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GIVEN :

C.P of canvas = Rs. 500

M.P of canvas = Rs. 800

Discount = 10%

CONCEPT :

Discount & Marked price.

FORMULA USED :

1) Profit = S.P - C.P

2) Profit% = (Profit / C.P) × 100

3) S.P = {(100 - D%) / M.P} × 100

CALCULATION :

After allowing a discount of 10% we get,

⇒ 90% of 800

⇒ (90 / 100) × 800

⇒ 720

∴ S.P = Rs. 720

Profit = S.P - C.P

⇒ 720 - 500

⇒ 220

Profit% = (Profit / C.P) × 100

⇒ (220 / 500) × 100

⇒ 44%

Given:

MP of the article = Rs 840

Discount, (D) = 15%

Profit, (P) = 19%

Formula used:

Selling price, (SP) = Marked price × (100 – D%)/100

Cost price, (CP) = (Selling price × 100)/(100 + P%)

Concept used:

The discount is given on the marked price

Calculation:

SP = Marked price × (100 – D%)/100

⇒ 840 × (100 - 15)/100

⇒ 840 × (85)/100

⇒ 714

Now, CP = (Selling price × 100)/(100 + P%)

⇒ (714 × 100)/(100 + 19)

⇒ (714 × 100)/(119)

⇒ 600

∴ The cost price of the article is Rs 600

Given-

Discount% = 10%

Profit% = 25%

CP = 360

Formula Used-

Discount% = (MP – SP)/MP × 100  [where MP = Marked Price and SP = Selling Price] 

Profit% = (CP - SP)/CP × 100       [where CP = Cost Price and SP = Selling Price]

Calculation-

Let the Marked Price be 100x

SP = 100x(1 - 10/100)

⇒ 90x

According to Condition -

90x = 360 × (1 + 25/100)

⇒ 90x = 360 × 5/4

⇒ 90x = 450

⇒ x = 5

∴ Marked Price of the article = 100 × 5

⇒ Rs. 500 

Given:

Market price = cost price + 30% of cost price

Discount = 25%

Formula used:

Profit = selling price – cost price, if Selling price is more than cost price

Loss = cost price – selling price, if selling price is less than cost price

Selling price = market price – discount

Calculation:

Let the cost price is Rs. 100

Market price = Rs. 130

Selling price = Rs. 130 – 25% of 130

⇒ Rs. 97.5

Selling price is less than cost price. So, there is loss

Loss = Rs. 100 – Rs.97.5

GIVEN:

MP = 45% above CP

Discount = 20%

Concept used:

Selling Price =  Market price[(100 - Discount)/100]

CALCULATION:
let the CP of goods be RS 100

MP = Rs 145

SP after discount = 80% of 145 = RS 116

Profit = 116 - 100 = Rs16

profit% = 16/100 = 16%

∴ There is a profit of 16%

Given:

MP of TV = Rs. 56000

Service tax fee = 2% of the item purchased

Discounts of 50% and 10% are offered.

Formula Used:

Discount = Marked Price – Selling Price

Discount Percentage = (Discount/Marked price) x 100

Calculation:

After the successive discounts of 50% and 10%

Selling Price of TV will be = Rs.(56000 × 0.50 × 0.9)

⇒ Rs. 25200

After using E-pay and paying 2% service tax,

The Selling Price of TV = Rs. 25200(1+0.02)

⇒ Rs. 25704

∴ Akash has to pay Rs. 25,704 for the HD television

Given:

Mark up = 20%

Discount = 10%

Measure 100 grams more

Profit = Rs. 1.8/kg

Formula Used:

Marked price = Cost price × (100 + mark up%)/100

Selling price = Marked price × (100 - discount%)/100

Profit = SP - CP

Calculation:

Let the cost price of 1 kg of sugar be Rs. x

He uses a faulty balance which reads 100 grams more than actual weight

Instead of 1 kg he sells only 900 grams

Cost price of 900 grams of sugar = Rs. (900/1000) × x

⇒ Rs. 0.9x

He marked the price to 20%

Marked Price of 1 kg sugar = Cost price × (100 + mark up%)/100

⇒ x × (100 + 20)/100

⇒ Rs. 1.2x

He gives a discount of 10%

Selling price of 1 kg sugar = Marked price × (100 - discount%)/100

⇒ 1.2x × (100 - 10)/100

⇒ Rs. 1.08x

Profit = SP - CP

⇒ 1.8 = 1.08x - 0.9x

⇒ 0.18x = 1.8

⇒ x = 10

∴ Cost price of 1 kg of sugar is Rs. 10.

