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AMOUNT FOR SHRUTI = P*[1+ (R/100)]^NAMOUNT FOR SHRUTI = 12000*[1 + 10/100]^3= 12000*(11/10)^3= 12*11*11*11= Rs. 15972INTEREST FOR SHRUTI = AMOUNT - PRINCIPAL= 15972 - 12000= Rs. 3972

INTEREST FOR SHALINI = PRT/100= (12000*10*3)/100= Rs. 3600. SHRUTI PAYS MORE INTEREST BY ( 3972 - 3600) = Rs. 372

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greatest four digit number divisible by 15, 25, 40 &75 is =9999

11 because 15and15 is same answer it is 1and1 is same answer

1515 = 15*101 3737 = 37*101 so number of this form is exactly divisible by 101.

The greatest two digit number is 99if it is even number then 98 is the correct answer.

99 is the correct answer

The right answer 9996

9996 is the best answer

9996 is the answer of the following

9996 is the correct answer

Let S.P. = a

C.P. of 1st calculator= a x 100/( 100 - 15 ) = a x 100/85 = 20a/17

C.P. of 2nd calculator = a x 100/( 100 + 19 ) = a x 100/119 = 100a/119

a/q

20a/17 + 100a/119 = 4800

240a/119 = 4800

a = 2380

C.P. Of 1st cal. = 20 x 2380/17 = 2800

C.P. of 2nd cal. = 4800 - 2800 = 2000

C.P. of lower priced calculator = Rs. 2000

Assume ABC as any three digit no. which is multiple of 111 like - 777

Then,ABC+BCA+CAB=?777+777+777=2331which is a multiple of 111.

hence proved

Multiplicative inverse of 1 is 1

Mustard, Peas and Wheat are the Rabi crops. Rabi crops are sown after the Monsoon ends and harvested around April/ May. So Black gram is not a Rabi crop.

17 + 6p = 96p = 9 - 176p = -8p = -8/6p = -4/3

Factorizarion : 6p - 12q = 6(p-2q) = 2×3×(p-2q)

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By multiplicative inverse ,it means that it is a number which when multiplied to the given number the product comes to be one.

the multiplicative inverse is 5/3 is the correct answer

Any number whose sum of digits is divisible by 3 is divisible by 3. Ther are 30 such two digit numbers because (100-9)/3 = 30.

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The greatest no. is 999. Divide 999 by 10The remainder is 9So, 999-9=990Therefore, greatest 3 digit number divisible by 10 is 990

96 is the greatest two digit

96 is the greatest two digit number which is divisible by 6

the greatest 2 digit no. which is divisible by 6 is 96Then it is also divisible by 2,3because divisibility of 2 is last digit is always even no.and the divisibility of 3 is sum of number is come in the table of 3

9995 is the biggest 4 digit no. divisible by 5

10,000 is the smallest 5 digit no. multiple of 10

Abc is divisible by 3 => a+b+c = 3k for some integer k.

sum of threenumbersabc ,bca ,cab= 100 a+ 10 b+ c + 100 b+ 10c+ a+ 100 c+ 10 a+ b= 100 (a+b+c) + 10 (b+c+a) + (c+a+b) = (a+b+c) * 111 = 3 k * 3 * 37 = 9 * 37 * k

Hence, proved.

for a no. to be divisible by 3. the sum of digits should be multiple of 3. so. the sum = a+b+c . and in all the three below no. i.e abc , bca and cab the sum of no. will be same.. that is (a+ b+ c) + ( b+c+a) + ( c+a+b) = 3(a+b+c) .and since a+b+c is divisible by 3. so multiple pf a+b+c by 3, will be divisible by 3×3= 9.

thank you so much

(1) 2/5 (2) - 1/21

1)2/52)-1/213)1 is the correct answer of the given question

y=73 (linear opposite angles)z+153=180z=180-153z=27

x+y+z=18073+y+27=180100+y=180y=180-100y=80

(1)^2-2(1)(9n^2)+(9n^2)^2=(1-9n^2)^2

right answer sir............

1) x+y=14.......(1) x-y=4..........(2) (1)+(2) 2x=18; x=9 then y =5

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Observe the concept

A number ends with zero must 2 and 5 as its prime factors.Consider 2020 = 2 × 5 × 2 , 2 and 5 are present.

Now, we see that 6ⁿ can be expressed as(2×3)ⁿ. But it does not have 5 so, it is not possible that 6ⁿ will end with 0.

a= पहला पद=2/3दूसरा पद होता है= a+dso2/3+d= aअब हमें ज्ञात है कि तीसरा पद= a+2dहोता है so2/3+2d= 22d= 2-(2/3)2d= 4/3d= 4/6= 2/3so a= 2/3+2/3= 4/3

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the largest four digit number that is divisible by 16 is 9984

multiplicatove inverse of 3+2i=1/(3+2i)=(3-2i)/(3+2i)(3-2i)=(3-2i)/(9-4)=(3-2i)/5=3/5 -2i/5

18 is the answer of the above question

9=908=727=566=423=12this is the right answer of the following

18 is the right answer

18 is the correct answer

12 is correct answer .BCZ

4x²=16x²=16/4x²=4X=√4X=2 is the best answer.

x + 2 = 16subtracting 2 on both sides givesx + 2 -2 = 16 - 2x = 14

= Cross multiply

= 9 x * 1 = 15 * ( 7 - 6 x )

= 9 x = 105 - 90 x

= 99 x = 105

= x = 1.16.

