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The sum age of A and B after 4 years is 28. If the ratio of the present age of A to B is 3: 2, then find the average of their ages after 6 years.
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Ratio of present age of A and B = 3: 2
A’s age 4 yrs after = 3X + 4
B’s age 4 yrs after = 2X + 4
Sum of A’s and B’s age = 3X + 4 + 2X + 4 = 5X + 8 = 28
X = 4
Therefore, present age of A = 3 x 4 = 12 yrs
& present age of B = 2 x 4 = 8 yrs
Average of A & B after 6 yrs = (12 + 6) + (8 + 6) = 32/2 = 16 yrs
Shortcut:
Average age of A and B,
After 4 years = 28/2 = 14
After 6 years =14 + 2 = 16
Two trains start at the same time from P and Q and moves towards each other at the speed of 80 and 95 km/hr respectively. When they meet, it is found that fastest train covers 210 km more than the other. Find the distance between P and Q.
Distance = time * speed
Let the time taken by them to meet each other be x then,
95x – 85x = 210
x = 14 h
Required distance = (80 + 95) * 14 = 2450 km
A vessel of 90 litres is filled with spirit and water. 75% of spirit and 30% of water is taken out of the vessel. It is found that the vessel is emptied by 55%. Find the initial quantity of spirit.
Therefore, ratio of spirit to water in the vessel = 5: 4
Thus, spirit = 90/9 x 5 = 50
Water = 90/9 x 4 = 40
What least number must be added to 2211 to make it exactly divisible by 84?
On dividing 2211 by 84, we get 27 as the remainder.
Therefore the number to be added = 84 – 27 = 57
Three pipes A, B and C can fill a cistern in 6 hrs. After working together for 3 hrs, C is closed and A and B fill the cistern in 9 hrs. Then, find the time in which the cistern can be filled by pipe C.
After 3 h quantity of tank = 3/6 = 1/2
Time taken by A and B to fill the tank = 9/1 * 2 = 18 h
Time taken by C to fill the tank = 1 ÷ (1/6 – 1/18) = 9 h
In the following questions, find the value of two quantities given in the question and find the relation between them.
Quantity I: a2X x a3X ÷ aX = a8, then find value of X.
Quantity II: Y2 + 6Y + 9 = 0, then find the values of Y.
Quantity I:
a2X x a3X ÷ aX = a8
a2X + 3X – X = a8
2X + 3X – X = 8
X = 2
Quantity II:
Y2 + 6Y + 9 = 0
Y(Y + 3) + Y(Y + 3) = 0
Y = -3, -3
Therefore, Quantity I > Quantity II
Quantity I: Side of a square is 48 cm. Find its area.
Quantity II: Diameter of the circle is 70 cm. Find its area.
Area of the square = a2 = 482 = 2304 cm2
Diameter = 70
Radius = 35
Area of the circle = 22/7 x 35 x 35 = 3850cm2
Therefore, Quantity I < Quantity II
Quantity I: C.P of an article is Rs.720 and S.P is Rs. 840. Find the profit percent.
Quantity II: Marked price of an article is 60% more than the C.P of the article. After giving discount of 20% it is sold at S.P of Rs. 7680. Find the profit or loss%.
Profit % = (840 – 720)/720 x 100 = 16(2/3)%
Let the CP be 100
MP = 160/100 x 100 = 160
SP = 160 x 80/100 = 128
According to question,
128 = 7680
1 = 60
100 = 6000
Therefore, CP of the article = Rs. 6000
Profit % = (7680 – 6000)/6000 x 100 = 28%
Short cut:
Profit % = 28/100 * 100 = 28%
Quantity I < Quantity II
Quantity I: A car covers a distance of 150 km in 1(1/2) hrs. Find the speed of the car.
Quantity II: Speed of the train is 25 m/sec.
Speed = 150/3 x 2 = 100 km/hr
Speed of the train = 25 m/sec = 25 x 18/5 = 90 km/hr
Quantity I: The ratio of present age of A to B is 7: 8, and the sum of their age is 75 years. Find the present age of A.
Quantity II: Age of C is 35 yrs.
Sum of A + B = 15X = 75
X = 75/15 = 5
Present age of A = 5 x 7 = 35 yrs
Present age of C = 35 yrs
Therefore, Quantity I = Quantity II
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