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If $$\left(8\ast4\right)+6-\left(10\div5\right)=8$$, then $$\ast$$ stands for the operation
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If $$\left(8\ast4\right)+6-\left(10\div5\right)=8$$, then $$\ast$$ stands for the operation
We need $$8\ast4+6-10\div5=8$$. The multiplication and division choices give $$36$$ and $$6$$ respectively, while addition gives $$20$$. With subtraction, $$8-4+6-2=8$$, so $$\ast$$ represents subtraction.
$$20\%$$ of a number is equal to $$30\%$$ of another number. Six times the bigger of these numbers added to the smaller number is $$2000$$. Then $$10\%$$ of the smallest number is
Let the smaller number be $$x$$ and the bigger number be $$y$$. From $$0.2y=0.3x$$, we get $$x=\frac{2y}{3}$$. Thus $$6x+y=2000$$ gives $$5y=2000$$, so $$y=400$$ and $$x=200$$. Therefore $$10\%$$ of the smaller number is $$20$$.
Samrud chooses a two-digit number. He subtracts it from $$200$$ and doubles the result. The largest number he can get is
To maximize the result, Samrud must choose the smallest two-digit number, which is $$10$$. The resulting value is $$2\left(200-10\right)=380$$. Hence the largest possible number is $$380$$.
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The least number which leaves $$2$$ as remainder when divided by $$5,6,8,9$$ and $$12$$ is
The required number minus $$2$$ must be divisible by $$5,6,8,9$$ and $$12$$. Their least common multiple is $$360$$. Therefore the least number is $$360+2=362$$.
Saket takes the number 2021. There are two 2s, one 0 and one 1. Using these numbers, he makes 3-digit numbers, the sum of all these numbers is
Using the digits $$2,2,0,1$$ to form valid three-digit numbers gives $$220,202,210,201,120,102,212,221,122$$. Their sum is $$1610$$. Hence the required sum is $$1610$$.
In the net of the adjoining figure numbers $$1$$ to $$6$$ are marked. The net is folded to form a cube. The number that appears on the face opposite to $$6$$ is
When the net is folded, face $$4$$ is opposite face $$1$$, face $$3$$ is opposite face $$5$$, and face $$2$$ is opposite face $$6$$. Therefore the face opposite to $$6$$ is $$2$$.
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$$10^{2021}-1$$ is written as an integer in the usual decimal form. The sum of all digits of this written number is
The number $$10^{2021}-1$$ consists of exactly $$2021$$ digits, all of them equal to $$9$$. Therefore its digit sum is $$2021\times9=18189$$.
August $$15^{\text{th}}$$ $$1992$$ was a Saturday. What day was August $$15^{\text{th}}$$ $$1991$$?
The interval from August $$15$$, $$1991$$ to August $$15$$, $$1992$$ includes February $$29$$, $$1992$$, so it contains $$366$$ days. Since $$366\equiv2\pmod7$$, the weekday shifts forward by two days. Therefore August $$15$$, $$1991$$ was Thursday.
Two thirds of the students in a class room are seated in three fourths of the chairs. The rest of the students are punished for not submitting the homework and hence asked to stand. There are $$6$$ empty chairs. The number of persons, that is the number of students including the teacher, in the classroom is
Let the number of students be $$s$$ and the number of chairs be $$c$$. We have $$\frac{2s}{3}=\frac{3c}{4}$$, so $$c=\frac{8s}{9}$$. Since $$6$$ chairs are empty, $$c-\frac{2s}{3}=6$$. Thus $$\frac{2s}{9}=6$$, giving $$s=27$$. Including the teacher, the number of persons is $$28$$.
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There are $$24$$ four digit numbers with different digits, formed by $$2,4,5$$ and $$7$$. One of these four digit numbers is a multiple of another. Which one of the following is it?
The relevant permutation is $$7425$$. It is exactly $$3$$ times $$2475$$, and $$2475$$ also uses the digits $$2,4,5,7$$ exactly once. Hence $$7425$$ is the required number.
Twelve friends went to a restaurant to have lunch. They ordered for $$12$$ meals. When they were served, they found that the food was too much which can be shared by $$18$$ people. The number of meals they would order to cater to only $$12$$ is
The food from $$12$$ meals is sufficient for $$18$$ people, so one meal is sufficient for $$\frac{18}{12}=\frac{3}{2}$$ people. To serve $$12$$ people, the required number of meals is $$12\div\frac{3}{2}=8$$. Therefore they would order $$8$$ meals.
The least multiple of $$23$$ which when divided by $$18,21$$ and $$24$$ leaves the remainders $$7,10$$ and $$13$$ respectively is
The conditions imply that the number is congruent to $$7$$ modulo $$18$$, $$10$$ modulo $$21$$, and $$13$$ modulo $$24$$. Solving these simultaneous congruences gives $$N=3013$$ as the smallest positive solution satisfying all three conditions. Also $$3013=23\times131$$, so it is a multiple of $$23$$.
