Question 1
Find the sum:
4 + 16 + 24 + 36 + 44 + 56
Pilihan Jawaban
A. 160 B. 170 C. 180 D. 190 E. 200
✅ Lihat Solusi
Step 1 → 4 + 56 = 60
Step 2 → 16 + 44 = 60
Step 3 → 24 + 36 = 60
Total = 60 × 3 = 180
Jawaban: C. 180
Question 2
The sum obtained after we add double of 25 to one-third of 21 is subtracted from 100. What is the final number in the end?
Pilihan Jawaban
A. 60 B. 29 C. 43 D. 57 E. 71
✅ Lihat Solusi
🔹Operasi Campuran
100 − (2 × 25 + 21 ÷ 3)
Step 1 → 2 × 25 = 50
Step 2 → 21 ÷ 3 = 7
Step 3 → 50 + 7 = 57
Step 4 → 100 − 57 = 43
Jawaban: C. 43
Question 3
Suzy has five kids. Amy is older than Beth but younger than Carla.
David is older than Erwin but younger than Beth.
Who is the eldest child?
Pilihan Jawaban
A. Amy B. Beth C. Carla D. David E. Erwin
✅ Lihat Solusi
🔹Relasi: Carla > Amy > Beth > David > Erwin
Step 1 → Amy lebih tua dari Beth, tapi lebih muda dari Carla => Carla > Amy > Beth
Step 2 → David lebih tua dari Erwin, tapi lebih muda dari Beth => Beth > David > Erwin
Step 3 → Gabungan urutan: Carla > Amy > Beth > David > Erwin
Jawaban: C. Carla
Question 4
Sarah’s birthday falls on 6th September each year.
Today is 27th July and is a Sunday.
What day will be Sarah’s birthday this year?
Pilihan Jawaban
A. Thursday B. Saturday C. Sunday D. Monday E. Friday
✅ Lihat Solusi
🔹Operasi Hitung Hari
27 July (Sunday) → 6 September ( ? )
🔹 Langkah Perhitungan
Step 1 → Sisa hari di Juli: 31 − 27 = 4 hari
→ 27 Jul → 31 Jul = 4 hari
Step 2 → Tambah hari di Agustus: 31 hari
Total sementara = 4 + 31 = 35 hari
Step 3 → Tambah 6 hari di September
Total keseluruhan = 35 + 6 = 41 hari
Step 4 → 41 ÷ 7 = 5 minggu + sisa 6 hari
→ Geser 6 hari dari Minggu
Minggu + 6 hari = Sabtu ==>
Jawaban: B. Saturday
Question 5
On the number table of 1–100, we finally reached 50 after we added 72 and subtracted 36 from the original number.
What is the original number?
Pilihan Jawaban
A. 86 B. 14 C. 48 D. 50 E. 84
✅ Lihat Solusi
🔹 Operasi Campuran
Original + 72 − 36 = 50
🔹 Langkah Perhitungan
Step 1 → Gabungkan operasi: +72 −36 = +36
Step 2 → Original + 36 = 50
Step 3 → Original = 50 − 36 = 14
Jawaban: B. 14
Question 6
How many natural numbers are there which are less than 300 and their digits add up to 5?
Pilihan Jawaban
A. 18 B. 10 C. 12 D. 15 E. 14
✅ Lihat Solusi
🔹 Analisis
Cari bilangan < 300 dengan jumlah digit = 5
🔹 Langkah Perhitungan
Step 1 → Bilangan 1 digit: hanya angka 5 → {5}
Step 2 → Bilangan 2 digit: digit pertama + digit kedua = 5
Contoh: 14, 23, 32, 41, 50 → total 5 bilangan
Step 3 → Bilangan 3 digit (<300): digit pertama bisa
1 atau
2.
