46923.If x is a positive quantity such that $2^{x}={3^{log5}}^2$. then x is equal to
$log_5{8}$
$1+log3\left(\begin{array}{c}5\\ 3\end{array}\right)$
$log_5{9}$
$1+log5\left(\begin{array}{c}3\\ 5\end{array}\right)$
Explanation:
Given that:
$2^{x}={3^{log5}}^{2}$
$\Rightarrow2^{x}={2^{log5}}^{3}$
$\Rightarrow x={log_5}{3}$
$\Rightarrow x={log_5}{\dfrac{3*5}{5}}$
$\Rightarrow x={log_5}5+{log_5}{\dfrac{3}{5}}$
$\Rightarrow x=1+{log_5}{\dfrac{3}{5}}$. Hence, option D is the correct answer.
Given that:
$2^{x}={3^{log5}}^{2}$
$\Rightarrow2^{x}={2^{log5}}^{3}$
$\Rightarrow x={log_5}{3}$
$\Rightarrow x={log_5}{\dfrac{3*5}{5}}$
$\Rightarrow x={log_5}5+{log_5}{\dfrac{3}{5}}$
$\Rightarrow x=1+{log_5}{\dfrac{3}{5}}$. Hence, option D is the correct answer.
46924.In a circle, two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm. If the distance between these chords is 1 cm, then the radius of the circle, in cm, is
$\sqrt{13}$
$\sqrt{14}$
$\sqrt{11}$
$\sqrt{12}$
Explanation:
Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.
In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.
We can see that M and N are the mid points of AB and CD respectively.
AM = 3 cm and CD = 2 cm. Let OM be x cm.
In right angle triangle AMO,
$AO^{2}=AM^2+OM^2$
$\Rightarrow AO^{2}=3^2+ x^2$...(1)
In right angle triangle CNO,$CO^{2}=CN^2+ON^2$
$\Rightarrow CO^{2}=2^2+ (OM+MN)^2$
$\Rightarrow CO^{2}=2^2+ (x+1)^2$...(2)
We know that both AO and CO are the radius of the circle. Hence AO = CO
Therefore, we can equate equation (1) and (2)
$3^{2}+x^2=2^2+ (x+1)^2$
$ \Rightarrow x=2cm$
Therefore, the radius of the circle
$AO=\sqrt{{AM^2}+{OM^2}}$
$ \Rightarrow AO=\sqrt{{3^2}+{2^2}}=\sqrt{13}$
Hence, option A is the correct answer.
Given that two parallel chords on the same side of a diameter have lengths 4 cm and 6 cm.
In the diagram we can see that AB = 6 cm, CD = 4 cm and MN = 1 cm.
We can see that M and N are the mid points of AB and CD respectively.
AM = 3 cm and CD = 2 cm. Let OM be x cm.
In right angle triangle AMO,
$AO^{2}=AM^2+OM^2$
$\Rightarrow AO^{2}=3^2+ x^2$...(1)
In right angle triangle CNO,$CO^{2}=CN^2+ON^2$
$\Rightarrow CO^{2}=2^2+ (OM+MN)^2$
$\Rightarrow CO^{2}=2^2+ (x+1)^2$...(2)
We know that both AO and CO are the radius of the circle. Hence AO = CO
Therefore, we can equate equation (1) and (2)
$3^{2}+x^2=2^2+ (x+1)^2$
$ \Rightarrow x=2cm$
Therefore, the radius of the circle
$AO=\sqrt{{AM^2}+{OM^2}}$
$ \Rightarrow AO=\sqrt{{3^2}+{2^2}}=\sqrt{13}$
Hence, option A is the correct answer.
46925.Humans and robots can both perform a job but at different efficiencies. Fifteen humans and five robots working together take thirty days to finish the job, whereas five humans and fifteen robots working together take sixty days to finish it. How many days will fifteen humans working together (without any
robot) take to finish it?
robot) take to finish it?
45
36
32
40
Explanation:
Let the efficiency of humans be h and the efficiency of robots be r.
