Section A
Sec A MCQs — 15 Questions
(+4)
1
arithmetic
easy
What is the sum of the digits of 5654+6545+5465+4556?
A
9
B
8
C
16
D
18
E
10
Show answer
Answer: B
5654+6545=12199; 12199+5465=17664; 17664+4556=22220. The sum of the digits of 22220 is 2+2+2+2+0=8.
2
time
medium
Garry left his house for the park which is 10 minutes away on foot. He then played football for 45 minutes and was back after chatting with his friends for 15 minutes. If he got back home by 7.00 pm, then what time did he leave his house?
A
6.00 pm
B
5.50 pm
C
5.40 pm
D
8.10 pm
E
8.20 pm
Show answer
Answer: C
Garry's total time away from home is: 10 minutes (walk to park) + 45 minutes (football) + 15 minutes (chatting) + 10 minutes (walk back home) = 80 minutes = 1 hour 20 minutes. Since he got back at 7.00 pm, he must have left at 7.00 pm − 1 hour 20 minutes = 5.40 pm.
3
number-theory
hard
How many 3-digit numbers have the sum of the digits less than 5?
A
4
B
8
C
12
D
15
E
19
Show answer
Answer: E
3-digit numbers (a,b,c digits, a≥1) with digit sum s<5 must have s∈{1,2,3,4} (s=0 is impossible since a≥1). Counting by bucket: s=1 gives only 100 (1 number). s=2 gives 101, 110, 200 (3 numbers). s=3 gives 102, 111, 120, 201, 210, 300 (6 numbers). s=4 gives 103, 112, 121, 130, 202, 211, 220, 301, 310, 400 (10 numbers). Total = 1+3+6+10 = 20. (See this question's 'notes' field: the source's own worked solution undercounts the s=4 bucket by omitting 400, getting 19 instead of the correct 20; since 20 is not among the printed options, the source's own answer of 19 is retained here.)
4
calendar
medium
If the last day of May is a Friday, what day is the 1st of April?
A
Monday
B
Sunday
C
Saturday
D
Wednesday
E
Tuesday
Show answer
Answer: A
April has 30 days, so the number of days from April 1st to May 31st (inclusive of the gap) is 30 (rest of April after the 1st) + 31 (all of May, but we count up to the 31st) — concretely, 60 days separate April 1st from May 31st. 60÷7 = 8 remainder 4, so May 31st falls 4 days (in the weekly cycle) after April 1st. Since May 31st is a Friday, going back 4 days from Friday gives: Thursday, Wednesday, Tuesday, Monday. So April 1st was a Monday.
5
ratio-proportion
medium
If 5 postmen deliver 200 letters in 2 days. How many postmen will be needed to deliver 600 letters in 3 days?
A
5
B
10
C
15
D
20
E
8
Show answer
Answer: B
5 postmen deliver 200 letters in 2 days, so 1 postman delivers 200/(5×2)=20 letters per day. To deliver 600 letters in 3 days requires 600/3=200 letters per day, which needs 200/20=10 postmen.
6
arithmetic
medium
Fill in the operations to make the following equation true. (Note that we follow the order of operations from left to right and not BODMAS)
45 _ 54 _ 11 _ 2 = 18
A
÷×+
B
÷×-
C
+÷×
D
+×÷
E
+÷+
Show answer
Answer: C
Working strictly left to right: 45+54=99; 99÷11=9; 9×2=18. So the operations, in order, are +, ÷, ×.
7
number-theory
easy
What is the greatest 3-digit number divisible by 5 and 11?
A
899
B
990
C
992
D
995
E
999
Show answer
Answer: B
A number divisible by both 5 and 11 must be divisible by 5×11=55 (since 5 and 11 are distinct primes). The greatest 3-digit multiple of 55 is found from 999÷55=18 remainder 9, so the greatest multiple is 18×55=990.
8
arithmetic
easy
What should be added to the number formed by 5 hundred 3 tens and 8 ones so that the new number has the same digits but in the reversed order?
(E.g., if the number is 123, reversed order number is 321)
A
358
B
853
C
300
D
297
E
279
Show answer
Answer: D
5 hundred 3 tens 8 ones = 538. Reversing its digits gives 835. The amount to add is 835−538=297.
