STEP 1: Here, let p = 255, q = 204 and r = 68 Check whether p = q or p > q or p < q. Here p > q.
STEP 2: Let us find the HCF of 255 and 204 first. Now, subtract smaller number from larger number till p = q.

Now p = q, Hence, we conclude that the HCF of 255 and 204 is 51.
STEP 3: Now repeated same procedure for r – HCF (p, q) Now, subtract smaller number from larger number till HCF (p, q) = r.

Now HCF (p, q) = r, Hence, we conclude that the HCF of 255, 204 and 68 is 17. Comparing both the repeated division and repeated subtraction methods, in finding the HCF, we can conclude that the repeated subtraction, in one way is easier and gives the HCF faster that the repeated division and one would want to easely do subtraction rather than division. Isn’t it?

STEP 1: Here , take m = 144 and n = 120 Check whether m = n or m > n or m < n. Here m > n.
STEP 2: Subtract the smaller number from the larger number till m = n.

Now m = n , Hence, we conclude that the HCF of 144 and 120 is 24.

Here, we have to find HCF of 270, 384 and 588
STEP 1: First find the HCF of any two of the given numbers (follow the same step 1, 2 and 3 of the above example). Here, find HCF of (384, 588) first.
STEP 2: The HCF of the first two numbers which is 12 becomes the divisor and the third number 270 becomes the dividend.
STEP 3: Repeat this division process till the remainder becomes zero. The last divisor is the HCF. Here, 6 is the last divisor.
Hence, HCF of 270, 384 and 588 is 6. Therefore, we needs 6 containers so that each of them contains (270 ÷ 6 = 45) 45 ginger chocolates, (384 ÷ 6 = 64) 64 milk chocolates and (588 ÷ 6 = 98) 98 coconut chocolates.

Let the distance be ‘ x ’ km. (Recall that time = \(\frac { Distance }{ Speed } \))
Time taken to cover ‘ x ’ km at 35 km/hr: T₁ = \(\frac { x }{ 35 } hr\)
Time taken to cover ‘ x ’ km at 50 km/hr: T₂ = \(\frac { x }{ 50 } hr\)
According to the problem, the difference between two timings
= 4–(–5)
= 4+5 = 9 minutes
= \(\frac { 9 }{ 60 } \)hour (changing minutes to hour)
Given, T₁ – T₂ =\(\frac { 9 }{ 60 } \)
\(\frac { x }{ 35 } \)-\(\frac { x }{ 50 } \) = \(\frac { 9 }{ 60 } \)
\(\frac { 10x-7x }{ 350 } =\frac { 9 }{ 60 } \)
\(\frac { 3x }{ 350 } =\frac { 9 }{ 60 } \)
x \(=\frac { 9 }{ 60 } \times \frac { 350 }{ 3 } \)
The distance to his office x = 17\(\frac { 1 }{ 2 } \) km.

Let the original fraction be \(\frac { x }{ y } \)
Given that y = x + 3. (Denominator = Numerator + 3).
Therefore, the fraction can be written as \(\frac { x }{ x+3 } \).As per the given condition, \(\\ \frac { x+2 }{ (x+3)+9 } =\frac { 5 }{ 6 } \)
By cross multiplication, 6( x +2) = 5 ( x +3+9)
6 x +12 = 5( x +12)
6 x +12 = 5 x +60
6 x − 5 x = 60 − 12
x = 60 − 12
x = 48.
Therefore, the original fraction is \(\frac { x }{ x+3 } =\frac { 48 }{ 48+3 } =\frac { 48 }{ 51 } \).

Let us assume that the length of the first piece is x cm.
Th en the length of the second piece is (200cm – x cm) i.e., (200 − x) cm.
According to the given statement (change m to cm),
First piece = 40 less than twice the second piece.
x = 2× (200 − x) − 40
x = 400 − 2x − 40
x + 2x = 360
3x = 360
x = \(\frac { 360 }{ 3 } \)
x = 120
Thus the length of the first piece is 120cm and
the length of second piece is 200cm − 120cm = 80cm, which happens to be the smaller.

