101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Let the required number of units of work = k
According to the formula,
W1= m and W2 = k
M1T1D1W2 = M2T2D2W1
? m x m x m x k = n x n x n x m
? x= (m x n3) / m3 = n3/m2
Let the ages of mother, father and boys be M, F, B1 and B2, respectively.
The total age of four member
= 19 x 4
= 76 yr.
Given (B1 + B1)/2 = 11/2
? B1 + B2 = 11
M + F + B1 + B2 = 76 - 11
? M + F = 65 .......(i)
According to the question,
(B1 + B2 + F) / 3 = (B1 + B2 + M) / 3 + 3
? B1 + B2 + F) = B1 + B2 + M + 9
? F = M + 9
? F - M = 9 ............. (ii)
From Eqs (i) and (ii), we get
F = 37 yr and M = 28 yr
Given expression = a2+b2+2ab = (a+b)2
where a = 387 and b = 114
= (387 + 114 )2
= (501)2
=(500 +1 )2
=(500)2 + (1)2+ 2 x 500 x 1
= 250000 + 1 +1000 = 251001.
3 pumps, can empty a tank in 2 days by working 8 hours a day.
1 pumps, can empty a tank in 2 days by working 8 x 3 hours a day.
1 pumps, can empty a tank in 1 days by working 8 x 3 x 2hours a day.
4 pumps, can empty a tank in 1 days by working 8 x 3 x 2 / 4 hours a day.
4 pumps, can empty a tank in 1 days by working 2 x 3 x 2 hours a day.
4 pumps, can empty a tank in 1 days by working 12 hours a day.
Total number of letters = 5
Number of vowels = 2.
If we consider both vowel as a one letter then,
Required number = 4! 2! = 48.
S.I for 1 year = Rs. (925 - 850) = Rs. 75
S.I for 3 year = Rs. (75 x 3) = Rs. 225
? Sum = Rs. (850 - 225) = Rs. 625
Original area = ?(d/2)2
= (?d2) / 4
New area = ?(2d/2)2
= ?d2
Increase in area = (?d2 - ?d2/4)
= 3?d2/4
? Required increase percent
= [(3?d2)/4 x 4/(?d2) x 100]%
= 300%
Interval of changes = (L . C . M of 48, 72, 108) sec. = 432 sec.
So, the lights will simultaneously changes after every 432 seconds i,e 7 min. 12 sec.
So, the next simultaneously changes will take place at 8 : 27 : 12 hrs.
Here n1 = 19, a1 = 74, n2 = 38, a1 = 63
Total average weight = (n1a1 + n2 a2) / (n1 + n2)
= [(19) (74) + (38) (63)] / (19 + 38)
= 3800/57 ? 67 kg
84 = 22 x 3 x 7
∴ H.C.F. = 22 x 3 = 12.
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