Given:

Cost price of the article = Rs. 360

Profit = 25%

Discount = 10%

Formula used:

M.P/C.P = (100 + Profit%)/(100 – Discount%)

Calculation:

M.P/C.P = (100 + Profit%)/(100 – Discount%)

M.P/360 = (100 + 25)/(100 - 10)

⇒ M.P = (125 × 360)/90

⇒ M.P = Rs. 500

Calculation:

Let cost price of goods be Rs.100

Marked price of goods = 100 × 132/100 = Rs.132

Selling price when discount is 20% = 132 × 80/100 = Rs.105.6

Selling price when discount is 40% = 132 × 60/100 = Rs.79.2

Total selling price of goods = 3/5 × 132 + 1/5 × 105.6 + 1/5 × 79.2 = Rs.116.16

∴ Profit percentage = (116.16 - 100)/100 × 100 = 16.16%

Given:

Marked up % = 50%

Discount given on MP = 20%

Weight used = 950 gm

Concept:

Assume the original weight be 1000 gm and the CP of 1 gm be Rs. 1 and continue.

Formula used:

Profit % = [(SP – CP)/CP] × 100

Calculation:

Let the original weight = 1000 gm

Weight used by shopkeeper = 950 gm

Let the CP of 1 gm = Rs. 1.

∴ CP of 1000 gm = Rs. 1000

Marked up Price = (150/100) × 1000

= Rs. 1500

Discount offered = 20% of MP

= (20/100) × 1500

= Rs. 300

∴ Selling Price = Rs. 1500 - Rs. 300

= Rs. 1200

∵ He sold 950 gm only,

∴ CP of 950 gm = Rs. 950.

∴ Profit = SP - CP

= Rs. 1200 - Rs. 950

= Rs. 250

∴ Profit % = [(SP – CP)/CP] × 100

= (250/950) × 100

= 26(6/19)%

Calculation:

Let the cost price of the article be Rs.100

Then, the marked price of the article = 100 × (125/100) = Rs.125

Discount of 10% given on marked price 

Selling price of the article = 125 × (90/100) = Rs.112.5

Profit percentage = (SP – CP/CP) × 100 

∴ Profit percentage = {(112.5 – 100)/100} × 100 = 12.5%

Give

A shopkeeper marks his goods at 20% above the cost price.

He sells three - fourth of the goods at the marked price and the remaining at 30% discount on the marked price.

Formula used

MP(1 - D/100) = SP

MP = Marked Price

SP = Selling Price

D = Discount

Calculation

Let CP be Rs.x

⇒ MP = 1,2x

⇒ SP of 3/4 goods = (3/4) × 1.2x

⇒ 0.9x

⇒ SP of remaining 1/4 goods = (70/100) × (1/4) × 1.2x

⇒ 0.21x

⇒ Total SP = 0.9x + 0.21x = 1.11x

⇒ Gain = 1.11x - x = 0.11x

⇒ Gain% = .11x/x × 100

∴ Gain% is 11%

Given:

Discount, (D) = 20%

Formula used:

Selling price = Marked price × (100 – D%)/100

Profit % = (Profit/CP) × 100

Successive profit = P₁ + P₂ + (P₁P₂)/100

Here, P₁ and P₂ is profit

Calculation:

Let the cost price of the goods be 100x

Then, MP = CP + 50% of CP

⇒ 100x + 50x = 150x

Now, Selling price = Marked price × (100 – D%)/100

⇒ 150x × (100 – 20)/100 = 120x

Profit = SP – CP

⇒ 120x – 100x = 20x

Profit % = (20x/100x) × 100 = 20%

When the trader weights the goods 20% less then it means, instead of 1000 gram trade weights 800 grams

Trader earns profit of 200 grams in 800 grams then the profit % is

Profit % = (200/800) × 100 = 25%

Overall profit = 20 + 25 + (20 × 25)/100 = 50%

∴ The net profit earned by the trader is 50%

Given: 

An article is Marked 25% above the cost price.

