M4maths Previous Puzzles

03 Jun 2022 Geometry

If x and y are the lengths of two sides of a triangle such that the product xy=18, where x and y are integers, then how many such triangles are possible ?

OPtion

1) 7
2) 10
3) 4
4) 9
5) 5
6) 12
7) 8
8) 6
9) 13
10) None of these

Solution

The length of third side of a triangle must always be between 'Sum and the difference of other two sides '
Let x, y and z are sides of a triangle, where (x-y) < z < (x+y)
Given, product of two sides x and y = xy = 18
Different possible values of (x,y) are (1,18) (2,9) (3,6)
For x=1, y=18 : (18-1) < z < (18+1), z=18
For x=2, y=9 : (9-2) < z < (9+2), z=8, 9, 10
For x=3, y=6 : (6-3) < z < (6+3), z=4, 5, 6, 7, 8
Thus third side z can take 9 different values.
Correct Option 4)

Correct Option: 4 Best Solution (0)

02 Jun 2022 Average

In a class there are total 63 students. The average age of girls is 24 years and that of boys is 21 years. If the average age of whole class is 22 years and 8 months, then number of boys in a class is

OPtion

1) 42
2) 35
3) 36
4) 14
5) 52
6) 48
7) 56
8) 28
9) 44
10) None of these

Solution

Let there be 'x' boys and (63-x) girls in a class.
22 years & 8 months = 22 (8/12) years = 68/3 years
Average age of whole class = [ 21x + (63-x)*24 ] / 63 = 68/3
Solving, we get x=28
.'. Number of boys in a class is 28.

Correct Option: 8 Best Solution (1)

01 Jun 2022 Profit & Loss

A trader bought two articles for Rs. 17,000. He sold one at a profit of 20% and the other at loss of 25%. If the selling price of an article sold at 25% loss is 500/9% of the selling price of an article sold at 20% profit. The gain or loss percentage in the whole transaction is

OPtion

1) 1.17% Gain
2) 2.35% Gain
3) 3.82% Loss
4) 1.77% Loss
5) 3.82% Gain
6) 5.85% Gain
7) 1.17% Loss
8) 2.35% Loss
9) 5.85% Loss
10) None of these

Solution

Let the CP of two articles be Rs. x and Rs. (17000-x) respectively and first article is sold at 20% profit and second at 25% loss.
SP of an article at 20% gain = (120/100)*x
SP of an article at 25% loss = (75/100)*(17000-x)
Then, (75/100)*(17000-x) = (500/9)*(1/100)*(120/100)*x
Solving, we get x=9000
.'. CP of an articles are Rs. 9000 and Rs. 8000
Total SP = (120/100)*9000 + (75/100)*8000 = 16800
Loss = CP - SP = 17000 - 16800 = Rs. 200
% Loss = (200/17000)*100 = 1.17%

Correct Option: 7 Best Solution (0)

31 May 2022 Sequence & Series

Select the letter cluster that can replace the question mark (?) in the given series
BGFZ , FKJU , ? , OSRK , TWVF

OPtion

1) LOPP
2) LPNP
3) JONP
4) JNNP
5) KOOP
6) KNNP
7) LONP
8) KONP
9) KPNP
10) None of these

Solution

Observe the separate letter-wise pattern according to its position in the given series.
1st letters: B, F, ?, O, T ⇒ Letters positioned at 2, 6, __, 15, 20 i.e. with difference of 4, 5, 4, 5, .., so letter at postion 11=K
2nd letters: G, K, ? , S, W ⇒ Letters positioned at 7, 11, __, 19, 23 i.e. with difference of 4, so letter at position 15=O
3rd letters: F, J, ? , R, V ⇒ Letters positioned at 6, 10, __, 18, 22 i.e. with difference of 4, so letter at position 14=N
4th letters: Z, U, ?, K, F ⇒ Letters positioned at 26, 21, __, 11, 6 i.e. with difference of 5 in reverse order, so letter at position 16=P
.'. Required cluster of lettrs is KONP
Correct Option 8)

Correct Option: 8 Best Solution (0)

30 May 2022 Number System

If a + b = 40 and a² + b² = 872, then a³ + b³ = ?

OPtion

1) 8640
2) 2880
3) 11520
4) 10160
5) 28000
6) 20320
7) 15240
8) 40720
9) 19000
10) None of these

Solution

Given, a + b = 40 and a² + b² = 872
(a+b)² = a² + b² + 2ab
40² = 872 + 2ab
⇒ ab = 364
Now, a³ + b³ = (a+b) (a² + b² - ab) = 40*(872 -364) = 20320
Correct Option 6)

Correct Option: 6 Best Solution (1)

27 May 2022 Time Distance & Speed

Train A travelling at 81 kmph overtakes another train B, 230 metre long and completely passses it in 104 seconds. If the trains had been going in opposite directions, they would have passed each other in 13 seconds. The length (in metre) of A and the speed (in Kmph) of B are respectively

OPtion

1) 210 ; 42
2) 290 ; 35
3) 320 ; 35
4) 290 ; 17.5
5) 275 ; 42
6) 320 ; 63
7) 245 ; 35
8) 290 ; 63
9) 290 ; 42
10) None of these

