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Maths 3 Mark Question

Factors and Multiples · Maharashtra Board STD 6 (MSBSHSE - English Medium). 51 questions.

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Question 13 Marks
Show that the following pairs are co-primes:385, 621
Answer
Given numbers are 385 and 621.$\begin{array}{c|c}5&385\\\hline7&77\\\hline11&11\\\hline&1\end{array}$
$\begin{array}{c|c}3&621\\\hline3&207\\\hline3&69\\\hline23&23\\\hline&1\end{array}$
$385=5\times7\times11\times1$
$621=3\times3\times3\times23=3^3\times23\times1$
$\therefore$ HCF = 1
Hence, they are co-primes.
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Question 23 Marks
Test the divisibility of 5869473 by 11.
Answer
5869473A number is divisible by 11 if the the difference of the sums of the digits at the odd places and that at the even places (starting from ones place) is either 0 or a multiple of 11.
Sum of the digits at even places = 7 + 9 + 8
= 24
Sum of the digits in odd places = 3 + 4 + 6 + 5
= 18
Difference = 24 - 18
= 6
Since 6 is not divisible by 11, 5869473 is not divisible by 11.
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Question 33 Marks
Find the greatest number which divides 2011 and 2623, leaving remainders 9 and 5 respectively.
Answer
Clearly, we have to find the number which exactly divides (2011 - 9) and (2623 - 5).So, the required numbers is the HCF of 2002 and 2618.

$\therefore$ The required number is 154.
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Question 43 Marks
In the following numbers, replace * by the smallest number to make it divisible by 11:
26*5
Answer
26*5
Sum of the digits at odd places = 5 + 6 = 11
Sum of the digits at even places = * + 2
Difference = sum of odd terms - sum of even terms
= 11 - (* + 2)
= 11 - * - 2
= 9 - *
Now, (9 - *) will be divisible by 11 if * = 9.
i.e., 9 - 9 = 0
0 is divisible by 11.
$\therefore$ * = 9
Hence, the number is 2695.
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Question 53 Marks
Determine the longest tape which can be used to measure exactly the lengths 7m, 3m, 85cm and 12m, 95cm.
Answer
7m = 700cm
3m, 85cm = 385cm
12m, 95cm = 1395cm
The required lenths of the tape that can measure the lenths 700cm, 385cm and 1295cm will given by the HCF of 700cm, 385cm and 1295cm.
Evaluating the HCF of 700, 385 and 1295 using prime fractorisation method, we have:
$\begin{array}{c|c}2&700\\\hline2 &350\\\hline5&175\\\hline5&35\\\hline7&7\\\hline&1\end{array}$
$700=2\times2\times5\times5\times7=2^2\times5^2\times7$
$\begin{array}{c|c}5&1295\\\hline11&259\\\hline7&7\\\hline&1\end{array}$
$385=5\times11\times7$
$\begin{array}{c|c}5&1295\\\hline7&259\\\hline37&37\\\hline&1\end{array}$
$1295=5\times7\times37$
$\therefore$ Hence, the longest tape which can measure the lengths 7m, 3m, 85cm, and 12m, 95cm exactly is 35cm.
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Question 63 Marks
Pair of numbers, werify that their product = (HCF × LCM)
87, 145
Answer
87 and 145
$\begin{array}{c|c}3&87\\\hline29&29\\\hline&1\end{array}$
$\begin{array}{c|c}5&145\\\hline29&29\\\hline&1\end{array}$
We have:
$87=3\times29$
$145=5\times29$
HCF = 29
LCM $=29\times15\times1=435$
Now, HCF × LCM $=29\times435=12615$
Product of the two numbers $=87\times145=12615$
$\therefore$ HCF × LCM = Product of the two numbers.
Verified.
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Question 73 Marks
In the following numbers, replace * by the smallest number to make it divisible by 11:
467*91
Answer
467*91Sum of the digits at odd places 1 + * + 6 = 7 + *
Sum of the digits at even places 9 + 7 + 4 = 20
Difference = sum of odd terms - sum of even terms
= (7 + *) - 20
= * - 13
Now, (* - 13) will be divisible by 11 if * = 2.
i.e., 2 - 13 = -11
-11 is divisible by 11.
∴ * = 2
Hence, the number is 467291.
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Question 83 Marks
Find the HCF of the numbers in the following using the division method:399, 437
Answer
The given numbers are 399 and 437.We have:

