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The three-digit number $$3a4$$ is added to 278 to obtain another three-digit number $$6b2$$, where $$a$$ and $$b$$ are digits. If $$6b2$$ is divisible by 9, then $$a + b = $$
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The three-digit number $$3a4$$ is added to 278 to obtain another three-digit number $$6b2$$, where $$a$$ and $$b$$ are digits. If $$6b2$$ is divisible by 9, then $$a + b = $$
Since $$6b2$$ is divisible by 9, the digit sum $$6 + b + 2 = 8 + b$$ must be a multiple of 9, which forces $$b = 1$$. Then $$6b2 = 612$$ and $$3a4 = 612 - 278 = 334$$, so $$a = 3$$. Hence $$a + b = 3 + 1 = 4$$.
In an AMTI workshop, the ratio of number of boys to the number of girls was $$5 \colon 8$$. If there are total 65 students in the workshop, how many more girls were there in the workshop than boys?
The 65 students are divided into $$5 + 8 = 13$$ equal parts, so each part has $$\frac{65}{13} = 5$$ students. This gives $$5 \times 5 = 25$$ boys and $$8 \times 5 = 40$$ girls. The number of extra girls is $$40 - 25 = 15$$.
A person covers first 1.5 km at 5 km/h and next 3 km at 6 km/h, then his average speed is
The first stretch takes $$\frac{1.5}{5} = 0.3$$ hours and the second takes $$\frac{3}{6} = 0.5$$ hours, so the total time is $$0.8$$ hours for a total distance of $$4.5$$ km. The average speed is $$\frac{4.5}{0.8} = \frac{45}{8}$$ km/h.
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A triangle and a square have equal perimeters. The lengths of three sides of the triangle are 7.4 cm, 5.7 cm, 6.9 cm. Then area of the square is ______ square cm.
The perimeter of the triangle is $$7.4 + 5.7 + 6.9 = 20$$ cm, so the square also has perimeter 20 cm and side $$\frac{20}{4} = 5$$ cm. Its area is $$5^2 = 25$$ square cm.
Ramanujan Primary School has 1200 students. Each class has 30 students. Each student takes 5 sessions a day and each teacher teaches 4 sessions a day. What is the minimum number of teachers in the School?
There are $$\frac{1200}{30} = 40$$ classes, and each class needs 5 sessions a day, so $$40 \times 5 = 200$$ sessions must be taught daily. Since one teacher handles 4 sessions a day, at least $$\frac{200}{4} = 50$$ teachers are needed.
In a magic triangle of addition, each of six natural numbers from 21 to 26 is placed in one of the six empty circles such that the sum $$S$$ of the three numbers from the circles along each side would be the same. Then the maximum possible value of $$S$$ is
Adding the three side sums counts every number once and each of the three corner numbers twice, so $$3S = (21 + 22 + \cdots + 26) + (\text{sum of corners}) = 141 + C$$. To make $$S$$ largest take the largest numbers at the corners, $$C = 24 + 25 + 26 = 75$$, giving $$3S = 216$$ and $$S = 72$$. This is achievable, since 23, 21 and 22 placed between the pairs $$(24, 25)$$, $$(25, 26)$$ and $$(26, 24)$$ make every side add to 72.
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A ream of identical papers containing 500 sheets is 5 cm thick. How many sheets of this type of paper would be there in a 7.2 cm thick bundle?
A thickness of 5 cm holds 500 sheets, so 1 cm holds $$\frac{500}{5} = 100$$ sheets. Hence a 7.2 cm bundle holds $$100 \times 7.2 = 720$$ sheets.
$$n = 1111\ldots11$$ is a 2026-digit number with each digit as 1. If $$n$$ is divided by 1111, then the quotient is $$Q$$ and the remainder is $$R$$, then
Since $$2026 = 2 + 4 \times 506$$, write $$n = 11 \times 10^{2024} + \underbrace{11\ldots1}_{2024}$$, and the repunit with 2024 ones is a multiple of 1111. As $$10^4 = 9999 + 1$$ leaves remainder 1 on division by 1111, $$10^{2024}$$ also leaves remainder 1, so $$R = 11$$. Dividing a 2026-digit number beginning with 11 by the 4-digit number 1111 gives a quotient of $$2026 - 3 = 2023$$ digits.
A 'stair-step' figure is made of alternating black and white squares in each row. All rows begin and end with white square. First four figures of the stair-step figure are shown below. Fig-1 has 1 white square, Fig-2 has 3 white and 1 black squares, Fig-3 has 6 white and 3 black squares, and it continues. What is the total number of white squares in the Fig-20?
The white squares in successive figures are $$1, 3, 6, 10, \ldots$$, which are the triangular numbers, so Fig-$$n$$ has $$\frac{n(n + 1)}{2}$$ white squares. For $$n = 20$$ this gives $$\frac{20 \times 21}{2} = 210$$.
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$$m$$, $$n$$, $$p$$ are all different natural numbers each between 2 and 9. If $$\frac{m + n + p}{m + n}$$ is an integer, then the number of all possible values of $$\frac{m + n + p}{m + n}$$ is
Write $$\frac{m + n + p}{m + n} = 1 + \frac{p}{m + n}$$, so $$m + n$$ must divide $$p$$. As $$m$$ and $$n$$ are different numbers greater than 2, $$m + n \ge 7$$, while $$p \le 8$$, so the only possibility is $$p = m + n$$, for instance $$(m, n, p) = (3, 4, 7)$$ or $$(3, 5, 8)$$. In every such case the value of the expression is $$1 + 1 = 2$$, so only 1 value is possible.
$$1\frac{7}{20} \div 2.7 + 2.7 \div 1.35 + \left(0.4 \div 2\frac{1}{2}\right) \times \left(4.2 - 1\frac{3}{40}\right) = $$ ______.
