A beats B by 11 meters. When B completes the 11 meters, there is a lead of 80 meters to C
So, C must have travelled only 90 - 80 = 10 meters
When B travels 11 meters, C travels only 10 meters
Ratio of distance travelled by second and third horse are11x and 10 x respectively
We know that the second horse beats the third horse by 80 meters.
So, Length of the track = Distance travelled by the second horse = 11 x 80 = 880 meters
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport 1/3 of his total journey time, while Bimal took each mode of transport 1/3 of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to
Let the distance be = 60 Kms.
So, Bimal travels a distance of 20Kms in each mode and Amal travels 10, 20 and 30 Kms respectively in 1 hour each
Time taken by Bimal = Time taken to travel 20 kms in 10 Kmph + Time taken to travel 20 kms in 20 Kmph + Time taken to travel 20 kms in 30 Kmph
So, Percentage increase in time = 22
Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs 250?
We know, Interest income Amount invested
And Interest income Interest rate
Therefore, Interest income must be in the ratio of the product of their Amount invested and Interest rate.
So, Ratios of Interest incomes of Amala, Bimala and Gouri = 18: 20: 20
Bina’s Interest income exceeds Amala by 250 Rs.
So, 20 x - 18 x = 250
2x = 250
So, Total Interest income = 250 x (9 + 10 + 10) = 7250 Rs.
Total Interest Income = Rs. 7250
This question can be solved by comparing powers of 2 and 3
Comparing the powers in both sides,
3n + 12 = 4m
4 + 2m = n
Substituting the value of n we get,
3 (4 + 2m) + 12 = 4m
12 + 6m + 12 = 4m
24 = -2m
m = -12
A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is
In Liquid L1 - 1L = 1000 grams
So, 1000 mL = 1000 grams
1 mL = 1 gram
x mL = x grams
In Liquid L2 - 1L = 800 grams
1000 mL = 800 grams
1mL = 0.8 grams
(500-x) mL = (500-x) x 0.8 grams
Total mass = x + 400 - 0.8x = 480 grams
0.2x = 80 grams
x = 400 grams
Therefore, Liquid 1 has 400 mL and Liquid 2 has 500 - 400 = 100 Ml
Therefore, Percentage of Liquid 1 100 = 80%
If the rectangular faces of a brick have their diagonals in the ratio then the ratio of the length of the shortest edge of the brick to that of its longest edge is
Let S be the set of all points (x,y) in the x-y plane such that |x| + |y| ≤ 2 and |x| ≥ 1. Then, the area, in square units, of the region represented by S equals [TITA]
Construct the given data on a rough graph.
Required area = Sum of area of two smaller triangles [ (-2,0) (-1,1) (-1,-1) and (2,0) (1,-1) (1,1) ]
Required area = base x height
Required area = 1 x 2 = 2 Sq units