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Borrowed date : 17th August Returned date : 4th September do not have to include return date Number of days = 15 days of August + 3 days of September = 18 days.

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Area of a tile = 10×10 = 100cm²area of wall = 4×2.5 = 10m² = 1000cm²

now let the number of tiles used to cover the wall be n

therefore 1000 = n×1001000/100 = nn = 10

therefore 10 tiles are required to cover the wall.

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ans=U308} . .

U3O8 is the answer best

The answer is U3O8 .

u308 is the best answer

the correct answer is U308

please right the full question NB

what is hip.................

bhai dekhai nahi de raha he

nice

a.15b.21c.23d.200e.88f.91g.

Using Euclid's algorithm :- 12576 > 4052

4052) 12576 ( 3 12156 ---------- 420) 4052 ( 9 3780 ---------- 272 ) 420( 1 272 ------ 148 ) 272 ( 1 148 ------ 124 ) 148 ( 1 124 ----- 24) 124 ( 5 120 ----- 4 ) 24 ( 6 24 ---- 0

→ Hence the H.C.F of4052 and 12576 is 4

bro lcm

nice answer............

X:6:5::3 SO, X=(6*5)/3=10

x = 10

Step 1. Divide the larger number by the smaller one:147 ÷ 105 = 1 + 42;

Step 2. Divide the smaller number by the above operation's remainder:105 ÷ 42 = 2 + 21;Step 3. Divide the remainder from the step 1 by the remainder from the step 2:42 ÷ 21 = 2 + 0;At this step, the remainder is zero, so we stop:21 is the number we were looking for, the last remainder that is not zero.This is the greatest common factor (divisor).

gcf, gcd (105; 147) = 21;coprime numbers (relatively prime) (105; 147)? No.

Euclid division lemma:-

a = bq + r

0 ≤ r < b

a > b

(i)8624 and 21658

21658 = 8624(2) + 4410

8624 =4410(1) + 4214

4410 = 4214(1) + 196

4214 = 196(21) + 98

196 = 98(2) + 0

HCF of 8624 and 21658 is 98

Let a be a positive integer.

Then, by Euclid ’s division lemma, corresponding to the positive integers a and 5, there exist non-negative integers m and r such that

a = 5m + r, where 0 ≤ r < 5⇒ a2= (5m + r2) = 25mr + r2+ 10mr

[∵(a+b)2= a2+ 2ab + b2]

⇒ a2= 5(5m2+ 2mr) + r2...(i)where, 0 ≤ r < 5

Case I

When r = 0, then putting r = 0 in Eq.(i), we geta2= 5(5m2) = 5qwhere, q = 5m2is an integer.

Case II

When r = 1, then putting r = 1 is Eq.(i), we geta2= 5(5m2+ 2m) + 1⇒ q = 5q + 1where,q = (5m2+ 2m) is an integer.

Case III

When r = 2,then putting r = 2 in eq.(i),we geta2= 5(5m2+ 4m) + 4 = 5q + 4where, q = (5m2+ 4m) is an integer.

Case IV

When r = 3, then putting r = 3 in Eq.(i), we geta2= 5(5m2+ 6m) + 9 = 5 (5m2+ 6m) + 5 + 4= 5(5m2+ 6m + 1) + 4 = 5q + 4where, q = (5m2+ 8m + 3) is an integer.

Case V

When r = 4, then putting r = 4 in Eq.(i), we get

a2= 5(5m2+ 8m) + 16 = 5 (5m2+ 8m) + 15 + 1⇒ a2= 5(5m2+ 8m + 3) + 1 = 5q + 1

where, q = (5m2+ 8m + 3) is an integer.

Hence, the square of any positive integer cannot be of the form 5q + 2 or 5q + 3 for any integer q.

P(A)/(1-P(A))=7/1212P(A)=7-7P(A)so 19P(A)=7so P(A)=7/19

Let abe any positive integer.By Euclid's division lemma,a=bm+rwhere b= 5⇒a= 5m+rSo, rcan be any of 0, 1, 2, 3, 4∴ a= 5mwhen r= 0a= 5m+ 1 when r= 1a= 5m+ 2 when r= 2a= 5m+ 3 when r= 3a= 5m+ 4 when r= 4So, "a" is any positive integer in the form of 5m, 5m+ 1 , 5m+ 2 , 5m+ 3 , 5m+ 4 for some integerm.Case I :a= 5m⇒ a2= (5m)2= 25m2⇒ a2= 5(5m2)= 5q,where q= 5m2Case II :a= 5m+ 1 ⇒ a2= (5m+ 1)2= 25m2+ 10m+ 1 ⇒ a2= 5 (5m2+ 2m) + 1= 5q+ 1, whereq= 5m2+ 2mCase III :a= 5m+ 2⇒ a2= (5m+ 2)2= 25m2+ 20m+4= 25m2+ 20m+4= 5 (5m2+ 4m) + 4= 5q+ 4 whereq= 5m2+ 4mCase IV:a= 5m+ 3⇒a2=(5m+ 3)2= 25m2+ 30m+ 9=25m2+ 30m+ 5 + 4=5 (5m2+ 6m+ 1) + 4=5q+ 4 where q= 5m2+ 6m+ 1Case V: a= 5m+ 4⇒a2=(5m+ 4)2= 25m2+ 40m+ 16=25m2+ 40m+ 15 + 1=5 (5m2+ 8m+ 3) + 1=5q+ 1 where q= 5m2+ 8m+ 3From all these cases, it is clear that square of any positive integer can not be of the form 5m+ 2 or 5m+ 3