9
logic
medium
A, B, C, D and E were standing in a line. A was standing between B and D. E was not standing in the last place. B was standing first in line. Who was standing in the last place?
A
A
B
B
C
C
D
D
E
E
Show answer
Answer: C
Since B stands first and A stands between B and D, the order starting from the front must be B, A, D (positions 1, 2, 3), so that A is sandwiched directly between B and D. That leaves positions 4 and 5 for C and E. Since E is given not to stand last (position 5), E must be in position 4, leaving C in the last position (position 5).
10
combinatorics
hard
If Freddy has a few $1, $2, $5 and $10 coins, and he has to make a payment of $10, how many different ways are there that he can make the payment using one or more coins of any given value?
A
8
B
12
C
9
D
10
E
11
Show answer
Answer: E
Count the ways to make $10 using coins of $1, $2, $5, $10 (any number of each, order doesn't matter). Using one $10 coin: 1 way. Using zero $10 coins, make $10 from $1/$2/$5 coins: with two $5 coins, the rest is 0, giving 1 way; with one $5 coin, the remaining $5 must be made from $1/$2 coins, which can be done as 5=5(1)+0(2), 3(1)+1(2), 1(1)+2(2) — 3 ways; with zero $5 coins, $10 must be made from $1/$2 coins, using k many $2 coins for k=0..5 (each with a unique number of $1 coins) — 6 ways. That sub-total is 1+3+6=10 ways without any $10 coin. Adding the 1 way using a $10 coin gives 10+1=11 total ways.
11
arithmetic
easy
A total of 400 points were scored by 5 players. The top two players scored 100 each. The other two players scored 88 and 76. How much did the 5th player score?
A
100
B
46
C
146
D
36
E
64
Show answer
Answer: D
The first four players' total is 100+100+88+76=364. So the 5th player scored 400−364=36 points.
12
sequences
easy
What is the difference between the 15th and 18th terms of the following sequence?
44, 59, 74, 89, …
A
15
B
30
C
45
D
40
E
35
Show answer
Answer: C
The sequence has a common difference of 15 (59−44=15, 74−59=15, 89−74=15). Since the 18th term is 3 steps ahead of the 15th term, the difference between them is 15×3=45.
13
time
easy
What will be the time after 5400 seconds if the time now is 5 pm?
A
6.00pm
B
6.15 pm
C
6.30 pm
D
4.30 pm
E
3.30 pm
Show answer
Answer: C
1 hour = 3600 seconds. 5400−3600=1800 seconds = 30 minutes. So 5400 seconds = 1 hour 30 minutes. Adding this to 5 pm gives 6.30 pm.
14
number-theory
hard
In the following addition, different letters represent different digits. What is the sum of A, B and D?
7 A
3 B 5
1 C 7 0
+ D 2 0 9
8 0 2 6
A
15
B
12
C
16
D
18
E
11
Show answer
Answer: A
Writing the sum by place value: 7A=70+A, 3B5=305+10B, 1C70=1070+100C, D209=1000D+209. Their sum is 1654+(A+10B+100C+1000D)=8026, so A+10B+100C+1000D=6372. Since D is a single digit and 1000D≤6372, try D=6 (1000×6=6000), leaving A+10B+100C=372; with C=3 (300), leaving A+10B=72, giving B=7, A=2. Checking: A=2, B=7, C=3, D=6, all distinct digits, and indeed 72+375+1370+6209=8026. So A+B+D=2+7+6=15.
15
fractions
easy
Sheila has to write 48 pages. She finished writing only one-fourth of the pages. How many more pages are yet to be written?
A
12
B
24
C
36
D
32
E
28
Show answer
Answer: C
One-fourth of 48 is 48÷4=12 pages written. The remaining pages to be written are 48−12=36.
Section B
Sec B Integers — 10 Questions
(+5)
16
age-problems
medium
Five years ago, the sum of a couple's ages was 76. Fifteen years later, the man will be 4 years older than the woman. What will their total ages be 30 years later?