Here $$1\frac{7}{20} = 1.35$$, so the first term is $$\frac{1.35}{2.7} = 0.5$$ and the second term is $$\frac{2.7}{1.35} = 2$$. In the third term $$0.4 \div 2.5 = 0.16$$ and $$4.2 - 1.075 = 3.125$$, whose product is $$0.5$$. The total is $$0.5 + 2 + 0.5 = 3$$.
$$a$$ and $$b$$ are natural numbers such that $$a < b$$ and $$a \times b = 2002$$. Then the minimum value of $$b - a$$ is ______.
Factorising, $$2002 = 2 \times 7 \times 11 \times 13$$, and the difference $$b - a$$ is least when the two factors are as close as possible to $$\sqrt{2002} \approx 44.7$$. The closest divisor pair is $$26 \times 77$$, since the neighbouring pairs $$22 \times 91$$ and $$14 \times 143$$ are further apart. Hence the minimum value is $$77 - 26 = 51$$.
In the adjoining figure, measure of $$\angle PQR = 120^\circ$$, measure of $$\angle PET = 40^\circ$$, and measure of $$\angle PTS = 110^\circ$$, then the measure of $$\angle PAQ = $$ ______.
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$$a$$, $$b$$, $$c$$, $$d$$, $$e$$ are five consecutive natural numbers. Another natural number $$n$$ always divides $$(a \times b \times c \times d \times e)$$ completely. Then the greatest possible value of $$n$$ is ______.
Among any five consecutive natural numbers there is a multiple of 5, at least one multiple of 3, and enough even numbers to supply a factor of 8, so the product is always divisible by $$5! = 120$$. The product $$1 \times 2 \times 3 \times 4 \times 5 = 120$$ shows that no larger number can always divide it. Hence the greatest such $$n$$ is 120.
An army general wanted to make a square arrangement of his soldiers. There were 2350 soldiers. After completing the square arrangement, he found that 46 soldiers were left over. Hence the number of soldiers in each row of the square arrangement is ______.
The soldiers actually used in the square number $$2350 - 46 = 2304$$. Since $$2304 = 48^2$$, each row of the square arrangement has 48 soldiers.
ABGH is a rectangle, CDFG is a square. B is the midpoint of CG and E is the midpoint of DF. The length of the rectangle ABGH is three times of its breadth. The area of the shaded region is $$80\,\text{cm}^2$$. Then the perimeter of ABCDFGHIA is ______ cm.
TO BE FILLED - figure required
80% of $$a$$ is equal to 16% of $$b$$. $$b$$ is $$x\%$$ of $$a$$, then 20% of $$x$$ is ______.
From $$0.8a = 0.16b$$ we get $$b = 5a$$, so $$b$$ is 500% of $$a$$ and $$x = 500$$. Therefore 20% of $$x$$ is $$\frac{20}{100} \times 500 = 100$$.
In the adjoining figure, $$\triangle ABC$$ is isosceles triangle with $$CA = CB$$. $$\angle ABC = 50^\circ$$, point $$D$$ is on side $$AB$$ such that $$\angle BCD = 46^\circ$$. Let $$\angle CAB = y$$, $$\angle ADC = x$$, $$\angle FCE = z$$, then $$x - y + 2z = $$ ______ degrees.
Since $$CA = CB$$, the base angles are equal, so $$y = \angle CAB = 50^\circ$$ and $$\angle ACB = 180^\circ - 50^\circ - 50^\circ = 80^\circ$$. In $$\triangle BCD$$, $$\angle BDC = 180^\circ - 50^\circ - 46^\circ = 84^\circ$$, so $$x = \angle ADC = 180^\circ - 84^\circ = 96^\circ$$. As $$CE$$ and $$CF$$ are the extensions of the sides $$CA$$ and $$CB$$ beyond $$C$$, the angle $$z$$ is vertically opposite $$\angle ACB$$ and equals $$80^\circ$$, giving $$x - y + 2z = 96 - 50 + 160 = 206$$.
A man drove from 7.45 am to 9.15 am (on the same day) and covered 84 kilometer. Then his driving speed is ______ km/h.
The drive lasted from 7.45 am to 9.15 am, which is 1 hour 30 minutes, or $$1.5$$ hours. The speed is therefore $$\frac{84}{1.5} = 56$$ km/h.
In the addition of two 2-digit numbers, each blank space, including those in the answer, is to be filled with one of the digits 0, 1, 2, 3, 4, 5, 6, each used exactly once. The unit's place digit of the sum is ______.
Seven digits are used in all, so the sum has three digits, and being the total of two 2-digit numbers it must start with 1 followed by 0. If the units digit of the sum is $$g$$, then the four digits of the two numbers add to $$20 - g$$, and comparing with the value $$100 + g$$ of the sum gives $$9 \times (\text{sum of the tens digits}) = 80 + 2g$$. Only $$g = 5$$ makes this a whole number, and $$42 + 63 = 105$$ shows it works, so the units digit of the sum is 5.
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