$\therefore$ The HFC of 19.
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Question 93 Marks
Find the HCF and LCM of:693, 1078
Answer
The given numbers are 693 and 1078.We have:
$\begin{array}{c|c}3&693\\\hline3&231\\\hline7&77\\\hline11&11\\\hline&1\end{array}$
$\begin{array}{c|c}2&1078\\\hline7&539\\\hline7&77\\\hline11&11\\\hline&1\end{array}$
Now, we have:
$693=3\times3\times7\times11$
$1078=2\times7\times7\times11$
$\therefore$ HCF $=7\times11=77$
Also, LCM $=2\times3\times3\times7\times7\times11=9702$
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Question 103 Marks
Three pieces of timber, 42-m, 49-m and 63-m long, have to be devided into planks of the same length. What is the greatest possible length of each plank?
Answer
The lengths of the three pieces of timber are 42-m, 49-m and 63-m.The greatest possible length of each plank will be given by the HCF of 42, 49 and 63.
Firstly, we will find the HCF of 42 and 49 by division method.

$\therefore$ The HCF of 42 and 49 is 7.
Now, we will find the HCF of 7 and 63.

$\therefore$ The HCF of 7 and 63 is 7.
Therefore, HCF of all three numbers is 7.
Hence, the greatest possible length of each plank is 7-m.
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Question 113 Marks
Find the HCF and LCM of:234, 572
Answer
The given numbers are 234 and 572.We have:
$\begin{array}{c|c}2&234\\\hline3&117\\\hline3&39\\\hline13&13\\\hline&1\end{array}$
$\begin{array}{c|c}2&572\\\hline2&286\\\hline13&143\\\hline11&11\\\hline&1\end{array}$
Now, we have:
$234=2\times3\times3\times13$
$572=2\times2\times13\times11$
$\therefore$ LCM $=13\times2\times2\times11\times9$
$=5148$
Also, HCF $=13\times2=26$
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Question 123 Marks
Show that the following pairs are co-primes:847, 1014
Answer
Given numbers are 847 and 1014.$\begin{array}{c|c}7&847\\\hline11&121\\\hline11&11\\\hline&1\end{array}$
$\begin{array}{c|c}2&1014\\\hline3&507\\\hline13&169\\\hline13&13\\\hline&1\end{array}$
$847=7\times11\times11\times1=7\times11^2\times1$
$1014=2\times3\times13\times13\times1$
$\therefore$ HCF = 1
Hence, 847 and 1014 are co-primes.
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Question 133 Marks
On dividing 5035 by 31, the remainder is 13. Find the quotient.
Answer
Remainder is 13
$\therefore$ Number exactly divisible by 31 = 5035 - 13
= 5022

So, the required quotient is 162.
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Question 143 Marks
Find the LCM of the numbers given below:28, 36, 45, 60
Answer
The given numbers are 28, 36, 45 and 60.We have:
$\begin{array}{c|c}2&28,36,45,60\\\hline2&14,18,45,30\\\hline3&7,9,45,15\\\hline3&7,3,15,5\\\hline5&7,1,5,5\\\hline7&7,1,1,1\\\hline&1,1,1,1\end{array}$
$\therefore$ LCM = 2 × 2 × 3 × 3 × 5 × 7
= 1260
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Question 153 Marks
Three measuring rods are 45cm, 50cm and 75cm in length. What is the least length (in metres) of a rope that can be measured by the full length of each of these three rods?
 
