Simplication of $3\frac{1}{5}\times10\frac{1}{2}$ gives us
- A$\frac{166}{5}$
- B$\frac{167}{5}$
- ✓$\frac{168}{5}$
- D$\frac{161}{5}$
Answer: C.
View full solution →Question types
Playing With Numbers question types
540 questions across 8 question groups — pick any mix to generate a MATHS paper with step-by-step answer keys.
540
Questions
8
Question groups
5
Question types
01
M.C.Q. [1 Marks Each]
292 Q→02Assertion (A) & Reason (B) MCQ
30 Q→03TRUE-FLASE [1 Marks Each]
38 Q→04BLANKS [1 Marks Each]
18 Q→051 Marks Question
49 Q→062 Marks Questions
64 Q→073 Marks Question
27 Q→085 Marks Questions
22 Q→Sample Questions
Playing With Numbers questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Which of the following is $NOT$ a positive multiple of $12?$
- ✓$3$
- B$12$
- C$24$
- D$48$
Answer: A.
View full solution →Mark the correct alternative in the following:
The $LCM$ of $100$ and $101$ is:
The $LCM$ of $100$ and $101$ is:
- ✓$10100$
- B$1001$
- C$10101$
- DNone of these.
Answer: A.
View full solution →Mark the correct alternative in the following:
Every counting number has an infinite number of
Every counting number has an infinite number of
- AFactors
- ✓Multiples
- CPrime factors
- DNone of these
Answer: B.
View full solution →$\frac{11}{7}$ can be expressed in the form.
- A$7\frac{1}{4}$
- B$4\frac{1}{7}$
- ✓$1\frac{4}{7}$
- D$11\frac{1}{7}$
Answer: C.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A): The multiples of $7$ are $7, 14, 21, 28,…$
Reason (R): every multiple of a number is greater than or equal to that number
Assertion (A): The multiples of $7$ are $7, 14, 21, 28,…$
Reason (R): every multiple of a number is greater than or equal to that number
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A$
- C$A$ is true but $R$ is false
- D$A$ is false but $R$ is true
Answer: A.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A): $48$ is divisible by $8.$
Reason (R): a number with $4$ or more digits is divisible by $8,$ if the number formed by the last three digits is divisible by $8.$
Assertion (A): $48$ is divisible by $8.$
Reason (R): a number with $4$ or more digits is divisible by $8,$ if the number formed by the last three digits is divisible by $8.$
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A$
- C$A$ is true but $R$ is false
- D$A$ is false but $R$ is true
Answer: A.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A): The $LCM$ of $20, 25$ and $30$ is $20.$
Reason (R): The Lowest Common Multiple $(LCM)$ of two or more given numbers is the lowest (or smallest or least) of their common multiples.
Assertion (A): The $LCM$ of $20, 25$ and $30$ is $20.$
Reason (R): The Lowest Common Multiple $(LCM)$ of two or more given numbers is the lowest (or smallest or least) of their common multiples.
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A$
- C$A$ is true but $R$ is false
- ✓$A$ is false but $R$ is true
Answer: D.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A): $60$ is not divisible by $3$ and $5$ which are co-primes.
Reason (R): If a number is divisible by two co-prime numbers then it is divisible by their product also.
Assertion (A): $60$ is not divisible by $3$ and $5$ which are co-primes.
Reason (R): If a number is divisible by two co-prime numbers then it is divisible by their product also.
- ABoth $A$ and $R$ are true and $R$ is the correct explanation of $A$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A$
- C$A$ is true but $R$ is false
- ✓$A$ is false but $R$ is true
Answer: D.
View full solution →Directions: In the following questions, the Assertions $(A)$ and Reason(s) $(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion (A): $7$ is a multiple of itself
Reason (R): every number is a multiple of itself
Assertion (A): $7$ is a multiple of itself
Reason (R): every number is a multiple of itself
- ✓Both $A$ and $R$ are true and $R$ is the correct explanation of $A$
- BBoth $A$ and $R$ are true but $R$ is not the correct explanation of $A$
- C$A$ is true but $R$ is false
- D$A$ is false but $R$ is true
Answer: A.
View full solution →Sum of two prime numbers is always even.
Prime numbers do not have any factors.
All prime numbers are odd.
If an even number is divided by $2$, the quotient is always odd.
View full solution →The product of three odd numbers is odd.
The smallest even number is ________.
The smallest composite number is _______.
The smallest prime number is ________.
Here are two different factor trees for $60.$ Write the missing numbers.
View full solution →A number is divisible by $3$ and $5.$ So it is also divisible by$………. [(5+3), (5 x 3). (5-3)]$
View full solution →What is the HCF of two consecutive odd numbers$?$
View full solution →What is the $HCF$ of two consecutive even numbers$?$
View full solution →What is the $HCF$ of two consecutive numbers$?$
View full solution →In the given expression, prime factorisation has been done or not$?\ 54 = 2 \times 3 \times 9$
View full solution →In the given expression, prime factorisation has been done or not$?\ 70 = 2 \times 5 \times 7$
View full solution →$HCF$ of co-prime numbers $4$ and $15$ was found as follows:
$4 = 2$ $\times$ $2$ and $15 = 3$ $\times$ $5$
since there is no common factor, so $H.C.F.$ of $4$ and $15$ is $0$. Is the answer correct? If not, what is the correct $H.C.F$.
View full solution →$4 = 2$ $\times$ $2$ and $15 = 3$ $\times$ $5$
since there is no common factor, so $H.C.F.$ of $4$ and $15$ is $0$. Is the answer correct? If not, what is the correct $H.C.F$.
Find the $H.C.F$ of the numbers: $18, 54, 81.$
View full solution →Find the H.C.F of the numbers: $91, 112, 49.$
View full solution →Find the $H.C.F$ of the numbers: $70, 105, 175$
View full solution →Find the $H.C.F$ of the numbers: $34, 102.$
View full solution →Find the smallest four digit number which is divisible by $18, 24$ and $32.$
View full solution →Find the least number which when divided by $6, 15$ and $18$ leave remainder $5$ in each case.
View full solution →Three tankers contain $403$ litres, $434$ litres and $465$ litres of diesel, respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
View full solution →The traffic lights at three different road crossings change after every $48$ seconds, $72$ seconds and $108$ seconds respectively. If they change simultaneously at $7 a.m$., at what time will they change simultaneously again?
View full solution →Determine the largest $3-$digit number exactly divisible by $8, 10$ and $12.$
View full solution →The product of three consecutive numbers is always divisible by $6.$ Verify this statement with the help of some examples.
View full solution →Here are two different factor trees for $60$. Write the missing numbers.
View full solution →Using divisibility tests, determine which of the following numbers are divisible by $2;$ by $3;$ by $4;$ by $5;$ by $6;$ by $8;$ by $9;$ by $10;$ by $11 ($say, yes or no$):$
View full solution →| Number | Divisible by | ||||||||
| $2$ | $3$ | $4$ | $5$ | $6$ | $8$ | $9$ | $10$ | $11$ | |
| $128$ | |||||||||
| $990$ | |||||||||
| $1586$ | |||||||||
| $275$ | |||||||||
| $6686$ | |||||||||
| $639210$ | |||||||||
| $429714$ | |||||||||
| $2856$ | |||||||||
| $3060$ | |||||||||
| $406839$ | |||||||||
Find the $LCM$ of $12$ and $18.$
View full solution →Find the least number which when divided by $12, 16, 24$ and $36$ leaves a remainder $7$ in each case.
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