Question 13 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 of each would you need to buy?
AnswerNumber of color pencils in one pack = 24
No of crayons in pack = 32
$\therefore$ The least number of both colors to be purchased
= LCM of 24 and 32
= 2 × 2 × 2 × 2 × 2 × 3
$\therefore$ Number of packs of pencils to be bought $= \frac{96}{24} = 4$
And number of packs of crayon to be bought $= \frac{96}{32} = 3$
View full question & answer→Question 23 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 of the room exactly.
AnswerLength of room = 8m 25cm = 825cm
Breadth of room = 6m 75cm = 675cm
Height of room = 4m 50cm = 450cm
$\therefore$ The required longest rod
= HCF of 825, 675 and 450
First consider 675 and 450
By applying Euclid’s division lemma
675 = 450 × 1 + 225
450 = 225 × 2 + 0
$\therefore$ HCF of 675 and 450 = 225
Now consider 225 and 825
By applying Euclid’s division lemma
825 = 225 × 3 + 150
225 = 150 × 1 + 75
150 = 75 × 2 + 0
H.C.F of 825, 675 and 450 = 75
View full question & answer→Question 33 Marks
Prove that the square of any positive integer is of the form $5q, 5q + 1, 5q + 4$ for some integer q.
AnswerLet a be the positive integer, and
Let a = 5m, then
$a^2 = (5m)^2 = 25m^2$
$= 5(5m^2) = 5q$
Where $q = 5m^2$
and $a = (5m + 1) then$
$a^2 = (5m + 1)^2$
$= 25m^2 + 10m + 1$
$= 5(5m^2 + 2m) + 1$
$= 5q + 1 where q = 5m^2 + 2m$
and let $a = 5m + 4, then$
$a^2 = (5m + 4)^2 = 25m^2 + 40m + 16$
$= 25m^2 + 40m + 15 + 1$
$= 5(5m^2 + 8m + 3) + 1$
$= 5q + 1 where q = 5m^2 + 8m + 3$
and $a = 5m + 2, then$
$a^2 = (5m + 2)^2$
$= 25m^2 + 20m + 4$
$= 5(5m^2 + 4m) + 4$
$= 5q + 4 where q = 5m^2 + 4m$
and $a = 5m + 3, then$
$a^2 = (5m + 3)^2 = 25m^2 + 30m + 9$
$= 25m^2 + 30m + 5 + 4$
$= 5(5m^2 + 6m + 1) + 1$
$= 5q + 4 where q = 5m^2 + 6m + 1$
Hence proved.
View full question & answer→Question 43 Marks
Find the HCF of the following pairs of integers and express it as a linear combination of them.
963 and 657
AnswerBy applying Euclid’s division lemma 963 = 657 × 1 + 306 …..(i)
Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306
657 = 306 × 2 + 45 …..(ii)
Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 4.
306 = 45 × 6 + 36 …..(iii)
Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36.
45 = 36 × 1 + 9 …..(iv)
Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9.
36 = 9 × 4 + 0
Therefore, H.C.F = 9
Now 9 = 45 – 36 × 1 [from (iv)]
= 45 - [306 – 45 × 6] × 1 [from (iii)]
= 45 - 306 × 1 + 45 × 6
= 45 × 7 - 306 × 1
= 657 × 7 - 306 × 14 - 306 × 1 [from (ii)]
= 657 × 7 - 306 × 15
= 657 × 7 - [963 – 657 × 1] × 15 [from (i)]
= 657 × 22 - 963 × 15
View full question & answer→Question 53 Marks
Show that the following numbers are irrational.
$3-\sqrt{5}$
AnswerLet assume that $(3-\sqrt{5})$ is rational than there exist co-prime a and b such that$(3-\sqrt{5})=\frac{\text{a}}{\text{b}}\ \dots(1)$
$\Rightarrow\ \sqrt{5}=3-\frac{\text{a}}{\text{b}}$
$\Rightarrow\ \sqrt{5}=\frac{3\text{b}-\text{a}}{\text{b}}\ \dots(2)$
Education (2) shows that $\sqrt{5}$ is rational. This is centradiction
Thus, $(3-\sqrt{5})$ is irrational.
View full question & answer→Question 63 Marks
Prove that $2-3\sqrt{5}$ is an irrational number.
AnswerLet us assume that $2-3\sqrt{5}$ is rational.
Then, there exist positive co primes a and b such that,
$2-3\sqrt{5}=\frac{\text{a}}{\text{b}}$
$2-\frac{\text{a}}{\text{b}}=3\sqrt{5}$
$\frac{2\text{b}-\text{a}}{\text{b}}=3\sqrt{5}$
$\frac{2\text{b}-\text{a}}{3\text{b}}=\sqrt{5}$
This contradicts the fact that $\sqrt{5}$ is an irrational number.