Answer
The length of the required rope must be such that it is exactly divisible by 45, 50 and 75. The least length will be given by the LCM of 45, 50 and 75.
$\begin{array}{c|c}2&45,50,75\\\hline3&45,25,75\\\hline3&15,25,25\\\hline5&5,25,25\\\hline5&1,5,5\\\hline&1,1,1\end{array}$
Required length of the rope that can be measured by the full length of each of the three rods is 450cm.
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Question 163 Marks
In the following numbers, replace * by the smallest number to make it divisible by 11:
1723*4
Answer
1723*4Sum of the digits at odd places 4 + 3 + 7 = 14
Sum of the digits at even places * + 2 + 1 = 3 + *
Difference = sum of odd terms - sum of even terms.
= 14 - (3 + *)
= 11 - *
Now, (11 - *) will be divisible by 11 if * = 0.
i.e., 11 - 0 = 11
11 is divisible by 11.
$\therefore$ * = 0
Hence, the number is 172304.
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Question 173 Marks
Give the prime factorization of the following number:
13915
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}5&13915\\\hline11&2783\\\hline11&253\\\hline23&23\\\hline&1\end{array}$
$\therefore13915=5\times11\times11\times23$
$=3\times11^2\times23$
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Question 183 Marks
Find the LCM of the numbers given below:
$48, 64, 72, 96, 108$
Answer
The given numbers are 48, 64, 72, 96 and 108.

We have:

$\begin{array}{c|c}2&48,64,72,96,108\\\hline2&24,32,36,48,54\\\hline2&12,16,18,24,27\\\hline2&6,8,9,12,27\\\hline3&3,4,9,6,27\\\hline2&1,4,3,2,9\\\hline2&1,2,3,1,9\\\hline3&1,1,3,1,9\\\hline3&1,1,1,1,3\\\hline&1,1,1,1,1\end{array}$

$\therefore \text { LCM }=2^6 \times 3^3$

$=1728$
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Question 193 Marks
Find the LCM of the numbers given below:36, 40, 126
Answer
The given numbers are 36, 40 and 126.We have:
$\begin{array}{c|c}2&36,40,126\\\hline3&18,20,60\\\hline3&6,20,21\\\hline2&2,20,7\\\hline2&1,10,7\\\hline5&1,5,7\\\hline7&1,1,7\\\hline&1,1,1\end{array}$
$\therefore$ LCM = 2 × 3 × 3 × 2 × 2 × 5 × 7
= 2520
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Question 203 Marks
Find the HCF of the numbers in the following using the prime factorization method:144, 252, 630
Answer
The given numbers are 144, 252 and 630.We have:
$\begin{array}{c|c}2&144\\\hline2&72\\\hline2&36\\\hline2&18\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$\begin{array}{c|c}2&252\\\hline2&126\\\hline3&63\\\hline3&21\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}2&630\\\hline3&315\\\hline3&105\\\hline5&35\\\hline7&7\\\hline&1\end{array}$
Now, $144=2\times2\times2\times2\times3\times3$
$252=2\times2\times3\times3\times7$
$630=2\times3\times3\times5\times7$
$\therefore$ HCF $=2\times3\times3=18$
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Question 213 Marks
Find the LCM of the numbers given below:16, 28, 40, 77
Answer
The given numbers are 16, 28, 40 and 77.We have:
$\begin{array}{c|c}2&16,28,40,77\\\hline7&8,14,20,77\\\hline2&8,2,20,11\\\hline2&4,1,10,11\\\hline2&2,1,5,11\\\hline5&1,1,5,11\\\hline11&1,1,1,11\\\hline&1,1,1,1\end{array}$
$\therefore$ LCM = 2 × 7 × 2 × 2 × 2 × 5 × 11
= 6160
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Question 223 Marks
Find the HCF and LCM of:2923, 3239
Answer
HCF of 2923 and 3239:

$\therefore$ HCF = 79
We know that product of two numbers = HCF × LCM
$\Rightarrow\text{LCM}=\frac{\text{Product of two numbers}}{\text{HCF}}$
$\Rightarrow\text{LCM}=\frac{2923\times3239}{79}$
$\therefore\text{LCM}=119843$
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Question 233 Marks
Show that the following pairs are co-primes:161, 192
Answer
The given numbers are 161 and 192.
We have:
$\begin{array}{c|c}2&161\\\hline23&23\\\hline&1\end{array}$
$\begin{array}{c|c}2&192\\\hline2&96\\\hline2&48\\\hline2&24\\\hline2&12\\\hline2&6\\\hline3&3\\\hline&1\end{array}$
Now, $161=7\times23\times1$
$192=2\times2\times2\times2\times2\times2\times3=2^6\times3\times1$
$\therefore$ HCF = 1
Hence, 161 and 192 are co-primes.
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Question 243 Marks
Three boys step off together from the same place. If their steps measure 36cm, 48cm and 54cm, at what distance from the starting point will they again step together?
Answer
From the starting point, they will step together again when they travel a distance that is exactly divisible by the lengths of their steps.
The least distance from the starting point where they will step together will be given by the LCM of 36, 48 and 54.
$\begin{array}{c|c}2&36,48,54\\\hline2&18,24,27\\\hline3&9,12,27\\\hline3&3,4,9\\\hline3&1,4,3\\\hline2&1,4,1\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
The required distance = 2 × 2 × 3 × 3 × 3 × 2 × 2
= 16 × 27
= 432cm
They will step together again at a distance of 432cm from the starting point
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Question 253 Marks
Find the least number which when divided by 16, 36 and 40 leaves 5 as remainder in each case.
Answer
On subtracting 5 from each number:
16 - 5 = 11
36 - 5 = 31
40 - 5 = 35
The required number will be the least common multiple of 11, 31 and 35.
LCM of 11, 31 and 35 = 11 × 31 × 35
= 11935
This is because they do not have any factor in common.
So, 11935 is the required number.
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Question 263 Marks
An electronic device makes a beep after every 15 minutes. Another device makes a beep after every 20 minutes. They beeped together at 6 a.m. At what time will they next beep together?
Answer
The LCM of the time intervals of the beeps will give the time when the electronic devices will beep together.
LCM of 15 and 20.
$\begin{array}{c|c}5&15,20\\\hline3&3,4\\\hline2&1,4\\\hline2&1,2\\\hline&1,1\end{array}$
Required time 5 × 3 × 2 × 2
= 60 min
So, they will beep simultaneously after 60 min or 1h.
$\therefore$ They will beep together again at 7:00 a.m.
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Question 273 Marks
Find the HCF of the numbers in the following using the prime factorization method:72, 108, 180
Answer
The given numbers are 72, 108 and 180.We have:
$\begin{array}{c|c}2&72\\\hline2&36\\\hline2&18\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$\begin{array}{c|c}2&108\\\hline2&54\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
$\begin{array}{c|c}2&180\\\hline2&90\\\hline3&45\\\hline3&15\\\hline5&5\\\hline&1\end{array}$
Now, $72=2\times2\times2\times3\times3=2^3\times3^2$
$108=2\times2\times3\times3\times3=2^2\times3^3$
$180=2\times2\times3\times3 \times5=2^2\times3^2\times5$
$\therefore$ HCF $=2^2\times3^2=36$
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Question 283 Marks
Find the HCF of the numbers in the following using the division method:1045, 1520
Answer
The given numbers are 1045 and 1520.We have:

$\therefore$ The HFC of 1045 and 15202 is 95.
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Question 293 Marks
Find the HCF and LCM of:145, 232
Answer
The given numbers are 145 and 232.We have:
$\begin{array}{c|c}5&145\\\hline29&29\\\hline&1\end{array}$
$\begin{array}{c|c}2&232\\\hline2&116\\\hline2&58\\\hline29&29\\\hline&1\end{array}$
Now, we have:
$145=5\times29$
$232=2\times2\times2\times29$
$\therefore$ HCF = 29
Also, LCM $=29\times2\times2\times2\times5=1160$
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Question 303 Marks
Find the HCF and LCM of:861, 1353
Answer
The given numbers are 861 and 1353.We have:
$\begin{array}{c|c}3&861\\\hline7&287\\\hline{41}&41\\\hline&1\end{array}$
$\begin{array}{c|c}3&1353\\\hline11&451\\\hline41&41\\\hline&1\end{array}$
Now, we have:
$861=3\times41\times7$
$1353=41\times11\times3$
$\therefore$ HCF $=41\times3=123$
Also, LCM $=41\times3\times11\times7=9471$
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Question 313 Marks
In the following numbers, replace * by the smallest number to make it divisible by 11:
86*72
Answer
86*72Sum of the digits at odd places 2 + * + 8 = 10 + *
Sum of the digits at even places 6 + 7 = 13
Difference = sum of odd terms - sum of even terms.
= 10 + * - 13
= * - 3
Now, (* - 3) will be divisible by 11 if * = 3.
i.e., 3 - 3 = 0
0 is divisible by 11.
$\therefore$ * = 3
Hence, the number is 86372.
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Question 323 Marks
In the following numbers, replace * by the smallest number to make it divisible by 11:
9*8071
Answer
9*8071Sum of the digits at odd places 1 + 0 + * = 1 + *
Sum of the digits at even places 7 + 8 + 9 = 24
Difference = sum of odd terms - sum of even terms.
=1 + * - 24
= * - 23
Now, (* - 23) will be divisible by 11 if * = 1.
i.e., 1 - 23 = -22
-22 is divisible by 11.
$\therefore$ * = 1
Hence, the number is 918071.
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Question 333 Marks
In the following numbers, replace * by the smallest number to make it divisible by 11:
39*43
Answer
39*43Sum of the digits at odd places = 3 + * + 3 = 6 + *
Sum of the digits at even places = 4 + 9 = 13
Difference = sum of odd terms - sum of even terms
= 6 + * – 13
= * - 7
Now, (* - 7) will be divisible by 11 if * = 7.
i.e., 7 - 7 = 0
0 is divisible by 11.
$\therefore$ * = 7
Hence, the number is 39743.
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Question 343 Marks
Find the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each case.
Answer
25, 40 and 60 exactly divides the least number that is equal to their LCM.So, the required number that leaves 9 as a remainder will be LCM + 9.
Finding the LCM.
$\begin{array}{c|c}2&25,40,60\\\hline2&25,20,30\\\hline2&25,10,15\\\hline3&25,5,15\\\hline5&25,5,5\\\hline5&5,1,1\\\hline&1,1,1\end{array}$
$\text { LCM }=2^3 \times 3 \times 5^2=600$
$\therefore$ Required number $=600+9=609$
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Question 353 Marks
The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds. If they start changing simultaneously at 8 a.m., after how much time will they change again simultaneously?
Answer
The time when the lights will change simultaneously again will be quantity which is exactly divisible by 48, 72 and 108. The least time when they change simultaneously will be given by their LCM.
$\begin{array}{c|c}2&48,72,108\\\hline2&24,36,54\\\hline2&12,18,27\\\hline2&6,9,27\\\hline3&3,9,27\\\hline3&1,3,9\\\hline3&1,1,3\\\hline&1,1,1\end{array}$
$\text { Required time }=2^4 \times 3^3$
$=432 \text { seconds }$
$=7 \text { min. } 12 \text { second }$
So, the lights will changes simultaneouslyat 8:07:12 a.m.
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Question 363 Marks
Show that the following pairs are co-primes:343, 432
Answer
The given numbers are 343 and 432.
We have:
$\begin{array}{c|c}7&343\\\hline7&49\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}2&432\\\hline2&216\\\hline2&108\\\hline2&54\\\hline3&27\\\hline3&9\\\hline3&3\\\hline&1\end{array}$
Now, $343=7\times7\times7\times1=7^3\times1$
$432=2\times2\times2\times2\times3\times3\times3\times1$
$\therefore$ HCF = 1
Hence, 343 and 432 are co-primes.
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Question 373 Marks
Pair of numbers, werify that their product = (HCF × LCM)
186, 403
Answer
186 and 403
$\begin{array}{c|c}2&186\\\hline3&93\\\hline31&31\\\hline&1\end{array}$
$\begin{array}{c|c}13&403\\\hline31&31\\\hline&1\end{array}$
We have:
$186=2\times3\times31$
$40=31\times13$
HCF = 31
LCM $=31\times13\times6=2418$
Now, HCF × LCM $=31\times2418=74958$
Product of the two numbers $=168\times403=74958$
$\therefore$ HCF × LCM = Product of the two numbers.
Verified.
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Question 383 Marks
Reduce the following fractions to the lowest terms:$\frac{517}{799}$
Answer
$\frac{517}{799}$
To reduce the given fraction to its lowest terms, we will divide the numerator and the denominators by their HCF.
Now, we will find the HCF of 517 and 799.

