Aptitude
1.0.0
Di dalam air, arah di sepanjang sungai disebut hilir. Dan, arah terhadap aliran disebut hulu.
Jika kecepatan perahu di air diam adalah U km/jam dan kecepatan alirannya adalah V km/jam, maka:
Kecepatan hilir = (u + v) km/jam.
Kecepatan hulu = (u - v) km/jam.
Jika kecepatan hilir adalah km/jam dan kecepatan hulu adalah b km/jam, maka:
Kecepatan dalam air diam = 1/2*(a + b) km/jam.
Laju aliran = 1/2*(a - b) km/jam.
Laba, p = sp - cp; SP> CP
Kehilangan, L = CP - SP; CP> sp
P% = (p/cp) x 100
L% = (l/cp) x 100
Sp = {(100 + p%)/100} x cp
Sp = {(100 - l%)/100} x cp
Cp = {100/(100 + p%)} x sp
Cp = {100/(100 - l%)} x sp
Diskon = MP - SP
Sp = mp -discou
Untuk bobot palsu , persentase laba akan menjadi P% = (berat sejati - berat palsu/ berat palsu) x 100 .
Ketika ada dua keuntungan yang berhasil mengatakan M% dan n%, maka laba persentase bersih sama dengan (m+n+mn)/100
Ketika laba adalah M% dan rugi adalah N%, maka laba atau rugi bersih akan menjadi: (Mn-Mn)/100
Jika suatu produk dijual dengan laba M% dan kemudian dijual lagi dengan laba N% maka harga biaya aktual produk adalah: CP = [100 x 100 x P/(100+m) (100+N)] . Dalam hal kerugian , cp = [100 x 100 x p/(100-m) (100-N)]
Jika P% dan L% sama, maka p = L dan % kehilangan = p^2/100
Untuk memeriksa kalikan angka dengan 5 atau 2 dan dari angka itu tambahkan (jika Mul dengan 5) lain jika Mul dengan 2 (kurangi) melalui nomor kiri
misalnya: 532 div dengan 7
Langkah 1- Mul 2x5 = 10 Langkah 2- Abaikan digit unit dan tambahkan ke nomor kiri 53 | 2+10 = 63
periksa nomor yang dapat dibagi jika ya daripada nomor asli juga dapat dibagi.
Similary kelipat dengan 2 dan sebagai gantinya penambahan kurangi.
Kalikan unit digit dengan 5 dan kurangi nilai yang berlipat ganda dengan digit istirahat jika jumlah multile ini dari angka ini dapat dibagi dengan 17.
misalnya: 272 iv pada 17
Langkah 1-
2 Mul*5 = 10 ** Langkah 2- **
27-10 = 17 17 dibagi dengan 17 karenanya 272 dibagi dengan 17.
Wajah atau dial jam adalah lingkaran yang kelilingnya dibagi menjadi 60 bagian yang sama, bernama Minute Spaces.
Tangan jam dan menit jam, satu jam memiliki dua tangan. Tangan yang lebih kecil disebut tangan jam atau tangan pendek dan yang lebih besar disebut tangan kecil atau tangan panjang.
Dalam 60 menit, tangan menit mendapatkan ruang 55 menit dari tangan.
(Dalam 60 menit, jam jam akan menggerakkan spasi 5 menit sementara tangan menit akan menggerakkan spasi 60 menit. Pada dasarnya gain ruang dari tangan menit sehubungan dengan tangan jam akan 60 - 5 = 55 menit.)
Kedua tangan jam bertepatan sekali dalam setiap jam.
Tangan jam berada di garis lurus yang sama ketika mereka bertepatan atau berlawanan satu sama lain.
Ketika dua tangan jam berada di sudut kanan , mereka terpisah 15 menit .
Ketika tangan jam berada di arah yang berlawanan, mereka terpisah 30 menit.
Sudut dilacak dengan tangan jam dalam 12 jam = 360 °
Angle dilacak dengan tangan kecil dalam 60 menit. = 360 ° .
