A paper is in the form of a rectangle ABCD where AB = 22 cm and BC = 14 cm. A semicircular portion with BC as diameter is removed. Find the remaining area of the paper [in cm^{2}].

Solution
Remaining area of the paper
\(=22times 14frac{1}{2}times frac{22}{7}times left ( frac{14}{2} right )^{2}\)
= 308  11 × 7
= 308  77 = 231 cm^{2}
Three pipes A, B and C can fill a cistern in 10, 12 and 15 hours respectively, while working alone. If all the three pipes are opened together, then find. the time taken to fill the cistern

Solution
∵ Part of the cistern filled by (A+ B + C) pipes
in 1 hour =\(frac{1}{10}+frac{1}{12}+frac{1}{15}=frac{6+5+4}{60}=frac{15}{60}=frac{1}{4}\)
∴ Reqd. time taken to fill the cistern = 4 hours
If ^{1}⁄_{3} of A = 75% of B = 0.6 of C, then A : B : C IS

Solution
∵ ^{1}⁄_{3}3 × A = 75% × B = 0.6 × C
\(Rightarrow frac{1}{3}A=frac{3}{4}B=frac{3}{5}C\)
∴ A:B:C = A:^{4}⁄_{9}:^{9}⁄_{5}=9:4:5
A shopkeeper offered a discount of 9% for an article, but he marked it at 25% higher than the cost price. Find his profit percentage

Solution
Let the cost price of the article be Rs x Marked price of the article
\(=left ( frac{100+25}{100} right )times x=Rsfrac{5}{4}x\)
Selling price. of the article
\(=left ( frac{1009}{100} right )times frac{5}{4}x=Rsfrac{91}{80}x\)
∴ Reqd. profit percentage = \(=frac{frac{91}{80}xx}{x}times 100\)%
\(=frac{frac{91}{80}xx}{x}times 100\)%
=^{55}⁄_{4}%= 13.75%
The number of bricks required for a wall which is 8 m long, 6 m high and 22.5 em thick, if each brick measures 25 cm × 11.25 cm × 6 cm, is

Solution
Reqd. number of bricks
\(=frac{Volume ; of ; the ; wall}{Volume ; of ; one ; brick} = frac{800 times 600 times 22.5}{25 times 11.25 times 6}\)
\(=frac{8times 6times 225times 100000}{25 times 1125 times 6}=6400\)
Cauvery water sharing is a dispute between
The Magsaysay Award for social service was instituted by
In how many denominations is Indian paper currency printed at present?
Tundras are
Tides in the sea are caused by
If a = 1, b = 2 and c = – 3, then the value of\(\frac{a^{3}+ b^{3}+ c^{3} 3ab}{ab + be + ca – (a^{2} + b^{2} + c^{2})}\) is

Solution
∴ a = 1, b = 2 and c =  3
∴ a + b + C = 1 + 2  3 = 0
∴ a^{3} + b^{3} + c^{3} = 3 abc
∴ Given expression=\(frac{a^{3}+ b^{3}+ c^{3} 3ab}{ab + be + ca  (a^{2} + b^{2} + c^{2})}\)
\(=frac{0}{ab + be + ca  (a^{2} + b^{2} + c^{2})}\)
When a = ^{4}⁄_{3} the value of 27a^{3} – 108a^{2} + 144a 317 is

Solution
Given expression = 27a^{3} 108a^{2} + 144a  317
\(left [ a=frac{4}{3} right ]\)
\(=27left ( frac{4}{3} right )^{3}108left ( frac{4}{3} right )^{2}+144frac{4}{3}317\)
= 64  192 + 192  317 =  253
The circumference of a circle is equal to the perimeter of a square of side 22 cm. The area of the circle is

Solution
Let the radius of the circle be R cm.
∵ Circumference of circle = Perimeter of a square⇒ 2 × ^{22}⁄_{7} × R = 4 × 22 ⇔ R = 2 × 7 = 14 cm
∴ Area of the circle = πR^{2}= π× (14)^{2}= 196 πcm^{2}
A refrigerator listed at ₹ 4000. Due to the festival season a shopkeeper announces a discount of 5%. Then the selling price of refrigerator (in ₹) is

Solution
Selling price of refrigerator =\(frac{(1005)}{100}× Rs. 4000\)
\(=frac{19}{20}times 4000 = Rs.3800\)
ABCD is a rectangle with AB = h^{2} and AD = 3p. If h is doubled and p is halved, then the

Solution
∴ Area of rectangle ABCD = AB × AD = h^{2} × 3p
If h is doubled and p is halved, then
New area of rectangle =\(left ( 2h right )^{2}times 3left ( frac{p}{2} right )\)
\(=4h^{2}times frac{3p}{2}\)
= 2 ^{2} (h2 × 3p)
= Area is doubled
The planet nearest to the Sun is

Solution
Mercury is the closest planet to the Sun. It is extremely hot planet. Mercury planet has no protective blanket like Ozone around it to prevent from harmful radiations. It has no satellite.
In winter, when water freezes due to cold, fishes and other aquatic animals

Solution
In winter, when water freezes due to cold, fishes and other aquatic animals can live because only the upper layer of the water freezes.
HIV is a

Solution
HIV is a viral disease. ELISA(Enzyme Linked Immune Sorvent Assay) is a test for HIV
virus.
For a living organism, greatest available energy is from

Solution
The greatest available ertergy is from ATP. 2 Atom of ATP equals to 16000 calorie (2×8000) energy
Haemophilia is a kind of disease which is

Solution
Haemophilia is a kind of disease which is hereditary.