Let the number of passengers having Rs.8 tickets be y. Then, the number of passengers with Rs.10 tickets is (56−y).
Total money received from the passengers = Rs.500
That is, y × Rs.8 + (56 − y) × Rs.10 = 500
8y +560 −10y = 500
8y−10y = 500 − 560
− 2y = − 60
y = \(\frac { 60 }{ 2 } \)
y = 30
Hence, the number of passengers having,
(i) Rs.8 tickets = 30
(ii) Rs.10 tickets = 56−30 = 26

Given:
DE = 8 cm, ES = 5.5 cm, KS = 5 cm, KD = 6 cm and \(\overset { \_ \_ }{ DE } \) || \(\overset { \_ \_ }{ KS } \)


Steps:
1. Draw a line segment DE = 8cm.
2. Mark the point A on DE such that DA = 5 cm.
3. With A and E as centres, draw arcs of radii 6 cm and 5.5 cm respectively. Let them cut at S. Join AS and ES.
4. With D and S as centres, draw arcs of radii 6 cm and 5 cm respectively. Let them cut at K. Join DK and KS.
5. DESK is the required trapezium.
Calculation of area:
Area of the trapezium DESK = \(\frac { 1 }{ 2 } \) x h x (a + b) sq. units
= \(\frac { 1 }{ 2 } \) x 5.5 x (8 + 5) = 35.75 sq. cm

Given:
CA = 9 cm, ㄥCAR = 70⁰ AR = 6 cm, and CD = 7 cm and \(\overset { \_ }{ CA } \) || \(\overset { \_ }{ DR } \)


Steps:
1. Draw a line segment CA = 9 cm.
2. Construct an angle ㄥCAX = 70⁰ at A.
3. With A as centre, draw an arc of radius 6 cm cutting AX at R.
4. Draw RY parallel to CA.
5. With C as centre, draw an arc of radius 7 cm cutting RY at D.
6. Join CD. CARD is the required trapezium.
Calculation of area:
Area of the trapezium CARD = \(\frac { 1 }{ 2 } \) x h x(a + b) sq.units
= \(\frac { 1 }{ 2 } \) x 5.6 x (9 + 11) = 56 sq. cm

Cost of the food = Rs. 350
Extra 5% GST is equally shared by Central and State Governments at 2.5% each
∴ CGST = SGST = 350 x \(\frac { 2.5 }{ 100 } \) = Rs. 8.75 

Let the number of eggs bought by her be x.
Then, Cost Price = \(\frac { 18 }{ 4 } \times x= \frac { 9x }{ 2 } \)
Selling Price = Rs.\(\frac { 24 }{ 5 } \times x=\frac { 24x }{ 5 } \)
∴ Gain = S.P – C.P = \( \frac { 24x }{ 5 } -\frac { 9x }{ 2 } \)
= \(\frac { 48x-45x }{ 10 } =\frac { 3x }{ 10 } \)
Given, gain = Rs. 90
That is,\(\frac { 3x }{ 10 } \) = Rs. 90
∴ x = \(\frac { 90\times 10 }{ 3 } \) = 300

Given,
Discount = Marked Price –Selling Price
= 1060 – 901
= Rs. 159
∴ Discount = \(\frac { 159 }{ 1060 } \)× 100%
= 15%

Let the C.P of one strawberry box be Rs.  x.
Then C.P of 20 strawberry boxes = 20 x and
S.P of 20 strawberry boxes = C.P of 16 strawberry boxes = 16 x
Thus, S.P < C.P, hence there is a loss.
Loss = C.P –S.P = 20 x − 16 x = 4 x
∴ Loss % =\(\left( \frac { Loss }{ C.P } \times 100 \right) \)%
=\(\left( \frac { 4x }{ 20x } \times 100 \right) \)%
= 20 %

Total C.P of the washing machine
= C.P + Overhead Expenses
= 16150 + 1350 = Rs. 17500
S.P = Rs. 19250
Therefore, we find S.P > C.P.
Gain % = \(\left( \frac { Gain }{ C.P } \times 100 \right) %\)% = \(\left( \frac { 19250-17500 }{ 17500 } \times 100 \right) %\)
=\(\frac { 1750 }{ 17500 } \times 100\) = 10%

Let the C.P of the LED TV be rs x.
∴ S.P = \(\frac { 5 }{ 4 } \)x
Profit = S.P – C.P = \(\frac { 5 }{ 4 } \)x - x = \(\frac { x }{ 4 } \)
∴Profit % =\(\left( \frac { Profit }{ C.P } \times 100 \right) %\)
=\(\left( \frac { { x }/{ 4 } }{ x } \times 100 \right) %\)
=\(\left( \frac { 1 }{ 4 } \times 100 \right) %\) = 25%