Hence $2-3\sqrt{5}$ is irrational.
View full question & answer→Question 73 Marks
A circular field has a circumference of $360\ km$. Three cyclists start together and can cycle $48, 60$ and $72\ km$ a day, round the field. When will they meet again?
AnswerWe need to find LCM of no. of days taken by cyclist to cover 360 days.
We will find the time taken by each cyclist in covering the distance in order to calculate the time when they meet again?
Number of days $1^{st}$ cyclist took to cover $360\text{km}= \frac{\text{Total distance}}{\text{Distance covered in 1 day}}$
$=\frac{360}{48} = \frac{15}{2}$
Number of days $2^{\text {nd }}$ cyclist took to cover $360 km=\frac{360}{60}=6$
Number of days $3^{\text {rd }}$ cyclist took to cover $360 km=\frac{360}{72}=5$
LCM of (15/2, 6 and 5) $= \frac{\text{LCM of numerators}}{\text{HCF of denominators}}$
$= \frac{30}{1} = 30$
Thus, all of them will meet after 30 days.
View full question & answer→Question 83 Marks
A positive integer is of the form $3q + 1$, q being a natural number. Can you write its square in any form other than $3m + 1, 3m$ or $3m + 2$ for some integer m? Justify your answer.
AnswerNo, by Euclid's Lemma, $b=a q+r, 0 \leq r<a$
Here, $b$ is any positive integer
$a=3, b=3 q+r \text { for } 0 \leq r<3$
So, this must be in the form $3 q, 3 q+1$ or $3 q+2$
Now, $(3q)^2 = 9q^2 = 3m$ [here, $m = 3q^2]$
and $(3q + 1)^2 = 9q^2 + 6q + 1$
$= 3(3q^2 + 2q) + 1 = 3m + 1$
[where, $m = 3q^2 + 2q]$
Also, $(3q + 2)^2 = 9q^2 + 12q + 4$
$= 9q^2 + 12q + 3 + 1$
$= 3(3q^2 + 4q + 1) + 1$
$= 3m + 1 [here, m = 3q^2 + 4q + 1]$
Hence, square of a positive integer is of the form $3q + 1$ is always in the form $3m + 1$ for some integer m.
View full question & answer→Question 93 Marks
Show that the square of an odd positive integer is of the form $8q + 1$, for some integer q.
AnswerTo Prove: that the square of an odd positive integer is of the form $8q + 1$, for some integer q.
Proof: Since any positive integer n is of the form $4m + 1 and 4m + 3$
if $n = 4m + 1
\Rightarrow n^2 = (4m + 1)^2
\Rightarrow n^2 = (4m)^2 + 8m + 1
\Rightarrow n^2 = 16m^2 + 8m + 1
\Rightarrow n^2 = 8m(2m + 1) + 1
\Rightarrow n^2 = 8q + 1(q = m(2m + 1))
If n = 4m + 3
\Rightarrow n^2 = (4m + 3)^2
\Rightarrow n^2 = 16m^2 + 24m + 9
\Rightarrow n^2 = 8(2m^2 + 3m + 1) + 1
\Rightarrow n^2 = 8q + 1(q = (2m^2 + 3m + 1))
$ Hence $n^2$ integer is of the form 8q + 1, for some integer q. View full question & answer→Question 103 Marks
Show that the following numbers are irrational.
$\frac{1}{\sqrt{2}}$
AnswerLet, us assume that $\frac{1}{\sqrt{2}}$ is rational.
Then, there exist positive co primes a and b such that,
$\frac{1}{\sqrt{2}}=\frac{\text{a}}{\text{b}}$
$\frac{1}{\sqrt{2}}=\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$\Rightarrow\ \frac{1}{2}=\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow\ \text{b}^2=2\text{a}^2$
$\Rightarrow\ 2|\text{b}^2(\because2|2\text{a}^2)$
$\Rightarrow\ 2|\text{b}$
$\Rightarrow\ \text{b}=2\text{c}$ for some positive integer c
$\Rightarrow\ 2\text{a}^2=\text{b}^2$
$\Rightarrow\ 2\text{a}^2=4\text{c}^2\ (\because\text{a}=\text{pc})$
$\Rightarrow\ \text{a}^2=2\text{c}^2$
$\Rightarrow\ 2|\text{a}^2\ (\because2|2\text{c}^2)$
$\Rightarrow\ 2|\text{b}$
$\Rightarrow\ 2|\text{b and 2|b}$
This contradicts the fact that a and b are co-primes.
Hence $\frac{1}{\sqrt{2}}$ is irrational
View full question & answer→Question 113 Marks
Find the smallest number which when increased by $17$ is exactly divisible by both $520$ and $468.$
Answergiven that the smallest member which when increased by $17$ is (exactly) by both 520 and 468.