After giving x% discount still makes a profit of 5.5%.

Concept: 

S.P = [(100 + gain%)/100] × C.P

S.P = [M.P - (M.P × Discount%)]

Calculation:

Let the cost price of the article is 100 unit.

According to the question

The selling price of the article is 

⇒ [(100 + 5.5)/100] × 100

⇒ 105.5 units 

The Marked price of the article is 

⇒ [100 + (100 × 25%)]

⇒ 100 + 25

⇒ 125 units 

Also,

The selling price of the article is 

⇒ [125 - (125 × x%)] = 105.5

⇒ 125 - 5x/4 = 105.5

⇒ 5x/4 = 19.5

⇒ x = 19.5 × 4/5

⇒ x = 15.6

∴ The required value of x is 15.6.

Given:

11- digit number = 765y88436x6 

Divisor = 72

Concept used:

First, we should have to know the factors of divisor if all the factors are divided the given number, then the number is divisibly by 72

Divisibility rule of 8: If the last three digits are divisible by then the number is divisible by 8

Divisibility of 9:  If the sum of all digits is divisible by 9 then the number is divisible by 9

2 and 4 are the factors of 8 so, if the number is divisible by 8 is also divisible by 2 and 4

3 is the factor of 9 so, if the number is divisible by 9 then the number is divisible by 3

2 and 3 are the factors of 6 

Calculations:

Factors of 72 = 1 , 2, 3, 4, 6, 8, 9

To verify the number is divisible by 8 last three digits = 6x6

If the 6x6 is divisible by 8  then the number must be 616 or 656 or 696

Hence, x = 1 or 5 or 9

To verify the number is divisible by 9 sum of digits = 7 + 6 + 5 + y + 8 + 8 + 4 + 3 + 6 + x + 6

⇒ 53 + x+ y

53 + x +y is divisible by 9 it must be either 63 

53 + x + y = 63

⇒ x + y = 63 - 53

⇒ x + y = 10 

If we substitute x = 1

1 + y = 10

⇒ y = 10 - 1

⇒ y = 9

Here, the condition is x > y so  x = 1 doesnot satisfy this condition 

If we substitute x = 5

5 + y = 10 

⇒ y = 10 -5

⇒ y = 5

x = y condition doesn't satify 

x = 9

9 + y = 10 

⇒ y = 10 - 9 

⇒ y = 1

x > y so condition satisfies

x - y = 9 - 1 

⇒ x - y = 8

∴ The required value is 8

Let the original cost price be Rs.100

Then the marked price = 50% of 100 + 100

⇒ Marked price = 150

Selling price = 150 - x% of 150                 1

%Profit = 9.85%

⇒ Selling price = 9.85% of 100 + 100       2

Equating 1 and 2 

⇒ 109.85 = 150 - x% of 150

⇒ x% of 150 = 150 - 109.85

⇒ 3/2 x = 40.15

⇒ x = 40.15 × 2/3

⇒ x = 26.76%

∴ The required answer = 26.76%

Given:

Number = 45x97y

Concept Used:

A number is divisible by 8 if its last 3 digits are divisible by 8.

A number is divisible by 9 if the sum of digits is divisible by 9.

Calculations:

45x97y is divisible by 72

⇒ 97y must be divisible by 8

⇒ y = 6

⇒ Number = 45x976

Sum of digits = (4 + 5 + x + 9 + 7 + 6)

⇒ Sum of digits = (31 + x)

Number is divisible by 9

⇒ Sum of digits is divisible by 9

⇒ x = 5

Now, (2x – y) = 2 × 5 – 6

⇒ (2x – y) = 4

∴ The value of (2x – y) is 4.

Given:

Discount = 15%

Gains = 12%

Formula used:

Discount % = (M.P – S.P)/M.P × 100

Calculation:

Let C.P of an article be 100 unit

⇒ For 12% profit S.P = 112 unit

15% discount on mark price

⇒ S.P = 85% = 112 unit

⇒ M.P = 100% = ?

