3 Mark Question - Maths STD 6 Questions - Vidyadip

Questions

3 Mark Question

Take a timed test

54 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks

For the following pairs of numbers, verify the property:
Product of the number = Product of their H.C.F and L.C.M
87,145

Answer

Given numbers are 87 and 145.
Prime factorization of 87 = 3 × 29
Prime factorization of 145 = 5 × 29
HCF of 87 and 145 = 29
LCM of 87 and 145 = 3 × 5 × 29 = 435
Product of the given number = 87 × 145 = 12615
Product of their HCF and LCM = 29 × 435 = 12615
Therefore, Product of the number = Product of their HCF and LCM (Verified)

View full question & answer→

Question 23 Marks

Show that the following pairs are co-prime:
875,1859

Answer

We know that two numbers are co-primes if their HCF is 1.
875 and 1859
Here, dividend = 1859 and divisor = 875

Clearly, the last divisor is 1.
Hence, the given numbers are co-prime.

View full question & answer→

Question 33 Marks

A merchant has 120 litres of oil of one kind, 180 liters of another kind and 240 litres of third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?

Answer

The maximum capacity of the required tin is the HCF of the three quantities of oil.
Prime factorization of 120 = 2 × 2 × 2 × 3 × 5
Prime factorization of 180 = 2 × 2 × 3 × 3 × 5
Prime factorization of 240 = 2 × 2 × 2 × 2 × 3 × 5
Therefore, HCF of 120, 180, and 240 = 2 × 2 × 3 × 5 = 60
Hence, the required greatest capacity of the tin must be 60 liters.

View full question & answer→

Question 43 Marks

In the following numbers. replace * by the smallest number to make it divisible by 11:
9 × 8071

Answer

Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.
9 × 8071
Sum of the digits at the odd places = 9 + 8 + 7 = 24
Sum of the digits at the even places = missing number + 0 + 1 = missing number + 1
Difference = 24 - [missing number + 1] = 23 - missing number
According to rule, 23 - missing number = 22 [Because 22 is a multiple of 11 and the missing number is a single digit]
Thus, missing number = 1
Hence, the smallest required number is 1.

View full question & answer→

Question 53 Marks

During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many each would you need to buy?

Answer

To find the required number of pencils and crayons, we need to find the LCM of 24 and 32.
Prime factorization of 24 = 2 × 2 × 2 × 3
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
Required LCM of 24 and 32 =2 × 2 × 2 × 2 × 3 = 96
Thus, number of pencils and crayons needed to be bought is 96 each, i.e. 96 ÷ 24 = 4 packs of color pencils and 96 ÷ 32 = 3 packs of crayons.

View full question & answer→

Question 63 Marks

Determine the longest tape which can be used to measure exactly the lengths 7m, 3m 85cm and 12m 95cm.

Answer

Given: Length of the first tape = 7m = 700cm
Length of the second tape = 3m 85cm = 385cm
Length of the third tape = 12m 95cm = 1,295cm
The length of the longest tape will be the HCF of 700, 385, and 1,295.
Prime factorization of 700 = 2 × 2 × 5 × 5 × 7
Prime factorization of 385 = 5 × 7 × 11
Prime factorization of 1,295 = 5 × 7 × 37 I-ICF of 700, 385, and 1,295 = 5 × 7 = 35
Required length of the longest tape = 35cm

View full question & answer→

Question 73 Marks

Define a perfect number. Write two perfect numbers.

Answer

A perfect number is a positive number that equals the sum of its divisors, excluding itself.
Divisors of 6 = 1, 2, 3
Divisors of 28 = 1, 2, 4, 7, 14, 28
In 6, the sum of divisors except itself, 1 + 2 + 3 is 6.
In 28, the sum of divisors except itself, 1 + 2 + 4 + 7 + 14 is 28.
Two perfect numbers are 6 and 28.

View full question & answer→

Question 83 Marks

Can two numbers have 12 as their HCF and 512 as their LCM? Justify your answer.

