Check the option. 32 and 128 are nither of a square and nor cube root. 64 is the square of 8 and a cube of 4

1369 = 37 × 37 It means there are 37 rows and 37 columns.
And each row will have 37 students, Therefore, the last row also have 37 students.

$$\eqalign{ & {\text{Let}}\,{\text{two}}\,{\text{numbers}}\,{\text{are}}\,x\,{\text{and}}\,y. \cr & {\text{Now,}}\,{\text{Sum}}\,{\text{of}}\,{\text{the}}\,{\text{numbers}}, \cr & x + y = 12 \cr & {\text{Product}}\,{\text{of}}\,{\text{the}}\,{\text{number}}, \cr & xy = 35 \cr & {\text{Then}}\,{\text{sum}}\,{\text{of}}\,{\text{the}}\,{\text{reciprocals}} \cr & = {\frac{1}{x}} + {\frac{1}{y}} \, \cr & = \frac{{ {x + y} }}{{xy}} \cr & = \frac{{12}}{{35}} \cr} $$

$$\eqalign{ & {\text{Let}}\,{\text{the}}\,{\text{fraction}}\,{\text{be}}\,\frac{{3k}}{{4k}} \cr & \therefore \frac{{ {3k - 9} }}{{4k}} = {\frac{2}{3}} \cr & \therefore 9k - 27 = 8k \cr & \therefore k = 27 \cr } $$

$$\eqalign{ & {\text{Let}}\,{\text{total}}\,{\text{property}}\,{\text{of}}\,{\text{Ram}} = x \cr & {\text{His}}\,{\text{widow}}\,{\text{gets}} \cr & \frac{1}{3}\,of\,x = \frac{x}{3} \cr & {\text{His}}\,{\text{daughter}}\,{\text{gets}}\, \cr & = \frac{3}{5}\,of\,\left[ {x - {\frac{x}{3}} } \right] \cr & = \frac{3}{5} \times {\frac{{2x}}{3}} \cr & = \frac{{2x}}{5} \cr & \cr & {\text{His}}\,{\text{son}}\,{\text{gets}}\,{\text{his}}\,{\text{rest}}\,{\text{property}} \cr & {\text{His}}\,{\text{son's}}\,{\text{property}} \cr & = x - \left[ { {\frac{x}{3}} + {\frac{{2x}}{5}} } \right] \cr & 6400 = x - {\frac{{11x}}{{15}}} \cr & 6400 = \frac{{ {15x - 11x} }}{{15}} \cr & 4x = 6400 \times 15 \cr & x = 1600 \times 15 \cr & x = 24000 \cr & {\text{Total}}\,{\text{Property}} = 24000 \cr} $$

5(x + 3) = 25(3x - 4)
Or, 5(x + 3) = 52 × (3x - 4)
Now, comparing the powers,
(x + 3) = 2 × (3x - 4)
Or, x + 3 = 6x - 8
Or, 6x - x = 8 + 3
Or, 5x = 11
Hence, x = $$\frac{{11}}{5}$$

LCM of 5, 6, 7 and 8 = 840 Hence the number can be written in the form (840k + 3) which is divisible by 9 If k = 1, number = (840 × 1) + 3 = 843 which is not divisible by 9 If k = 2, number = (840 × 2) + 3 = 1683 which is divisible by 9 Hence, 1683 is the least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder

$$\eqalign{ & X = \frac{{ {693} }}{{ {9 \times 11} }} \cr & X = 7 \cr} $$

LCM of 252, 308 and 198 = 2772 Hence, they all will be again at the starting point after 2772 seconds. or 46 minutes 12 seconds.

Ratio of the numbers = x : y HCF of the numbers = z So, z is the common factor of the numbers Then, First number = xz Second Number = yz Now, First Number × Second Number = HCF and LCM of the numbers xzyz = z × LCM LCM = xyz

HCF of 551 and 1073 is 29. So second number is 29
First number = $$\frac{{551}}{{29}}$$ = 19
Third number = $$\frac{{1073}}{{29}}$$  = 37
So,
Required Sum
= 19 + 29 + 37
= 85

