Given Expression 112/?196 x ?579/12 x ?256/8
=(112/14) x (24/12) x (16/8)
= 32
?y/169= 54/39
? y/169 = (54/39) x (54/39)
? y= (54/39) x (54/39) x 169 = 324
In 19683, 19 lies between 23 and 33, so left digit is 2 and 683 ends with 3, so right digit is 7.
thus 27 is a cube root of 19683.
Total number of students = 480
percentage of total students passed = 75% of total student
= (75 x 480)/100 = 360 students
Now, using the condition from the question.
Let the number of boys be N
Then, 70% of N + 85% of (480 - N) = 360
? [(70 x N)/100] + [85 x (480 - N)]/100 = 360
? 70N - 85N + 40800 = 36000
? 40800 - 36000 = 85N - 70N
? 4800 = 15N
? N = 4800/15 = 320
? There are 320 boys who appeared for the examination,
Let us assume that Arun uses N units of petrol everyday. so the amount of of petrol in the tank when it is full will be 10N. If he starts using 25% more petrol everyday. then the units of petrol he now use everyday will be
N(1 + 25/100) = 1.25N
So, the number of days his petrol will now last will be equal to (amount of petrol in tank) / (number of units used everyday)
= (10N) / (1.25N) = 10 / 1.25
= 8 days
In 1998
Let number of students in x = 3k
Number of students in y = 5k and
Number of students in z = 6k
Next year,
number of students in x = 3k + 20% of 3k = 18k/5
Number of students in y = 5k + 10% of 5k = 11k/2
Number of students in z = 6k + 20% of 6k = 36k/5
According to the question,
(18k/5) / (36k/5) = 1/2
Thus, data is insufficient.
?4096 + ?40.96 + ?.004096
= ?4096 + ?4096/100 + ?4096/1000000
=?4096 + ?4096/10 + ?4096/1000
= 64 + 64/10 + 64/1000
= 70.464
?.04 = ?4/100
= 2/10
= 0.2
?288 / ?128 = ?9/4
= 3/2
?10 x ?15
= ?10 x 15
= ?150
= ?25 x 6
= ?25 x ?6
= 5?6
Let √x / 200 = 0.02
Then, √x = 200 x 0.02 = 4.
So, x = 16
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