Answer

HCF of two numbers is a factor of the LCM of those numbers
Thu, we cannot have two numbers whose HCF is 12 and LCM is 512.
Because, when we divide 512 by 12, we get a remainder of 42.68
Thus, 12 is not a factor of 512.
Hence, we cannot have two numbers of whose HCF is 12 and LCM is 512.

View full question & answer→

Question 93 Marks

For the following pairs of numbers, verify the property:
Product of the number = Product of their H.C.F and L.C.M
117,221

Answer

Given numbers are 117 and 221.
Prime factorization of 117 = 3 × 3 × 13
Prime factorization of 221 = 13 × 17
HCF of 117 and 221 = 13
LCM of 117 and 221 = 3 × 3 × 13 × 17 = 1, 989
Product of the given number = 117 × 221 = 12, 857
Product of their HCF and LCM = 13 × 1,989 = 12,857
Therefore, Product of the number = Product of their HCF and LCM (verified)

View full question & answer→

Question 103 Marks

The L.C.M and H.C.F of two numbers are 180 and 6 respectively. If ones of the numbers is 30, find the other number.

Answer

Given: HCF of two numbers = 6
LCM of two numbers = 180
One of the given number = 30
Product of the two numbers = Product of their HCF and LCM
Therefore, 30 x other number = 6 × 180
Other number = 6 × 18030 = 36
Thus, the required number is 36.

View full question & answer→

Question 113 Marks

Find the greatest number that can divide 101 and 115 leaving remainders 5 and 7 respectively.

Answer

First we will subtract 5 and 7 from 101 and 115 respectively.
Now, we have $101-5=96$ and $115-7=108$
$96=2 \times 2 \times 2 \times 2 \times 2 \times 3=2^5 \times 3^1$
$108=2 \times 2 \times 3 \times 3 \times 3=2^2 \times 3^3$
HCF of 96 and $108=2^2 \times 3^1=12$
Hence, the greatest number that can divide 101 and 115 leaving remainders 5 and 7 respectively is 12.

View full question & answer→

Question 123 Marks

What is the H.C.F of two consecutive numbers?

Answer

The HCF of two consecutive numbers is 1.
Example:
D = 4 and d= 5 are two consecutive numbers.
Here, we have dividend = 5 and divisor = 4

Clearly, the last divisor is 1.
Hence, HCF of 4 and 5 is 1.

View full question & answer→

Question 133 Marks

Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.

Answer

We have to find the greatest number which divides (285 - 9) and (1,249 - 7) exactly.
The required number will be given by the HCF of 276 and 1242.
Resolving 276 and 1242 into prime factors, we have:
276 =2 × 2 × 3 × 23
1242 =2 × 3 × 3 × 3 × 23
HCF of 276 and 1242 is 2 × 3 × 23 = 138.

View full question & answer→

Question 143 Marks

Find the prime factors of 1729. Arrange the factors in ascending order, and find the relation between two consecutive prime factors.

Answer

The given number is 1729.
We have:

7	1729
13	247
19	19

Thus, the number 1729 can be expressed in the form of its prime factors ass 7 × 13 × 19.
Relation between its two consecutive prime factors:
The consecutive prime factors of the given number are 7, 13 and 19.
Clearly, 13 - 7 = 6 and 19 - 13 =6
Here, in two consecutive prime factors, the latter is 6 more than the previous one.

View full question & answer→

Question 153 Marks

A school bus picking up children in a colony of flats stops at every sixth block of flats. Another school bus starting from the same place stops at every eighth blocks of flats. Which is the first bus stop at which both of them will stop?

Answer

First bus stop at which both the buses will stop together = LCM of 6th block and 8th block
Prime factorization of 6 = 2 × 3
Prime factorization of 8 = 2 × 2 × 2
Therefore, Required LCM = 2 × 2 × 2 × 3 = 24
Hence, the first bus stop at which both the buses will stop together will be at the 24th block.

