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In a village, the ratio of number of males to females is 5 : 4. The ratio of number of literate males to literate females is 2 : 3. The ratio of the number of illiterate males to illiterate females is 4 : 3. If 3600 males in the village are literate, then the total number of females in the village is
In the question, it is given that the ratio of number of literate males to literate females is 2 : 3.
Given, the number of literate males = 3600
The number of literate females = $$\frac{3600}{2}\times3$$ = 5400
Males : Females = 5 : 4
Let the number of males be 5y and the number of females be 4y
Illiterate males = 5y - 3600
Illiterate females = 4y - 5400
It is given,
$$\ \frac{\ 5y-3600}{4y-5400}=\frac{4}{3}$$
15y - 10800 = 16y - 21600
y = 10800
Number of females = 4y = 4*10800 = 43200
The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is
Let the original number of students be 'n' whose average weight is 'x'
Let the number of students added be 'm' and the average weight will be x + 3
We need to find the value of n : m
It is given, average weight of students in a class increased by 0.6 after new students are added.
Therefore,
$$\ \frac{\ nx+m\left(x+3\right)}{n+m}=x+0.6$$
$$\ \ nx+mx+3m=mx+nx+0.6n+0.6m$$
$$2.4m=0.6n$$
$$4m=n$$
$$\frac{n}{m}=\frac{4}{1}$$
The answer is option C.
For any natural number n, suppose the sum of the first n terms of an arithmetic progression is $$(n + 2n^2)$$. If the $$n^{th}$$ term of the progression is divisible by 9, then the smallest possible value of n is
It is given,
$$S_n=2n^2+n$$
$$S_{n-1}=2\left(n-1\right)^2+\left(n-1\right)$$
$$S_{n-1}=2n^2-3n+1$$
$$T_n=S_n-S_{n-1}=2n^2+n-2n^2+3n-1=4n-1$$
$$T_n=4n-1$$
The terms are 3, 7, 11, 15, 19, 23, 27,......
27 is the first term in the series divisible by 9.
27 is the 7th term.
Therefore, the least possible value of n is 7.
The answer is option C.
Let $$0 \leq a \leq x \leq 100$$ and $$f(x) = \mid x - a \mid + \mid x - 100 \mid + \mid x - a - 50\mid$$. Then the maximum value of f(x) becomes 100 when a is equal to
x>=a, so |x-a| = x-a
x<100, so |x-100| = 100-x
f(x) = (x-a) + (100-x) + |x-a-50| =100
or, |x-a-50| = a
From the graph we can can see that when x=a then
|x-a-50|=a
or, a= 50
Similarly when x=a+100
|x-a-50|=a
or, a= 50
So value of a is 50 when f(x) is 100.
Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is
M - First meeting point
Let the speeds of trains A and B be 'a' and 'b', respectively.
$$\dfrac{x}{a}=\ \dfrac{\ D-x}{b}$$
It is given,
$$\dfrac{D}{a}=10$$ and $$\dfrac{x}{b}=9$$
$$\dfrac{x}{\dfrac{D}{10}}=\ \dfrac{\ D-x}{\dfrac{x}{9}}$$
$$\dfrac{10x}{D}=\ \dfrac{\ 9D-9x}{x}$$
$$10x^2=\ \ 9D^2-9Dx$$
$$10x^2+9Dx-9D^2=\ 0$$
Solving, we get $$x=\dfrac{3D}{5}$$
$$\dfrac{x}{b}=9$$
$$\dfrac{3D}{b\times5}=9$$
$$\dfrac{D}{b}=15$$
The total time taken by train B to travel from station Y to station X is 15 minutes.
The answer is option B
A trapezium $$ABCD$$ has side $$AD$$ parallel to $$BC, \angle BAD = 90^\circ, BC = 3$$ cm and $$AD= 8$$ cm. If the perimeter of this trapezium is 36 cm, then its area, in sq. cm, is
CD = $$\sqrt{\ y^2+25}$$
$$11+y+\sqrt{y^2+25}=36$$
$$\sqrt{y^2+25}=25-y$$
$$y^2+25=25^2+y^2-50y$$
2y = 24
y = 12
Area of trapezium = $$3y+\frac{5y}{2}=\frac{11y}{2}=\frac{11}{2}\left(12\right)=66$$
Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is
It is given,
7C = 30P = 9A and Ankita bought 4C, 14P and 6A.
