M4maths Previous Puzzles

03 Jan 2019 Number of Cans

Chintu works at a soft drink company and was arranging some particular number of cylindrical aluminium soft drink cans in a square box but the box became full and there were 8 cans remaining to be put in. Then, he started arranging all Cans in a rectangular box where he could arrange 6 cans more along the length than the breadth after putting all the cans in the rectangular box. He found out that there was still space left for another 11 cans. What is the number of cans Chintu had if it is known that the number of cans were more than 15 and no can was put on top of another can?

OPtion

1) 25
2) 33
3) 34
4) 36
5) 55
6) 44
7) 45
8) 54
9) 56
10) None of these

Solution

In a square box the number of cans arranged along the length and breadth would be same.
In case of rectangular box, lets assume that number of cans along breadth is x then number of boxes along the length would be x+6
Total number of cans the rectangular box can take = x*(x+6)
Total number of cans Chintu was able to put in the rectangular box (as some empty space was remaining ) = x*(x+6) - 6 ---- (i)


Given that the number of cans are more than 15, hence 2 & 3 cans along the length are ruled out

Lets try with 4 cans along length of square box, total cans in box would be 16
Total number of cans which Chintu has = 16+8 = 24 ---- (ii)
(i = (ii)
x*(x+6) - 11 = 24
=> x^2 + 6x = 35
Trying to solve we realize that there is no such whole number which x can take

Lets try with 5 cans along length of square box, total cans in box would be 25
Total number of cans which Chintu has = 25+8 = 33 ---- (ii)
(i = (ii)
x*(x+6) - 11 = 33
=> x^2 + 6x = 44
Trying to solve we realize that there is no such whole number which x can take


Lets try with 6 cans along length of square box, total cans in box would be 36
Total number of cans which Chintu has = 36+8 = 44 ---- (ii)
(i = (ii)
x*(x+6) - 11 = 44
=> x^2 + 6x = 55
Solving we get x = 5

So total number of cans which Chintu had = x*(x+6) - 11 = 5(5+6) - 11 = 5*11 -11 = 44

Correct Option: 6 Best Solution (1)

G.S.Kantharao   3 Months AGO

Let the side of the Square=x
x^2>15=> x^2=16/25/36/...
Length and breadths are l and b where l=b+6
l*b=b(b+6)=16+8+11/25+8+11/36+8+11/...
b^2+6b=35/44/55/...
b^2+6b-55=0(55 gives real solution)
(b-5)(b+11)=0
Breadth =5
Length=11
Area =55
Chintu had at first =55-11=44
My option is 6👈

Like? Yes (7) | No (2)  

02 Jan 2019 Profit and Loss

A vendor bought 1300 toffees for 5850 rupees. How many toffees in 1053 rupees must he sell to gain 56%?

OPtion

1) 133.33
2) 134
3) 140
4) 150
5) 135
6) 155
7) 160
8) 250
9) 210
10)None of these

Solution

CP of 1 toffees = 5850/1300 = Rs. 4.5
SP of 1 toffees for 56% profit = 4.5*1.56 = Rs. 7.02
So in 1053 Rupees number of toffees to be sold = 1053/7.02 = 150

Option 4)

Correct Option: 4 Best Solution (10)

G.S.Kantharao   3 Months AGO

C.P.of 1300=₹5850
C.P.of each=₹4.5
%Profit =56.
S.P of each =4.5*(156/100)=₹7.02
No of Toffees=1053/7.02=150
Hence the right option is 4👈

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Devendra Marghade   3 Months AGO

CP of 1 toffee=5850/1300=Rs. 9/2
SP of 1 toffee with 56% gain=(9/2)*(156/100)=Rs. 351/50
So in Rs. 1053, quantities to be sold=1053/(351/50)= 150
Option 4)

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Kariamal   3 Months AGO

C.P of 1 toffee=5850/1300=4.5
Selling a toffee gain of 56%=4.5*1.56=7.02
So, number of toffees sold for rs.1053=1053/7.02=150

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RAMADURAI NATESAN   3 Months AGO

P.P.of 1 toffee =5850/1300 = Rs 4.5.
At 56% profit,S.P.of 1 toffee =1.56*4.5 = 7.02.
Hence in Rs 1053, he mast sell 1053/7.02 = 150 toffees

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Sowmya Reddy   3 Months AGO

cost price of 1 toffee is 4.5
sp=(100+56)%4.5
sp=7.02
sp of 1 toffee to obtain 56% gain is 7.02
for 1053 rs we can sell 150 toffees