View full question & answer→

Question 163 Marks

What is the smallest number that both 33 and 39 divide leaving remainders of 5?

Answer

We have to find prime factorization of 33 and 39.
Prime factorization of 33 = 3 × 11
Prime factorization of 39 = 3 × 13
Therefore, Required LCM = 3 × 11 × 13 = 429
Thus, 429 is the smallest number exactly divisible by 33 and 39.
To get the remainder as 5: Smallest number = 429 + 5 = 434
Thus, the required number is 434.

View full question & answer→

Question 173 Marks

Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case.

Answer

First we subtract the required remainder from 615 and 963.
Thus, we will get 609 and 957.
$609=3 \times 7 \times 29=3^1 \times 7^1 \times 29^1$
$165=3 \times 11 \times 29=3^1 \times 11^1 \times 29^1$
HCF $=3^1 \times 29^1=87$
Hence, the greatest number which divides 615 and 963 , leaving the remainder 6 in each case is 87 .

View full question & answer→

Question 183 Marks

In a morning walk three persons step off together. Their steps measure 80cm, 85cm, and 90cm respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?

Answer

We have to find the LCM of 80cm, 85cm, and 90cm.
Prime factorization of 80 = 2 × 2 × 2 × 2 × 5
Prime factorization of 85 = 5 × 17
Prime factorization of 90 = 2 × 3 × 3 × 5
Therefore, Required LCM = 2 × 2 × 2 × 2 × 3 ×3 × 5 × 17 = 12,240
Therefore, Required minimum distance = LCM of 80cm, 85cm, and 90cm
= 12,240cm
= 122m 40cm (since 1m =100cm)

View full question & answer→

Question 193 Marks

The HCF of two numbers is 4 and their LCM is 400. How many pairs of values can the numbers assume?

Answer

HCF of two numbers is 4
So, the numbers can be 4a and 4b.
Now, 4a × 4b = 4 × 400
⇒ ab = 100
So, we can get the pairs
a = 25 and b = 4
a = 4 and b = 25
Thus, the numbers are 4 × 25 = 100 and 4 × 4 = 16.
Also, we can get the other pair 4 × 1 = 4 and 400.
Hence, there are two pairs.

View full question & answer→

Question 203 Marks

Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).

Answer

To find the required least number, we have to find the LCM of the numbers from 1 to 10.
We know that 2, 3, 5, and 7 are prime number.
Prime factorization of 4 = 2 × 2
Prime factorization of 6 = 2 × 3
Prime factorization of 8= 2 × 2 × 2
Prime factorization of 9 = 3 × 3
Prime factorization of 10 = 2 × 5
Therefore, Required least number = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 2,520

View full question & answer→

Question 213 Marks

For the following pairs of numbers, verify the property:
Product of the number = Product of their H.C.F and L.C.M
25,65

Answer

Given numbers are 25 and 65
Prime factorization of 25 = 5 × 5
Prime factorization of 65 = 5 × 13
HCF of 25 and 65 = 5
LCM of 25 and 65= 5 × 5 × 13 = 325
Product of the given numbers = 25 × 65 = 1,625
Product of their HCF and LCM = 5 × 325 = 1,625
Therefore, Product of the number = Product of their HCF and LCM (Verified)

View full question & answer→

Question 223 Marks

Find the least 5-digit number which is exactly divisible by 20, 25 and 30.

Answer

$\quad 20=1 \times 2 \times 2 \times 5=2^2 \times 5^1$
$25=1 \times 5 \times 5 \times 31=5^2$
$30=1 \times 2 \times 3 \times 5=2^1 \times 3^1 \times 5^1$
LCM of 20,25 and $30=2^2 \times 3^1 \times 5^2=300$
Least five digit number is 10000
Now, if we divide 10000 by 60 , we will get 33.33 as quotient.
The integer just greater than 33.33 is 34
$\therefore$ Required number $=300 \times 34=10200$
Hence, the least 5-digit number which is exactly divisible by $20,25,30$ is 10200.

