Let us assume the first installment is a and difference between two consecutive installments is d.
According to question,
Sum of 40 Installments = 3600
Apply the formula Sum of first n terms in an Arithmetic Progression = Sn = n/2[ 2a + (n?1)d ]
where a = the first term, d = common difference, n = number of terms.
n/2[ 2a + (n?1)d ] = 3600
Put the value of a, n and d from question,
40/2[ 2a + (40 ?1)d ] = 3600
20[ 2a + 39d ] = 3600
[ 2a + 39d ] = 3600/20 = 180
2a + 39d = 180...........................(1)
Again according to given question,
After paying the 30 installments the unpaid amount = 1/3(total unpaid amount) = 3600 x 1/3
After paying the 30 installments the unpaid amount = 1200
So After paying the 30 installments the paid amount = 3600 - 1200 = 2400
Sum of 30 installments = 2400
30/2[ 2a + (30 ?1)d ] = 2400
[ 2a + (30 ?1)d ] = 2400 x 2/30
2a + 29d = 80 x 2 = 160
2a + 29d = 160..........................(2)
Subtract the Eq. (2) from Eq. (1), we will get
2a + 39d - 2a - 29d = 180 - 160
10d = 20
d = 2
Put the value of d in Equation (1), we will get
2a + 39 x 2 = 180
2a = 180 - 78
a = 102/2
a = 51
The value of first installment = a = 51
The HCF of m and n is 1, so m and n are prime number.
Let m = 7 and n = 5 ? m + n = 12
HCF of(m + n)and m = HCF of 12 and 7 = 1
Similarly, HCF of (m - n) and n = 1
Let be the ten's digit be x and unit's digit be y.
So the two digit number = 10x + y ( where x > y )
According to question
x + y = 14 ....(i)
x - y = 2 ....(ii)
Solving Eqs. (i) and (ii), we get
x = 8 and y = 6
? Required product = 8 x 6 = 48
Increase in 3 years over 100 = 100 x {1 + 7/(2 x 100)}3
= (100 x 207/200 x 207/200 x 207/200)
= (200 + 7)3 / 80000
= 88869743 / 80000 = 110.8718
Required increase % = (110.8 - 100)% = 10.8%
? 2log4x = 1 + log4(x-1)
? log4x2 = log44 + log4(x-1)
? x2 = 4(x-1)
? x2 - 4x + 4 = 0
? (x-2)2 = 0
? x = 2
Let the rate of interest allowed by bank be r%
According to the question,
[(12000 x 5 x 10)/100] - [(12000 x 3 x r)/100] = 3320
? 6000 - 360r = 3320
? 360r = 6000 - 3320 = 2680
? r = 2680/360 = 74/9%
Given log90 = 1.9542
? log(32 x 10) = 1.9542
? 2log 3 + log 10 = 1.9542
? log 3 = 0.9542 / 2 = 0.4771
A leap year has 366 days = 52 weeks + 2 days. These 2 days can be (Sunday, Monday, Wednesday) .... or (Saturday, Sunday). Out of these total 7 out comes there are 2 cases favourable to the desired event i, e. (Sunday, Monday) and (Monday, Tuesday)
? Required probability = 2/7
Let C(O, r) is a circle and let M is a point within it where O is the centre and r is the radius of the circle.
Let CD is another chord passes through point M.
We have to prove that AB < CD.
Now join OM and draw OL perpendicular to CD.
In right angle triangle OLM,
OM is the hypotenuse.
So OM > OL
? chord CD is nearer to O in comparison to AB.
? CD > AB
? AB < CD
So all chords of a circle of a circle at a given point within it, the smallest is one which is bisected at that point.
Hence required answer will be option D .
From given figure , we can see that
Let AB = 16 m , OM = 6 m , AM = BM = 8 m and OA = r
In triangle AMO ,
By pythagoras theorem ,
OA2 = AM2 + OM2
OA2 = 82 + 62 = 64 + 36
OA2 = 100 ? OA = ? 100 = 10
? OA = r = 10 m
The radius of the circular lawn = 10 m
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