In the first case,
Total work = (15h + 5r) * 30......(i)
In the second case,
Total work = (5h + 15r) * 60......(ii)
On equating (i) and (ii), we get
(15h + 5r) * 30 = (5h + 15r) * 60
Or, 15h + 5r = 10h + 30r
Or, 5h = 25r
Or, h = 5r
Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h
Time taken by 15 humans = $\dfrac{480h}{50h}$days= 32 days.
Hence, option C is the correct answer.
Let the efficiency of humans be h and the efficiency of robots be r.
In the first case,
Total work = (15h + 5r) * 30......(i)
In the second case,
Total work = (5h + 15r) * 60......(ii)
On equating (i) and (ii), we get
(15h + 5r) * 30 = (5h + 15r) * 60
Or, 15h + 5r = 10h + 30r
Or, 5h = 25r
Or, h = 5r
Total work = (15h + 5r) * 30 = (15h + h) * 30 = 480h
Time taken by 15 humans = $\dfrac{480h}{50h}$days= 32 days.
Hence, option C is the correct answer.
46926.Let ABCD be a rectangle inscribed in a circle of radius 13 cm. Which one of the following pairs can represent, in cm, the possible length and breadth of ABCD?
24, 10
25, 9
25, 10
24, 12
Explanation:
We know that AC is the diameter and $\angle$ABC = 90°. AC = 2*13 = 26 cm
In right angle triangle ABC,
$AC^{2}=AB^2+BC^2$
$\Rightarrow AB^2+BC^2=26^2$
$\Rightarrow AB^2+BC^2=676$
Let us check with the options.
Option (A): $24^{2}+10^2=676$. Hence, this is a possible answer.
Option (B): $25^{2}+9^2=706\neq676$. Hence, this is an incorrect pair.
Option (C): $25^{2}+10^2=725\neq676$. Hence, this is an incorrect pair.
Option (D): $25^{2}+12^2=720\neq676$. Hence, this is an incorrect pair.
Therefore, we can say that option A is the correct answer.
We know that AC is the diameter and $\angle$ABC = 90°. AC = 2*13 = 26 cm
In right angle triangle ABC,
$AC^{2}=AB^2+BC^2$
$\Rightarrow AB^2+BC^2=26^2$
$\Rightarrow AB^2+BC^2=676$
Let us check with the options.
Option (A): $24^{2}+10^2=676$. Hence, this is a possible answer.
Option (B): $25^{2}+9^2=706\neq676$. Hence, this is an incorrect pair.
Option (C): $25^{2}+10^2=725\neq676$. Hence, this is an incorrect pair.
Option (D): $25^{2}+12^2=720\neq676$. Hence, this is an incorrect pair.
Therefore, we can say that option A is the correct answer.
46927.Points E, F, G, H lie on the sides AB, BC, CD, and DA, respectively, of a square ABCD. If EFGH is also a square whose area is 62.5% of that of ABCD and CG is longer than EB, then the ratio of length of EB to that of CG is
3 : 8
2 : 5
4 : 9
1 : 3
Explanation:
It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1.
Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.
Let us assume that x+1 is the length of side of square ABCD.
Area of square ABCD = $(x+1)^{2}$sq. units.Therefore, area of square EFGH =$\dfrac{62.5}{100}*(x+1)^{2}=\dfrac{5(x+1)^2}{8}...(1)$
In right angle triangle EBF,
$EF^2 = EB^2 + BF^2$
$\Rightarrow EF=\sqrt{{1^2}+{x^2}}$
Therefore, the area of square EFGH =$EF^2=x^2+1..(2)$
By equating (1) and (2),
x^2+1= $\dfrac{5(x+1)^2}{8}$
$\Rightarrow 8x^{2}+8=5x^2+10x+5$
$\Rightarrow 3x^{2}-10x+3=0$
$\Rightarrow (x-3)(3x-1)=0$
$\Rightarrow x=3 or 1/3$
The ratio of length of EB to that of CG = 1 : x
EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer.