View full question & answer→

Question 233 Marks

The length, breadth and height of a room are 1050 cm, 750 cm and 425 cm respectively. Find the length of the longest tape which can measure the three dimensions of the room exactly.

Answer

$1050=1 \times 2 \times 3 \times 5 \times 5 \times 7=2^1 \times 3^1 \times 5^2 \times 7^1$
$750=1 \times 2 \times 3 \times 5 \times 5 \times 5=2^1 \times 3^1 \times 5^3$
$425=1 \times 5 \times 5 \times 17=5^2 \times 17^1$
$30=1 \times 2 \times 3 \times 5=2^1 \times 3^1 \times 5^1$
HCF of 1050,750 , and $425=5^2=25$
Hence, the length of the longest tape which can measure the three dimensions of the room exactly is 25 cm

View full question & answer→

Question 243 Marks

Show that the following pairs are co-prime:
59,97

Answer

We know that two numbers are co-primes if their HCF is 1.
59 and 97
Here, dividend = 97 and divisor = 59

Clearly, the last divisor is 1.
Hence, the given numbers are co-primes.

View full question & answer→

Question 253 Marks

Three tankers contain 403 liters, 434 liters and 465 liters of diesel respectively. Find the maximum capacity of a container that can measure the diesel of three containers exact number of times.

Answer

The maximum capacity of three containers is equal to the HCF of 403,434 and 465.
$403=1 \times 13 \times 31=13^1 \times 31^1$
$434=1 \times 2 \times 7 \times 31=2^1 \times 7^1 \times 31^1$
$465=1 \times 3 \times 5 \times 31=3^1 \times 5^1 \times 31^1$
HCF of 403,434 and $465=31^1=31$
Hence, the maximum capacity of a container that can measure the diesel of three containers exact number of times is 31 liters.

View full question & answer→

Question 263 Marks

Find the largest number which 615 and 963 leaving remainder 6 in each case.

Answer

We have to find the largest number which divides (615 - 6) and (963 - 6) exactly.
Therefore, the required number = HCF of 609 and 957
Resolving 609 and 957 into prime factors, we have:
609 = 3 × 7 × 29
957 = 3 × 11 × 29
Therefore, HCF of 609 and 957 = 29 × 3 = 87
Hence, the required largest number is 87.

View full question & answer→

Question 273 Marks

The H.C.F of two numbers is 16 and their product is 3072. Find their L.C.M.

Answer

Given: HCF of two numbers = 16
Product of these two numbers = 3,072
Product of the two numbers = Product of their HCF and LCM
Therefore, 3,072 = 16 × LCM
LCM = 307216 = 192
Thus, the required LCM is 192.

View full question & answer→

Question 283 Marks

In the following numbers. replace * by the smallest number to make it divisible by 11:
467 × 91

Answer

Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.
467 × 91
Sum of the digits at the odd places = 4 + 7 + 9 = 20
Sum of the digits at the even places = 6 + missing number + 1 = missing number + 7 Difference = 20 - [missing number + 7] = 13 - missing number
According to rule, 13 - missing number = 11 [Because the missing number is a single digit]
Thus, missing number = 2
Hence, the smallest required number is 2.

View full question & answer→

Question 293 Marks

Reduce the following fractions to the lowest terms:
$\frac{196}{481}$

Answer

$\frac{196}{481}$
For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF.
Now, we have to find the HCF of 296 and 481.
Prime factorization of 296 = 2 × 2 × 2 × 37
Prime factorization of 481 = 13 × 37
Therefore, HCF of 296 and 481 = 37
Now, 296 + 37481 + 37 = 813
Hence, 813 is the required fraction.

View full question & answer→

Question 303 Marks

Two brands of chocolates are available in packs of 24 and 15 respectively. If i need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind i would need to buy?

