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URVASHI GARG

Two pipes A and B can fill a tank in 36 minutes and 45 minutes respectively. Another pipe C can empty the tank in 30 minutes. First A and B are opened. After 7 minutes, C is also opened. The tank is filled up in

(1) 39 minutes (2) 46 minutes (3) 40 minutes (4) 45 minutes Solution: Part of the tank filled by pipes A and B in 1 minute = 1/36+1/45= 5+4/180 = 9/180 = 1/20 Part of the tank filled by these pipes in 7 minutes =7/20 Remaining unfilled part = 1-7/20= 20-7/20= 13/20 When all three […]

Two pipes A and B can fill a tank in 36 minutes and 45 minutes respectively. Another pipe C can empty the tank in 30 minutes. First A and B are opened. After 7 minutes, C is also opened. The tank is filled up in Read More »

The sum of length, breadth and height of a rectangular parallelopiped is 20 cm.and its whole surface area is 264 sq. cm. Find the area of the square whose side is equal to the length of the diagonal of the parallelopiped.

(1) 136 square cm. (2) 120 square cm. (3) 125 square cm. (4) 100 square cm. Solution: Let the length, breadth and height of the parallelopiped be a, b and c cm respectively. :. a+ b+ c= 20 2 (ab + bc + ca) = 264 :. (a+b+c)2= a2+b2+c2+2(ab + bc+ ca) ⇒400= a2+b2+c2+264 ⇒a2+b2+c2=

The sum of length, breadth and height of a rectangular parallelopiped is 20 cm.and its whole surface area is 264 sq. cm. Find the area of the square whose side is equal to the length of the diagonal of the parallelopiped. Read More »

A moving train passes a platform 50 metres long in 14 seconds and a lamp-post in 10 seconds. The speed of the train is

(1) 24 km/hr. (2) 36 km/hr. (3) 40 km/hr. (4)45 km/hr. Solution: ⇒Suppose length of train be x  According to question x+50/14= x/10 ⇒ 14x = 10x+500 ⇒ 4x= 500 ⇒ x= 500/4=125m Therefore, speed = 125/10*18/5= 45kmph

A moving train passes a platform 50 metres long in 14 seconds and a lamp-post in 10 seconds. The speed of the train is Read More »

By walking 5/3 of usual of speed a student reaches school 20 minutes earlier. Find his usual time.

(1) 45 Minutes (2) 50 Minutes (3) 60 Minutes (4) None of these Solution: 5/3 of usual 3/5 of usual time as he reaches earlier. = 3/5 usual time + 20 Minutes= Usual time 20 minutes = (1-3/5) usual time = 2/5 usual time Usual time = 20*5/2= 50 Minutes Then answer is 50 minutes.

By walking 5/3 of usual of speed a student reaches school 20 minutes earlier. Find his usual time. Read More »