Step 3a → Ratusan = 1 → puluhan + satuan = 4
Kombinasi: 104, 113, 122, 131, 140
Total = 5
Step 3b → Ratusan = 2 → puluhan + satuan = 3
Kombinasi: 203, 212, 221, 230
Total = 4
1 (satu digit) + 5 (dua digit) + (5 + 4) (tiga digit) = 15 ==>
Jawaban: D. 15
Question 7
Roy walks 10 m to his school and then continues the remaining three-fourths of the whole journey by bus.
How far is Roy’s school from his house?
Pilihan Jawaban
A. 134 m B. 100 m C. 30 m D. 40 m E. 400 m
✅ Lihat Solusi
🔹 Analisis
Roy berjalan 10 m = 1/4 perjalanan
Sisa 3/4 perjalanan ditempuh dengan bus
🔹 Langkah Perhitungan
Step 1 → 10 m = 1/4 perjalanan
Step 2 → Total perjalanan = 10 × 4 = 40 m
Step 3 → Jadi jarak rumah ke sekolah = 40 m
🔹 Kesimpulan
Jawaban: D. 40 m
Question 8
A square \(ABCD\) has an area of \(72 \,\text{m}^2\).
The square is divided into 9 smaller squares of equal area.
Find the total area of 5 of these smaller squares.
Pilihan Jawaban
A. -36 B. 40 C. 45 D. 50 E. 52
✅ Lihat Solusi
🔹 Analisis
Total luas persegi besar = 72 m²
Dibagi menjadi 9 persegi kecil dengan luas sama
🔹 Langkah Perhitungan
Step 1 → Luas tiap persegi kecil = 72 ÷ 9 = 8 m²
Step 2 → Luas 5 persegi kecil = 5 × 8 = 40 m²
🔹 Kesimpulan
Jawaban: B. 40
Question 9
A 1‑liter bottle was filled up using a 150‑ml and a 50‑ml jug.
Both jugs were poured thrice into the empty 1‑liter bottle.
How much more liquid can be poured into the bottle to fill it to the brim?
Pilihan Jawaban
A. 600 ml B. 650 ml C. 450 ml D. 400 ml E. 500 ml
✅ Lihat Solusi
🔹 Analisis
Total kapasitas botol = \(1000 \,\text{ml}\) |
Isi dari 150 ml jug + 50 ml jug | masing‑masing dituang 3 kali.
🔹 Langkah Perhitungan
Step 1 → Isi dari 150 ml jug = \(150 \times 3 = 450 \,\text{ml}\)
Step 2 → Isi dari 50 ml jug = \(50 \times 3 = 150 \,\text{ml}\)
Step 3 → Total isi = \(450 + 150 = 600 \,\text{ml}\)
Step 4 → Sisa kapasitas = \(1000 - 600 = 400 \,\text{ml}\)
🔹 Kesimpulan
Jawaban: D. 400 ml
Question 10
If all alphabets are represented by a number starting with
\(A = 1, B = 2, C = 3, \ldots, Z = 26\),
find the sum of the digits represented by the word SMART.
(Note: DOG is written as 4157)
Pilihan Jawaban
A. 20 B. 24 C. 26 D. 28 E. 30
✅ Lihat Solusi
🔹 Analisis
Konversi huruf ke angka sesuai urutan alfabet, lalu jumlahkan digitnya.
🔹 Langkah Perhitungan
S = 19 → \(1+9 = 10\)
M = 13 → \(1+3 = 4\)
A = 1 → \(1\)
R = 18 → \(1+8 = 9\)
T = 20 → \(2+0 = 2\)
🔹 Total
\[
10 + 4 + 1 + 9 + 2 = 26
\]
🔹 Kesimpulan
Jawaban: C. 26
Question 11
How many times do the hour and minute hand cross an even number on the clock from 1 a.m. to 1 p.m.?
Pilihan Jawaban
A. 70 B. 72 C. 78 D. 76 E. 80
✅ Lihat Solusi
🔹 Analisis
Jam analog memiliki angka 1–12.
Angka genap = {2, 4, 6, 8, 10, 12} → total 6 angka genap.
🔹 Langkah Perhitungan
Step 1 → Dalam 1 jam, jarum menit melewati semua 12 angka sekali penuh.