$\therefore$ HCF = 47
Dividing the numerator and the denominators by the HCF, we get:
$\frac{517\div47}{799\div47}=\frac{11}{17}$
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Question 393 Marks
Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case.
Answer
Because the remainder is 6, we have to find the number that exactly devides (615 - 6) and (963 - 6).
Required number = HCF of 609 and 957.

Therefore, the required number is 87.
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Question 403 Marks
Pair of numbers, werify that their product = (HCF × LCM)
490, 1155
Answer
490 and 1155
$\begin{array}{c|c}2&490\\\hline5&525\\\hline7&49\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}5&1155\\\hline7&231\\\hline3&33\\\hline11&11\\\hline&1\end{array}$
$490=7\times7\times2\times5$
$1155=5\times7\times3\times11$
HCF $=7\times5=35$
LCM $=7\times5\times7\times2\times3\times11=16170$
Now, HCF × LCM $=35\times16170=565950$
Product of the two numbers $=490\times1155=565950$
$\therefore$ HCF × LCM = Product of the two numbers.
Verified.
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Question 413 Marks
Find the HCF of the numbers in the following using the division method:58, 70
Answer
We have:
$\therefore$ The HFC of 58 and 70 is 2.
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Question 423 Marks
Find the HCF of the numbers in the following using the prime factorization method:1197, 5320, 4389
Answer
The given numbers are 1197, 5320 and 4389.We have:
$\begin{array}{c|c}3&1197\\\hline3&399\\\hline7&133\\\hline19&19\\\hline&1\end{array}$
$\begin{array}{c|c}2&5320\\\hline2&2660\\\hline2&1330\\\hline5&665\\\hline7&133\\\hline19&19\\\hline&1\end{array}$
$\begin{array}{c|c}3&4389\\\hline7&1463\\\hline19&209\\\hline11&11\\\hline&1\end{array}$
Now, $119=3\times3\times7\times19=3^2\times7\times19$
$5320=2\times2\times2\times5\times7\times19=2^3\times5\times7\times19$
$4389=3\times7\times19\times11$
$\therefore$ Required HCF $=19\times7=133$
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Question 433 Marks
Find the HCF and LCM of:117, 221
Answer
The given numbers are 117 and 221.We have:
$\begin{array}{c|c}3&117\\\hline3&39\\\hline13&13\\\hline&1\end{array}$
$\begin{array}{c|c}13&221\\\hline17&17\\\hline&1\end{array}$
Now,
$ 177=3\times3\times13$
$221=13\times17$
$\therefore$ HCF = 13 × 1
Now, LCM $=13\times17\times3\times3$
$=1989$
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Question 443 Marks
Find the LCM of the numbers given below:
$144,180,384$
Answer
The given numbers are 144, 180 and 384.We have:
$\begin{array}{c|c}2&144,180,384\\\hline2&72,90,192\\\hline2&36,46,96\\\hline2&18,45,48\\\hline3&9,45,24\\\hline3&3,15,8\\\hline2&1,5,8\\\hline2&1,5,4\\\hline2&1,5,2\\\hline5&1,5,1\\\hline&1,1,1\end{array}$
$\therefore LCM=2^7 \times 3^2 \times 5$
$=5760$
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Question 453 Marks
Find the HCF of the numbers in the following using the prime factorization method:84, 120, 138
Answer
The given numbers are 84, 120 and 138.We have:
$\begin{array}{c|c}2&84\\\hline4&42\\\hline3&21\\\hline7&7\\\hline&1\end{array}$
$\begin{array}{c|c}2&120\\\hline2&60\\\hline2&30\\\hline3&15\\\hline5&5\\\hline&1\end{array}$
$\begin{array}{c|c}2&138\\\hline3&69\\\hline23&23\\\hline&1\end{array}$
Now, $84=2\times2\times3\times7$
$120=2\times2\times2\times3\times5$
$138=2\times3\times23$
$\therefore$ HCF $=2\times3=6$
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Question 463 Marks
Find the HCF of the numbers in the following using the prime factorization method:106, 159, 371
Answer
The given numbers are 106, 159 and 371.We have:
$\begin{array}{c|c}2&106\\\hline53&53\\\hline&1\end{array}$
$\begin{array}{c|c}3&159\\\hline53&53\\\hline&1\end{array}$
$\begin{array}{c|c}7&371\\\hline53&53\\\hline&1\end{array}$
Now, $106=2\times53$
$159=3\times53$
$371=7\times53$
$\therefore$ HCF $=53$
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Question 473 Marks
Find the least numbres divisible by 15, 20, 24, 32 and 36.
Answer
The given numbers are $15, 20, 24, 32$ and $36.$
The smallest number divisible by the numbers given above will be their LCM.'
$\begin{array}{c|c}2&15,20,24,31,36\\\hline2&15,10,12,16,18\\\hline5&5,10,4,16,2\\\hline2&1,2,4,16,6\\\hline2&1,1,2,8,3\\\hline2&1,1,1,4,3\\\hline2&1,1,1,2,3\\\hline3&1,1,1,1,3\\\hline&1,1,1,1,1\end{array}$
LCM = $2^5 \times 3^2 \times 5$
$= 1440$
$\therefore$ The least numbers divisible by $15, 20, 24, 32$ and $36$ is $1440.$
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Question 483 Marks
Reduce the following fractions to the lowest terms:$\frac{296}{481}$
Answer
$\frac{296}{481}$
To reduce the given fraction to its lowest terms, we will divide the numerator and the denominators by their HCF.
Now, we will find the HCF of 296 and 481.