Answer

Let the brand ‘A’ contain 24 chocolates in one packet and brand ‘B’ contain 14 chocolates in one packet.
Equal number of chocolates of each kind can be found out by taking LCM of the number of chocolates in each packet.
Therefore, LCM of 15 and 24 is:
$\begin{array}{c|c} 2 & 15, 24 \\ \hline 2 & 15,12 \\ \hline2&15,6\\ \hline3&15,3\\ \hline5&5,1\\ \hline&1,1 \end{array}$
Required LCM = 2 × 2 × 2 × 3 × 5 =120
Therefore, minimum 12o chocolates of each kind should be purchased.
Number of boxes of brand ‘A’ which needs to be purchased = 120 ÷ 24 = 5
Number of boxes of brand ‘B’ which needs to be purchased = 120 ÷ 15 = 8

View full question & answer→

Question 313 Marks

For the following pairs of numbers, verify the property:
Product of the number = Product of their H.C.F and L.C.M
490,1155

Answer

Given numbers are 490 and 1155.
Prime factorization of 490 = 2 × 5 × 7 × 7
Prime factorization of 1155 = 3 × 5 × 7 × 11
HCF of 490 and 1155 = 35
LCM of 490 and 1155 = 2 × 3 × 3 × 5 × 7 × 7 × 11 = 16710
Product of the given number = 490 × 1155 = 5,65,950
Product of their HCF and LCM = 35 × 16,170 = 5,65,950
Therefore, Product of the number = Product of their HCF and LCM (Verified)

View full question & answer→

Question 323 Marks

For the following pairs of numbers, verify the property:
Product of the number = Product of their H.C.F and L.C.M
35,40

Answer

Given numbers are 35 and 40.
Prime factorization of 35 = 5 × 7
Prime factorization of 40 = 2 × 2 × 2 × 5
HCF of 35 and 40 = 5
LCM of 35 and 40 = 2 × 2 × 2 × 5 × 7 = 280
Product of the given number = 35 × 40 = 1400
Product of their HCF and LCM = 5 × 280 = 1400
Therefore, Product of the number = Product of their HCF and LCM (Verified)

View full question & answer→

Question 333 Marks

Find the least three digit number which when divided by $20,30,40$ and 50 leaves remainder 10 in each case.

Answer

$20=1 \times 2 \times 2 \times 5=2^2 \times 5^1$
$30=1 \times 2 \times 3 \times 5=2^1 \times 3^1 \times 5^1$
$40=1 \times 2 \times 2 \times 2 \times 5=2^3 \times 5^1$
$50=1 \times 2 \times 5 \times 5=2^1 \times 5^2$
LCM of $20,30,40$ and $50=2^3 \times 3^1 \times 5^2=600$
$\therefore$ Required number $=600+10=610$.
Hence, the least three digit number which when divided by $20,30,40$ and 50 leaves remainder 10 in each case is 610.

View full question & answer→

Question 343 Marks

The H.C.F of two numbers is 145, their L.C.M is 2175. If one number is 725, find the other.

Answer

HCF of two numbers = 145
LCM of two numbers = 2,175
One of the given numbers = 725
Product of the given two numbers = Product of their LCM and HCF
Therefore, 725 x other number = 145 × 2,175
Other number = 145 × 2175725 = 435
Thus, the required number is 435.

View full question & answer→

Question 353 Marks

Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Answer

First, we have to find the LCM of 28 and 32.
Prime factorization of 28 = 2 × 2 × 7
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
Therefore, Required LCM = 2 × 2 × 2 × 2 × 2 × 7 = 224
It is given that when we divide the number by 28, the remainder is 8 and when we divide the number by 32, the remainder is 12.
We observe:
28 - 8 = 20
32 - 12 = 20
Therefore, Required number = 224 - 20 = 204

View full question & answer→

Question 363 Marks

Reduce the following fractions to the lowest terms:
$\frac{161}{207}$

Answer

$\frac{161}{207}$
For reducing the given fraction to the lowest terms, we have to divide its numerator and denominator by their HCF.
Now, we have to find the HCF of 161 and 207.
Prime factorization of 161 = 7 × 23
Prime factorization of 207 = 3 × 3 × 23
Therefore, HCF of 161 and 207 = 23
Now, 161 + 23207 + 23 = 79
Hence, 79 is the required fraction.

