Let each literature text book cost be Rs. X
Then, cost of mathematics text book = Rs. (X + 2)
ATQ,
5X = 38 + 3(X + 2)
5X = 38 + 3X + 6
2X = 44
X = 22.
Therefore, cost of each literature text book = Rs. X = Rs. 22.
And that of masthematics = X + 2 = Rs. 24.
To find the largest 4 digit number exactly divisible by 88,
We should divide the largest possible 4 digit number by 88, and if we get any remainder than subtract it from that largest number.
The largest possible 4 digit number is 9999
Now,
88) 9999 (113
88
_______
119
88
_______
319
264
_______
55
_______
Therefore, the largest 4 digit number exactly divisible by 88 is given by
9999 - 55 = 9944.
Let the middle digit be x.
Then, 2x = 10 or x = 5. So, the number is either 253 or 352
Since the number increases on reversing the digits, so the hundred's digit is smaller than the unit's digit.
Hence, required number = 253.
Two painters can complete two rooms in two hours.So 18 rooms can be painted in 6 hrs by 6 painters
Let t be the usual time. If the usual rate is 1 unit, then the unusual rate is 6/7.
Balance by the distance,
1 x t = (6/7)(t + 30)
7t = 6(t + 30)
we get t as,
t = 180 minutes = 3 hrs.
Let the number be x. Then,
15x - x = 56
x = 4
In mathematics, a geometric progression, also known as a geometric sequence, is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio.
Herein the given sequence, 4, 2, 1, 0.5, 0.25, 0.125,...
Common Ratio r = 2/4 = 1/2 = 0.5/1 = 0.25/0.5 = 0.125/0.25 == 0.5.
Let the number be v. Then, 3 (2x + 9) = 75.
2x + 9 = 25
=> 2x = 16
=> x = 8
Let the fraction be x/y. Then, x /(y+1) = 1/2
=> 2x - y = 1 ..........(i)and
(x + 1) / y = 1 => x - y = 1 .......(ii)
Solving (i) and (ii), we get : x = 2, y = 3. Hence, the required fraction is 2/3..
Let the ten's digit be x and unit's digit be y. Then, (10x + y) - (x + y) = 9 or x = 1.
From this data we cannot find y, the unit's digit. So the data is inadequate..
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