Let 7C = 30P = 9A = 630k
C = 90k, P = 21k, and A = 70k
Cost price of 4C, 14P and 6A = 4(90k)+14(21k)+6(70k) = 1074k
Marked up price = 1074k + 1752
S.P = $$\frac{1}{6}\left(1074k+1752\right)+\left(\frac{4}{5}\right)\left(\frac{5}{6}\right)\left(1074k+1752\right)$$ = $$\frac{5}{6}\left(1074k+1752\right)$$
S.P - C.P = profit
$$1460-\frac{1074k}{6}=744$$
$$\frac{1074k}{6}=716$$
k = 4
Money spent on buying almonds = 420k = 420*4 = Rs 1680
The answer is option A.
Let A be the largest positive integer that divides all the numbers of the form $$3^k + 4^k + 5^k$$, and B be the largest positive integer that divides all the numbers of the form $$4^k + 3(4^k) + 4^{k + 2}$$ , where k is any positive integer. Then (A + B) equals
A is the HCF of $$3^k+4^k+5^k$$ for different values of k.
For k = 1, value is 12
For k = 2, value is 50
For k = 3, value is 216
HCF is 2. Therefore, A = 2
$$4^k+3\left(4^k\right)+4^{k+2}=4^k\left(1+3\right)+4^{k+2}=4^{k+1}+4^{k+2}=4^{k+1}\left(1+4\right)=5\cdot4^{k+1}$$
HCF of the values is when k = 1, i.e. 5*16 = 80
Therefore, B = 80
A + B = 82
Let a, b, c be non-zero real numbers such that $$b^2 < 4ac$$, and $$f(x) = ax^2 + bx + c$$. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be
$$b^2 < 4ac$$ means that the discriminant is less than 0. Therefore, f(x)>0 for all x if the coefficient of $$x^2$$ is positive, and f(x)<0 for all x if the coefficient of $$x^2$$ is negative.
We are given that f(m)<0 and m is an integer.
So the set containing values of m will either be empty if the coefficient of $$x^2$$ is positive, or it will be a set of all integers if the coefficient of $$x^2$$ is negative.
The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is
Let the number of balloons each child received be 2a, 2b, 2c and 2d
2a + 2b + 2c + 2d = 20
a + b + c + d = 10
Each of them should get more than zero balloons.
Therefore, total number of ways = $$(n-1)_{C_{r-1}}=(10-1)_{C_{4-1}}=9_{C_3}=84$$
Let a and b be natural numbers. If $$a^2 + ab + a = 14$$ and $$b^2 + ab + b = 28$$, then $$(2a + b)$$ equals
a(a + b + 1) = 14 ...... (1)
b(a + b + 1) = 28 ...... (2)
$$\frac{a}{b}=\frac{1}{2}$$
b = 2a
Substituting in (1), we get
a(3a + 1) = 14
$$3a^2+a-14=0$$
$$3a^2-6a+7a-14=0$$
$$3a\left(a-2\right)+7\left(a-2\right)=0$$
Given, a and b are natural numbers.
Therefore, a = 2 and b = 4
2a + b = 2(2) + 4 = 8
The answer is option A.
Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg and makes an overall profit of 64%. Then, Amal’s cost price for syrup, in rupees per kg, is
Total syrup - 110 kg
Total juice - 120 kg
It is given, cost price of syrup is 20% less than the cost price of juice.
Let the cost price of juice per kg be 10CP
Cost price of syrup per kg is 8CP
10kg syrup -> cost price = 80CP
It is given, 10kg syrup is sold at 10% profit. This implies selling price = 1.1*80CP = 88CP
20kg juice -> cost price = 200CP
It is given, 20kg juice is sold at 20% profit. This implies selling price = 1.2*200CP = 240CP
It is given, Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg
Selling price of the remaining mixture = 308.32*200 = Rs 61664
Total S.P = 61664 + 328CP
Total C.P = 880CP + 1200CP = 2080CP
Overall profit = 64%
$$61664+328CP=\frac{164}{100}\left(2080CP\right)$$
Solving, we get CP = 20
Cost price for syrup per kg = 8CP = 8*20 = Rs 160
All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is
$$A=lb$$
$$l^2+b^2=4r^2$$
$$P\ =2\left(l+b\right)$$
$$\frac{P}{2}=l+b$$
Squaring on both the sides, we get
$$\frac{P^2}{4}=l^2+b^2+2lb$$
$$\frac{P^2}{4}=4r^2+2lb$$
$$\frac{P^2}{8}-2r^2=lb$$
The answer is option B.
The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is
It is given that average of three numbers is 13.