No of 5 cm cubes = 15*25*30/5*5*5

= 15*6

= 90 cubes

20*10*4=n*2*2*2n=20*10*4/2*2*2n=10*5*2n=100

Option c is correct

2dm = 0.2 mvolume of cube = 0.008m^3volume of bigger cube 3*3*3= 27m^3number of cubes = 27/0.0083cubes approximately

Probability of failure = 1/2Probability of success = 1/2ATQProbability of success is = 2(1/2) = 1

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P = Rs 300 r = 12% per annum = 1% per monthSum on maturity= Rs 8,100. let n be number of months after which the amount become mature.

the first installment earns interest for n months, the second one for n-1 months, and the last installment earns interest for 1 month. The sum is a geometric series with ratio 1+r/100.

Sum = P (1 + r/100)^n + P (1+r/100)^n-1 + P(1+r/100)^n-2+......+P(1+r/100)^18,100 = 300 [ 1.01^n + 1.01^n-1 + 1.01^n-2 +.... + 1.01^2 + 1.01 ]27 = 1.01 [ 1.01^(n+1) - 1 ] / [ 1.01 - 1] 27 = 101 [ 1.01^(n+1) - 1 ] 1.01^(n+1) = 1 + 27/101 = 128/101 (n+1) Log 1.01 = Log 128/101 n+ 1 = [ log 128/101 ] / Lo g 1.01 = 23.81

n = 22.81 months or, 1.984 years

is the correct answer

thx a lot

Installment per month (P) = 600Number of months (n) = 20Rate of Interest (r) = 10%

Then,SI = P * n(n+1)/2*12 * r/100 = 600 * 20*21/24 * 10/100 = 600 * 420/24 * 10/100 = 1050

The amount Manish will get at the time of Maturity = 600*20 + 1050= 12000 + 1050= Rs 13050

For recurring deposit interest can be calculated using the formula

SI = P * n(n+1)/2*12 * r/100

Where SI is simple interest, P is the the money deposited per month, n is number of months for which money deposited and r is rate of interest per annum.

Therefore,SI = P * 18(18 + 1)/2*12 * 10/100= P * (18*19)/(2*12) * 1/10= P * (3*19)/40= 57P/40

A = P*n + SI1554 = 18P + 57P/401554 = (720 + 57)P/40P = 1554*40/777P = 2*40 = 80

Therefore, Sudhir deposit every month Rs. 80

Number of riped mangoes is 1/2 × 60 30 mangoes are riped

Let the apple be denoted as 'x' and mangoes be denoted by 'y', and the money he carrying be 'm'

Now, man can purchase 16x=10y=m

A/c to the statement, he bought 4x+5y=m-20

4*(m/16)+5*(m/10)=m-20m/4+m/2=m-208m-6m=160m=80 Rupees

Therefore, for apples 16x=mi.e. x=80/16One apple cost x=5 Rs.

Similarly for mangoes 10y=mi.e. 10y=80Price of 1 mango is y=8 Rs.

And the difference between them is Rs.3 .................Answer

C.P of 1000 pencils=Rs.3000C.P of one pencil= 3000/1000 =Rs.3

Manish sold 200 pencils at the gain of 5%C.P of 200 pencils= 3(200) =Rs.600Profit= 5% of CP of 200 pencils= 5*600/100 = Rs.30

S.P of 200 pencils= C.P of 200 pencils + Profit = 600+30 =Rs.630

Manish wants to make a profit of 15% of whole transaction

Profit of whole transaction= 15*3000/100 =Rs.45

Profit to be made for 2nd transaction= Profit of whole transaction - Profit of 1st transaction

= 450-30

= Rs.420

C.P of remaining pencils= 800*3 =Rs.2400

Let 'x' be the profit percentage of 2nd transaction

Profit of 2nd transaction= C.P of remaining pencils * (x/100)

=> 420 = 2400x/100=> 420 = 24x=> x = 420/24=> x = 17.5

Answer is 17.5

Let sp of on mango be rs1

Then sp of 125mangoes = 125 × 1

= rs 125

now he gains sp of 5 mangoes = 5×1

= rs5

so,

cp = sp - gain

cp = 125-5

cp=120rs

Now gain% = gain * 100 / cp

= 5 * 100/120

Gain% = 4.16666

Gain% = 4.167 %

1 mango = 15/4 rupees

For 210 rupees,

15/4 * x = 210

x = 210 * 4/15

x = 4*14

x = 56 mangoes

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by basic addition837+208+363=1408

So the answer is 40 mangoes

please shootcut method solve

the cube of a number n is its third power. the result of the number multiplied by itself twice. Cube of n is defined asn3 = n × n × n