Prime factors of 520 and 468 are
$520 = 2^3 \times 5 \times 13$
$468 = 2^2 \times 3^2 \times 13$
$L.C.M (520, 468) = 2^3 \times 3^2 \times 5 \times 13$
$= 8 \times 9 \times 65$
$= 4680$
Number is increased by $17$ so, smallest number will be
$= L.C.M (520, 468) - 17$
$= 4680 - 17$
$= 4663$
Thus, the required number is $4663.$
View full question & answer→Question 123 Marks
An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
AnswerWe are given that an army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. We need to find the maximum number of columns in which they can march.
Members in army = 616
Members in band = 32.
Therefore,
Maximum number of columns = H.C.F of 616 and 32.
By applying Euclid’s division lemma
616 = 32 × 19 + 8
32 = 8 × 4 + 0
Therefore, H.C.F = 8
Hence, the maximum number of columns in which they can march is 8.
View full question & answer→Question 133 Marks
144 cartons of Coke Cans and 90 cartons of Pepsi Cans are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have?
AnswerGiven that 144 cartons of coke cans and 90 cartons of Pepsi cans are to be stacked in a canteen. If each stack is of the same height and contains cartons of the same drink We need to find the greatest number of cartons, each stack would have.
Given that,
Number of cartons of coke cans = 144.
Number of cartons of Pepsi cans = 90.
Therefore, the greatest number of cartons in one stack = H.C.F. of 144 and 90.
By applying Euclid’s division lemma
144 = 90 × 1 + 54
90 = 54 × 1 + 36
36 = 18 × 2 + 0
H.C.F = 18.
Hence, the greatest number cartons in one stack 18.
View full question & answer→Question 143 Marks
Prove that following numbers are irrationals:
$\frac{2}{\sqrt{7}}$
AnswerLet us assume that $\frac{2}{\sqrt{7}}$ is rational.
Then, there exist positive co primes a and b such that,
$\frac{2}{\sqrt{7}}=\frac{\text{a}}{\text{b}}$
$\sqrt{7}=\frac{2\text{b}}{\text{a}}$
$\sqrt{7}$ is rational number which is a contradication.
Hence $\frac{2}{\sqrt{7}}$ is irrational
View full question & answer→Question 153 Marks
If the HCF of 657 and 963 is expressible in the form 657x + 963x - 15, find x.
Answer9 = 45 - 36
= 45 - (306 - 45 × 6)
= 45 - 306 + 45 × 6
= 45 × 7 - 306 = [657 - (306 × 2)] × 7 - 306
= 657 × 7 - 306 × 14 - 306
= 657 × 7 - 306 × 15
= 657 × 7 - (963 - 657) × 15
= 657 × 7 - 963 × 15 + 657 × 15
= 657 × 22 - 963 × 15
= 657 × 22 + 963 × (-15)
= 657 × x + 963 × (-15)
Comparing, we get
x = 22
View full question & answer→Question 163 Marks
A merchant has $120$ litres of oil of one kind, $180$ litres 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?
AnswerQuantity of oil A = 120 litres
Quantity of oil B = 180 litres
Quantity of oil A = 240 litres
we want to fill oils A, B, and C in tins of the some capacity.
Therefore The greatest capacity of the in the tins at can hold oil. $A, B,$ and $C = H.C.F.$ of $120, 180$ and $240$.
By fundamental Theorem of Arithematic,
$120 = 2^3 \times 3 \times 5$
$180 = 2^2\times 3^2\times 5$
$240 = 2^4 \times 3 \times 5$
$H.C.F = 2^2 \times 3 \times 5 = 4 \times 3 \times 5 = 60$ litres.
The greatest capacity of tin $= 60$ litres.
View full question & answer→Question 173 Marks
For any positive integer n, prove that $n^3 - n$ is divisible by $6.$
AnswerLet $n = 6q or 6q + 1, 6q + 2, 6q + 3 ... 6q + 5$
If $n = 6q$, then
Then $n^3 - n = (6q)^3 - 6q = 216q^3 - 6q$
$= 6 (36q^3 - q)$
Which is divisible by 6
If $n = 6q + 1$, then
$n^3 - n = (6q + 1)^3 - (6q + 1)$
$= 216q^3 + 108q^2 + 18q + 1 - 6q - 1$
$= 216q^3 + 108q^2 + 12q$
$= 6(36q^3 + 18q^2 + 2q)$
Which is also divisible by 6
If $n = 6q + 2,$ then
$n^3 - n = (6q + 2)^3 - (6q + 2)$
$= 216q^3 + 216q^2 + 72q + 8 - 6q - 2$
$= 216q^3 + 216q^2 + 66q + 6$
$= 6(36q^3 + 36q^2 + 11q + 1)$
Which is divisible by $6$
Hence we can similarly, prove that $n^2 - n$ is divisible by $6$ for any positive integer n.