It is given that EFGH is also a square whose area is 62.5% of that of ABCD. Let us assume that E divides AB in x : 1.
Because of symmetry we can se that points F, G and H divide BC, CD and DA in x : 1.
Let us assume that x+1 is the length of side of square ABCD.
Area of square ABCD = $(x+1)^{2}$sq. units.Therefore, area of square EFGH =$\dfrac{62.5}{100}*(x+1)^{2}=\dfrac{5(x+1)^2}{8}...(1)$
In right angle triangle EBF,
$EF^2 = EB^2 + BF^2$
$\Rightarrow EF=\sqrt{{1^2}+{x^2}}$
Therefore, the area of square EFGH =$EF^2=x^2+1..(2)$
By equating (1) and (2),
x^2+1= $\dfrac{5(x+1)^2}{8}$
$\Rightarrow 8x^{2}+8=5x^2+10x+5$
$\Rightarrow 3x^{2}-10x+3=0$
$\Rightarrow (x-3)(3x-1)=0$
$\Rightarrow x=3 or 1/3$
The ratio of length of EB to that of CG = 1 : x
EB : CG = 1 : 3 or 3 : 1. Hence, option D is the correct answer.
46928.Point P lies between points A and B such that the length of BP is thrice that of AP. Car 1 starts from A and moves towards B. Simultaneously, car 2 starts from B and moves towards A. Car 2 reaches P one hour after car 1 reaches P. If the speed of car 2 is half that of car 1, then the time, in minutes, taken by car 1 in reaching P from A is
14
16
13
12
Explanation:
Let the distance between A and B be 4x.
Length of BP is thrice the length of AP.
$\Rightarrow$ AP = x and BP = 3x
Let the speed of car 1 be s and the speed of car 2 be 0.5s.
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.
$\Rightarrow$ x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60x/s = 12
Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.
Let the distance between A and B be 4x.
Length of BP is thrice the length of AP.
$\Rightarrow$ AP = x and BP = 3x
Let the speed of car 1 be s and the speed of car 2 be 0.5s.
Car 2 reaches P one hour (60 minutes) after Car 1 reaches P.
$\Rightarrow$ x/s + 60 = 3x/0.5s
x/s + 60 = 6x/s
5x/s = 60x/s = 12
Time taken by car 1 in reaching P from A = x/s = 12 minutes.
Therefore, 12 is the correct answer.
46929.Let f(x) = min $(2x^2, 52-5x)$ where x is any positive real number. Then the maximum possible value of f(x) is
32
33
28
25
Explanation:
f(x) = min $(2x^2, 52-5x)$
The maximum possible value of this function will be attained at the point in which $2x^2$ is equal to 52-5x.
$2x^2=52-5x$
$2x^2+5x-52=0$
$(2x+13)(x-4)=0$
$\Rightarrow x=\dfrac{-13}{2}$ or x=4
It has been given that is a positive real number. Therefore, we can eliminate the case $x=\dfrac{-13}{2}$.
x=4 is the point at which the function attains the maximum value. is not the maximum value of the function.
Substituting in the original function, we get, $2x^2 = 2 * 4^2 = 32
$
f(x) = 32.
Therefore, 32 is the right answer.
f(x) = min $(2x^2, 52-5x)$
The maximum possible value of this function will be attained at the point in which $2x^2$ is equal to 52-5x.
$2x^2=52-5x$
$2x^2+5x-52=0$
$(2x+13)(x-4)=0$
$\Rightarrow x=\dfrac{-13}{2}$ or x=4
It has been given that is a positive real number. Therefore, we can eliminate the case $x=\dfrac{-13}{2}$.
x=4 is the point at which the function attains the maximum value. is not the maximum value of the function.
Substituting in the original function, we get, $2x^2 = 2 * 4^2 = 32
$
f(x) = 32.
Therefore, 32 is the right answer.