Theta (alias derajat) = 30 x h- (11/2) xm
Sudut reflaks = 360 - theta
1. Bilangan alami (n) = 1, 2, 3 ,. . . . 2. Bilangan seluruh (W) = 0, 1, 2, 3 ,. . . . 3. Intezers (Z) = −∞. . . −2, −1, 0, 1, 2, 3 ,. . . 4. Bilangan rasional (q) = Angka bentuk p⁄Q di mana q ≠ 0. misalnya: 1⁄5, 0,46, 0,333333 mereka mengakhiri, mengulangi . 5. Bilangan irasional (i) = Angka dari bentuk x1⁄n ≠ intezer. Juga π dan e juga bilangan irasional mereka tidak terminasi, tidak berulang .
Jenis Angka Lainnya: a. Bahkan angka: Angka yang persis dapat dibagi dengan 2. Angka -angka ini berada dalam format 2n. B. Angka ganjil: Angka yang memberikan sisa 1 ketika dibagi dengan 2. Angka -angka ini dalam format 2n ± 1. C. Bilangan prima: Angka yang dapat dibagi dengan 1 dan angka itu sendiri adalah bilangan prima. Yang paling utama adalah 2. d. Angka Komposit: Jumlah yang dapat dibagi dengan lebih dari 2 angka.
ganjil ± ganjil = bahkan;
bahkan ± bahkan = bahkan;
bahkan ± ganjil = ganjil
ganjil × ganjil = ganjil;
bahkan × bahkan = bahkan;
bahkan × ganjil = bahkan.
ganjil (angka apapun) = ganjil
bahkan (angka apapun) = bahkan
(A - B) 2 = (A2 + B2 - 2AB)
(a + b) 2 = (a2 + b2 + 2ab)
(a + b) (a - b) = (a2 - b2)
(A3 + B3) = (A + B) (A2 - AB + B2)
(A3 - B3) = (A - B) (A2 - AB + B2)
(A + B + C) 2 = A2 + B2 + C2 + 2 (AB + BC + CA)
(A3 + B3 + C3 - 3ABC) = (A + B + C) (AR> 2 + B2 + C2 - AB - BC - AC)
Tip dan trik cepat
Langkah1: Saat menyederhanakan ekspresi yang diberikan, tanda kurung pertama harus dihapus dalam urutan: '––', () ',' {} ',' []
Langkah2: Operasi harus dilakukan secara ketat dalam urutan: Divisi, Perkalian, Penambahan dan Pengurangan
Bodmas adalah jalan pintas yang digunakan untuk mengingat prosedur penyederhanaan.
B: Braket
O: Pesanan (yaitu kekuatan, akar kuadrat dll.)
D: Divisi (kiri ke kanan)
M: Perkalian (kiri ke kanan)
A: Penambahan (kiri ke kanan)
S: pengurangan (kiri ke kanan)
Langkah1: Saat menyederhanakan ekspresi yang diberikan, tanda kurung pertama harus dihapus dalam urutan: (), {}, []
Langkah2: Operasi harus dilakukan secara ketat dalam urutan: Divisi, Perkalian, Penambahan dan Pengurangan
** Formula penting: **
Formula penting
Jumlah kuadrat n bilangan alami = n (n+1) (2n+1)/2
Jumlah N angka alami = n (n+1)/2
Jumlah dalam ap = nx (a+l)/2
Rata -rata angka ganjil n ganjil = n
Rata -rata n even number = n+1
Rata -rata n naturat number = (n+1)/2
Rata -rata angka * jumlah total = jumlah
Pertanyaan berbasis populasi
Rasio dan proporsi
Kode/decode
Log
Pertanyaan verbal amcat
Sisa negatif
Tes kelayakan
735/2
735/704
3/704
3/735
Larutan
Option A
Find the LCM of the numerators.
LCM (147, 30) = 1470
Step 2: Find the HCF of denominators.