⇒ M.P = (100 × 112)/85

⇒ M.P = 2240/17

If C.P increased by 10%

⇒ New S.P = 110 unit

⇒ For 12% profit new S.P = (110 × 112)/100

⇒ 616/5

Discount % = (M.P – S.P)/M.P × 100

⇒ (2240/17 – 616/5)/(2240/17) × 100

⇒ 6.5%

∴ The discount percentage should be allowed now on the same marked price is 6.5%

Given:

MP = 2 × CP

Profit% = 15%

Formula used:

Gain = SP – CP

Gain% = Gain/CP × 100

Discount% = (MP – SP)/MP × 100

Calculation:

Let the CP be Rs. x

And the MP be Rs. 2x.

Gain% = 15%

SP = Rs. (115x/100) = Rs. 23x/20

Discount% = {(2x – 23x/20)/2x × 100}% = 42% (approx)

∴ required discount is 42% (approx).

Ans: 3

Given:

5x2y6z is divisible by 7, 11 and 13

Concept used:

7 × 11 × 13 = 1001, when any 3 digit number multiplied by 1001 then it repeats itself and always completely divisible by 7,11 and 13.

Calculation:

Let 3 digit number be pqr.

⇒ pqr × 1001 = 5x2y6z

⇒ pqrpqr = 5x2y6z

Compare both sides

⇒ p = 5, q = 6 and r = 2

Hence, that number is 562562.

⇒ 5x2y6z = 562562

by comparing both numbers, x = 6, y = 5, and z = 2 

Required value = (x – y + 3z)

⇒ 6 – 5 + 3 × 2

⇒ 7

∴ The required value is 7.

GIVEN:

If the six-digit number 879xyz is exactly divisible by 7, 11 and 13

CONCEPT:

We know that, xyzxyz = xyz × 1001 = xyz × 7 × 11 × 13

CALCULATION:

Therefore, the six-digit number 879xyz exactly divisible by 7, 11, and 13.

⇒ The number will be 879879.

∴ x = 8, y = 7 and z = 9

According to question –

⇒ {(x + y) ÷ z} = {(8 + 7) ÷ 9}

⇒ 15/9

⇒ 5/3

Given :

A shopkeeper gained 25% while allowing 20% discount 

Formula used :

Selling price = Marked price - discount on marked price 

Selling price = cost price + profit on cost price 

Calculations :

Let the initial cost price be 100x 

Selling price = 100x + 25% of 100x 

⇒ 125x 

Selling price = Marked price - 20% of marked price 

Marked price = 625x/4 = 156.25x

Now Cost price got decreased by 20%, so new cost price 

New cost price = 80x 

New selling price = 80x + 25% of 80x 

⇒ 100x 

Discount per cent on marked price to get this selling price 

Discount per cent = (Marked price - selling price)/(Marked price) × 100 

⇒ (156.25x - 100x)/(156.25x) × 100 

⇒ 36% 

∴ Discount should be 36% 

Given :

The marked price of the article = Rs.2400

The selling price of the article = Rs.1542.24

Three successive discounts are 15%,10% and x%

Formula used :

Selling price = Marked price - Discount

Calculation :

⇒ 1542.24 = 2400 × (85/100) × (90/100) × (\( { {100-x} \over 100}\))

⇒ 1542.24 = {1836 × (100 - x)}/100

⇒ 154224 = 183600 - 1836x

⇒ 1836x = 29376

⇒ x = 16

∴ The value of 'x' is 16

Given:

Six digit number is 479xyz

Divisible by 7, 11, and 13.

Concept Used:

Any number is divisible by 7, 11, and 13 

If the number is in the form of xyzxyz (repeating form)

Calculation:

479xyz = 479479 

⇒ x = 4, y = 7, and z = 9

⇒ x × y × z = 4 × 7 × 9

⇒ x × y × z = 252

∴ The product of xyz is 252.

Given:

The number is 10, 12, 15 and 20

Concept used:

The greatest number of 3 digits is 999.

Calculation:

LCM of 10, 12, 15 and 20

10 = 2 × 5

12 = 2² × 3

15 = 3 × 5

20 = 2² × 5

LCM = 2² × 3 × 5 = 60

The greatest number of 3 digits = 999

⇒ 999 ÷ 60

⇒ Remainder = 39

⇒ 3 digit greatest numbers = 999 - 39 = 960

∴ The greatest number of 3 digits is 960.