46930.While multiplying three real numbers, Ashok took one of the numbers as 73 instead of 37. As a result, the product went up by 720. Then the minimum possible value of the sum of squares of the other two numbers is
45
48
40
41
Explanation:
We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20Product of 2 real numbers is 20.
We have to find the minimum possible value of the sum of the squares of the 2 numbers.
Let x=a*b
It has been given that a*b=20
The least possible sum for a given product is obtained when the numbers are as close to each other as possible.
Therefore, when a=b, the value of a and b will be $\sqrt{20}.$
Sum of the squares of the 2 numbers = 20 + 20 = 40.
Therefore, 40 is the correct answer.
We know that one of the 3 numbers is 37.
Let the product of the other 2 numbers be x.
It has been given that 73x-37x = 720
36x = 720
x = 20Product of 2 real numbers is 20.
We have to find the minimum possible value of the sum of the squares of the 2 numbers.
Let x=a*b
It has been given that a*b=20
The least possible sum for a given product is obtained when the numbers are as close to each other as possible.
Therefore, when a=b, the value of a and b will be $\sqrt{20}.$
Sum of the squares of the 2 numbers = 20 + 20 = 40.
Therefore, 40 is the correct answer.
46931.Train T leaves station X for station Y at 3 pm. Train S, traveling at three quarters of the speed of T, leaves Y for X at 4 pm. The two trains pass each other at a station Z, where the distance between X and Z is three-fifths of that between X and Y. How many hours does train T take for its journey from X to Y?
18
11
15
20
Explanation:
Train T starts at 3 PM and train S starts at 4 PM.
Let the speed of train T be t.
=> Speed of train S = 0.75t.
When the trains meet, train t would have traveled for one more hour than train S.
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours.
Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t
We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities.
=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25x = 9 hours.
Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3)= 15 hours.
Therefore, 15 is the correct answer.
Train T starts at 3 PM and train S starts at 4 PM.
Let the speed of train T be t.
=> Speed of train S = 0.75t.
When the trains meet, train t would have traveled for one more hour than train S.
Let us assume that the 2 trains meet x hours after 3 PM. Trains S would have traveled for x-1 hours.
Distance traveled by train T = xt
Distance traveled by train S = (x-1)*0.75t = 0.75xt-0.75t
We know that train T has traveled three fifths of the distance. Therefore, train S should have traveled two-fifths the distance between the 2 cities.
=> (xt)/(0.75xt-0.75t) = 3/2
2xt = 2.25xt-2.25t
0.25x = 2.25x = 9 hours.
Train T takes 9 hours to cover three-fifths the distance. Therefore, to cover the entire distance, train T will take 9*(5/3)= 15 hours.
Therefore, 15 is the correct answer.
46932.Two types of tea, A and B, are mixed and then sold at Rs. 40 per kg. The profit is 10% if A and B are mixed in the ratio 3 : 2, and 5% if this ratio is 2 : 3. The cost prices, per kg, of A and B are in the ratio
17 : 25
18 : 25
19 : 24
21 : 25
Explanation:
The selling price of the mixture is Rs.40/kg.
Let a be the quantity of tea A in the mixture and b be the quantity of tea B in the mixture.
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x.
It has been given that 1.1x = 40
x = 40/1.1$\dfrac{3a+2b}{5}=\dfrac{40}{1.1}$
3.3a + 2.2b = 200...(1)
The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
$\dfrac{2a+3b}{5}=\dfrac{40}{1.05}$
2.1a + 3.15b = 200...(2)
Equating (1) and (2), we get,
3.3a + 2.2b = 2.1a + 3.15b
1.2a = 0.95b
$\dfrac{a}{b}=\dfrac{0.95}{1.2}$
$\dfrac{a}{b}=\dfrac{19}{24}$
Therefore, option C is the right answer.
The selling price of the mixture is Rs.40/kg.
Let a be the quantity of tea A in the mixture and b be the quantity of tea B in the mixture.
It has been given that the profit is 10% if the 2 varieties are mixed in the ratio 3:2
Let the cost price of the mixture be x.
