Solution of SEAMO 2016 Paper D

SEAMO 2016 PAPER D

SEAMO 2016 Paper D No. 1

A flight of stairs has twelve steps. To go up the stairs, Mary takes either one or two steps at a time. The number of ways for her to go up the stairs is ….
A. 34
B. 55
C. 89
D. 144
E. 233

Solution: Show/Hide

$1^{st}$ step: 1, 1 way
$2^{nd}$ step: 1 – 1, 2, 2 ways
$3^{rd}$ step: 1-1-1, 1-2, 2-1, 3 ways
$4^{th}$ step: 1-1-1-1-1, 1-1-2, 1-2-1, 2-1-1, 2-2, 5 ways
…
1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, …
It is a Fibonacci Sequence
Answer: E

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SEAMO 2016 Paper D No. 2

If the side of a square is increased by 30%, by how many percent does its are increase?
A. 30%
B. 54%
C. 69%
D. 144%
E. None of the above

Solution: Show/Hide

Let the side be $a$.
$\begin{align}\frac{(1.3a)^2-a^2}{a^2}\times 100\% &= 0,69\times 100\% \\ &= 69\% \end{align}$
Answer: C

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SEAMO 2016 Paper D No. 3

A coin is tossed 4 times. Find the probability that ‘Heads’ appears twice.

A. $\frac{2}{8}$
B. $\frac{3}{8}$
C. $\frac{3}{16}$
D. $\frac{5}{6}$
E. None of the above

Solution: Show/Hide

The possible outcomes:
HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, THTH, TTHH, HTTH, TTTH, TTHT, THTT, HTTT, TTTT.
$P(2\,heads)=\frac{6}{16}=\frac{3}{8}$
Answer: B

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SEAMO 2016 Paper D No. 4

Find the value of $x$ in the following number sequence.
2, 9, $-4$, 3, $-10$, $-3$, $-16$, $x$, $-22$, $-15$, ….
A. $-9$
B. $-8$
C. $-7$
D. $-6$
E. None of the above

Solution: Show/Hide

2 + 7 = 9
9 – 13 = -4
-4 + 7 = 3
3 – 13 = -10
-10 + 7 = -3
-3 – 13 = -16
-16 + 7 = -9 = x
Answer: A

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SEAMO 2016 Paper D No. 5

$m$ and $n$ are two prime numbers such that $m+n=21$. Find possible values of $\frac{2n+1}{m}$.
A. $\frac{5}{19}$, $\frac{39}{2}$
B. $\frac{9}{17}$, $\frac{17}{9}$
C. $\frac{17}{13},\frac{13}{17}$
D. $\frac{23}{10},\frac{21}{10}$
E. None of the above

Solution: Show/Hide

When $m=19$, $n=2$ values of $\frac{2n+1}{m}=\frac{5}{19}$.
When $m=2$, $n=19$ values of $\frac{2n+1}{m}=\frac{39}{2}$.
Answer: A

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SEAMO 2016 Paper D No. 6

What is the sum of all negative factors of 105?
A. $-105$
B. $-106$
C. $-144$
D. $-170$
E. $-192$

Solution: Show/Hide

(-1) + (-3) + (-5) + (-7) + (-15) + (-21) + (-35) + (-105)
= -192
Answer: E

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SEAMO 2016 Paper D No. 7

Tae Kwon multiplies the month of his birthday by 31. He the multiplies the day of his birthday by 12. The sum of the two products is 256. When is Tae Kwon’s birthday?
A. $10^{th}$ of March
B. $1^{st}$ of April
C. $7^{th}$ of April
D. $4^{th}$ of July
E. $9^{th}$ of August

Solution: Show/Hide

Let,
$\begin{align}31m+12d &= 265 \\ 12d &= 265-31m \\ d &= \frac{265-31m}{12} \end{align}$
When $m=7$
$d=4$
$4^{th}$ of July
Answer: D

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SEAMO 2016 Paper D No. 8

Find the value of
$2015\times 20162016-2016\times 20152015$
A. $-2$
B. $-1$
C. 0
D. 1
E. 2

Solution: Show/Hide

$\begin{align}\overline{abcdabcd} &= \overline{abcd0000}+\overline{abcd} \\ &= \overline{abcd}\times 10000+\overline{abcd} \\ \overline{abcdabcd} &= \overline{abcd}\times 10001 \end{align}$
$2015\times 20162016-2016\times 20152015$
= 2015 $\times$ 2016 $\times$ 10001 $-$ 2016 $\times$ 2015 $\times$ 10001
= 2016 $\times$ 2015 $\times$ 10001 $-$ 2016 $\times$ 2015 $\times$ 10001
= 0
Answer: C

