = ____ lb.
- 77 t = ____ lb.
- 98 lb = ____ t.
- 25 lb = ____ t.
- 30 t = ____ lb.
Show answer
Answer:
- 55 yd = 1.980 in
- 43 ft = 14 yd 1 ft
- 31 in = 2 ft 7 in
- 29 ft = 348 in
- 72 in = 6 ft
- 13 ft = 4 yd 1 ft.
- 54 lb = 0,027 t
- 26 t = 52.000 lb
- 77 t = 154.000 lb
- 98 lb = 0,049 t
- 25 lb = 0?0125 t
- 30 t = 60.000 lb.
Ratios and percentages for the 6th-8th grades
It is another area that children can find quite difficult. Let’s look at simple examples of how to find percentages and ratios.
1. A chess club has 25 members, of which 13 are males, and the rest are females. What is the ratio of males to all club members?
2. A group has 8 boys and 24 girls. What is the ratio of girls to all children?
3. A pattern has 4 red triangles for every 12 yellow triangles. What is the ratio of red triangles to all triangles?
4. An English club has 21 members, of which 13 are males, and the rest are females. What is the ratio of females to all club members?
5. Dan drew 1 heart, 1 star, and 26 circles. What is the ratio of circles to hearts?
6. Percentages of whole numbers:
- 50% of 60 = …
- 100% of 70 = …
- 90% of 70 = …
- 20% of 30 = …
- 40% of 10 = …
- 70% of 60 = …
- 100% of 20 = …
- 80% of 90 = …
Show answer
Answer:
- 50% of 60 = 30
- 100% of 70 = 70
- 90% of 70 = 63
- 20% of 30 = 6
- 40% of 10 = 4
- 70% of 60 = 42
- 100% of 20 = 20
- 80% of 90 = 72.
Probability and data relationships for the 8th grade
1. John ‘s probability of winning the game is 60%. What is the probability of John not winning the game?
2. The probability that it will rain is 70%. What is the probability that it won’t rain?
3. There is a pack of 13 cards with numbers from 1 to 13. What is the probability of picking a number 9 from the pack?
4. A bag had 4 red toy cars, 6 white cars, and 7 blue cars. When a car is picked from this bag, what is the probability of it being red or blue?
5. In a class, 22 students like orange juice, and 18 students like milk. What is the probability that a schooler likes juice?
Geometry for the 7th grade
The following task is to write out equations and find the angles. Complementary angles are two angles that sum up to 90 degrees, and supplementary angles are two angles that sum up to 180 degrees.
1. The complement of a 32° angle = …
2. The supplement of a 10° angle = …
3. The complement of a 12° angle = …
4. The supplement of a 104° angle = …
Variables/ equation word problems for the 5th grades
1. The park is 𝑥 miles away from Jack’s home. Jack had to drive to and from the beach with a total distance of 36 miles. How many miles is Jack’s home away from the park?
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Answer: 2𝑥 = 36 → 𝑥 = 18 miles.
2. Larry bought some biscuits which cost $24. He paid $x and got back $6 of change. Find x.
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Answer: x = 24 + 6 = $30.
3. Mike played with his children on the beach for 90 minutes. After they played for x minutes, he had to remind them that they would be leaving in 15 minutes. Find x.
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Answer: x = 90 – 15 = 75 minutes.
4. At 8 a.m., there were x people at the orchard. Later at noon, 27 of the people left the orchard, and there were 30 people left in the orchard. Find x.
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Answer: x = 30 + 27 = 57 people
Travel time word problems for the 5th-7th grades
1. Tony sprinted 22 miles at 4 miles per hour. How long did Tony sprint?
Show answer
Answer: 22 miles divided by 4 miles per hour = 5.5 hours.
2. Danny walked 15 miles at 3 miles per hour. How long did Danny walk?
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Answer: 15 miles divided by 3 miles per hour = 5 hours.
3. Roy sprinted 30 miles at 6 miles per hour. How long did Roy sprint?
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Answer: 30 miles divided by 6 miles per hour = 5 hours.
4. Harry wandered 5 hours to get Pam’s house. It is 20 miles from his house to hers. How fast did Harry go?
Show answer
Answer: 20 miles divided by 5 hours = 4 miles per hour.
STEM subjects for kids
STEM courses for kids ages 7-13 in physics, chemistry, math and logic in interactive game format
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This lesson is part of a series of practice test questions for the quantitative reasoning section of the GRE revised General Test.
Example 1:
This is a multiple-choice question, select only one answer from a list of five choices.
For the first 5 hours of a trip, a plane averaged 120 kilometers per hour. For the remainder of the trip, the plane travelled an average speed of 180 kilometers per hour. If the average speed for the entire trip was 170 kilometers per hour, how many hours long was the entire trip?
(A) 15
(B) 20
(C) 25
(D) 30
(E) 35
- Show Step-by-step Solutions
These are the directions for the Quantitative Comparison Questions.
Directions: Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices:
(A) Quantity A is greater.
(B) Quantity B is greater.
(C) The two quantities are equal.
(D) The relationship cannot be determined from the information given.
Example 2:
Cleve is 4 times as old as Al. Bob is 3 years younger than Al. The sum of their ages is 81.
Quantity A: Al’s age
Quantity B: 13
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Example 3:
The speed of light is approximately 3 × 105 kilometers per second.
Quantity A: Approximate number of kilometers that light can travel in 1 hour.
Quantity B: 1.08 × 108
- Show Step-by-step Solutions
Example 4:
In 12 years, Murray will be 4 times as old as he is now.
Quantity A: Number of years until Murray is 8 times as old as he is now.
Quantity B: 24
- Show Step-by-step Solutions
Example 5:
The population of bacteria doubles every 30 minutes. At 3:30 pm on Monday, the population was 240.
Quantity A: The bacteria population at 2:00 pm on Monday.
Quantity B: 40
- Show Step-by-step Solutions
Try the free Mathway calculator and
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