It has been given that 1.1x = 40
x = 40/1.1$\dfrac{3a+2b}{5}=\dfrac{40}{1.1}$
3.3a + 2.2b = 200...(1)
The profit is 5% if the 2 varieties are mixed in the ratio 2:3.
$\dfrac{2a+3b}{5}=\dfrac{40}{1.05}$
2.1a + 3.15b = 200...(2)
Equating (1) and (2), we get,
3.3a + 2.2b = 2.1a + 3.15b
1.2a = 0.95b
$\dfrac{a}{b}=\dfrac{0.95}{1.2}$
$\dfrac{a}{b}=\dfrac{19}{24}$
Therefore, option C is the right answer.
46933.Given an equilateral triangle T1 with side 24 cm, a second triangle T2 is formed by joining the midpoints of the sides of T1. Then a third triangle T3 is formed by joining the midpoints of the sides of T2. If this process of forming triangles is continued, the sum of the areas, in sq cm, of infinitely many such triangles T1, T2, T3,... will be
$188\sqrt{3}$
$248\sqrt{3}$
$164\sqrt{3}$
$192\sqrt{3}$
Explanation:
We can see that $T_2$ is formed by using the mid points of $T_1$ . Hence, we can say that area of triangle of $T_2$ will be (1/4)th of the area of triangle T_1$.
Area of triangle $T_1=\dfrac{\sqrt{3}}{4}*(24)^2=144\sqrt{3}sq.cm$
Area of triangle $T_2=\dfrac{144\sqrt{3}}{4}=36\sqrt{3} sq.cm$
Sum of the area of all triangles = $T_1 + T_2 + T_3 + ...
$\Rightarrow T_1+ \dfrac{T_2}{4}+ \dfrac{T_3}{4^2}+...$
$\Rightarrow \dfrac{T_1}{1-0.25}$
$\Rightarrow \dfrac{4}{3}*T_1$
$\Rightarrow \dfrac{4}{3}*144\sqrt{3}$
$\Rightarrow192\sqrt{3}$
Hence, option D is the correct answer.
We can see that $T_2$ is formed by using the mid points of $T_1$ . Hence, we can say that area of triangle of $T_2$ will be (1/4)th of the area of triangle T_1$.
Area of triangle $T_1=\dfrac{\sqrt{3}}{4}*(24)^2=144\sqrt{3}sq.cm$
Area of triangle $T_2=\dfrac{144\sqrt{3}}{4}=36\sqrt{3} sq.cm$
Sum of the area of all triangles = $T_1 + T_2 + T_3 + ...
$\Rightarrow T_1+ \dfrac{T_2}{4}+ \dfrac{T_3}{4^2}+...$
$\Rightarrow \dfrac{T_1}{1-0.25}$
$\Rightarrow \dfrac{4}{3}*T_1$
$\Rightarrow \dfrac{4}{3}*144\sqrt{3}$
$\Rightarrow192\sqrt{3}$
Hence, option D is the correct answer.
46934.If $log_{12}{81}=p, then 3\left(\begin{array}{c}4-p\\ 4+p\end{array}\right) $is equal to
$log_{4}{16}$
$log_{6}{16}$
$log_{2}{8}$
$log_{6}{8}$
Explanation:
Given that: $log_{12}{81}=p$
$\Rightarrow log_{81}{12}=\dfrac{1}{p}$
$\Rightarrow 4log_3{3*4}=\dfrac{1}{p}$
$\Rightarrow 1+log_3{4}=\dfrac{4}{p}$
Using Componendo and Dividendo,
$\Rightarrow \dfrac{1+log_3{4}-1}{1+log_3{4}+1}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{log_3{4}}{2+log_3{4}}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{log_3{4}}{log_3{9}+log_3{4}}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{log_3{4}}{log_3{36}}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{4-p}{3*4+p}=\dfrac{3log_3{4}}{log_3{36}}$
$\Rightarrow \dfrac{4-p}{3*4+p}=\dfrac{log_3{64}}{log_3{36}}$
$\Rightarrow \dfrac{4-p}{3*4+p}=log_{36}{64}$
$\Rightarrow \dfrac{4-p}{3*4+p}=log_6^2{8^2}=log_6{8}$. Hence, option D is the correct answer.