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SEAMO 2016 Paper D No. 9

Find the value of
$100^2-99^2+98^2-97^2+...+2^2-1^2$
A. 100
B. 500
C. 1000
D. 2000
E. None of the above

Solution: Show/Hide

Remember: $a^2-b^2=(a+b)(a-b)$
$100^2-99^2+98^2-97^2+...+2^2-1^2$
= $(100+99)(100-99)$ + $(98+97)(98-97)$ +...+ $(2+1)(2-1)$
= 100 + 99 + 98 + 97 + … + 2 + 1
= $\frac{100(100+1)}{2}$
= 5050
Answer: E

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SEAMO 2016 Paper D No. 10

The number of digits in the product of $3\times 4^{22}\times 5^{45}$ is ….
A. 44
B. 45
C. 46
D. 47
E. None of the above

Solution: Show/Hide

$\begin{align}3\times 4^{22}\times 5^{45} &= 3\times (2^2)^{22}\times 5^{45} \\ &= 3\times 2^{44}\times 5^{44}\times 5 \\ &= 3\times 5\times (2\times 5)^{44} \\ &= 15\times (10)^{44} \\ &= 15\times 1\underbrace{00..00}_{44\,digits} \\ &= 15\underbrace{00..00}_{44\,digits} \end{align}$
There are 46 digits.
Answer: C

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SEAMO 2016 Paper D No. 11

It is Wednesday today. On which day of the week is 20152015 days later?
A. Wednesday
B. Thursday
C. Friday
D. Saturday
E. None of the above

Solution: Show/Hide

There are 7 days in a week.
Remainder 0 = Wednesday
Remainder 1 = Thursday
Remainder 2 = Friday
Remainder 3 = Saturday
Remainder 4 = Sunday
Remainder 5 = Monday
Remainder 6 = Tuesday
$20152015=\text{2878859}\times 7+2$
20152015 divided by 7 remainder 2 = Friday
Answer: C

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SEAMO 2016 Paper D No. 12

Mr. Bond left Follonica for Arezza at 8:00 AM at a constant speed of 150 km/h. he completed his mission in Arezza with in 30 minutes, the immediately mad his way back to Follonica at a constant speed of 100 km/h. How far was Arezza from Fellonica, if he arrived at Follonica at 11:00 AM?
A. 110
B. 120
C. 130
D. 140
E. 150

Solution: Show/Hide

Total travelling time:
= (11:00 AM – 08:00 AM) – 30 minutes
= 3 h – 0.5 h
= 2.5 h
$t_1+t_2=2.5$ h
$t_2=2.5-t_1$
$\begin{align}v_1\times t_1 &= v_2\times t_2 \\ 150\times t_1 &= 100\times t_2 \\ 150t_1 &= 100(2.5-t_1) \\ 150t_1 &= 250-100t_1 \\ 250t_1 &= 250 \\ t_1 &= 1 \end{align}$
Thus, the distance Arezza from Fellonica is:
= $v_1\times t_1$
= $150\times 1$
= 150 km
Answer: E

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SEAMO 2016 Paper D No. 13

Find the value of
$1+2^1+2^2+2^3+...+2^8$
A. 488
B. 511
C. 513
D. 516
E. None of the above

Solution: Show/Hide

Let,
$S=1+2^1+2^2+2^3+...+2^8\,....\,(1)$
$2S=2^1+2^2+2^3+...+2^8+2^9\,....\,(2)$
(2) – (1)
$\frac{\begin{align}2S &= 2^1+2^2+2^3+...+2^8+2^9 \\ S &= 1+2^1+2^2+2^3+...+2^8 \end{align}}{\begin{align}S &= 2^9-1 \\ S &= 512-1 \\ S &= 511 \end{align}}-$
Answer: B

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SEAMO 2016 Paper D No. 14

How many ways are there to fill each circle with a number from 1 to 7, without repetition, such that the sum of three numbers along each line is equal?

