4 Marks Questions

Question 14 Marks

Show that any number of the form $4^n, n \in N$ can never end with the digit $0$.

Answer

We have $4^n=(2 \times 2)^n=2^n \times 2^n$
The prime factors of $4^n$ are $2^n$
For any $n$, the number $2^n$ would with a zero digit if number is divisible by $5$
So, the prime factorisation should contain at least one prime factor $5$
But $4 n=2^n \times 2^n$ which contains the prime factors $2$
So, by the uniqueness of the fundamental theorem, no other prime factorisation of $4^n$ exists.
Thus, any number of the form $4^n$ can never end with the digit $0$

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Question 24 Marks

Use Euclid's algorithm to find HCF of 1190 and 1445. express the HCF in the form 1190m + 1445n.

Answer

Using Euclid's algorithm
HCF of (1190 and 1445).
1445 = 1190 × 1 + 255
1190 = 255 × 4 + 170
255 = 170 × 1 + 85
170 = 85 × 2 + 0
HCF = 85
Now,
85 = 255 - 170
= (1445 - 1190 × 1) - (1190 - 225 × 4)
= (1445 - 1190 × 1) - (1150 - (1445 - 1190 × 1) × 4)
= 1445 - 190 × 1 - 1190 + 1445 × 4 - 1190 × 4
= 1445 × 5 - 1190 × 6
= 85

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Question 34 Marks

The traffic lights at three different road crossing change after every $48$ secons, $72$ seconds and $108$ seconds respectively. If they all change simultaneously at $8 a.m.$ then at what time will they again change simultaneously?

Answer

Let us find the LCM of $48, 72$ and $108$ through prime factorisation:
$\begin{array}{c|c} 2 & 48 \\ \hline 2 & 24\\ \hline2&12\\ \hline2&6\\ \hline&3 \end{array}$ $\begin{array}{c|c} 2 & 72 \\ \hline 2 & 36\\ \hline2&18\\ \hline3&9\\ \hline&3 \end{array}$ $\begin{array}{c|c} 2 & 108 \\ \hline 2 & 54\\ \hline3&27\\ \hline3&9\\ \hline&3 \end{array}$
$48 = 2 \times 2 \times 2 \times 2 \times 3 = 2^4 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3 = 2^3 \times 3^2$
$108 = 2 \times 2 \times 3 \times 3 \times 3 = 2^2 \times 3^3$
LCM of $48, 72, 108$ is $2^4 \times 3^3$
$= 16 \times 27\ sec$
$= 432\ sec$
$=7$ min $12$ sec
Three bells toll together after $7$ min $12$ sec.

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Question 44 Marks

Find the leash number of square tiles required to pave the ceiling of a room 15m 17cm long and 9m 2cm broad.

Answer

Length of room = 15m 17cm= 1500cm + 17cm = 1517cm
Breadth of room = 9m 2cm
= 900cm + 2cm = 902cm
Now, H.C.F. of 1517 and 902 is 41
Thus, area of each tile = (41 × 41) = 1681 sq. cm
Hence, required number of tiles $=\frac{1517\times902}{1681}=814$

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Question 54 Marks

Three measuring rods are $64\ cm,$ $80\ cm$ and $96\ cm$ in length. Find the least length of cloth that can be measured an exact number of times, using any of the rods.

Answer

Let us find the LCM of 64, 80 and 96 through prime factorization:
$\begin{array}{c|c} 2 & 64 \\ \hline 2 & 32\\ \hline2&16\\ \hline2&8\\ \hline2&4\\\hline&2 \end{array}$ $\begin{array}{c|c} 2 & 80 \\ \hline 2 & 40\\ \hline2&20\\ \hline2&10\\ \hline&5 \end{array}$ $\begin{array}{c|c} 2 & 96 \\ \hline 2 & 48\\ \hline2&24\\ \hline2&12\\ \hline2&6\\\hline&3 \end{array}$
$64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^6$
$80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5$
$96 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 = 2^5 \times 3$
L.C.M. of $64, 80$ and $96$
$= 2^6 \times 5 \times 3 = 64 \times 15$
$= 960\ cm = 9.6\ m$
Therefore, the least length of the cloth that can be measured an exact number of times by the rods of $64\ cm, 80\ cm$ and $96\ cm = 9.6\ m$

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Question 64 Marks

In a seminar, the number of participants in Hindi, English and mathematics are $60, 84$ and $108$ respectively. Find the minimum number of rooms required, if in each room, the same number of participants are to be seated and all of them being in the same subject.