Given that: $log_{12}{81}=p$
$\Rightarrow log_{81}{12}=\dfrac{1}{p}$
$\Rightarrow 4log_3{3*4}=\dfrac{1}{p}$
$\Rightarrow 1+log_3{4}=\dfrac{4}{p}$
Using Componendo and Dividendo,
$\Rightarrow \dfrac{1+log_3{4}-1}{1+log_3{4}+1}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{log_3{4}}{2+log_3{4}}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{log_3{4}}{log_3{9}+log_3{4}}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{log_3{4}}{log_3{36}}=\dfrac{4-p}{4+p}$
$\Rightarrow \dfrac{4-p}{3*4+p}=\dfrac{3log_3{4}}{log_3{36}}$
$\Rightarrow \dfrac{4-p}{3*4+p}=\dfrac{log_3{64}}{log_3{36}}$
$\Rightarrow \dfrac{4-p}{3*4+p}=log_{36}{64}$
$\Rightarrow \dfrac{4-p}{3*4+p}=log_6^2{8^2}=log_6{8}$. Hence, option D is the correct answer.
46935.John borrowed Rs. 2,10,000 from a bank at an interest rate of 10% per annum, compounded annually. The loan was repaid in two equal instalments, the first after one year and the second after another year. The first instalment was interest of one year plus part of the principal amount, while the second was the rest
of the principal amount plus due interest thereon. Then each instalment, in Rs., is
of the principal amount plus due interest thereon. Then each instalment, in Rs., is
1,22,780
1,25,660
1,21,000
2,00,000
Explanation:
We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years.
John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years.
Let the amount paid as installment every year be Rs.x.
John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid.
$\Rightarrow$1.1x + x = 2,54,100
2.1x = 2,54,100
$\Rightarrow$ x = Rs. 1,21,000.
Therefore, 121000 is the correct answer.
We have to equate the installments and the amount due either at the time of borrowing or at the time when the entire loan is repaid. Let us bring all values to the time frame in which all the dues get settled, i.e, by the end of 2 years.
John borrowed Rs. 2,10,000 from the bank at 10% per annum. This loan will amount to 2,10,000*1.1*1.1 = Rs.2,54,100 by the end of 2 years.
Let the amount paid as installment every year be Rs.x.
John would pay the first installment by the end of the first year. Therefore, we have to calculate the interest on this amount from the end of the first year to the end of the second year. The loan will get settled the moment the second installment is paid.
$\Rightarrow$1.1x + x = 2,54,100
2.1x = 2,54,100
$\Rightarrow$ x = Rs. 1,21,000.
Therefore, 121000 is the correct answer.
46937.How many numbers with two or more digits can be formed with the digits 1,2,3,4,5,6,7,8,9, so that in every such number, each digit is used at most once and the digits appear in the ascending order?
507
502
503
505
Explanation:
It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number. There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $9C_2$ ways.
3 digit numbers can be formed in $9C_3$ ways.
9 digit number can be formed in 9C9 ways.
We know that $nC_0+nC_1+nC_2+...+nC_n=2^n$
$\Rightarrow 9C_0+9C_1+9C_2+...+9C_9=2^9$
$9C_0+9C_1+...+9C_9=519$
We have to subtract $9C_0$ and $9C_1$ from both the sides of the equations since we cannot form single digit numbers.
$\Rightarrow 9C_2+9C_3+...+9C_9=512 -1-9$
$ 9C_2+9C_3+...+9C_9=502$
Therefore, 502 is the right answer.
It has been given that the digits in the number should appear in the ascending order. Therefore, there is only 1 possible arrangement of the digits once they are selected to form a number. There are 9 numbers (1,2,3,4,5,6,7,8,9) in total.
