M4maths Previous Puzzles

29 Nov 2021 Milk Dealer

A Milk dealer has 125 liters, 150 liters and 275 liters of three different kinds of milk. Find the least number of containers of equal size required to store all the milk without mixing.

OPtion

1. 22
2. 28
3. 51
4. 48
5. 35
6. 37
7. 53
8. 43
9. 42
10. None of the above

Solution

Since minimum number of containers are required, the size of the containers is greatest. Also, the containers in three cases are of equal size.
The size of the containers is the H.C.F. of 125, 150 and 275 which is 25.
Now, the number of containers required for storing the milk = (125+150+275) ÷ 25 = 22

Correct Option: 1 Best Solution (4)

26 Nov 2021 AP

Find the 26th term of the sequence 5, 2, - 1, - 4, ....

OPtion

1. -25
2. -27
3. -55
4. -52
5. -67
6. -65
7. -70
8. -73
9. -77
10. None of the above

Solution

Clearly, the given sequence is an A.P. with a = 5 and d = -3.
∴a11 = a+(n−1)d
a26 = 5+ (26 – 1)-3
= 5 + 25(-3)
= 5 -75
= -70

Correct Option: 7 Best Solution (0)

25 Nov 2021 Speed

A and B take part in a 200 m race. A run at 5.4 Km/h. A gives B a start of 12 m and still beat him by 6 sec. Speed of B is:

OPtion

1. 5.15 Km/h
2. 4.14 Km/h
3. 4.25 Km/h
4. 4.40 Km/h
5. 4.51 Km/h
6. 5.20 Km/h
7. 4.85 Km/h
8. 4.95 Km/h
9. 5.11 Km/h
10. None of the above

Solution

A’s speed = (5.4x 5/18) m/sec = 27/18 m/sec
Time taken by A to cover 200 m
= (200 x 18/27) sec = 133.33 seconds.
∴ B covers 188 m in 133.33 + 6 = 139.33 seconds.
B’s speed = {(188/139.33) x (18/5)} Km/h
= 4.85 Km/h

Correct Option: 7 Best Solution (0)

24 Nov 2021 Machine

The value of a machine is ₹7000. It decreases by 10% during the first year, 15% during the second year and 20% during the third year. What will be the value of the machine after 3 years?

OPtion

1. ₹3150
2. ₹3500
3. ₹3550
4. ₹4200
5. ₹4250
6. ₹4400
7. ₹4284
8. ₹4244
9. ₹4310
10. None of the above

Solution

Here, A = 7000, x = –10, y = –15 and z = –20.
∴ Value of the machine after 3 years
= 7000 (1- 10/100)(1-15/100)(1-20/100)
= (7000 x 90 x 85 x 80)/100 x 100 x 100
= 4,28,40,00,000 / 10,00,000
= ₹4284

Correct Option: 7 Best Solution (4)

23 Nov 2021 Income

If Anil’s income is 40% less than Satish’s income, then how much per cent is Satish’s income more than Anil’s income?

OPtion

1. 55.55%
2. 60.50%
3. 66.66%
4. 70.55%
5. 70.00%
6. 60.00%
7. 50.00%
8. 55.00%
9. 80.55%
10. None of the above

Solution

Here x = 40%
Required answer = {(x/(100-x)x100}%
= {40/(100-40)x100)%
= 66.66%

Correct Option: 3 Best Solution (2)

22 Nov 2021 Number

Two numbers are respectively 30% and 80% more than a third number. What per cent is the first of the second?

OPtion

1. 80%
2. 65%
3. 90%
4. 75%
5. 70%
6. 60%
7. 68%
8. 72%
9. 85%
10. None of the above

Solution

Here, x = 30 and y = 80.
∴ First number = {(100 + ×)/ (100 + y)} x 100% of the second
= {(100 + 30)/ (100 + 80)} x 100% of the second
= (130 / 180)/100
= 72%

Correct Option: 8 Best Solution (3)

19 Nov 2021 Age

Age of X is thrice the age of Y. 3 year before age of X was 4 times that of Y. Sum of ages of X and Y in present will be

OPtion

1) 24
2) 28
3) 30
4) 32
5) 40
6) 48
7) 56
8) 36
9) 90
10) none of these

Solution

let age of x is a and y is b
now age of x is thrice age of y
a=3b;
now before three years (a-3)=4(b-3)
=>3b-3=4b-12
b=9
=>a=3*9=27
sum of ages is 27+9=36

Correct Option: 8 Best Solution (3)

18 Nov 2021 X

x % of x % of x % of x % 0f x is equal to x, so x will be
(given that x is a real number)

OPtion

1) 25
2) 50
3) 100
4) 200
5) 500
6) 800
7) 1000
8) data are insufficient to solve the problem
9) question is wrong
10) none of these

Solution

(x/100)*(x/100)*(x/100)*(x/100)*x=x
or,x^4=100000000
or,x=100

Correct Option: 3 Best Solution (2)

17 Nov 2021 Cards

In a game Romit first throws a coin. If he gets Tail then he puts 1 card from a deck of 52 cards. Find the probability to get a face card?

OPtion

P(A)= probability of getting tail : 1/2
P(B)= probability of getting a face card is : 12/52 = 3/13

so total probability P(A and B) = P(A).P(B)=
(1/2)*(3/13)= 3/26

Solution

1) 1/2
2) 1/4
3) 1/8
4) 1/26
5) 1/52
6) 3/26
7) 3/4
8) 1/10
9) incomplete data
10) none of these

Correct Option: 6 Best Solution (0)

16 Nov 2021 station

The distance between two stations is 100 km. 2 cars start from A and B at the same time and return to the original station with speed 48 km/hr and 64 km/hr. The time gap between their meeting twice will be

OPtion

1) 10/7 hr
2) 20/17 hr
3) 25/14 hr
4) 30/7 hr
5) 27/14 hr
6) 3 hr
7) 20/7 hr
8) 4 hr
9) 5 hr
10) none of these

Solution

Total distance covered between their meetings is 200 KM travelled together by 2 cars.
total combined speed = 64+48 = 112
Time take to cover 200KM with 112 KM/H speed is
200/112
=25/ 14 Hour

Correct Option: 3 Best Solution (1)