View full question & answer→

Question 373 Marks

Telegraph poles occur at equal distances of 220m along a road and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole?

Answer

We have to find the LCM of 220 m and 300 m.
Prime factorization of 220 = 2 × 2 × 5 × 11
Prime factorization of 300 = 2 × 2 × 3 × 5 × 5
Therefore, Required LCM = 2 × 2 × 3 × 5 × 5 × 11 = 3,300
Hence, 3,300 m far is the next heap that lies at the foot of a pole.

View full question & answer→

Question 383 Marks

If $x$ is prime, $y$ is a composite number such that $x+y=240$ and their LCM is 4199 . Find $x$ and $y$.

Answer

We know that the LCM of a prime number and a composite number is equal to their product.
So, $x y=4199$
Now, $x+y=240$
$\Rightarrow y=240-x$
Substituting the value of $y$ in $x y=4199$, we will get
$x(240-x)=4199$
$\Rightarrow 240 x-x^2=4199$
$\Rightarrow x^2-240 x+4199=0$
$\Rightarrow x^2-19 x-221 x+4199=0$
$\Rightarrow x(x-19)-221(x-19)=0$
$\Rightarrow(x-19)(x-221)=0$
$\Rightarrow(x-19)=0 \text { or }(x-221)=0$
$\Rightarrow x=19,221$
$\therefore x=19 \text { and } y=221 \text { or } x=221 \text { and } y=19$

View full question & answer→

Question 393 Marks

Define:
Multiple

Answer

Multiple:When a number ‘a’ is multiplied by another number ‘b’, the product is the multiple of both the numbers ‘a’ and ‘b’.
Examples of multiples:
6 is a multiple of 2 because 2 × 3 = 6
8 is a multiple of 4 because 4 × 2 = 8
12 is a multiple of 6 because 6 × 2 = 12
21 is a multiple of 7 because 7 × 3 = 21

View full question & answer→

Question 403 Marks

Find the smallest 5-digit number which is exactly divisible by $20,25,30$.

Answer

$20=1 \times 2 \times 2 \times 5=2^2 \times 5^1$
$25=1 \times 5 \times 5 \times 31=5^2$
$30=1 \times 2 \times 3 \times 5=2^1 \times 3^1 \times 5^1$
LCM of 20,25 and $30=2^2 \times 3^1 \times 5^2=300$
Smallest five digit number is 10000
Now, if we divide 10000 by 60, we will get 33.33 as quotient.
The integer just greater than 33.33 is 34
$\therefore$ Required number $=300 \times 34=10200$
Hence, the smallest 5-digit number which is exactly divisible by $20,25,30$ is 10200.

View full question & answer→

Question 413 Marks

What is the largest number that divides 626,3127 and 15628 and leaves remainders of 1,2 and 3 respectively?

Answer

We have to find the largest number which divides (626 - 1), (3,127 - 2), and (15,628 - 3) exactly.
The required number will be given by the HCF of 625, 3,125 and 15,625.
Resolving 625, 3125, and 15625 into prime factors, we have:
625 = 5 × 5 × 5 × 5 3,
125 = 5 × 5 × 5 × 5 × 5 15,
625 = 5 × 5 × 5 × 5 × 5 × 5
Therefore, HCF of 625, 3125 and 15625 = 5 × 5 × 5 × 5 = 625 Hence, the required largest number is 625.

View full question & answer→

Question 423 Marks

105 goats, 140 donkeys and 175 cows have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip.

Answer

We have to find the largest possible number of animals. Thus, we will have to find the HCF of 105, 140, and 175.
Prime factorization of 105 = 3 × 5 × 7
Prime factorization of 140 = 2 × 2 × 5 × 7
Prime factorization of 175 = 5 × 5 × 7
Required HCF = 5 × 7 = 35 Hence, 35 animals went in each trip.