Step 2 → Jadi dalam 1 jam, jarum menit “cross” 6 angka genap.
Step 3 → Dari 1 a.m. sampai 1 p.m. = 12 jam.
Total crossing jarum menit = \(12 \times 6 = 72\).
Step 4 → Dari 1 a.m. sampai 1 p.m. = 12 jam.
Total crossing jarum jam = 6.
🔹 Kesimpulan
Total crossing = 72 + 6 = 78 => Jawaban: C. 78
Question 12
Count the number of blocks in the following picture.
Pilihan Jawaban
A. -15 B. 18 C. 20 D. 21 E. 24
✅ Lihat Solusi
🔹 Analisis
Struktur kubus terdiri dari beberapa lapisan bertingkat.
Kita hitung jumlah kubus per lapisan.
🔹 Langkah Perhitungan
Lapisan bawah = 9 kubus
Lapisan tengah = 7 kubus
Lapisan atas = 5 kubus
Total = \(9 + 7 + 5 = 21\)
🔹 Kesimpulan
Jawaban: D. 21
Question 13
A fan has three blades. How many blades will there be in 150 such fans?
Pilihan Jawaban
A. -450 B. 150 C. 500 D. 50 E. 45
✅ Lihat Solusi
🔹 Analisis
Setiap kipas memiliki 3 baling‑baling.
Soal menanyakan jumlah baling‑baling pada 150 kipas.
🔹 Langkah penyelesaian
Step 1: 1 kipas = 3 baling‑baling
Step 2: 150 kipas × 3 baling‑baling = 450
🔹 Kesimpulan
Jawaban: 450 → tidak ada di opsi, tapi seharusnya A. -450 adalah salah ketik dan maksudnya 450.
Question 14
Count the number of triangles in the picture.
Pilihan Jawaban
A. 5 B. 6 C. 4 D. 3 E. 9
✅ Lihat Solusi
🔹 Analisis
Gambar terdiri dari segitiga besar di kiri, segitiga di kanan, dan garis‑garis yang membentuk segitiga kecil tambahan.
Soal menanyakan total segitiga yang terbentuk dari kombinasi garis tersebut.
🔹 Langkah penyelesaian
Step 1: Identifikasi segitiga besar utama (2 buah).
Step 2: Hitung segitiga kecil yang terbentuk dari garis dalam (beberapa segitiga tambahan).
Step 3: Gabungkan segitiga kecil menjadi segitiga lebih besar (komposit).
Total: Setelah dihitung semua kombinasi, jumlah segitiga = 5.
🔹 Kesimpulan
Jawaban: A. 5
Question 15
What number has 2 hundreds, 5 more tens than 20 and 1 less than 7?
Pilihan Jawaban
A. 251 B. 272 C. 276 D. 271 E. 254
✅ Lihat Solusi
🔹 Analisis
Soal menanyakan bilangan dengan komposisi:
2 hundreds | 5 tens lebih banyak dari 20 | 1 lebih kecil dari 7
🔹 Langkah
Step 1: 2 hundreds = 200
Step 2: 5 tens lebih banyak dari 20 → 50 + 20 = 70
Step 3: 7 − 1 = 6
Total: 200 + 70 + 6 = 276
🔹 Kesimpulan
Jawaban: C. 276
Question 16
Harry scored 3 more marks than Alwin.
Sophia scored the same as Harry.
The sum of their scores was 51.
How many marks did Alwin score?
Pilihan Jawaban
(Open – Ended Questions)
✅ Lihat Solusi
🔹 Analisis
Diketahui:
Harry = Alwin + 3 |
Sophia = Harry |
Total (Alwin + Harry + Sophia) = 51
*Ditanya: nilai Alwin.*
🔹 Langkah
Step 1: Misalkan Alwin = A
Step 2: Harry = A + 3
Step 3: Sophia = Harry = A + 3
Step 4: Total = A + (A + 3) + (A + 3) = 3A + 6
Step 5: 3A + 6 = 51 → 3A = 45 → A = 15
🔹 Kesimpulan
Alwin scored 15 marks.