HCF (64, 44) = 4
So, the LCM of 147/64 and 30/44 is (LCM of Numerators) / (HCF of Denominators) = 1470 / 4 = 735/2
A) 5 mph
B) 10 mph
C) 12 mph
D) 20 mph
Larutan
Option C
Let the speed of the river=x mph, then
Time taken row 30 miles upstream and 30 miles downstream = 30/(15-x) + 30/(15+x) = 9/2
= 10/(15-x) + 10/(15+x) = 3/2
= 2[10(15+x) + 10(15-x)] = 3(15-x)²
= 300 + 20x + 300 – 20x = 675 -3x²
x² = 25 or x=5
A) 3 hari
B) 4 hari
C) 4,5 hari
D) 5,4 hari
Larutan
option A
Working 5 hours a day, A can complete the work in 8 days = 5*8 = 40 hours
Working 6 hours a day, B can complete the work in 10 days = 6*10 = 60 hours
(A+B)’s 1 hour’s work = (1/40+1/60)
=(3+2)/120
= 1/24
Hence, A and B can complete the work in 24 hours which is equal to 3 days.
A) 6,5 liter
B) 5 liter
C) 4 liter
D) 7,5 liter
Larutan
Option B
A mixture of 40 liters of milk = 36 liters of Milk and 4 liters of water = 90:10 ratio
Now the new mixture should be in the ratio of 80:20
Hence 80% is equivalent to 36 liters (no addition of milk is done)
100% is (36/80)*100 = 45 liters of milk is present in the new mixture
Thus water shall be added= (45 – 36 – 4) = 5 liters of water
**or**
A) 12 tengah malam
B) 3 pagi
C) 6 pagi
D) 9 pagi
Larutan
Option D
Four different devices beep after every 30 mins, 60 mins, 90 mins and 105 mins.
LCM of 30,60,90 and 105 is 1260.
Which means the devices beep together after every 1260 mins = 1260/60 = 21 hours
Hence 12 noon + 21 hours = 9 a.m
A) 21 m
B) 26 m
C) 28 m
D) 29 m
Larutan
Option C
When A travels 100 m, B travels 75 m. Hence A:B = 100:75
When B travels 100 m, C travels 96 m. Hence B:C = 100:96
When B Travels 75 m, C travels (96 x 75)/100 = 72 m
Hence B:C = 75:72.
Therefore, A:B:C = 100:75:72.
So, when A Travels 100 m, C travels 72 m.
Therefore, A beat C by 28 m
A) 62
B) 68
C) 66
D) 69
Larutan
Option A
70% students passed in physics = 30% failed in Physics.
65% students passed in Chemistry = 35% failed in Chemistry
Percentage of students failed in both subject = 27%.
Percentage of students failed = 35 + 30 – 27
= 38%.
Percentage of students passed = 100 – 38% = 62%
A) 40
B) 60
C) 100
D) 160
Larutan
Option D
15 oxen take 80 days so, 6 oxen take x days
x = 15*80/6 = 200 days
20 oxen also take 80 day. So, 2 cows take y days
y = 20*80/2 = 800 days
Together work will be done in 800*200/(800+200) = 160 days
A) 297/10377
B) 188/121
C) 21/34
D) 33/163
Larutan
Solution: Option D
A) 6c4
B) 6p4
C) 4^6
D) 6^4
Larutan
C is correct because all 4 can get 4 ticket one by one
A) 550
B) 450
C) 350
D) 150
Larutan
Option B
Going through the options,
Taking Cost Price as Rs 450.
Profit = 650 – 450 = 200
Loss = 350 – 450 = 100
Clearly profit is twice the loss incurred.
Hence, Rs 450 is the correct option.