Given:

Selling price of article = 575

Marked price of article = 850

Successive discounts = 20% and X%

Formula used:

S.P = markup – discount

Percent discount = discount/M.P × 100

Calculation:

M.P of article = 850

Value of article after discount of 20% = 850 – 20% of 850 = 850 – 170 = 680

Now, new marked price of an article = 680

⇒ S.P = 575

Discount = 680 – 575 = 105

Discount% = 105/680 × 100 = 15.4%

Given:

The marked Price of an article is Rs. 5000.

Three successive discounts of 15%, 10% and 18%, are given on its marked price.

Formula Used:

Selling price = Marked price × (100 - Discount%)/100

Calculation:

The marked Price of an article is Rs. 5000.

The Selling Price = 5000 × (100 - 15)/100 × (100 - 10)/100 × (100 - 18)/100

⇒ 5000 × 85/100 × 90/100 × 82/100

⇒ Rs.3136.5

∴ The Selling price of the article is Rs.3136.5.

Given:

Two successive discounts of 20% and 25%.

SP of the article = Rs. 600

Formula used:

Calculation:

Let the marked price be Rs. x

After first discount of 20%, price = Rs.(x - x × 20/100) = Rs. (4x)/5

After second discount of 25%, price = Rs.[(4x)/5 - (4x)/5 × 25/100) = Rs. (3x)/5

According to problem,

⇒ (3x)/5 = 600

⇒ x = 1000

∴ The Marked price of the article is Rs. 1000.

Given:

Sum of selling price = Rs. 1995

Discount on 1^st case = 35 percent

Discount on 2^nd case (Successive) = 20 percent & 15 percent

Formula used:

Selling price = [(discount)/marked price] × 100

Or Selling price = [(100 – discount percent)/100] × marked price

Calculation:

Let marked price be 100x

[{(100 - 35)/100} × 100x] + [{(100-20)/100} × {(100-15)/100} × 100x] = 1995

⇒ 65x + [(80 × 85)/100]x = 1995

⇒ 65x + 68x = 1995

⇒ 133x = 1995

⇒ x = 15

⇒ 100x = 1500

∴ The marked price of the article is Rs. 1500

Given:

Marked price: Rs. 1,250

Discount offered 5%, 15% and 20%

Formula used:

S.P= M.P × {(100 - d1)/100} × {(100 - d2)/100} × ......× {(100 - dn)/100}

M.P = Marked price

S.P = Selling price

d1,d2....dn = discount%

Calculation:

We need to find S.P

S.P = M.P × {(100 - d1)/100} × {(100 - d2)/100} × ......× {(100 - dn)/100}

⇒ S.P = 1250 × {(100 - 5)/100} × {(100 - 15)/100} × {(100 - 20)/100}

⇒ S.P = 1250 × (95/100) × (85/100) × (80/100) 

⇒ S.P = 807.50

∴ The selling price is Rs. 807.50

CONCEPT:

In this question, we will divide the largest five-digit number with 93 and find out the remainder because once the remainder is subtracted from the largest five-digit number we get the number which is divisible by 93.

Detailed Solution:

The largest 5 digit number = 99,999

When we divide this number by 93 we get a remainder of 24 And we can find the required number by subtracting the remainder from 99,999.

Hence, required numbers = 99,999 – 24 = 99975

Short Trick: 

Go through the options one by one by dividing by 93 and starting with the largest number.

Given:

the 7-digit number 2y6810x is divisible by 88.

Concept used:

Divisibility

Calculation:

If a number is divisible by 88 means it is divisibly by 11 and 8

⇒ 11 × 8 = 88

To check the divisibility of 8 check the last three digits

In 10x let the x be 4

104 is completely divisible by 8

∴ x = 4

Divisibility of 11

⇒ (sum of digits at odd places - Sum of digits at even places) = 0 or 11

⇒ (4 + 1 + 6 + 2) - (y + 8) = 0

⇒ 13 - y - 8 = 0

∴  y = 5

Product of x and y 

⇒  4 × 5 = 20

Given:

Number = 179x091y 

Concept used:

1) If the last three digits of a number are divisible by 8, then the number is completely divisible by 8

2) If the difference of the sum of alternative digits of a number is divisible by 11, then that number is divisible by 11 completely.