A. 1
B. 2
C. 3
D. 4
E. 5

Solution: Show/Hide

1 + 2 + 3 + 4 + 5 + 6 + 7 = 28
Let the middle number be $x$, $(28+2x)$ is divisible by 3.
$x$ = 1, 4, 7
There are 3 ways to arrange the numbers.
Answer: C

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SEAMO 2016 Paper D No. 15

$\overline{a8819b}$ is divisible by 12. How many such six-digit numbers are there?
A. 3
B. 4
C. 5
D. 6
E. None of the above

Solution: Show/Hide

$12=3\times 4$
Case 1. $b=2$
$a+8+8+1+9+2=a+28$
$(a+28)|3$ so $a$ = 2, 5, 8
Case 2. $b=6$
$a+8+8+1+9+6=a+32$
$(a+32)|3$ so $a$ = 1, 4, 7
Thus, there are 6 numbers that satisfy.
Answer: D

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SEAMO 2016 Paper D No. 16

A circle of circumference 2 m rolls around the equilateral triangle of a perimeter of 6 m. How many turns does the circle make as it rolls around triangle once, without slipping?

A. 3
B. 4
C. 5
D. 6
E. None of above

Solution: Show/Hide

Circumference of the circle = 2 meters
Perimeter of the equilateral triangle = 6 meters
Number of turns does the circle make as it rolls around the triangle once, without slipping is $\frac{6}{2}=3$
Answer: A

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SEAMO 2016 Paper D No. 17

How long will it be before the minute hand next lies directly over the hour hand?

A. 1 hours 15 minutes
B. 2 hours 40 minutes
C. 4 hours
D. 12 hours
E. None of the above

Solution: Show/Hide

For every lap made by the minute hand, the hour make $\frac{1}{12}$ laps.
Ratio of speed = 12 : 1
$\frac{1}{60}-\frac{1}{720}=\frac{11}{720}$
The minute hand catches up by $\frac{11}{720}$ laps every minute.
At 8:00 o’clock, the minute hand is $\frac{2}{3}$ laps behind the hour hand.
Thus,
$\frac{2}{3}\div \frac{11}{720}=\frac{2}{3}\times \frac{720}{11}=\frac{480}{11}=43\frac{7}{11}$
The minute hand will take $43\frac{7}{11}$ minutes to catch up to the hour hand.
Answer: E

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SEAMO 2016 Paper D No. 18

There are 9 white, 5 red and 6 black balloons. 10 balloons are picked at random such that there are at least 2 but not more than 8 while, there are at least 2 red and there are not more than 3 black balloons. How many ways are there to do this?
A. 4
B. 5
C. 16
D. 18
E. None of the above

Solution: Show/Hide

There are 16 ways.
Answer: C

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SEAMO 2016 Paper D No. 19

Candle A took 3 hours to finish burning. Candle B took 5 hours to finish burning. Candle B wa shorter and thicker than candle A. they were lit at the same time and had the same height two hours later. What was the ratio their heights first?

A. 8 : 5
B. 9 : 5
C. 8 : 3
D. 5 : 3
E. None of the above

Solution: Show/Hide

Let the initial heights be $m$, $n$ respectively.
Amount burnt = $m\times \frac{120}{180}=\frac{2m}{3}$
Amount burnt = $n\times \frac{120}{300}=\frac{2n}{3}$
Remained: $\frac{m}{3}$, $\frac{3n}{5}$
$\frac{m}{3}:\frac{3n}{5}=1:1$
$m:n=9:5$
Answer: B

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SEAMO 2016 Paper D No. 20

Find the largest $n$ such that there is only one whole number $k$ that satisfies $\frac{9}{17} < \frac{n}{n+k} < \frac{8}{15}$.
A. 72
B. 127
C. 144
D. 255
E. None of the above

Solution: Show/Hide

$\frac{9}{17} < \frac{n}{n+k} < \frac{8}{15}$
$\frac{15}{8} < \frac{n+k}{n} < \frac{17}{9}$
$\frac{15}{8} < 1+\frac{k}{n} < \frac{17}{9}$
$\frac{15}{8}-1 < 1-1+\frac{k}{n} < \frac{17}{9}-1$
$\frac{7}{8} < \frac{k}{n} < \frac{8}{9}$
$\frac{63}{72} < \frac{k}{n} < \frac{64}{72}$
$\frac{126}{144} < \frac{k}{n} < \frac{128}{144}$
$n=144$
Answer: C

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SEAMO 2016 Paper D No. 21

Fill in each circle with numbers from 1 to 10, without repetition, such that the average of any group of 5 adjacent numbers is the minimum.