A) 7 jam 30 menit
B) 8 jam
C) 8 jam 15 menit
D) 8 jam 25 menit
Larutan
Option C
In 1hr, Ronald types = 32/6 pages and Elan types = 40/5 pages
If they work together they will type = 32/6 + 40/5 = 40/3 pages in 1 hr
Time needed to complete the assignment is = (3 x 110)/40 = 33/4 = 8hrs 15mins
Hence, the time required is 8 hrs 15 mins.
A) menurun sebesar 20%
B) meningkat sebesar 20%
C) meningkat sebesar 10%
D) menurun sebesar 10%
Larutan
Option D
Let the initial Price = Rs.100 and initial sales = 100
So, the initial revenue = Rs. 10000
Now, the price is reduced to 25% which is equal to Rs.75 and Sales is increased by 20% which is equal to 120.
Now new revenue = 120 x 75 = Rs. 9000
Change in revenue = (10000 – 9000) = Rs.1000 decrease
% decrease = (1000/10000) x 100 = 10%
Hence, the correct option is decrease of 10%.
A) 19/21
B) 3/7
C) 2/21
D) 1/3
Larutan
Option A
Probability of getting atleast one nestle chocolate = [(10C1 x 5C1) + 10C2] / 15C2
[(10 x 5) + (10 x 9)/2] / [(15 x 14)/2] = 19/21.
Hence, the required probability is 19/21.
A) Rs 7490
B) Rs 7350
C) Rs 8250
D) Rs 8530
E) tidak satupun dari ini
Larutan
Option B
Solution:
Share of Anil : Share of Ruhi : Share of Teena is
2000×8 + 2600×4 : 2800×8 + 3200×4 : 4200×4
33 : 44 : 21
so share of Teena = 21/(33+44+21) × 34300 = Rs 7350
A) Rs 3200
B) Rs 4500
C) Rs 3800
D) Rs 3500
E) Rs 2800
Larutan:
Option C
Solution:
Rs 3800
Solution:
(7000-x)*8*4/100 = x [ (1 + 10/100)2 – 1] + 226
70*8*4 – 32x/100 = 21x/100 + 226
2240 – 226 = 53x/100
2014 = 53x/100
So, x = Rs 3800
A) 16
B) 18
C) 12
D) 10
E) 22
Larutan:
Option A
Solution:
20 men in 8 days so 16 men in 20 × 8/16 = 10 days and
25 women in 12 days so 10 women in 25 × 12/10 = 30 days
So in 3 days, they complete (1/10 + 1/30) × 3 = 2/5
So remaining work = 1 – 2/5 = 3/5
20 m 1 work in 8 days and x men 3/5 work in 6 days
So 20 × 8 × 3/5 = x × 6 × 1
So, x = 16 men
A) 3/5
B) 2/9
C) 1/8
D) 3/7
E) tidak satupun dari ini
Larutan:
Option D
Solution
Number of multiples of 3 in 140 = 140/3 = 46
Number of multiples of 7 in 140 = 140/7 = 20
Number of multiples of 3×7= 21 in 140 = 140/21 = 6
So required probability = (46+20 – 6)/140 = 60/140 = 3/7
A. Ibu
B. Sister
C. keponakan
D. bibi ibu
Solusi c
A A
B) b
C) c
D) e
Larutan:
E
Solution:
D is father of A and grandfather of F. So, A is father of F.
Thus. D and A are the two fathers. C is the sister of F So. C is the daughter of A.
Since there is only one mother, it is evident that E is the wife of A and hence the mother of C and F.
So, B is brother of A There are three brothers. So. F is the brother of C.
Clearly, A is E's Husband.
A) Ibu
B) saudari
C) bibi
D) Nenek
Larutan:
C
Solution:
Only son of Amar's mother's father — Amar's maternal uncle.
So, the girl's maternal uncle is Arnar's maternal uncle. Thus, the girl's mother is Amar's aunt.
A) 12
B) 13
C) 14
D) 15
Larutan:
A
Solution:
A+B=B+C+12
so
A=12
Larutan:
P=(3*2*5)/1=30
Q=(4*2*5)/1=40
Larutan:
(150+x)/15=16.
