Method 1 to solve this question.
Let us assume the numbers of days of tour before extending the tour are D and daily expenses is E.
According to question,
Total expenses on tour = per day expenses x total number of days
Total expenses on tour = E x D = ED
360 = ED
? ED = 360
? E = 360/D .................(1)
Again according to question,
After extending the tour by 4 days, Number of days of tour = D + 4
Expenses will be reduced by 3 rupees, then everyday expenses = E - 3
so total expenses = (D + 4) x (E - 3)
360 = (D + 4) x (E - 3)
(D + 4) x (E - 3) = 360 .............................................(2)
Put the value of E in Equation (2). we will get,
(D + 4) x (360/D - 3) = 360
? (D + 4) x ( (360 - 3D)/D ) = 360
? (D + 4) x ( (360 - 3D) ) = 360 x D
? (D + 4) x (360 - 3D) = 360 x D
After multiplication by algebra law,
? 360 x D - 3D x D + 4 x 360 - 4 x 3D = 360D
? 360 x D - 3D
? - 3
D
? - 3
D
? 3
D
?
D
?
D
?
D
?
D(
D + 24) - 20(
D + 24) = 0
? (
D + 24) (
D - 20) = 0
Either (
D + 24) = 0 or (
D - 20) = 0
So
D = - 24 or
D = 20
But days cannot be negative so D = 20 days.
Method 2 to solve this question.
Let us assume the original day of tour is d days.
Given that his tour is extended for 4 days
Hence daily expenses per days = 360/(d + 4)
Therefore, According to question,
360/d - 360/(d + 4) = 3
? (360(d + 4) - 360d)/d x (d + 4) = 3
? (360(d + 4) - 360d)/d x (d + 4) = 3
? 360(d + 4) - 360d = 3d x (d + 4)
? 360d + 1440 ? 360d = 3(d
2 + 4d)
? 1440 = 3d
2 + 12d
? 3d
2 + 12d ? 1440 = 0
? d
2 + 4d ? 480 = 0
? d
2 + 24d ? 20d ? 480 = 0
? d(d + 24) ? 20(d + 24) = 0
? (d + 24)(d ? 20) = 0
? (d + 24) = 0 or (d ? 20) = 0
? d = ?24 or d = 20
Since days cannot be negative, d = 20
Hence his original duration of the tour is 20 days.
5% of A + 4% of B = | 2 | (6% of A + 8% of B) |
3 |
⟹ | 5 | A + | 4 | B | = | 2 | ❨ | 6 | A + | 8 | B | ❩ |
100 | 100 | 3 | 100 | 100 |
⟹ | 1 | A + | 1 | B | = | 1 | A + | 4 | B |
20 | 25 | 25 | 75 |
⟹ | ❨ | 1 | - | 1 | ❩ A = | ❨ | 4 | - | 1 | ❩ B |
20 | 25 | 75 | 25 |
⟹ | 1 | A = | 1 | B |
100 | 75 |
A | = | 100 | = | 4 | . |
B | 75 | 3 |
∴ Required ratio = 4 : 3
Speed = | ❨ | 240 | ❩m/sec = 10 m/sec. |
24 |
∴ Required time = | ❨ | 240 + 650 | ❩sec = 89 sec. |
10 |
Odd days in 1600 years = 0
Odd days in 300 years = (5 x 3) ≡ 1
97 years has 24 leap years + 73 ordinary years.
Number of odd days in 97 years ( 24 x 2 + 73) = 121 = 2 odd days.
Jan. Feb. March April May June (31 + 28 + 31 + 30 + 31 + 17) = 168 days
∴ 168 days = 24 weeks = 0 odd day.
Total number of odd days = (0 + 1 + 2 + 0) = 3.
Given day is Wednesday.
Other number = | ❨ | 11 x 7700 | ❩ | = 308. |
275 |
Let 1872/?N = 234.
Then ?N = 1872/234 = 8
? N = 8 x 8 = 64
Let distance covered by dog in 1 leap is x and
Distance covered by cat in 1 leap is y
Then, 3x = 4y
? x= 4/3y
Now, Ratio of speed of dog and cat = Ratio of distance covered by them in the same time = 4x : 5y
= 16/3y : 5y
= 16 : 15
Total volume of cuboid = (10 x 5 x 2) cm3
= 100 cm3
Volume curved = 1/3 x (22/7) x 3 x 3 x 7 cm3
= 66 cm3
% of wood wasted = (100 - 66)% = 34 %
Series pattern
12, 32, 52, 72, 92, 112
So missing term = 112 = 121
Angle of 360° is covered by hour hand in 12 h .
So, angle of 135° is covered by hour hand in 12/360° x 135° = 4.5 h
Thus, required time = (3 + 4.5)h = 7.5 h = 7 : 30
Try yourself. It was Wednesday.
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