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.
Distance travelled = | ❨ | 18 x | 12 | ❩km = 3.6 km. |
60 |
Let the age of father = 5k
and the age of son = 2k
According to the question,
5k x 2k = 1000
? 10k2 = 1000
? k2 = 100
? k = ?100 = 10
? Father's age after 10 yr = (5k + 10)
= 5 x 10 + 10
= 60 yr
Here , x = a ? b, y = b ? c, z = c ? a
We have x + y + z = a ? b + b ? c + c ? a = 0
? x3 + y + z ? 3xyz = 0
Hence , the numerical value of algebraic expression is 0 .
Part filled by (A + B) in x min. + Part filled by A in (30 -x) min. = 1.
∴ x | ❨ | 2 | + | 1 | ❩ | + (30 - x). | 2 | = 1 |
75 | 45 | 75 |
⟹ | 11x | + | (60 -2x) | = 1 |
225 | 75 |
⟹ 11x + 180 - 6x = 225.
⟹ x = 9.
Weight of new student = Total weight of all 20 students - Total weight of initial 19 students
= (20 x 14.8 - 19 x 15) kg
= 11 kg.
Let their shares be Rs. 4x and Rs. 3x.
Then 3x = 2400
? x = 800
? Total amount = 7x = Rs. 5600
47619 x 7 = 333333.
Volume of rectangular box = 10 x 8 x 6 = 480 cm3
Volume of cubes = 6240 cm3
? Required boxes = Volume of cubes / Volume of rectangular box
= 6240/480 = 13
Hence, 13 boxes are needed.
Let man be represented by m and woman be represented by w.
? 2m + 1w = 1/14
? 14 x (2m + 1w ) = 1 ...(i)
and 4w + 2m = 1/8
8 x (4w + 2m) = 1 ...(ii)
On equating Eqs. (i) and (ii), we get
14 (2m + 1w) = 8 (4w + 2m)
? 28m + 14w = 32w + 16m
? 28m - 16m = 32w - 14w
? 12m = 18w
? m/w = 18/12 = 3/2
So, efficiency of 1 man and 1 woman is 3 : 2.
So, their wages must be in the same ratio
90/x = 3/2
[here, x = wages of a woman ]
? x = 90 x 2 / 3 = ? 60
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