3) If any number is completely divisible by 88 then the number is also completely divisible by 8 and 11

Calculation:

Given number is completely divisible by 88 then it is also divisible by 11 and 8

When the number is divisible by 8 then,

91y is divisible by 8 when y = 2      ----(1)

When the number is divisible by 11 then,

(7 + x + 9 + y) – (1 + 9 + 0 + 1) 

⇒ 16 + x + y – 11 

⇒ x + y + 5 

⇒ x + 2 + 5      ----(From eq (1))

⇒ x + 7

Now, (x + 7) is divisible by 11 when x = 4 then,

x – y = 4 – 2 = 2

∴ The required difference is 2 

Calculation:

The largest known 4 digit number is 9999

Divide 9999 by 88

it leaves reminder 55 

Subtracting 55 from 9999

9999 - 55 = 9944

∴The largest 4 digit number exactly divisible by 88 is 9944

Given:

90457416xy is divisible by 9 

x - y = 3

Concept Used:

If sum of all digits of a number is divisible by 9 then the number should also be divisible by 9

Calculation:

(9 + 0 + 4 + 5 + 7 + 4 + 1 + 6 + x + y)/9 = (36 + x + y)/9

For 10 digit number to divisible by 9  then (36 + x + y) is completely divisible by 9.

⇒ x + y = 9

Also given x - y = 3

Solving,

x = 6 and y = 3

5x + 3y = 5 × 6 + 3 × 3 = 39

∴ Required value of (5x + 3y) is 39.

Given:

The number 15785xy3 is divisible by 9.

Concept used:

A number is divisible by 9 when sum of it’s digits is divisible by 9.

Calculation:

Sum of all digits = 1 + 5 + 7 + 8 + 5 + x + y + 3

⇒ Sum of digits = 29 + x + y

As the number is divisible by 9, so 29 + x + y is divisible by 9

Multiple of 9 nearest to 29 is 36

29 + x +y = 36

⇒ x + y = 36 – 29 = 7

But this is not the maximum value of x and y

Next nearest multiple of 9 is 45

 29 + x +y = 45

⇒ x + y = 16

For 2x + 3y to be maximum y should be greater than x

For y = 9, x = 16 – 9 = 7

2x + 3y = 2 × 7 + 3 × 9 = 41

∴ Maximum value of 2x + 3y is 41

Given:

24357x8y0 is divisible by 9

Concept:

If the sum of all the digits of a number is divisible by 9, the number is also divisible by 9.

Calculation:

Sum of all the digits =  2 + 4 + 3 + 5 + 7 + x + 8 +y + 0

⇒ 29 + x + y

As the number is divisible by 9, so 29 + x + y is also divisible by 9

Multiple of 9 nearest to 29 is 36

⇒ 29 + x + y = 36

⇒ x + y = 7

Squaring both the sides

⇒ (x + y)² = 7²

⇒ (x + y)² = 49

Given:

Our given number is 67x2y

Concept used:

If the sum of the digits of the number is divisible by 9 then the number is divisible by 9

Calculation:

Our given number is 67x2y

Sum of digits of number = 15 + x + y

The value of x + y ≥ 3

And x + y ≤ 4

Possible numbers of pair are (0, 3), (3, 0), (1, 2), (2, 1)

∴ The number of possible pairs is 4

Concept

Given number is divisible by 9, if the sum of the digits of the number is divisible by 9

Explanation

Sum of the digits (A + B + 4 + 0) should be divisible by 9.

For A + B + 4 = 9 

The value of A + B = 9 - 4 = 5

A and B could be (5, 0), (2, 3), (3, 2), (1, 4), (4, 1)

Note:- We cannot take A as 0, because at A = 0, the number is not a 4 digit number.

For A + B + 4 = 18

The value of A + B = 18 - 4 = 14

So, value of A and B could be (7, 7); (6, 8); (8, 6); (9, 5)and (5, 9)

This number could be 5040; 2340; 3240; 1440; 4140; 7740; 6840; 5940; 8640 and 9540

∴ 10 pairs are possible.

Note: 

Pair (0,5) is not considered, because if we 0 used as the first position

Then we get 0540, 0540 is not the 4 digit value, it's a three-digit value.

That's why we have not considered this pair. 

