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In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016mm. What is the total thickness of the stack?

Solution ∵ Thickness of one book = 20 mm∴ Thickness of 5 books = 5 × 20 mm = 100 mm Again,Thickness of 1 paper sheet = 0.016 mm∴ Thickness of 5 paper sheets = 5 × 0.016 mm= 0.080m∴ Total thickness = 100 mm + 0.080 mm= 100.08 mm = 1.0008 × 10² mm

In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016mm. What is the total thickness of the stack? Read More »

In an AP, the sum of the first 3 terms is -36 and that of the last 3 is 27. If there are 10 terms, what are the 1 st term and the common difference respectively? A. 15,3 B. -15,3 C.15,-3 D.-15,-3

Solution The AP can be expressed as a, (a + d),-, (a + 9d). The sum of the first 3 terms is (3a + 3d) = -36 and the sum of the last 3 terms is (3a + 24d) = 27. Solving these two equations, we get a = -15 and d = 3.

In an AP, the sum of the first 3 terms is -36 and that of the last 3 is 27. If there are 10 terms, what are the 1 st term and the common difference respectively? A. 15,3 B. -15,3 C.15,-3 D.-15,-3 Read More »

A boy walks from his house at 4 km per hr. and reaches his school 9 minutes late. If his speed had been 5 km per hr. he would have reached his school 6 minutes earlier. How far his school from house?

(1) 6.5 km. (2) 5.5 km. (3) 6 km. (4)5 km. Solution: Let time taken to reach school at 4 kmph be x hrs.Then time taken to reach school at 5 kmph = (x+15/60)hrs Since, distance is equal. ,’. 4x=5(x+15/60) x= 5/4 hrs. Hence, distance between school & house = 4*5/4km= 5km Then the answer

A boy walks from his house at 4 km per hr. and reaches his school 9 minutes late. If his speed had been 5 km per hr. he would have reached his school 6 minutes earlier. How far his school from house? Read More »

On dividing p(x) by a polynomial x – 1 – x2, the quotient and remainder were (x – 2) and 3 respectively. Find p(x).

Solution Here,dividend = p(x)Divisor, g(x) = (x – 1 – x2)Quotient, q(x) = (x – 2)Remainder, r(x) = 3∵ Dividend = [Divisor × Quotient] + Remainder∴ P(x) = [g(x) × q(x)] + r(x)= [(x – 1 – x²) (x – 2)] + 3= [x² – x – x³ – 2x + 2 + 2x²] +

On dividing p(x) by a polynomial x – 1 – x2, the quotient and remainder were (x – 2) and 3 respectively. Find p(x). Read More »

The scores of a batsman in 10 cricket matches are as follows: 38, 45, 50, 60, 65, 70, 75, 80, 85, 90. Calculate the mean score.

Solution To find the mean score, add up all the scores and divide by the number of matches.Total sum of scores = 38 + 45 + 50 + 60 + 65 + 70 + 75 + 80 + 85 + 90 = 658.Number of matches = 10.Mean score = Total sum of scores / Number

The scores of a batsman in 10 cricket matches are as follows: 38, 45, 50, 60, 65, 70, 75, 80, 85, 90. Calculate the mean score. Read More »