Hence proved. View full question & answer→Question 183 Marks
Show that the square of an odd positive integer can be of the form $6 q+1$ or $6 q+3$ for some integer?
AnswerWe know that any positive integer can be of the form $6 m, 6 m+1,6 m+2,6 m+3,6 m+4$ or $6 m+5$, for some integer $m$. Thus, an odd positive integer can be of the form $6 m+1,6 m+3$, or $6 m+5$ Thus we have: $(6 m+1)^2=36 m^2+12 m+1=6\left(6 m^2+2 m\right)+1=6 q+1, q$ is an integer. $(6 m+3)^2=36 m^2+36 m+9=6\left(6 m^2+6 m+1\right)+3=6 q+3, q$ is an integer. $(6 m+5)^2=36 m^2+60 m+25=6\left(6 m^2+10 m+4\right)+1=6 q+1, q$ in an integer.
Thus, the square of an odd positive integer can be of the form $6 q+1$ or $6 q+3$.
View full question & answer→Question 193 Marks
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
AnswerWe need to find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
The required number when divides 280 and 1245, leaves remainder 4 and 3, this means 280 - 4 = 276 and 1245 - 3 = 1242 are completely divisible by the number.
Therefore, the required number = H.C.F. of 276 and 1242.
By applying Euclid’s division lemma.
1242 = 276 × 4 + 138
276 = 138 × 2 + 0
Therefore, H.C.F. = 138.
Hence, the required number is 138.
View full question & answer→Question 203 Marks
Show that the following numbers are irrational.
$7\sqrt{5}$
AnswerLet us assume that $7\sqrt{5}$ is rational.
Then , there exist positive co primes a and b such that,
$7\sqrt{5}=\frac{\text{a}}{\text{b}}$
$\sqrt{5}=\frac{\text{a}}{7\text{b}}$
We know that $\sqrt{5}$ is an irrational number.
Here we see that $\sqrt{5}$ is a rational number which is a contradiction.
View full question & answer→Question 213 Marks
Show that the following numbers are irrational.
$6+\sqrt{2}$
AnswerLet us assume that $6+\sqrt{2}$ is a rational number.
$\therefore\ 6+\sqrt{2}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-primes numbers.
$\Rightarrow\sqrt{2}=\frac{\text{a}}{\text{b}}-6$
$\Rightarrow\sqrt{2}=\frac{\text{a}-6\text{b}}{\text{b}}$
We know that $\sqrt{2}$ is an irrational number.
This contradicts our assumption that $6+\sqrt{2}$ is a rational number.
Hence, $\sqrt{2}$ must be irrational.
View full question & answer→Question 223 Marks
A rectangular courtyard is $18\ m 72\ cm$ long and $13\ m 20\ cm$ broad. it is to be paved with square tiles of the same size. Find the least possible number of such tiles.
AnswerLength of the yard $= 18m 72cm = 1800cm + 72cm = 1872cm$
Breadth of the yard $= 13m 20cm = 1300cm + 20cm = 1320cm$
Area of the yard $= 1872 \times 1320 = 2471040$
The size of the square tile of same size needed to the pave the rectangular yard is equal the HCF of the length and breadth of the rectangular yard.
Prime factorisation of length and breadth
$1872 = 2^4 \times 3^2 \times 13$
$1320 = 2^3 \times 3 \times 5 \times 11$
HCF of 1872 and$ 1320 = 2^3 \times 3 = 24$
Therefore, length of side of the square tile = 24cm
Area of the tile $= 24 \times 24 = 576cm^2$^
Number of tiles required $=\frac{\text{Area of courtyard}}{\text{Area of each tile}}$
$= \frac{2471040}{576} = 4290.$
View full question & answer→Question 233 Marks
Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.
AnswerThe require number when divides 285 and 1249, leaves remainder 9 and 7, this means 285 - 9 = 276 and 1249 - 7 = 1242 are completely divisible by the number
$\therefore$ The required number = HCF of 276 and 1242
1242 = 276 × 4 + 138
276 = 138 × 2 + 0
$\therefore$ H.C.F = 138
Hence remainder is = 0
Hence required number is 138
View full question & answer→Question 243 Marks
Show that $5-2\sqrt{3}$ is an irrational number.
AnswerLet us assume that $5-2\sqrt{3}$ is a rational number.
$\therefore\ 5-2\sqrt{3}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-prime number.
