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?

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 two digit number be xy (i.e., ten’s digit is x, ones digit is y)
Its value can be expressed as 10 x + y.
Given, x + y = 8 which gives y = 8 − x
Therefore its value is 10 x + y
= 10x + 8 − x
= 9x + 8.
The new number is yx with value is 10y + x
= 10(8 − x) + x
= 80 – 9x
Given, when 18 is added to the given number (xy) gives new number (yx)
(9x + 8) + 18 = 80 – 9x
This simplifies to 9x + 9x = 80 – 8 – 18
18x = 54
x = 3 ⇒ y = 8 – 3 = 5
The two digit number is xy = 35

Given condition: After two years, Mother’s age = 4 times of Daughter's age
5 x +2 = 4 ( x +2)
5 x +2 = 4 x +8
5 x − 4x = 8 − 2
x = 6
Hence daughter’s present age = 6 years;
and mother’s present age = 5 x = 5 × 6 = 30 years 

Let the breadth of the field be ‘ x ’ metres; then its length (x + 9) metres.
Perimeter of the P = 2(length + breadth) = 2(x + 9 + x) = 2(2x + 9)
Given that, 2(2x + 9) = 154.
4x + 18 = 154
4x = 154 − 18
4x = 136
x = 34
Th us, Breadth of the rectangular fi eld = 34m
Length of the rectangular fi eld = x + 9 = 34 + 9 = 43m

Let the smaller number be x and the greater number be x + 8
Given: the sum of two numbers = 36
x + (x+8) = 36
2 x + 8 = 36
2 x = 36 − 8
2 x = 28
x = \(\frac { 28 }{ 2 } \)
x = 14
The smaller number, x = 14
The greater number, x + 8 = 14 + 8 = 22

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:
DE = 7 cm, EA = 6.5 cm ㄥEDN = 100⁰ and ㄥDEA = 70⁰ and \(\overset { \_ \_ }{ DE } \) || \(\overset { \_ \_ }{ NA} \)


Steps:
1. Draw a line segment DE = 7cm.
2. Construct an angle ㄥDEX = 70⁰ at E.
3. With E as centre draw an arc of radius 6.5cm cutting EX at A.
4. Draw AY parallel to DE.
5. Construct an angle ㄥEDZ = 100⁰ at D cutting AY at N.
6. DEAN is the required trapezium.
Calculation of area:
Area of the trapezium DEAN = \(\frac { 1 }{ 2 } \) x h x (a+b) sq. units
= \(\frac { 1 }{ 2 } \) x 6.1 x (7 + 5.8) = 39.04 sq. units

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

Given:
BO = 7cm, OA = 6cm, BA = 10cm,
TA = 6 cm and \(\overset { - }{ BO } \) || \(\overset { - }{ TA } \)


​​​​​​Steps:
1. Draw a line segment BO = 7 cm.
2. With B and O as centres, draw arcs of radii 10cm and 6cm respectively and let them cut at A.
3. Join BA and OA.
4. Draw AX parallel to BO
5. With A as centre, draw an arc of radius 6cm cutting AX at T.
6. Join BT. BOAT is the required trapezium.
Calculation of area:
Area of the trapezium BOAT = \(\frac { 1 }{ 2 } \) x h x (a+b) sq units
= \(\frac { 1 }{ 2 } \) x 5.9 x (7+6) = 38.35 sq. cm

Let the marked price be Rs x.
Now, x + \(\frac { 18x }{ 100 } \) = 10502
\(\frac { 118x }{ 100 } \) = 10502
∴ Marked price, x = Rs. 8900.
GST at 18% = 8900 x \(\frac { 18 }{ 100 } \)
= Rs. 10502 – Rs. 8900
= Rs.1602

Let the marked price of an article be Rs. 100.
First discount of 25% = 100 x \(\frac { 25 }{ 100 } \)= Rs. 25
∴ Price aft er fi rst discount = 100 − 25 = Rs. 75.
Second discount of 20% = 75 ×\(\frac { 20 }{ 100 } \) = Rs. 25.
∴ Price aft er second discount = 75 − 15 = Rs. 60.
Net selling price = Rs. 60.
∴ Single discount equivalent to two given successive discounts = (100-60)% = 40%.

Marked price = Rs. 550 and Discount = 8%
∴ Discount = \(\frac { 8 }{ 100 } \)x 550 = Rs. 44
∴ Selling Price = Marked price – Discount
= Rs. 550 − 44
= Rs. 506
∴ The selling price of the tube light is Rs. 506

Total C.P of the car
= C.P + Overhead Expenses
= 24000 + 15000 +8500
= Rs. 263500
S.P = Rs. 258230
As S.P < C.P , there is a loss.
∴ Loss % = \(\left( \frac { loss }{ C.P } \times 100 \right) %\)%
= \(\left( \frac { 263500-258230 }{ 263500 } \times 100 \right) %\)%
= \(\left( \frac { 5270 }{ 263500 } \times 100 \right) %\)%
= 2%
∴ Loss percentage = 2%

S.P of the bicycle = Rs. 4275
Loss = 5%
∴ C.P =\(\frac { 100 }{ 100-loss% } \times S.P\)
= \(\frac { 100 }{ 95 } \times 4275\)
= Rs. 4500
Now,
C.P = Rs. 4500 and the desired profi t = 5%
∴ Desired S.P = \(\frac { 100+gain% }{ 100 } \times C.P\)
=\(\frac { 100+5 }{ 100 } \times 4500\)
= 105 x 45
= Rs. 4725
Hence, the desired selling price is Rs. 4725.