Concept used:

If the difference of the alternating sum of the digits of a number is a multiple of 11, then the number is divisible by 11

Calculation:

Checking above given digits 656656 or 486486

Adding digits alternately then we got (6 + 6 + 5) and (5 + 6 + 6)

The difference between the number is 0 which is multiple of 11

Checking other given number

Adding digits alternately then we got (4 + 6 + 8) and (8 + 4 + 6)

The difference between the number is 0 which is divisible by 11

∴ The above pattern of the number is always divisible by 11

Concept used:

Divisibility rule of 11: If the difference of the alternating sum of the digits of a number is a multiple of 11, then the number is divisible by 11

Divisibility rule of 7: Take the last digit of the number, double it then subtract the result from the rest of the number, If the resulting number is divisible by 7 then original number is divisible by 7

Calculation:

Checking above given digits xyzxyz

Adding digits alternately then we got (x + z + y) and (y + x + z)

The difference between the number is 0 which is multiple of 11

The number formed may or may not be divisible by 3

The number formed is always divisible by 7

Taking any number 142142, 343343 etc these number is not divisible by 3 but divisible by 7

∴ The above pattern of the number is always divisible by 7 and 11

An ion is an atom or group of atoms in which the number of electron s is different from the number of proton s. If the number of electrons is less than the number of protons, the particle is a positive ion, also called a cation. If the number of electrons is greater than the number of protons, the particle is a negative ion, also called an anion.

Given:

The given expression is \({\rm{A}} = {\rm{}}({11^{12}} + {11^{14}} + {11^{16}})\)

Formula Used:

Basic concept of exponent

Calculation:

From the given expression we take 11¹² common

∴ A = 11¹² (1 + 11² + 11⁴) = 11¹² × 14763

So, the given number is divisible by 7, 3 as well as 19

Hence, option (4) is correct

Given:

The given six digit number is 5678ab which is divisible by 4

Concept Used:

The number is divisible by 4 if the last two digits of the number are also divisible by 4

Calculation:

In the given question the last two digits are ab along with the condition

∴ Possible values of ab will be 60, 40, 32, 24, 20, 12, and 04 only these digits satisfy the given condition

So, there are 7 possible pairs

Hence, option (4) is correct

Given:

The given expression is (pq + 1/(pq)^-1)

Calculation:

The given expression can be written as (pq + 1/(pq)^-1) = (pq + pq) = 2pq

Now, we know that product of two odd numbers is always a odd number and sum and difference of any two odd number is always even

So, the given expression is divisible by p, q and 2

Hence, option (1) is correct

Correct option is C) until

18,Laxmi Bai Park,

Sadar Bazar, Jhansi (U.P.)

22nd April., 2012

Dear Hasan,

Thanks for the letter which I received yesterday. I was very glad to read that you have not forgotten your promise to come and spend a few days of your holidays with me. The summer vacation starts from 11th of May to 21st June. By the time this letter reaches you, your examination will be over. How have you done in the examination? I hope you will secure good division.

It will be really such a great pleasure to me and my family if you come over here and stay with us during the holidays. I assure you that you will certainly enjoy your visit and the beauty of this place. There are many historical monuments, palaces and gardens. Jhansi Fort, Rani Laxmi Bai Park, Zoo, Museums, Radha - Krishan Temple and a number of others are very famous places. They all are worth seeing. Besides, there are many places of outing. We will also visit beauty of mountains, temples, springs, etc. I have a number of friends here. You will really enjoy their company.

Please do reply and let me know when you will come here. I wish you to convey my regards to your parents and love to Shahni.

Rest is o. k. More when we see.

Sincerely Yours

Rahul Kumar

                                                                                                                                     Krishna

                                                                                                                No.23 fort Road, Gorur

                                                                                                                  Date : ……….

Dear Ramesh,

I am quite good in my health. I think you and your family members are fine.

No I am writing this letter to inform you about my sister’s marriage. My sister’s marriage falk on 25th May at Sri. Venkataramana temple, Gorur. We have invited only a few relatives and friends. The bridegroom is a teacher and he did not expect anything (dowry) from us. As I could not invite you personally, I am inviting you through this letter. I cordially invite you on the occasion of my sister’s marriage. You should be here by 20th May. Apply for leave and come as early as possible. I hope you will not disappoint me.Convey this and my regards to one and all in your house.

                                                                                                                    Yours loving friend

                                                                                                                          Krishna

To,

Ramesh

No. 40,

Church Road

Hassan