Let Price of each capsule were Rs. X, originally. So, Number of capsule purchased originally, = $$ \frac{{176}}{{\text{X}}}$$ If price of capsule is Rs. 3 less then 6 more capsule were purchased. In this case, the number of capsules, = $$ \frac{{176}}{{{\text{X}} - 3}}$$ According to question, $$\eqalign{ & \frac{{176}}{{{\text{X}} - 3}} - \frac{{176}}{{\text{X}}} = 6 \cr & \frac{{176{\text{X}} - 176{\text{X}} + 528}}{{{\text{X}} \times \left( {{\text{X}} - 3} \right)}} = 6 \cr & 6{{\text{X}}^2} - 18{\text{X}} - 528 = 0 \cr & {{\text{X}}^2} - 3{\text{X}} - 88 = 0 \cr & {{\text{X}}^2} - 11{\text{X}} + 8{\text{X}} - 88 = 0 \cr & \left( {{\text{X}} - 11} \right)\left( {{\text{X}} + 8} \right) = 0 \cr} $$ Either, X = 11 Or, X = -8 (Price cannot be negative) Thus, X = Rs. 11 Hence, Number of the capsule purchased = $$\frac{{176}}{{11}}$$ = 16

Given that ab + ac + ad = 0
Hence, out of these three terms, some are positive and some are negative. Since all of a, b, c, and d are negative integers, only that term which has an even power can be positive. Hence, the absolute value of at least one of a, b, c or d is even. Hence, a × b × c × d is also even

LCM of 4, 8, 10, 12, 15 and 20 = 120 120 seconds = 2 minutes Hence, all the six bells will ring together in every 2 minutes Hence, number of times they will ring together in 60 minutes = $$1 + \frac{{60}}{2}$$
= 31

Length of the room
= 15 m 17 cm
= 15 × 100 + 17
= 1517 cm.
Breadth of the room
= 9 m 2 cm
= 902 cm.
The HCF of the 1517 and 902 will be size of square tiles. HCF 1517 and 902 = 41 cm.
Area of the room
= length × breadth
= 1517 × 902 cm2
Area of tiles = 41 × 41 cm2 So, number of tiles required
= $$\frac{{1517 \times 902}}{{41 \times 41}}$$
= 814 tiles

Since the numbers are co-prime, their HCF = 1 Product of first two numbers = 119 Product of last two numbers = 391 The middle number is common in both of these products. Hence if we take HCF of 119 and 391, we get the common middle number. HCF of 119 and 391 = 17 So, Middle Number = 17 First Number = $$\frac{{119}}{{17}}$$ = 7 Last Number = $$\frac{{391}}{{17}}$$ = 23 Sum of the three numbers
= 7 + 17 + 23
= 47

Lcm of 8, 12,20,30 is 120. As they leave a remainder 4 So it should be 120 + 4. But when being divided by 21, remainder is 7 So the number will be exactly divisible by (21 + 7) = 28. The number will be in the form of (120k + 4) which is divisible by 28 Where k is any positive integer. Now, if we divide 120 by 28, the remainder is 8 We need a remainder of 24, because (24 + 4) = 28 Now (8 × 3) = 24 Thus, k = 3 Hence, the number would be (120 × 3) + 4 = 364

Here we have to verify the given two conditions for the given options.
Condition :1
As given, after finished his purchase he had the rupees as much as the paise when he went.
Condition :2
After finishing purchase, he had Half as many rupees as he had paise when he went.

We solve it through option checking method.
Case :1
Now suppose the answer is option c
Then he spent Rs. $$\frac{{99.98}}{2}$$ = Rs. 49.99
Condition :1
Here we have 99 paise so the rupees should be 99 when he went. He initially had 99 Rs. Hence condition 1 is satisfied.
Condition :2
He had 98 paise to begin with. So the rupees he spent should be $$\frac{{98}}{2}$$ = Rs. 49 as per condition 2.
It turns out that the rupees actually he spent was 49.
Therefore, condition 2 is satisfied as well.

Given, HCF = 44 LCM = 264 First Number = When divided by 2, the quotient is 44
So, First Number = 2 × 44 = 88 Second number = N. (let). Now, using formula, HCF × LCM of two numbers = First number × Second number 44 × 264 = 88 × N
N = 132
Second Number = 132 NOTE : Dividend = Divisor × Quotient + Remainder

Let the Original number is (X + 10Y) According to question, Sum of the digit of the number = 12 X + Y = 12 ----------- (1) When digit interchanged, number become 36 more than original number, so, 10X + Y = 10Y + X + 36 X - Y = 4 ------------------- (2) On solving equation (1) and (2), we get X = 8 Y = 4 So, Original number, = X + 10Y = 8 + 10 × 4 = 48