$\therefore$ HCF = 37
Dividing the numerator and the denominators by the HCF, we get:
$\frac{296\div37}{481\div37}=\frac{8}{13}$
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Question 493 Marks
Three bells toll at intervals of 9, 12, 15 minutes. If they start tolling together, after what time will they next toll together?
Answer
Three bells toll intervals of 9, 12 and 15 minutes.The time when they will toll together again is given by the LCM of 9, 12 and 15.
$\begin{array}{c|c}3&9,12,15\\\hline3&3,4,5\\\hline5&1,4,5\\\hline2&1,4,1\\\hline2&1,2,1\\\hline&1,1,1\end{array}$
Required time = $2^3 \times 3^2 \times 5$
= 180 minutes
= 3h
If they start tolling together, they will toll together again after 3h.
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Question 503 Marks
Reduce the following fractions to the lowest terms:$\frac{161}{207}$
Answer
$\frac{161}{207}$
To reduce the given fraction to its lowest terms, we will divide the numerator and the denominators by their HCF.
Now, we will find the HCF of 161 and 207.

$\therefore$ HCF = 23
Dividing the numerator and the denominators by the HCF, we get:
$\frac{161\div23}{207\div23}=\frac79$
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Question 513 Marks
Give the prime factorization of the following number:
8712
Answer
We will use the didvision method as shown below:
$\begin{array}{c|c}2&8712\\\hline2&4356\\\hline2&2178\\\hline3&1089\\\hline3&363\\\hline11&121\\\hline11&11\\\hline&1\end{array}$
$\therefore8712=2\times2\times2\times3\times3\times11\times11$
$=2^3\times3^2\times11^2$
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