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If x = √3-√2/√3+√2 and y = √3+√2/√3-√2 find the value of x3+y3 :

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(1) 807 (2) 907 (3) 970 (4)870 Solution: Rationallsing the denominators. we have x=√3-√2/√3+√2*√3-√2/√3-√2 =(√3-√2)2/(√3)2-(√2)2 = 3+2-2√3*√2/3-2 = 5-2√6 and,  y=√3+√2/√3-√2*√3+√2/√3+√2 = (√3+√2)2/(√3)2-(√2)2 = 3+2+2√3*√2/3-2 = 5+2√6 :.  x+ y = 5-2√6+5+2√6 = 10 and,  xy= (5-2√6) (5+2√6) = 52-(2√6)2 =25-24=1 x3+y3 =(x+y)3-3xy (x+ y) = 103– 3 *10 = 1000 – 30 =970 Then the option is […]

If x = √3-√2/√3+√2 and y = √3+√2/√3-√2 find the value of x3+y3 : Read More »

In an AP, the ratio of the 2nd term to the 7th term is 1/3. If the 5th term is 11, what is the 15th term? A. 28B. 31C. 33D. 36

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Solution The 2nd and the 7th terms are (a + d) and (a + 6d) respectively. The ratio of these terms is 1/3. Solving this ratio, we get 2a = 3d. The 5th term is (a + 4d) = 11. Substituting for a, we get a = 3 and d = 2. Therefore, the 15th

In an AP, the ratio of the 2nd term to the 7th term is 1/3. If the 5th term is 11, what is the 15th term? A. 28B. 31C. 33D. 36 Read More »

The average age of 30 students of a class is 14 years 4 months. After admission of 5 new students in the class the average becomes 13years 9 months. The youngest one of the five new students is 9 years 11 months old. The average age of the remaining 4 new students is:

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(1) 11 years 2 months (2) 13 years 6 months (3) 12 years 4 months (4) 10 years 4 months Solution: Total age of initial 30 students = 14years 4 months * 30 = 430 years Total ageof 35 students = 13years 9 months * 35 = (455 + 26)years 3 months = 481 years

The average age of 30 students of a class is 14 years 4 months. After admission of 5 new students in the class the average becomes 13years 9 months. The youngest one of the five new students is 9 years 11 months old. The average age of the remaining 4 new students is: Read More »

My grandfather was 9 times older than me 16years ago. He will be 3 times of my age 8 years from now. Eight years ago, the ratio of my age to that of my grandfather was:  (1) 3: 8 (2) 2: 5 (3) 1: 2 (4) 1: 5

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Solution: 16 years ago, My age = x years My grandfather’s age = 9x years After 8 years from the present, 9x+ 16+8=3(x+8+ 16) ⇒9x+ 24 = 3x+ 24 + 48 ⇒9x+24=3x+72 ⇒9x- 3x= 72 -24  ⇒6x= 48 ⇒x=48/6 = 8 Required ratio 8 years ago, = (x+ 8) : (9x+ 8) = (8 +

My grandfather was 9 times older than me 16years ago. He will be 3 times of my age 8 years from now. Eight years ago, the ratio of my age to that of my grandfather was:  (1) 3: 8 (2) 2: 5 (3) 1: 2 (4) 1: 5 Read More »

From the sum of 3a−b+9 and −b−9 , subtract 3a−b−9

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Solution Given: expressions 3a−b+9 , −b−9 , 3a−b−9We need to subtract 3a−b−9 from the sum of 3a−b+9 and −b−9The sum of the first two terms, −b−9 and 3a−b+9 will be3a−b+9+(−b−9)=3a−b+9−b−9=3a−2bNow subtracting 3a−b+9 from 3a−2b , we get3a−2b−(3a−b−9)=3a−2b−3a+b=9=−b+9

From the sum of 3a−b+9 and −b−9 , subtract 3a−b−9 Read More »

The ratio of the ages of two persons is 4 : 7 and the age of one of them is greater than that of the other by 30 years. The sum of their ages (in years) is

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(1) 110 (2) 100 (3) 70 (4) 40k Solution: Ages of the persons = 4x and 7x years. :. 7x-4x=30 ⇒3x=30 ⇒ x=10 :. Sum of their ages = 4x+7x = 11x years = 11*10= 110 years Then the option is (1) 110

The ratio of the ages of two persons is 4 : 7 and the age of one of them is greater than that of the other by 30 years. The sum of their ages (in years) is Read More »