=)150+x=240
=x=90
Rata -rata usia = jumlah/angka =) 90/5 = 18
A. 154 °
B. 170 °
C. 160 °
D. 180 °
Larutan:
We know that angle traced by hour hand in 12 hrs = 360°
From 8 to 2, there are 6 hours.
Angle traced by the hour hand in 6 hours = 6×360/12= 180°
A. 120 °
B. 125 °
C. 130 °
D. 135 °
Larutan:
C
Solution:
Angle traced by hour hand in 12 hrs. = 360°.
Angle traced by it in 11/3 hrs = (360/12 x 11/3)° = 110°.
Angle traced by minute hand in 60 min. = 360°.
Angle traced by it in 40 min. = (360/60x40)°= 240°.
Required angle (240 - 110)° = 130°.
A. 3.6
B. 7.2
C. 8.4
D. 10
Larutan:
B
Solution:
Speed =600/(5 x 60)= 2 m/sec.
Converting m/sec to km/hr (see important formulas section)
= (2 x18/5)km/hr= 7.2 km/hr.
A. 10
B. 8
C. 6
D. 4
Larutan:
C
Opposite direction
speed=60+6=66km/h
time=distance/speed=110/66=5/3 km/h
in m/s 5/3x18/5=6
A. 14.4 detik
B. 15,5 detik
C. 18.8 detik
D. 20.2 detik
Solusi:
A
Solution :
Let length of each train be x meter.
Then, speed of 1st train = x/18m/sec
Speed of 2nd train = x/12 m/sec
Now,
When both trains cross each other, time taken
=2x/(x/18+x/12)=2x/(2x+3x)/36=(2x X 36)/5x=725=14.4seconds
767 495 359 291 257?
A. 230 B. 240 C. 250 D. 280 E. 260
Soluton:
B
Solution:
797-495=272
495-359=136
So which means it is half of previous diffrences
272/2=136
291-257=32
34/2=17
So subtract 17
257-17=240
50 67 33 84 16?
A. 101 B. 109 C. 107 D. 103 E. 201
Larutan:
A
Solution:
17 is the gap once increase and than decrease follow this order you will get the answer
192
10
38
2
3
Larutan:
3
Solution:
Logic is 2×1 + 1 = 3, 3 × 2 + 4 =10, 10 × 3 + 9 = 39, 39 × 4 + 16 = 172…. So in place of 38, it should be 39.
999980
999990
999984
Tidak ada satupun dari ini
Larutan:
Greatest six-digit number is 999999. Divide this number by 12 and get remainder as 3. Since the remainder is 3, if you subtract 3 from the number, the remaining number will be a multiple of 12. So the greatest such number will be 999999 – 3 =999996.
5000
4950
4980
4900
Tidak ada satupun dari ini
Larutan:
Multiples of 3 between 100 and 200 are 102, 105, 108,… ,198.
Here, the first term = 102
last term = 198
Let the number of Multiples of 3 between 100 and 200 = n
W.K.T: Arithmetic Progression Formula:
an = a1 + (n - 1)d
Where, an = last term = 198
a1 = first term = 102
d = common difference = 105 - 102 = 3
---> 198 = 102 + (n - 1) * 3
---> 198 - 102 = (n - 1) * 3
---> 96 = (n - 1) * 3
---> (n - 1) = 96/3 = 32
---> n = 32 + 1
---> n = 33
Formula:
Sum of n terms = Sn = (n/2) * (a + l)
where n = number of elements = 33
a = first term = 102
l = last term = 198
Thus, using the above formula, Sum of all natural numbers between 100 and 200 which are multiples of 3 = (33/2) * (102 + 198)
= (33/2) * 300
= 33 * 150
= 4950
A. 5, 15, 25
B. 12, 15, 18
C. 10, 15, 20
D. -10, -15, -20
Larutan:
Assuming that the numbers are (a – d), a, (a + d) and their sum is 45, we get the middle number as 15. Now, the product (a – d) (a + d) = 200. Solving, we get d = 5. Therefore, the numbers are 10, 15 and 20.