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lakshman   3 Months AGO

CP=5850/1300
SP=1053/N

Therefore,
156/100*5850/1300=1053/N
N=150

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Alivia Pramanik   3 Months AGO

CP of 1 toffee = (5850/1300)
SP of 1 toffee = (5850/1300)*(156/100)
SP of 1053 toffee = 1053*(1300/5850)*(100/150)

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ushap   3 Months AGO

for 1 toff sp is 1.56×4.5=7,so for 1053 rupees 150 toff can be sold to get 56% profit

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Ramesh Muthukrishnan   3 Months AGO

1 tofee= 4.5Rs
S.P = 156/100 × 4.5
S.P of 1 toffee =Rs. 351/50
N toffees = 1053/351 ×50
N=150

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minnu   3 Months AGO

56% of 5850/1300=2.52
5850/1300=4.5
1053/(4.5+2.52)=150

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01 Jan 2019 Decode

If VERY= 7369, HAPPY= 54889 and NEW = 231 then code for YEAR would be

OPtion

1) 9345
2) 9846
3) 9316
4) 9376
5) 9046
6) 9348
7) 9346
8) 9340
9) 9356
10) None of these

Solution

AS VERY is coded as "7369", HAPPY is coded as "54889" and NEW as 231
So Y is coded as 9, E as 3, A as 4, R as 6

So code for YEAR would be = 9346

Correct Option: 7 Best Solution (12)

G.S.Kantharao   3 Months AGO

VERY=7369
HAPPY=54889
NEW=231
YEAR=9346
Hence the right option is 7👈

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Ritul rana   3 Months AGO

select the digit for every letter from the given words..

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Ayush Garg   3 Months AGO

E represents 3, Y represents 9, A represents 4, R represents 6

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Aritra Banerjee   3 Months AGO

9346

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Siddesh K C   3 Months AGO

Letters given in problem statement coded as
V E R Y H A P N W
7 3 6 9 5 4 8 2 1
Then YEAR can be coded as 9346

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Hema   3 Months AGO

y=9
E=3
A=4
R=6

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Sowmya Reddy   3 Months AGO

y is coded as 9 in happy and very where as E is coded as 3 in very and new

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Priyojeet Sarkar   3 Months AGO

In the word HAPPY and VERY Y=9 , in VERY and NEW E=3 , in HAPPY A=4 , in VERY R=6

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Abhijit   3 Months AGO

y=9
e=3
a=4
r=6
so year=9346

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Kariamal   3 Months AGO

YEAR=9346

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RAMADURAI NATESAN   3 Months AGO

From the code of VERY which is 7369 , we get
Y=9 , E=3 , and R = 6. From the code of HAPPY, which is
54889, we get A =4. Hence code for YEAR is 9346

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Kavya   3 Months AGO

By arranging the alphabets in order

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31 Dec 2018 Series

Find the next term of the series:
6, 11, 16, 21, 26, ?

OPtion

1) 27
2) 28
3) 29
4) 34
5) 35
6) 36
7) 31
8) 32
9) 30
10) None of these

Solution

Next term of series is derived
4*(n-1) + n+1 where n=1

So 4*1 + 2 = 6
4*2 + 3 = 11
4*3 + 4 = 16
4*4 + 5 = 21
4*5 + 6 = 26
4*6 + 7 = 31

Correct Option: 7 Best Solution (10)

G.S.Kantharao   3 Months AGO

6,11,16,21,26,... Common difference is 5
So Next term is 31.
My option is 7👈

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Varun   3 Months AGO

5 difference makes it 26+5 =31
= Option (7)

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vignesh   3 Months AGO

The difference between each term is 5 so next term will be 31

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sruthi pujari   3 Months AGO

next number in the series is calculated by adding +5 to previous number. so, 26+ 5 = 31
Hence, option>> 7

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Kariamal   3 Months AGO

6+5=11
11+5=16
16+5=21
21+5=26
26+5=31

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ARINJOY GHOSH   3 Months AGO

It's an Arithmetic Progression with first term and common difference 5. Next Term is =26+5=31
Ans. option(7)

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Ramana   3 Months AGO

6+5=11
11+5=16
16+5=21
21+5=26
26+5=31

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Sowmya Reddy   3 Months AGO

difference between 2 terms is 5

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Ritul rana   3 Months AGO

for each next term you have to add 5 to the present one i.e6+5=11,then 11+5=16 just like that 26+5=31.