2 digit numbers can be formed in $9C_2$ ways.
3 digit numbers can be formed in $9C_3$ ways.
9 digit number can be formed in 9C9 ways.
We know that $nC_0+nC_1+nC_2+...+nC_n=2^n$
$\Rightarrow 9C_0+9C_1+9C_2+...+9C_9=2^9$
$9C_0+9C_1+...+9C_9=519$
We have to subtract $9C_0$ and $9C_1$ from both the sides of the equations since we cannot form single digit numbers.
$\Rightarrow 9C_2+9C_3+...+9C_9=512 -1-9$
$ 9C_2+9C_3+...+9C_9=502$
Therefore, 502 is the right answer.
46938.A right circular cone, of height 12 ft, stands on its base which has diameter 8 ft. The tip of the cone is cut off with a plane which is parallel to the base and 9 ft from the base. With $\pi$= 22/7, the volume, in cubic ft, of the remaining part of the cone is
197
200
198
190
Explanation:
We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft
In triangle OAB and OCD
$\dfrac{OA}{AB}=\dfrac{OC}{CD}$
$\Rightarrow AB=\dfrac{3*4}{12}=1ft$
Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone
$\Rightarrow \dfrac{1}{3}*\pi*4^2*12-\dfrac{1}{3}*\pi*1^2*3$
$\Rightarrow \dfrac{1}{3}*\pi*189$
$\Rightarrow \dfrac{22}{7*3*189}$
$\Rightarrow 198 cubic ft$
We are given that diameter of base = 8 ft. Therefore, the radius of circular base = 8/2 = 4 ft
In triangle OAB and OCD
$\dfrac{OA}{AB}=\dfrac{OC}{CD}$
$\Rightarrow AB=\dfrac{3*4}{12}=1ft$
Therefore, the volume of remaining part = Volume of entire cone - Volume of smaller cone
$\Rightarrow \dfrac{1}{3}*\pi*4^2*12-\dfrac{1}{3}*\pi*1^2*3$
$\Rightarrow \dfrac{1}{3}*\pi*189$
$\Rightarrow \dfrac{22}{7*3*189}$
$\Rightarrow 198 cubic ft$
46939.Raju and Lalitha originally had marbles in the ratio 4:9. Then Lalitha gave some of her marbles to Raju. As a result, the ratio of the number of marbles with Raju to that with Lalitha became 5:6. What fraction of her original number of marbles was given by Lalitha to Raju?
$\dfrac{1}{5}$
$\dfrac{6}{19}$
$\dfrac{1}{4}$
$\dfrac{7}{33}$
Explanation:
Let the number of marbles with Raju and Lalitha initially be 4x and 9x.
Let the number of marbles that Lalitha gave to Raju be a.
It has been given that (4x+a)/(9x-a) = 5/6
24x + 6a = 45x - 5a
11a = 21x
a/x = 21/11
Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).
a/9x = 21/99
= 7/33.
Therefore, option D is the right answer.
Let the number of marbles with Raju and Lalitha initially be 4x and 9x.
Let the number of marbles that Lalitha gave to Raju be a.
It has been given that (4x+a)/(9x-a) = 5/6
24x + 6a = 45x - 5a
11a = 21x
a/x = 21/11
Fraction of original marbles given to Raju by Lalitha = a/9x (Since Lalitha had 9x marbles initially).
a/9x = 21/99
= 7/33.
Therefore, option D is the right answer.
46940.Each of 74 students in a class studies at least one of the three subjects H, E and P. Ten students study all three subjects, while twenty study H and E, but not P. Every student who studies P also studies H or E or both. If the number of students studying H equals that studying E, then the number of students studying H is
52
53
55
57
Explanation:
Let us draw a Venn diagram using the information present in the question.
It is given that the number of students studying H equals that studying E.
Let x be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.
Therefore,
x + 20 + 10 + x = 74
x = 22
Hence, the number of students studying H = 22 + 10+ 20 = 52
Let us draw a Venn diagram using the information present in the question.
