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Directions: The following pie chart shows the marks scored by a student in different subjects viz. Physics (Ph), Chemistry (Ch), Mathematics (M), Social Science (SS) and English (E) in an examination. Assuming that total marks obtained for the examination is 810, answer the questions given below.
The marks obtained in English, Physics and Social Science exceed the marks obtained in Mathematics and Chemistry by
The difference of marks between Physics and Chemistry is same as that between
As we know share of marks will be proportional to share of angles in pie chart
Hence Difference of marks in physics and chemistry is proportional to 15
and difference of marks in chemistry and social science is also proportional to 15 too.
So answer will be D
Study the above bar graph showing the production of food grains (in million tons). What is the ratio between the maximum production and the minimum production during the given period?
Graph should be drawn
Clearly, in the year 2007, maximum production of 100 mil tonnes is observed.
While in the year 2008, minimum production of 40 mil tonnes is observed.
Hence, the ratio is 100 : 40 = 5 : 2
Rs. 700 is divided among A, B, C in such a way that the ratio of the amounts of A and B is 2 : 3 and that of B and C is 4 : 5. Find the amounts in Rs. each received, in order A, B, C.
A : B = 2 : 3
B : C = 4 : 5
A : B : C = 8 : 12 : 15
amount received by A = 8/(8 + 12 + 15) * 700 = 160/
amount received by B = 12/(8 + 12 + 15) * 700 = 240/
amount received by C = 15/(8 + 12 + 15) * 700 = 300/
so the answer is option B.
ABC is a right angled triangle, B being the right angle. Midpoints of BC and AC are respectively B’ and A’. The ratio of the area of the quadrilateral AA’ B’B to the area of the triangle ABC is
let A', B', C' are the midpoints of AC, BC, AB.
areas of all 4 small triangles are same, let it be x.
area of quadrilateral AA'B'B = 3x
area of ABC = 4x
RATIO = 3x : 4x
= 3 : 4
so the answer is option C.
The ratio of the volumes of water and glycerine in 240cc of a mixture is 1 : 3. The quantity of water (in cc that should be added to the mixture so that the new ratio of the volume of water and glycerine becomes 2 : 3 is
Let's say we added x cc of water.
According to previous ratio amount of water in mixture is 240 × 1/4 = 60
So after adding amount of water = 60 + x
amount of glycerine = 240 × 3/4 = 180
Hence new ratio will be =
or x will be 60.
If the distance between two point (0, 5) and (x, 0) is 13 unit, then x =
Distance^{2} = diff. of x cordinates^{2} + diff. of y coordinates^{2}
13^{2} = x^{2} + 5^{2}
Hence x = ±12
A and B together can do a works in 12 days. B and C together do it in 15 days. If A's efficiency is twice that of C, then the days required for B alone to finish the work is
A and B will do in 1 day, 1/12 amount of work
B and C will do in 1 day, 1/15 amount of work
Hence adding up above two equations
A + 2B + C will do in 1 day, 9/60 amount of work
as it is mentioned A = 2C (i.e. efficiency of A is twice of C)
So 2C + 2B + C will do in 1 day 9/60 amount of work
or 2(B+C) + C = 9/60
or C = 1/60 (as B + C = 1/15)
So B alone will do work in 20 days.
A drum of kerosene is 3/4 full. When 30 litres of kerosene is drawn from it, it remains 1/2 full. The capacity of the drum is
Drum is 3/4 full.
When 30 liters are drawn out of it, it becomes 1/2 full.
Therefore 3/4  1/2 of drum = 30
1/4 of drum = 30
Total capacity of drum = 30 × 4 = 120 litres
A rectangular sheet of metal is 40 cm by 15 cm. Equal squares of side 4 cm are cut off at the corners and the remainder is folded up to form an open rectangular box. The volume of the box is
After cutting off squares of sides 4cm. from rectangular sheet of 40cm × 15cm and making a rectangular box
New dimentions of rectangular will be:
Base = 32cm × 7cm
Height = 4 cm
Hence volume will be 32 × 7 × 4 = 896 cm^{3}
Water flows into a tank which is 200m long and 150m wide, through a pipe of crosssection 0.3m × 0.2m at 20 km/hour. Then the time (in minutes) for the water level in the tank to reach 8cm is
Volume of water in time t to reach at 8 cm. height will be equal to volume of tank till height 8 cm.
hence volume of water in time t to reach at height 8 cm. = .3 × .2 × 2000/6 × (speed in per minute)
which is equal to volume of tank = 200 × 150 × 8/100
equating both and after solving we will get t = 120 min.
Taking cosθ outside in numerator and in denominator and making tanθ
hence eq will be
As it is given that 5tanθ = 4
after putting values and solving we will get the equation reduced to 1/7.
For asin θ + bcos θ + c,
for sin θ + cos θ , a = 1, b = 1, c = 0
so the answer is option B.
If x = cosecθ − sinθ and y = secθ − cosθ, then the value of x^{2}y^{2}(x^{2} + y^{2} + 3)
Similarly y = sinθcosθ
x^{2} + y^{2} + 3 = (sec^{2}θ + cosec^{2}θ)
Now putting above values in given equation, and after solving it will be reduced to 1
The value of sin^{2} 22° + sin^{2} 68° + cot^{2 }30° is
Expression: sin^{2} 22° + sin^{2} 68° + cot^{2 }30°
We know that sin(90° − θ) = cosθ
=> sin22° = sin(90° − 68°) = cos68°
=> cos^{2}68° + sin^{2}68° + (√3)^{2}
=> 1 + 3 = 4
If the number of vertices, edges and faces of a rectangualr parallelopiped are denoted by v, e and f respectively, the value of (v  e + f) is
Euler's polyhedral formula states that the number of vertices V, faces F and edges E in a polyhedron satisfy:
V + F  E = 2
=> Ans  (B)
The average age of a jury of 5 is 40. If a member aged 35 resigns and man aged 25 becomes a member, then the average age of the new jury is
Initially, sum of ages of the jury = 200,
then man aged 35 resigns, sum of ages = 200  35
then man aged 25 joins, sum of ages = 200  35 + 25
Final number of members in the jury = 5
So, average age of the jury = (200  35 + 25)/5 = 38
I am three times as old as my son. 15 years hence, I will by twice as old as my son. The sum of our ages is
Let's say son's age is x
hence father's present age will be 3x
after 15 years son's age will be x + 15 and father's age will be 3x + 15 and it is twice the age of son
so 3x + 15 = 2 (x + 15)
solve for x
If the cube root of 79507 is 43, then the value of is
=> 4.3 + 0.43 + 0.043 = 4.773
It is given that √6 x √15 = x√10
√6 x √15 = √(3x2) x √(3 x 5) = √(9 x 10) = 3 √10
Hence x = 3
If x + y = 15, then (x − 10)^{3} + (y − 5)^{3} is
given that x + y = 15
So rearranging the above equation
(x  10 ) + (y  5) =0
Now , x  10 = (y  5)
On cubing both sides
(x − 10)^{3} =  (y − 5)^{3}
And hence
(x − 10)^{3} + (y − 5)^{3} = 0
We need to find value of
= 0.519 ~ 0.52
It is given that
x = 16
If a + 1/a = 1, then the value of a^{2} + 1/a^{2 }is:
it is given that a + 1/a = 1
and we need to find value of
= 1^{2}  2 = 1
It is given that 5% of √2x is 0.01
x = 0.02
If a * b = 2a + 3b  ab, then the value of (3 * 5 + 5 * 3) is
For 3 * 5 put a = 3 and b = 5 in given equation
and for 5 * 3 put a = 5 and b = 3 in equation
now add both values
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