Question 17
Alice picked two cards from deck such that the sum of the numbers on the two cards was 15.
(A = 1, J = 11, Q = 12, K = 13, lainnya 2–10)
Pilihan Jawaban
(Open – Ended Questions)
✅ Lihat Solusi
🔹 Analisis
Soal menanyakan berapa banyak kombinasi 2 kartu dari 1 deck standar (tanpa suit dibedakan) yang jumlah nilainya = 15.
🔹 Langkah
Step 1: Cari pasangan nilai yang jumlahnya 15:
(2, 13)
- (3, 12)
- (4, 11)
- (5, 10)
- (6, 9)
- (7, 8)
Step 2: Total ada 6 pasangan nilai berbeda.
Step 3: Karena setiap kartu punya 4 suit, maka setiap pasangan nilai menghasilkan 4 × 4 = 16 kombinasi kartu.
Step 4: Total kombinasi = 6 × 16 = 96
🔹 Kesimpulan
Alice dapat memilih 96 kombinasi kartu yang jumlahnya 15.
Question 18
Five students entered a contest where they had to guess the number of candies in a box. Alan
guessed 25 candies while Beth guessed 30. Charles guessed 27 and David guessed 35. Roma
was declared as the winner. One of the students guessed 4 more than Roma’s number while
another student guessed 4 less than hers. What was the number guessed by Roma?
Pilihan Jawaban
(Open – Ended Questions)
✅ Lihat Solusi
🔹 Analisis
Roma harus punya nilai X sehingga ada satu tebakan = X + 4 dan satu tebakan = X − 4.
Kita cek tebakan siswa lain: 25, 30, 27, 35.
🔹 Langkah
Roma = 31 → Charles = 27 (−4), David = 35 (+4)
🔹 Kesimpulan
Jawaban: 31
Question 19
Ayer and Kumar tossed a coin. Ayer wins if it is a head while Kumar wins if it is a tail.
The winner gets 3 marbles from the loser.
Both had 20 marbles each in the beginning.
They played 10 rounds and Ayer won 4 rounds.
What is the total number of marbles Kumar had in the end?
✅ Solusi
🔹 Analisis
Awalnya Ayer = 20, Kumar = 20.
Setiap ronde: pemenang +3, kalah −3.
Total 10 ronde → Ayer menang 4, Kumar menang 6.
Kita hitung perubahan marbles untuk Kumar.
🔹 Langkah
Ayer menang 4 × 3 = 12 → Kumar −12
Kumar menang 6 × 3 = 18 → Kumar +18
Total perubahan Kumar = −12 + 18 = +6
Awal 20 → Akhir = 26
🔹 Kesimpulan
Kumar memiliki 26 marbles di akhir.
Question 20
The place cards shown are folded along the dotted line so that only a number or letter is visible.
Chrissy enters the room and sees all five place cards, with at least 2 number cards being shown.
The sum of numbers that she sees is less than 8.
How many different sets of numbers are there?
✅ Solusi
🔹 Analisis
Ada 5 kartu: angka 1–5.
Chrissy melihat minimal 2 angka.
Jumlah angka yang terlihat < 8.
Kita cari kombinasi angka yang memenuhi syarat.
🔹 Langkah
2 angka:
- {1,2}=3
- {1,3}=4
- {1,4}=5
- {1,5}=6
- {2,3}=5
- {2,4}=6
- {2,5}=7
- {3,4}=7
→ total 8 set
3 angka:
- {1,2,3}=6
- {1,2,4}=7
→ total 2 set
4 angka:
- {1,2,3,4}=10 (>8) → tidak masuk
Total kombinasi valid = 8 + 2 = 10
🔹 Kesimpulan
Ada 10 set angka yang mungkin.
Question 21
If the digits of a two-digit number are reversed to form a new number,
the difference between the two numbers is 45.
How many different such pairs are there?