Larutan:
x/(y+1)=1/2
and
(x+1)/y=1
2x-y=1 ....eq (1)
x=y-1 ....eq (2)
by solving eq 1 and 2 we get
x=2 and y=3
A. 80
B. 75
C. 42
D. 53
Menjawab:
Option D
Solution:
As the Number gives a remainder of 4 when it is divided by 7, then the number must be in form of (7x + 4)
The same gives remainder 1 when it is divided 4, so the number must be in the form of {4 × (7x + 4) + 1}
Also, the number when divided by 3 gives remainder 2, thus number must be in form of [3 × {4 × (7x + 4) + 1} + 2]
Now, On simplifying,
[3 × {4 × (7x + 4) + 1} + 2]
= 84x + 53
We get the final number 53 more than a multiple of 84 Hence, if the number is divided by 84,
The remainder will be 53
A) 55/601
B) 601/55
C) 11/120
D) 120/11
Larutan:
Then, a + b = 55 and ab = 5 x 120 = 600.
The required sum =1a+1b = a+bab= 55600=11120
Larutan:
5
2, 3, 5, 8, 13, 20, 34
Larutan:
first+second=third
follow this order you get **20** as a answer
196, 169, 144, 121, 100, 80
Larutan:
80
First - Second=2
A. cmn
B. Uji
C. VIJ
D. ijt
Larutan:
C
Olnnie
Onilen
Noilen
Lnoeni
Onnlie
Larutan:
1
A. 9 tahun
B. 10 tahun
C. 13 tahun
D. 15 tahun
Larutan:
Correct Option: (c)
We have to find the population of cities A and B after x years.
Step 1: Population of city A = 68000, decreases at the rate of 1200/year
68000 – 1200x
Step 2: Population of city B = 42000, increases at the rate of 800/year
42000 + 800x
Step 3: Find after how many population of cities A and B are equal.
Population of city A = Population of city B
68000 – 1200x = 42000 + 800x
68000 – 42000 = 1200x + 800x
26000 = 2000x
x = 13
A. 28 hari
B. 30 hari
C. 34 hari
D. 40 hari
Larutan:
Correct Option: (b)
Step 1: Number of days worked by the worker = 60 and he remained idle for x days. Therefore, number of days worked = (60 – x)
Step 2: Each day he was getting paid Rs. 20. Therefore, the payment received for working days = (60 – x) 20
Step 3: After subtracting the amount which he forfeited, he receives Rs. 300.
Therefore,
(60 – x) 20 – 10x = 300
1200 – 20x – 10x = 300
900 = 30x
x = 30 days
A. 11
B. 15
C. 16
D. 18
Larutan:
Correct Option: (b)
Let’s the number of farmers be y.
Step 1: Find number of heads
= (50 hens + 45 goats + 8 horses + y farmers)
= (103 + y)
Step 2: Number of feet
= [(Hens 2 × 50) + (45 × 4) + (8 × 4) + (y × 2)]
= [100 + 180 + 32 + 2y]
= 312 + 2y
Step 3: Find number of farmers
(312 + 2y) – (103 + y) = 224
312 + 2y – 103 – y = 224
y = 15
A) 49500
B) 49950
C) 45000
D) 49940
Larutan:
The Correct answer is (B)
Answer with explanation:
The digit 5 has two place values in the numeral, 5 * 105 = 50,000 and 5 * 101 = 50.
∴Required difference = 50000 - 50 = 49950
A) Rs. 180
B) Rs. 204
C) Rs. 210
D) Rs. 220
Larutan:
Option B
CI=P(1+r/100)^t
CI=2500*(1+4/100)^2
CI=2704
So the diffrenece is 204
A) Rs. 5222.2
B) Rs. 5777.7
C) Rs. 6222.2
D) Rs. 6777.7
Larutan:
Option B
80000*12/65000*6=32/13
113/32*20000=5777.7rs
A) Kamis
B) Rabu
C) Jumat
D) Minggu
Larutan:
The correct option is (B)
Explanation:
The year 1996 is divisible by 4, so it is a leap year with 2 odd days.