Solution: Show/hide

10, 7, 6, 3, 2, 9, 8, 5, 4, 1
Answer: 10, 7, 6, 3, 2, 9, 8, 5, 4, 1

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SEAMO 2016 Paper D No. 22

Evaluate
$\frac{1}{1+2}$+$\frac{1}{1+2+3}$+$\frac{1}{1+2+3+4}$+...+$\frac{1}{1+2+3+....+99+100}$

Solution: Show/Hide

$\frac{1}{1+2}$+$\frac{1}{1+2+3}$+$\frac{1}{1+2+3+4}$+...+$\frac{1}{1+2+3+....+99+100}$
= $\sum\limits_2^{100}{\frac{1}{\frac{n(n+1)}{2}}}$
= $\sum\limits_2^{100}{\frac{2}{n(n+1)}}$
= $\sum\limits_2^{100}{\left( \frac{2}{n}-\frac{2}{n+1} \right)}$
= $\frac{2}{2}-\frac{2}{3}+\frac{2}{3}-\frac{2}{4}+...+\frac{2}{100}-\frac{2}{101}$
= $1-\frac{2}{101}$
= $\frac{99}{101}$
Answer: 99/101

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SEAMO 2016 Paper D No. 23

The figure shows a $4\times 4$ grid. The sum of 4 numbers in each row, column and diagonal is 2016. Find $(a+b+c+d)$.

Solution: Show/Hide

$a+e+i+c=2016\,....\,(1)$
$b+h+i+d=2016\,....\,(2)$
$a+f+k+d=2016\,....\,(3)$
$b+g+j+c=2016\,....\,(4)$
(1) + (2) + (3) + (4),
$2\times (a+b+c+d)$ + $(e+f+g+h)$ + $(i+j+k+l)$ = $2016\times 4$
$\begin{align}2\times (a+b+c+d)+2016+2016 &= 2016\times 4 \\ 2\times (a+b+c+d)+2016\times 2 &= 2016\times 4 \\ 2\times (a+b+c+d) &= 2016\times 2 \\ a+b+c+d &= 2016 \end{align}$
Answer: a + b + c + d = 2016

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SEAMO 2016 Paper D No. 24

The figure below shows a right-angled triangle with semicircles A, B and C constructed using its sides as diameters. The circumference of semicircle A is $13\pi $. The area of semicircle B is $12.5\pi $. What is the radius of semicircle C?

Solution: Show/Hide

Let the radius of the semicircle A be $r_1$
$\begin{align}\frac{1}{2}\times 2\pi r &= 13\pi \\ r_1 &= 13 \end{align}$
$\begin{align}PR &= 2\times r_1 \\ &= 2\times 13 \\ PR &= 26 \end{align}$
Let the radius of semicircle B be $r_2$
$\begin{align}\frac{1}{2}\times \pi r_2^2 &= 12.5\pi \\ r_2^2 &= 25 \\ r_2 &= 5 \end{align}$
$\begin{align}PQ &= 2\times r_2 \\ &= 2\times 5 \\ PQ &= 10 \end{align}$
$\begin{align}QR &= \sqrt{PR^2-PQ^2} \\ &= \sqrt{26^2-10^2} \\ QR &= 24 \end{align}$
Radius of semicircle C = $\frac{1}{2}\times QR=\frac{1}{2}\times 24=12$.
Answer: 12

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SEAMO 2016 Paper D No. 25

$AE\bot BE$, $AF\bot CD$ in the parallelogram shown below. Given that $\angle EAF=60^\circ $, BE = 2 cm and DF = 3 cm, find $\angle ABC$ and the length of CD.

Solution: Show/Hide

$\angle AEC$ + $\angle ECF$ + $\angle CFA$ + $\angle EAF$ = $360^\circ$
$90^\circ $ + $\angle ECF$ + $90^\circ$ + $60^\circ$ = $360^\circ $
$\angle ECF=120^\circ$
$\angle BCD=\angle ECF=120^\circ$
$\begin{align}\angle ABC+\angle BCD &= 180^\circ \\ \angle ABC+120^\circ &= 180^\circ \\ \angle ABC &= 60^\circ \end{align}$
$\angle ABE=\angle ABC=60^\circ $
Consider triangle ABE:
$\begin{align}\cos \angle ABE &= \frac{BE}{AB} \\ \cos 60^\circ &= \frac{2}{AB} \\ \frac{1}{2} &= \frac{2}{AB} \\ AB &= 4 \end{align}$
$CD=AB$ = 4 cm
Answer: 4 cm
Answer: $\angle ABC=60^\circ $ and the length of CD is 4 cm

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