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SAYAN CHATTERJEE   3 Months AGO

The swreis in AP with common difference 5,
So, next term is 26+5= 31.
Option 7.

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28 Dec 2018 ODD Number

How many factors of 3622184564625 are odd numbers?

OPtion

1) 600
2) 188
3) 144
4) 180
5) 288
6) 576
7) 128
8) 256
9) 96
10) None of these

Solution

3622184564625 = 37^2 * 19^1 * 11^3 * 5^3 * 3^2 * 31^1

Any factor of this number should be of the form 37^a * 19^b * 11^c * 5^d * 3^e * 31^f

All the factors of this number would be an odd number as there is no even factor

a can take values 0, 1, 2
b can take values 0, 1,
c can take values 0, 1, 2, 3
d can take values 0, 1, 2, 3
e can take values 0, 1, 2, 3
and f can take value 0, 1

Total number of odd factors = 3*2*4*4*4*2= 768

Option 10)

Correct Option: 10 Best Solution (0)

27 Dec 2018 Mixture

A mixture contains sugar solution and colored water in the ratio 3 : 5. If 16 liters of colored water is added to the mixture, the ratio becomes 4 : 7. Find the initial quantity of colored water in the given mixture (in liters).

OPtion

1) 192
2) 120
3) 240
4) 160
5) 180
6) 320
7) 360
8) 150
9) 200
10) None of these

Solution

The initial ratio is 3 : 5
Let ‘k’ be the common ratio.
=> Initial quantity of sugar solution = 3 k
=> Initial quantity of colored water = 5 k
=> Final quantity of sugar solution = 3 k
=> Final quantity of colored water = 5 k + 16

Final ratio = 3 k : 5 k + 16 = 4 : 7
=> 21 k = 20k + 64
=> k = 64
Therefore, initial quantity of colored water in the given mixture (in liters) = 5 * 64 = 320 liter

Correct Option: 6 Best Solution (7)

G.S.Kantharao   3 Months AGO

Let Sugar solution : Coloured water =x : y
X:Y=3:5=>5x=3y=> x=3y/5
X:(Y+16)=4:7=>7x=4y+64
7(3y/5)=4y+64
0.2y=64=>y=320
My option is 6👈

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Siddesh K C   3 Months AGO

Initial Ratio --Sugar Solution : Color Water =3:5
Ratio after adding 16 litres of colored water,
Final Ratio --Sugar solution : Color Water =4:7
This can be written as
3x : 5x+16=4x:7x
21x^2=20x^2+64x
x^2=64x
x=64
Initial quantity of Colored water in mixture is 5x=5*64 =320 litres.

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Kariamal   3 Months AGO

3x/5x+16=4/7
x=64
Initial quantity 5x=320

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SAYAN CHATTERJEE   3 Months AGO

3x/(5x+16)= 4/7
So, x= 64
So, colored water is 64*5= 320.
Option 6.

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RAMADURAI NATESAN   3 Months AGO

Let initial mixture be 3x sugar and 5x col.water.After adding 16 l col.water,the mixture will be 3x sugar and 5x+16 col water.3x : (5x+16) :: 4:7 . 21x =4(5x+16) . x= 64.
Initial quantity of col water = 5x = 5*64 =320

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Soniya Chhatri   3 Months AGO

3x/5x+16=4x/7x
X=64
5x=320

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Veluprabhakaran.K   3 Months AGO

x/y=3/5,x/(y+16)=4/7
5x-3y=0
7x-4y=64
x=192
y=320

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26 Dec 2018 Race

In a race of 500 m, A can beat B by 30 m and C by 20 m. In a race of 1200 m, C will beat B by:

OPtion

1) 75
2) 25
3) 100
4) 10
5) 15
6) 350
7) 150
8) 50
9) 175
10) None of these

Solution

When A covers 500m B Covers 470m
When A Covers 500m C covers 480

A : B = 500 : 470
A : C = 500 : 480

C/B = (C/A*A/B) = (480/470) = 480:470
When C covers 480 m, B covers 470 m.
When C covers 1200 m, B covers (470/480)*1200)m = 1175m
Therefore, C beats B by (1200 - 1175) m = 25 m.

Option 2)

Correct Option: 2 Best Solution (0)

25 Dec 2018 Solve

Given (73-a)(73-b)(73-c)(73-d)(73-e) = 910.