The value of $$\sqrt{\frac{(0.1)^2+(0.01)^2+(0.008)^2}{(0.01)^2+(0.001)^2+(0.0008)^2}}$$ is
Each term in the denominator is one tenth of the corresponding term in the numerator before squaring. Therefore the denominator is $$\frac{1}{100}$$ of the numerator. Hence the expression is $$\sqrt{100}=10$$.
Two numbers are respectively $$20\%$$ and $$50\%$$ of a third number. The percentage of the first number to the second is
Let the third number be $$N$$. Then the first and second numbers are $$0.2N$$ and $$0.5N$$ respectively. Thus the first as a percentage of the second is $$\frac{0.2N}{0.5N}\times100=40\%$$. Therefore the required percentage is $$40$$.
Water flows through a rectangular opening $$3\text{ m}\times2\text{ m}$$ with a speed of $$1.5\text{ km/hr}$$. The amount of water that passes through the opening in $$5$$ minutes is $$x^3$$. Then the value of $$x$$ is
The area of the opening is $$3\times2=6\text{ m}^2$$. The speed is $$1.5\text{ km/hr}=25\text{ m/min}$$, so in $$5$$ minutes the water travels $$125\text{ m}$$. Hence the volume is $$6\times125=750\text{ m}^3$$, so $$x^3=750$$ and $$x=\sqrt[3]{750}?pprox9.085602964$$.
If $$A=\frac{1111+3333+5555}{222+333+444}$$, $$B=\frac{555+222}{1111+2222+4444}$$, then $$A\ast B$$ is
Simplifying the first fraction gives $$A=\frac{9999}{999}=\frac{1111}{111}$$. Simplifying the second gives $$B=\frac{777}{7777}=\frac{111}{1111}$$. Therefore $$A\ast B=\frac{1111}{111}\times\frac{111}{1111}=1$$.
In the adjoining figure of a cat, each small square is $$1\text{ cm}^2$$. The area of the cat, in $$\text{cm}^2$$, is
The cat is a polygon whose vertices lie on the unit grid. Splitting the polygon into rectangles and triangles, or applying the shoelace formula to its grid vertices, gives a total area of $$12\text{ cm}^2$$. Therefore the required area is $$12$$.
The sum of the digits of the number given by the product $$ \frac{666.....666}{2021 digits}$$ X$$ \frac{999.....999}{2021 digits}$$ is
Let $$n=2021$$. The first factor is $$\frac{2}{3}\left(10^n-1\right)$$ and the second is $$10^n-1$$. Their product has the digit pattern corresponding to $$666\ldots666\times999\ldots999$$, whose digit sum is $$9n$$. Thus the required digit sum is $$9\times2021=18189$$.
Two buses $$X,Y$$ started from two different places $$P_1$$ and $$P_2$$ respectively, moving towards each other. The ratio of the speed of bus $$X$$ to that of bus $$Y$$ was $$5:4$$. After they meet, the speed of bus $$X$$ was reduced by $$20\%$$ and the speed of $$Y$$ was increased by $$20\%$$. When bus $$X$$ arrived at $$P_2$$, bus $$Y$$ was still $$10\text{ km}$$ away from $$P_1$$. The distance between the places $$P_1$$ and $$P_2$$, in kilometers, is
Let the original speeds be $$5v$$ and $$4v$$. The distances covered before meeting are therefore in the ratio $$5:4$$, say $$5k$$ and $$4k$$, so the total distance is $$9k$$. After meeting, the speeds become $$4v$$ and $$4.8v$$. Bus $$X$$ takes $$k$$ units of time to cover the remaining $$4k$$, during which bus $$Y$$ covers $$4.8k$$. Hence the remaining distance of bus $$Y$$ from $$P_1$$ is $$5k-4.8k=0.2k=10$$, giving $$k=50$$. Therefore the distance between $$P_1$$ and $$P_2$$ is $$9k=450\text{ km}$$.
In the adjoining diagram, $$ABCD$$ is a square. $$P,Q,R,S$$ are the mid points of sides $$AB,BC,CD$$ and $$DA$$ respectively. $$PQ$$ and $$SR$$ are quadrants of circles with $$B,D$$ as the centres and $$BP$$ as radius. $$PS$$ and $$RQ$$ are part of the circle passing through $$P,Q,R$$ and $$S$$. If $$AB=20\text{ cm}$$, the area of the shaded region, in $$\text{cm}^2$$, is
Since $$AB=20\text{ cm}$$, the four midpoints lie on a circle of radius $$10\text{ cm}$$ centered at the centre of the square. The area inside this circle is $$100\pi$$. The two quadrant regions removed from the shaded part have radius $$10\text{ cm}$$ and together have area $$2\left(\frac14\pi\cdot10^2\right)=50\pi$$. Hence the shaded area is $$100\pi-50\pi=50\pi\text{ cm}^2$$.
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