$\Rightarrow\ 5-\frac{\text{a}}{\text{b}}=2\sqrt{3}$
$\Rightarrow\ \frac{5\text{b}-\text{a}}{\text{b}}=2\sqrt{3}$
$\Rightarrow\ \frac{5\text{b}-\text{a}}{2\text{b}}=\sqrt{3}$
We know that $\sqrt{3}$ is an irrational number.
This contradicts our assumption that $5-2\sqrt{3}$ is a rational number.
Hence, $5-2\sqrt{3}$ must be irrational.
View full question & answer→Question 253 Marks
Find the smallest number which leaves remainders $8$ and $12$ when divided by $28$ and $32$ respectively.
AnswerTo Find: The smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
L.C.M of 28 and 32.
$28=2^2 \times 7$
$32=2^5$
L.C.M of 28 , and $32=2^5 \times 7$
$=224$
Hence 224 is the least number which exactly divides 28 and 32 i.e. we will get a remainder of 0 in this case. But we need the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
Therefore,
$=224-8-12$
$=204$
Hence $=204$ is the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.
View full question & answer→Question 263 Marks
Prove that for any prime positive integer $\text{p},\sqrt{\text{p}}$ is an irrational number.
AnswerGiven that p is a prime positive integer.
Let us assume p is rational number and there exist two positive integer a and b which are co-prime such that
$\sqrt{\text{p}}=\frac{\text{a}}{\text{b}}\ \dots(1)$
squaring both sides,
$\Rightarrow\ \text{p}=\frac{\text{a}^2}{\text{b}^2}$
$\Rightarrow\ \text{pb}^2=\text{a}^2\ \dots(2)$
$\Rightarrow\ \text{p|a}^2$
$\Rightarrow\ \text{p|a}\ \dots(3)$
Let a = pc some positive integer c.
put a = pc in equation (2)
$\Rightarrow\ \text{pb}^2=\text{p}^2\text{c}^2$
$\Rightarrow\ \text{b}^2=\text{pc}^2$
$\Rightarrow\ \text{p|b}^2$
$\Rightarrow\ \text{p|b}\ \dots(4)$
From equation (3) and (4), p is common factor of a and b.
This is the constraction of our assumption because a and b are co-prime (no common factor other than.)
Thus, $\sqrt{\text{p}}$ is an irrational numbers.
View full question & answer→Question 273 Marks
Prove that the square of any positive integer of the form $5q + 1$ is of the same form.
AnswerLet $n = 5q + 1$ where q is a positive integer
$\therefore$ $n^2 = (5q + 1)^2$
$= 25q^2 + 10q + 1$
$= 5(5q^2 + 2q) + 1$
$= 5m + 1$, where m is some integer
Hence, the square of any positive integer of the form $5q + 1$ is of the same form.
View full question & answer→Question 283 Marks
Prove that following numbers are irrationals:
$4+\sqrt{2}$
AnswerLet us assume that $4+\sqrt{2}$ is rational number.
$\therefore\ 4+\sqrt{2}=\frac{\text{a}}{\text{b}}$ where, a and b are positive co-prime numbers.
$\Rightarrow\ \sqrt{2}=\frac{\text{a}}{\text{b}}-4$
$\Rightarrow\ \sqrt{2}=\frac{\text{a}-4\text{b}}{\text{b}}$
We know that $\sqrt{2}$ in an irratonal.
This contradicts our assumption that $4+\sqrt{2}$ is a rational number.
Hence, $4+\sqrt{2}$ must be irrational.
View full question & answer→Question 293 Marks
Find the $LCM$ and $HCF$ of the following integer by applying the prime factorisation method.
$84, 90$ and $120$
Answer$84, 90$ and $120$
Prime factor of 84, 90 and 120 are,
$84 = 2 \times 2 \times 3 \times 7 = 2^2 \times 3 \times 7$
$90 = 2 \times 3 \times 3 \times 5 = 2 \times 3^2 \times 5$
$120 = 2 \times 2 \times 2 \times 3 \times 5 = 2^3 \times 3 \times 5$
For H.C.F:
|
Common prime factor
|
Least component
|
|
2
|
1
|
|
3
|
1
|
H.C.F $(84, 90, 120) = 2 × 3 = 6$
For L.C.M:
|
Prime factor of 84, 90, 120
|
Greatest component
|
|
2
|
3
|
|
3
|
2
|
|
5
|
1
|
|
7
|
1
|
$L.C.M (84, 90, 120) = 2^3 \times 3^2 \times 5 \times 7$
$= 72 \times 35$
$= 2520$
Thus, H.C.F = 6 and L.C.M = 2520 View full question & answer→Question 303 Marks
Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.
AnswerThe required number when divides 2011 and 2623 leaves remainders 9 and 5 this means 2011 - 9 = 2002 and 2623 - 5 = 2618 are completely divisible by the number.