As per the question, the first day of the year 1996 was Monday, so the first day of the year 1997 must be two days after Monday. So, it was Wednesday.
A) 1,5 km/jam
B) 2 km/jam
C) 2,5 km/jam
D) 1 km/jam
Larutan:
The correct answer is B
Answer with explanation:
Let the speed of stream = X km/hr
Speed of boat = 5 km/hr
Speed upstream = 3km/hr
Apply formula: Speed upstream = speed of boat - speed of stream
∴ 3 = 5 - X
X = 5 - 3 = 2 km/hr
A) 24
B) 22
C) 23
D) 21
Larutan:
The Correct answer is (B)
Explanation:
The hands of a clock coincide only once between 11 O' clock and 1 O' clock, so in every 12 hours, the hands of a clock will coincide for 11 times.
∴ In a day or 24 hours, the hands of a clock will coincide for 22 (11+11) times.
A) 10 jam
B) 12 jam
C) 14 jam
D) 16 jam
Larutan:
Option B
Water enter in 1 hr=1/6
Water empty in 1 hr=1/12
net=1/6-1/12=1/12
or 12hr
A) 8,5 km/s
B) 7,5 km/s
C) 9,5 km/s
D) 6,5 km/s
Larutan:
Option B
sec=4*60=240s
speed=500/240=25/12m/s
in km/s speed is =25/12*18/5=7.5km/s.
Larutan:
Surface area of cube=6a^2
600=6*a^2
a=10
diagonal of cube=sqrt(3)*a
ans=sqrt(3)*10
HCF dari dua angka adalah 23 dan dua faktor lainnya dari LCM mereka adalah 13 dan 14. Yang lebih besar dari dua angka adalah:
A. 276
B. 299
C. 322
D. 345
Larutan:
Answer: Option C
Explanation:
Clearly, the numbers are (23 x 13) and (23 x 14).
Larger number = (23 x 14) = 322.
Enam lonceng mulai berbunyi bersama dan tol pada interval 2, 4, 6, 8 10 dan 12 detik masing -masing. Dalam 30 menit, berapa kali mereka bertanding bersama?
A. 4
B. 10
C. 15
D. 16
Larutan:
Answer: Option D
Explanation:
L.C.M. of 2, 4, 6, 8, 10, 12 is 120.
So, the bells will toll together after every 120 seconds(2 minutes).
In 30 minutes, they will toll together 30 + 1 = 16 times.
2
Misalkan N menjadi angka terbesar yang akan membagi 1305, 4665 dan 6905, meninggalkan sisa yang sama dalam setiap kasus. Maka jumlah digit di n adalah:
A. 4
B. 5
C. 6
D. 8
Larutan:
Answer: Option A
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
A. 9000
B. 9400
C. 9600
D. 9800
Larutan
Answer: Option C
Explanation:
Greatest number of 4-digits is 9999.
L.C.M. of 15, 25, 40 and 75 is 600.
On dividing 9999 by 600, the remainder is 399.
Required number (9999 - 399) = 9600.
A) 50 hari
B) 60 hari
C) 84 hari
D) 9.333 hari
Larutan:
Answer: C
Explanation:
Let 5 men can reap a field in x days
So, put the same quantities on the same side.
Men: Days
Now, Men and Days are inversely proportional to each other. If we increase the number of men fewer days will be required to complete the work.
Inversely proportional means Apti Chain Rule
Apti Chain Rule
i.e., 5: 15 = 28: x
Or, x = (28*15)/ 5
Or, x = 84 days
Hence, 5 men can reap a field in 84 days.