Solution

Speed of train A = 81 kmph = 81*5/18 = 45/2 m/s. Let the speed of train B = x m/s
Relative speed in same direction=(45/2 - x) m/s
If the length of train A be 'y' metres, then time taken to cross in same direction = (y+230)/(45/2 - x) = 104 sec.
⇒ 104x + y = 2110 ---(i)
Relative speed in opposite direction=(45/2 + x) m/s
Time taken to cross in opposite direction = (y+230)/(45/2 + x) = 13 sec.
⇒ 26x - 2y = -125 ---(i)
Solving eqn. (i) & (ii), we get x=17.5 , y=290
Speed of train B in kmph = 17.5*18/5 = 63 kmph.

Correct Option: 8 Best Solution (1)

26 May 2022 Algebra

If x + 1/x = 5/2, then possible value of x^5 + 1/x^2 is

OPtion

1) 16.03125
2) 4.0625
3) 32.125
4) 4.125
5) 32.25
6) 8.03125
7) 4.25
8) 16.125
9) 8.125
10) None of these

Solution

x + 1/x = 5/2
2x² - 5x + 2 = 0
(x-2) (2x-1) = 0
⇒ x=2 or 1/2
For x=2, x^5 + 1/x^2 = 2^5 + 1/2^2 = 32 + 0.25 = 32.25
For x=1/2, (1/2)^5 + 1/(1/2)^2 = 0.03125 + 4 = 4.03125
So, from the given options 32.25 is correct

Correct Option: 5 Best Solution (1)

25 May 2022 Time and Work

A can work 4/3 as fast as B and C together. A and B together can work 6 times as fast as C. If all three of them complete a job in 100/7 days, then how long would B alone take to complete the same work (in days)?

OPtion

1) 75
2) 60
3) 40
4) 25
5) 90
6) 80
7) 45
8) 50
9) 100
10) None of these

Solution

A, B and C together complete a job in 100/7 days
.'. 1/A + 1/B + 1/C = 7/100 ----(i)
A and B together can work 6 times as fast as C
1/A + 1/B = 6/C ----(ii)
A can work 4/3 as fast as B and C together
1/A = (4/3)*(1/B + 1/C) ⇒ 1/B + 1/C = (3/4)(1/A) ----(iii)
Substituting 1/A + 1/B = 6/C from eqn (ii) in eqn. (i), we get 6/C + 1/C = 7/100
⇒ 1/C=1/100
Substituting 1/B + 1/C = (3/4)(1/A) from eqn (iii) in eqn (i), we get 1/A + (3/4)(1/A) = 7/100
⇒ 1/A=1/25
.'. 1/25 + 1/B + 1/100 = 7/100
⇒ 1/B=1/50
Therefore, B would take 50 days to complete the work alone.

Correct Option: 8 Best Solution (0)

24 May 2022 Average

In a class there are certain number of students and their average weight is 15 kg. If 6 students with average weight of 18 kg join or 9 students with average weight of 14 kg leaves, then in both the cases average weight of students remains same. How many students are there initially in a class ?

OPtion

1) 48
2) 36
3) 42
4) 39
5) 24
6) 30
7) 33
8) 21
9) 27
10) None of these

Solution

Initially if there are 'x' students in a class, then their total weight = 15x kg.
When 6 students of average weight 18 kg joins, then average of all students = (15x + 18*6) / (x+6) Kg
When 9 students of average weight 14 kg leaves, then average of all students = (15x - 14*9) / (x-9) Kg
As both cases,the averages are same .'. (15x + 18*6) / (x+6) = (15x - 14*9) / (x-9)
(5x + 36) / (x+6) = (5x - 42) / (x-9)
5x² - 45x + 36x - 324 = 5x² - 42x + 30x - 252
3x = 72
x = 24
Hence, there are 24 students initially in a class.

Correct Option: 5 Best Solution (2)

23 May 2022 Ratio & Proportion

To celebrate party, Akshay placed an order for 9 pizzas which includes brand A and brand B pizzas. Price of brand A pizza is double that of brand B. When the order was delivered he found that the number of pizzas of the brands were interchanged and this increased the bill by 25%. The ratio of brand A and brand B pizzas in original order was

OPtion

1) 3:1
2) 2:7
3) 2:1
4) 1:3
5) 5:4
6) 7:2
7) 4:5
8) 1:2
9) 1:8
10) None of these

Solution

Let the number of pizzas of brand B=n and brand A=(9-n)
If price of one pizza of brand B = Rs. x and brand A=2x
Then price of original order = Rs. [nx + (9-n)*2x]
Bill after interchanging the number of two brands = Rs. [n*2x + (9-n)*x]
Given, new bill is 25% more than the original
⇒ [n*2x + (9-n)*x] = (125/100)*[nx + (9-n)*2x]
n + 9 = (5/4)*(18-n)
n = 6
.'. Number of brand A=(9-n)=3 and brand B=6
Required ratio = A : B = 3:6 = 1:2
Correct Option 8)

Correct Option: 8 Best Solution (1)