Answer

To find the minimum number of rooms required first find the maximum number of participants that can be accomodated in each room such that the number of participants in each room is same.
This can be determined by finding the HCF of $60, 84$ and $108$.
$60 = 2^2 \times 3 \times 5$
$84 = 2^2 \times 3 \times 7$
$108 = 2^2 \times 3^2$^
HCF $= 2^2 \times 3$
$= 12$
So, the minimum number of rooms required
$=\frac{\text{Total number of participants}}{12}$
$=\frac{60+84+108}{12}$
$=21$

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Question 74 Marks

Find the largest number which divides $546$ and $764$, leaving remainders 6 and 8 respectively.

Answer

$546$ and $764$ are divided by the largest number leaving remainers $6$ and $8$ respectively.
$546 - 6 = 540$
$764 - 8 = 756$
So, 540 and 756 are exactly divisible by the required number.
Thus, the required number is the HCF of 540 and 756
$540 = 2^2 \times 3^3 \times 5$
$756 = 2^2 \times 3^3 \times 7$
HCF$(540, 756) = 2^2 \times 3^3 = 108$
which is the required number.

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Question 84 Marks

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10, 12 minutes respectively. In 30 hours, how many times do they toll together?

Answer

2 = 2 × 1
4 = 2 × 2
6 = 2 × 3
8 = 2 × 2 × 2
10 = 2 × 5
12 = 2 × 2 × 3
L.C.M. of 2, 4, 6, 8, 10, 12 minutes
= 2 × 2 × 2 × 3 × 5 = 120 minutes
= 2 hours
After every 2 hours they toll together,
Required number of times $=\Big(\frac{30}{2}+1\Big)$
$$= 16 times

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Question 94 Marks

Find the HCF and LCM of $\frac{8}{9},\frac{10}{27}$ and $\frac{16}{81}.$

Answer

$\text{HCF}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{\text{HCF of the numerators}}{\text{LCM of the denominators}}$
$=\frac{\text{HCF}(8,10,16)}{\text{LCM}(9,27,81)}$
and
$\text{LCM}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{\text{LCM of the numerators}}{\text{HCF of the denominators}}$
$=\frac{\text{LCM}(8,10,16)}{\text{HCF}(9,27,81)}$
Consider,
$8 = 2^3$
$10 = 2 \times 5$
$16 = 2^4$
So, the HCF$(8, 10, 16) = 2$ and LCM$(8, 10, 16) = 2^4 \times 5 = 80$
$9 = 3^2$
$27 = 3^3$
$81 = 3^4$
So, the HCF$(9, 27, 81) = 3^2 = 9$ and LCM$(9, 27, 81) = 3^4 = 81$
$\Rightarrow\text{HCF}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{2}{81}$ and $\text{LCM}\Big(\frac{8}{9},\frac{10}{27},\frac{16}{81}\Big)=\frac{80}{9}$

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Question 104 Marks

An electronic device makes a beep after every 60 seconds. Another device makes a beep after 62 seconds. They beeped together at 10 a.m. At what time will they beep together at the earliest?

Answer

Interval of beeping together = LCM (60 seconds, 62 seconds)
The prime factorization of 60 and 62:
60 = 30 × 2, 62 = 31 × 2
$\therefore$ L.C.M of 60 and 62 is 30 31 × 2 = 1860 sec = 31min
$\therefore$ electronic device will beep after every 31 minutes
After 10 a.m., it will beep at 10hrs 31 minutes

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Question 114 Marks

Find the maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and the same number of pencils.

Answer

Total pens = 1001
Total pencils = 910
We need to find maximum no.of students among whom 1001 pens and 910 pencils can be distributed in such a way that each students get same no.of pens and pencils.
Then we need to find HCF of 1001 and 910
Prime factorization of,
1001 = 7 × 11 × 13
910 = 2 × 5 × 7 × 13
HCF=product of commom prime factor of least power
HCF = 7 × 13 = 91
Here HCF of 1001 and 910 is 91.
Hence among 91 students 1001 pens and 910 pencils can be distributed such that each student get same no. of pens and pencils.

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