Find a+b+c+d+e= ?

OPtion

1) 210
2) 220
3) 410
4) 110
5) 840
6) 820
7) 910
8) 920
9) 420
10) None of these

Solution

Note: There is some issue with the points allotted to participant, we would be correcting it soon.

Lets try to simplify:
910 = 13*7*5*2*1 = (73-60)*(73-66)*(73-68)*(73-71)*(73-72)

So a+b+c+d+e = 60 + 66 + 68 + 71 + 72 = 337

Correct Option: 10 Best Solution (6)

G.S.Kantharao   4 Months AGO

910=1*2*5*7*13
=(73-72)(73-71)(73-68)(73-66)(73-60)
A+b+c+d+e=72+71+68+66+60=337
Right option is 10👈

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RAMADURAI NATESAN   4 Months AGO

Factorising 910 =13*10*7=13*5*2*1*7=13*7*5*2*1
=(73-60)(73-66)(73-68)(73-71)(73-72)
60+66+68+71+72=337

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Karan   4 Months AGO

a,b,c,d, and e are 72,71,68,66,60 respectively.
72+71+68+66+60= 337

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Kariamal   4 Months AGO

Factors of 910=2*5*7*13*1
a=71, b=68, c=66, d=60, e=72
a+b+c+d+e=337

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Kethineni Vinodh   4 Months AGO

910 = 1*2*5*7*13
i.e., (73-72)(73-71)(73-68)(73-66)(73-60) sum = 337

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SAYAN CHATTERJEE   4 Months AGO

910= 13*5*2*7*1
a+b+c+d+e= (60+68+71+67+72)= 338.
Option 10.

Like? Yes | No (1)  

24 Dec 2018 Probability

A letter is randomly picked from the following longest word in the dictionary: pneumonoultramicroscopicsilicovolcanoconiosis
what is the probability that it is a vowel?

OPtion

1) 1/3
2) 16/45
3) 17/45
4) 2/5
5) 19/45
6) 4/9
7) 7/15
8) 22/45
9) 1/4
10) None of these

Solution

Total number of letters in pneumonoultramicroscopicsilicovolcanoconiosis = 45
Number of vowels a, e, i, o, and u = 20

Probability = 20/45 = 4/9

Option 6)

Correct Option: 6 Best Solution (4)

Kariamal   4 Months AGO

Total number of words =45
Total number of vowels=20
Probability=20/45
=4/9

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Divya   4 Months AGO

20/45=4/9

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MUTHU   4 Months AGO

number of outcomes=20
divided by
total number=45

20/45=4/9

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MANISHA JADHAV   4 Months AGO

Total no of alphabet=45
Total no of vowels=20
Probability that piked alphabet is vowel=20/45=4/9

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21 Dec 2018 RANK

Find the rank of the word WORD in the dictionary of words formed by the letters of the word WORD

OPtion

1) 10
2) 11
3) 12
4) 13
5) 14
6) 20
7) 21
8) 22
9) 24
10) None of these

Solution

Lets arrange the letters alphabetically DORW

Words starting with D or O or R = 3*3! = 3*6 = 18
Words Starting with WD = 2! = 2
Words starting with WOD = 1 =1
Next word is WORD = 1 =1

Rank = 18+2+1+1 = 22

Correct Option: 8 Best Solution (3)

G.S.Kantharao   4 Months AGO

Word:WORD
Writing in order=D,O,R,W
Words which start with D=3!
With O=3!
With R=3!
With WD=2!
With WOD=1!
WORD=1
Rank=3!+3!+3!+2!+1+!+1=22
My option goes with 8👈

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AASHRAY AJAY SHARMA   4 Months AGO

Assigning the alphabets according to their rank and position in dictionary,
W--4
O--2
R--3
D--1
first letter D, before it there are 3 letters so 3!
second letter O, before it 3 letters so 3!
third letter R, before it 3 letters so 3!
fourth letter WO, before it 2 letters so 2!
Then, letter WOD, before it only 1 letter so 1!

Therefore, rank = (3!*3) + (2!*1) + (1!*1) +1= 18+2+1+1=22
i.e., Rank of the word 'WORD' = 22 (Option 8)

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K.L.DEDEEPYA   4 Months AGO

alphabetical order of WORD is D,O,R,W
D _ _ _=3!
O _ _ _=3!
R _ _ _=3!
W D _ _=2!
W O D _=1!
W O R D=1
3!+3!+3!+2!+1!+1=22

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