Show answer
Answer: 0146
The age difference between two people never changes over time, so the '4 years older, 15 years later' fact does not affect the sum of their ages — it is extra information not needed for this calculation. Five years ago their combined age was 76, so right now (5 years later) their combined age is 76+2×5=86 (each of the two people has aged 5 years). Thirty years from now, their combined age will increase by 2×30=60 more, giving 86+60=146.
17
algebra
easy
The sum of the three consecutive numbers is 81. Find the smallest number.
Show answer
Answer: 0026
Let the three consecutive numbers be x, x+1, x+2. Their sum is 3x+3=81, so x=26. The smallest number is 26.
18
measurement
easy
11 lamps were installed from one end to another along the road of a farm which was 110 metres long. If they are placed at equal distance, what is the distance between any two lamps, in metres?
Show answer
Answer: 0011
With 11 lamps placed from one end of the road to the other, there are 11−1=10 equal gaps between consecutive lamps spanning the full 110 metres. So each gap is 110÷10=11 metres.
19
geometry
easy
If two identical squares with sides of 16 cm each are joined to make a rectangle, what will be the perimeter of the rectangle formed?
Show answer
Answer: 0096
Joining two 16 cm squares edge-to-edge forms a rectangle with length 16+16=32 cm and breadth 16 cm. Its perimeter is 2×(32+16)=2×48=96 cm.
20
logic
medium
A basket contains 30 red balls. Karla takes 5 red balls and replaces them with 6 green balls. She continues to exchange the balls in the same way until all the red balls are replaced by green balls. How many balls are there in the basket now?
Show answer
Answer: 0036
30 red balls are removed in groups of 5, requiring 30÷5=6 exchanges. Each exchange removes 5 balls and adds 6, a net gain of 1 ball per exchange. Starting from 30 balls, after 6 exchanges the basket has 30+6×1=36 balls.
21
arithmetic
medium
If a fruit vendor bought 4 dozen apples at $60 a dozen and sold each apple for $6. How much was his gain, in $?
Show answer
Answer: 0048
The cost per apple is $60÷12=$5 (since a dozen is 12 apples). Each apple is sold for $6, giving a gain of $6−$5=$1 per apple. With 4 dozen = 48 apples sold, the total gain is 48×$1=$48.
22
number-theory
medium
Mahesh wrote numbers from 31 to 80 in his notebook. How many times did he write the digit 4 at any place value?
Show answer
Answer: 0015
Count occurrences of the digit 4 in the tens place: the numbers 40 through 49 contribute 10 occurrences. Count occurrences in the ones place within 31-80: 34, 44, 54, 64, 74 contribute 5 occurrences. Total occurrences of the digit 4: 10+5=15.
23
combinatorics
medium
Two players were playing a dice game. Each player rolled the dice 3 times and noted down the numbers they got. If the first player got a sum of 13 in such a way that the product of numbers is the highest, what is the sum of the possible numbers he got in his turn?
Show answer
Answer: 0013
Each die shows a value from 1 to 6. Listing the ways three such values can sum to 13 and computing each product: (6,6,1)→36, (6,5,2)→60, (6,4,3)→72, (5,5,3)→75, (5,4,4)→80. The highest product, 80, occurs for the triple (5,4,4). The question asks for the sum of the numbers he got, which is stated to be 13 in the problem itself (and indeed 5+4+4=13), so the answer is 13.
24
algebra
medium
Luna had 56 bows. Her friend said: "If you give me 18 bows, I will have double the number of bows you will have". How many bows did her friend have?
Show answer
Answer: 0058
After Luna gives away 18 bows, she has 56−18=38 bows left. Her friend will then have double that: 38×2=76 bows (after receiving the 18). So before receiving them, her friend had 76−18=58 bows.
25
ratio-proportion
medium
If 9 lemons weigh the same as 3 peaches and 6 peaches weigh the same as 2 mangoes. How many lemons will weigh the same as 5 mangoes?
Show answer
Answer: 0045
From 9 lemons = 3 peaches, dividing by 3 gives 3 lemons = 1 peach. From 6 peaches = 2 mangoes, dividing by 2 gives 3 peaches = 1 mango. Substituting, 1 mango = 3 peaches = 3×(3 lemons) = 9 lemons. So 5 mangoes = 5×9=45 lemons.