View full question & answer→

Question 433 Marks

Find the largest number that divides 59 and 54 leaving remainders 3 and 5 respectively

Answer

First we will subtract 3 and 5 from 59 and 54 recpectively.
Now, we have $59-3=56$ and $54-5=49$
$56=2 \times 2 \times 2 \times 7=2^3 \times 7^1 \\
49=7 \times 7=7^2$
HCF of 56 and $49=7^1=7$
Hence, the largest number that divides 59 and 54 leaving remainders 3 and 5 respectively is 7 .

View full question & answer→

Question 443 Marks

What is the smallest number which when divided by 24, 36 and 54 gives a remainder of 5 each time?

Answer

We have to find prime factorization of 24, 36, and 54.
Prime factorization of 24 = 2 × 2 × 2 × 3
Prime factorization of 36 = 2 × 2 × 3 × 3
Prime factorization of 54 = 2 × 3 × 3 × 3
Therefore, Required LCM=2 × 2 × 2 × 3 × 3 × 3 = 216
Thus, 216 is the smallest number exactly divisible by 24, 36, and 54.
To get the remainder as 5:
Smallest number = 216 + 5 = 221
Thus, the required number is 221.

View full question & answer→

Question 453 Marks

In the following numbers. replace * by the smallest number to make it divisible by 11:
86 * 72

Answer

Rule: A number is divisible by 11 if the difference of the sums of the alternate digits is either 0 or a multiple of 11.
86 × 72
Sum of the digits at the odd places = 8 + missing number + 2 = missing number + 10
Sum of the digits at the even places = 6 + 7 = 13
Difference = [missing number + 10 ] - 13 = Missing number – 3
According to the rule, missing number - 3 = 0 [Because the missing number is a single digit]
Thus, missing number = 3
Hence, the smallest required number is 3.

View full question & answer→

Question 463 Marks

A number is divisible by both 7 and 16. By which other number will that number be always divisible?

Answer

Since the number is divisible by 7 and 16, they are the factors of that number.
So, the number will be divisible by the common factor of 7 and 16.
The factors of 7 are 1 and 7.
The factors of 16 are 1, 2, 4, 8, and 16.
Therefore, the common factor of 7 and 16 is 1 and the number is divisible by 1.

View full question & answer→

Question 473 Marks

Show that the following pairs are co-prime:
288,1375

Answer

We know that two numbers are co-primes if their HCF is 1.
288 and 1375
Here, dividend = 288 and divisor = 1375

Clearly, the last divisor is 1.
Hence, the given numbers are co-prime.

View full question & answer→

Question 483 Marks

Find the least number of five digits which is exactly divisible by each of $8,12,18,40$ and 45

Answer

$2 \times 2 \times 3=2^2 \times 3^1$
$18=1 \times 2 \times 3 \times 3=2^1 \times 3^2$
$40=1 \times 2 \times 2 \times 2 \times 5=2^3 \times 5^1$
$45=1 \times 3 \times 3 \times 5=3^2 \times 5^1$
LCM of $8,12,18,40$ and $45=2^3 \times 3^2 \times 5^1=360$
Smallest five digit number is 10000
Now, if we divide 10000 by 360 , we will get 27.78 as quotient.
The integer just greater than 27.78 is 28
$\therefore$ Required number $=360 \times 28=10080$
Hence, the least number of five digits which is exactly divisible by each of $8,12,18,40$ and 45 is 10080 .

View full question & answer→

Question 493 Marks

What is the smallest number that, when divide by 35, 56 and 91 leaves remainders of 7 in each case?

Answer

We have to find the prime factorization of 35, 56, and 91.
Prime factorization of 35 = 5 × 7
Prime factorization of 56 = 2 × 2 × 2 × 7
Prime factorization of 91 = 7 × 13
Therefore, Required LCM = 2 × 2 × 2 × 5 × 7 × 13 = 3,640
Thus, 3,640 is the smallest number exactly divisible by 35, 56, and 91.
To get the remainder as 7:
Smallest number = 3,640 + 7 = 3,647
Thus, the required number is 3,647.