A) 16
B) 13/5
C) -16/3
D) 12
Larutan:
Answer: C
Explanation:
Let log2√2 [1/256] = x
We know that loga y = x is similar to ax = y
So, we can write it as [1/256] = (2√2) x
Or, (2√2) x = [1/28]
Or, [21 * 21/2]x = 1/28
Or, 23x/2 = 2-8
Therefore, 3x/2 = -8
Hence, x = (-8 * 2)/ 3 = -16/3
Larutan:
(6!)/(2!)(2!)=180
Larutan:
You can see number less than 3! are not divisible by 8 so it decide your output
(1!+2!+3!)=9
9%8=1
1 is the answer
A. 2400
B. 2000
C. 1904
D. 1906
E. Tidak satu pun dari ini
Larutan:
106 x 106 - 94 x 94 = (106)2 - (94)2
= (106 + 94)(106 - 94) [Ref: (a2 - b2) = (a + b)(a - b)]
= (200 x 12)
= 2400.
Perbedaan dua angka adalah 1365. Pada membagi angka yang lebih besar dengan yang lebih kecil, kita mendapatkan 6 sebagai hasil bagi dan 15 sebagai sisa. Berapa angka yang lebih kecil?
A. 240
B. 270
C. 295
D. 360
Jawaban: Opsi b
Penjelasan:
Let the smaller number be x. Then larger number = (x + 1365).
x + 1365 = 6x + 15
5x = 1350
x = 270
Smaller number = 270
A. 1035
B. 1280
C. 2070
D. 2140
Jawaban: Opsi a
Penjelasan:
Let Sn =(1 + 2 + 3 + ... + 45). This is an A.P. in which a =1, d =1, n = 45.
Sn = n [2a + (n - 1)d] = 45 x [2 x 1 + (45 - 1) x 1] = 45 x 46 = (45 x 23)
2 2 2
= 45 x (20 + 3)
= 45 x 20 + 45 x 3
= 900 + 135
= 1035.
Shorcut Method:
Sn = n(n + 1) = 45(45 + 1) = 1035.
2 2
[A] 7 jam 30 menit
[B] 8 jam
[C] 8 jam 15 menit
[D] 8 jam 25 menit
Larutan)
C)
Ronald 1 hr work = 32/6=16/3
Elan 1 hr work = 40/5=8
Show both work in an hr=8+16/3=40/3
Show for 110 pages it will take 110/(40/3) or (110 x 3)/40=33/4hr
Since: convert it into hr 4*8=32 1 left in 1 hr 60 min 60/4=15min
Show final answer is 8hr 15 mon
Larutan:
tautan ke solusi
4^x/4^x + 6^x/4^x = 9^x/4^x
Now,
1 + (3/2)^x=(3/2)^(2x)
Consider (3/2)^x=u
Then,
1 + u = u^2
Simplifying this
0 = u^2 -u -1
By solving we get
u = (1 + sqrt(5))/2
and this equal to
(1 + sqrt(5))/2 = (3/2)^x
Take log both side
and you get 1.187 approx value.
Solution:
circumference of an wheel=πd
=22/7×98
22×14
=308cm =1 revolution
distance covered
1540×100=154000
now,154000÷308
500 rotations
A. 2%
B. 2,02%
C. 4%
D. 4,04%
E. Tidak satu pun dari ini
Jawaban: Opsi d
100 cm is read as 102 cm.
∴ A1 = (100 x 100) cm2 and A2 (102 x 102) cm2
(A2 - A1) = [(102)2 - (100)2]
= (102 + 100) x (102 - 100)
= 404 cm2
∴ Percentage error
=(404100×100×100)%=4.04%
Area of the square=484cm
side-22cm
perimeter=22*4=88cm
circumfrence of cirlce is 2*pi*r=88
r=14cm
area=pi*r*r=616cm^2
Larutan
P(A)=1/9
P(B)=1/6
P(C)=26/36=13/18
Apply GP
you get 2/5 ans