It is given that the number of students studying H equals that studying E.
Let x be the total number of students who studied H, and H and P but mot E.We can also say that the same will be the number of students who studied E, and E and P but not H.
Therefore,
x + 20 + 10 + x = 74
x = 22
Hence, the number of students studying H = 22 + 10+ 20 = 52
46941.A wholesaler bought walnuts and peanuts, the price of walnut per kg being thrice that of peanut per kg. He then sold 8 kg of peanuts at a profit of 10% and 16 kg of walnuts at a profit of 20% to a shopkeeper. However, the shopkeeper lost 5 kg of walnuts and 3 kg of peanuts in transit. He then mixed the remaining
nuts and sold the mixture at Rs. 166 per kg, thus making an overall profit of 25%. At what price, in Rs. per kg, did the wholesaler buy the walnuts?
96
98
86
84
Explanation:
Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer
Let the price of peanuts be Rs. 100x per kg
Then, the price of walnuts = Rs. 300x per kg
Cost price of peanuts for the shopkeeper = Rs. 110x per kg
Cost price of walnuts for the shopkeeper = Rs. 360x per kg
Total cost incurred to the shopkeeper while buying = Rs.(8 * 110x + 16 * 360x) = Rs. 6640x
Total selling price that the shopkeeper got = Rs. (166 * 16) = Rs. 2656
Profit = 25%
So, cost price = Rs. 2124.80
Therefore, 6640x = 2124.80On solving, we get x = 0.32
Therefore, price of walnuts = Rs. (300 * 0.32) = Rs. 96 per kg.
Hence, option A is the correct answer
46942.If among 200 students, 105 like pizza and 134 like burger, then the number of students who like only burger can possibly be
23
26
96
93
Explanation:
It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values.
The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too.
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.
The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible.
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95.
The number of students who like only burger should lie between 29 and 95 (both the values are included). 93 is the only value among the given options that satisfies this condition and hence, option D is the right answer.
It has been given that among 200 students, 105 like pizza and 134 like burger.
The question asks us to find out the number of students who can be liking only burgers among the given values.
The least number of students who like only burger will be obtained when everyone who likes pizza likes burger too.
In this case, 105 students will like pizza and burger and 134-105 = 29 students will like only burger. Therefore, the number of students who like only burger cannot be less than 29.
The maximum number of students who like only burger will be obtained when we try to separate the 2 sets as much as possible.
There are 200 students in total. 105 of them like pizza. Therefore, the remaining 95 students can like only burger and 134-95 = 39 students can like both pizza and burger. As we can see, the number of students who like burger cannot exceed 95.
The number of students who like only burger should lie between 29 and 95 (both the values are included). 93 is the only value among the given options that satisfies this condition and hence, option D is the right answer.
46943.If f(x+2)=f(x)+f(x+1) for all positive integers x, and f(11)=91, f(15)=617, then f(10) equals
50
54
56
53
Explanation:
f(x+2)=f(x)+f(x+1)
As we can see, the value of a term is the sum of the 2 terms preceding it.
It has been given that f(11)=91and f(15)=617.
We have to find the value of f(10).Let f(10)= b
f(12) = b + 91
f(13) = 91 + b + 91 = 182 + b
f(14) = 182+b+91+b = 273+2b
f(15) = 273+2b+182+b = 455+3b
It has been given that 455+3b = 617
3b = 162
$\Rightarrow $b = 54
Therefore, 54 is the correct answer.
f(x+2)=f(x)+f(x+1)
As we can see, the value of a term is the sum of the 2 terms preceding it.
It has been given that f(11)=91and f(15)=617.
We have to find the value of f(10).Let f(10)= b
f(12) = b + 91
f(13) = 91 + b + 91 = 182 + b
f(14) = 182+b+91+b = 273+2b
f(15) = 273+2b+182+b = 455+3b
It has been given that 455+3b = 617
3b = 162
$\Rightarrow $b = 54
Therefore, 54 is the correct answer.