The magnitude of a vector is never

Solution
The physical quantities, which possess both magnitude and direction are called vecors. Displacement, velocity. acceleration, force, momentum, gravitation field, electric field etc. are a few
In Bayer’s process the Bauxite ore (for purification) is digested in

Solution
In Bayer process the Bauxite ore is digested in NaOH
Which compound (given below) is not a peroxide?

Solution
A peroxide is a compound contatntng an oxygenoxygen single bond. PbO_{2} is not a peroxide
Bronze is an alloy. It’s constituents are

Solution
Bronze is an alloy of copper, zinc and tin.
What is the formula of Sodium Zincate?

Solution
The formula of sodium zmcate is Na_{2}ZnO_{2}
The ratio of the ages of X and y three years ago was 4 : 5 and that after three years will be 5 : 6. Then the sum ofthe ages of X and Y is

Solution
3 years ago,
X's age = 4x years
Y's age = 5x years
After 3 years of present age\(\frac{4\times +6}{5\times +6}=\frac{5}{6}\)
⇒ 25x+ 30 = 24x+ 36
⇒ x=3630=6
Sum of the present ages of X and Y
=4x+3 + 5x+ 3
= 9x + 6 = 9 x 6 + 6 = 60 years
The diagonal of a rectangular field is 50m and one of the sides is 48m. If the cost of cutting the grass of the field is Rs. 24 per square metre then the total cost of cutting all grass of the rectangular field is

Solution
AC = 50 metre
AB = 48 metre\(∴ BC=\sqrt{AC^{2}AB^{2}}\)
\(=\sqrt{50^{2}48^{2}}\)
\(=\sqrt{(50 + 48)(50  48)}\)
\(=\sqrt{98\times 2}\)
= 14 metre
∴ Area of field = 48 × 14
= 672 sq. metre
∴ Expenditure on cutting grass = 672 × 24=Rs.16128
The smallest positive integer n with 24 divisors, considering 1 and n as divisors, is

Solution
360 = 2^{3} × 3^{2} × 5
∴ Number of divisors
= (3 + 1) (2 + 1) (1 + 1)
=4×3×2=24
If 16 sheep or 12 horses eat the grass of a field in 20 days then in how many days will 5 sheep and 4 horses eat it ?

Solution
∵ 16 sheep ≡ 12 horses
∴ 5 sheep ≡^{12}⁄_{16}×5=^{15}⁄_{4}horses
∴5 sheep + 4 horses=^{15}⁄_{4}+4
=^{31}⁄_{4}horses
By M_{1}D_{1} =M_{2}D_{2}
12×20 =^{31}⁄_{4}×D_{2}\(\Rightarrow D_{2}=\frac{12\times 20\times 4}{31}\)
\(=\frac{960}{31}=30\frac{30}{31}\: days\)
The average height of 35 students in a class is 4’2″. Three students of average height 4′ 10″ moved to new section while 6 students of total height 33’4″ joined the class. The average height of the students in the class is now

Solution
Total height of 35 students = 50" × 35 = 1750"
Total height of 3 outgoing students = 58 × 3 = 174"
Total height of 32 students = 1750174= 1576"
∴ Total height of 38 students = 1576 + 33 × 12 + 4 = 1976"
Required average height\(=\frac{1976}{38}\)
=52"=4'4"