View full question & answer→

Question 503 Marks

In a school there are two sections - section A and section B of Class VI. There are 32 students in section A and 36 in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.

Answer

We have to find the LCM of 32 and 36.
Prime factorization of 32 = 2 × 2 × 2 × 2 × 2
Prime factorization of 36 = 2 × 2 × 3 × 3
Required LCM = 2 × 2 × 2 × 2 × 2 × 3 × 3 = 288
Therefore, Minimum number of books required = LCM of 32 and 36 = 288 books

View full question & answer→

Question 513 Marks

Define:
Factor

Answer

Factor:A factor of a number is an e×act divisor of that number.
For e×ample, 4 e×actly divide 32. Therefore, 4 is a factor of 32.
Examples of factors are:
2 and 3 are factors of 6 because 2 × 3 = 6
2 and 4 are factors of 8 because 2 × 4 = 8
3 and 4 are factors of 12 because 3 × 4 = 12
3 and 5 are factors of 15 because 3 × 5 = 15

View full question & answer→

Question 523 Marks

A rectangular courtyard is 20m 16cm long and 15m 60cm broad. It is to be paved with square stones of the same size. Find the least possible number of such stones.

Answer

Length of the rectangular courtyard = 20m 16cm = 2,016cm
Breadth of the rectangular courtyard = 15m 60cm = 1,560cm
Least possible side of the square stones used to pave the rectangular courtyard = HCF of (2,016 and 1,560)
Prime factorization of 2,016 =2 × 2 × 2 × 2 × 2 × 3 × 3 × 7
Prime factorization of 1,560 = 2 × 2 × 2 × 3 × 5 × 13 HCF of (2,016, 1,560) = 2 × 2 × 2 × 3= 24
Least possible side of square stones used to pave the rectangular courtyard is 24 cm. Number of square stones used to pave the rectangular courtyard
= Area of rectangular courtyard Area of square stone = 2016cm × 1560cm (24cm) 2 = 5460 Thus, the least number of square stones used to pave the rectangular courtyard is 5,460.

View full question & answer→

Question 533 Marks

The length, breadth and height of a room are 8m 25cm, 6m 75cm and 4m 50cm, respectively. Determine the longest rod which can measure the three dimensions if the room exactly.

Answer

Length of the room = 8m 25cm = 825cm
Breadth of the room = 6m 75cm = 675cm
Height of the room = 4m 50cm = 450cm
The longest rod will be given by the HCF of 825, 675 and 450.
Prime factorization of 825 = 3 × 5 × 5 × 11
Prime factorization of 675 = 3 × 3 × 3 × 5 × 5
Prime factorization of 450 = 2 × 3 × 3 × 5 × 5 Therefore, HCF of 825, 675 and 450 = 3 × 5 × 5 = 75
Thus, the required length of the longest rod is 75cm.

View full question & answer→

Question 543 Marks

Find the greatest number of four digits which is exactly divisible by each of $8,12,18$ and 30 .

Answer

$8=1 \times 2 \times 2 \times 2=2^3$
$12=1 \times 2 \times 2 \times 3=2^2 \times 3^1 \\
18=1 \times 2 \times 3 \times 3=2^1 \times 3^2 \\
30=1 \times 2 \times 3 \times 5=2^1 \times 3^1 \times 5^1$
LCM of $8,12,18$, and $30=2^3 \times 3^2 \times 5^1=360$
Largest 4-digit number is 9999
Now, if we divide 9999 by 360 , we will get 27.78 as quotient.
The integer just less than 27.78 is 27
$\therefore$ Required number $=360 \times 27=9720$
Hence, the greatest number of four digits which is exactly divisible by each of $8,12,18$ and 30 is 9720 .

View full question & answer→

Generate Paper Free All Playing with numbers topics