Ayush's present age = 10 years.
His mother's present age = (10 + 20) years = 30 years.
Ayush's father's present age = (30 + 5) years = 35 years.
Ayush's father's age at the time of Ayush's birth = (35 - 10) years = 25 years.
Therefore Ayush's father's age at the time of marriage = (25 - 2) years = 23 years.
Let Varun's age today = x years.
Then, Varun's age after 1 year = (x + 1) years.
Therefore x + 1 = 2 (x - 12) => x + 1 = 2x - 24 => x = 25.
Number of cuts made to cut a roll into 10 pieces = 9.
Therefore, Required number of rolls = (45 x 24)/9 = 120.
Let Rs. x be the fare of city B from city A and Rs. y be the fare of city C from city A.
Then, 2x + 3y = 77 ...(i) and
3x + 2y = 73 ...(ii)
Multiplying (i) by 3 and (ii) by 2 and subtracting, we get: 5y = 85 or y = 17.
Putting y = 17 in (i), we get: x = 13.
Clearly, From 1 to 100, there are ten numbers with 3 as the unit's digit - 3, 13, 23, 33, 43, 53, 63, 73, 83, 93 and ten numbers with 3 as the ten's digit - 30, 31, 32, 33, 34, 35, 36, 37, 38, 39.
So, required number = 10 + 10 = 20.
Almost all the options are in support with the passage. As B underlines the main cause, it is the strongest argument among all the given options.
Let R, G and B represent the number of balls in red, green and blue boxes respectively.
Then,
R + G + B = 108 ...(i),
G + R = 2B ...(ii)
B = 2R ...(iii)
From (ii) and (iii), we have G + R = 2x 2R = 4R or G = 3R.
Putting G = 3R and B = 2R in (i), we get:
R + 3R + 2R = 108 => 6R = 108 => R = 18.
Therefore Number of balls in green box = G = 3R = (3 x 18) = 54.
Let the number of boys and girls participating in sports be 3x and 2x respectively.
Then, 3x = 15 or x = 5.
So, number of girls participating in sports = 2x = 10.
Number of students not participating in sports = 60 - (15 + 10) = 35.
Let number of boys not participating in sports be y.
Then, number of girls not participating in sports = (35 -y).
Therefore (35 - y) = y + 5 <=> 2y<=> 30<=> y = 15.
So, number of girls not participating in sports = (35 - 15) = 20.
Hence, total number of girls in the class = (10 + 20) = 30.
Since one of the numbers on the dial of a telephone is zero, so the product of all the numbers on it is 0.
Let the numbers be 3x, 3x + 3 and 3x + 6.
Then,
3x + (3x + 3) + (3x + 6) = 72
9x = 63
x = 7
Largest number = 3x + 6 = 27.
=> Second largest number = 27 - 3 = 24
Clearly, we have :
A = B - 3 ...(i)
D + 5 = E ...(ii)
A+C = 2E ...(iii)
B + D = A+C = 2E ...(iv)
A+B + C + D + E=150 ...(v)
From (iii), (iv) and (v), we get: 5E = 150 or E = 30.
Putting E = 30 in (ii), we get: D = 25.
Putting E = 30 and D = 25 in (iv), we get: B = 35.
Putting B = 35 in (i), we get: A = 32.
Putting A = 32 and E = 30 in (iii), we get: C = 28.
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