✅ Solusi
🔹 Analisis
Misalkan angka = 10x + y (x = puluhan, y = satuan).
Angka terbalik = 10y + x.
Selisih = |(10x + y) − (10y + x)| = 9(x − y).
Diketahui selisih = 45 → 9(x − y) = 45 → x − y = 5.
🔹 Langkah
x − y = 5
Kemungkinan pasangan digit:
(x,y) = (6,1), (7,2), (8,3), (9,4)
Total ada 4 pasang angka dua digit yang memenuhi.
🔹 Kesimpulan
Jawaban: 4 pairs
Question 22
Harry wrote 1, 2, 3, 11, 22, 33, 111, 222, 333, … until he got 17 numbers.
What is the sum of the digits of the last number he wrote?
✅ Solusi
🔹 Analisis
Urutan angka:
- 1 digit: 1,2,3 (3 angka)
- 2 digit: 11,22,33 (3 angka)
- 3 digit: 111,222,333 (3 angka)
- 4 digit: 1111,2222,3333 (3 angka)
- 5 digit: 11111,22222,33333 (3 angka)
→ total 15 angka.
Lanjut ke 6 digit: 111111, 222222, 333333 → jadi 18 angka.
Maka angka ke‑17 = 222222.
🔹 Langkah
Angka ke‑17 = 222222
Jumlah digit = 2+2+2+2+2+2 = 12
🔹 Kesimpulan
Jawaban: 12
Question 23
A jar of 30 chocolates is shared among Ann, Alice, and Alisha.
Alice takes more than anyone else.
What is the least number of chocolates she could have taken?
✅ Solusi
🔹 Analisis
Total = 30.
Alice harus lebih banyak dari Ann dan Alisha.
Untuk minimum, bagiannya harus sedikit lebih besar dari yang lain.
Sebenarnya cari langsung rata-ratanya (dimana mereka punya jumlah cokelat yang sama)= 30 : 3 = 10 => agar syarat terpenuhi ditambah 1 saja => 10 + 1 = 11
🔹 Langkah
Jika Alice = 11 → sisa 19.
Bisa dibagi Ann = 9, Alisha = 10.
Alice (11) > 9 dan 10 → valid.
Jika Alice < 11 → tidak mungkin, karena salah satu orang lain akan sama atau lebih besar.
🔹 Kesimpulan
Jawaban: Alice paling sedikit bisa ambil 11 chocolates.
Question 24
Few kids were playing cricket.
Each of them carried 4 balls.
While playing, 17 balls were hit outside the ground and lost.
In the end, they were left with a total of 55 balls.
How many kids were playing together?
✅ Solusi
🔹 Analisis
Awalnya setiap anak bawa 4 bola.
Jika jumlah anak = n → total awal = 4n.
Setelah hilang 17 bola → sisa = 4n − 17.
Diketahui sisa = 55.
🔹 Langkah
Persamaan: 4n − 17 = 55
→ 4n = 72
→ n = 18
🔹 Kesimpulan
Ada 18 kids yang bermain bersama.
Question 25
Patrick started jumping every day. He jumped 10 times on Day 1.
He jumped 8 more times than Day 1 on Day 2.
He jumped 8 more times than Day 2 on Day 3.
He continued to jump 8 more times than the previous day every day.
Which day will be the first day when he would have jumped at least 100 times in one day?
✅ Solusi
🔹 Analisis
Setiap hari bertambah 8 dari hari sebelumnya.
Ini barisan aritmetika:
a₁ = 10, d = 8.
Rumus hari ke‑n: aₙ = 10 + (n−1)×8.
Cari n saat aₙ ≥ 100.
🔹 Langkah
Rumus: aₙ = 10 + (n−1)×8
Syarat: 10 + (n−1)×8 ≥ 100
(n−1)×8 ≥ 90
n−1 ≥ 11.25 → n ≥ 12.25
Maka n = 13
🔹 Kesimpulan
Patrick pertama kali ≥100 lompatan pada Day 13.