Therefore, the required number = H.C.F. of 2002 and 2618
By applying Euclid’s division lemma
2618 = 2002 × 1 + 616
2002 = 616 × 3 + 154
616 = 154 × 4 + 0
H.C.F of 2002 and 2618 = 154
Hence, the required number is 154.
View full question & answer→Question 313 Marks
Find the LCM and HCF of the following integer by applying the prime factorisation method.
$40, 36$ and $126$
AnswerLet us first find the factors of $40, 36$ and $126$
$40 = 2^3 \times 5$
$36 = 2^2 \times 3^2$
$126 = 2 \times 3^2 \times 7$
$L.C.M$ of $40, 36$ and $126 = 2^3 \times 3^2 \times 5 \times 7$
$L.C.M$ of $40, 36$ and $126 = 2520$
$H.C.F$ of $40, 36$ and $126 = 2$
View full question & answer→Question 323 Marks
Can two numbers have 16 as their H.C.F. and 380 as their L.C.M? Give reason.
AnswerH.C.F. of two numbers = 16 and their L.C.M. = 380
We know the H.C.F. of two numbers is a factor of their L.C.M. but 16 is not a factor of 380 or 380 is not divisible by 16.
It can not be possible.
View full question & answer→Question 333 Marks
Prove that following numbers are irrationals:
$\frac{3}{2\sqrt{5}}$
AnswerLet us assume that $\frac{3}{2\sqrt{5}}$ is rational. Then, there exist positive co-primes a and b such that, $\frac{3}{2\sqrt{5}}=\frac{\text{a}}{\text{b}}$ $\sqrt{5}=\frac{3}{2}\frac{\text{b}}{\text{a}}$ $\sqrt{5}$ is rational. This is a contradication.Thus $\frac{3}{2\sqrt{5}}$ is irrational.
View full question & answer→Question 343 Marks
Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
AnswerThe required number when divides 445, 572 and 699 leaves remainders 4, 5 and 6
This means 445 - 4 = 441, 572 - 5 = 567 and 699 - 6 = 693 are completely divisible by the number
$\therefore$ The required number = HCF of 441, 567 and 693
First consider 441 and 567
By applying Euclid’s division lemma
567 = 441 × 1 + 126
441 = 126 × 3 + 63
126 = 63 × 2 + 0
$\therefore$ HCF of 441 and 567 = 63
Now consider 63 and 693
By applying Euclid’s division lemma
693 = 63 × 11 + 0
$\therefore$ HCF of 441, 567 and 693 = 63
Hence required number is 63.
View full question & answer→Question 353 Marks
Find the $LCM$ and $HCF$ of the following integer by applying the prime factorisation method.
$24, 15$ and $36$
AnswerLet us first find the factors of $24, 15$ and $36$.
$24 = 2^3 \times 3$
$15 = 3 \times 5$
$36 = 2^2 \times 3^2$
$L.C.M$ of $24, 15$ and $36 = 2^3 \times 3^2 \times 5$
$L.C.M$ of $24, 15$ and $36 = 360$
$H.C.F$ of $24, 15$ and $36 = 3$
View full question & answer→Question 363 Marks
Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
AnswerWe need to express the H.C.F. of 468 and 222 as 468x + 222y
Where x, y are integers in two different ways.
Given integers are 468 and 222, where 468 > 222
By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24.
Since the remainder ≠ 0, so apply division lemma on divisor 222 and remainder 24.
222 = 24 × 9 + 6.
Since the remainder ≠ 0, so apply division lemma on divisor 24 and remainder 6.
24 = 6 × 4 + 0.
We observe that remainder is 0. So the last divisor 6 is the H.C.F. of 468 and 222 from we have
6 = 222 - 24 × 9
⇒ 6 = 222 - (468 - 222 × 2) × 9 [Substiuting 24 = 468 - 222 × 2]
⇒ 6 = 222 - 468 × 9 + 222 × 18
⇒ 6 = 222 × 19 - 468 × 9
⇒ 6 = 222y + 468x, where x = -9 and y = 19
View full question & answer→Question 373 Marks
A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required?
AnswerA mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10ft. by 8ft. We need to find the size in inches of the tile required that has to be cut and number of such tiles are required.
Size of bathroom = 10ft by 8ft
= (10 × 12) inch by (8 × 12) inch
= 120 inch by 96 inch
The largest size of tile required = H.C.F. of 120 and 96.
By applying Euclid’s division lemma
120 = 96 × 1 + 24
96 = 24 × 4 + 0
Therefore, H.C.F. = 24.
Thus, largest size of tile required = 24 inches.