Sum = 3*13 = 39
It is given, $$\ \frac{\ 39+n}{4}$$ is a odd number.
Minimum value $$\ \frac{\ 39+n}{4}$$ can take such that n is a natural number is 11
$$\ \frac{\ 39+n}{4}=11$$
n = 5
The answer is option C.
A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is
Lemon juice : sugar syrup in the mixture is 1:1, i.e. 50% Lemon juice and 50% sugar syrup.
In sugar syrup, 100% is sugar syrup.
These two are mixed in the ratio 1:3.
Lemon juice = $$\ \dfrac{\ 1\left(50\%\right)}{1+3}$$
Sugar syrup = $$\ \dfrac{\ 1\left(50\%\right)+3\left(100\%\right)}{1+3}=\dfrac{350}{4}$$
Required ratio = 50:350 = 1:7
The answer is option D.
The largest real value of a for which the equation $$\mid x + a \mid + \mid x - 1 \mid = 2$$ has an infinite number of solutions for x is
In the question, it is given that the equation $$\mid x + a \mid + \mid x - 1 \mid = 2$$ has an infinite number of solutions for any value of x. This is possible when x in |x+a| and x in |x-1| cancels out.
Case 1:
x + a < 0, x - 1 $$\ge\ $$ 0
- a - x + x - 1 = 2
a = -3
Case 2:
x + a $$\ge\ $$ 0 and x - 1 < 0
x + a - x + 1 = 2
a = 1
Largest value of a is 1.
The answer is option C.
In a class of 100 students, 73 like coffee, 80 like tea and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is
Let n, s, d and t be the number of students who likes none of the drinks, exactly one drink, exactly 2 drinks and all three drinks, respectively.
It is given,
n + s + d + t = 100 ...... (1)
s + 2d + 3t = 73 + 80 + 52
s + 2d + 3t = 205 ...... (2)
(2)-(1), we get
d + 2t - n = 105
Maximum value t can take is 52, i.e. t = 52, d = 1 and n = 0
Minimum value t can take is 5, i.e. t = 5, d = 95 and n = 0 (This also satisfies equation (1))
Difference = 52 - 5 = 47
The answer is option A.
Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (−2, 8), respectively. Then, the coordinates of the vertex D are
In a parallelogram, two diagonals of parallelogram bisects each other, which concludes that mid-point of both diagonals are the same.
Midpoint of AC = $$\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$
Let the coordinates of vertex D be (x,y)
$$\left(\ \frac{\ x+3}{2},\ \frac{\ y+4}{2}\right)=\left(\ \frac{\ 1-2}{2},\ \frac{\ 1+8}{2}\right)$$
x = -4 and y = 5
The answer is option D.
For natural numbers x, y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is
It is given, y(x + z) = 19
y cannot be 19.
If y = 19, x + z = 1 which is not possible when both x and z are natural numbers.
Therefore, y = 1 and x + z = 19
It is given, z(x + y) = 51
z can take values 3 and 17
Case 1:
If z = 3, y = 1 and x = 16
xyz = 3*1*16 = 48
Case 2:
If z = 17, y = 1 and x = 2
xyz = 17*1*2 = 34
Minimum value xyz can take is 34.
Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is
Let the savings invested in first part and second part be 'x' and 'y', respectively.
It is given,
$$\ \frac{\ x\times15\times4}{100}=\ \frac{\ y\times12\times3}{100}$$
60x = 36y
5x = 3y
Required percentage = $$\frac{x}{x+y}\times100=\frac{3}{3+5}\times100=37.5\%$$
The answer is option A.
Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is
Let the number of persons standing ahead and behind of Pinky be 3a and 5a.
Total number of persons = 3a + 5a + 1(including pinky) = 8a + 1
8a + 1 < 300
8a < 299
a < 37.375
Maximum value a can take is 37.
The maximum possible number of persons standing ahead of Pinky = 3a = 3*37 = 111
For any real number x, let [x] be the largest integer less than or equal to x. If $$\sum_{n=1}^N \left[\dfrac{1}{5} + \dfrac{n}{25}\right] = 25$$ then N is
It is given,
$$\Sigma_{n=1}^N\ \left[\dfrac{1}{5}+\dfrac{n}{25}\right]=25$$
$$\Sigma_{n=1}^N\ \left[\dfrac{5+n}{25}\right]=25$$
For n = 1 to n = 19, value of function is zero.
For n = 20 to n = 44, value of function will be 1.
44 = 20 + n - 1
n = 25 which is equal to given value.
This implies N = 44
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