15(7-6x)= 9x105-90x=9x99x=105x=105/99x= 1.06

आधार समान। हेतु ghate जुड़ जायगीx-2= 4x= 6please like the solution 👍 ✔️👍

thanks

kya aap mere aur questions ko solve karne me meri help karenge please sir

not do so long i have very short method for you from the IIT BOOKS 1_ take last digit i.e _ 2 .which is cube root of 8 hence the last digit of cube root is 8 2_now after strikking three digit we have _ 21 .and find below (less)cube root from 21 ..i.e 2×2×2= 8 which is less than 21 but 3×3×3=27 which is largest from 21 hence tens digit is 2 3_ cube root 21952= 28

Given ratio of sum of n terms of two AP’s = (7n+1):(4n+27)Let’s consider the ratio these two AP’s mth terms as am : a’m →(2)Recall the nth term of AP formula, an = a + (n – 1)dHence equation (2) becomes,am : a’m = a + (m – 1)d : a’ + (m – 1)d’On multiplying by 2, we getam : a’m = [2a + 2(m – 1)d] : [2a’ + 2(m – 1)d’]= [2a + {(2m – 1) – 1}d] : [2a’ + {(2m – 1) – 1}d’]= S2m – 1 : S’2m – 1= [7(2m – 1) + 1] : [4(2m – 1) +27] [from (1)]= [14m – 7 +1] : [8m – 4 + 27]= [14m – 6] : [8m + 23]Thus the ratio of mth terms of two AP’s is [14m – 6] : [8m + 23].

Let us assume that the two APs are:

a1 + (n-1)d1

And

a2 + (n-1)d2

Now the sum of these respective APs are

n/2(2*a1 + (n-1)d1)

n/2(2*a2 + (n-1)d2)

Taking the ratio we get

(2*a1 + (n-1)d1) : (2*a2 + (n-1)d2) = (7n-5) : (5n+17)

Comparing the coefficients of n on the RHS with those in both the terms on the LHS, we get

d1= 7 and d2= 5

Substituting these values in the equation and comparing the constants we get,

a1= 1 and a2= 11

Therefore the series are

1,8,15,22,29,36,43,50,57,64,71….

And

11,16,21,26,31,36,41,46,51,56,61,66,71,76…

So the series of numbers common in both APs is

36,71,106,141,176,211,….

Or (35n+1) where n is a natural number. 35 because it's the LCM of 7 and 5, the common differences of the APs.

So the number of equal terms are infinite.

Consider cosec theta - sin theta = a³⇒ !/sin theta - sin theta = a³⇒ 1 - sin² theta/sin theta = a³cos² theta/ sin theta = a³→ (1)⇒ (cos² theta/sin theta)²/³ = (a³)²/³⇒ cos⁴/³ theta/sin²/³ theta = a²→ (2)Now consider, sec theta - cos theta = b³⇒ 1/cos theta - cos theta = b³⇒ 1 - cos²theta/cos theta = b³⇒ sin² theta/cos theta = b³→ (3)⇒ (sin² theta/cos theta)²/³ = (b³)²/³⇒ sin⁴/³ theta/cos²/³ theta = b²→ (4)Multiply (2) and (4), we get(cos⁴/³ theta/sin²/³ theta)× (sin⁴/³ theta/cos²/³ theta) = a²b²→ (5)a² + b² =(cos⁴/³ theta/sin²/³ theta) + (sin⁴/³ theta/cos²/³ theta)(cos² theta + sin² theta)/(sin²/³ theta cos²/³ theta)= 1/sin²/³ thetacos²/³ thetaConsider, a²b²(a²+b²) =(sin²/³ theta cos²/³ theta)× 1/sin²/³ theta cos²/³ theta= 1 Hence proved.

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The answer is not fully correct

Given number is 17576.

1) Split the number from right into the group of three digit numbers. We have 2 parts 17 and 576.

2)the unit digit of 576 is 6. Hence the unit digit of required cube root is 6.

3) Now, We have to estimate the cube root of 17.

It lies between 8 and 27 i.e cube of 2 and 3. (2^3 = 8 and 3^3 = 27)

Take the units digit of the smallest number i.e 2.

So, The cube root of 17576 is 26.

0.2 aayega iska sir ji

78 in to fraction in lowest from

Cp 1600cp×120/100 =mp1600×120/100 =mpmp=1920sp = 1920×84/100=1612.8 cp=1600 sp = 1612.8 profit= 12.8profit% = 12.8/1600 ×100=8/10=.8%