Therefore,
No, of tiles required $= \frac{\text{Area of bathroom}}{\text{Area of 1 tile}}$
$=\frac{120 \times 96}{24 \times 24}$
$= 5 \times 4$
$= 20 \text{ tiles}.$
View full question & answer→Question 383 Marks
Write the exponent of $2$ in the price factorization of $144.$
AnswerUsing the factor tree for prime factorization, we have:
$144 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 144 = 2^4 \times 3^2$
Hence the exponent of 2 in 144 is 4. View full question & answer→Question 393 Marks
Show that the square of any positive integer cannot be of the form $3 m+2$, where m is a natural number.
AnswerBy Euclid's lemma, $b=a q+r, 0 \leq r<a$.
Here, b is a positive integer and $a =3$.
$\therefore b=3 q+r \text {, for } 0 \leq r<3$
This must be in the form $3 q, 3 q+1$ or $3 q+2$.
Now,
$(3 q)^2=9 q^2=3 m$, where $m=3 q^2$
$(3 q+1)^2=9 q^2+6 q+1=3\left(3 q^2+2 q\right)+1=3 m+1$, where $m=3 q^2+2 q$
$(3 q+2)^2=9 q^2+12 q+4=3\left(3 q^2+4 q+1\right)+1=3 m+1$, where $m=3 q^2+4 q+1$
Therefore, the square of a positive integer cannot be of the form $3 m+2$, where $m$ is a natural number.
View full question & answer→Question 403 Marks
Find the HCF of the following pairs of integers and express it as a linear combination of them.
506 and 1155
AnswerGiven that two positive integers are 506 and 1155 and 1155 > 506.
So. applying Euclid's division algorithem
1155 = 506 × 2 + 143 …..(i)
Here, remainder 143 ≠ 0, so again apply euclid's division algorithem on divisor 506 and remainder 143.
506 = 143 × 3 + 77 …..(ii)
Here, remainder 77 ≠ 0, so again apply euclid's division algorithem on divisor 143 and remainder 77.
143 = 77 × 1 + 66 …..(iii)
Here, remainder 66 ≠ 0, so again apply euclid's division algorithem on divisor 77 and remainder 66.
77 = 66 × 1 +11 …..(iv)
Here, remainder 11 ≠ 0, so again apply euclid's division algorithem on divisor 66 and remainder 11.
66 = 11 × 6 + 0 …..(v)
Now, remainder = 0.
So, tha division of this stage and remainder of previous stage i.e., 11 is HCF of 506 and 1155.
HCF (506, 1155) = 11
Linear Combination:
11 = 16 × 506 + (-7) × 1155
View full question & answer→Question 413 Marks
Show that $2-\sqrt{3}$ is an irrational number.
AnswerLet us assume that $2-\sqrt{3}$ is rational.
Then, there exist positive co primes a and b such that,
$2-\sqrt{3}=\frac{\text{a}}{\text{b}}$
$\sqrt{3}=2-\frac{\text{a}}{\text{b}}$
This implies, $\sqrt{33}$ is a rational number, which is a contradiction.
Hence, $2-\sqrt{3}$ is irrational number
View full question & answer→Question 423 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?
AnswerNumber of chocolates of $1^{st}$ brand in one pack $= 24$
Number of chocolates of $2^{nd}$ b and in one pack $= 15$
$\therefore$ The least number of chocolates 1 need to purchase
$24 = 2 \times 2 \times 2 \times 3$
$15 = 3 \times 5$
So, that $LCM$ of $24$ and $15$
$= 2 \times 2 \times 2 \times 3 \times 5$
$= 120$
$\therefore$ The number of packet of 1st brand $=\frac{120}{24}=5$
And the number of packet of 2nd brand $=\frac{120}{15}=8$
View full question & answer→Question 433 Marks
Prove that $4-5\sqrt{2}$ is an irrational number.
AnswerLet $4-5\sqrt{2}$ is not are irrational number.
and let $4-5\sqrt{2}$ is a rational number.
and $4-5\sqrt{2}=\frac{\text{a}}{\text{b}}$ where a and b are positive prime integers,
$\Rightarrow\ 4-\frac{\text{a}}{\text{b}}=5\sqrt{2}$
$\Rightarrow\ \frac{4\text{b}-\text{a}}{\text{b}}=5\sqrt{2}$
$\Rightarrow\ \frac{4\text{b}-\text{a}}{5\text{b}}=\sqrt{2}$
$\sqrt{2}$ is a rational number.
But $\sqrt{2}$ is an irrational number.
Our supposition is wrong.
$4-5\sqrt{2}$ is an irrational number.
View full question & answer→Question 443 Marks
15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain?
AnswerNumber of pastries = 15
Number of biscuit packets = 12
$\therefore $ The required no of boxes to contain equal number = HCF of 15 and 12
By applying Euclid’s division lemma
15 = 12 × 1 + 3
12 = 3 × 4 + 0
$\therefore $ No. of boxes required = 3
Hence each box will contain $\frac{15}{3}=5$ pastries and $\frac{2}{3}$ biscuit packets.
View full question & answer→Question 453 Marks
Prove that $\sqrt{2}+\sqrt{3}$ is an irrational number.
AnswerLet us suppose that $\sqrt{2}+\sqrt{3}$ is rational.
Let $\sqrt{2}+\sqrt{3}=\text{a},$ where a is rational.
Therefore, $\sqrt{2}=\text{a}-\sqrt{3}$
Squaring on both sides, we get
$2=\text{a}^2+3-2\text{a}\sqrt{3}$
Therefore,
$\sqrt{3}=\frac{\text{a}^2+1}{2\text{a}}$
which is a contradiction as the right hand side is a rational number while $\sqrt{3}$ is irrational.
Hence, $\sqrt{2}+\sqrt{3}$ is irrational.
View full question & answer→Question 463 Marks
Find the HCF of the following pairs of integers and express it as a linear combination of them.
592 and 252
AnswerWe need to find the H.C.F. of 592 and 252 and express it as a linear combination of 592 and 252.
By applying Euclid’s division lemma
592 = 252 × 2 + 88
Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 88.
252 = 88 × 2 + 76
Since remainder ≠ 0, apply division lemma on divisor 88 and remainder 76.
88 = 76 × 1 + 12
Since remainder ≠ 0, apply division lemma on divisor 76 and remainder 12.
76 = 12 × 6 + 4
Since remainder ≠ 0, apply division lemma on divisor 12 and remainder 4.
12 = 4 × 3 + 0
Therefore, H.C.F. = 4.
Now,
4 = 76 - 12 × 6
= 76 - [88 - 76 × 1] × 6
= 76 - 88 × 6 + 76 × 6
= 76 × 7 - 88 × 6
= (252 - 88 × 2) × 7 - 88 × 6
= 252 × 7 - 88 × 14 - 88 × 6
= 252 × 7 - 88 × 20
= 252 × 7 - [592 - 252 × 2] × 20
= 252 × 7 - 592 × 20 + 252 × 40
= 252 × 47 - 592 × 20
= 252 × 47 + 592 × (-20)
View full question & answer→Question 473 Marks
What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.
AnswerThe reqwired number when divises 626, 3127 and 157628, leaves remainder 1, 2 and 3.
This meams 626 - 1 = 625, 3127 - 2 = 3125 and 15628 - 3 = 15625 are completely divisible by the number
$\therefore$ The required number = H.C.F of 625, 3125 and 15625
First consider 625 and 3125
By applying Euclid’s division lemma
3125 = 625 × 5 + 0
H.C.F of 625 and 3125 = 625
Now consider 625 and 15625
By applying Euclid’s division lemma
15625 = 625 × 25 + 0
$\therefore$ H.C.F of 625, 3125 and 15625 = 625
Hence required number is 625.
View full question & answer→Question 483 Marks
Find the greatest number of $6$ digits exactly divisible by $24, 15$ and $36.$
Answergreatest number of 6 digit = 999999
Prime factor of 24, 15 and 36 are
$24 = 2^3 \times 3$
$15 = 3 \times 5$
$36 \times 2^2 \times 3^2$
$L.C.M (24, 15, 36) = 2^3 \times 3^2 \times 5$
$= 8 \times 9 \times 5$
$= 360$
Required number must be divisible by LCM of 24, 15 and 36 i.e., 360.
Required number = 999999 - Remainder when 999999 is divided by 360.
$= 999999 - (999999 mod 360)$
$= 999999 - 279$
$= 999720$
View full question & answer→Question 493 Marks
Prove that following numbers are irrationals:
$5\sqrt{2}$
AnswerLet us assume that $5\sqrt{2}$ is rational.
Then, there exist positive co-primes a and b such that,
$5\sqrt{2}=\frac{\text{a}}{\text{b}}$
$\sqrt{2}=\frac{\text{a}}{5\text{b}}$
$\sqrt{2}$ is a rational number which is a contradication.
Hence $5\sqrt{2}$ is irrational.
View full question & answer→Question 503 Marks
Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
AnswerWe need to find the largest number which divides 615 and 963 leaving remainder 6 in each case.
The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.
Therefore,
The required number = H.C.F. of 609 and 957.
By applying Euclid’s division lemma
957 = 609 × 1 + 348
609 = 348 × 1 + 261
348 = 216 × 1 + 87
261 = 87 × 3 + 0.
Therefore, H.C.F. = 87.
Hence, the required number is 87.
View full question & answer→